Buffa, Anthony J. - Wilson, Jerry D. - College Physics-Addison-Wesley (2010)

College
Physics
College
Physics
SEVENTH EDITION
Jerry D. Wilson
LANDER UNIVERSITY
GREENWOOD, SC
Anthony J. Buffa
CALIFORNIA POLYTECHNIC STATE UNIVERSITY
SAN LUIS OBISPO, CA
Bo Lou
FERRIS STATE UNIVERSITY
BIG RAPIDS, MI
ADDISON-WESLEY
SAN FRANCISCO BOSTON NEW YORK
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Library of Congress Cataloging-in-Publication Data
Wilson, Jerry D.
College physics / Jerry D. Wilson, Anthony J. Buffa, Bo Lou. -- 7th ed.
p. cm.
ISBN 978-0-321-60183-4 -- ISBN 978-0-321-59277-4
1. Physics--Textbooks. I. Buffa, Anthony J. II. Lou, Bo. III. Title.
QC21.3.W35 2010
530--dc22
2008051063
ISBN: 978-0-321-60183-4 (student copy)

ISBN: 978-0-321-59277-4 (professional copy)
Copyright © 2010, 2007, 2003 Pearson Education, Inc., publishing as Pearson Addison-Wesley, 1301 Sansome St., San Francisco, CA
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About the Authors
J E R R Y D . W I L S O N , a native of A N T H O N Y J . B U F F A received B O L O U is currently Professor of

Ohio, is Emeritus Professor of his B.S. degree in physics from Physics at Ferris State University in
Physics and former chair of the Rensselaer Polytechnic Institute (RPI) Michigan. His primary teaching
Division of Biological and Physical in Troy, New York, and M.S. and responsibilities are undergraduate
Sciences at Lander University in Ph.D. degrees in physics from the introductory physics lectures and
Greenwood, South Carolina. He University of Illinois, laboratories. Professor Lou
received a B.S. degree from Ohio Urbana–Champaign. In 1970, emphasizes the importance of
University, an M.S. degree from Professor Buffa joined the faculty at conceptual understanding of the
Union College, and, in 1970, a Ph.D. California Polytechnic State basic laws and principles of physics
from Ohio University. He earned his University, San Luis Obispo. and their practical applications to the
M.S. degree while employed as a Recently retired, he teaches at Cal real world. He is also an enthusiastic
Materials Behavior physicist. Poly as an Emeritus Professor. advocate of using technology in
As a doctoral graduate student, During his career he was involved in teaching and learning.
Professor Wilson held the faculty nuclear physics research at several Professor Lou received his B.S.
rank of Instructor and began national laboratories. On campus, he and M.S. degrees in optical
teaching physical science courses. was a research associate in the engineering from Zhejiang
During this time, he coauthored a department’s radioanalytical facility. University (China) in 1982 and 1985,
physical science text that is now in its During his tenure at Cal Poly, he has respectively, and a Ph.D. in
twelfth edition. In conjunction with taught courses from introductory condensed matter physics from
his teaching career, Professor Wilson physical science to quantum Emory University in 1989.
continued his writing and has mechanics, developed and revised Dr. Lou, his wife Lingfei, and
authored or coauthored six titles. many laboratory experiments, and their daughter Alina reside in Big
Having retired from full-time taught physics to local K-12 teachers Rapids, Michigan. The family enjoys
teaching, he continues to write, at NSF workshops. Combining travel, nature, and tennis.
producing, among other works, The physics with interests in art and
Curiosity Corner, a weekly column for architecture, Dr. Buffa develops his
local newspapers that can also be own artwork and sketches, which he
found on the Internet. He and his uses to increase his effectiveness in
wife Sandy are lake dwellers in teaching. In addition to teaching,
Greenwood, SC. during his (partial) retirement, he
and his wife intend to travel more
and enjoy their future grandkids.
v
ActivPhysics™ OnLine Activities www.masteringphysics.com
1.1 Analyzing Motion Using 6.1 Momentum and Energy 9.7 Releasing a Vibrating 13.4 Magnetic Force on a
Diagrams Change Skier II Particle
1.2 Analyzing Motion Using 6.2 Collisions and Elasticity 9.8 One- and Two-Spring 13.5 Magnetic Force on a Wire
Graphs 6.3 Momentum Conservation Vibrating Systems 13.6 Magnetic Torque on a
1.3 Predicting Motion from and Collisions 9.9 Vibro-Ride Loop
Graphs 6.4 Collision Problems 9.10 Pendulum Frequency 13.7 Mass Spectrometer
1.4 Predicting Motion from 6.5 Car Collision: Two 9.11 Risky Pendulum Walk 13.8 Velocity Selector
Equations Dimensions 9.12 Physical Pendulum 13.9 Electromagnetic
1.5 Problem-Solving 6.6 Saving an Astronaut 10.1 Properties of Mechanical Induction
Strategies for Kinematics 6.7 Explosion Problems Waves 13.10 Motional emf
1.6 Skier Races Downhill 6.8 Skier and Cart 10.2 Speed of Waves on a 14.1 The RL Circuit
1.7 Balloonist Drops 6.9 Pendulum Bashes Box String 14.2 The RLC Oscillator
Lemonade 6.10 Pendulum Person- 10.3 Speed of Sound in a Gas 14.3 The Driven Oscillator
1.8 Seat Belts Save Lives Projectile Bowling 10.4 Standing Waves on 15.1 Reflection and Refraction
1.9 Screeching to a Halt 7.1 Calculating Torques Strings 15.2 Total Internal Reflection
1.10 Pole-Vaulter Lands 7.2 A Tilted Beam: Torques 10.5 Tuning a Stringed 15.3 Refraction Applications
1.11 Car Starts, Then Stops and Equilibrium Instrument: Standing 15.4 Plane Mirrors
1.12 Solving Two-Vehicle 7.3 Arm Levers Waves 15.5 Spherical Mirrors: Ray
Problems 7.4 Two Painters on a Beam 10.6 String Mass and Standing Diagrams
1.13 Car Catches Truck 7.5 Lecturing from a Beam Waves 15.6 Spherical Mirror: The
1.14 Avoiding a Rear-End 7.6 Rotational Inertia 10.7 Beats and Beat Frequency Mirror Equation
Collision 7.7 Rotational Kinematics 10.8 Doppler Effect: 15.7 Spherical Mirror: Linear
2.1.1 Force Magnitudes 7.8 Rotoride: Dynamics Conceptual Introduction Magnification
2.1.2 Skydiver Approach 10.9 Doppler Effect: Problems 15.8 Spherical Mirror:
2.1.3 Tension Change 7.9 Falling Ladder 10.10 Complex Waves: Fourier Problems
2.1.4 Sliding on an Incline 7.10 Woman and Flywheel Analysis 15.9 Thin-Lens Ray Diagrams
2.1.5 Car Race Elevator: Dynamics 11.1 Electric Force: Coulomb’s 15.10 Converging Lens
2.2 Lifting a Crate Approach Law Problems
2.3 Lowering a Crate 7.11 Race Between a Block 11.2 Electric Force: 15.11 Diverging Lens Problems
2.4 Rocket Blasts Off and a Disk Superposition Principle 15.12 Two-Lens Optical
2.5 Truck Pulls Crate 7.12 Woman and Flywheel 11.3 Electric Force Systems
2.6 Pushing a Crate Up a Elevator: Energy Superposition Principle 16.1 Two-Source Interference:
Wall Approach (Quantitative) Introduction
2.7 Skier Goes Down a Slope 7.13 Rotoride: Energy 11.4 Electric Field: Point 16.2 Two-Source Interference:
2.8 Skier and Rope Tow Approach Charge Qualitative Questions
2.9 Pole-Vaulter Vaults 7.14 Ball Hits Bat 11.5 Electric Field Due to a 16.3 Two-Source Interference:
2.10 Truck Pulls Two Crates 8.1 Characteristics of a Gas Dipole Problems
2.11 Modified Atwood 8.2 Maxwell-Boltzmann 11.6 Electric Field: Problems 16.4 The Grating: Introduction
Machine Distribution: Conceptual 11.7 Electric Flux and Qualitative
3.1 Solving Projectile Motion Analysis 11.8 Gauss’s Law Questions
Problems 8.3 Maxwell-Boltzmann 11.9 Motion of a Charge in an 16.5 The Grating: Problems
3.2 Two Balls Falling Distribution: Electric Field: 16.6 Single-Slit Diffraction
3.3 Changing the x-Velocity Quantitative Analysis Introduction 16.7 Circular Hole Diffraction
3.4 Projectile x- and y- 8.4 State Variables and Ideal 11.10 Motion in an Electric 16.8 Resolving Power
Accelerations Gas Law Field: Problems 16.9 Polarization
3.5 Initial Velocity 8.5 Work Done by a Gas 11.11 Electric Potential: 17.1 Relativity of Time
Components 8.6 Heat, Internal Energy, Qualitative Introduction 17.2 Relativity of Length
3.6 Target Practice I and First Law of 11.12 Electric Potential, Field, 17.3 Photoelectric Effect
3.7 Target Practice II Thermodynamics and Force 17.4 Compton Scattering
4.1 Magnitude of Centripetal 8.7 Heat Capacity 11.13 Electrical Potential 17.5 Electron Interference
Acceleration 8.8 Isochoric Process Energy and Potential 17.6 Uncertainty Principle
4.2 Circular Motion Problem 8.9 Isobaric Process 12.1 DC Series Circuits 17.7 Wave Packets
Solving 8.10 Isothermal Process (Qualitative) 18.1 The Bohr Model
4.3 Cart Goes Over Circular 8.11 Adiabatic Process 12.2 DC Parallel Circuits 18.2 Spectroscopy
Path 8.12 Cyclic Process: Strategies 12.3 DC Circuit Puzzles 18.3 The Laser
4.4 Ball Swings on a String 8.13 Cyclic Process: Problems 12.4 Using Ammeters and 19.1 Particle Scattering
4.5 Car Circles a Track 8.14 Carnot Cycle Voltmeters 19.2 Nuclear Binding Energy
4.6 Satellites Orbit 9.1 Position Graphs and 12.5 Using Kirchhoff’s Laws 19.3 Fusion
5.1 Work Calculations Equations 12.6 Capacitance 19.4 Radioactivity
5.2 Upward-Moving 9.2 Describing Vibrational 12.7 Series and Parallel 19.5 Particle Physics
Elevator Stops Motion Capacitors 20.1 Potential Energy
5.3 Stopping a Downward- 9.3 Vibrational Energy 12.8 RC Circuit Time Diagrams
Moving Elevator 9.4 Two Ways to Weigh Constants 20.2 Particle in a Box
5.4 Inverse Bungee Jumper Young Tarzan 13.1 Magnetic Field of a Wire 20.3 Potential Wells
5.5 Spring-Launched Bowler 9.5 Ape Drops Tarzan 13.2 Magnetic Field of a Loop 20.4 Potential Barriers
5.6 Skier Speed 9.6 Releasing a Vibrating 13.3 Magnetic Field of a
5.7 Modified Atwood Skier I Solenoid
Machine
vi
Brief Contents
PART ONE: Mechanics PART FIVE: Optics
1 Measurement and Problem Solving 1 22 Reflection and Refraction of Light 751
2 Kinematics: Description of Motion 33 23 Mirrors and Lenses 777
3 Motion in Two Dimensions 67 24 Physical Optics: The Wave Nature
of Light 810
4 Force and Motion 103
25 Vision and Optical Instruments 844
5 Work and Energy 141
6 Linear Momentum and Collisions 180
7 Circular Motion and Gravitation 222 PART SIX: Modern Physics
8 Rotational Motion and Equilibrium 266 26 Relativity 875
9 Solids and Fluids 311 27 Quantum Physics 910
28 Quantum Mechanics
and Atomic Physics 938
PART TWO: Thermodynamics
29 The Nucleus 965
10 Temperature and Kinetic Theory 355
30 Nuclear Reactions
11 Heat 386 and Elementary Particles 1001
12 Thermodynamics 417
APPENDICES
PART THREE: Oscillations and Wave Motion I Mathematical Review (with Examples)
for College Physics A-1
13 Vibrations and Waves 455 II Kinetic Theory of Gases A-6
14 Sound 489 III Planetary Data A-7
IV Alphabetical Listing
of the Chemical Elements A-7
PART FOUR: Electricity and Magnetism
V Properties of Selected Isotopes A-8
15 Electric Charge, Forces, and Fields 529
VI Answers to Follow-Up Exercises A-10
16 Electric Potential, Energy,
VII Answers to Odd-Numbered Exercises A-18
and Capacitance 560
17 Electric Current and Resistance 596
18 Basic Electric Circuits 623 Photo Credits P-1
19 Magnetism 657 Index I-1
20 Electromagnetic Induction and Waves 696

21 AC Circuits 729
vii
Learn By Drawing
Cartesian Coordinates and From Cold Ice To Hot Steam 396 Graphical Relationship between Electric
One-Dimensional Displacement 37 Leaning on Isotherms 430 Field Lines and Equipotentials 571
Signs of Velocity and Acceleration 44 Representing Work in Thermal Electric Circuit Symbols and Circuits 599
Make a Sketch and Add Them Up 79 Cycles 437 Kirchhoff Plots: A Graphical
Forces on an Object on an Inclined Plane Oscillating in a Parabolic Interpretation of Kirchhoff’s Loop
and Free-body Diagram 117 Potential Well 460 Theorem 636
Work: Area under Using the Superposition Principle to Tracing the Reflected Rays 755
the F-versus-x Curve 144 Determine the Electric Field A Mirror Ray Diagram 783
Determining the Sign of Work 145 Direction 541 A Lens Ray Diagram 792
Energy Exchanges: Sketching Electric Lines of Force Three Polarizers (see Integrated
A Falling Ball 163 for Various Point Charges 544 Example 24.6) 830
The Small-Angle Approximation 225 ¢V Is Independent of the Reference The Photoelectric Effect and Energy
Thermal Area Expansion 369 Point 562 Conservation 915
Applications (Insights appear in boldface, and “(bio)” indicates a biomedical application)
CHAPTER 1 Leg traction (bio) 120 CHAPTER 7

Why Study Physics? 2 On your toes (bio) 120 Measuring angular distance 223
Global Positioning System (GPS) 6 Burning in (racer tires) 122 Merry-go-round and rotational speeds
Capillary system (bio) 15 Airfoils 127 228
Is Unit Conversion Important? 17 Sky diving and terminal velocity 128 Compact discs (CDs) and angular
Drawing blood (bio) 24 Aerobraking 129 acceleration 229
How many red cells in blood (bio) 25 Down force and race cars 134 The Centrifuge: Separating Blood
Capillary length (bio) 15, 29 Racing tires versus passenger-car tires Components (bio) 231
Circulatory system (bio) 29 134 Centrifuge speed 232
Heartbeat (bio) 29 Down-slope run 139 Driving on a curved road 234
Red blood cells (bio) 29 Cooking evenly in a microwave oven
CHAPTER 5
Nutrition labels (bio) 31 238
White cells and platelets (bio) 31 People Power: Using Body Energy (bio) Geosynchronous satellite orbit 242
Human hair (bio) 31 158 Space Exploration: Gravity Assists 246
Hybrid Energy Conversion 166 Satellite orbits 251
CHAPTER 2 Power ratings of motors 167 ”Weightlessness” (zero gravity) and
Galileo Galilei and the Leaning Tower Weightlifting (bio) 173 apparent weightlessness 252
of Pisa 52 ”Weightlessness”: Effects on the Human
CHAPTER 6
Reaction time (bio) 53 Body (bio) 254
Free-fall on the Moon 55 How to catch a fastball 188 Space colonies and artificial gravity 254
Tower 63 Impulse force and body injury (bio) 188 Thrown outward when driving on a
Free-fall on Mars 65 Following through in sports 189 curved road 259
Taipei 65 The Automobile Air Bag and Martian Banked roads 259
Air Bags 190 Space walks 259
CHAPTER 3 Center of mass of a high jumper 208
Air resistance and range 86 Squid jet propulsion (bio) 208 CHAPTER 8
The longest jump (bio) 87 Recoil of a rifle 209 Muscle torque (bio) 270
Air-to-air refueling 89 Rocket thrust 209 My aching back (bio) 271
Reverse thrust of jet aircraft 210 Back pain (bio) 271
CHAPTER 4 Karate chop 213 No net torque: the Iron Cross (bio) 276
g’s of Force and Effects on the Human Propulsion of fan boats 214 Gymnast and balance (bio) 276
Body (bio) 109 Flamingo on one leg (bio) 214 Low bases and center of gravity of race
Sailing into the Wind—Tacking 115 Bird catching fish (bio) 219 cars 278
viii
APPLICATIONS ix
The center-of-gravity challenge (bio) 278 Human Body Temperature (bio) 361 CHAPTER 13
Stabilizing the Leaning Tower of Pisa 279 Warm-Blooded Versus Cold-Blooded A pendulum-driven clock 465
Stability in Action 282 (bio) 361 Damping: bathroom scales, shock
Yo-yo torque 287 Expansion gaps 370 absorbers, and earthquake protection
Slide or Roll to a Stop? Antilock Brakes Why lakes freeze at top first 371 467
292 Osmosis and kidneys (bio) 375 Surf 471
Angular momentum in diving and Physiologal Diffusion in Life Processes Earthquakes, Seismic Waves, and
skating (bio) 295 (bio) 376 Seismology 472
Tornadoes and Hurricanes 295 Highest and lowest recorded Destructive interference: active noise
Throwing a spiraling football 296 temperatures 382 cancellation headphones 474
Gyrocompass 296 Cooling in open-heart surgery (bio) 382 Stringed musical instruments 478
Precession of the Earth’s axis 297 Lung capacity (bio) 383 Pushing a swing in resonance 479
Helicopter rotors 298 Gaseous diffusion and the atomic bomb
Tightrope walkers 302 The collapse of the “Galloping Gertie”
385 480
Falling cat (bio) 303
Muscle force (bio) 304 CHAPTER 11 Tuning a guitar 481
Russell traction (bio) 304 Radio frequencies 486
Working off that birthday cake (bio) 389
Knee physical therapy (bio) 305 Specific heat and burning your mouth CHAPTER 14
Roller coaster loop-the-loop 308 (bio) 390
Infrasonic and ultrasonic hearing in
Cooking at Pike’s Peak 397
CHAPTER 9 animals (bio) 491
Keeping organs ready for transplant (bio)
Bone (femur) extension (bio) 315 Sonar 491
398
Osteoporosis and Bone Mineral Density Ultrasound in Medicine (bio) 492
Physiological Regulation of Body
(BMD) (bio) 319 Low-frequency fog horns 497
Temperature (bio) 399
Hydraulic brakes, shock absorbers, lifts, The Physiology and Physics of the Ear
Copper-bottomed pots 400
and jacks 322 and Hearing (bio) 497
Thermal insulation: Helping prevent heat
Manometers, tire gauges, and barometers Protect your hearing (bio) 502
loss 401
324 Beats and stringed instruments 506
Physics, the Construction Industry, and
An Atmospheric Effect: Possible Traffic radar 510
Energy Conservation 403
Earaches (bio) 325 Sonic booms 511
R-values 403
Blood Pressure and Intraocular Pressure Crack of a whip 511
Day–night atmospheric convection cycles
(bio) 326 Doppler Applications: Blood Cells and
404
An IV: gravity assist (bio) 327 Raindrops (bio) 512
Forced convection in refrigerators,
Fish swim bladders or gas bladders (bio) Pipe organs 514
heating and cooling systems, and the
332 Wind and brass instruments 515
body (bio) 404
Tip of the iceberg 333 Ultrasound in medical diagnosis (bio)
Polymer-foam insulation 405
Blood flow: cholesterol and plaque (bio) 523
335 The Greenhouse Effect (bio) 406
Thermography (bio) 407 Ultrasound and dolphins (bio) 524
Speed of blood in the aorta (bio) 336 Speed of sound in human tissue (bio)
Chimneys, smokestacks, and the Saving fruit trees from frost (bio) 408
Dressing for the desert 408 524
Bernoulli effect 337 Size of eardrum (bio) 524
Airplane lift 337 Thermal bottle 408
Passive solar design 408 Fundamental frequency of ear canal (bio)
Water strider (bio) 339
Bridge ices before road 412 527
The Lungs and Baby’s First Breath (bio)
Solar collectors for heating 415 Helium and “Donald Duck” sound (bio)
340
Cycling and perspiration (bio) 416 527
Motor oils and viscosity 341
Poiseuille’s law: a blood transfusion (bio) CHAPTER 15
CHAPTER 12
342
Energy balancing: Exercising using Uses of semiconductors 533
A bed of nails (bio) 347
Shape of water towers 347 physics (bio) 422 Application of electrostatic charging 536
Pet water dispenser 347 How not to recycle a spray can 427 Lightning and Lightning Rods 546
Plimsoll mark for depth loading 347 Perpetual-motion machines 432 Electric Fields in Law Enforcement and
Perpetual motion machine 347 Global Warming: Some Inconvenient Nature (bio) 547
Indy race cars and Venturi tunnel 348 Facts 435 Safety in lightning storms (Question
Speed of blood flow (bio) 348 Thermal efficiency of engines 437 15.20) (bio) 555
Zeppelins 352 Internal Combustion Engines and the
Otto Cycle 438 CHAPTER 16
Blood flow in the pulmonary artery (bio)
353 Thermodynamics and the Human Body Creation of X-rays 565
Blood transfusion (bio) 353 (bio) 440 The water molecule: the molecule of life
Drawing blood (bio) 353 Refrigerators as thermal pumps 441 (bio) 567
Air conditioner/heat pump: Thermal Cardiac defibrillators (bio) 577
CHAPTER 10 switch hitting 442 Electric Potential and Nerve Signal
Thermometers and thermostats 358 Compression ignition 449 Transmission (bio) 578
x APPLICATIONS
Computer monitor operation (Ex. 16.19) CHAPTER 20 The Rainbow 769

592 Wind farm generation of electrical energy
Nerve signal transmission (Exs. 16.63 and CHAPTER 23
697
16.64) (bio) 595 Electromagnetic Induction at Work: Coating of mirrors 778
Flashlights and Antiterrorism 702 Plane mirrors 778
CHAPTER 17 It’s All Done with Mirrors 780
Induced currents and equipment hazards
Battery operation 597 703 Spherical mirrors 782
Electrical hazards in the house 603 Electric generators 705 Store-monitoring diverging mirror 782
The “Bio-generation” of High Voltage Electromagnetic Induction at Play: Spherical aberration of mirrors 783
(bio) 604 Hobbies and Transportation 707 Converging lenses 790
Bioelectrical Impedance Analysis (BIA) Renewable electric energy generation: Diverging lenses 790
(bio) 607 new and old 708 Camera, slide projector, magnifying glass
An electrical thermometer 608 dc motors and black emf 709 795
Applications of superconductivity 608 Transformers 710 Combination of lenses 796
Power requirements of appliances 610 Fresnel Lenses 797
Eddy currents in braking rapid-transit
Caution during repair of electric Lens power and optometry (bio) 799
railcars 712
appliances 611 Lens aberrations 800
Electric energy transmission 713
Electrical energy efficiency and natural Making lens to correct nearsightedness
Radiation pressure and space exploration
resources 612 802
716
Various appliance applications in Day–night rearview mirror 804
Power waves and electrical noise 717
exercises 617–622 Backward lettering on emergency
Radio and TV waves 717
CHAPTER 18 Microwaves 718 vehicles 805
IR radiation: heat lamps and the Spoon as concave and convex mirror 805
Strings of Christmas tree lights 629 Dual mirrors for driving 805
greenhouse effect (bio) 718
Applications of RC Circuits to Cardiac Compound-microscope geometry 808
Visible light and eyesight (bio) 719
Medicine (bio) 639
UV light, ozone layer, sunburn and skin Autocollimation 809
Flash photography 640
cancer (bio) 719
Blinker (flashing) circuit operation 640 CHAPTER 24
UV light and photogray sunglasses (bio)
Ammeter design 641
719 Measuring the wavelength of light 813
Voltmeter design 642
X-rays, TV tube, medical applications Peacock feathers (bio) 816
Multimeter design 643
and CT scans (bio) 719 Interference of oil and soap films 816
Household circuit wiring 644
Microwave Ovens (Ex. 20.44) 727 Nonreflecting Lenses 817
Fuses and circuit breakers 645
Optical flats 818
Electrical safety and grounding (bio) 646 CHAPTER 21 Newton’s rings 819
Polarized plugs 646
British versus U.S. electrical systems 732 Diffraction of water around natural
Electricity and Personal Safety (bio) 647
Converters and adapters 733 barriers 820
Various circuit applications to medicine
Oscillator Circuits: Broadcasters of Diffraction around a razor blade 820
and safety in exercises (bio) 650–656
Electromagnetic Radiation 743 Diffraction and radio reception 822
CHAPTER 19 Resonance circuits and radio tuning 743 Diffraction gratings 823
Magnetically levitated trains 657 AM versus FM radio broadcasting 744 Compact-disc diffraction 824
Cathode-ray tubes, oscilloscopes, TV Spectrometers 825
CHAPTER 22 X-ray diffraction 826
screens and monitors 664
Mass spectrometer operation 664 How we see things 753 Polaroid™ and dichroism 828
Submarine propulsion using Diffuse reflection and seeing illuminated Polarizing sunglasses and glare reduction
magnetohydrodynamics 666 objects 753 (bio) 831
Galvanometer operation in ammeters A Dark, Rainy Night 754 Glare reduction 831
and voltmeters 671 The human eye: Refraction and Birefringent crystals 832
dc motor operation 671 wavelength (bio) 760 Optical activity and stress 833
Electronic balance 672 Mirages 761 LCDs and Polarized Light 834
Electromagnets and magnetic materials Not where it should be 761 The blue sky 835
679 Refraction and depth perception 761 Red sunrises and sunsets 836
The Magnetic Force in Future Medicine Atmospheric effects 762 Mars, the red planet 836
(bio) 680 Negative Index of Refraction and the Optical Biopsy (bio) 836
The Earth’s magnetic field and Superlens 763
Prisms in binoculars 764 CHAPTER 25
geomagnetism 682
Magnetism in Nature (bio) 683 Diamond brilliance and fire 765 The human eye (bio) 845
Navigating with compasses 683 Optical networking and communication Cameras 846
The aurorae 684 767 Nearsightedness and corrective lenses
Doorbell and chime operation (CQ. 19.17) Fiber Optics: Medical Applications (bio) (bio) 848
689 767 Farsightedness and corrective lenses (bio)
Charged pions and cancer treatment (Ex. Endoscopy and cardioscopes (bio) 767 848
19.15) (bio) 691 Prisms and spectrum 768 Bifocals (bio) 849
APPLICATIONS xi
Cornea “Orthodontics” and Surgery Photographic light meters 917 Radioactive dating (bio) 978
(bio) 850 Solar-energy conversion 917 Carbon-14 dating of bones (bio) 978
Astigmatism and corrective lenses (bio) Photocell applications, electric eyes and Radiation detectors 986
851 garage door safety 917 Biological radiation hazards (bio) 987
Visual actuity (bio) 852 Gas discharge tubes 920 Radiation dosage (bio) 988
The magnifying glass (bio) 852 Fluorescence in nature and mineral Biological effects and medical
The compound microscope (bio) 854 detection 926 applications of radiation exposure
Refracting telescopes 857 Lasers 926 (bio) 988
Prism binoculars 858 Phosphorescent materials 927 Radiation Dosage for Thyroid Cancer
Reflecting telescopes 859 Industrial lasers 929 Treatment (bio) 989
Giant Magellan Telescope (GMT) 860 CD and DVD Systems 929 Biological and Medical Applications of
Hubble space telescope 861 Lasers in Modern Medicine (bio) 930 Radiation (bio) 989
Telescopes Using Nonvisible Radiation Laser tattoo removal (bio) 930 Radioactive tracers in medicine (bio) 991
861 Laser varlcose vein treatment (bio) 930 SPET and PET (bio) 991
Automobile’s headlights resolution 863 Holography 930 Domestic and industrial applications of
Viewing from space: The Great Wall of radiation 991
China (bio) 864 CHAPTER 28 Smoke detectors 991
Color vision (bio) 865 Radioactive tracers in industry 991
Crystallography using electron
Paint and mixing of pigments (bio) 866 Neutron activation in screening for
diffraction 942
Photographic filters 867 bombs 992
The Electron Microscope (bio) 943
Oil-immersion lenses 868 Gamma radiation for sterilizing food
The Scanning Tunneling Microscope
A camera’s f-stops 874 (bio) 992
(STM) 946
CHAPTER 26 Magnetic Resonance Imaging (MRI) CHAPTER 30
Relativity and space travel 888 (bio) 948 Energy from fission: the power reactor
Relativity in Everyday Living 896 Structure of the Elements, Chemistry, and 1008
Gravitational lensing 896 the Periodic Table 952 Energy from fission: the breeder reactor
Black holes 897 Molecular binding 954 1009
Black Holes, Gravitational Waves, and Cloud chamber 958 Nuclear electrical generation 1010
LIGO 898 Nuclear-reactor safety 1010
CHAPTER 29 Fusion as an energy source 1011
CHAPTER 27 Bone scans (bio) 965 Energy from fusion: magnetic
Laser induced nuclear fusion 910 Plusses and minuses of radiation (bio) confinement 1013
Blackbody radiation, star color and 973 Energy from fusion: inertial confinement
temperature 912 Thyroid treatment with I-131 (bio) 977 1014
Preface
We believe there are two basic goals in any introductory THE SEVENTH EDITION
physics course: (1) to impart an understanding of the We have added new material to further student
basic concepts of physics and (2) to enable students to understanding and make physics more relevant,
use these concepts to solve a variety of problems. interesting, and memorable for students.
These goals are linked. We want students to apply
their conceptual understanding as they solve problems. Learning Path To provide students with a clear
Unfortunately, students often begin the problem-solving overview of the key concepts that they will be expected
process by searching for an equation. There is the temp- to learn as the chapter progresses, we have incorporated
tation to try to plug numbers into equations before visu- a flow chart that shows the learning path that students
alizing the situation or considering the physical concepts will take. It is reinforced throughout the chapter to keep
that could be used to solve the problem. In addition, stu- students focused on the key concepts as they proceed.
dents often do not check their numerical answer to see if Physics Facts. Each chapter begins with four to six
it matches their understanding of the relevant physical interesting facts about discoveries or everyday phenom-
concept. ena applicable to the chapter.
We feel, and users agree, that the strengths of this text-
book are as follows: Learning Path Review. Each end-of-chapter summary
includes visual representations of the key concepts from the
chapter to serve as a reminder for students as they review.
Conceptual Basis. Giving students a secure grasp of
physical principles will almost invariably enhance their Biological Applications. The number and scope of bio-
problem-solving abilities. We have organized discus- logical and biomedical applications make them a popu-
sions and incorporated pedagogical tools to ensure that lar feature. Examples of biological applications include
conceptual insight drives the development of practical “g’s of Force and Effects on the Human Body,” “People
skills. Power: Using Body Energy,” “Osteoporosis and Bone
Concise Coverage. To maintain a sharp INSIGHT 9.1 Osteoporosis and Bone Mineral Density (BMD)
focus on the essentials, we have avoided Bone is a living, growing tissue. Your body is continuously tak-
ing up old bone (resorption) and making new bone tissue. In
puted, commonly in grams per cubic centimeter, after the bone
is weighed to determine mass. If you burn the bone, weigh the
topics of marginal interest. We do not the early years of life, bone growth is greater than bone loss.
This continues until a peak bone mass is reached as a young
remaining ash, and divide by the volume of the overall bone
(bone tissue), you get the bone tissue mineral density, which is
derive relationships when they shed no adult. After this, bone growth is slowly outpaced by bone loss.
Bones naturally become less dense and weaker with age.
commonly called the bone mineral density (BMD).
To measure the BMD of bones in vivo, types of radiation
additional light on the principle Osteoporosis (“porous bone”) occurs when bones deteriorate
to the point where they are easily fractured (Fig. 1).
transmission through the bone are measured, which is related
to the amount of bone mineral present. Also, a “projected”
area of the bone is measured. Using these measurements, a
involved. It is usually more important projected BMD is computed in units of mg>cm2. Figure 2
illustrates the magnitude of the effect of bone density loss
for students in this course to understand with aging.
The diagnosis of osteoporosis relies primarily on the mea-
what a relationship means and how it surement of BMD. The mass of a bone, measured by a BMD
test (also called a bone densitometry test), generally correlates to
can be used than to understand the the bone strength. It is possible to predict fracture risk, much
as blood pressure measurements can help predict stroke risk.
mathematical or analytical techniques Bone density testing is recommended for all women age 65
and older, and for younger women at an increased risk of
employed to derive it. 䉱 F I G U R E 1 Bone mass loss An X-ray micrograph of the osteoporosis. This testing also applies to men. Osteoporosis is
often thought to be a woman’s disease, but 20% of osteoporo-
bone structure of the vertebrae of a 50-year-old (left) and a
70-year-old (right). Osteoporosis, a condition characterized by sis cases occur in men. A BMD test cannot predict the certainty
bone weakening caused by loss of bone mass, is evident for of developing a fracture, but only predicts the degree of risk.
the vertebrae on the right. So how is BMD measured? This is where the physics comes
Applications. College Physics is known in. Various instruments, divided into central devices and
Osteoporosis and low bone mass affect an estimated 24 mil- peripheral devices, are used. Central devices are used primarily
for the strong mix of applications related lion Americans, most of whom are women. Osteoporosis to measure the bone density of the hip and spine. Peripheral
results in an increased risk of bone fractures, particularly of devices are smaller, portable machines that are used to mea-
to medicine, science, technology, and the hip and the spine. Many women take calcium supple- sure the bone density in such places as the heel or finger.
ments to help prevent this. The most widely used central device relies on dual energy
everyday life in its text narrative and To understand how bone density is measured, let’s first dis-
tinguish between bone and bone tissue. Bone is the solid mater-
X-ray absorptiometry (DXA), which uses X-ray imaging to
measure bone density. (See Section 20.4 for a discussion of
Insight boxes. The seventh edition con- ial composed of a protein matrix, most of which has calcified.
Bone tissue includes the marrow spaces within the matrix.
X-rays.) The DXA scanner produces two X-ray beams of dif-
ferent energy levels. The amount of X-rays that pass through
tinues to have a wide range of applica- (Marrow is the soft, fatty, vascular tissue in the interior cavi-
ties of bones and is a major site of blood cell production.) The
a bone is measured for each beam; the amounts vary with the
density of bone. The calculated bone density is based on the
tions we have also increased the number marrow volume varies with the bone type.
If the volume of an intact bone is measured (for example, by
difference between the two beams. The procedure is nonin-
trusive and takes 10–20 min, and the X-ray exposure is usu-
of biological and biomedical applications water displacement), then the bone tissue density can be com- ally about one-tenth of that of a chest X-ray (Fig. 3).
in recognition of the high percentage of 䉴 F I G U R E 2 Bone density loss with aging

An illustration of how normal bone density
premed and allied health majors who loss for a female hip bone increases with age
(scale on right). Osteopenia refers to
decreased calcification or bone density. A per-
take this course. A complete list of appli- son with osteopenia is at risk for developing
osteoporosis, a condition that causes bones to
cations, with page references, is found on become brittle and prone to fracture.
pages viii–xi.
xii
PREFACE xiii
Mineral Density (BMD),” and “The Magnetic Force in ■ Examples that illustrate the detailed problem-solving
Future Medicine.” process, showing how the general procedure is
applied in practice
We have enhanced the following pedagogical features in the
seventh edition:
Problem-Solving Strategies and Hints. The initial
Learn by Drawing Boxes. Visualization is one of the treatment of problem solving is followed throughout
most important steps in problem solving. In many cases, with an abundance of suggestions, tips, cautions, short-
if students can make a sketch of a problem, they can cuts, and useful techniques for solving specific kinds of
solve it. “Learn by Drawing” features offer students spe- problems. These strategies and hints help students apply
cific help on making certain types of sketches and general principles to specific contexts as well as avoid
graphs that will provide key insights into a variety of common pitfalls and misunderstandings.
physical situations.
Learning Path Questions (LPQs) and Did You Learn? Conceptual Examples. These Examples ask students to
(DYL). The usual section objectives have been replaced think about a physical situation and conceptually solve a
by LPQs at the beginning of each chapter section. The question or choose the correct prediction out of a set of
LPQs are two to three general questions on important possible outcomes on the basis of an understanding of
section topics. They alert the student to important con- relevant principles. The discussion that follows (“Rea-
cepts covered in the section. The DYL at the end of each soning and Answer”) explains clearly how the correct
section is a new feature. They are general statements of answer can be identified, as well as why the other
items that should have been learned in the section. answers are wrong.
Demonstrations. Six new demonstrations with photos

have been added to the seventh edition. Examples Worked Examples. We have tried to make in-text
Examples as clear and detailed as possible. The aim is
include “Tension in a String: Action and Reaction
not merely to show students which equations to use, but
Forces” and “Buoyancy and Density.”
to explain the strategy being employed
and the role of each step in the overall
plan. Students are encouraged to learn the
DEMONSTRATION 4 Simple Harmonic Motion (SHM) and Sinusoidal Oscillation “why” of each step along with the “how.”
A demostration to show that SHM can be represented by a
sinusoidal function. A “graph” of the function is generated
Our goal is to provide a model for stu-
with an analogue of a strip chart recorder. dents to use as they solve problems. Each
worked Example includes the following:
■ Thinking It Through focuses students
on the critical thinking and analysis they
should undertake before beginning to use
equations.
A salt-filled funnel oscillates, suspended from two strings.
■ Given and Find are provided as the first
part of every Solution to remind students
the importance of identifying what is
known and what needs to be solved.
■ Follow-Up Exercises at the end of each
Conceptual Example and each worked
Example further reinforce the importance
Away we go. The salt falls on a black-painted poster board The salt trail traces out a plot of displacement versus time, or
of conceptual understanding and offer
that will be pulled in a direction perpendicular to the plane
of the funnel’s oscillation.
y = A sin1vt + d2. Note that in this case the phase constant
is about d = 90° and y = A cos vt. (Why?) additional practice. (Answers to Follow-
Up Exercises are given at the back of the
text.)
Suggested Problem-Solving Procedure. Section 1.7 Integrated Examples. In order to further emphasize the
provides a framework for thinking about problem solv- connection between conceptual understanding and
ing. This section includes the following: quantitative problem solving, we have developed Inte-
grated Examples for each chapter. These Examples work
■ An overview of problem-solving strategies through a physical situation both qualitatively and
■ A six-step procedure that is general enough to apply quantitatively. The qualitative portion is solved by con-
to most problems in physics, but is easily used in ceptually choosing the correct answer from a set of
specific situations possible answers. The quantitative portion involves a
xiv PREFACE
mathematical solution related to the conceptual part, Integrated Exercises. Like the Integrated Examples in
demonstrating how conceptual understanding and the chapter, Integrated Exercises (IE) ask students to
numerical calculations go hand in hand. solve a problem quantitatively as well as answer a con-
ceptual question dealing with the exercise. By answering
Pulling It Together Examples. These worked examples both parts, students can see if their numerical answer
show students how to work problems that involve mul- matches their conceptual understanding.
tiple concepts from the chapter and provide a bridge
from the individual worked examples in the chapter to Pulling It Together: Multiconcept Exercises. To ensure
the end-of-chapter comprehensive Pulling It Together that students can synthesize concepts, each chapter con-
problems. cludes with a section of comprehensive exercises drawn
from all sections of the chapter and per-
haps basic principles from previous
PULLING IT TOGETHER Ideal Gas Law, Thermodynamics, and Thermal Efficiency
chapters.
Assume you have 0.100 mol of an ideal monatomic gas that energy, heat, and the ideal gas law. Care, however, needs to be
follows the cycle given in Fig. 12.14b and that the pressure taken because heat exchanges can occur during more than
and temperature at the lower left-hand corner of that figure one of the processes in the cycle. To determine heat input dur-
are 1.00 atm and 20 °C, respectively. Further assume that the ing the isobaric expansion, the change in internal energy and
pressure doubles during the isometric process and the vol- thus the change in temperature are needed. So it seems likely ABSOLUTELY ZERO TOLERANCE
ume also doubles during the isobaric expansion. What would that the temperatures at all four corners of the cycle will be
be the thermal efficiency of this cycle? needed. These can be calculated using the ideal gas law. The FOR ERRORS CLUB (AZTEC)
four thermodynamic processes involved are two isobaric and
THINKING IT THROUGH. This example combines thermal effi- two isometric processes.
ciency (Eq. 12.12), thermal dynamic processes, work, internal We have continued to ensure accuracy
SOLUTION. The four corners are labeled with numbers as shown in Figure 12.14b. Listing the data given and converting to SI units, through the Absolutely Zero Tolerance for
Given: p4 = p3 = 1.00 atm = 1.01 * 105 N>m2
n = 0.100 mol
Find: e (thermal efficiency)
Errors Club (AZTEC). Tony Buffa of Cal
T4 = 20 °C = 293 K
p1 = p2 = 2.00 atm = 2.02 * 105 N>m2
Poly San Luis Obispo headed the AZTEC
V2 = V3 = 2V4 = 2V1
team and was supported by the text’s co-
First, the volumes and temperatures at the corners are computed, using the ideal gas law:
nRT1 10.100 mol238.31 J>1mol # K241293 K2

authors as well as by Wayne Anderson and
V4 = V1 = = 2.41 * 10-3 m3
p1
=
1.01 * 105 N>m2 Sen-Ben Liao. Every end-of-chapter Exer-
Therefore, cise was worked by three of the five team
V2 = V3 = 2V1 = 4.82 * 10-3 m3
members individually and independently.
During isometric processes, temperature (absolute in kelvins) is directly proportional to pressure (p>T = constant), and during
isobaric processes, temperature is directly proportional to volume (V>T = constant). Therefore, The results were collected and discrepan-
T1 = 2T4 = 586 K cies were resolved by a team discussion.
T2 = 2T1 = 1172 K
T3 = 12 T2 = 586 K While there has never been a text that was
Now the heat transfers can be calculated. W = 0 during the 4–1 process, and for a monatomic gas, ¢U = 32 nR¢T. Therefore, absolutely free of errors, that was our goal;
Q41 = ¢U41 = 32 nR¢T41 = 32 10.100 mol238.31 J>1mol # K241586 K - 293 K2 = + 365 J we worked very hard to make the book
During the 1–2 process, the gas expands and its internal energy increases. The work done by the gas is
error-free.
W12 = p1 ¢V12 = 12.02 * 105 N>m2214.82 * 10-3 m3 - 2.41 * 10-3 m32 = + 487 J
Since work was done and the internal energy increased,

The seventh edition is supplemented by
Q12 = ¢U12 + W12 = 32 nR¢T12 + 487 J a media and print ancillary package devel-
oped to address the needs of both students
and instructors.
End-of-Chapter Exercises. Each section of the end-of- FOR THE INSTRUCTOR

chapter material begins with multiple-choice questions
Instructor Solutions Manual, prepared by Wayne
(MC) to allow students a quick self-test for that section.
Anderson, this manual is available online at the Instruc-
These are followed by short-answer conceptual ques-
tor Resource Center: www.pearsonhighered.com/
tions (CQ) that test students’ conceptual understanding
educator. It includes complete, worked-out solutions to
and ask students to reason from principles. Quantitative
all end-of-chapter exercises.
problems round out the Exercises in each section. College
Physics provides short answers to all odd-numbered Instructor Resource Manual with Notes on ConcepTest
Exercises (quantitative and conceptual) at the back of the Questions Available at the Instructor Resource Center,
text, so students can check their understanding. www.pearsonhighered.com/educator, this online man-
ual has two parts. The first part, prepared by Katherine
Paired Exercises. To encourage students to work Whatley and Judith Beck (both of University of North
problems on their own, most sections include at least Carolina, Asheville), contains sample syllabi, lecture out-
one set of Paired Exercises that deal with similar situa- lines, notes, demonstration suggestions, readings, and
tions. The first problem in a pair is solved in the additional references and resources. The second part,
Student Study Guide and Solutions Manuals; the second prepared by Cornelius Bennhold and Gerald Feldman
problem, which explores a situation similar to that (both of George Washington University), contains an
presented in the first problem, has only an answer at overview of the development and implementation of
the back of the book. ConcepTests, as well as instructor notes for each
PREFACE xv
ConcepTest found on the Instructor Resource Center and encourage students to confront misconceptions, reason
available on the Instructor Resource DVD. qualitatively, and learn to think critically. They cover all
topics from mechanics to electricity and magnetism and
Test Bank Available at the Instructor Resource Center, from optics to modern physics. The ActivePhysics OnLine
www.pearsonhighered.com/educator, the test bank was companion workbooks help students work through com-
fully revised by Delena Gatch (University of North plex concepts and understand them more clearly.
Alabama). This online, cross-platform test bank offers
more than 2800 multiple-choice, true>false, and short- FOR THE STUDENT
answer>essay questions, approximately 50% conceptual.
The questions are organized and referenced by chapter Student Study Guide and Selected Solutions Manual
section and by question type. by Bo Lou (Ferris State University); Volume 1: 0-321-
59274-3; Volume 2: 0-321-59278-6 This guide presents
Instructor Resource DVD (ISBN 0-321-59273-5) This chapter-by-chapter reviews, chapter summaries and dis-
cross-platform DVD provides virtually every electronic cussions, mathematical summary, additional worked
asset you’ll need in and out of the classroom. The DVD is examples, practice quizzes, and solutions to paired and
organized by chapter and includes all text illustrations selected exercises.
and tables from College Physics, seventh edition in jpeg
format. The IRDVD also contains ConcepTest “Clicker” MasteringPhysicsTM Available at www.mastering-
Questions in PowerPoint, the eleven Physics You Can physics.com, this homework, tutorial, and assess-
See demonstration videos, and pdf files of the Instructor ment system is based on years of research into how
Resource Manual with Notes on ConcepTest Questions. students work physics problems and precisely where
they need help. Studies show that students who use
MasteringPhysicsTM Available at www.master- MasteringPhysics significantly increase their final scores
ingphysics.com, this homework, tutorial, and compared to those who use handwritten homework. Mas-
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the end-of-chapter problems from the text are available when they get stuck, and partial credit for their method(s)
in MasteringPhysics. This system provides instructors used. This individualized, 24>7 Socratic tutoring is recom-
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myeBook The interactive myebook is available through
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upgrade online. Allowing students access to the text
wherever they have acces to the Internet, myeBook com-
ActivPhysics OnlineTM Accessed through the Self Study
prises the full text, including figures that can be enlarged
area of www.masteringphysics.com, ActivPhysics
for better viewing. Within myeBook, students are able to
Online provides students with a suite of highly regarded
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applet-based self-study tutorials (see description on pre-
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vious page). The following workbooks provide a range
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ActivPhysics OnlineTM Accessed through the Self Study
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xvi PREFACE
Pearson Tutor Services (www.pearsontutorservices. be connected to highly qualified e-structors who provide
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Acknowledgments
The members of AZTEC—Wayne Anderson and Sen- the work on this edition. As always, several colleagues of
Ben Liao—as well as accuracy reviewer Todd Pedlar mine at Cal Poly gave of their time for fruitful discus-
deserve more than a special thanks for their tireless, sions. Among them are Professors Joseph Boone, Ronald
timely, and extremely thorough review of this book. Brown, and Theodore Foster. My family—my wife,
Dozens of other colleagues, listed in the upcoming Connie, and daughters, Jeanne and Julie—was, as
section, helped us identify ways to make the seventh always, a continuous and welcomed source of support. I
edition a better learning tool for students. We are also acknowledge the support of my father, Anthony
indebted to them, as their thoughtful and constructive Buffa, Sr. Lastly, I thank the students in my classes who
suggestions benefited the book greatly. contributed excellent ideas over the past few years.
We owe many thanks to the editorial and production Finally, we would like to urge anyone using the
team at Addison-Wesley, including Nancy Whilton, book—student or instructor—to pass on to us any sug-
Executive Editor, and Chandrika Madhavan, Project Edi- gestions that you have for its improvement. We look for-
tor. In particular, the authors wish to acknowledge the ward to hearing from you.
outstanding performance of Simone Lukashov, Produc-
—Jerry D. Wilson
tion Editor. His courteous, conscientious, and cheerful
jwilson@greenwood.net
manner made for an efficient and enjoyable production
process. —Anthony J. Buffa
In addition, I (Tony Buffa) once again extend many abuffa@calpoly.edu
thanks to my co-authors, Jerry Wilson and Bo Lou, for —Bo Lou
their cheerful helpfulness and professional approach to loub@ferris.edu
REVIEWERS OF PREVIOUS EDITIONS
David Aaron Raymond D. Benge Michael Browne

South Dakota State University Tarrant County College University of Idaho
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E. Daniel Akpanumoh William Berres Debra L. Burris
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ACKNOWLEDGMENTS xvii
Lattie F. Collins Allen Grommet Mark Lindsay

East Tennessee State University East Arkansas Community College University of Louisville
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xviii ACKNOWLEDGMENTS
Donald S. Presel Terry Scott Gabriel Umerah

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Contents
3.3 Projectile Motion 80
1 Measurement
and Problem Solving 1
3.4 Relative Velocity
LEARNING PATH REVIEW
88
94 EXERCISES 96
INSIGHT: 1.1 Why Study Physics? 2

1.1 Why and How We Measure 2
1.2 SI Units of Length, Mass, and Time 3
I N S I G H T : 1.2 Global Positioning System (GPS) 6 4 Force and Motion 103
1.3 More about the Metric System 8 4.1 The Concepts of Force and Net Force 104
1.4 Unit Analysis 12 4.2 Inertia and Newton’s First Law of Motion 105
1.5 Unit Conversions 14 4.3 Newton’s Second Law of Motion 107
I N S I G H T : 1.3 Is Unit Conversion Important? 17 I N S I G H T : 4.1 g’s of Force and Effects
1.6 Significant Figures 17

on the Human Body 109
4.4 Newton’s Third Law of Motion 113
1.7 Problem Solving 21
I N S I G H T : 4.2 Sailing into the Wind—Tacking 115
LEARNING PATH REVIEW 26 EXERCISES 28
4.5 More on Newton’s Laws: Free-Body Diagrams
and Translational Equilibrium 116
Forces on an Object on an Inclined
2
LEARN BY DRAWING:
Kinematics: Plane and Free-body Diagrams 117
Description of Motion 33 4.6 Friction 121

2.1 Distance and Speed: Scalar Quantities 34
2.2 One-Dimensional Displacement and Velocity:
Vector Quantities 36
L E A R N B Y D R A W I N G : Cartesian Coordinates and One-
2.3 Acceleration 42
Dimensional Displacement 37 5 Work and Energy 141
5.1 Work Done by a Constant Force 142
L E A R N B Y D R A W I N G : Signs of Velocity L E A R N B Y D R A W I N G : Work: Area under
and Acceleration 44 the F-versus-x Curve 144
2.4 Kinematic Equations (Constant Acceleration) 46 L E A R N B Y D R A W I N G : Determining the Sign
2.5 Free Fall 50 of Work 145
I N S I G H T : 2.1 Galileo Galilei and the Leaning Tower 5.2 Work Done by a Variable Force 147
of Pisa 52 5.3 The Work–Energy Theorem: Kinetic Energy 150
LEARNING PATH REVIEW 57 EXERCISES 60 5.4 Potential Energy 154
5.5 Conservation of Energy 157
I N S I G H T : 5.1 People Power: Using Body Energy
3 Motion in Two Dimensions 67 L E A R N B Y D R A W I N G : Energy Exchanges:

A Falling Ball 163
158
3.1 Components of Motion 68

I N S I G H T : 5.2 Hybrid Energy Conversion 166
3.2 Vector Addition and Subtraction 72
5.6 Power 166
L E A R N B Y D R A W I N G : Make a Sketch
and Add Them Up 79 LEARNING PATH REVIEW 171 EXERCISES 174
xix
xx CONTENTS
6 Linear Momentum
and Collisions 180
8 Rotational Motion
and Equilibrium 266
6.1 Linear Momentum 181 8.1 Rigid Bodies, Translations, and Rotations 267
6.2 Impulse 186 8.2 Torque, Equilibrium, and Stability 270

6.3 Conservation of Linear Momentum 189 8.3 Rotational Dynamics 280
I N S I G H T : 8.1 Stability in Action 282
I N S I G H T : 6.1 The Automobile Air Bag
and Martian Air Bags 190 8.4 Rotational Work
and Kinetic Energy 288
6.4 Elastic and Inelastic Collisions 195
8.5 Angular Momentum 291
6.5 Center of Mass 203
I N S I G H T : 8.2 Slide or Roll to a Stop?
6.6 Jet Propulsion and Rockets 208
Antilock Brakes 292
9 Solids and Fluids 311

9.1 Solids and Elastic Moduli 312
9.2 Fluids: Pressure and Pascal’s Principle 317
I N S I G H T : 9.1 Osteoporosis and Bone
Mineral Density (BMD) 319
I N S I G H T : 9.2 An Atmospheric Effect:
Possible Earaches 325
I N S I G H T : 9.3 Blood Pressure
and Intraocular Pressure 326
9.3 Buoyancy and Archimedes’ Principle 328
9.4 Fluid Dynamics and Bernoulli’s Equation 333
*9.5 Surface Tension, Viscosity,
and Poiseuille’s Law 338
I N S I G H T : 9.4 The Lungs and Baby’s First Breath 340
7 Circular Motion and Gravitation 222 10 Temperature and Kinetic Theory 355
7.1 Angular Measure 223 10.1 Temperature and Heat 356
L E A R N B Y D R A W I N G : The Small-Angle
10.2 The Celsius and Fahrenheit
Approximation 225 Temperature Scales 358
INSIGHT: 10.1 Human Body Temperature 361
7.2 Angular Speed and Velocity 226
I N S I G H T : 10.2 Warm-Blooded
7.3 Uniform Circular Motion
versus Cold-Blooded 361
and Centripetal Acceleration 229
10.3 Gas Laws, Absolute Temperature, and the Kelvin
I N S I G H T : 7.1 The Centrifuge: Separating
Temperature Scale 362
Blood Components 231
10.4 Thermal Expansion 368
7.4 Angular Acceleration 236
L E A R N B Y D R A W I N G : Thermal Area Expansion 369
7.5 Newton’s Law of Gravitation 238
10.5 The Kinetic Theory of Gases 372
I N S I G H T : 7.2 Space Exploration: Gravity Assists 246 I N S I G H T : 10.3 Physiological Diffusion
7.6 Kepler’s Laws and Earth Satellites 247 in Life Processes 376
I N S I G H T : 7.3 “Weightlessness”: *10.6 Kinetic Theory, Diatomic Gases,
Effects on the Human Body 254 and the Equipartition Theorem 376
LEARNING PATH REVIEW 256 EXERCISES 260 LEARNING PATH REVIEW 379 EXERCISES 382
CONTENTS xxi
11 Heat 386 13 Vibrations and Waves 455

11.1 Definition and Units of Heat 387 13.1 Simple Harmonic Motion 456
11.2 Specific Heat and Calorimetry 389 13.2 Equations of Motion 459
11.3 Phase Changes and Latent Heat 393 LEARN BY DRAWING: Oscillating in a Parabolic
LEARN BY DRAWING: From Cold Ice to Hot Steam 396 Potential Well 460
I N S I G H T : 11.1 Physiological Regulation 13.3 Wave Motion 468
of Body Temperature 399 INSIGHT: 13.1 Earthquakes, Seismic Waves,
11.4 Heat Transfer 400 and Seismology 472
I N S I G H T : 11.2 Physics, the Construction Industry, 13.4 Wave Properties 473
and Energy Conservation 403 13.5 Standing Waves and Resonance 477
I N S I G H T : 11.3 The Greenhouse Effect 406 LEARNING PATH REVIEW 481 EXERCISES 484
14 Sound 489
14.1 Sound Waves 490
INSIGHT: 14.1 Ultrasound in Medicine 492
14.2 The Speed of Sound 494
I N S I G H T : 14.2 The Physiology and Physics of the Ear
and Hearing 497
14.3 Sound Intensity and Sound Intensity Level 498
14.4 Sound Phenomena 503
14.5 The Doppler Effect 507
I N S I G H T : 14.3 Doppler Applications:
Blood Cells and Raindrops 512
14.6 Musical Instruments
and Sound Characteristics 514
12 Thermodynamics 417
12.1 Thermodynamic Systems, States,
and Processes 418
12.2 The First Law of Thermodynamics 420
12.3 Thermodynamic Processes for an Ideal Gas 424
L E A R N B Y D R A W I N G : Leaning on Isotherms 430
12.4 The Second Law of Thermodynamics
and Entropy 431
I N S I G H T : 12.1 Global Warming:
Some Inconvenient Facts 435
12.5 Heat Engines and Thermal Pumps 436
L E A R N B Y D R A W I N G : Representing Work
in Thermal Cycles 437
I N S I G H T : 12.2 Thermodynamics
and the Human Body 440
12.6 The Carnot Cycle and Ideal Heat Engines 443

xxii CONTENTS
15 Electric Charge, Forces, and Fields 529

15.1 Electric Charge 530
15.2 Electrostatic Charging 532
15.3 Electric Force 536
15.4 Electric Field 540
LEARN BY DRAWING: Using the Superposition Principle
to Determine the Electric
Field Direction 541
L E A R N B Y D R A W I N G : Sketching Electric Lines
of Force for Various
Point Charges 544
I N S I G H T : 15.1 Lightning and Lightning Rods 546
I N S I G H T : 15.2 Electric Fields in Law Enforcement
INSIGHT: 17.2 Bioelectrical Impedance
and Nature: Stun Guns and Electric Fish 547
Analysis (BIA) 607
15.5 Conductors and Electric Fields 548
17.4 Electric Power 609
*15.6 Gauss’s Law for Electric Fields:
A Qualitative Approach 550 LEARNING PATH REVIEW 616 EXERCISES 619
18 Basic Electric Circuits 623

16 Electric Potential, Energy,
and Capacitance 560
18.1 Resistances in Series, Parallel,
and Series—Parallel Combinations 624
18.2 Multiloop Circuits and Kirchhoff’s Rules 631
16.1 Electric Potential Energy
L E A R N B Y D R A W I N G : Kirchhoff Plots: A Graphical
and Electric Potential Difference 561
Interpretation of Kirchhoff’s
LEARN BY DRAWING: ≤V Is Independent Loop Theorem 636
of the Reference Point 562
18.3 RC Circuits 637
16.2 Equipotential Surfaces
I N S I G H T : 18.1 Applications of RC Circuits
and the Electric Field 568
to Cardiac Medicine 639
L E A R N B Y D R A W I N G : Graphical Relationship between
18.4 Ammeters and Voltmeters 641
Electric Field Lines
and Equipotentials 571 18.5 Household Circuits and Electrical Safety 644
16.3 Capacitance 575 I N S I G H T : 18.2 Electricity and Personal Safety 647
I N S I G H T : 16.1 Electric Potential LEARNING PATH REVIEW 649 EXERCISES 652

and Nerve Signal Transmission 578
16.4 Dielectrics 579
16.5 Capacitors in Series and in Parallel 582

19 Magnetism 657
19.1 Permanent Magnets, Magnetic Poles,
and Magnetic Field Direction 658
17 Electric Current and Resistance 596

19.2
19.3
Magnetic Field Strength and Magnetic Force
Applications: Charged Particles
660
17.1 Batteries and Direct Current 597 in Magnetic Fields 664

LEARN BY DRAWING: Electric Circuit Symbols 19.4 Magnetic Forces on Current-Carrying Wires 667
and Circuits 599 19.5 Applications: Current-Carrying Wires
17.2 Current and Drift Velocity 600 in Magnetic Fields 671
17.3 Resistance and Ohm’s Law 602 19.6 Electromagnetism:
I N S I G H T : 17.1 The “Bio-Generation” The Source of Magnetic Fields 673
of High Voltage 604 19.7 Magnetic Materials 678
CONTENTS xxiii
INSIGHT: 19.1 The Magnetic Force

in Future Medicine 680
*19.8 Geomagnetism: The Earth’s Magnetic Field 682
23 Mirrors and Lenses 777
23.1 Plane Mirrors 778
INSIGHT: 19.2 Magnetism in Nature 683
INSIGHT: 23.1 It’s All Done with Mirrors 780
LEARNING PATH REVIEW 686 EXERCISES 690 23.2 Spherical Mirrors 782
L E A R N B Y D R A W I N G : A Mirror Ray Diagram 783
23.3 Lenses 790
20 Electromagnetic
and Waves
Induction
696
L E A R N B Y D R A W I N G : A Lens Ray Diagram
I N S I G H T : 23.2 Fresnel Lenses 797
792
23.4 The Lens Maker’s Equation 798

20.1 Induced emf: Faraday’s Law and Lenz’s Law 697 *23.5 Lens Aberrations 800
INSIGHT: 20.1 Electromagnetic Induction at Work: LEARNING PATH REVIEW 803 EXERCISES 806
Flashlights and Antiterrorism 702
20.2 Electric Generators and Back emf 705
I N S I G H T : 20.2 Electromagnetic Induction at Play:
Hobbies and Transportation 707
20.3 Transformers and Power Transmission 710
20.4 Electromagnetic Waves 714
21 AC Circuits 729
21.1 Resistance in an AC Circuit 730
21.2 Capacitive Reactance
21.3 Inductive Reactance 735
733
24 Physical Optics:
The Wave Nature Of Light 810
21.4 Impedance: RLC Circuits 737
24.1 Young’s Double-Slit Experiment 811
21.5 Circuit Resonance 742
24.2 Thin-Film Interference 815
INSIGHT: 21.1 Oscillator Circuits: Broadcasters
INSIGHT: 24.1 Nonreflecting Lenses 817
of Electromagnetic Radiation 743
24.3 Diffraction 819
24.4 Polarization 827
LEARN BY DRAWING: Three Polarizers
(see Integrated Example 24.6) 830
22 Reflection
of Light
and Refraction
751
*24.5 Atmospheric Scattering of Light 833
I N S I G H T : 24.2 LCDs and Polarized Light 834
I N S I G H T : 24.3 Optical Biopsy 836
22.1 Wave Fronts and Rays 752 LEARNING PATH REVIEW 838 EXERCISES 840
22.2 Reflection 753

INSIGHT: 22.1 A Dark, Rainy Night 754
LEARN BY DRAWING: Tracing the Reflected Rays 755
25 Vision and Optical Instruments 844
25.1 The Human Eye 845
22.3 Refraction 756
INSIGHT: 25.1 Cornea “Orthodontics” and Surgery 850
I N S I G H T : 22.2 Negative Index of Refraction
25.2 Microscopes 852
and the Superlens 763
25.3 Telescopes 856
22.4 Total Internal Reflection and Fiber Optics 764
I N S I G H T : 25.2 Telescopes Using
I N S I G H T : 22.3 Fiber Optics: Medical Applications 767 Nonvisible Radiation 861
22.5 Dispersion 768 25.4 Diffraction and Resolution 862
I N S I G H T : 22.4 The Rainbow 769 *25.5 Color 865
LEARNING PATH REVIEW 771 EXERCISES 773 LEARNING PATH REVIEW 868 EXERCISES 871
xxiv CONTENTS
26 Relativity 875
INSIGHT: 28.2 The Scanning Tunneling
Microscope (STM) 946
I N S I G H T : 28.3 Magnetic Resonance Imaging (MRI) 948
26.1 Classical Relativity and the Michelson–Morley
Experiment 876 28.4 The Heisenberg Uncertainty Principle 955
26.2 The Special Relativity Postulate and the Relativity 28.5 Particles and Antiparticles 958
of Simultaneity 878
26.3 The Relativity of Length and Time: Time Dilation
and Length Contraction 882
26.4 Relativistic Kinetic Energy, Momentum, Total
Energy, and Mass—Energy Equivalence 890 29 The Nucleus 965
26.5 The General Theory of Relativity 893 29.1 Nuclear Structure and the Nuclear Force 966
I N S I G H T : 26.1 Relativity in Everyday Living 896 29.2 Radioactivity 969
I N S I G H T : 26.2 Black Holes, Gravitational Waves, 29.3 Decay Rate and Half-Life 975
and LIGO 898 29.4 Nuclear Stability and Binding Energy 981
*26.6 Relativistic Velocity Addition 899 29.5 Radiation Detection, Dosage,
and Applications 986
I N S I G H T : 29.1 Biological and Medical Applications
of Radiation 989
27 Quantum Physics 910 LEARNING PATH REVIEW 994 EXERCISES 997
27.1 Quantization: Planck’s Hypothesis 911

27.2 Quanta of Light: Photons
and the Photoelectric Effect 914 30 Nuclear Reactions
and Elementary Particles 1001
L E A R N B Y D R A W I N G : The Photoelectric Effect and Energy
Conservation 915 30.1 Nuclear Reactions 1002
27.3 Quantum “Particles”: The Compton Effect 918 30.2 Nuclear Fission 1006
27.4 The Bohr Theory of the Hydrogen Atom 920 30.3 Nuclear Fusion 1011
27.5 A Quantum Success: The Laser 926 30.4 Beta Decay and the Neutrino 1014
I N S I G H T : 27.1 CD and DVD Systems 929
30.5 Fundamental Forces and Exchange Particles 1016
I N S I G H T : 27.2 Lasers in Modern Medicine 930
30.6 Elementary Particles 1019
LEARNING PATH REVIEW 932 EXERCISES 935 30.7 The Quark Model 1021
30.8 Force Unification Theories, the Standard Model,
28 Quantum Mechanics
and Atomic Physics 938
and the Early Universe 1023
I N S I G H T : 30.1 The Large Hadron Collider 1025
28.1 Matter Waves: The de Broglie Hypothesis 939

INSIGHT: 28.1 The Electron Microscope 943 APPENDIX I Mathematical Review (with Examples)
28.2 The Schrödinger Wave Equation 944 for College Physics A-1
28.3 Atomic Quantum Numbers APPENDIX II Kinetic Theory of Gases A-6
and the Periodic Table 945 APPENDIX III Planetary Data A-7
APPENDIX IV Alphabetical Listing
of the Chemical Elements A-7
APPENDIX V Properties of Selected Isotopes A-8
APPENDIX VI Answers to Follow-Up Exercises A-10
APPENDIX VII Answers to Odd-Numbered
Exercises A-18
Photo Credits P-1

Index I-1
Measurement and
CHAPTER 1 LEARNING PATH
1 Problem Solving †
1.1 Why and how
we measure (2)
1.2 SI units of length, mass,

and time (3)
1.3 More about

the metric system (8)
■ metric prefixes and the liter
1.4 Unit analysis (12)
1.5 Unit conversions (14)

PHYSICS FACTS
I
■ conversion factors
✦ Tradition holds that in the twelfth
s it first and ten in the chapter-
century King Henry I of England opening photo? A measurement
decreed that the yard should be
1.6 Significant figures (17) the distance from the tip of his is needed, as with many other
■ estimating uncertainty royal nose to the thumb of his out-
stretched arm. (Had King Henry’s
things in our lives. Length measure-
arm been 3.37 inches longer, the ments tell us how far it is between
yard and the meter would be
1.7 Problem solving (21) equal in length.) cities, how tall you are, and as in the
■ suggested procedure ✦ The abbreviation for the pound, lb, photo, if it’s first and ten (yards to
comes from the Latin word libra,
which was a Roman unit of weight go). Time measurements tell you
approximately equal to a pound.
The word pound comes from the
how long it is until the class ends,
Latin pondero, “to weigh.” Libra is when the semester or quarter
also a sign of the zodiac and is
symbolized by a set of scales (used begins, and how old you are. Drugs
for weight measurement).
taken because of illnesses are given
✦ Is the old saying “A pint’s a pound
the world around” true? It depends in measured doses. Lives depend
on what you are talking about. The
saying is a good approximation for on various measurements made by
water and other similar liquids.
doctors, medical technologists, and
Water weighs 8.3 pounds per gal-
†
The mathematics needed in this chapter lon, so one-eighth of that, or a pint, pharmacists in the diagnosis and
involves scientific (powers-of-10) notation weighs 1.04 lb.
and trigonometry relationships. You may treatment of disease.
want to review these topics in Appendix I.
2 1 MEASUREMENT AND PROBLEM SOLVING
Measurements enable us to compute quantities and solve problems. Units of

measurement are also important in measurements and problem solving. For
example, in finding the volume of a rectangular box, if you measure its dimensions
in inches, the volume would have units of in3 (cubic inches); if measured in cen-
timeters, then the units would be cm3 (cubic centimeters). Measurement and
problem solving are part of our lives. They play a particularly central role in our
attempts to describe and understand the physical world, as will be seen in this
chapter. For some reasons why one should study physics, see Insight 1.1.
1.1 Why and How We Measure

LEARNING PATH QUESTIONS
➥ What does physics attempt to do?

➥ How does a unit become a standard?
➥ What is a system of units?
Imagine that someone is giving you directions to her house. Would you find it
helpful to be told, “Drive along Elm Street for a little while, and turn right at one
of the lights. Then keep going for quite a long way”? Or would you want to deal
with a bank that sent you a statement at the end of the month saying, “You still
have some money left in your account. Not a great deal, though.”
Measurement is important to all of us. It is one of the concrete ways in which
we deal with our world. This concept is particularly true in physics. Physics is con-
cerned with the description and understanding of nature, and measurement is one of its
most important tools.
INSIGHT 1.1 Why Study Physics?

The question “Why study physics?” occurs to many stu- More than likely you are a life or biological science major (such as
dents at some time during their college careers. The truth is biology, premedicine, preveterinary, medical technology, or
that there are probably as many answers as there are stu- physical therapy). In this case, physics can provide a back-
dents, much as with any other subject. However, the ground understanding of the principles involved in your work.
answers can usually be arranged into several general Although the applications of the laws of physics may not be
groups, as follows. immediately obvious, understanding them can be a valuable
You are probably not a physics major, but for these stu- tool. For example, if you are a medical professional, it may be
dents, the answer is obvious. Introductory physics provides necessary to evaluate MRI (magnetic resonance imaging) results,
the foundation of their careers. The fundamental goal of a procedure that is now commonplace. Would you be surprised
physics is to discover and understand the rules (“laws”) to know that MRI scans are based on a physical phenomenon
that govern observed phenomena. These students will use called nuclear magnetic resonance, first discovered by physicists
their knowledge of physics continually in various fields. As and still used for measuring nuclear and solid-state properties?
an example of an application, consider the development of If you are a student in a nonscience major, the physics
the laser in the 1960s. It currently plays an important role in requirement is intended to provide a well-rounded education,
various fields—medicine, industry, music (CD-DVD play- that is, the ability to evaluate technology in the context of
ers), and so on. societal needs. For example, you may be called on to vote on
You are also probably not an engineering “applied physics” tax benefits for an energy production source, and you may
major. For these students, physics provides the basis of the want to evaluate the pros and cons of the process. Or you may
engineering principles used to solve technological (applied be tempted to vote for an official who has strong views on
and practical) problems. Some of these students may not use nuclear waste disposal. Are these views scientifically justified?
physics directly, but a good understanding of physics is cru- To fully evaluate them, a knowledge of physics is necessary.
cial to the problem solving needed in technological So as you can see, there is no one answer to the question
advances. For example, after the discovery of the transistor “Why study physics?” However, there is one overriding
by physicists, engineers then developed uses for it. Decades theme: Knowledge of the laws of physics can provide an
later it evolved into the modern computer chip, which is an excellent background for understanding of the world around
electrical computing network containing millions of tiny you, or it can simply help make you a better and more well-
transistor elements. rounded citizen.
1.2 SI UNITS OF LENGTH, MASS, AND TIME 3
There are ways of describing the physical world that do not involve measure-
ment. For instance, we might talk about the color of a flower or a dress. But the
perception of color is subjective; it may vary from one person to another. Indeed,
many people are color-blind and cannot tell certain colors apart. Light received by
our eyes can be described in terms of wavelengths and frequencies. Different
wavelengths are associated with different colors because of the physiological
response of our eyes to light. But unlike the sensations or perceptions of color,
wavelengths can be measured. They are the same for everyone. In other words,
measurements are objective. Physics attempts to describe and understand nature in an
objective way through measurement.
STANDARD UNITS
Measurements are expressed in terms of unit values, or units. As you are probably
aware, a large variety of units are used to express measured values. Some of the
earliest units of measurement, such as the foot, were originally referenced to parts
of the human body. Even today, the hand is still used as a unit to measure the
height of horses. One hand is equal to 4 inches (in.). If a unit becomes officially
accepted, it is called a standard unit. Traditionally, a government or international
body establishes standard units.
A group of standard units and their combinations is called a system of units.
Two major systems of units are in use today—the metric system and the British
system. The latter is still widely used in the United States, but has virtually disap-
peared in the rest of the world, having been replaced by the metric system.
Different units in the same system or units of different systems can be used to
describe the same thing. For example, your height can be expressed in inches, feet,
centimeters, meters—or even miles, for that matter (although this unit would not
be very convenient). It is always possible to convert from one unit to another, and
such conversions are sometimes necessary. However, it is best, and certainly most
practical, to work consistently within the same system of units, as will be seen.
DID YOU LEARN?

➥ Physics attempts to describe and understand nature through measurement.
➥ A government or international body establishes measurement standards.
➥ A group of standard units and their combinations form a system of units.The two
major systems of units in use today are the metric system and the British system.
1.2 SI Units of Length, Mass, and Time

➥ What is the difference between base and derived units?

➥ How are the meter (m), the kilogram (kg), and the second (s) currently defined?
Length, mass, and time are fundamental physical quantities that are used to
describe a great many quantities and phenomena. In fact, the topics of mechanics
(the study of motion and force) covered in the first part of this book require only
these physical quantities. The system of units used by scientists to represent these
and other quantities is based on the metric system.
Historically, the metric system was the outgrowth of proposals for a more uni-
form system of weights and measures in France during the seventeenth and eigh-
teenth centuries. The modern version of the metric system is called the
International System of Units, officially abbreviated as SI (from the French
Système International des Unités).
The SI includes base quantities and derived quantities, which are described by
base units and derived units, respectively. Base units, such as the meter (m), the
kilogram (kg), and the second (s) are defined by standards. Other quantities that
are expressed in terms of combinations of base units are called derived units.
䉴 F I G U R E 1 . 1 The SI length stan- North Pole

dard: the meter (a) The meter was 0°
originally defined as 1>10 000 000 of
the distance from the North Pole to Dunkirk
the equator along a meridian run- Paris
ning through Paris, of which a por-
LENGTH: METER
tion was measured between
Dunkirk and Barcelona. A metal bar 330° Barcelona
75°
(called the Meter of the Archives) 345° 10 000 000 m 60°
was constructed as a standard. 0°
15° 30° 45°
(b) The meter is currently defined in 1m
terms of the speed of light. See the E quator
1 m = distance traveled by light in a
text for description. vacuum in 1/299 792 458 s
(a) (b)
(Think of how we commonly measure the length of a trip in miles (mi) and the
amount of time the trip takes in hours (h). To express how fast, or the rate we
travel, the derived unit of miles per hour (mi>h) is used, which represents distance
traveled per unit of time, or length per time.)
LENGTH
Length is the base quantity used to measure distances or dimensions in space. We
commonly say that length is the distance between two points. But the distance
between any two points depends on how the space between them is traversed,
which may be in a straight or a curved path.
The SI unit of length is the meter (m). The meter was originally defined as
1>10 000 000 of the distance from the North Pole to the equator along a meridian
running through Paris (䉱 Fig. 1.1a).* A portion of this meridian between Dunkirk,
France, and Barcelona, Spain, was surveyed to establish the standard length,
which was assigned the name metre, from the Greek word metron, meaning “a
measure.” (The American spelling is meter.) A meter is 39.37 in.—slightly longer
than a yard (3.37 in. longer).
The length of the meter was initially preserved in the form of a material stan-
dard: the distance between two marks on a metal bar (made of a platinum–iridium
alloy) that was stored under controlled conditions in France and called the Meter of
the Archives. However, it is not desirable to have a reference standard that changes
with external conditions, such as temperature. In 1983, the meter was redefined in
terms of a more accurate standard, an unvarying property of light: the length of the
path traveled by light in a vacuum during an interval of 1>299 792 458 of a second
(Fig. 1.1b). Light travels 299 792 458 m in a second, and the speed of light in a vac-
uum is c = 299 792 458 m>s. (c is the common symbol for the speed of light.) Thus,
light travels 1 m in 1>299 792 458 s. Note that the length standard is referenced to
time, which can be measured with great accuracy.
MASS
Mass is the base quantity used to describe amounts of matter. The more massive
an object, the more matter it contains. The SI unit of mass is the kilogram (kg). The
kilogram was originally defined in terms of a specific volume of water, that is, a
cube 0.10 m (10 cm) on a side (thereby associating the mass standard with the
*Note that this book and most physicists have adopted the practice of writing large numbers with a
thin space for three-digit groups—for example, 10 000 000 (not 10,000,000). This is done to avoid confu-
sion with the European practice of using a comma as a decimal point. For instance, 3.141 in the United
States would be written as 3,141 in Europe. Large decimal numbers, such as 0.537 84, may also be sepa-
rated, for consistency. Spaces are generally used for numbers with more than four digits on either side
of the decimal point.
length standard). However, the kilogram is now referenced to a specific material MASS: KILOGRAM
standard: the mass of a prototype platinum–iridium cylinder kept at the Interna-
tional Bureau of Weights and Measures in Sèvres, France (䉴Fig. 1.2). The United
States has a duplicate of the prototype cylinder. The duplicate serves as a reference
for secondary standards that are used in everyday life and commerce. It is hoped
that the kilogram may eventually be referenced to something other than a material
standard.
0.10 m
You may have noticed that the phrase weights and measures is generally used
water
instead of masses and measures. In the SI, mass is a base quantity, but in the British
system, weight is used to describe amounts of mass—for example, weight in
pounds instead of mass in kilograms. The weight of an object is the gravitational 0.10 m
attraction that the Earth exerts on the object. For example, when you weigh yourself 0.10 m
on a scale, your weight is a measure of the downward gravitational force exerted on (a)
your mass by the Earth. Weight is a measure of mass in this way near the Earth’s
surface, because weight and mass are directly proportional to each other.
But treating weight as a base quantity creates some problems. A base quantity
should have the same value everywhere. This is the case with mass—an object has
the same mass, or amount of matter, regardless of its location. But this is not true of
weight. For example, the weight of an object on the Moon is less than its weight on
the Earth (one-sixth as much). This is because the Moon is less massive than the
Earth and the gravitational attraction exerted on an object by the Moon (the
object’s weight) is less than that exerted by the Earth. That is, an object with a
given amount of mass has a particular weight on the Earth, but on the Moon, the
same amount of mass will weigh only about one-sixth as much. Similarly, the
weight of an object would vary for different planets.
For now, keep in mind that in a given location, such as on the Earth’s surface,
weight is related to mass, but they are not the same. Since the weight of an object of a
certain mass can vary with location, it is much more practical to take mass as the
base quantity, as the SI does. Base quantities should remain the same regardless of
where they are measured, under normal or standard conditions. The distinction
between mass and weight will be more fully explained in a later chapter. Our dis- (b)
cussion until then will be chiefly concerned with mass. 䉱 F I G U R E 1 . 2 The SI mass stan-
dard: the kilogram (a) The kilogram
was originally defined in terms of a
TIME specific volume of water, that of a
Time is a difficult concept to define. A common definition is that time is the contin- cube 0.10 m (10 cm) on a side,
thereby associating the mass stan-
uous, forward flow of events. This statement is not so much a definition as an dard with the length standard.
observation that time has never been known to run backward, as it might appear (b) The standard kilogram is now
to do when you view a film run backward in a projector. Time is sometimes said to defined by a metal cylinder. The
be a fourth dimension, accompanying the three dimensions of space (x, y, z, t). international prototype of the kilo-
That is, if something exists in space, it also exists in time. In any case, events can be gram is kept at the French Bureau of
Weights and Measures. It was man-
used to mark time measurements. The events are analogous to the marks on a ufactured in the 1880s of an alloy of
meterstick used for measurements of length. [An old view: Time does not exist in 90% platinum and 10% iridium.
itself, but only through the perceived object, from which the concepts of past, of Copies have been made for use as
present, and of future ensue. Lucretis (c. 99 BC–c. 55 BC) ] 1-kg national prototypes, one of
The SI unit of time is the second (s). The solar “clock” was originally used to which is the mass standard for the
United States. (Shown in the photo.)
define the second. A solar day is the interval of time that elapses between two suc- It is kept at the National Institute of
cessive crossings of the same longitude line (meridian) by the Sun. A second was Standards and Technology (NIST) in
fixed as 1>86 400 of this apparent solar day 11 day = 24 h = 1440 min = 86 400 s2. Gaithersburg, MD. (Notice that the
However, the elliptical path of the Earth’s motion around the Sun causes apparent bell jar can be evacuated so the
solar days to vary in length. cylinder can be stored under partial
vacuum.)
As a more precise standard, an average, or mean, solar day was computed from
the lengths of the apparent solar days during a solar year. In 1956, the second was
referenced to this mean solar day. But the mean solar day is not exactly the same
for each yearly period because of minor variations in the Earth’s motions and a
very small, but steady, slowing of its rate of rotation due to tidal friction. So scien-
tists kept looking for something better.
One frequency oscillation
Cesium-133 1 s = 9 192 631 770 oscillations Radiation

(a) detector
䉱 F I G U R E 1 . 3 The SI time standard: the second The second

was once defined in terms of the average solar day. (a) It is now
defined by the frequency of the radiation associated with an
atomic transition. (b) The atomic fountain “clock” shown here, at
NIST, is the time standard for the United States. The variation of
this “timepiece” is less than 1 s per 20 million years. (b)
In 1967, an atomic standard was adopted as a better reference. The second was
defined by the radiation frequency of the cesium-133 atom. This “atomic clock”
used a beam of cesium atoms to maintain our time standard, with a variation of
about 1 s in 300 years. In 1999, another cesium-133 atomic clock was adopted, the
atomic fountain clock, which, as the name implies, is based on the radiation fre-
quency of a fountain of cesium atoms rather than a beam (䉱 Fig. 1.3). The variation
of this “timepiece” is less than 1 s per 20 million years!*
A modern practical application involving length and time in designating a
position or location on the Earth is the GPS. See Insight 1.2, Global Positioning
System (GPS)
*An even more precise clock, the all-optical atomic clock, is under development. It is so named
because it uses laser technology and measures a time interval of 0.000 01 s. This new clock does not use
cesium atoms, but rather a single cooled ion of liquid mercury linked to a laser oscillator. The
frequency of the mercury ion is 100 000 times the frequency of cesium atoms, hence the shorter, more
precise time interval.
INSIGHT 1.2 Global Positioning System (GPS)

The GPS consists of a network of two dozen satellites. These solar-
powered satellites circle the Earth at altitudes of about 20 000 km
(12 400 mi), making two complete orbits every day. The orbits are
arranged so that there are at least four satellites observable at any
time from anywhere on the Earth (Fig. 1).
Originally developed for the Department of Defense as a mili-
tary navigation system, the GPS is now available to everyone.
All you need is a GPS receiver to find your location anywhere
on Earth, except where the satellite radio signals cannot be
received such as in caves or underwater.
GPS receivers are becoming increasingly commonplace for
finding locations in navigation and other applications. They
are used by hunters, hikers, and boaters. GPSs are found in
automobiles to provide locations for roadside assistance,
F I G U R E 1 Global Positioning
System (GPS) An artist’s concep-
tion of GPS satellites.
along with sophisticated systems that can look up addresses passing student, you ask how far it is to the bell tower over
and give directions to a particular location. The accuracy of a there; the answer is one block. Drawing a circle with a one-
receiver depends on how much you want to spend. High-end block radius with the bell tower at the center, you know that
receivers have accuracies down to 1 m. Really expensive units you are somewhere on the circle (Fig. 2a).
can come within 1 cm! That doesn’t help much, so you ask another student how
So how does the GPS determine a position on the Earth (lat- far it is to the gym; the answer is two blocks. Drawing a circle
itude and longitude)? The electronics and so on are quite with a two-block radius with the gym at the center, you know
complicated, but the basic principles of locating a position can you are at either point A or B where the circles intersect (Fig.
be understood. The process involves triangulation. You have 2b). Doing the same for the campus gate, which you are told
probably seen one form of this on TV or in a movie where is three blocks away, you now know your location is at point
police are trying to locate a radio transmitter. One receiver A where the three circles intersect (Fig. 2c).
gets a “fix” or direction of the transmitter and a straight line is The same idea works in three dimensions on spheres. The
drawn on a map. Another receiver at another location does satellites send time radio signals to the receiver and its elec-
the same, and where the two directional lines cross is the loca- tronics interprets these in terms of the satellites’ distance. The
tion of the transmitter. Just to make sure, a third receiver is satellites carry highly accurate “atomic clocks” for time mea-
used for a three-line intersection. surements. For GPS to work, the clocks in orbit must be “in
However in the case of the GPS, it is distance rather than sync” with the corresponding clocks on the Earth. If not, the
direction that is used. Let’s consider a two-dimensional exam- travel time will be incorrect and the distances will be wrong.
ple of finding a location. Suppose you are at a big university Due to the satillites’ orbital speeds (several kilometers per
and want to find your location on a campus map. Stopping a second), there are special relativity time dilations to account
for, along with general relativity effects. (See the Chapter 26
Insight 26.1, Relativity in Everyday Living.)
The distance to a satellite is computed by a simple equa-
tion, d = vt (distance = speed * time, Section 2.1). Here,
radio waves, which travel at the speed of light, are used, so
Gym d = ct, where c = 3.0 * 108 m>s (186 000 mi>s). Then, analo-
gous to the previous two-dimensional example, the positions
and distances provide three circles on the globe, the intersec-
Two blocks tion of which is the receiver’s location (Fig. 3).*
A
B
One
block
Bell tower Bell tower
(a) (b)
Satellite 1
Distance from
satellite 1
Gym
A
Three blocks
B Satellite 2
College
gate
Bell tower Distance from
satellite 2
Distance from
satellite 3
(c)
Satellite 3
F I G U R E 2 Finding a location Triangulation can be used to
find a location. (a) You are somewhere on the circle. (b) You are F I G U R E 3 Location on Earth Satellite data pro-
at either point A or point B. (c) You are at point A, where all vide three circles on the globe, the intersection of
three circles intersect. See text for detailed description. which is the receiver’s location.
*Actually, the receiver’s altitude should also be supplied. By adding a fourth satellite, the receiver’s latitude, longitude, and altitude can be
determined.
TABLE 1.1 The Seven Base Units of the SI

Name of Unit (abbreviation) Property Measured
meter (m) length

kilogram (kg) mass
second (s) time
ampere (A) electric current
kelvin (K) temperature
mole (mol) amount of substance
candela (cd) luminous intensity
SI BASE UNITS
The SI has seven base units for seven base quantities, which are assumed to be mutu-
ally independent. In addition to the meter, kilogram, and second for (1) length,
(2) mass, and (3) time, SI units include (4) electric current (charge>second) in
amperes (A), (5) temperature in kelvins (K), (6) amount of substance in moles (mol),
and (7) luminous intensity in candelas (cd). See Table 1.1.
The foregoing quantities are thought to compose the smallest number of base
quantities needed for a full description of everything observed or measured in
nature.
DID YOU LEARN?

➥ Base units are defined by standards. Derived units are combinations of base units.
➥ The meter is defined in terms of the speed of light, the standard mass of 1 kg is
associated with a platinum–iridium cylinder, (the only SI standard unit referenced
to a material artifact), and the second is defined by the radiation frequency of the
cesium-133 atom in an “atomic clock.”
1.3 More about the Metric System

➥ What is the difference between the mks and cgs systems of units?
➥ What is the proper order, from smallest to largest, of the metric prefixes
kilo-, milli-, mega-, micro-, and centi- ?
➥ Why does 1 L of water have a mass of 1 kg?
The metric system involving the standard units of length, mass, and time, now incor-
porated into the SI, was once called the mks system (for meter–kilogram–second).
Another metric system that has been used in dealing with relatively small quantities
is the cgs system (for centimeter–gram–second). In the United States, the system still
generally in use is the British (or English) engineering system, in which the standard
units of length, mass, and time are foot, slug, and second, respectively. You may not
have heard of the slug, because as mentioned earlier, gravitational force (weight) is
commonly used instead of mass—pounds instead of slugs—to describe quantities of
matter. As a result, the British system is sometimes called the fps system (for
foot–pound–second).
The metric system is predominant throughout the world and is coming into
increasing use in the United States. Because it is simpler mathematically, the SI is
the preferred system of units for science and technology. SI units are used
throughout most of this book. All quantities can be expressed in SI units. How-
ever, some units from other systems are accepted for limited use as a matter of
practicality—for example, the time unit of hour and the temperature unit of
1.3 MORE ABOUT THE METRIC SYSTEM 9
TABLE 1.2 Some Multiples and Prefixes for Metric Units*

Multiple† Prefix (and Abbreviation) Pronunciation
1012 tera- (T) ter’a (as in terrace)

9
10 giga- (G) jig’a (jig as in jiggle, a as in about)
6
10 mega- (M) meg’a (as in megaphone)
103 kilo- (k) kil’o (as in kilowatt)
2
10 hecto- (h) hek’to (heck-toe)
10 deka- (da) dek’a (deck plus a as in about)
10 -1
deci- (d) des’i (as in decimal)
10-2 centi- (c) sen’ti (as in sentimental)
10 -3
milli- (m) mil’li (as in military)
10 -6
micro- 1m2 mi’kro (as in microphone)
10 -9
nano- (n) nan’o (an as in annual)
10-12 pico- (p) pe’ko (peek-oh)
10 -15
femto- (f) fem’to (fem as in feminine)
10 -18
atto- (a) at’toe (as in anatomy)
*For example, 1 gram (g) multiplied by 1000, or 103, is 1 kilogram (kg); 1 gram multiplied
by 1>1000, or 10-3, is 1 milligram (mg).
†
The most commonly used prefixes are printed in blue. Note that the abbreviations for the
multiples 106 and greater are capitalized, whereas the abbreviations for the smaller multiples are
lowercased.
degree Celsius. British units will sometimes be used in the early chapters for
comparison purposes, since these units are still employed in everyday activities
and many practical applications.
The increasing worldwide use of the metric system means that you should be
familiar with it. One of the greatest advantages of the metric system is that it is a
decimal, or base-10, system. This means that larger or smaller units may be
obtained by multiplying or dividing by powers of 10. A list of some multiples and
corresponding prefixes for metric units is given in Table 1.2.
For metric measurements, the prefixes micro-, milli-, centi-, kilo-, and mega- are the
ones most commonly used—for example, microsecond 1ms2, millimeter (mm), cen-
timeter (cm), kilogram (kg), and megabyte (MB) as for computer disk or CD storage
sizes. The decimal characteristics of the metric system make it convenient to change
measurements from one size of metric unit to another. In the British system, differ-
ent conversion factors must be used, such as 16 for converting pounds to ounces
and 12 for converting feet to inches, whereas in the metric system, the conversion
factors are multiples of 10. For example, 100 (102) to convert meters to centimeters
(1 m = 100 cm) and 1000 (103 ) to convert meters to millimeters (1 m = 1000 mm).
You are already familiar with one base-10 system—U.S. currency. Just as a
meter can be divided into 10 decimeters, 100 centimeters, or 1000 millimeters, the
“base unit” of the dollar can be broken down into 10 “decidollars” (dimes), 100
“centidollars” (cents), or 1000 “millidollars” (tenths of a cent, or mills, used in fig-
uring property taxes and bond levies). Since all the metric prefixes are powers of
10, there are no metric analogues for quarters or nickels.
The official metric prefixes help eliminate confusion. For example, in the
United States, a billion is a thousand million (109 ); in Great Britain, a billion is a
million million (1012 ). The use of metric prefixes eliminates any confusion, since
giga- indicates 109 and tera- stands for 1012. You will probably be hearing more
about nano-, the prefix that indicates 10-9, with respect to nanotechnology
(nanotech for short). In general, nanotechnology is any technology done on the
䉴 F I G U R E 1 . 4 Molecular Man
This figure was crafted by moving
28 molecules, one at a time. Each of
the gold-colored peaks is the image
of a carbon monoxide molecule. The
molecules rest on a single crystal
platinum surface. “Molecular Man”
measures 5 nm tall and 2.5 nm wide
(hand to hand). It would take about
16 000 such figures, linked hand to
hand, to span a single human hair.
The molecules in the figure were
positioned using a special micro-
scope at very low temperatures.
nanometer scale. A nanometer (nm) is one billionth 110-92 of a meter, about the
width of three to four atoms. Basically, nanotechnology involves the manufacture
1 cm3 = 1 mL = 1 cc or building of things one atom or molecule at a time, so the nanometer is the
appropriate scale. One atom or molecule at a time? That may sound a bit far-
(1 cm3) 1000 cm3 = 1 L fetched, but it’s not (see 䉱 Fig. 1.4).
The chemical properties of atoms and molecules are well understood. For
example, rearranging the atoms in coal can produce a diamond. (This is already
10 cm done without nanotechnology using heat and pressure.) Nanotechnology presents
the possibility of constructing novel molecular devices or “machines” with extra-
ordinary properties and abilities, for example, in medicine. Nanostructures might
be injected into the body to go to a particular site, such as a cancerous growth, and
deliver a drug directly. Other organs of the body would then be spared any effects
10 cm
10 cm of the drug. (This process might be considered nanochemotherapy.)
(a) Volume It is difficult for us to grasp or visualize the new concept of nanotechnology.
Even so, keep in mind that a nanometer is one billionth of a meter. The diameter of
Mass of an average human hair is about 40 000 nm—huge compared with the new
1 mL water = 1 g nanoapplications. The future should be an exciting nanotime.
Mass of
1 L water = 1 kg
VOLUME
In the SI, the standard unit of volume is the cubic meter (m3)–the three-dimensional
10 cm derived unit of the meter base unit. Because this unit is rather large, it is often more
convenient to use the nonstandard unit of volume (or capacity) of a cube 10 cm on a
Water
side. This volume was given the name litre, which is spelled liter (L) in the United
States. The volume of a liter is 1000 cm3 110 cm * 10 cm * 10 cm2. Since
1 L = 1000 mL (milliliters, mL) it follows that 1 mL = 1 cm3. See 䉳 Fig. 1.5a. [The
10 cm
10 cm cubic centimeter is sometimes abbreviated as cc, particularly in chemistry and biol-
(b) Mass ogy. Also, the milliliter is sometimes abbreviated as ml, but the capital L is preferred
(mL) so as not to be confused with the numeral one, 1.]
䉱 F I G U R E 1 . 5 The liter and the Recall from Fig. 1.2 that the standard unit of mass, the kilogram, was originally
kilogram Other metric units are
derived from the meter. (a) A unit of defined to be the mass of a cubic volume of water 10 cm, or 0.10 m, on a side, or
volume (capacity) was taken to be the mass of one liter (1 L) of water*. That is, 1 L of water has a mass of 1 kg (Fig. 1.5b).
the volume of a cube 10 cm, or Also, since 1 kg = 1000 g and 1 L = 1000 cm3 1= 1000 mL2, then 1 cm3 (or 1 mL)
0.10 m, on a side and was given the of water has a mass of 1 g.
name liter (L). (b) The mass of a liter
of water was defined to be 1 kg.
Note that the decimeter cube con-
tains 1000 cm3, or 1000 mL. Thus, *This is specified at 4 °C. A volume of water changes slightly with temperature (thermal expansion,
1 cm3, or 1 mL, of water has a mass Section 10.4). For our purposes here, a volume of water will be considered to remain constant under
of 1 g. normal temperature conditions.
1.3 MORE ABOUT THE METRIC SYSTEM 11
EXAMPLE 1.1 The Metric Ton (or Tonne): Another Unit of Mass
As discussed, the metric unit of mass was originally related to T H I N K I N G I T T H R O U G H . A cubic meter is a relatively large
length, with a liter (1000 cm3) of water having a mass of 1 kg. volume and holds a large amount of water (more than a cubic
The standard metric unit of volume is the cubic meter (m3) yard; why?). The key is to find how many cubic volumes
and this volume of water was used to define a larger unit of measuring 10 cm on a side (liters) are in a cubic meter. A large
mass called the metric ton (or tonne, as it is sometimes spelled). number would be expected.
A metric ton is equivalent to how many kilograms?
3
S O L U T I O N . Each liter of water has a mass of 1 kg, so we need to find out how many liters are in 1 m . Since there are 100 cm in a
meter, a cubic meter is simply a cube with sides 100 cm in length. Therefore, a cubic meter (1 m3) has a volume of
10 2 cm * 10 2 cm * 10 2 cm = 106 cm3. Since 1 L has a volume of 103 cm3, there must be 1106 cm32>1103 cm3>L2 = 1000 L in 1 m3.
Thus, 1 metric ton is equivalent to 1000 kg.
Note that this line of reasoning can be expressed very concisely in a single ratio:
1 m3 100 cm * 100 cm * 100 cm

= = 1000 or 1 m3 = 1000 L
1L 10 cm * 10 cm * 10 cm
F O L L O W - U P E X E R C I S E . What would be the length of the sides of a cube that contained a metric kiloton of water? (Answers to all
Follow-Up Exercises are given in Appendix VI at the back of the book.)
You are probably more familiar with the liter than you think. The use of the liter
is becoming quite common in the United States, as 䉴 Fig. 1.6 indicates.
Because the metric system is coming into increasing use in the United States,
you may find it helpful to have an idea of how metric and British units compare.
The relative sizes of some units are illustrated in 䉲 Fig. 1.7. The mathematical con-
version from one unit to another will be discussed shortly.
DID YOU LEARN?

➥ The mks (meter-kilogram-second) system has standard units.The cgs (centimeter-
gram-second) system, although not standard, is useful in measuring relatively small
quantities.
➥ The values of commonly used metric prefixes in ascending order are micro- (10-6),
milli- (10-3), centi- (10-2), kilo- (103), and mega- (106). 䉱 F I G U R E 1 . 6 Two, three, one,
➥ The kilogram was defined to be the mass of a cube of water 10 cm on a side.This and one-half liters The liter is now a
volume was taken to be a liter, so 1 L of water has a mass of 1 kg. common volume unit for soft drinks.
Volume
1 L = 1.06 qt
1L
1 qt = 0.947 L
1 qt
Length
Mass
1 cm 1 cm = 0.394 in.
1 kg weighs An object weighing 1 lb
1 in. 1 in. = 2.54 cm 1 kg 2.2 lb at the 1 lb at the Earth's surface has
Earth's surface a mass of 0.454 kg
1m 1 m = 1.09 yd 0 0
3.5 0.5 1.750 0.250
1 yd 1 yd = 0.914 m Pounds Kilograms
3.0 1.0 1.500 0.500
2.5 1.5 1.250 0.750

2.0 1.000
1 km 1 km = 0.621 mi
1 mi
1 mi = 1.61 km
䉱 F I G U R E 1 . 7 Comparison of some SI and British units The bars illustrate the relative magnitudes of each
pair of units. (Note: The comparison scales are different in each case.)
1.4 Unit Analysis

➥ How is unit analysis useful?

➥ What units should be used in working a problem?
➥ What does density represent?
The fundamental or base quantities used in physical descriptions are called

TABLE 1.3 Some Units of dimensions. For example, length, mass, and time are dimensions. You could mea-
Common sure the distance between two points and express it in units of meters, centime-
Quantities ters, or feet, but the quantity would still have the dimension of length.
Quantity Unit Dimensions provide a procedure by which the consistency of equations may be
checked. In practice, it is convenient to use specific units, such as m, kg, and s. (See
mass kg Table 1.3.) Such units can be treated as algebraic quantities and be canceled. Using
time s units to check equations is called unit analysis, which shows the consistency of
length m units and whether an equation is dimensionally correct.
You have used equations and know that an equation is a mathematical equality.
area m2
Since physical quantities used in equations have units, the two sides of an equation must
volume m3 be equal not only in numerical value (magnitude), but also in units (dimensions). For exam-
velocity (v) m>s ple, suppose you had the length quantities a = 3.0 m and b = 4.0 m. Inserting these
m>s2 values into the equation a * b = c gives 3.0 m * 4.0 m = 12 m2. Both sides of the
equation are numerically equal 13 * 4 = 122, and both sides have the same units,
acceleration
(a or g)
m * m = m2 = 1length22. If an equation is correct by unit analysis, it must be
dimensionally correct. Example 1.2 demonstrates the further use of unit analysis.
EXAMPLE 1.2 Checking Dimensions: Unit Analysis

A professor puts two equations on the board: (a) v = vo + at The equation is dimensionally correct, since the units on each
and (b) x = v>2a, where x is distance in meters (m); v and vo side are meters per second. (The equation is also a correct
are velocities in meters>second 1m>s2 ; a is acceleration in relationship, as will be seen in Chapter 2.)
(meters>second)>second, or meters>second2 1m>s22; and t is (b) Using unit analysis, the equation
time in seconds (s). Are the equations dimensionally correct?
Use unit analysis to find out. v
x =
2a
T H I N K I N G I T T H R O U G H . Simply insert the units for the quanti-
ties in each equation, cancel, and check the units on both sides. is
a b
SOLUTION. m
(a) The equation is s m s2 (not dimensionally
m = = * or m = s
correct)
a b
v = vo + at m s m
Inserting units for the physical quantities gives (Table 1.3) s2
+ a 2 * sb + a * sb
m m m m m m The meter (m) is not the same unit as the second (s), so in this
case, the equation is not dimensionally correct 1length Z time2,
= or =
s s s s s s * s
Notice that units cancel like numbers in a fraction. Then, and therefore is also not physically correct.
simplifying,
FOLLOW-UP EXERCISE. Is the equation ax = v 2 dimension-
m m m (dimensionally ally correct? (Answers to all Follow-Up Exercises are given in
= +
s s s correct) Appendix VI at the back of the book.)
Unit analysis will tell if an equation is dimensionally correct, but a dimension-

ally consistent equation may not correctly express the physical relationship of
quantities. For example, in terms of units, the equation
x = at2
is
m = 1m>s221s 22 = m
This equation is dimensionally correct 1length = length2. But, as will be learned in
Chapter 2, it is not physically correct. The correct form of the equation—both dimen-
1.4 UNIT ANALYSIS 13
sionally and physically—is x = 12 at2. (The fraction 12 has no dimensions; it is an exact

dimensionless number.) Unit analysis cannot tell you if an equation is physically cor-
rect, only whether or not it is dimensionally consistent.
MIXED UNITS
Unit analysis also allows you to check for mixed units. In general, when working
problems, you should always use the same system of units and the same unit for a given
dimension throughout an exercise.
Suppose you wanted to buy a rug to fit a rectangular floor area and you measure
the sides to be 4.0 yd * 3.0 m. The area of the rug would then be A = l * w =
4.0 yd * 3.0 m = 12 yd # m, which might cause a problem at the carpet store. Note
that this equation is dimensionally correct, 1length22 = 1length22, but the units are
inconsistent or mixed. So, unit analysis will point out mixed units. Note that it is possi-
ble for an equation to be dimensionally correct, even if the units are mixed.
Let’s look at mixed units in an equation. Suppose that you used centimeters
(cm) as the unit for x in the equation
v 2 = v 2o + 2ax
and the units for the other quantities as in Example 1.2. In terms of units, this
equation would give
m 2 m 2
a b = a b + a b
m * cm
s s s2
or
m2 m2 m * cm
2
= 2
+
s s s2
which is dimensionally correct, 1length22>1time22, on both sides of the equation.
But the units are mixed (m and cm). The value of x in centimeters needs to be con-
verted to meters to be used in the equation.
DETERMINING THE UNITS OF QUANTITIES

Another aspect of unit analysis that is very important in physics is the determina-
tion of the units of quantities from defining equations. For example, the density
(R) of an object (represented by the Greek letter rho, r) is defined by the equation
kg
a b
m
r = (1.1)
V m3
where m is the object’s mass and V its volume. (Density is the mass per unit vol-
ume and is a measure of the compactness of the mass of an object or substance.) In
SI units, mass is measured in kilograms and volume in cubic meters, which gives
the derived SI unit for density as kilograms per cubic meter 1kg>m32.
How about the units of p? The relationship between the circumference (c) and
the diameter (d) of a circle is given by the equation c = pd, so p = c>d. If the
lengths are measured in meters, then unitwise,
a b
c m
p =
d m
Thus, p has no units. It is unitless, or a dimensionless constant.
DID YOU LEARN?

➥ Unit analysis can tell if an equation is dimensionally correct, but not physically
correct.
➥ In working problems, the same system of units should be used throughout.
➥ Density is a measure of the compactness of the mass of an object (mass>volume).
1.5 Unit Conversions

➥ What is an equivalence statement?

➥ How are conversion factors written?
Because units in different systems, or even different units in the same system, can
be used to express the same quantity, it is sometimes necessary to convert the
units of a quantity from one unit to another. For example, we may need to convert
feet to yards or convert inches to centimeters. You already know how to do many
unit conversions. If a sidewalk is 12 ft long, what is its length in yards? Your
immediate answer is 4 yd.
How did you do this conversion? Well, you must have known a relationship
between the units of foot and yard. That is, you know that 3 ft = 1 yd. This is
what is called an equivalence statement. As was seen in Section 1.4, the numerical
values and units on both sides of an equation must be the same. In equivalence
statements, we commonly use an equal sign to indicate that 1 yd and 3 ft stand for
the same, or equivalent, length. The numbers are different because they stand for
different units of length.
Mathematically, to change units conversion factors are used, which are sim-
ply equivalence statements expressed in the form of ratios—for example,
1 yd>3 ft or 3 ft>1 yd. (The “1” is often omitted in the denominators of such ratios
for convenience—for example, 3 ft>yd.) To understand why such ratios are
useful, note the expression 1 yd = 3 ft in ratio form:
1 yd 3 ft 3 ft 1 yd
= = 1 or = = 1
3 ft 3 ft 1 yd 1 yd
As can be seen, a conversion factor has an actual value of unity or one—and you
can multiply any quantity by one without changing its value or size. Thus, a con-
version factor simply lets you express a quantity in terms of other units without changing
its physical value or size.
The manner in which 12 ft is converted to yards may be expressed mathemati-
cally as follows:
1 yd
12 ft * = 4 yd (units cancel)
3 ft
Using the appropriate conversion factor form, the units cancel, as shown by the
slash marks, giving the correct unit analysis, yd = yd.
Suppose you are asked to convert 12.0 in. to centimeters. You may not know
the conversion factor in this case, but it can be obtained from a table (such as the
one that appears inside the front cover of this book). The needed relationships
are 1 in. = 2.54 cm or 1 cm = 0.394 in. It makes no difference which of these
equivalence statements you use. The question, once you have expressed the
equivalence statement as a ratio conversion factor, is whether to multiply or
divide by that factor to make the conversion. In doing unit conversions, take advan-
tage of unit analysis—that is, let the units determine the appropriate form of con-
version factor.
Note that the equivalence statements can give rise to two forms of the conver-
sion factors: 1 in.>2.54 cm and 2.54 cm>in. When changing inches to centimeters,
the appropriate form for multiplying is either 2.54 cm>in. or 1 cm>0.394 in. When
changing centimeters to inches, use the form 1 in.>2.54 cm or 0.394 in.>cm. For
example,
2.54 cm
12.0 in. * = 30.5 cm
in.
0.394 in.
15.0 cm * = 5.91 in.
cm
1.5 UNIT CONVERSIONS 15
The multiplication of conversion factors in cancelling units is usually more conve-

nient than division. In general, the multiplication form of conversion factors will
be used throughout this book.*
A few commonly used equivalence statements are not dimensionally or physi-
cally correct; for example, consider 1 kg = 2.2 lb, which is used for quickly deter-
mining the weight of an object near the Earth’s surface given its mass. The
kilogram is a unit of mass, and the pound is a unit of weight. This means that 1 kg
is equivalent to 2.2 lb; that is, a 1-kg mass has a weight of 2.2 lb. Since mass and
weight are directly proportional, the dimensionally incorrect conversion factor
1 kg>2.2 lb may be used (but only near the Earth’s surface).
EXAMPLE 1.3 Converting Units: Use of Conversion Factors

A championship male pole vaulter goes over a bar set at 6.14 m. A championship
female vaulter clears a height of 4.82 m. What is the difference in these heights in feet?
THINKING IT THROUGH. After using the correct conversion factor, the rest is arithmetic.
SOLUTION.
From the conversion table, 1 m = 3.28 ft, so converting the heights to feet: (a)
3.28 ft
6.14 m * = 20.1 ft
m
3.28 ft
4.82 m * = 15.8 ft
m
And the difference in height is ¢h = 20.1 ft - 15.8 ft = 4.3 ft.
Another approach would be to subtract the heights in meters and have only a single
conversion:
3.28 ft
6.14 m - 4.82 m = 1.32 m * = 4.33 ft
m
The answers aren’t the same. Is there something wrong? No, as will be discussed in the
Section 1.6 Problem-Solving Hint, The “Correct” Answer, the difference is usually due to (b)
rounding differences. This may occur when working a problem by another method.
Another foot–meter conversion is shown in 䉴 Fig. 1.8a. Is it correct? 䉱 F I G U R E 1 . 8 Unit conversion
Signs sometimes list both the British
F O L L O W - U P E X E R C I S E . Rather than use a single conversion factor from the table, use and metric units, as shown here for
commonly known factors to convert a 30-day month to seconds. (Answers to all Follow- elevation (a) and speed (b). Note the
Up Exercises are given in Appendix VI at the back of the book.) highlighted km.
EXAMPLE 1.4 More Conversions: A Really Long Capillary System

Capillaries, the smallest blood vessels of the body, connect the 䉳 FIGURE 1.9
arterial system with the venous system and supply our tissues Capillary system
with oxygen and nutrients (䉴Fig. 1.9). It is estimated that if all Capillaries con-
of the capillaries of an average adult were unwound and nect the arterial
spread out end to end, they would extend to a length of about and venous sys-
64 000 km. (a) How many miles is this length? (b) Compare tems in our bod-
ies. They are the
this length with the circumference of the Earth.
smallest blood
T H I N K I N G I T T H R O U G H . (a) This conversion is straightfor- vessels, but
ward—just use the appropriate conversion factor. (b) How is their total
length is
the circumference of a circle or sphere calculated? There is an
impressive.
equation to do so, but the radius or diameter of the Earth must
be known. (If you do not remember one of these values, see the
solar system data table inside the back cover of this book.) (continued on next page)
*Quantities may also be divided by conversion factors. For example,
12 in. n a b = 12 in. * 12.54 cm>in.2 = 30 cm

1 in.
2.54 cm
Using the multiplication form saves the step of inverting the ratio.
SOLUTION. [To make a general comparison, p 1 = 3.14 Á 2 is rounded off

(a) From the conversion table, 1 km = 0.621 mi, so to 3. The L symbol means “approximately equal to.”]
So,
64 000 km * 0.621 mi capillary length 40 000 mi
= 40 000 mi (rounded off) L = 1.7
1 km Earth’ s circumference 24 000 mi
The capillaries of your body have a total length that would
(b) A length of 40 000 mi is substantial. To see how this length extend about 1.7 times around the world. Wow!
compares with the circumference (c) of the Earth, recall that the
radius of the Earth is approximately 4000 mi, so the diameter (d) F O L L O W - U P E X E R C I S E . Taking the average distance between
is 8000 mi. The circumference of a circle is given by c = pd the East Coast and West Coast of the continental United States
(Appendix I-C), and to be 4800 km, how many times would the total length of your
body’s capillaries cross the country? (Answers to all Follow-Up
c = pd L 3 * 8000 mi L 24 000 mi (rounded off) Exercises are given in Appendix VI at the back of the book.)
EXAMPLE 1.5 Converting Units of Area: Choosing the Correct Conversion Factor
A hall bulletin board has an area of 2.5 m2. What is this area (a) Then using the conversion factor explicitly squared:
(a) in square centimeters (cm2), and (b) square inches (in2)?
10 2 cm 2 104 cm2
THINKING IT THROUGH. This problem is a conversion of area 2.5 m2 * a b = 2.5 m2 * = 2.5 * 104 cm2
1m 1 m2
units, and we know that 1 m = 100 cm. So, some squaring
must be done to get square meters related to square centimeters.
(b) Using the cm2 result found in (a), by a similar procedure,
SOLUTION. A common error in such conversions is the use
0.394 in. 2
of incorrect conversion factors. Because 1 m = 100 cm, it is 2.54 * 104 cm2 a b =
sometimes assumed that 1 m2 = 100 cm2, which is wrong. The cm
correct area conversion factor may be obtained directly from
0.155 in2
the correct linear conversion factor, 100 cm>1 m, or 2.54 * 104 cm2 * a b = 3.94 * 103 in2
102 cm>1 m, by squaring the linear conversion factor: cm2
10 2 cm 2 104 cm2
a b = F O L L O W - U P E X E R C I S E . How many cubic centimeters are in
1m 1 m2 1 m3 ? (Answers to all Follow-Up Exercises are given in Appendix
Hence, 1 m2 = 104 cm2 1= 10 000 cm22. VI at the back of the book.)
EXAMPLE 1.6 The Better Deal

A grocery store has a sale on sodas. A 2-L bottle sells for $1.35, Then in terms of quarts, the base price of the liquid in the 2.0-
and the price of a half-gallon bottle is $1.32. Which is the bet- L bottle is
ter buy? $1.35 $0.64
=
2.1 qt qt
T H I N K I N G I T T H R O U G H . The answer is obtained by knowing
Similarly for the half-gallon volume:
the price per common volume. This means that liters must be
converted to quarts or vice versa. (The cancellation slashes $1.32 $0.66
=
will now be omitted as being understood.) 2.0 qt qt
So, the better buy is the 2-L bottle, even though its price for
SOLUTION. To get a common volume, let’s convert liters to
2 L is higher. (Keep in mind a liter is larger than a quart.)
quarts using the conversion factor given inside the front cover
of the book: F O L L O W - U P E X E R C I S E . Work the Example the other way—
changing quarts to liters—to see if the result is the same.
1.056 qt
2.0 L a b = 2.1 qt (Answers to all Follow-up Exercies are given in Appendix VI at the
L back of the book.)
Some examples of the importance of unit conversion are given in the accompa-
nying Insight 1.3, Is Unit Conversion Important?
DID YOU LEARN?

➥ 3 ft = 1 yd is not an equation, but an equivalence statement representing
equivalent lengths: 3 ft is the equivalent length of 1 yd.
➥ The conversion factor for foot and yard may be written 3 ft>1 yd or 1 yd>3 ft.
1.6 SIGNIFICANT FIGURES 17
INSIGHT 1.3 Is Unit Conversion Important?

The answer to this question is, you bet! Here are a couple of How could this have happened? Investigations showed
cases in point. In 1999, the $125 million Mars Climate Orbiter that the failure of the Orbiter was primarily a problem of a
was making a trip to the Red Planet to investigate its atmos- lack of unit conversion. At Lockheed Martin Astronautics,
phere (Fig. 1). The spacecraft approached the planet in Sep- which built the spacecraft, the engineers calculated the navi-
tember, but suddenly contact between the Orbiter and gational information in British units. When scientists at
personnel on Earth was lost, and the Orbiter was never heard NASA’s Jet Propulsion Laboratory received the data, they
from again. Investigations showed that the orbiter had assumed that the information was in metric units, as was
approached Mars at a far lower altitude than planned. Instead called for in the mission specifications. The unit conversions
of the Orbiter passing 147 km (91 mi) above the Martian sur- weren’t made, and a $125 million spacecraft was lost on the
face, tracking data showed that it was on a trajectory that Red Planet—causing more than a few red faces.
would have taken it as close as 57 km (35 mi) from the sur- Closer to Earth, in 1983 Air Canada Flight 143 was on
face. As a result, the spacecraft either burned up in the Mart- course from Montréal to Edmonton, Canada, with sixty-one
ian atmosphere or crashed into the surface. passengers in a new Boeing 767, at the time the most
advanced jetliner in the world. Almost halfway into the flight,
a warning light came on for a fuel pump, then for another,
and finally for all four pumps. The engines quit, and this
advanced plane was now a glider, about 100 mi from the
nearest major airport, at Winnipeg. Without engines, Flight
143’s descent would bring it down 10 mi short of the airport,
so it was diverted to an old Royal Canadian Air Force landing
field at Gimli. The pilot maneuvered the powerless plane to a
landing, stopping just short of a barrier. Did the plane, which
was dubbed “The Gimli Glider,” have bad fuel pumps? No—
it had run out of fuel!
This near-disaster was caused by another conversion prob-
lem. The fuel computers weren’t working properly so the
mechanics had used the old procedure of measuring the fuel
in the tanks with a dipstick. In this method, the length of the
stick that is wet is used to determine the volume of fuel by
means of conversion values in tables. Air Canada had for
years computed the amount of fuel in pounds, but the new
767’s fuel consumption was expressed in kilograms. Even
worse, the dipstick procedure gave the amount of fuel
onboard in liters instead of pounds or kilograms. The result
was that the aircraft was loaded with 22 300 lb of fuel instead
of the required 22 300 kg. Since 1 lb has a mass of 0.45 kg, the
plane had less than half the required fuel.
F I G U R E 1 Mars Climate Orbiter An artist’s conception of
These incidents underscore the importance of using appro-
the orbiter near the surface of Mars. The actual orbiter either priate units, making correct unit conversions, and working
burned up in the Martian atmosphere or crashed into the consistently in the same system of units. Several exercises at
surface. The cause was attributed to a mix-up in units, the end of the chapter will challenge you to develop your
resulting in the loss of a $125 million spacecraft. skills in accurate unit conversions.
1.6 Significant Figures

➥ Are there some numbers without uncertainty or error?

➥ What numbers generally have uncertainty and error?
➥ What determines the degree of accuracy or number of significant figures of a mea-
sured quantity?
Most of the time, you will be given numerical data when asked to solve a problem.
In general, such data are either exact numbers or measured numbers (quantities).
Exact numbers are numbers without any uncertainty or error. This category
includes numbers such as the 100 used to calculate a percentage and the 2 in the
equation r = d>2 relating the radius and diameter of a circle. Measured numbers
are numbers obtained from measurement processes and thus generally have some
degree of uncertainty or error.
When calculations are done with measured numbers, the uncertainty and/or
error of measurement is propagated, or carried along, by the mathematical opera-
tions. The question of how to report a result arises. For example, suppose that you
are asked to find time (t) from the equation x = vt and are given that x = 5.3 m
and v = 1.67 m>s. That is,
x 5.3 m
t = = = ?
v 1.67 m> s
Doing the division operation on a calculator yields a result such as 3.173 652 695 s
(䉳 Fig. 1.10). How many figures, or digits, should you report in the answer?
The uncertainty of the result of a mathematical operation may be computed by
statistical methods.* However, a simpler and more widely used procedure for esti-
mating uncertainty involves the use of significant figures (sf), sometimes called
significant digits. The degree of accuracy of a measured quantity depends on how
finely divided the measuring scale of the instrument is. For example, you might
measure the length of an object as 2.5 cm with one instrument and 2.54 cm with
another. The second instrument with a finer scale provides more significant fig-
ures and thus a greater degree of accuracy.
䉱 F I G U R E 1 . 1 0 Significant fig- Basically, the significant figures in any measurement are the digits that are known
ures and insignificant figures For with certainty, plus one digit that is uncertain. This set of digits is usually defined as
the division operation 5.3>1.67, a all of the digits that can be read directly from the instrument used to make the
calculator with a floating decimal measurement, plus one uncertain digit that is obtained by estimating the fraction
point gives many digits. A calcu-
lated quantity can be no more accu-
of the smallest division of the instrument’s scale.
rate than the least accurate quantity The quantities 2.5 cm and 2.54 cm have two and three significant figures,
involved in the calculation, so this respectively. This is rather evident. However, some confusion may arise when a
result should be rounded off to two quantity contains one or more zeros. For example, how many significant figures
significant figures—that is, 3.2. does the quantity 0.0254 m have? What about 104.6 m, or 2705.0 m? In such cases,
the following rules will be used to determine significant figures:
1. Zeros at the beginning of a number are not significant. They merely locate
the decimal point. For example,
0.0254 m has three significant figures (2, 5, 4)
2. Zeros within a number are significant. For example,
104.6 m has four significant figures (1, 0, 4, 6)
3. Zeros at the end of a number after the decimal point are significant. For
example,
2705.0 m has five significant figures (2, 7, 0, 5, 0)
4. In whole numbers without a decimal point that end in one or more zeros
(trailing zeros)—for example, 500 kg—the zeros may or may not be signifi-
cant. In such cases, it is not clear which zeros serve only to locate the decimal
point and which are actually part of the measurement. For example, if the
first zero after the 5 in 500 kg is the estimated digit in the measurement, then
there are only two significant figures. Similarly, if the last zero is the esti-
mated digit (500 kg), then there are three significant figures. This ambiguity
may be removed by using scientific (powers-of-10) notation:
5.0 * 102 kg has two significant figures
5.00 * 102 kg has three significant figures
This notation is helpful in expressing the results of calculations with the
proper numbers of significant figures, as will be seen shortly. (Appendix I
includes a review of scientific notation.)
(Note: To avoid confusion regarding numbers having trailing zeros used
as given quantities in text examples and exercises, the trailing zeros will be
*Measurement error can arise because of a miscalibrated instrument and/or a personal error in
reading the instrument.
1.6 SIGNIFICANT FIGURES 19
considered significant. For example, assume that a time of 20 s has two sig-
nificant figures, even if it is not written out as 2.0 * 101 s.)
It is important to report the results of mathematical operations with the proper
number of significant figures. This is accomplished by using rules for (1) multipli-
cation and division and (2) addition and subtraction. To obtain the proper number
of significant figures, the results are rounded off. Here are some general rules that
will be used for mathematical operations and rounding.
SIGNIFICANT FIGURES IN CALCULATIONS

1. When multiplying and dividing quantities, leave as many significant figures
in the answer as there are in the quantity with the least number of significant
figures.
2. When adding or subtracting quantities, leave the same number of decimal
places (rounded) in the answer as there are in the quantity with the least
number of decimal places.
RULES FOR ROUNDING*

1. If the first digit to be dropped is less than 5, leave the preceding digit as is.
2. If the first digit to be dropped is 5 or greater, increase the preceding digit
by one.
The rules for significant figures mean that the result of a calculation can be no
more accurate than the least accurate quantity used. That is, accuracy cannot
be gained performing mathematical operations. Thus, the result that should be
reported for the division operation discussed at the beginning of this section is
(2 sf)
5.3 m
= 3.2 s (2 sf)
1.67 m>s
(3 sf)
The result is rounded off to two significant figures. (See Fig. 1.10.)
Applications of these rules are shown in the following Examples.
EXAMPLE 1.7 Using Significant Figures in Multiplication

and Division: Rounding Applications
The following operations are performed and the results rounded off to the proper
number of significant figures:
Multiplication:
2.4 m * 3.65 m = 8.76 m2 = 8.8 m2 (rounded to two sf)

(2 sf) (3 sf)
Division:
(4 sf)
725.0 m
= 5800 m>s = 5.80 * 103 m>s (represented with three sf; why?)
0.125 s
(3 sf)
F O L L O W - U P E X E R C I S E . Perform the following operations, and express the answers in

the standard powers-of-10 notation (one digit to the left of the decimal point) with the
proper number of significant figures: (a) 12.0 * 105 kg210.035 * 102 kg2 and
(b) 1148 * 10-6 m2>10.4906 * 10-6 m2. (Answers to all Follow-Up Exercises are given in
Appendix VI at the back of the book.)
*It should be noted that these rounding rules give an approximation of accuracy, as opposed to the
results provided by more advanced statistical methods.
EXAMPLE 1.8 Using Significant Figures in Addition and Subtraction: Application of Rules
The following operations are performed by finding the num- Subtraction:
ber that has the least number of decimal places. (Units have The same rounding procedure is used. Here, 157 has the least
been omitted for convenience.) number of decimal places (none).
Addition: 157
In the numbers to be added, note that 23.1 has the least num- -5.5
ber of decimal places (one): 151.5 " 152
(rounding off)
23.1
0.546 F O L L O W - U P E X E R C I S E . Given the numbers 23.15, 0.546, and
1.058, (a) add the first two numbers and (b) subtract the last
+ 1.45
" 25.1 number from the first. (Answers to all Follow-Up Exercises are
25.096 (rounding off) given in Appendix VI at the back of the book.)
Suppose that you have to deal with mixed operations—multiplication and>or divi-
sion and addition and>or subtraction. What do you do in this case? Just follow the reg-
ular rules for order of algebraic operations, and observe significant figures as you go.*
The number of digits reported in a result depends on the number of digits in
the given data. The rules for rounding will generally be observed in this book.
However, there will be exceptions that may make a difference, as explained in the
following Problem-Solving Hint.
PROBLEM-SOLVING HINT: THE “CORRECT” ANSWER
When working problems, you naturally strive to get the correct answer and will proba-
bly want to check your answers against those listed in the Answers to Odd-Numbered
Exercises section in the back of the book. However, on occasion, your answer may differ
slightly from that given, even though you have solved the problem correctly. There are
several reasons why this could occur.
It is best to round off only the final result of a multipart calculation, but this practice
is not always convenient in elaborate calculations. Sometimes, the results of intermedi-
ate steps are important in themselves and need to be rounded off to the appropriate
number of digits as if each were a final answer. Similarly, Examples in this book are often
worked in steps to show the stages in the reasoning of the solution. The results obtained
when the results of intermediate steps are rounded off may differ slightly from those
obtained when only the final answer is rounded.
Rounding differences may also occur when using conversion factors. For example, in
changing 5.0 mi to kilometers using the conversion factor listed inside the front cover of
this book in different forms,
5.0 mi a b = 18.045 km2 = 8.0 km (two significant figures)

1.609 km
1 mi
and
5.0 mi a b = 18.051 km2 = 8.1 km (two significant figures)

1 km
0.621 mi
The difference arises because of rounding of the conversion factors. Actually,
1 km = 0.6214 mi, so 1 mi = 11>0.62142 km = 1.609 269 km L 1.609 km. (Try repeating
these conversions with the unrounded factors, and see what you get.) To avoid rounding
differences in conversions, the multiplication form of a conversion factor will generally
be used, as in the first of the foregoing equations, unless there is a convenient exact fac-
tor, such as 1 min>60 s.
Slight differences in answers may also occur when different methods are used to
solve a problem. Keep in mind that when solving a problem, if your answer differs from
that in the text in only the last digit, the disparity is most likely the result of a rounding difference
for an alternative method of solution being used.
*Order of operations: (1) calculations done from left to right, (2) calculations inside parentheses,
(3) multiplication and division, (4) addition and subtraction.
1.7 PROBLEM SOLVING 21
DID YOU LEARN?

➥ There are exact numbers, such as the numeral coefficients in an equation, for
example, d = 2r.
➥ Measured values generally have some degree of uncertainty or error.
➥ The more finely divided an instrument’s measurement scale, the greater the
accuracy and number of significant figures.
1.7 Problem Solving

➥ What is the first step in problem solving?

➥ What is a final step in problem solving?
➥ What is meant by an order of magnitude?
An important aspect of physics is problem solving. In general, this involves the

application of physical principles and equations to data from a particular situation
in order to find some unknown or wanted quantity. There is no universal method
for approaching problem solving that will automatically produce a solution.
However, although there is no magic formula for problem solving, there are some
sound practices that can be very useful. The steps in the following procedure are
intended to provide you with a framework that can be applied to solving most of
the problems you will encounter during your course of study. (Modifications may
be made to suit your own style.)
These steps will generally be used in dealing with the Example problems through-
out the text. Additional problem-solving hints will be given where appropriate.
GENERAL PROBLEM-SOLVING STEPS

1. Read the problem carefully, and analyze it. What is given, and what is wanted?
2. Where appropriate, draw a diagram as an aid in visualizing and analyzing the phys- 1. Read the problem carefully
ical situation of the problem. This step may not be necessary in every case, but and analyze it.
it is often useful.
3. Write down the given data and what is to be found. Make sure the data are expressed
in the same system of units (usually SI). If necessary, use the unit conversion 2. Where appropriate,
draw a diagram.
procedure learned earlier in the chapter. Some data may not be given explic-
itly. For example, if a car “starts from rest,” its initial speed is zero 1vo = 02;
in some instances, you may be expected to know certain quantities, such as
3. Write down the given data and
the acceleration due to gravity, g, or can look them up in tables. what is to be found.
(Make unit conversions if necessary.)
4. Determine which principle(s) and equation(s) are applicable to the situation, and
how they can be used to get from the information given to what is to be found. You
may have to devise a strategy that involves several steps. Also, try to sim-
4. Determine which principle(s)
plify equations as much as possible through algebraic manipulation. The and equation(s) are applicable.
fewer calculations you do, the less likely you are to make a mistake—so don’t
put in numbers until you have to.
5. Substitute the given quantities (data) into the equation(s) and perform calculations. 5. Perform calculations with
Report the result with the proper units and proper number of significant given data.
figures.
6. Consider whether the results are reasonable. Does the answer have an appropri-
6. Consider whether the results
ate magnitude? (This means, is it in the right ballpark?) For example, if a per- are reasonable.
son’s calculated mass turns out to be 4.60 * 102 kg, the result should be
questioned, since a mass of 460 kg has a weight of 1010 lb. (Also, in motion
䉱 F I G U R E 1 . 1 1 A flowchart for
problems, direction may be important.) 䉴 Fig. 1.11 summarizes the main the suggested problem-solving
steps in the form of a flowchart. procedure
TABLE 1.4 Types of Examples

Example—primarily mathematical in nature
Sections: Thinking It Through
Solution
Integrated Example—(a) conceptual multiple choice, (b) mathematical follow-up
Sections: (a) Conceptual Reasoning
(b) Quantitative Reasoning and Solution
Conceptual Example—in general, needs only reasoning to obtain the answer, although
some simple math may be required at times to justify the reasoning
Sections: Reasoning and Answer
Pulling It Together—at the end of chapter, these examples demonstrate the use of
several different concepts. Their purpose is to create a Learning Bridge from the chapter
Learning Path to the End of Chapter Exercises, particularly the multiconcept type.
Sections: Same as Example or Integrated Example depending on the question(s)
asked
In general, there are four types of examples in this book, as listed in Table 1.4. The
preceding steps are applicable to the three types, because they include calculations.
Conceptual Examples, in general, do not follow these steps, being primarily concep-
tual in nature. The chapter Putting It Together is a multiconcept example.
In reading the worked Examples and Integrated Examples, you should be able
to recognize the general application or flow of the preceding steps. This format
will be used throughout the text. Let’s take an Example and an Integrated Exam-
ple as illustrations. Comments will be made in these examples to point out the
problem-solving approach and steps that will not be made in the text Examples,
but should be understood. Since no physical principles have really been covered,
math and trig problems will be used, which should serve as a good review.
EXAMPLE 1.9 Finding the Outside Surface Area of a Cylindrical Container

A closed cylindrical container used to store material from a
manufacturing process has an outside radius of 50.0 cm and a
height of 1.30 m. What is the total outside surface area of the
container? r = 50.0
cm
T H I N K I N G I T T H R O U G H . (In this type of Example, the Think-
ing It Through section generally combines problem-solving
steps 1 and 2 given previously.)
It should be noted immediately that the length units are
given in mixed units, so a unit conversion will be in order. To
visualize and analyze the cylinder, drawing a diagram is h =1.30 m Ab
helpful (䉴 Fig. 1.12). With this information in mind, proceed to
finding the solution, using the expression for the area of a 䉳 FIGURE 1.12
cylinder (the combined areas of the circular ends and the A helpful step in
cylinder’s side). problem solving
Ae Drawing a diagram
cylinder axis helps you visualize
and better under-
A = 2A e + A b stand the situation.
SOLUTION. Writing what is given and what is to be found conversions are not obvious, so going through the unit con-
(step 3 in our procedure): version for illustration:
r = 50.0 cm a b = 0.500 m
1m
Given: r = 50.0 cm Find: A (the total outside surface
100 cm
h = 1.30 m area of the cylinder)
There are general equations for areas (and volumes) of
First, let’s tend to the mixed units. You should be able in this commonly shaped objects. The area of a cylinder can be easily
case to immediately write r = 50.0 cm = 0.500 m. But often looked up (given in Appendix I-C), but suppose you didn’t
have such a source. In this case, you may be able to figure it Then the total area A is
out for yourself. Looking at Fig. 1.12, note that the outside
surface area of a cylinder consists of that of two circular ends A = 2A e + A b = 2pr2 + 2prh
and that of a rectangle (the body of the cylinder laid out flat).
The data could be put into the equation, but sometimes an
Equations for the areas of these common shapes are generally
equation may be simplified to save some calculation steps.
remembered. So the area of the two ends would be
2A e = 2 * pr2 (2 times the area of the circular end; A = 2pr1r + h2 = 2p10.500 m210.500 m + 1.30 m2
area of a cicle = pr 2) = p11.80 m22 = 5.65 m2
and the area of the body of the cylinder is The result appears reasonable considering the cylinder’s
A b = 2pr * h (circumference of circular end dimensions.
times height)
F O L L O W - U P E X E R C I S E . If the wall thickness of the cylinder’s side and ends is 1.00 cm, what is the inside volume of the cylinder?
(Answers to all Follow-Up Exercises are given in Appendix VI at the back of the book.)
INTEGRATED EXAMPLE 1.10 Sides and Angles*

(a) A gardener has a rectangular plot measuring 3.0 m * 4.0 m.
She wishes to use half of this area to make a triangular flower
bed. Of the two types of triangles shown in 䉴 Fig. 1.13, which u2
should she use to do this: (1) the right triangle, (2) the isosce-
les triangle—two sides equal, or (3) either one? (b) In laying
out the flower bed, the gardener decides to use a right trian- r
gle. Wishing to line the sides with rows of stone, she wants to 4.0 m
know the total length (L) of the triangle sides. She would also
like to know the values of the acute angles of the triangle. Can
you help her so she doesn’t have to do physical measure- right
triangle
ments? (Appendix I includes a review of trigonometric rela-
tionships as well as the marginal note on the next page.)
u1
(A) CONCEPTUAL REASONING. The rectangular plot has a total
2
area of 3.0 m * 4.0 m = 12 m . It is obvious that the right tri- 3.0 m
angle divides the plot in half (Fig. 1.13). This is not as obvious
for the isosceles triangle. But with a little study you should
see that the outside areas could be arranged such that their
combined area would be the same as that of the shaded
isosceles triangle. So the isosceles triangle also divides the
plot in half and the answer is (3). [This could be proven math-
ematically by computing the areas of the triangles.
Area = 12 1altitude * base2.] 4.0 m
(B) QUANTITATIVE REASONING AND SOLUTION. To find the
total length of the sides, the length of the hypotenuse of the isosceles
triangle is needed. This can be done using the Pythagorean triangle
theorem, x 2 + y 2 = r2, and
䉳 FIGURE 1.13
r = 2x 2 + y 2 = 213.0 m22 + 14.0 m22 = 225 m2 = 5.0 m A flower bed project
Two types of triangles
(Or directly, you may have noticed that this is a 3–4–5 right 3.0 m for a new flower bed.
triangle.) Then,
L = 3.0 m + 4.0 m + 5.0 m = 12.0 m Similarly,
u2 = tan-1 a b = 37°
3.0 m
The acute angles of the triangle can be found by using
4.0 m
trigonometry. Referring to the angles in Fig. 1.13,
side opposite The two angles add to 90° as would be expected with the
right angle 153° + 37° = 90°2.
4.0 m
tan u1 = =
side adjacent 3.0 m
F O L L O W - U P E X E R C I S E . What are the total length of the sides
and
and the interior angles for the isosceles triangle in Fig. 1.13?
u1 = tan-1 a b = 53°
4.0 m (Answers to all Follow-Up Exercises are given in Appendix VI at
3.0 m the back of the book.)
*Here and throughout the text, angles will be considered exact, that is, they do not determine the
number of significant figures.
Basic trigonometric functions: These examples illustrate how the problem-solving steps are woven into find-
side adjacent ing the solution of a problem. You will see this pattern throughout the solved
a b
x
cos u = examples in the text, although not as explicitly explained. Try to develop your
r hypotenuse
problem-solving skills in a similar manner.
y side opposite
sin u = a b
r hypotenuse
APPROXIMATION AND ORDER-OF-MAGNITUDE
y side opposite
= a b
sin u CALCULATIONS
tan u =
cos u x side adjacent
At times when solving a problem, you may not be interested in an exact answer,
but want only an estimate or a “ballpark” figure. Approximations can be made by
rounding off quantities so as to make the calculations easier and, perhaps, obtain-
able without the use of a calculator. For example, suppose you want to get an idea
of the area of a circle with radius r = 9.5 cm. Then, rounding 9.5 cm L 10 cm, and
p L 3 instead of 3.14,
A = pr2 L 3110 cm22 = 300 cm2
(Note that significant figures are not a concern in calculations involving approxi-
mations.) The answer is not exact, but it is a good approximation. Compute the
exact answer and see.
Powers-of-10, or scientific, notation is particularly convenient in making esti-
mates or approximations in what are called order-of-magnitude calculations.
Order of magnitude means that a quantity is expressed to the power of 10 closest to
the actual value. For example, in the foregoing calculation, approximating
9.5 cm L 10 cm is expressing 9.5 as 101, and we say that the radius is on the order of
10 cm. Expressing a distance of 75 km L 102 km indicates that the distance is on
the order of 102 km. The radius of the Earth is 6.4 * 103 km L 104 km, or on the
order of 104 km. A nanostructure with a width of 8.2 * 10-9 m is on the order of
10-8 m, or 10 nm. (Why an exponent of - 8?)
An order-of-magnitude calculation gives only an estimate, of course. But this
estimate may be enough to provide you with a better grasp or understanding of a
physical situation. Usually, the result of an order-of-magnitude calculation is pre-
cise within a power of 10, or within an order of magnitude. That is, the number (pre-
fix) multiplied by the power of 10 is somewhere between 1 and 10. For example,
if a length result of 105 km were obtained, it would be expected that the exact
answer was somewhere between 1 * 105 km and 10 * 105 km.
EXAMPLE 1.11 Order-of-Magnitude Calculation: Drawing Blood

A medical technologist draws 15 cc of blood from a patient’s practicality when dealing with small, whole-number quanti-
vein. Back in the lab, it is determined that this volume of blood ties in some situations. The cc abbreviation is commonly
has a mass of 16 g. Estimate the density of the blood in SI units. used in the medical and chemistry fields for cm3. Density
1r2 is mass per unit volume, where r = m>V (Section 1.4).
T H I N K I N G I T T H R O U G H . The data are given in cgs
(centimeter-gram-second) units, which are often used for
SOLUTION.
First, changing to SI standard units:
1 kg
Given: m = 16 g a b = 1.6 * 10-2 kg L 10-2 kg Find: estimate of r (density)
1000 g
3
V = 15 cm3 a b = 1.5 * 10-5 m3 L 10-5 m3
1m
102 cm
So, we have
m 10-2 kg
r = L = 103 kg>m3
V 10-5 m3
This result is quite close to the average density of whole blood, 1.05 * 103 kg>m3.
F O L L O W - U P E X E R C I S E . A patient receives 750 cc of whole standard units. (Answers to all Follow-Up Exercises are given in
blood in a transfusion. Estimate the mass of the blood, in Appendix VI at the back of the book.)
EXAMPLE 1.12 How Many Red Cells Are in Your Blood?

The blood volume in the human body varies with a person’s cells cells
age, body size, and sex. On average, this volume is about 5.0 L. cells>volume = 5.0 * 106 L 107
mm3 mm3
A typical value of red blood cells (erythrocytes) per volume is Then, changing to cubic meters,
5 000 000 15.0 * 1062 cells per cubic millimeter. Estimate how
cells 103 mm 3
a b L 1016
many red blood cells you have in your body. cells cells
M 107
volume mm3 1m m3
T H I N K I N G I T T H R O U G H . The red blood cell count in cells per
cubic millimeter is sort of a red blood cell “number density.” (Note: The conversion factor for liters to cubic meters was
Multiplying this figure by the total volume of blood obtained directly from the conversion tables, but there is no
[1cells>volume2 * total volume] will give the total number of conversion factor given for converting cubic millimeters to
cells. But note that the volumes must have the same units. First cubic meters, so a known conversion factor is cubed.) Then,
let’s start by converting 5.0 L to cubic meters (m3): 1 L = 10 - 3 m3.
a b 1total volume2 L a1016 b 110-2 m32
cells cells
(See inside front cover.)
volume m3
SOLUTION. = 1014 red blood cells
Given: V = 5.0 L Find: the approxi-
That’s a bunch of cells. Red blood cells (erythrocytes) are
3 mate number
= 5.0 L a 10 b
-3 m one of the most abundant cells in the human body.
of red blood
L
cells in the
= 5.0 * 10-3 m3 M 10-2 m3 body
F O L L O W - U P E X E R C I S E . The average number of white blood cells (leukocytes) in human blood is normally 5000 to 10 000 cells
per cubic millimeter. Estimate the number of white blood cells you have in your body. (Answers to all Follow-Up Exercises are
given in Appendix VI at the back of the book.)
DID YOU LEARN?

➥ It is essential to initially understand a problem.
➥ Answers should be checked to ensure that they are reasonable in magnitude, and
in some cases, direction.
➥ For an order of magnitude or a “ball park” figure, a number may be approximated
by expressing it to the closest power of ten.
PULLING IT TOGETHER Painting by Pythagoras

A painter uses a 10.0 ft ladder to reach the top area of a wall. (a) From the Pythagorean theorem, L2 = h2 + d2. Thus
d = 2L2 - h2 = 2110.0 ft22 - 18.50 ft22 = 5.27 ft
The ladder is to be placed so that its top supports are at a
height of 8.50 ft above the floor. (a) How far out from the wall
are the bottom ladder feet? (b) What angle does the ladder (b) From a sketch the angle is given by
make with respect to the vertical? (c) If the painter wants to
u = tan-1 a b = tan-1 a b = 31.8°
d 5.27 ft
adjust the ladder so that the top supports are 60.0 cm below h 8.50 ft
the initial 8.50 ft height, how much further (in cm) must the
floor feet be moved from the wall? (c) Since the answer is asked for in centimeters, let’s convert
all the dimensions into cm:
T H I N K I N G I T T H R O U G H . In part (a), if you make a quick 12.0 in. 2.54 cm
sketch, you can see that the third side of a right triangle is L = 10.0 ft * * = 305 cm
ft in.
being asked for, so the Pythagorean theorem applies. In (b), a 12.0 in. 2.54 cm
trig function is clearly appropriate. (c) This is a repeat of (a) h = 8.50 ft * * = 259 cm
ft in.
but a conversion of units is needed.
12.0 in. 2.54 cm
SOLUTION. d = 5.27 ft * * = 161 cm
ft in.
Given: L = 10.0 ft Find: (a) d (distance ladder Then the new distance of the top supports up the wall is
(triangle hypotenuse) feet from wall)
h¿ = h - ¢h = 259 cm - 60.0 cm = 199 cm
h = 8.50 ft (b) angle of ladder
Applying the Pythagorean theorem one more time to find the
(initial height of top lad- from vertical
new distance d¿ out from the wall and the change in that distance:
der supports on wall)
d¿ = 2L2 - 1h¿22 = 21305 cm22 - 1199 cm22 = 231 cm
(c) ¢d (distance
¢h = 60.0 cm ladder feet
(distance top ladder move out) and
supports move down)
¢d = d¿ - d = 231 cm - 161 cm = 70 cm
Learning Path Review
■ SI units of length, mass, and time. The meter (m), the kilo- ■ Unit analysis. Unit analysis can be used to determine the
gram (kg), and the second (s), respectively. consistency of an equation, that is, if the equation is dimen-
LENGTH: METER
sionally correct, but not if physically correct. Unit analysis
can also be used to find the unit of a quantity.
■ Significant figures (digits). The digits that are known with
certainty, plus one digit that is uncertain, in a measured
1m
1 m = distance traveled by light in a
value.
vacuum in 1/299 792 458 s
■ Problem solving. Problems should be worked using a con-
MASS: KILOGRAM
sistent procedure. Order-of-magnitude calculations may be
done when an estimated value is desired.
Suggested Procedure for Problem Solving:
1. Read the problem carefully and analyze it.
0.10 m 2. Where appropriate draw a diagram.
water 3. Write down the given data and what is to be found.
(Make unit conversions if necessary.)
0.10 m 4. Determine which principle(s) and equation(s) are
0.10 m applicable.
One frequency oscillation
5. Perform calculations with given data.
6. Consider whether the results are reasonable.
■ Density (R). The mass per unit volume of an object or sub-
stance, which is a measure of the compactness of the mater-
Cesium-133 1 s = 9 192 631 770 oscillations Radiation ial it contains:
detector
a b
m mass
r =
■ Liter (L). A volume of 10 cm * 10 cm * 10 cm = 1000 cm3 V volume
or 1000 mL. A liter of water has a mass of 1 kg or 1000 g.
Therefore, 1 cm3 or 1 mL has a mass of 1 gram.
Volume
1 L = 1.06 qt
1L
1 qt = 0.947 L
1 qt
Learning Path Questions and Exercises For instructor-assigned homework, go to www.masteringphysics.com

MULTIPLE CHOICE QUESTIONS
1.2 SI UNITS OF LENGTH, MASS, 7. A new technology is concerned with objects the size of
AND TIME what metric prefix: (a) nano-, (b) micro-, (c) mega-, or
(d) giga-?
1. How many base units are there in the SI: (a) 3, (b) 5, (c) 7,
or (d) 9? 8. Which of the following has the greatest volume: (a) 1 L,
2. The only SI standard represented by material standard (b) 1 qt, (c) 2000 mL, or (d) 2000 mL?
or artifact is the (a) meter, (b) kilogram, (c) second, 9. Which of the following metric prefixes is the smallest:
(d) electric charge. (a) micro-, (b) centi-, (c) nano-, or (d) milli-?
3. Which of the following is the SI base unit for mass:
(a) pound, (b) gram, (c) kilogram, or (d) ton?
1.4 UNIT ANALYSIS
4. Which of the following is not related to a volume of
water: (a) kilogram, (b) pound, (c) gram, or (d) tonne? 10. Both sides of an equation are equal in (a) numerical
value, (b) units, (c) dimensions, (d) all of the preceding.
1.3 MORE ABOUT THE METRIC SYSTEM 11. Unit analysis of an equation cannot tell you if (a) the
equation is dimensionally correct, (b) the equation is
5. The prefix giga- means (a) 10-9, (b) 109, (c) 10-6, (d) 106. physically correct, (c) the numerical value is correct,
6. The prefix micro- means (a) 106, (b) 10-6, (c) 103, (d) 10-3. (d) both b and c.
CONCEPTUAL QUESTIONS 27
1.5 UNIT CONVERSIONS 17. In a multiplication and>or division operation involving

the numbers 15 437, 201.08, and 408.0 * 105, the result
12. A good way to ensure proper unit conversion is to
should have how many significant figures: (a) 3, (b) 4,
(a) use another measurement instrument, (b) always
(c) 5, or (d) any number?
work in the same system of units, (c) use unit analysis,
(d) have someone check your math.
1.7 PROBLEM SOLVING
13. You often see 1 kg = 2.2 lb. This expression means that
(a) 1 kg is equivalent to 2.2 lb, (b) this is a true equation, 18. An important step in problem solving before mathemati-
(c) 1 lb = 2.2 kg, (d) none of the preceding. cally solving an equation is (a) checking units, (b) check-
14. You have a quantity of water and wish to express this in ing significant figures, (c) checking with a friend,
volume units that give the largest number. Which of the (d) checking to see if the result will be reasonable.
following units should be used: (a) in3, (b) mL, (c) mL, or 19. An important final step in problem solving before report-
(d) cm3? ing an answer is (a) saving your calculations, (b) reading
the problem again, (c) seeing if the answer is reasonable,
1.6 SIGNIFICANT FIGURES (d) checking your results with another student.
15. Which of the following has the greatest number of signif- 20. In order-of-magnitude calculations, you should (a) pay
icant figures: (a) 103.07, (b) 124.5, (c) 0.09916, or close attention to significant figures, (b) work primarily
(d) 5.408 * 105? in the British system, (c) get results within a factor of 100,
(d) express a quantity to the power of 10 closest to the
16. Which of the following numbers has four significant fig-
actual value.
ures: (a) 140.05, (b) 276.02, (c) 0.004 006, or (d) 0.073 004?
CONCEPTUAL QUESTIONS
1.2 SI UNITS OF LENGTH, MASS, 13. Does it make any difference whether you multiply or
AND TIME divide by a conversion factor? Explain.
14. A popular saying is “Give him an inch and he’ll take a
1. Why are there not more SI base units?
mile.” What would be the equivalent numerical values
2. Why is weight not a base quantity? and units in the metric system?
3. What replaced the original definition of the second and
why? Is the replacement still used? 1.6 SIGNIFICANT FIGURES
4. Give a couple of major differences between the SI and 15. What is the purpose of significant figures?
the British system.
16. Are all the significant figures reported for a measured
value accurately known? Explain.
1.3 MORE ABOUT THE METRIC SYSTEM 17. How are the number of significant figures determined
5. If a fellow student tells you he saw a 3-cm-long ladybug, for the results of calculations involving (a) multiplica-
would you believe him? How about another student tion, (b) division, (c) addition, and (d) subtraction?
saying she caught a 10-kg salmon? 18. Why is 5 chosen to be the major digit for rounding?
6. Explain why 1 mL is equivalent to 1 cm3.
7. Explain why a metric ton is equivalent to 1000 kg. 1.7 PROBLEM SOLVING
19. What are the main steps in the problem-solving proce-
1.4 UNIT ANALYSIS dure suggested in this chapter?
8. Can unit analysis tell you whether you have used the 20. When you do order-of-magnitude calculations, should
correct equation in solving a problem? Explain. you be concerned about significant figures? Explain.
21. When doing an order-of-magnitude calculation, how
9. The equation for the area of a circle from two sources is
given as A = pr 2 and A = pd2>2. Can unit analysis tell
accurate can you expect the answer to be? Explain.
you which is correct? Explain. 22. The largest organ of the human body is the skin. The
total external skin area of the average human covers an
10. How might unit analysis help determine the units of a area of approximately 2.0 m2. If you were asked to com-
quantity? pute the approximate skin area, how would you go
11. Why is p unitless? about it? (Hint: see Example 8.18.)
23. Is the following statement reasonable? It took 300 L of
gasoline to fill the car’s tank. (Justify your answer.)
1.5 UNIT CONVERSIONS
24. Is the following statement reasonable? A car traveling
12. Are an equation and an equivalence statement the same? 30 km>h through a school speed zone exceeds the speed
Explain. limit of 25 mi>h. (Justify your answer.)
EXERCISES*
Integrated Exercises (IEs) are two-part exercises. The first part typically requires a conceptual answer choice based
on physical thinking and basic principles. The following part is quantitative calculations associated with the con-
ceptual choice made in the first part of the exercise.
Many exercise sections include “paired” exercises. These exercise pairs, identified with red numbers, are
intended to assist you in problem solving and learning. In a pair, the first exercise (even numbered) is worked out in
the Study Guide so that you can consult it should you need assistance in solving it. The second exercise (odd num-
bered) is similar in nature, and its answer is given in Appendix VII at the back of the book.
1.3 MORE ABOUT THE METRIC SYSTEM 14. ●●The units for pressure (p) in terms of SI base units are
kg
1. ● The metric system is a decimal (base-10) system, and known to be . For a physics class assignment, a
the British system is, in part, a duodecimal (base-12) sys- m # s2
student derives an expression for the pressure exerted by
the wind on a wall in terms of the air density 1r2 and wind
tem. Discuss the ramifications if our monetary system
had a duodecimal base. What would be the possible val-
ues of our coins if this were the case? speed (v) and her result is p = rv2. Use SI unit analysis to
show that her result is dimensionally consistent. Does this
2. ● (a) In the British system, 16 oz = 1 pt and 16 oz = 1 lb.
prove that this relationship is physically correct?
Is something wrong here? Explain. Here’s an old one: A
15. ● ● Is the equation for the area of a trapezoid,
A = 12 a1b1 + b22, where a is the height and b1 and b2 are
pound of feathers weighs more than a pound of gold.
How can that be? [Hint: Look up ounce in the dictionary.]
the bases, dimensionally correct? (䉲 Fig. 1.14.)
3. ● Convert the following: (a) 40 000 000 bytes to MB,
(b) 0.5722 mL to L, (c) 2.684 m to cm, and (d) 5 500 bucks b1 䉳 F I G U R E 1 . 1 4 The
to kilobucks. area of a trapezoid See
4. ● ● A sailor tells you that if his ship is traveling at 25 Exercise 15.
knots (nautical miles per hour), it is moving faster than a
the 25 mi>h your car travels. How can that be?
5. ● ● A rectangular container measuring 25 cm * 35 cm * b2
55 cm is filled with water. What is the mass of this volume
of water? 16. ● ● ● Newton’s second law of motion (Section 4.3) is
6. ● ● What size cube (in centimeters) would have a vol- expressed by the equation F = ma, where F represents
ume equal to that of a quart? force, m is mass, and a is acceleration. (a) The SI unit of
7. ● ● (a) What volume in liters is a cube 20 cm on a side? force is, appropriately, called the newton (N). What are
(b) If the cube is filled with water, what is the mass of the the units of the newton in terms of base quantities?
water? (b) An equation for force associated with uniform circu-
lar motion (Section 7.3) is F = mv2>r, where v is speed
and r is the radius of the circular path. Does this equa-
1.4 UNIT ANALYSIS
tion give the same units for the newton?
8. ● Show that the equation x = xo + vt, where v is velocity,
17. ● ● ● The angular momentum (L) of a particle of mass m
x and xo are lengths, and t is time, is dimensionally correct. moving at a constant speed v in a circle of radius r is given
9. ● If x refers to distance, vo and v to velocities, a to accelera- by L = mvr (Section 8.5). (a) What are the units of angular
tion, and t to time, which of the following equations is momentum in terms of SI base units? (b) The units of
dimensionally correct: (a) x = vo t + at3, (b) v2 = v2o + 2at, kg # m2
(c) x = at + vt2, or (d) v2 = v2o + 2ax? kinetic energy in terms of SI base units are .
s2
10. ● ● Use SI unit analysis to show that the equation Using SI unit analysis, show that the expression for the
A = 4pr2, where A is the area and r is the radius of a kinetic energy of this particle in terms of its angular
sphere, is dimensionally correct. L2
momentum, K = , is dimensionally correct. (c) In
11. ● ● The general equation for a parabola is
2mr2
y = ax 2 + bx + c, where a, b, and c are constants. What the previous equation, the term mr2 is called the moment
are the units of each constant if y and x are in meters? of inertia of the particle in the circle. What are the units of
12. ● ● You are told that the volume of a sphere is given by moment of inertia in terms of SI base units?
V = pd3>4, where V is the volume and d is the diameter 18. ● ● ● Einstein’s famous mass–energy equivalence
of the sphere. Is this equation dimensionally correct? is expressed by the equation E = mc 2, where E is energy,
(Use SI unit analysis to find out.) m is mass, and c is the speed of light. (a) What are the SI
13. ● ● The correct equation for the volume of a sphere is base units of energy? (b) Another equation for energy is
V = 4pr3>3, where r is the radius of the sphere. Is the E = mgh, where m is mass, g is the acceleration due to
equation in Exercise 12 correct? If not, what should it be gravity, and h is height. Does this equation give the same
when expressed in terms of d? units as in part (a)?
*Keep in mind here and throughout the text that your answer to an odd-numbered exercise may differ slightly from that given in Appendix VII at
the back of the book because of rounding. See the Problem-Solving Hint: The “Correct” Answer in this chapter.
EXERCISES 29
1.5 UNIT CONVERSION 30. ● ● An automobile speedometer is shown in 䉲 Fig. 1.16.
(a) What would be the equivalent scale readings (for

19. ● Figure 1.8 (top) shows the elevation of a location in
each empty box) in kilometers per hour? (b) What would
both feet and meters. Is the conversion correct?
be the 70-mi>h speed limit in kilometers per hour?
20. IE ● (a) If you wanted to express your height with the
largest number, which units would you use: (1) meters, meters per hour
(2) feet, (3) inches, or (4) centimeters? Why? (b) If you are Kilo
6.00 ft tall, what is your height in centimeters?
50
21. ● If the capillaries of an average adult were unwound and 40 per h 60
spread out end to end, they would extend to a length over iles ou
SPEED
30 M 70
r
40 000 mi (Fig. 1.9). If you are 1.75 m tall, how many times LIMIT
your height would the capillary length equal? 20 80 70 mi/h
22. IE ● (a) Compared with a 2-L soda bottle, a half-gallon 10 90 km/h

soda bottle holds (1) more, (2) the same amount of,
(3) less soda. (b) Verify your answer for part (a). 0 mi/h 100
23. ● (a) A football field is 300 ft long and 160 ft wide. What km/h 0
are the field’s dimensions in meters? (b) A football is 11.0
to 1114 in. long. What is its length in centimeters? 䉱 F I G U R E 1 . 1 6 Speedometer readings See Exercise 30.
24. ● Suppose that when the United States goes completely
metric, the dimensions of a football field are established 31. ●● A person weighs 170 lb. (a) What is his mass in kilo-
as 100 m by 54 m. Which would be larger, the metric grams? (b) Assuming the density of the average human
football field or a current football field (see Exercise 23a), body is about that of water (which is true), estimate his
and what would be the difference between the areas? body’s volume in both cubic meters and liters. Explain
why the smaller unit of the liter is more appropriate
25. ●Water is sold in pint bottles. What is the mass of the (convenient) for describing a volume of this size.
water in a full bottle?
32. ●● If the components of the human circulatory system
26. ● How many (a) quarts and (b) gallons are there in 10.0 L? (arteries, veins, and capillaries) were completely
27. ● A submarine is submerged 175 fathoms below the sur- extended and placed end to end, the length would be on
face. What is its depth in meters? (A fathom is an old nau- the order of 100 000 km. Would the length of the circula-
tical measurement equal to 2 yd.) tory system reach around the circumference of the
Moon? If so, how many times?
28. ●● Driving a jet-powered car, Royal Air Force pilot Andy
Green broke the sound barrier on land for the first time 33. ●● The human heartbeat, as determined by the pulse
and achieved a record land speed of more than 763 mi>h rate, is normally about 60 beats>min. If the heart pumps
in Black Rock Desert, Nevada, on October 15, 1997 75 mL of blood per beat, what volume of blood is
(䉲Fig. 1.15). (a) What is this speed expressed in m>s? pumped in one day in liters?
(b) How long would it take the jet-powered car to travel 34. ●● Some common product labels are shown in 䉲Fig. 1.17.
the length of a 300-ft football field at this speed? (Hint: From the units on the labels, find (a) the number of milli-
v = d>t.) liters in 2 fl. oz and (b) the number of ounces in 100 g.
䉱 F I G U R E 1 . 1 7 Conversion factors See Exercise 34.
䉱 F I G U R E 1 . 1 5 Record run See Exercise 28. 35. ●● 䉲 Fig. 1.18 is a picture of red blood cells seen under a
scanning electron microscope. Normally, women possess
about 4.5 million of these cells in each cubic millimeter of
29. IE ● ● (a) Which of the following represents the greatest blood. If the blood flow to the heart is 250 mL>min, how
speed: (1) 1 m>s, (2) 1 km>h, (3) 1 ft>s, or (4) 1 mi>h? many red blood cells does a woman’s heart receive each
(b) Express the speed 15.0 m>s in mi>h. second?
42. ● Using a meterstick, a student measures a length and

reports it to be 0.8755 m. What is the smallest division on
the meterstick scale?
43. ● Determine the number of significant figures in the fol-
lowing measured numbers: (a) 1.007 m, (b) 8.03 cm,
(c) 16.272 kg, (d) 0.015 ms (microseconds).
44. ● Express each of the numbers in Exercise 43 with two
significant figures.
45. ● Round the following numbers to two significant fig-
ures: (a) 95.61, (b) 0.00208, (c) 9438, (d) 0.000344
䉳 FIGURE 1.18 46. ● Which of the following quantities has three significant
Red blood cells figures: (a) 305.0 cm, (b) 0.0500 mm, (c) 1.000 81 kg,
See Exercise 35. (d) 8.06 * 104 m2?
47. ●● The cover of your physics book measures 0.274 m
36. ●● A student was 18 in. long when she was born. She is long and 0.222 m wide. What is its area in square
now 5 ft 6 in. tall and 20 years old. How many centime- meters?
ters a year did she grow on average?
48. ●● The interior storage compartment of a restaurant
37. ●●● How many minutes of arc does the Earth rotate in
refrigerator measures 1.3 m high, 1.05 m wide, and 67 cm
1 min of time?
deep. Determine its volume in cubic feet.
38. ●●● The density of metal mercury is 13.6 g>cm3.
49. IE ● ● The top of a rectangular table measures 1.245 m by
(a) What is this density as expressed in kilograms per
0.760 m. (a) The smallest division on the scale of the
cubic meter? (b) How many kilograms of mercury
measurement instrument is (1) m, (2) cm, (3) mm. Why?
would be required to fill a 0.250-L container?
(b) What is the area of the tabletop?
39. ● ● ● The Roman Coliseum used to be flooded with water
to re-create ancient naval battles. Assuming the circular 50. IE ● ● The outside dimensions of a cylindrical soda can
floor be 250 m in diameter and the water to have a depth are reported as 12.559 cm for the diameter and 5.62 cm
of 10 ft, (a) how many cubic meters of water are for the height. (a) How many significant figures will the
required? (b) How much mass would this water have in total outside area have: (1) two, (2) three, (3) four, or
kilograms? (c) How much would the water weigh in (4) five? Why? (b) What is the total outside surface area
pounds? of the can in square centimeters?
40. ● ● ● In the Bible, Noah is instructed to build an ark 300

51. ●● Express the following calculations using the proper
cubits long, 50.0 cubits wide, and 30.0 cubits high number of significant figures: (a) 12.634 + 2.1,
(䉲Fig. 1.19). Historical records indicate a cubit is equal to (b) 13.5 - 2.134, (c) p10.25 m22, (d) 12.37>3.5
half a yard. (a) What would be the dimensions of the ark 52. IE ● ● ● In doing a problem, a student adds 46.9 m and
in meters? (b) What would be the ark’s volume in cubic 5.72 m and then subtracts 38 m from the result. (a) How
meters? To approximate, assume that the ark is to be many decimal places will the final answer have: (1) zero,
rectangular. (2) one, or (3) two? Why? (b) What is the final answer?
53. ● ● ● Work this exercise by the two given procedures as
directed, commenting on and explaining any difference

50.0 in the answers. Use your calculator for the calculations.
cubits Compute p = mv, where v = x>t, given x = 8.5 m,
30.0
cubits t = 2.7 s, and m = 0.66 kg. (a) First compute v and then
p. (b) Compute p = mx>t without an intermediate step.
300 cubits (c) Are the results the same? If not, why?
1.7 PROBLEM SOLVING

54. ● A corner construction lot has the shape of a right
triangle. If the two sides perpendicular to each other
are 37 m long and 42.3 m long, what is the length of the
hypotenuse?
55. ●● The lightest solid material is silica aerogel, which has
䉱 F I G U R E 1 . 1 9 Noah and his ark See Exercise 40. a typical density of only about 0.10 g>cm3. The molecu-
lar structure of silica aerogel is typically 95% empty
space. What is the mass of 1 m3 of silica aerogel?
1.6 SIGNIFICANT FIGURES
56. ●● A cord of wood is a volume of cut wood equal to a
41. ● Express the length 50 500 mm (micrometers) in centime- stack 8.0 ft long, 4.0 ft wide, and 4.0 ft high. How many
ters, decimeters, and meters, to three significant figures. cords are there in 3.0 m3?
EXERCISES 31
57. ●● Nutrition Facts labels now appear on most foods. 63. ●●Two students go into Tony’s Pizza Palace and order a
An abbreviated label concerned with fat is shown in 12-in. (diameter) pizza. Shortly thereafter, the waitress
䉲 Fig. 1.20. When burned in the body, each gram of fat brings an 8-in. pizza special. She explains that the 12-in.
supplies 9 Calories. (A food Calorie is really a kilocalo- pizza was given to someone else by mistake and they
rie, as will be learned in Chapter 11.) (a) What percent- could have the 8-in. now and she would bring another
age of the Calories in one serving is supplied by fat? 8-in. shortly to make up for the missing 12-in. pizza. Was
(b) You may notice that our answer doesn’t agree with this a good deal?
the listed Total Fat percentage in Fig. 1.20. This is 64. ● ● In 䉲 Fig. 1.22, which black region has the greater area,
because the given Percent Daily Values are the percent- the center circle or the outer ring?
ages of the maximum recommended amounts of nutri-
ents (in grams) contained in a 2000-Calorie diet. What
are the maximum recommended amounts of total fat
and saturated fat for a 2000-Calorie diet?
3.32 cm
1.28 cm
Nutrition Facts
Serving Size: 1 can
Calories: 310 3.56 cm
Amount Per Serving % Daily Value*
Total Fat 18 g 28% 䉱 F I G U R E 1 . 2 2 Which black area is greater? See Exercise 64.
Saturated Fat 7g 35%
65. ●● The Channel Tunnel, or “Chunnel,” which runs
under the English Channel between Great Britain and
France, is 31 mi long. (There are actually three separate
䉳 FIGURE 1.20
* Percent Daily Values are based on a 2,000 tunnels.) A shuttle train that carries passengers through
Calorie diet. Nutrition Facts
See Exercise 57. the tunnel travels with an average speed of 75 mi>h. On
average, how long, in minutes, does the shuttle take to
58. ●● The thickness of the numbered pages of a textbook is make a one-way trip through the Chunnel?
measured to be 3.75 cm. (a) If the last page of the book is 66. ●● Human adult blood contains, on average, 7000>mm3
numbered 860, what is the average thickness of a page? white blood cells (leukocytes) and 250 000>mm3 platelets
(b) Repeat the calculation by using order-of-magnitude (thrombocytes). If a person has a blood volume of 5.0 L,
calculations. estimate the total number of white cells and platelets in
the blood.
59. ●● The mass of the Earth is 5.98 * 1024 kg. What is the
average density of the Earth in standard units? 67. ●● The average number of hairs on the normal human
scalp is 125 000. A healthy person loses about 65 hairs
60. IE ● ● To go to a football stadium from your house, you
per day. (New hair from the hair follicle pushes the old
first drive 1000 m north, then 500 m west, and finally
hair out.) (a) How many hairs are lost in one month?
1500 m south. (a) Relative to your home, the football sta-
(b) Pattern baldness (top-of-the-head hair loss) affects
dium is (1) north of west, (2) south of east, (3) north of
about 35 million men in the United States. If an average
east, (4) south of west. (b) What is the straight-line dis-
of 15% of the scalp is bald, how many hairs are lost per
tance from your house to the stadium?
year by one of these “bald is beautiful” people?
61. ●● Two chains of length 1.0 m are used to support a 68. IE ● ● A car is driven 13 mi east and then a certain dis-
lamp, as shown in 䉲 Fig. 1.21. The distance between the tance due north, ending up at a position 25° north of east
two chains along the ceiling is 1.0 m. What is the vertical of its initial position. (a) The distance traveled by the car
distance from the lamp to the ceiling? due north is (1) less than, (2) equal to, (3) greater than 13
mi. Why? (b) What distance due north does the car travel?
1.0 m
69. IE ● ● ● At the Indianapolis 500 time trials, each car
makes four consecutive laps, with its overall or average
speed determining that car’s place on race day. Each lap
1.0 m 1.0 m covers 2.5 mi (exact). During a practice run, cautiously
and gradually taking his car faster and faster, a driver
records the following average speeds for each successive
lap: 160 mi>h, 180 mi>h, 200 mi>h, and 220 mi>h. (a) Will
his average speed be (1) exactly the average of these
speeds 1190 mi>h2, (2) greater than 190 mi>h, or (3) less
than 190 mi>h? Explain. (b) To corroborate your concep-
䉱 F I G U R E 1 . 2 1 Support the lamp See Exercise 61. tual reasoning, calculate the car’s average speed.
70. ● ● ● Approximately 118 mi wide, 307 mi long, and aver-
62. ●● Tony’s Pizza Palace sells a medium 9.0-in. (diameter) aging 279 ft in depth, Lake Michigan is the second-
pizza for $7.95, and a large 12-in. pizza for $13.50. Which largest Great Lake by volume. Estimate its volume of
pizza is the better buy? water in cubic meters.
71. IE ● ● ● In the Tour de France, a bicyclist races up two

successive (straight) hills of different slope and length.
The first is 2.00 km long at an angle of 5° above the hori- Island
zontal. This is immediately followed by one 3.00 km
long at 7°. (a) What will be the overall (net) angle from
start to finish: (1) smaller than 5°, (2) between 5° and 7°, ?
or (3) greater than 7°? (b) Calculate the actual overall
30° 40°
(net) angle of rise experienced by this racer from start to 50 m
finish, to corroborate your reasoning in part (a).
72. ● ● ● A student wants to determine the distance from the Beach
lakeshore to a small island (䉴Fig. 1.23). He first draws a
50-m line parallel to the shore. Then, he goes to the ends of
the line and measures the angles of the lines of sight from
the island relative to the line he has drawn. The angles are 䉱 F I G U R E 1 . 2 3 Measuring with lines of sight See Exercise 72
30° and 40°. How far is the island from the shore?
PULLING IT TOGETHER: MULTICONCEPT EXERCISES
The Multiconcept Exercises require the use of more than one fundamental concept for their understanding and
solution from this chapter.
73. A farmer owns a piece of land in the shape of an equilat- cylindrical hole is completely filled with plastic (with a
eral triangle, 200 m on a side, which is totally fenced in. density twice that of water), determine the overall (aver-
He wishes to construct an additional fence parallel to the age) density of the brick>plastic combination after fabri-
side fronting the road (䉲 Fig. 1.24) so that the area cation is complete.
fronting the road takes up one-third of the total area.
This area will be for his horses. On the remaining two- 75. A spherical shell is formed by taking a solid sphere of
thirds he plans to construct his dream home. How far radius 20.0 cm and hollowing out a spherical section from
back from the road (shown as the distance h) should the the shell’s interior. Assume the hollow section and the
fence be located? sphere itself have the same center location (that is, they are
concentric). (a) If the hollow section takes up 90.0 percent
of the total volume, what is its radius? (b) What is the ratio
of the outer area to the inner area of the shell?
76. Two separate seismograph stations receive indication of
an earthquake in the form of a wave traveling to them in
20
0m
a straight line from the epicenter and shaking the

0m
20
ground at their locations. Station B is 50 km due east of

station A. The epicenter is located due north of station A
and 30° north of due west from station B. (a) Draw a
h sketch and use it to determine the distance from the epi-
center to A. (b) Determine the distance from the epicen-
ter to B. (c) Station C is located an additional 20 km east
of B. At what angle does C report the direction of the epi-
center to be?
(Top view)
77. You are sailing a radio-controlled model powerboat on a
䉱 F I G U R E 1 . 2 4 Don’t fence me in See Exercise 73. perfectly circular pool of water. The boat travels at a con-
stant 0.500 m>s. It takes 30.0 s to make the trip from one
side of the pool, through the center, to the other side.
74. In a radioactivity experiment, a solid lead brick (with
(a) How long would it take the boat to travel completely
same measurements as a patio brick, 2.00 in. * 4.00 in.
around the edge of the pool? (b) If the pool is uniformly
* 8.00 in., except with a density that is 11.4 times that of
1.50 m deep, how many gallons of water does it hold?
water) is to be modified to hold a solid cylindrical piece
of plastic. To accomplish this, the machinists are told to 78. A certain material has a density of 9.0 g>cm3. It is formed
drill a cylindrical hole 2.0 cm in diameter through into a solid rectangular brick with dimensions
the center of the brick parallel to the longest side of the 1.0 cm * 2.0 cm * 4.0 cm. (a) What is its mass in kilo-
brick. (a) What is the mass of lead (in kilograms) grams? (b) If you wanted to make a cube of this same
removed from the brick? (b) What percentage of the material containing twice the mass of this brick, what
original lead remains in the brick? (c) Assuming the would be the length of one side of the cube?
Kinematics:
2 Description of Motion †
2.1 Distance and speed:
scalar quantities (34)
■ magnitude only
2.2 One-dimensional
displacement and velocity:
vector quantities (36)
■ magnitude and
direction
2.3 Acceleration (42)

■ time rate of change
of velocity
2.4 Kinematic equations

PHYSICS FACTS
T
(constant acceleration) (46)
■ description of motion ✦ “Give me matter and motion and I
will construct the universe.” Rene he cheetah is running at full
Descartes (1640)
stride in the chapter-opening
✦ Nothing can exceed the speed of
2.5 Free fall (50) light (in vacuum), photo. This fastest of all land ani-
solely under the influence 3.0 * 108 m>s (186 000 mi>s).
■
mals is capable of attaining speeds
of gravity ✦ A bullet from a high-powered rifle
travels at a speed of about up to 113 km>h, or 70 mi>h. The
2900 km>h (1800 mi>h).
✦ NASA’s X-43A uncrewed jet flew at
sense of motion in the photograph
a speed of 7700 km>h (4800 mi>h) is so strong that you can almost feel
—faster than a speeding bullet.
✦ Electrical signals between your
the air rushing by. And yet this
brain and muscles travel at about sense of motion is an illusion.
435 km>h (270 mi>h).
✦ A person at the equator is traveling
Motion takes place in time, but the
at a speed of 1600 km>h (1000 mi>h) photo can “freeze” only a single
due to the Earth’s rotation.
✦ Aristotle thought heavy objects fall instant. You’ll find that without the
faster than lighter ones. Galileo
dimension of time, motion cannot
wrote, “Aristotle says that an iron
ball falling from a height of one be described at all.
hundred cubits reaches the
† ground before a one-pound ball The description of motion
The mathematics needed in this chapter
has fallen a single cubit. I say they
involves general algebraic equation manipu- involves the representation of a
lation. You may want to review this in arrive at the same time.”
Appendix I. restless world. Nothing is ever
34 2 KINEMATICS: DESCRIPTION OF MOTION
perfectly still. You may sit, apparently at rest, but your blood flows, and air moves
into and out of your lungs. The air is composed of gas molecules moving at differ-
ent speeds and in different directions. And while experiencing stillness, you, your
chair, the building you are in, and the air you breathe are all rotating and revolving
through space with the Earth, part of a solar system in a spiraling galaxy in an
expanding universe.
The branch of physics concerned with the study of motion and what produces
and affects motion is called mechanics. The roots of mechanics and of human
interest in motion go back to early civilizations. The study of the motions of heav-
enly bodies, or celestial mechanics, grew out of measuring time and location. Sev-
eral early Greek scientists, notably Aristotle, put forth theories of motion that were
useful descriptions, but were later proved to be incomplete or incorrect. Our cur-
rently accepted concepts of motion were formulated in large part by Galileo
(1564–1642) and Isaac Newton (1642–1727).
Mechanics is usually divided into two parts: (1) kinematics and (2) dynamics.
Kinematics deals with the description of the motion of objects, without considera-
tion of what causes the motion. Dynamics analyzes the causes of motion. This
chapter covers kinematics and reduces the description of motion to its simplest
terms by considering linear motion, that is, motion in a straight line. You’ll learn to
analyze changes in motion—speeding up, slowing down, and stopping. Along the
way, a particularly interesting case of accelerated motion will be presented: free fall
(motion under the influence of gravity only).
2.1 Distance and Speed: Scalar Quantities

State ➥ Why is distance a scalar quantity?
Podunk University
➥ What is the difference between average speed and instantaneous speed?
48 km (30 mi) ➥ When are the average and instantaneous speeds equal?
i)
m
50 DISTANCE
81 km (
97
Motion is observed all around us. But what is motion? This question seems sim-
km
60 ple; however, you might have some difficulty giving an immediate answer. (And,
(
m it’s not fair to use forms of the verb to move to describe motion.) After a little
i)
thought, you should be able to conclude that motion (or moving) involves changing
Hometown position. Motion can be described in part by specifying how far something travels in
changing position—that is, the distance it travels. Distance is simply the total path
䉱 F I G U R E 2 . 1 Distance—total length traversed in moving from one location to another. For example, you may
path length In driving to State Uni- drive to school from your hometown and express the distance traveled in miles or
versity from Hometown, one stu-
dent may take the shortest route kilometers. In general, the distance between two points depends on the path trav-
and travel a distance of 81 km eled (䉳 Fig. 2.1).
(50 mi). Another student takes a Along with many other quantities in physics, distance is a scalar quantity. A
longer route in order to visit a friend scalar quantity is one with only magnitude or size. That is, a scalar has only a
in Podunk before returning to numerical value, such as 160 km or 100 mi. (Note that the magnitude includes
school. The longer trip is in two seg-
ments, but the distance traveled is units.) Distance tells you the magnitude only—how far, but not the direction.
the total length, 97 km + 48 km = Other examples of scalars are quantities such as 10 s (time), 3.0 kg (mass), and
145 km 190 mi2. 20 °C (temperature). Some scalars may have negative values, for example, -10 °F.
2.1 DISTANCE AND SPEED: SCALAR QUANTITIES 35
SPEED
When something is in motion, its position changes with time. That is, it moves a
certain distance in a given amount of time. Both length and time are therefore
important quantities in describing motion. For example, imagine a car and a
pedestrian moving down a street and traveling a distance of one block. You would
expect the car to travel faster and thus to cover the same distance in a shorter time
than the person. A length–time relationship can be expressed by using the rate at
which distance is traveled, or speed.
Average speed (sq ) is the distance d traveled, that is, the actual path length,
divided by the total time ¢t elapsed in traveling that distance:
distance traveled
average speed = (2.1)
total time to travel that distance
d d
qs = =
¢t t2 - t1
SI unit of speed: meters per second (m>s)
A symbol with a bar over it is commonly used to denote an average. The Greek let-
ter delta, ¢ , is used to represent a change or difference in a quantity, in this case
the time difference between the beginning (t1) and end (t2) of a trip, or the elapsed
total time.
The SI standard unit of speed is meters per second (m>s, length>time),
although kilometers per hour (km>h) is used in many everyday applications.
The British standard unit is feet per second (ft>s), but a commonly used unit is
miles per hour (mi>h). Often, the initial time is taken to be zero, t1 = 0, as in
resetting a stopwatch, and thus the equation is written qs = d>t, where it is
understood that t is the total time.
Since distance is a scalar (as is time), speed is also a scalar. The distance does not
have to be in a straight line (see Fig. 2.1). For example, you probably have com-
puted the average speed of an automobile trip by using the distance obtained
from the starting and ending odometer readings. Suppose these readings were
17 455 km and 17 775 km, respectively, for a 4.0-h trip. Subtracting the readings
gives a total traveled distance d of 320 km, so the average speed of the trip is
d>t = 320 km>4.0 h = 80 km>h (or about 50 mi>h).
Average speed gives a general description of motion over a time interval ¢t.
In the case of the auto trip with an average speed of 80 km>h, the car’s speed
wasn’t always 80 km>h. With various stops and starts on the trip, the car must
have been moving more slowly than the average speed at various times. It there-
fore had to be moving more rapidly than the average speed another part of the
time. With an average speed, you don’t know how fast the car was moving at
any particular instant of time during the trip. By analogy, the average test score
of a class doesn’t tell you the score of any particular student.
On the other hand, instantaneous speed tells how fast something is moving
at a particular instant of time. That is, when
¢t : 0 (the time interval approaches zero),
䉳 F I G U R E 2 . 2 Instantaneous
which represents an instant of time. The speed The speedometer of a car
speedometer of a car gives an approximate gives the speed over a very short
instantaneous speed. For example, the interval of time, so its reading
speedometer shown in 䉴 Fig. 2.2 indicates a approaches the instantaneous
speed of about 44 mi>h, or 70 km>h. If the car speed. Note the speeds are given in
mi>h and km>h. (MPH is a nonstan-
travels with constant speed (so the dard abbreviation.)
speedometer reading does not change), then
the average and instantaneous speeds will
be equal. (Do you agree? Think of the previ-
ous average test score analogy. What if all of
the students in the class got the same score?)
EXAMPLE 2.1 Slow Motion: Rover Moves Along

In January 2004, a Mars Exploration Rover touched down on T H I N K I N G I T T H R O U G H . (a) Because the average speed and
the surface of Mars and rolled out for exploration (䉲 Fig. 2.3). the distance are known, the time can be computed from the
The average speed of the Rover on flat, hard ground is 5.0 cm>s. equation for average speed (Eq. 2.1). (b) Here, to calculate the
(a) Assuming the Rover traveled continuously over this terrain average speed, the total time, which includes stops, must be
at its average speed, how much time would it take to travel used.
2.0 m nonstop in a straight line? (b) However, in order to
SOLUTION.
ensure a safe drive, the Rover was equipped with hazard
avoidance software that caused it to stop and assess its location Listing the data in symbol form (cm>s is converted directly to
every few seconds. It was programmed to drive at its average m>s):
speed for 10 s, then stop and observe the terrain for 20 s before Given: Find:
moving onward for another 10 s and repeating the cycle. Tak- (a) qs = 5.0 cm>s = 0.050 m>s (a) ¢t (time to travel
ing its programming into account, what would be the Rover’s d = 2.0 m distance)
average speed in traveling the 2.0 m? (There were actually two
(b) cycles of 10-s travel, (b) qs (average speed)
Rovers on this mission, named Spirit and Opportunity. At the
20-s stops
time of this writing, Spring 2008, both Rovers are still function-
ing after over four years on the Red Planet.)
d
(a) Rearranging Eq. 2.1, qs = , to solve for time,
¢t
d 2.0 m
¢t = = = 40 s
qs 0.050 m>s
So it takes the Rover 40 s to travel a path length of 2.0 m.

(b) Here, the total time for the 2.0-m distance is needed. In
each 10-s interval, a distance of 0.050 m>s * 10 s = 0.50 m
would be traveled. So, the total time would be four 10-s inter-
vals for actual travel and three 20-s intervals of stopping, giv-
ing ¢t = 4 * 10 s + 3 * 20 s = 100 s. Then
d d 2.0 m
qs = = = = 0.020 m>s
¢t t2 - t1 100 s
F O L L O W - U P E X E R C I S E . Suppose the Rover’s programming
䉱 F I G U R E 2 . 3 A Mars exploration rover Twin Rovers was for 5.0 s of travel and for 10-s stops. How long would it
landed on opposite sides of the Martian planet in search of take to travel the 2.0 m in this case? (Answers to all Follow-Up
answers about the history of water on Mars. Exercises are given in Appendix VI at the back of the book.)
DID YOU LEARN?

➥ Distance is the total path length and has only magnitude.
➥ Average speed is speed over a period of time; instantaneous speed is speed at a
particular instant of time.
➥ If the speed is constant, the average and instantaneous speeds are equal.
2.2 One-Dimensional Displacement and Velocity:

Vector Quantities
➥ How are displacement and velocity related?

➥ When are average velocity and instantaneous velocity related for linear motion?
DISPLACEMENT
For straight-line, or linear, motion, it is convenient to specify position by using the
familiar two-dimensional Cartesian coordinate system with x- and y-axes at right
angles. A straight-line path can be in any direction relative to the axes, but for con-
venience, the coordinate axes are usually oriented so that the motion is along one
of them. (See the accompanying Learn by Drawing 2.1, Cartesian Coordinates and
One-Dimensional Displacement.)
2.2 ONE-DIMENSIONAL DISPLACEMENT AND VELOCITY: VECTOR QUANTITIES 37
As was discussed in the previous section, distance is a scalar quantity with only
magnitude (and units). However, to more completely describe motion, more
information can be given by adding a direction. This information is particularly
easy to convey for a change of position in a straight line. Displacement is defined LEARN BY DRAWING 2.1
as the straight-line distance between two points, along with the direction directly
from the starting point to the final position. Unlike distance (a scalar), displace- Car tesian coordinates
ment can have either positive or negative values, with the signs indicating the
directions along a coordinate axis.
and one-dimensional
As such, displacement is a vector quantity. In other words, a vector has both displacement
magnitude and direction. For example, when describing the displacement of an
airplane as 25 km north, this is a vector description (magnitude and direction).
Other vector quantities include velocity and acceleration, as will be learned later
in the chapter.
There is an algebra that applies to vectors, which involves how to specify and
deal with the direction part of the vector. This is done relatively easily in one
dimension by using + and - signs to indicate directions. To illustrate this with
displacements, consider the situation shown in 䉲 Fig. 2.4, where x1 and x2 indicate
the initial and final positions, respectively, on the x-axis as a student moves in a
straight line from his locker to the physics lab. As can be seen in Fig. 2.4a, the
scalar distance traveled is 8.0 m. To specify displacement (a vector) between x1 and
x2, we use the expression (a) A two-dimensional Cartesian
coordinate system. A displacement
¢x = x2 - x1 (2.2) vector d locates a point (x, y).
where ¢ is again used to represent a change in a quantity. Then, as in Fig. 2.4b,

¢x = x2 - x1 = + 9.0 m - 1+ 1.0 m2 = + 8.0 m
where the + signs indicate the positions on the positive x-axis. Hence, the stu-
dent’s displacement (magnitude and direction) is 8.0 m in the positive x-direction,
as indicated by the positive 1+2 result in Fig. 2.4b. (As in “regular” mathematics,
the plus sign is often omitted, as being understood, so this displacement can be (b) For one-dimensional, or
straight-line, motion, it is convenient
written as ¢x = 8.0 m instead of ¢x = + 8.0 m.) to orient one of the coordinate axes
along the direction of motion.
PHYSICS
LAB
x1 x2
x
1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 (meters)
8.0 m
(a) Distance (magnitude or numerical value only)
x1 x2
x
1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 (meters)
∆x = x2 − x1 = 9.0 m − 1.0 m = +8.0 m
(b) Displacement (magnitude and direction)
䉱 F I G U R E 2 . 4 Distance (scalar) and displacement (vector) (a) The distance (straight-

line path) between the student on the left and the physics lab is 8.0 m and is a scalar
quantity. (b) To indicate displacement, x1 and x2 specify the initial and final positions,
respectively. The displacement is then ¢x = x2 - x1 = 9.0 m - 1.0 m = + 8.0 m—that
is, 8.0 m in the positive x-direction.
Vector quantities in this book are usually indicated by boldface

B
type with an over-
arrow; for example, a displacement vector is indicated by d or xB, and a velocity vector
B
is indicated by v . However, when working in one dimension, this notation is not
needed. Instead, plus and minus signs can be used to indicate the only two possible
directions. The x-axis is commonly used for horizontal motions, and a plus 1 +2 sign
is taken to indicate the direction to the right, or in the “positive x-direction,” and a
minus 1 - 2 sign indicates the direction to the left, or in the “negative x-direction.”
Keep in mind that these signs only “point” in particular directions. An object
moving along the negative x-axis toward the origin would be moving in the posi-
tive x-direction. How about an object moving along the positive x-axis toward the
origin? If you said in the negative x-direction, you are correct.
Suppose the other student in Fig. 2.4 walks from the physics lab (her initial
position is different, x1 = + 9.0 m) to the end of the lockers (the final position is
now x2 = + 1.0 m). Her displacement would be
¢x = x2 - x1 = + 1.0 m - 1+ 9.0 m2 = - 8.0 m
The minus sign indicates that the direction of the displacement was in the nega-
tive x-direction or to the left in the figure. In this case, we say that the two stu-
dents’ displacements are equal (in magnitude) and opposite (in direction).
VELOCITY
As has been learned, speed, like the distance it incorporates, is a scalar quantity—it
has magnitude only. Another more descriptive quantity used to describe motion is
velocity. Speed and velocity are often used synonymously in everyday conversa-
tion, but the terms have different meanings in physics. Speed is a scalar, and
velocity is a vector—velocity has both magnitude and direction. Unlike speed,
one-dimensional velocities can have both positive and negative values, indicating
the only two possible directions (as with displacement).
Velocity tells how fast something is moving and in which direction it is moving.
And just as there are average and instantaneous speeds, there are average and
instantaneous velocities involving vector displacements. The average velocity is
the displacement divided by the total travel time. In one dimension, this involves
just motion along one axis, which is taken to be the x-axis. In this case,
displacement
average velocity = (2.3)*
total travel time
¢x x2 - x1
vq = =
¢t t2 - t1
SI unit of velocity meters per second (m>s)

In the case of more than one displacement (such as for successive displace-
ments), the average velocity is equal to the total or net displacement divided by the
total time. The total displacement is found by adding the displacements alge-
braically according to the directional signs.
You might be wondering whether there is a relationship between average
speed and average velocity. A quick look at Fig. 2.4 will show that if all the motion
*Another common form of this equation is

¢x 1x2 - x12 1x - xo2 1x - xo2
1t2 - t12 1t - to2
vq = = = =
¢t t
or, after rearranging,
x = xo + vqt (2.3)
where xo is the initial position, x is the final position, and ¢t = t with to = 0. See Section 2.3 for more
on this notation.
is in one direction, that is, there is no reversal of direction, the distance is equal to
the magnitude of the displacement. Then the average speed is equal to the magni-
tude of the average velocity. However, be careful. This set of relationships is not true
if there is a reversal of direction, as Example 2.2 shows.
EXAMPLE 2.2 There and Back: Average Velocities

A jogger jogs from one end to the other of a straight 300-m T H I N K I N G I T T H R O U G H . The average velocities are com-
track in 2.50 min and then jogs back to the starting point in puted from the defining equation. Note that the times given
3.30 min. What was the jogger’s average velocity (a) in jog- are the ¢t’s associated with the particular displacements.
ging to the far end of the track, (b) coming back to the starting
point, and (c) for the total jog?
SOLUTION.
Given: ¢x1 = + 300 m (taking the initial direction as positive) Find: Average velocities for
¢x2 = - 300 m (taking the direction of the return trip (a) the first leg of the jog,
as negative) (b) the return jog,
¢t1 = 2.50 min160 s>min2 = 150 s (c) the total jog
¢t2 = 3.30 min160 s>min2 = 198 s
(a) The jogger’s average velocity for the trip down the track is The total or net displacement for this case could have been
found using Eq. 2.3: found by simply taking ¢x = xfinal - xinitial = 0 - 0 = 0,
where the initial and final positions are taken to be the origin,
¢x1 +300 m but it was done in parts here for illustration purposes.
vq1 = = = + 2.00 m>s
¢t1 150 s
(b) Similarly, for the return trip,
¢x2 -300 m
vq2 = = = - 1.52 m>s
¢t2 198 s
(c) For the total trip, there are two displacements to consider,
down and back, so these are added together to get the total
displacement, and then divided by the total time:
¢x1 + ¢x2 300 m + 1- 300 m2

vq3 = = = 0 m>s
¢t1 + ¢t2 150 s + 198 s
The average velocity for the total trip is zero! Do you see
䉱 F I G U R E 2 . 5 Back home again! Despite having covered
why? Recall from the definition of displacement that the mag-
nearly 110 m on the base paths, at the moment the runner slides
nitude of displacement is the straight-line distance between through the batter’s box (his original position) into home plate,
two points. The displacement from one point back to the his displacement is zero—at least, if he is a right-handed batter.
same point is zero; hence the average velocity is zero. (See No matter how fast he ran the bases, his average velocity for
䉴 Fig. 2.5.) the round trip is zero.
F O L L O W - U P E X E R C I S E . Find the jogger’s average speed for each of the cases in this Example, and compare these with the mag-
nitudes of the respective average velocities. [Will the average speed for part (c) be zero?] (Answers to all Follow-Up Exercises are
As Example 2.2 shows, average velocity provides only an overall description of

motion. One way to take a closer look at motion is to take smaller time intervals,
that is, to let the observation time interval 1¢t2 become smaller and smaller. As
with speed, when ¢t approaches zero, an instantaneous velocity is obtained,
which describes how fast something is moving and in which direction at a particu-
lar instant of time.
䉴 F I G U R E 2 . 6 Uniform linear Distance

motion—constant velocity In uni- 50 km 100 km 150 km
form linear motion, an object travels
at a constant velocity, covering the 0 1.0 h 2.0 h 3.0 h
same distance in equal time inter-
Time
vals. (a) Here, a car travels 50 km
each hour. (b) An x-versus-t plot is a
straight line, since equal distances ∆ x (km) ∆t (h) ∆ x/∆t
are covered in equal times. The
numerical value of the slope of the 50 1.0 50 km/1.0 h = 50 km/h
line is equal to the magnitude of the 100 2.0 100 km/2.0 h = 50 km/h
velocity, and the sign of the slope 150 3.0 150 km/3.0 h = 50 km/h
gives its direction. (The average
velocity equals the instantaneous (a)
velocity in this case. Why?)
x
200 Slope = v (= v)
∆ x 50 km
Slope = = = 50 km/h
∆t 1.0 h
Position (km)
150
100 x2
∆ x = x2 − x1 = 100 − 50 = 50 km
50 x1
∆t = t2 − t1 = 2.0 − 1.0 = 1.0 h
t1 t2
0 t
1.0 2.0 3.0 4.0
Time (h)
Uniform velocity
(b)
Instantaneous velocity is defined mathematically as

¢x
v = lim (2.4)
¢t: 0 ¢t
This expression is read as “the instantaneous velocity is equal to the limit of ¢x>¢t
as ¢t goes to zero.” The time interval does not ever equal zero (why?), but
approaches zero. Instantaneous velocity is technically still an average velocity, but
over such a very small ¢t that it is essentially an average “at an instant in time,”
which is why it is called the instantaneous velocity.
Uniform motion means motion with a constant or uniform velocity (constant
magnitude and constant direction). As a one-dimensional example of this, the car
in 䉱 Fig. 2.6 has a uniform velocity. It travels the same distance and experiences the
same displacement in equal time intervals (50 km each hour) and the direction of
its motion does not change. Hence, the magnitudes of the average velocity and
instantaneous velocity are equal in this case. The average of a constant is equal to
that constant.
GRAPHICAL ANALYSIS
Graphical analysis is often helpful in understanding motion and its related quan-
tities. For example, the motion of the car in Fig. 2.6a may be represented on a plot
of position versus time, or x versus t. As can be seen from Fig. 2.6b, a straight line
is obtained for a uniform, or constant, velocity on such a graph.
Recall from Cartesian graphs of y versus x that the slope of a straight line is
given by ¢y>¢x. Here, with a plot of x versus t, the slope of the line, ¢x>¢t, is
x x
Slope = –v = ∆ x = –50 km = –50 km/h

200 ∆t 1.0 h
x2
Position (km)
∆ x = x2 – x1 = 100 – 150 = –50 km

150 x1
100 x2
∆ t = t2 – t1 = 2.0 – 1.0 = 1.0 h
50
Position
t1 t2 v
0 t ∆x e
=
1.0 2.0 3.0 4.0 op
Sl
Time (h)
䉱 F I G U R E 2 . 7 Position-versus-time graph for an object in (5) ∆x

uniform motion in the negative x-direction A straight line on (2)
an x-versus-t plot with a negative slope indicates uniform (3)
∆t
motion in the negative x-direction. Note that the object’s x1 (1)
(4)
location changes at a constant rate. At t = 4.0 h, the object is ∆t
at x = 0. How would the graph look if the motion continues t
for t 7 4.0 h? t1 t2
Time
䉱 F I G U R E 2 . 8 Position-versus-
therefore equal to the average velocity vq = ¢x>¢t. For uniform motion, this value time graph for an object in nonuni-
form linear motion For a
is equal to the instantaneous velocity. That is, vq = v. (Why?) The numerical value
nonuniform velocity, an x-versus-t
of the slope is the magnitude of the velocity, and the sign of the slope gives the plot is a curved line. The slope of
direction. A positive slope indicates that x increases with time, so the motion is in the line between two points is the
the positive x-direction. (The plus sign is often omitted as being understood, average velocity between those
which will be done in general from here on.) positions, and the instantaneous
velocity at any time t is the slope of
Suppose that a plot of position versus time for a car’s motion is a straight line
a line tangent to the curve at that
with a negative slope, as in 䉱 Fig. 2.7. What does this indicate? As the figure shows, point. Five tangent lines are shown,
the position (x) values get smaller with time at a constant rate, indicating that the with the intervals for ¢x> ¢t in the
car is traveling in uniform motion, but now in the negative x-direction which cor- fifth. Can you describe the object’s
relates with the negative value of the slope. motion in words?
In most instances, the motion of an object is nonuniform, meaning that different
distances are covered in equal intervals of time. An x-versus-t plot for such motion
in one dimension is a curved line, as illustrated in 䉱 Fig. 2.8. The average velocity
of the object during any interval of time is the slope of a straight line between the
two points on the curve that correspond to the starting and ending times of
the interval. In the figure, since vq = ¢x>¢t, the average velocity of the total trip is
the slope of the straight line joining the beginning and ending points of the curve.
The instantaneous velocity is equal to the slope of the tangent line to the curve
at the time of interest. Five typical tangent lines are shown in Fig. 2.8. At (1), the
slope is positive, and the motion is therefore in the positive x-direction. At (2), the
slope of a horizontal tangent line is zero, so there is no motion for an instant. That
is, the object has instantaneously stopped 1v = 02 at that time. At (3), the slope is
negative, so the object is moving in the negative x-direction. Thus, the object
stopped in the process of changing direction at point (2). What is happening at
points (4) and (5)?
By drawing various tangent lines along the curve, it can be seen that their
slopes vary, in magnitude and direction (sign), indicating that the instantaneous
velocity is changing with time. An object in nonuniform motion can speed up,
slow down, or change direction. How nonuniform motion is described in the topic
of Section 2.3.
DID YOU LEARN?

➥ Velocity is the time rate of change of displacement.
➥ For linear motion in one direction, average speed is equal to the magnitude of the
average velocity. (This is not true if there is a reversal in direction.)
2.3 Acceleration
➥ What is evidence of an acceleration?

➥ What is required for a deceleration?
➥ Is a negative acceleration always a deceleration?
The basic description of motion involving the time rate of change of position (and
direction) is called velocity. Going one step further, we can consider how this rate of
change itself changes. Suppose an object is moving at a constant velocity and then
the velocity changes. Such a change in velocity is called an acceleration. The gas
pedal on an automobile is commonly called the accelerator. When you press down
on the accelerator, the car speeds up; when you let up on the accelerator, the car
slows down. In either case, there is a change in velocity with time. Acceleration is
defined as the time rate of change of velocity.
Analogous to average velocity, the average acceleration is defined as the
change in velocity divided by the time taken to make the change:
change in velocity
average acceleration = (2.5)
time to make the change
¢v
qa =
¢t
v2 - v1 v - vo
= =
t2 - t1 t - to
SI unit of acceleration: meters per second squared 1m>s22

Note that the initial and final variables have been changed to a more commonly
used notation. That is, vo and to are the initial or original velocity and time, respec-
tively, and v and t are the general velocity and time at some point in the future,
such as when you want to know the velocity v at a particular time t. (This may or
may not be the final velocity of a particular situation. There may be an acceleration
after this time.)
From ¢v>¢t, the SI units of acceleration can be seen to be meters per second
1¢v2 per second 1¢t2, that is, 1m>s2>s or m>1s # s2, which is commonly expressed as
meters per second squared 1m>s 22. In the British system, the units are feet per second
squared 1ft>s22.
Because velocity is a vector quantity, so is acceleration, as acceleration repre-
sents a change in velocity. Since velocity has both magnitude and direction, a
change in velocity may involve changes in either or both of these factors. Thus an
acceleration may result from a change in speed (magnitude), a change in direction,
or a change in both, as illustrated in 䉴 Fig. 2.9.
For straight-line, linear motion, plus and minus signs will be used to indicate
the directions of velocity and acceleration, as was done for linear displacements.
Eq. 2.5 is commonly simplified and written as
v - vo
qa = (2.6)
t
where to is taken to be zero. (vo may not be zero, so it cannot generally be omitted.)
Analogous to instantaneous velocity, instantaneous acceleration is the acceler-
ation at a particular instant of time. This quantity is expressed mathematically as
¢v
a = lim (2.7)
¢t:0 ¢t
The conditions of the time interval approaching zero are the same here as
described for instantaneous velocity.
2.3 ACCELERATION 43
a a
Acceleration Deceleration
(velocity magnitude increases) (velocity magnitude
decreases)
v (20 km/h) v (30 km/h) v (40 km/h) v=0

0 t = 2.0 s t = 4.0 s t = 6.0 s
(a) Change in velocity magnitude but not direction
v2
(80
km )
/h) /h
t=
1.0 km
s 0
(6
v1 t=
0 2.
t= 0
s
v1 (80 km/h)
v2
(4
0
km
/h
)
t=0
(b) Change in velocity direction but not magnitude (c) Change in velocity magnitude and direction
䉱 F I G U R E 2 . 9 Acceleration—the time rate of change of velocity Since velocity is a vector quantity, with magnitude and direction,
an acceleration can occur when there is (a) a change in magnitude, but not direction; (b) a change in direction, but not magnitude; or
(c) a change in both magnitude and direction.
EXAMPLE 2.3 Slowing It Down: Average Acceleration

A couple in a sport-utility vehicle (SUV) are traveling at [Here, the instantaneous velocities are assumed to be in the pos-
90 km>h on a straight highway. The driver sees an accident in itive direction, and conversions to standard units (meters per
the distance and slows down to 40 km>h in 5.0 s. What is the second) are made right away, since it is noted that the speed is
average acceleration of the SUV? given in km/h. In general, standard units should be used.]
T H I N K I N G I T T H R O U G H . To find the average acceleration, the
Given the initial and final velocities and the time interval,
variables as defined in Eq. 2.6 must be given, and they are. the average acceleration can be found by using Eq. 2.6:
SOLUTION. Listing the data and converting units, v - vo 11 m>s - 125 m>s2
qa = = = - 2.8 m>s2
t 5.0 s
Given: vo = 190 km>h2a b
0.278 m>s
Find: qa (average
1 km>h The minus sign indicates the direction of the (vector) accelera-
acceleration)
= 25 m>s tion. In this case, the acceleration is opposite to the direction
v = 140 km>h2a b
0.278 m>s of the motion and the car slows. Such an acceleration is some-
1 km>h times called a deceleration, since the car is slowing. (Note: This
= 11 m>s is why vo cannot arbitrarily be set to zero, because as shown
t = 5.0 s here there may be motion, and vo Z 0 at to = 0.)
F O L L O W - U P E X E R C I S E . Does a negative acceleration necessarily mean that a moving object is slowing down (decelerating) or
that its speed is decreasing? [Hint: See the accompanying Learn by Drawing 2.2, Signs of Velocity and Acceleration.] (Answers to
all Follow-Up Exercises are given in Appendix VI at the back of the book.)
CONSTANT ACCELERATION
Although acceleration can vary with time, our study of motion will generally be
restricted to constant accelerations for simplicity. (An important constant accelera-
LEARN BY DRAWING 2.2 tion is the acceleration due to gravity near the Earth’s surface, which will be consid-
ered in the next section.) Since for a constant acceleration, the average acceleration is
signs of velocity and equal to the constant value 1aq = a2, the bar over the acceleration in Eq. 2.6 may be
acceleration omitted. Thus, for a constant acceleration, the equation relating velocity, accelera-
tion, and time is commonly written (rearranging Eq. 2.6) as follows:
a positive Result: v = vo + at (constant acceleration only) (2.8)

Faster in
+x direction
v positive
(Note that the at term represents the change in velocity, since at = v - vo = ¢v.)
–x +x
a negative Result:
Slower in EXAMPLE 2.4 Fast Start, Slow Stop: Motion with
+x direction Constant Acceleration
v positive
–x +x A drag racer starting from rest accelerates in a straight line at a constant rate of 5.5 m>s2
for 6.0 s. (a) What is the racer’s velocity at the end of this time? (b) If a parachute
a positive Result: deployed at this time causes the racer to slow down uniformly at a rate of 2.4 m>s2, how
Slower in long will the racer take to come to a stop?
–x direction
v negative
T H I N K I N G I T T H R O U G H . The racer first speeds up and then slows down, so close atten-
–x +x tion must be given to the directional signs of the vector quantities. Choose a coordinate
system with the positive direction in the direction of the initial velocity. (Draw a sketch
Result: of the situation for yourself.) The answers can then be found by using the appropriate
a negative Faster in equations. Note that there are two different parts to the motion, with two different
–x direction accelerations. Let’s distinguish these phases with subscripts of 1 and 2.
v negative
–x +x SOLUTION. Taking the initial motion to be in the positive direction, we have the fol-
lowing data:
Given: (a) vo = 0 1at rest2 Find: (a) v1 (final velocity for
a 1 = 5.5 m>s 2 first part of the
t1 = 6.0 s motion)
(b) vo = v1 [from part 1a2] (b) t2 (time for second
v2 = 0 1comes to stop2 part of the
a 2 = - 2.4 m>s21opposite direction of vo2 motion)
The data have been listed in two parts. This practice helps avoid confusion with sym-
bols. Note that the final velocity v1 that is to be found in part (a) becomes the initial
velocity vo for part (b).
(a) To find the final velocity v1, Eq. 2.8 may be used directly:
v1 = vo + a 1t1 = 0 + 15.5 m>s2216.0 s2 = 33 m>s
(b) Here we want to find time, so solving Eq. 2.6 for t2 and using vo = v1 = 33 m>s from
part (a),
v2 - vo 0 - 133 m>s2
t2 = = = 14 s
a2 -2.4 m>s2
Note that the time comes out positive, as it should. Why?
F O L L O W - U P E X E R C I S E . What is the racer’s instantaneous velocity 10 s after the para-
chute is deployed? (Answers to all Follow-Up Exercises are given in Appendix VI at the back
of the book.)
Motions with constant accelerations are easy to represent graphically by plot-

ting instantaneous velocity versus time. In this case, the v-versus-t plot is a straight
line, the slope of which is equal to the acceleration, as illustrated in 䉴Fig. 2.10. Note
that Eq. 2.8 can be written as v = at + vo, which, as you may recognize, has the
form of an equation of a straight line, y = mx + b (slope m and intercept b).
2.3 ACCELERATION 45
vo
Sl
op
Velocity
v e –at
v = vo + at =
–a
Velocity
= +a vo
pe
Slo at
vo v v = vo – at
vo
0 t Time 0 t Time
(a) Motion in positive direction—speeding up (b) Motion in positive direction—slowing down
t
0 vo
–vo Time
–vo Sl vo
Velocity
op –at
Velocity
Slo e
pe 0 =
=– –at t1 –a t2
a Time
–v = vo – at2
–v –v = –vo –at –v
(c) Motion in negative direction—speeding up (d) Changing direction
䉱 F I G U R E 2 . 1 0 Velocity-versus-time graphs for motions with constant accelerations The slope of a v-versus-t plot is the
acceleration. (a) A positive slope indicates an increase in the velocity in the positive direction. The vertical arrows to the
right indicate how the acceleration adds velocity to the initial velocity, vo. (b) A negative slope indicates a decrease in the ini-
tial velocity, vo, or a deceleration. (c) Here a negative slope indicates a negative acceleration, but the initial velocity is in the
negative direction, -vo, so the speed of the object increases in that direction. (d) The situation here is initially similar to that
of part (b) but ends up resembling that in part (c). Can you explain what happened at time t1?
In Fig. 2.10a, the motion is in the positive direction, and the acceleration term adds
to the velocity after t = 0, as illustrated by the vertical arrows at the right of the
graph. Here, the slope is positive 1a 7 02. In Fig. 2.10b, the negative slope 1a 6 02
indicates a negative acceleration that produces a slowing down, or deceleration.
However, Fig. 2.10c illustrates how a negative acceleration can speed things up
(for motion in the negative direction). The situation in Fig. 2.10d is slightly more
complex. Can you explain what is happening there?
When an object moves at a constant acceleration, its velocity changes by the
same amount in each equal time interval. For example, if the acceleration is
10 m>s 2 in the same direction as that of the initial velocity, the object’s velocity
increases by 10 m>s each second. As an example of this, suppose that the object
has an initial velocity vo of 20 m>s in a particular direction at to = 0. Then, for
t = 0, 1.0, 2.0, 3.0, and 4.0 s, the magnitudes of the velocities are 20, 30, 40, 50, and
60 m>s, respectively.
The average velocity may be computed in the regular manner (Eq. 2.3), or you
may recognize that the uniformly increasing series of numbers 20, 30, 40, 50, and 60
has an average value of 40 (the midway value of the series) and vq = 40 m>s. Note
that the average of the initial and final values also gives the average of the series—
that is, 120 + 602>2 = 40. Only when the velocity changes at a uniform rate because
of a constant acceleration is vq then the average of the initial and final velocities:
v + vo
vq = (constant acceleration only) (2.9)
2
EXAMPLE 2.5 On the Water: Using Multiple Equations

A motorboat starting from rest on a lake accelerates in a lem then involves finding vq . Then by Eq. 2.9, vq = 1v + vo2>2,
straight line at a constant rate of 3.0 m>s2 for 8.0 s. How far and with vo = 0, only the final velocity v is needed to solve
does the boat travel during this time? the problem. Equation 2.8, v = vo + at, can be used to calcu-
late v from the given data. So it follows that:
T H I N K I N G I T T H R O U G H . We have only one equation for dis-
The velocity of the boat at the end of 8.0 s is
tance (Eq. 2.3, x = xo + vqt), but this equation cannot be used
directly. The average velocity must first be found, so multiple v = vo + at = 0 + 13.0 m>s2218.0 s2 = 24 m>s
equations and steps are involved.
The average velocity over that time interval is
SOLUTION.
Reading the problem, summarizing the given data, and iden- v + vo 24 m>s + 0
vq = = = 12 m>s
tifying what is to be found (assuming the boat to accelerate in 2 2
the +x-direction) gives the following:
Finally, the magnitude of the displacement, which in this case
Given: xo = 0 Find: x (distance) is the same as the distance traveled, is given by Eq. 2.3 (choos-
vo = 0 ing the boat’s initial location at the origin, so xo = 0):
a = 3.0 m>s2
t = 8.0 s x = vqt = 112 m>s218.0 s2 = 96 m
(Note that all of the units are standard.) F O L L O W - U P E X E R C I S E . (Sneak preview.) In Section 2.4, the
In analyzing the problem, one might reason: To find x, following equation will be derived: x = vo t + 12 at2. Use the
Eq. 2.3 needs to be used in the form x = xo + vqt. (The average data in this Example to see if this equation gives the distance
velocity vq must be used because the velocity is changing and traveled. (Answers to all Follow-Up Exercises are given in
thus not constant.) With time given, the solution to the prob- Appendix VI at the back of the book.)
DID YOU LEARN?

➥ A change in velocity is evidence of an acceleration.
➥ A deceleration requires a velocity in the opposite direction of the acceleration.
➥ A negative acceleration for motion in the negative direction increases the velocity.
2.4 Kinematic Equations (Constant Acceleration)

➥ How is the velocity affected for a constant acceleration?

➥ What is necessary for a moving object to come to a stop?
The description of motion in one dimension with constant acceleration requires

only three basic equations. From previous sections, these equations are
x = xo + vqt (2.3)
v + vo
vq = (constant acceleration only) (2.9)
2
v = vo + at (constant acceleration only) (2.8)
(Keep in mind that the first equation, Eq. 2.3, is general and is not limited to situa-
tions in which there is constant acceleration, as are the latter two equations.)
However, as Example 2.5 showed, the description of motion in some instances
requires multiple applications of these equations, which may not be obvious at
first. It would be helpful if there were a way to reduce the number of operations in
solving kinematic problems, and there is—by combining equations algebraically.
For instance, suppose an expression that gives location x in terms of time and
acceleration is wanted, rather than one in terms of time and average velocity
(as in Eq. 2.3). Eliminating vq from Eq. 2.3 by substituting for vq from Eq. 2.9 into
Eq. 2.3,
v + vo
x = xo + vqt (2.3) and vq = (2.9)
2
2.4 KINEMATIC EQUATIONS (CONSTANT ACCELERATION) 47
and on substituting,
x = xo + 12 1v + vo2t (constant acceleration only) (2.10)
Then, substituting for v from Eq. 2.8 (v = vo + at) gives

x = xo + 12 1vo + at + vo2t
Simplifying,
x = xo + vo t + 12 at2 (constant acceleration only) (2.11)
Essentially, this series of steps was done in Example 2.5. The combined equation
allows the displacement of the motorboat in that Example to be computed
directly:
x - xo = ¢x = vo t + 12 at2 = 0 + 12 13.0 m>s2218.0 s22 = 96 m
Much easier, isn’t it?
We may want an expression that gives velocity as a function of position x rather
than time (as in Eq. 2.8). In this case, t can be eliminated from Eq. 2.8 by using Eq.
2.10 in the form
1x - xo2
v + vo = 2
t
Then, multiplying this equation by Eq. 2.8 in the form 1v - vo2 = at gives
1v + vo21v - vo2 = 2a1x - xo2
and using the relationship v2 - v2o = 1v + vo21v - vo2,
v2 = v2o + 2a1x - xo2 (constant acceleration only) (2.12)
PROBLEM-SOLVING HINT
Students in introductory physics courses are sometimes overwhelmed by the various

kinematic equations. Keep in mind that equations and mathematics are the tools of
physics. As any mechanic or carpenter will tell you, tools make your work easier as long as
you are familiar with them and know how to use them. The same is true for physics tools.
Summarizing the equations for linear motion with constant acceleration:
v = vo + at (2.8)
2 1v + vo2t
1
x = xo + (2.10)
1 2
x = xo + vo t + 2 at (2.11)
2
v = v2o + 2a1x - xo2 (2.12)
This set of equations is used to solve the majority of kinematic problems. (Occasionally
there may be interest in average speed or velocity, and for that Eq. 2.3 can be used.)
Note that each of the equations in the list has four or five variables. All but one of the
variables in an equation must be known in order to be able to solve for what you are try-
ing to find. Generally, an equation with the unknown or wanted quantity is chosen. But,
as pointed out, the other variables in the equation must be known. If they are not, then
the wrong equation was chosen or another equation must be used to find the variables.
(Another possibility is that not enough data are given to solve the problem, but that is
not the case in this textbook.)
Always try to understand and visualize a problem. Listing the data as described in
the suggested problem-solving procedure in Section 1.7 may help you decide which
equation to use, by determining the known and unknown variables. Remember this
approach as you work through the remaining Examples in the chapter. Also, don’t over-
look any implied data or restrictive conditions, as illustrated in the following examples.
CONCEPTUAL EXAMPLE 2.6 Something Is Wrong!

A student working a problem with a constantly accelerating object wants to find v, and
is given that vo = 0 and t = 3.0 s, but is not given the acceleration a. He observes the
kinematic equations and decides, using v = at and x = 12 at2 (with xo = vo = 0), that the
unknown a can be eliminated. With a = v>t and a = 2x>t2 and equating,
v 2x
= 2
t t
but x is not known, so he decides to use x = vt to eliminate it, and
v 2vt
= 2
t t
Simplifying,
v = 2v or 1 = 2!
What’s wrong here?
REASONING AND ANSWER. Obviously something is big-time wrong, and it goes back
to the problem-solving procedure given in Section 1.7. Step 4 there states: Determine
which principle(s) and equation(s) are applicable to the situation. Since only equations
were used, one equation must not apply to the situation. On inspection and analyzing,
this can be seen to be x = vt, which applies only to nonaccelerated motion, and hence
doesn’t apply to the problem.
FOLLOW-UP EXERCISE. Given only vo and t, is there any way to find v using the given
kinematic equations? Explain. (Answers to all Follow-Up Exercises are given in Appendix VI
at the back of the book.)
EXAMPLE 2.7 Moving Apart: Where Are They Now?

Two riders on dune buggies sit 10 m apart on a long, straight SOLUTION. Listing the data:
track, facing in opposite directions. Starting at the same time,
Given: x oA = 0 Find: separation
both riders accelerate at a constant rate of 2.0 m>s2. How far
aA = - 2.0 m>s2 distance at
apart will the dune buggies be at the end of 3.0 s?
t = 3.0 s t = 3.0 s
T H I N K I N G I T T H R O U G H . The dune buggies are initially 10 m xoB = 10 m
apart, so they can be positioned anywhere on the x-axis. It is
aB = 2.0 m>s2
convenient to place one at the origin so that one initial position
(xo) is zero. A sketch of the situation is shown in 䉲Fig. 2.11.
– +
A B
A B
x=0 x = 10 m
Initial
separation
= 10 m
Final separation = ?
䉱 F I G U R E 2 . 1 1 Away they go! Two dune buggies accelerate away from each other. How far apart are they at a
later time?
2.4 KINEMATIC EQUATIONS (CONSTANT ACCELERATION) 49
The displacement of each vehicle is given by Eq. 2.11 [the only And for buggie B with a nonzero xo,
displacement 1¢x2 equation with acceleration (a)]: xB = xoB + voBt + 12 aBt2
x = xo + vot + 12 at2. But there is no vo in the Given list. Some
implied data must have been missed. It should be quickly = 10 m + 0 + 1212.0 m>s2213.0 s22 = 19 m
noted that vo = 0 for both vehicles, so Hence vehicle A is 9.0 m to the left of the origin on the - x-axis,
1 2 whereas vehicle B is at a position of 19 m to the right of the ori-
xA = xoA + voAt + 2 a At gin on the +x-axis. And so, the separation distance between
= 0 + 0 + 121- 2.0 m>s2213.0 s22 = - 9.0 m the two dune buggies is 19 m + 9 m = 28 m.
F O L L O W - U P E X E R C I S E . Would it make any difference in the separation distance if vehicle B had been initially put at the origin
instead of vehicle A? Try it and find out. (Answers to all Follow-Up Exercises are given in Appendix VI at the back of the text.)
EXAMPLE 2.8 Putting On the Brakes: Vehicle Stopping Distance

The stopping distance of a vehicle is an important factor in stopping distance x is proportional to the square of the initial
road safety. This distance depends on the initial speed (vo) speed. Doubling the initial speed therefore increases the stop-
and the braking capacity, which produces the deceleration, a, ping distance by a factor of 4 (for the same deceleration). That
assumed to be constant. Express the stopping distance x in is, if the stopping distance is x1 for an initial speed of v1, then
terms of these quantities. for a twofold increase in the initial speed 1v2 = 2v12, the stop-
ping distance would increase fourfold:
T H I N K I N G I T T H R O U G H . The signs of the velocity and accel-
eration are taken to be plus and minus respectively, indicating v21
x1 =
they are in opposite directions so the car comes to a stop. 2a
12v122
Again, a kinematic equation is required, and the appropriate
v22 v21
one may be better determined by listing what is given and x2 = = = 4 a b = 4x1
what is to be found. Notice that the distance x is wanted and 2a 2a 2a
time is not involved. The same result can be obtained by using ratios:
SOLUTION. Here the quantities are variables and repre- v22 v2 2
= 2 = a b = 22 = 4
x2
sented in symbol form:
x1 v1 v1
Given: +vo Find: stopping distance
-a x (in terms of the Do you think this consideration is important in setting
v = 0 (car comes to stop) given variables) speed limits, for example, in school zones? (The driver’s reac-
xo = 0 (car taken to be ini- tion time should also be considered. A method for approxi-
tially at the origin) mating a person’s reaction time is given in Section 2.5.)
Again, it is helpful to make a sketch of the situation, particu- F O L L O W - U P E X E R C I S E . Tests have shown that the Chevy
larly when vector quantities are involved (䉲 Fig. 2.12). Since Blazer has an average braking deceleration of 7.5 m>s2,
Eq. 2.12 has the variables we want, it should allow us to find while that of a Toyota Celica is 9.2 m>s2. Suppose these two
the stopping distance x. Expressing the negative acceleration vehicles are being driven down a straight, level road at
explicitly and assuming xo = 0 gives 97 km>h (60 mi/h), with the Celica in front of the Blazer. A
v2 = v2o - 2ax cat runs across the road ahead of them, and both drivers
apply their brakes at the same time and come to safe stops
Since the vehicle comes to a stop 1v = 02, solving for x: (not hitting the cat). Assuming constant acceleration and the
v2o same reaction times for both drivers, what is the minimum
x = safe tailgating distance for the Blazer so that there won’t be a
2a
rear-end collision with the Celica when the two vehicles
This equation gives x expressed in terms of the vehicle’s come to a stop? (Answers to all Follow-Up Exercises are given in
initial speed and stopping acceleration. Notice that the Appendix VI at the back of the book.)
䉳 F I G U R E 2 . 1 2 Vehicle
– + stopping distance A sketch to
vo help visualize the situation.
Car stopped
v=0
a
x=?
xo = 0 (Stopping distance)
GRAPHICAL ANALYSIS OF KINEMATIC EQUATIONS
v As was shown in Fig. 2.10, plots of v versus t give straight-line graphs where the
slopes are values of the constant accelerations. There is another interesting aspect
of v-versus-t graphs. Consider the one shown in 䉳 Fig. 2.13a, particularly the
Velocity
shaded area under the curve. Suppose we calculate the area of the shaded triangle,
where, in general, A = 12 ab C Area = 12 1altitude21base2 D .
A For the graph in Fig. 2.13a, the altitude is v and the base is t, so A = 12 vt. But
from the equation v = vo + at, we have v = at, where vo = 0 (zero intercept on
graph). Therefore,
A = 12 vt = 12 1at2t = 12 at2 = ¢x
t
Time
(a) Hence, ¢x is equal to the area under a v-versus-t curve.
Now look at Fig. 2.13b. Here, there is a nonzero value of vo at t = 0, so the
v object is initially moving. Consider the two shaded areas. We know that the area
of the triangle is A 2 = 12 at2, and the area of the rectangle can be seen (with xo = 0)
to be A 1 = vo t. Adding these areas to get the total area yields
A 1 + A 2 = vo t + 12 at2 = ¢x
Velocity
A2 This is just Eq. 2.11, which is equal to the area under the v-versus-t curve.
vo DID YOU LEARN?

➥ The velocity changes linearly as a function of time for a constant acceleration and
gives a straight line on a v versus-t graph.
A1
➥ An acceleration in the opposite direction of the motion is needed for a moving
object to come to a stop.
t
Time
(b)
2.5 Free Fall
䉱 F I G U R E 2 . 1 3 v-versus-t graphs,
one more time (a) In the straight-line LEARNING PATH QUESTIONS
plot for a constant acceleration, the ➥ What is required for an object to be in free fall?
area under the curve is equal to ¢x, ➥ What is different about free fall on the Moon compared to that on the Earth?
the distance covered. (b) If vo is not
➥ A heavy object and a light object are in free fall, having been dropped from equal
zero, the distance is still given by the
area under the curve ¢x, but here heights.Which object strikes the ground first?
divided into two parts, areas A1
and A2. One of the more common cases of constant acceleration is the acceleration due to grav-
ity near the Earth’s surface. When an object is dropped, its initial velocity (at the instant
it is released) is zero. At a later time while falling, it has a nonzero velocity. There has
been a change in velocity and thus, by definition, an acceleration. This acceleration
due to gravity (g) near the Earth’s surface has an approximate magnitude of
g = 9.80 m>s2 (acceleration due to gravity)
(or 980 cm>s2) and is directed downward (toward the center of the Earth). In
British units, the value of g is about 32.2 ft>s2.
The values given here for g are only approximate because the acceleration due
to gravity varies slightly at different locations as a result of differences in elevation
and regional average mass densities of the Earth. These small variations will be
ignored in this book unless otherwise noted. (Gravitation is studied in more detail
in Section 7.5.) Air resistance is another factor that affects (reduces) the accelera-
tion of a falling object, but it too will be ignored here for simplicity. (The frictional
effect of air resistance will be considered in Section 4.6.)
Objects in motion solely under the influence of gravity are said to be in free fall. The
words free fall may bring to mind dropped objects. However, the term applies to
any motion under the sole influence of gravity. Objects released from rest, thrown
upward or downward, are all in free fall once they are released. That is, after t = 0
(the time of release), only gravity is influencing the motion. (Even when an object
projected upward is traveling upward, it is still accelerating downward.) Thus, the
set of equations for motion in one dimension with constant acceleration given in
the last section can be used to describe free fall.
2.5 FREE FALL 51
䉳 F I G U R E 2 . 1 4 Free fall and air

resistance (a) When dropped
simultaneously from the same
height, a feather falls more slowly
than a coin, because of air resis-
tance. But when both objects are
dropped in an evacuated container
with a good partial vacuum, where
air resistance is negligible, the
feather and the coin both have the
same constant acceleration. (b) An
actual demonstration with multi-
flash photography: An apple and a
feather are released simultaneously
through a trap door into a large vac-
uum chamber, and they fall
together—almost. Because the
chamber has only a partial vacuum,
there is still some air resistance.
(a) (b) (Can you tell?)
The acceleration due to gravity, g, has the same value for all free-falling objects,
regardless of their mass or weight. It was once thought that heavier bodies acceler-
ate faster than lighter bodies. This concept was part of Aristotle’s theory of
motion. You can easily observe that a coin accelerates faster than a sheet of paper
when dropped simultaneously from the same height. But in this case, air resis-
tance plays a noticeable role. If the paper is crumpled into a compact ball, it gives
the coin a much better race. Similarly, a feather “floats” down much more slowly
than a coin falls. However, in a near-vacuum, where there is negligible air resis-
tance, the feather and the coin have the same acceleration—the acceleration due to
gravity (䉱 Fig. 2.14).
Astronaut David Scott performed a similar experiment on the Moon in 1971 by
simultaneously dropping a feather and a hammer from the same height. He did
not need a vacuum pump. The Moon has no atmosphere and therefore no air
resistance. The hammer and the feather reached the lunar surface together, but
both had a smaller acceleration and fell at a slower rate than on Earth. The acceler-
ation due to gravity near the Moon’s surface is only about one-sixth of that near
the Earth’s surface 1gM L g>62.
Currently accepted ideas about the motion of falling bodies are due in large
part to Galileo. He challenged Aristotle’s theory and experimentally investigated
the motion of such objects. Legend has it that Galileo studied the accelerations of
falling bodies by dropping objects of different weights from the top of the Leaning
Tower of Pisa. (See the accompanying Insight 2.1, Galileo Galilei and the Leaning
Tower of Pisa.)
It is customary to use y to represent the vertical direction and to take upward as
positive (as with the vertical y-axis of Cartesian coordinates). Because the accelera-
tion due to gravity is always downward, it is in the negative y-direction. This neg-
ative acceleration, a = - g = - 9.80 m>s2, should be substituted into the equations
of motion. However, the relationship a = - g may be expressed explicitly in the
equations for linear motion for convenience:
v = vo - gt (2.8¿)
(free-fall equations with
y = yo + vo t - 12 gt2 (2.11¿)
a y = - g expressed explicity)
v 2 = v 2o - 2g1y - yo2 (2.12¿)
Equation 2.10 applies to free fall as well, but it does not contain g:
y = yo + 121v + vo2t (2.10¿)

INSIGHT 2.1 Galileo Galilei and the Leaning Tower of Pisa

Galileo Galilei (䉴 Fig. 1) was born in Pisa, Italy, in 1564 during F I G U R E 1 Galileo
the Renaissance. Today, he is known throughout the world by Galilei is alleged to have
his first name and is often referred to as the father of modern performed free-fall
science and experimental physics, which attests to the magni- experiments by dropping
objects off the Leaning
tude of his scientific contributions.
Tower of Pisa.
One of Galileo’s greatest contributions to science was the
establishment of the scientific method—that is, investigation
through experiment. In contrast, Aristotle’s approach was
based on deduction. By the scientific method, for a theory to
be valid, it must correctly predict, or agree with, experimental
results. If it doesn’t, it is invalid or requires modification.
Galileo said, “I think that in the discussion of natural prob-
lems we ought not to begin at the authority of places of Scrip-
ture, but at sensible experiments and necessary
demonstrations.”* There is some debate as to whether Galileo actually did this,
Probably the most popular and well-known legend about but there is little doubt that he questioned Aristotle’s view on
Galileo is that he performed experiments with falling bodies the motion of falling objects. In 1638, Galileo wrote,
by dropping objects from the Leaning Tower of Pisa (䉲 Fig. 2).
Aristotle says that an iron ball of one hundred pounds
falling from a height of one hundred cubits reaches the
ground before a one-pound ball has fallen a single cubit. I
say that they arrive at the same time. You find, on making
the experiment, that the larger outstrips the smaller by two
finger-breadths, that is, when the larger has reached the
ground, the other is short of it by two finger-breadths; now
you would not hide behind these two fingers the ninety-
nine cubits of Aristotle.†
The experiments at the Tower of Pisa supposedly took
place around 1590. In his writings of about that time,
Galileo mentions dropping objects from a high tower, but
never specifically names the Tower of Pisa. A letter written
to Galileo from another scientist in 1641 describes the
dropping of a cannonball and a musket ball from the
Tower of Pisa. The first account of Galileo doing a similar
experiment was written a dozen years after his death by
Vincenzo Viviani, his last pupil and first biographer. It is
not known whether Galileo told this story to Viviani in his
declining years or Viviani created this picture of his former
teacher.
The important point is that Galileo recognized (and proba-
bly experimentally showed) that free-falling objects fall with
the same acceleration regardless of their mass or weight. (See
Fig. 2.14.) Galileo gave no reason as to why all objects in free
F I G U R E 2 The Leaning Tower of Pisa The tower, con- fall have the same acceleration, but Newton did, as will be
structed as a belfry for a nearby cathedral, was built on shift- learned in a later chapter.
ing subsoil. Construction began in 1173, and the tower started
to shift one way and then the other before inclining to its pre-
sent direction. Today, the tower leans about 5 m (16 ft) from *From Growth of Biological Thought: Diversity, Evolution & Inheritance,
the vertical at the top. It was closed in 1990, and efforts were by F. Meyr (Cambridge, MA: Harvard University Press, 1982).
†
made to stabilize and correct the leaning. Some improvement From Aristotle, Galileo, and the Tower of Pisa, by L. Cooper (Ithaca,
was made and the tower is now open to the public. NY: Cornell University Press, 1935).
The origin 1y = 02 of the frame of reference is usually taken to be at the initial

position of the object. Writing -g explicitly in the equations is a reminder of its
direction. Then the value of g is simply inserted as 9.80 m>s 2.
The equations can also be written with a = g, for example, v = vo + gt, with the
directional minus sign associated directly with g. In this case, a value of
g = - 9.80 m>s2 must be substituted for g each time. However, either method works,
and the choice is arbitrary. Your instructor may prefer one method over the other.
2.5 FREE FALL 53
Note that you must be explicit about the directions of vector quantities. The
location y and the velocities v and vo may be positive (up) or negative (down), but
the acceleration due to gravity is always downward.
The use of these equations and the sign convention (with -g explicitly
expressed in the equations) are illustrated in the following Examples. (This con-
vention will be used throughout the text.)
EXAMPLE 2.9 A Stone Thrown Downward: The Kinematic Equations Revisited

A boy on a bridge throws a stone vertically downward with Notice that g is listed as a positive number, since by our con-
an initial speed of 14.7 m>s toward the river below. If the vention the directional minus sign has already been put into
stone hits the water 2.00 s later, what is the height of the the previous equations of motion.
bridge above the water? Which equation(s) will provide the solution using the
given data? It should be evident that the distance the stone
T H I N K I N G I T T H R O U G H . This is a free-fall problem, but note
travels in an amount of time t is given directly by Eq. 2.11¿.
that the initial velocity is downward, which is taken as the
Taking yo = 0:
negative direction. It is important to express this factor explic-
itly. Draw a sketch to help you analyze the situation if needed. y = vo t - 12 gt2
SOLUTION. As usual, first writing what is given and what is = 1- 14.7 m>s212.00 s2 - 1
2 A 9.80 m>s2 B 12.00 s22
to be found:
= - 29.4 m - 19.6 m = - 49.0 m
Given: vo = - 14.7 m>s Find: y (bridge height
t = 2.00 s above water) The minus sign indicates that the rock’s displacement is
g 1 = 9.80 m>s22 downward, as it should be. Thus the height bridge is 49.0 m.
F O L L O W - U P E X E R C I S E . How much longer would it take for the stone to reach the river if the boy in this Example had dropped
the ball rather than thrown it? (Answers to all Follow-Up Exercises are given in Appendix VI at the back of the book.)
Reaction time is the time it takes a person to notice, think, and act in response to
a situation—for example, the time between first observing and then responding to
an obstruction on the road ahead by applying the brakes. Reaction time varies
with the complexity of the situation (and with the individual). In general, the
largest part of a person’s reaction time is spent thinking, but practice in dealing
with a given situation can reduce this time. The following Example gives a simple
method for measuring reaction time.
EXAMPLE 2.10 Measuring Reaction Time: Free Fall

A person’s reaction time can be measured by hav-
ing another person drop a ruler (without warning)
even with and through the first person’s thumb
and forefinger, as shown in 䉴 Fig. 2.15. After
observing the unexpected release, the first person
grasps the falling ruler as quickly as possible, and
the length of the ruler below the top of the finger is
noted. Suppose the ruler descends 18.0 cm before it
is caught. What is the person’s reaction time?
T H I N K I N G I T T H R O U G H . Both distance and time
are involved. This observation indicates which
kinematic equation should be used.
䉴 F I G U R E 2 . 1 5 Reaction time
A person’s reaction time can be mea-
sured by having the person grasp a
dropped ruler.
(continued on next page)
SOLUTION. Notice that only the distance of fall is given. Then solving for t,
However, a couple of other things are known, such as vo 2y 21- 0.180 m2
and g. So, taking yo = 0: t = = = 0.192 s
A -g C -9.80 m>s2
Given: y = - 18.0 cm = - 0.180 m Find: t (reaction time) Try this experiment with a fellow student and measure
vo = 0 your reaction time. Why do you think another person besides
you should drop the ruler?
g 1= 9.80 m>s22
F O L L O W - U P E X E R C I S E . A popular trick is to substitute a crisp
dollar bill lengthwise for the ruler in Fig. 2.15, telling the person
(Note that the distance y has been converted to meters. Why?)
that he or she can have the dollar if able to catch it. Is this proposal
It can be seen that Eq. 2.11¿; applies here (with vo = 0), giving
a good deal? (The length of a dollar is 15.7 cm.) (Answers to all
y = - 12 gt2 Follow-Up Exercises are given in Appendix VI at the back of the book.)
Here are some interesting facts about free-fall motion of an object thrown upward
in the absence of air resistance. First, if the object returns to its launch elevation, the
times of flight upward and downward are the same. Similarly, note that at the very
top of the trajectory, the object’s velocity is zero for an instant, but the acceleration
(even at the top) remains a constant 9.8 m>s2 downward. It is a common misconcep-
tion that at the top of the trajectory the acceleration is zero. If this were the case, the
object would remain there, as if gravity had been turned off!
Finally, the object returns to the starting point with the same speed as that at
which it was launched. (The velocities have the same magnitude, but are opposite
in direction.)
EXAMPLE 2.11 Free Fall Up and Down: Using Implicit Data

A worker on a scaffold in front of a billboard throws a ball v=0
straight up. The ball has an initial speed of 11.2 m>s when it y = ymax
leaves the worker’s hand at the top of the billboard
(䉴 Fig. 2.16). (a) What is the maximum height the ball reaches g
relative to the top of the billboard? (b) How long does it take
the ball to reach this height? (c) What is the position of the ball
v
at t = 2.00 s?
T H I N K I N G I T T H R O U G H . In part (a), only the upward part of ymax
g v g
the motion has to be considered. Note that the ball stops (zero
instantaneous velocity) at the maximum height, which allows
this height to be determined. In part (b), knowing the maxi-
mum height allows the determination of the upward time of
vo = 11.2 m/s
flight. In part (c), the distance–time equation (Eq.2.11¿) applies
for any time and therefore allows calculation of the position (y)
yo = 0
of the ball relative to the launch point at t = 2.00 s.
EWTON'S g v g
䉴 F I G U R E 2 . 1 6 Free fall up and down

Note the lengths of the velocity and acceler-
ation vectors at different times. (The
upward and downward paths of the ball
are horizontally displaced for illustration
purposes.)
SOLUTION. It might appear that all that is given is the initial velocity vo at time to. However, a couple of other pieces of informa-
tion are implied that should be recognized. One, of course, is the acceleration g, and the other is the velocity at the maximum
height where the ball stops. Here, in changing direction, the velocity of the ball is momentarily zero, so (again taking yo = 0):
Given: vo = 11.2 m>s Find: (a) ymax (maximum height above launch point)
g 1 = 9.80 m>s22 (b) tu (time upward)
v = 0 (at ymax) (c) y (at t = 2.00 s)
t = 2.00 s [for part (c)]
2.5 FREE FALL 55
(a) Notice that the height 1yo = 02 is referenced to the top of (c) The height of the ball at t = 2.00 s is given directly by
the billboard. For this part of the problem, we need be con- Eq. 2.11¿:
cerned with only the upward motion—a ball is thrown y = vo t - 12 gt2
upward and stops at its maximum height ymax. With v = 0 at
this height, ymax may be found directly from Eq. 2.12¿, = 111.2 m>s212.00 s2 - 12 19.80 m>s2212.00 s22
= 22.4 m - 19.6 m = 2.8 m
v 2 = 0 = v2o - 2gymax
Note that this height is 2.8 m above, or measured upward
So, from, the reference point 1yo = 02. The ball has reached its
v2o 111.2 m>s22 maximum height in 1.14 s and is on the way back down.
ymax = = 6.40 m
219.80 m>s22
2g
= Considered from another reference point, the situation in
part (c) can be analyzed by imagining dropping a ball from a
relative to the top of the billboard (yo = 0; see Fig. 2.16). height of ymax above the top of the billboard with vo = 0 and
asking how far it falls in a time t = 2.00 s - tu = 2.00 s -
(b) The time the ball travels upward to its maximum height is 1.14 s = 0.86 s. The answer is (this time with yo = 0 at the
designated tu. This is the time it takes for the ball to reach maximum height)
ymax, where v = 0. Since vo and v are known, the time tu can
be found directly from Eq. 2.8¿, y = vo t - 1
2 gt2 = 0 - 1
2 19.80 m>s2210.86 s22 = - 3.6 m
v = 0 = vo - gtu This height is the same as the position found previously, but
is measured with respect to the maximum height as the refer-
So, ence point; that is,
vo 11.2 m>s ymax - 3.6 m = 6.4 m - 3.6 m = 2.8 m
tu = = = 1.14 s
g 9.80 m>s 2 above the starting point.
F O L L O W - U P E X E R C I S E . At what height does the ball in this Example have a speed of 5.00 m>s? [Hint: The ball attains this height
twice—once on the way up, and once on the way down.] (Answers to all Follow-Up Exercises are given in Appendix VI at the back of
the book.)
When working vertical projectile problems involving motions up and down, it is often
convenient to divide the problem into two parts and consider each part separately. As
seen in Example 2.11, for the upward part of the motion, the velocity is zero at the maxi-
mum height. A quantity of zero usually simplifies the calculations. Similarly, the down-
ward part of the motion is analogous to that of an object dropped from a height where
the initial velocity can be taken as zero.
However, as Example 2.11 shows, the appropriate equations may be used directly for
any position or time of the motion. For instance, note in part (c) that the height was
found directly for a time after the ball had reached the maximum height. The velocity of
the ball at that time could also have been found directly from Eq. 2.8¿, v = vo - gt.
Also, note that the initial position was consistently taken as yo = 0. This assumption
is taken for convenience when the situation involves only one object (then yo = 0 at
to = 0). Using this convention can save a lot of time in writing and solving equations.
The same is true with only one object in horizontal motion: You can usually take
xo = 0 at to = 0. There are a couple of exceptions to this case, however. The first is if the
problem specifies the object to be initially located at a position other than xo = 0, and
the second is if the problem involves two objects, as in Example 2.7. In the latter case, if
one object is taken to be initially at the origin, the other’s initial position is not zero.
EXAMPLE 2.12 Lunar Landing

A Lunar Lander makes a descent toward a level plain on T H I N K I N G I T T H R O U G H This appears to be analogous to a
the Moon. It descends slowly by using retro (braking) rock- simple free-fall problem of throwing an object downward—
ets. At a height of 6.0 m above the surface, the rockets are and it is, but the situation takes place on the Moon. It was
shut down with the Lander having a downward speed of noted previously that the acceleration due to gravity on the
1.5 m>s. What is the speed of the Lander just before touch- Moon, gM, is one-sixth of that on the Earth, gE. (No problem
ing down? with air resistance on the Moon—it has no atmosphere.)
SOLUTION. So,
v 2 = 21 m2>s 2 and v = 221 m2>s 2 = ; 4.6 m>s

Given: y = - 6.0 m Find: v (just before
vo = - 1.5 m>s touching down)
gM = gE >6 This is the velocity, which we know is downward, so the neg-
= 19.8 m>s22>6 ative root is selected, and v = - 4.6 m>s, and the speed is
= 1.6 m>s2 4.6 m>s.
Then Eq. 2.12¿ can be used: F O L L O W - U P E X E R C I S E . From the 6.0-m height, how long did
v2 = v2o - 2gM y = 1 - 1.5 m>s22 - 211.6 m>s221 -6.0 m2 the Lander’s descent take? (Answers to all Follow-Up Exercises
= 21 m2>s 2 are given in Appendix VI at the back of the book.)
DID YOU LEARN?

➥ When gravitational attraction is the sole influence on an object, it is in free fall.This
includes objects dropped from rest, projected upward, or thrown downward.
➥ The gravitational attraction on the Moon is one-sixth of that on the Earth, hence
objects in free fall on the Moon fall more slowly than on the Earth.
➥ Heavy and light objects in free fall dropped from equal heights strike the ground at
the same time because the acceleration due to gravity is the same for both.
PULLING IT TOGETHER Learning Kinematics by Playing

A student drops a tennis ball from a dormitory window 13.0 m T H I N K I N G I T T H R O U G H . This example uses kinematic equa-
above the ground. At that instant, a student on the ground tions in free-fall situations—down and up. (a) Since both balls
launches another ball straight up directly toward the dropped experience constant acceleration, the curves will be parabolas,
ball with an initial speed of 15.0 m>s in order to hit the dropped intersecting at some height and time to be determined from
ball. The upward ball is launched from shoulder height of 1.0 m the mathematics. Sign conventions are crucial. (b) When the
above the ground. (a) Sketch the location of each ball as a func- balls collide, their heights above the launch point are the
tion of time. You should show two curves, labeled 1 for the same. Setting these equal should enable the determination of
dropped ball and 2 for the thrown ball. (b) How long does it take the time. (c) Once the time is known, the height of the balls
before they hit? (c) How far above the ground are the balls when can be determined from kinematics. (d) The sign of the
they hit? (d) Which way is the thrown ball moving up or down thrown ball’s velocity at this time will tell which way it is
when they hit, or is it at rest at that instant? moving on collision.
SOLUTION.
Given: 13.0 m - 1.0 m = 12.0 m. (Initial distance between Find: (a) each ball’s location variation with time
the balls.) The height of 1.0 m will be taken as yo = 0 (b) t (time to hit)
for convenience, making the window at y = 12.0 m, (c) y (distance above ground when they hit)
relative to this reference point. (d) direction of thrown ball at collision
vo1 = 0 (dropped ball initial velocity)
vo2 = 15.0 m>s (thrown ball initial velocity)
yo1 = 12.0 m (dropped ball initial location)
yo2 = 0 m (thrown ball initial location)
(a) The dropped ball’s trajectory has zero slope to start since +y
its initial velocity is zero, and it is a downward curving 12.0 m
parabola starting at y = 12.0 m (䉴 Fig. 2.17). The thrown ball
also will be a downward-curving parabola, but due to its ini- 1
tial upward velocity, it starts with an upward slope starting at
y = 0 m. The intersection point represents the balls’ common
location and the time when they collide.
(b) Each ball’s location follows the general one-dimensional
free fall equation y = yo + vo t - 12 gt2. By putting in the num-
bers for each ball, an equation for each ball’s location as a
function of time is obtained. [All times are in seconds (s), 2
velocities in meters per second 1m>s2 and accelerations in
meters per second squared 1m>s22, but units are omitted here
for convenience.]
0
y1 = yo1 + vo1 t - 12 gt2 = 12.0 + 0 - 4.90 t2 t
䉱 F I G U R E 2 . 1 7 Ball location versus time.

LEARNING PATH REVIEW 57
and Then the height above the ground is

1 2 1.0 m + y1 = 1.0 m + 8.86 m = 9.9 m.
y2 = yo2 + vo2 t - 2 gt = 0 + 15.0t - 4.90t2
Setting y1 = y2 yields 12.0 - 4.90t2 = 15.0t - 4.90t2, which (d) The general equation for velocity in one-dimensional free
can be solved for t since it becomes 12.0 = 15.0t. Thus the col- fall is v = vo - gt. Applying this to ball 2:
lision time is t = 0.80 s. v2 = vo2 - gt = 15.0 - 9.8010.802 = + 7.16 m>s
(c) Either location equation can be used to determine the
So from the plus sign for the velocity, it is clear that ball 2 is
height at collision. Ball 1 has a simpler equation, hence
rising when the balls collide.
y1 = 12.0 + 0 - 4.9010.8022 = 8.86 m
(You should check to see that y2 = 8.86 m also.)
■ Motion involves a change of position; it can be described in ¢v v2 - v1

qa = = (2.5)
terms of the distance moved (a scalar) or the displacement ¢t t2 - t1
(a vector).
■ A scalar quantity has magnitude (value and units) only; a
vector quantity has magnitude and direction. v v = vo + at
Velocity
a
e =+
Slop at
vo
vo
0 t Time
Motion in positive direction—speeding up
x1 x2
x
1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 (meters)
8.0 m t
0
–vo Time
–vo
■ Average speed (sq) (a scalar) is the distance traveled divided
Velocity
Slo
pe
by the total time: =–
a –at
distance traveled –v = –vo –at

average speed = –v
total time to travel that distance
d d
qs = = (2.1) Motion in negative direction—speeding up
¢t t2 - t1
■ Average velocity (a vector) is the displacement divided by
the total travel time: ■ The kinematic equations for constant acceleration:
displacement v + vo
average velocity = vq = (2.9)
total travel time 2
¢x x2 - x1 v = vo + at (2.8)
vq = = or x = xo + vqt (2.3)
x = xo + 12 1v + vo2t
¢t t2 - t1
(2.10)
Distance
x = xo + vo t + 12 at2 (2.11)
50 km 100 km 150 km
v 2 = v2o + 2a1x - xo2 (2.12)
0 1.0 h 2.0 h 3.0 h
Time
a positive Result:
■ Instantaneous velocity (a vector) describes how fast some- Faster in
+x direction
thing is moving and in what direction at a particular instant v positive
of time. –x +x
■ Acceleration is the time rate of change of velocity and hence Result:
a negative
is a vector quantity: Slower in
+x direction
change in velocity v positive
average acceleration = –x +x
time to make the change
■ An object in free fall has a constant acceleration of magni- ■ Expressing a = - g in the kinematic equations for constant
tude g = 9.80 m>s2 (acceleration due to gravity) near the acceleration in the y-direction yields the following:
surface of the Earth. v = vo - gt (2.8¿)
y = yo + 12 1v + vo2t (2.10¿)
y = yo + vo t - 12 gt2 (2.11¿)
2
v = v2o - 2g1y - yo2 (2.12¿)

2.1 DISTANCE AND SPEED: SCALAR 10. For a constant acceleration, what changes uniformly?
QUANTITIES (a) acceleration, (b) velocity, (c) displacement, (d) distance.
AND 11. Which one of the following is true for a deceleration?
2.2 ONE-DIMENSIONAL DISPLACEMENT (a) The velocity remains constant. (b) The acceleration is
AND VELOCITY: VECTOR QUANTITIES negative. (c) The acceleration is in the direction opposite
to the velocity. (d) The acceleration is zero.
1. A scalar quantity has (a) only magnitude, (b) only direc-
12. A car accelerates from 80 km>h to 90 km>h, while a
tion, (c) both magnitude and dirrection.
moped accelerates from 0 to 20 km>h in twice the time.
2. Which of the following is always true about the magni- Which of the following is true: (a) The car has the greater
tude of the displacement: (a) It is greater than the dis- acceleration; (b) the moped has the greater acceleration; or
tance traveled; (b) it is equal to the distance traveled; (c) they both have the same magnitude of acceleration?
(c) it is less than the distance traveled; or (d) it is less
than or equal to the distance traveled?
3. A vector quantity has (a) only magnitude, (b) only direc- 2.4 KINEMATIC EQUATIONS (CONSTANT
tion, (c) both direction and magnitude. ACCELERATION)
4. What can be said about average speed relative to the 13. For a constant linear acceleration, the velocity-versus-time
magnitude of the average velocity? (a) greater than, graph is (a) a horizontal line, (b) a vertical line, (c) a non-
(b) equal to, (c) both a and b. horizontal and nonvertical straight line, (d) a curved line.
5. Distance is to displacement as (a) centimeters is to 14. For a constant linear acceleration, the position-versus-
meters, (b) a vector is to a scalar, (c) speed is to velocity, time graph would be (a) a horizontal line, (b) a vertical
(d) distance is to time. line, (c) a nonhorizontal and nonvertical straight line,
(d) a curve.
15. An object accelerates uniformly from rest for t seconds.
2.3 ACCELERATION
The object’s average speed for this time interval is
6. On a position-versus-time plot for an object that has a (a) 12 at, (b) 12 at2, (c) 2at, (d) 2at2.
constant acceleration, the graph is (a) a horizontal line,
(b) a nonhorizontal and nonvertical straight line, (c) a
vertical line, (d) a curve. 2.5 FREE FALL
7. An acceleration may result from (a) an increase in speed, 16. An object is thrown vertically upward. Which of the fol-
(b) a decrease in speed, (c) a change of direction, (d) all of lowing statements is true: (a) Its velocity changes nonuni-
the preceding. formly; (b) its maximum height is independent of the
8. A negative acceleration can cause (a) an increase in initial velocity; (c) its travel time upward is slightly greater
speed, (b) a decrease in speed, (c) either a or b. than its travel time downward; or (d) its speed on return-
9. The gas pedal of an automobile is commonly referred to ing to its starting point is the same as its initial speed?
as the accelerator. Which of the following might also be 17. The free-fall motion described in this section applies to
called an accelerator: (a) the brakes, (b) the steering (a) an object dropped from rest, (b) an object thrown ver-
wheel, (c) the gear shift, or (d) all of the preceding? tically downward, (c) an object thrown vertically
Explain. upward, (d) all of the preceding.
18. A dropped object in free fall (a) falls 9.8 m each second, 20. When an object is thrown vertically upward, it is accelerat-
(b) falls 9.8 m during the first second, (c) has an increase ing on (a) the way up, (b) the way down, (c) both a and b.
in speed of 9.8 m>s each second, (d) has an increase in
acceleration of 9.8 m>s2 each second.
19. An object is thrown straight upward. At its maximum
height, (a) its velocity is zero, (b) its acceleration is zero,
(c) both a and b.
2.1 DISTANCE AND SPEED: SCALAR 10. An object traveling at a constant velocity vo experi-
QUANTITIES ences a constant acceleration in the same direction for
AND a period of time t. Then an acceleration of equal mag-
2.2 ONE-DIMENSIONAL DISPLACEMENT nitude is experienced in the opposite direction of vo for
the same period of time t. What is the object’s final
AND VELOCITY: VECTOR QUANTITIES
velocity?
1. Can the displacement of a person’s trip be zero, yet the
distance involved in the trip be nonzero? How about the 11. Car A is in a straight-line distance d from a starting line,
reverse situation? Explain. and Car B is a distance of 2 d from the line. Accelerating
uniformly from rest, it is desired that both cars cross the
2. You are told that a person has walked 750 m. What can
starting line at the same speed. If so, which car has the
you safely say about the person’s final position relative
greater acceleration, and how much greater?
to the starting point?
3. If the displacement of an object is 300 m north, what can
you say about the distance traveled by the object? 2.4 KINEMATIC EQUATIONS (CONSTANT
4. Speed is the magnitude of velocity. Is average speed the ACCELERATION)
magnitude of average velocity? Explain.
12. If an object’s velocity-versus-time graph is a horizontal
5. The average velocity of a jogger on a straight track is
line, what can you say about the object’s acceleration?
computed to be +5 km>h. Is it possible for the jogger’s
instantaneous velocity to be negative at any time during 13. In solving a kinematic equation for x, which has a nega-
the jog? Explain. tive acceleration, is x necessarily negative?
14. How many variables must be known to solve a kine-
2.3 ACCELERATION matic equation?
15. Consider Eq. 2.12, v2 = v 2o + 2a(x - xo). An object starts
6. A car is traveling at a constant speed of 60 mi>h on a cir-
from rest (vo = 0) and accelerates. Since v is squared and
cular track. Is the car accelerating? Explain.
therefore always positive, can the acceleration be nega-
7. Does a fast-moving object always have higher acceleration tive? Explain.
than a slower object? Give a few examples, and explain.
8. A classmate states that a negative acceleration always
means that a moving object is decelerating. Is this state-
2.5 FREE FALL
ment true? Explain. 16. When a ball is thrown upward, what are its velocity and
9. Describe the motions of the two objects that have the acceleration at its highest point?
velocity-versus-time plots shown in 䉲 Fig. 2.18. 17. If the instantaneous velocity of an object is zero, is the
acceleration necessarily zero?
v (a) 18. Imagine you are in space far away from any planet, and
(b) you throw a ball as you would on the Earth. Describe the
ball’s motion.
19. A person drops a stone from the window of a building.
One second later, she drops another stone. How does the
Velocity
distance between the stones vary with time?

20. How would free fall on the Moon differ from that on the
Earth?
0 t
Time
䉱 F I G U R E 2 . 1 8 Description of motion See Conceptual

Question 9.
EXERCISES
Many exercise sections will include “paired” exercises. These exercise pairs, identified with red numbers, are
bered) is similar in nature, and its answer is given at in Appendix VII the back of the book.
2.1 DISTANCE AND SPEED: SCALAR caught at the initial height 2.4 s after being thrown,
QUANTITIES (a) what is the ball’s average speed, and (b) what is its
AND average velocity?
2.2 ONE-DIMENSIONAL DISPLACEMENT 13. ● ● An insect crawls along the edge of a rectangular
AND VELOCITY: VECTOR QUANTITIES swimming pool of length 27 m and width 21 m
(䉲 Fig. 2.19). If it crawls from corner A to corner B in
1. ● What is the magnitude of the displacement of a car 30 min, (a) what is its average speed, and (b) what is the
that travels half a lap along a circle that has a radius of magnitude of its average velocity?
150 m? How about when the car travels a full lap?
2. ● A motorist travels 80 km at 100 km>h, and 50 km at A
75 km>h. What is the average speed for the trip?
3. ● An Olympic sprinter can run 100 yd in 9.0 s. At the same 21 m
rate, how long would it take the sprinter to run 100 m?
4. ● A senior citizen walks 0.30 km in 10 min, going
around a shopping mall. (a) What is her average speed

27 m
in meters per second? (b) If she wants to increase her
average speed by 20% when walking a second lap, what B
would her travel time in minutes have to be?
5. ● ● A hospital patient is given 500 cc of saline by IV.
䉱 F I G U R E 2 . 1 9 Speed versus velocity See Exercise 13.
If the saline is received at a rate of 4.0 mL>min, how long (Not drawn to scale; insect is displaced for clarity.)
will it take for the half liter to run out?
6. ● ● A hospital nurse walks 25 m to a patient’s room at the 14. ● ● A plot of position versus time is shown in 䉲 Fig. 2.20
end of the hall in 0.50 min. She talks with the patient for 4.0 for an object in linear motion. (a) What are the average
min, and then walks back to the nursing station at the velocities for the segments AB, BC, CD, DE, EF, FG, and
same rate she came. What was the nurse’s average speed? BG? (b) State whether the motion is uniform or nonuni-
7. ● ● A train makes a round trip on a straight, level track. form in each case. (c) What is the instantaneous velocity
The first half of the trip is 300 km and is traveled at a at point D?
speed of 75 km>h. After a 0.50 h layover, the train returns x
the 300 km at a speed of 85 km>h. What is the train’s
(a) average speed and (b) average velocity? 10.0
8. IE ● ● A car travels three-quarters of a lap on a circular D
9.0
track of radius R. (a) The magnitude of the displacement
is (1) less than R, (2) greater than R, but less than 2R, 8.0
(3) greater than 2R. (b) If R = 50 m, what is the magni- 7.0 C E
tude of the displacement?
Position (m)
6.0
9. ● ● The interstate distance between two cities is 150 km.
(a) If you drive the distance at the legal speed limit of 5.0
65 mi>h, how long would the trip take? (b) Suppose on 4.0
the return trip you pushed it up to 80 mi>h (and didn’t
get caught). How much time would you save? 3.0
F G
10. IE ● ● A race car travels a complete lap on a circular 2.0
track of radius 500 m in 50 s. (a) The average velocity of A B
1.0
the race car is (1) zero, (2) 100 m>s, (3) 200 m>s, (4) none
of the preceding. Why? (b) What is the average speed of t
0 2.0 4.0 6.0 8.0 10.0 12.0
the race car?
Time (s)
11. IE ● ● A student runs 30 m east, 40 m north, and 50 m
west. (a) The magnitude of the student’s net displace- 䉱 F I G U R E 2 . 2 0 Position versus time See Exercise 14.
ment is (1) between 0 and 20 m, (2) between 20 m and
40 m, (3) between 40 m and 60 m. (b) What is his net 15. ●● In demonstrating a dance step, a person moves in
displacement? one dimension, as shown in 䉴 Fig. 2.21. What are (a) the
12. ● ● A student throws a ball vertically upward such that average speed and (b) the average velocity for each
it travels 7.1 m to its maximum height. If the ball is phase of the motion? (c) What are the instantaneous
EXERCISES 61
velocities at t = 1.0 s, 2.5 s, 4.5 s, and 6.0 s? (d) What is How long will it take for the runners to meet, and at what
the average velocity for the interval between t = 4.5 s position will they meet if they maintain these speeds?
and t = 9.0 s? [Hint: Recall that the overall displacement
is the displacement between the starting point and the 4.50 m/s 3.50 m/s
ending point.]
4.0
100 m
3.0
䉱 F I G U R E 2 . 2 2 When and where do they meet?
Position (m)
2.0 See Exercise 22.

1.0
2.3 ACCELERATION
0 t
2.0 4.0 6.0 8.0 10.0 23. ● An automobile traveling at 15.0 km>h along a straight,
–1.0 level road accelerates to 65.0 km>h in 6.00 s. What is the
magnitude of the auto’s average acceleration?
–2.0
Time (s) 24. ● A sports car can accelerate from 0 to 60 mi>h in 3.9 s.
What is the magnitude of the average acceleration of the
䉱 F I G U R E 2 . 2 1 Position versus time See Exercise 15. car in meters per second squared?
25. ● If the sports car in Exercise 24 can accelerate at a rate
16. ●● A high school kicker makes a 30.0-yd field goal of 7.2 m>s2, how long does the car take to accelerate from
attempt (in American football) and hits the crossbar at a 0 to 60 mi>h?
height of 10.0 ft. (a) What is the net displacement of the 26. IE ● ● A couple is traveling by car down a straight high-
football from the time it leaves the ground until it hits way at 40 km>h. They see an accident in the distance, so
the crossbar? (b) Assuming the football took 2.50 s to the driver applies the brakes, and in 5.0 s the car uni-
hit the crossbar, what was its average velocity? formly slows down to rest. (a) The direction of the accel-
(c) Explain why you cannot determine its average speed eration vector is (1) in the same direction as, (2) opposite
from these data. to, (3) at 90° relative to the velocity vector. Why? (b) By
17. ● ● The location of a moving particle at a particular time how much must the velocity change each second from
is given by x = at - bt2, where a = 10 m>s and the start of braking to the car’s complete stop?
b = 0.50 m>s2. (a) Where is the particle at t = 0?
27. ●● A paramedic drives an ambulance at a constant
(b) What is the particle’s displacement for the time inter-
speed of 75 km>h on a straight street for ten city blocks.
val t1 = 2.0 s and t2 = 4.0 s?
Because of heavy traffic, the driver slows to 30 km>h in
18. ● ● The displacement of an object is given as a function
6.0 s and travels two more blocks. What was the average
of time by x = 3t2 m. What is the magnitude of the acceleration of the vehicle?
average velocity for (a) ¢t = 2.0 s - 0 s, and
(b) ¢t = 4.0 s - 2.0 s? 28. ●● During liftoff, a hot-air balloon accelerates upward at a
rate of 3.0 m>s2. The balloonist drops an object over the side
19. ● ● Short hair grows at a rate of about 2.0 cm>month.
of the gondola when the speed is 15 m>s. (a) What is the
A college student has his hair cut to a length of 1.5 cm.
object’s acceleration after it is released (relative to the
He will have it cut again when the length is 3.5 cm. How
ground)? (b) How long does it take to hit the ground?
long will it be until his next trip to the barber shop?
20. ● ● ● A student driving home for the holidays starts at
29. ●● A new-car owner wants to show a friend how fast
8:00 AM to make the 675-km trip, practically all of which her sports car is. The friend gets in his car and drives
is on nonurban interstate highways. If she wants to down a straight, level highway at a constant speed of
arrive home no later than 3:00 PM, what must be her min- 60 km>h to a point where the sports car is waiting. As
imum average speed? Will she have to exceed the the friend’s car just passes, the sports car accelerates at a
65-mi>h speed limit? rate of 2.0 m>s2. (a) How long does it take for the sports
car to catch up to the friend’s car? (b) How far down the
21. ● ● ● A regional airline flight consists of two legs with an
road does the sports car catch up to the friend’s car?
intermediate stop. The airplane flies 400 km due north
(c) How fast is the sports car going at this time?
from airport A to airport B. From there, it flies 300 km
due east to its final destination at airport C. (a) What is 30. ●● After landing, a jetliner on a straight runway taxis to
the plane’s displacement from its starting point? (b) If a stop at an average velocity of - 35.0 km>h. If the plane
the first leg takes 45 min and the second leg 30 min, what takes 7.00 s to come to rest, what are the plane’s initial
is the average velocity for the trip? (c) What is the aver- velocity and acceleration?
age speed for the trip? (d) Why is the average speed not 31. ●● A train on a straight, level track has an initial speed
the same as the magnitude for the average velocity? of 35.0 km>h. A uniform acceleration of 1.50 m>s2 is
22. ● ● ● Two runners approaching each other on a straight applied while the train travels 200 m. (a) What is the
track have constant speeds of 4.50 m>s and 3.50 m>s, speed of the train at the end of this distance? (b) How
respectively, when they are 100 m apart (䉴Fig. 2.22). long did it take for the train to travel the 200 m?
32. A hockey puck sliding along the ice to the left hits the
●● 36. ●●● A train normally travels at a uniform speed of
boards head-on with a speed of 35 m>s. As it reverses 72 km>h on a long stretch of straight, level track. On a
direction, it is in contact with the boards for 0.095 s, particular day, the train must make a 2.0-min stop at a
before rebounding at a slower speed of 11 m>s. Deter- station along this track. If the train decelerates at a uni-
mine the average acceleration the puck experienced form rate of 1.0 m>s 2 and, after the stop, accelerates at a
while hitting the boards. Typical car accelerations are rate of 0.50 m>s 2, how much time is lost because of stop-
5.0 m>s2. Comment on the size of your answer, and why ping at the station?
it is so different from this value, especially when the
puck speeds are similar to car speeds.
33. ● ● What is the acceleration for each graph segment in 2.4 KINEMATIC EQUATIONS (CONSTANT
䉲 Fig. 2.23? Describe the motion of the object over the
ACCELERATION)
total time interval. 37. ● At a sports car rally, a car starting from rest accelerates
v uniformly at a rate of 9.0 m>s2 over a straight-line dis-
tance of 100 m. The time to beat in this event is 4.5 s.
Does the driver beat this time? If not, what must the
10.0 minimum acceleration be to do so?
(4.0, 8.0) (10.0, 8.0)
Velocity (m/s)
8.0 38. ● A car accelerates from rest at a constant rate of 2.0 m>s2
6.0 for 5.0 s. (a) What is the speed of the car at the end of that
time? (b) How far does the car travel in this time?
4.0
39. ● A car traveling at 25 mi>h is to stop on a 35-m-long
2.0 shoulder of the road. (a) What is the required magni-
t tude of the minimum acceleration? (b) How much time
0
4.0 8.0 12.0 16.0 will elapse during this minimum deceleration until the
Time (s) car stops?
40. ● A motorboat traveling on a straight course slows uni-
䉱 F I G U R E 2 . 2 3 Velocity versus time See Exercises 33 and 51. formly from 60 km>h to 40 km>h in a distance of 50 m.
What is the boat’s acceleration?
34. ●● 䉲 Figure 2.24 shows a plot of velocity versus time for
an object in linear motion. (a) Compute the acceleration 41. ●● The driver of a pickup truck going 100 km>h applies
for each phase of motion. (b) Describe how the object the brakes, giving the truck a uniform deceleration of
moves during the last time segment. 6.50 m>s2 while it travels 20.0 m. (a) What is the speed of
the truck in kilometers per hour at the end of this dis-
v tance? (b) How much time has elapsed?
10.0 42. ●● A roller coaster car traveling at a constant speed of
8.0
20.0 m>s on a level track comes to a straight incline
with a constant slope. While going up the incline, the
6.0 car has a constant acceleration of 0.750 m>s2 in magni-
4.0 tude. (a) What is the speed of the car at 10.0 s on the
Velocity (m/s)
incline? (b) How far has the car traveled up the incline
2.0
at this time?
0 t 43. ●● A rocket car is traveling at a constant speed of
2.0 4.0 6.0 8.0 10.0 12.0
–2.0 250 km>h on a salt flat. The driver gives the car a reverse
thrust, and the car experiences a continuous and con-
– 4.0
stant deceleration of 8.25 m>s2. How much time elapses
–6.0 until the car is 175 m from the point where the reverse
–8.0 thrust is applied? Describe the situation for your answer.
44. ●● Two identical cars capable of accelerating at 3.00 m>s2
–10.0
are racing on a straight track with running starts. Car A
–12.0 has an initial speed of 2.50 m>s; car B starts with speed of
Time (s)
5.00 m>s. (a) What is the separation of the two cars after
10 s? (b) Which car is moving faster after 10 s?
䉱 F I G U R E 2 . 2 4 Velocity versus time See Exercises 34 and 55.
45. ●●According to Newton’s laws of motion (which will
35. ●● A car initially traveling to the right at a steady speed be studied in Chapter 4), a frictionless 30° incline should
of 25 m>s for 5.0 s applies its brakes and slows at a con- provide an acceleration of 4.90 m>s2 down the incline. A
stant rate of 5.0 m>s2 for 3.0 s. It then continues traveling student with a stopwatch finds that an object, starting
to the right at a steady but slower speed with no addi- from rest, slides down a 15.00-m very smooth incline in
tional braking for another 6.0 s. (a) To help with the cal- exactly 3.00 s. Is the incline frictionless?
culations, make a sketch of the car’s velocity versus 46. IE ● ● An object moves in the +x-direction at a speed of
time, being sure to show all three time intervals. 40 m>s. As it passes through the origin, it starts to experi-
(b) What is its velocity after the 3.0 s of braking? (c) What ence a constant acceleration of 3.5 m>s 2 in the
was its displacement during the total 14.0 s of its -x-direction. (a) What will happen next? (1) The object
motion? (d) What was its average speed for the 14.0 s? will reverse its direction of travel at the origin; (2) the
EXERCISES 63
object will keep traveling in the +x-direction; (3) the 57. ● ● ● A car accelerates horizontally from rest on a level
object will travel in the + x-direction and then reverses road at a constant acceleration of 3.00 m>s2. Down the
its direction. Why? (b) How much time elapses before road, it passes through two photocells (“electric eyes”
the object returns to the origin? (c) What is the velocity of designated by 1 for the first one and 2 for the second
the object when it returns to the origin? one) that are separated by 20.0 m. The time interval to
47. ●● A rifle bullet with a muzzle speed of 330 m>s is fired travel this 20.0-m distance as measured by the electric
directly into a special dense material that stops the bullet eyes is 1.40 s. (a) Calculate the speed of the car as it
in 25.0 cm. Assuming the bullet’s deceleration to be con- passes each electric eye. (b) How far is it from the start to
stant, what is its magnitude? the first electric eye? (c) How long did it take the car to
get to the first electric eye?
48. ●● The speed limit in a school zone is 40 km>h (about
58. ● ● ● An automobile is traveling on a long, straight high-
25 mi>h). A driver traveling at this speed sees a child run
onto the road 13 m ahead of his car. He applies the way at a steady 75.0 mi>h when the driver sees a wreck
brakes, and the car decelerates at a uniform rate of 150 m ahead. At that instant, she applies the brakes
8.0 m>s2. If the driver’s reaction time is 0.25 s, will the (ignore reaction time). Between her and the wreck are
car stop before hitting the child? two different surfaces. First there is 100 m of ice, where
the deceleration is only 1.00 m>s2. From then on, it is dry
49. ●● Assuming a reaction time of 0.50 s for the driver in
concrete, where the deceleration is a more normal
Exercise 48, will the car stop before hitting the child?
7.00 m>s2. (a) What was the car’s speed just after leaving
50. ●● A bullet traveling horizontally at a speed of 350 m>s the icy portion of the road? (b) What is the total distance
hits a board perpendicular to its surface, passes through her car travels before it comes to a stop? (c) What is the
and emerges on the other side at a speed of 210 m>s. If total time it took the car to stop?
the board is 4.00 cm thick, how long does the bullet take
to pass through it?
51. ●● (a) Show that the area under the curve of a velocity- 2.5 FREE FALL
versus-time plot for a constant acceleration is equal to 59. ● A student drops a ball from the top of a tall building;
the displacement. [Hint: The area of a triangle is ab>2, or the ball takes 2.8 s to reach the ground. (a) What was the
one-half the altitude times the base.] (b) Compute the ball’s speed just before hitting the ground? (b) What is
distance traveled for the motion represented by Fig. 2.23. the height of the building?
52. IE ● ● An object initially at rest experiences an acceleration 60. IE ● The time it takes for an object dropped from the top
of 2.00 m>s2 on a level surface. Under these conditions, it of cliff A to hit the water in the lake below is twice the
travels 6.00 m. Let’s designate the first 3.00 m as phase 1 time it takes for another object dropped from the top of
with a subscript of 1 for those quantities, and the second cliff B to reach the lake. (a) The height of cliff A is (1) one-
3.00 m as phase 2 with a subscript of 2. (a) The times for half, (2) two times, (3) four times that of cliff B. (b) If it
traveling each phase should be related by which condition: takes 1.80 s for the object to fall from cliff A to the water,
(1) t1 6 t2, (2) t1 = t2, or (3) t1 7 t2? (b) Now calculate the what are the heights of cliffs A and B?
two travel times and compare them quantitatively.
61. ● For the motion of a dropped object in free fall, sketch
53. IE ● ● A car initially at rest experiences loss of its park-
the general forms of the graphs of (a) v versus t and (b) y
ing brake and rolls down a straight hill with a constant
versus t.
acceleration of 0.850 m>s 2, traveling a total of 100 m.
Let’s designate the first half of the distance as phase 1 62. ● You can perform a popular trick by dropping a dollar
with a subscript of 1 for those quantities, and the sec- bill (lengthwise) through the thumb and forefinger of a
ond half as phase 2 with a subscript of 2. (a) The car’s fellow student. Tell your fellow student to grab the dol-
speeds at the end of each phase should be related lar bill as fast as possible, and he or she can have the dol-
by which condition (1) v1 6 12 v2 , (2) v1 = 12 v2 , or lar if able to catch it. (The length of a dollar is 15.7 cm,
(3) v1 7 12 v2? (b) Now calculate the two speeds and and the average human reaction time is about 0.20 s.
compare them quantitatively. See Fig. 2.15.) Is this proposal a good deal? Justify your
answer.
54. ● ● An object initially at rest experiences an acceleration
of 1.5 m>s2 for 6.0 s and then travels at that constant 63. ● A juggler tosses a ball vertically a certain distance.
velocity for another 8.0 s. What is the object’s average How much higher must the ball be tossed so as to spend
velocity over the 14-s interval? twice as much time in the air?
55. ● ● ● Figure 2.24 shows a plot of velocity versus time for 64. ● A boy throws a stone straight upward with an initial
an object in linear motion. (a) What are the instantaneous speed of 15.0 m>s. What maximum height will the stone
velocities at t = 8.0 s and t = 11.0 s? (b) Compute the reach before falling back down?
final displacement of the object. (c) Compute the total 65. ● In Exercise 64, what would be the maximum height of
distance the object travels. the stone if the boy and the stone were on the surface of
56. IE ● ● ● (a) A car traveling at a speed of v can brake to an the Moon, where the acceleration due to gravity is only
emergency stop in a distance x. Assuming all other driving one-sixth of that of the Earth’s?
conditions are similar, if the traveling speed of the car dou- 66. ●● The Petronas Twin Towers in Malaysia and the
bles, the stopping distance will be (1) 22x, (2) 2x, (3) 4x. Chicago Sears Tower have heights of about 452 m and
(b) A driver traveling at 40.0 km>h in a school zone can 443 m, respectively. If objects were dropped from the top
brake to an emergency stop in 3.00 m. What would be the of each, what would be the difference in the time it takes
braking distance if the car were traveling at 60.0 km>h? the objects to reach the ground?
67. ●● In an air bag test, a car traveling at 100 km>h is

remotely driven into a brick wall. Suppose an identical
car is dropped onto a hard surface. From what height
would the car have to be dropped to have the same
impact as that with the brick wall?
68. ●● You throw a stone vertically upward with an initial
speed of 6.0 m>s from a third-story office window. If the
1.35 m
window is 12 m above the ground, find (a) the time the
stone is in flight and (b) the speed of the stone just before
it hits the ground.
69. IE ● ● A Super Ball is dropped from a height of 4.00 m.
Assuming the ball rebounds with 95% of its impact 䉳 FIGURE 2.26
speed, (a) the ball would bounce to (1) less than 95%, From where did it
(2) equal to 95%, or (3) more than 95% of the initial come? See Exercise 73.
height? (b) How high will the ball go?
74. ● ● ● A tennis ball is dropped from a height of 10.0 m. It
70. ●● In 䉲Fig. 2.25, a student at a window on the second
floor of a dorm sees his math professor walking on the rebounds off the floor and comes up to a height of only
sidewalk beside the building. He drops a water balloon 4.00 m on its first rebound. (Ignore the small amount of
from 18.0 m above the ground when the professor is time the ball is in contact with the floor.) (a) Determine the
1.00 m from the point directly beneath the window. If ball’s speed just before it hits the floor on the way down.
the professor is 1.70 m tall and walks at a rate of (b) Determine the ball’s speed as it leaves the floor on its
0.450 m>s, does the balloon hit her? If not, how close way up to its first rebound height. (c) How long is the ball
does it come? in the air from the time it is dropped until the time it
reaches its maximum height on the first rebound?
75. ● ● ● A pollution-sampling rocket is launched straight
upward with rockets providing a constant acceleration of
12.0 m>s2 for the first 1000 m of flight. At that point the
rocket motors cut off and the rocket itself is in free fall.
Ignore air resistance. (a) What is the rocket’s speed when
the engines cut off? (b) What is the maximum altitude
reached by this rocket? (c) What is the time it takes to get
to its maximum altitude?
18.0 m 76. ● ● ● A test rocket containing a probe to determine the com-
0.450 m/s position of the upper atmosphere is fired vertically upward
from an initial position at ground level. During the time t
while its fuel lasts, the rocket ascends with a constant
1.70 m
upward acceleration of magnitude 2g. Assume that the
rocket travels to a small enough height that the Earth’s
gravitational force can be considered constant. (a) What are
1.00 m
the speed and height, in terms of g and t, when the rocket’s
fuel runs out? (b) What is the maximum height the rocket
䉱 F I G U R E 2 . 2 5 Hit the professor See Exercise 70. (This reaches in terms of g and t? (c) If t = 30.0 s, calculate the
figure is not drawn to scale.)
rocket’s maximum height.
77. ● ● ● A car and a motorcycle start from rest at the same time
71. ●● A photographer in a helicopter ascending vertically on a straight track, but the motorcycle is 25.0 m behind the
at a constant rate of 12.5 m>s accidentally drops a camera car (䉲 Fig. 2.27). The car accelerates at a uniform rate of
out the window when the helicopter is 60.0 m above the 3.70 m>s2 and the motorcycle at a uniform rate of 4.40 m>s 2.
ground. (a) How long will the camera take to reach the (a) How much time elapses before the motorcycle overtakes
ground? (b) What will its speed be when it hits? the car? (b) How far will each have traveled during that
time? (c) How far ahead of the car will the motorcycle be
72. IE ● ● The acceleration due to gravity on the Moon is 2.00 s later? (Both vehicles are still accelerating.)
about one-sixth of that on the Earth. (a) If an object were
dropped from the same height on the Moon and on the 25.0 m
Earth, the time it would take to reach the surface on the HD
Moon is (1) 16, (2) 6, or (3) 36 times the time it would d

take on the Earth. (b) For a projectile with an initial HD
velocity of 18.0 m>s upward, what would be the maxi- Start

mum height and the total time of flight on the Moon and
on the Earth?
73. ●●● It takes 0.210 s for a dropped object to pass a win-
dow that is 1.35 m tall. From what height above the top 䉱 F I G U R E 2 . 2 7 A tie race See Exercise 77. (This figure
of the window was the object released? (See 䉴Fig. 2.26.) is not drawn to scale.)
PULLING IT TOGETHER: MULTICONCEPT EXERCISES 65
solution. The concepts may be from this chapter, but may include those from previous chapters.
78. Two joggers run at the same average speed. Jogger A speed elevators that service it reach a peak speed of
cuts directly north across the diameter of the circular 1008 m>min on the way up and 610 m>min on the way
track, while jogger B takes the full semicircle to meet down. Assuming these peak speeds are reached at the
his partner on the opposite side of the track. Assume midpoint of the run and that the accelerations are con-
their common average speed is 2.70 m>s and the track stant for each leg of the runs, (a) what are the accelera-
has a diameter of 150 m. (a) How many seconds ahead tions for the up and down runs? (b) How much longer
of jogger B does jogger A arrive? (b) How do their is the trip down than the trip up?
travel distances compare? (c) How do their displace-
81. From street level, Superman spots Lois Lane in trouble—
ments compare? (d) How do their average velocities
the evil villain, Lex Luthor, is dropping her from near the
compare?
top of the Empire State Building. At that very instant, the
79. Many highways with steep downhill areas have “run- Man of Steel starts upward at a constant acceleration to
away truck” inclined paths just off the main roadbed. attempt a midair rescue of Lois. Assuming she was
These paths are designed so that if a vehicle’s braking dropped from a height of 300 m and that Superman can
system gives out, the driver can steer it onto this accelerate straight upward at 15.0 m>s2, determine
incline (usually composed of loose gravel or sand). The (a) how far Lois falls before he catches her, (b) how long
idea is that the vehicle can then roll up the incline and Superman takes to reach her, and (c) their speeds at the
come permanently and safely to rest with no need of a instant he reaches her. Comment on whether these
braking system. In one region of Hawaii the incline speeds might be a danger to Lois, who, being a mere
distance is 300 m and provides a (constant) decelera- mortal, might get hurt running into the impervious Man
tion of 2.50 m>s2. (a) What is the maximum speed that of Steel if the speeds are too great.
a runaway vehicle can have as it enters the incline?
82. In the 1960s there was a contest to find the car that could
(b) How long would such a vehicle take to come to rest?
do the following two maneuvers (one right after the
(c) Suppose another vehicle moving 10 mi>h (4.47 m/s)
other) in the shortest total time: First, accelerate from rest
faster than the maximum value enters the incline. What
to 100 mi>h (45.0 m/s), and then brake to a complete
speed will it have as it leaves the gravel-filled area?
stop. (Ignore the reaction time correction that occurs
80. The Taipei 101 Tower in Taipei, Taiwan is a 509-m between the speeding-up and slowing-down phases and
(1667-ft), 101-story building (䉲Fig. 2.28). The outdoor assume that all accelerations are constant.) For several
observation deck is on the 89th floor, and two high- years, the winner was the “James Bond car,” the Aston
Martin. One year it won the contest when it took a total
of only 15.0 seconds to perform these two tasks! Its brak-
ing acceleration (deceleration) was known to be an excel-
lent 9.00 m>s2. (a) Calculate the time it took during the
braking phase. (b) Calculate the distance it traveled dur-
ing the braking phase. (c) Calculate the car’s acceleration
during the speeding-up phase. (d) Calculate the distance
it took to reach 100 mi>h.
83. Let’s investigate a possible vertical landing on Mars that
includes two segments: free fall followed by a parachute
deployment. Assume the probe is close to the surface, so
the Martian acceleration due to gravity is constant at
3.00 m>s2. Suppose the lander is initially moving verti-
cally downward at 200 m>s at a height of 20 000 m above
the surface. Neglect air resistance during the free-fall
phase. Assume it first free falls for 8000 m. (The para-
chute doesn’t open until the lander is 12 000 m from the
surface. See 䉲 Fig. 2.29.) (a) Determine the lander’s speed
at the end of the 8000-m free-fall drop. (b) At 12 000 m
above the surface, the parachute deploys and the lander
immediately begins to slow. If it can survive hitting the
surface at speeds of up to 20.0 m>s, determine the mini-
䉱 F I G U R E 2 . 2 8 A tall one The Taipei 101 Tower in mum constant deceleration needed during this phase.
Taipei, Taiwan, is a tall building with 101 stories. It has a (c) What is the total time taken to land from the original
height of 509 m (1671 ft). See Exercise 80. height of 20 000 m?
䉴 FIGURE 2.29 Suddenly that car speeds up and passes you, traveling at
Down she comes a constant acceleration until it is 40.0 m in front of you
See Exercise 83. v1 7.00 s later. (a) Qualitatively sketch the location-versus-
time graphs for both cars on the same axes, letting t = 0
be the start of the acceleration, and x = 0 be the location
Free fall of the other car at that time. (b) Determine the other car’s
for 8000 m acceleration. (c) How far did each of you travel during
the passing procedure? (d) What is the other car’s speed
at the end of the passing procedure?
v2 85. A car is traveling on a straight, level road under wintry
conditions. Seeing a patch of ice ahead of her, the driver
of the car slams on her brakes and skids on dry pave-
ment for 50 m, decelerating at 7.5 m>s2. Then she hits the
icy patch and skids another 80 m before coming to rest. If
Parachute her initial speed was 70 mi>h, what was the deceleration
slowdown for on the ice?
the last 12 000 m 86. On a water slide ride, you start from rest at the top of a
45.0-m-long incline (filled with running water) and
accelerate down at 4.00 m>s2. You then enter a pool of
water and skid along the surface for 20.0 m before stop-
ping. (a) What is your speed at the bottom of the
incline? (b) What is the deceleration caused by the
water in the pool? (c) What was the total time for you to
stop? (d) How fast were you moving after skidding the
v3 first 10.0 m on the water surface?
87. A toy rocket is launched (from the ground) vertically
upward with a constant acceleration of 30.0 m>s2. After
traveling 1000 m, its engines stop. When it reaches the
very top of its motion, it falls for 0.500 s before a para-
chute deploys and it descends safely to the ground at the
speed it has at that time. (a) What is the maximum alti-
v4
tude reached by the rocket? (b) How long does the rocket
take to get to its maximum altitude? (c) How long does
the total trip, from launch to ground impact, take?
Just above the Martian surface
88. A Superball is dropped from a height of 2.5 m and
rebounds off the floor to a height of 2.1 m. If the ball is in
84. You are driving slowly in the right lane of a straight contact with the floor for 0.70 ms, determine (a) the
country road. For a while, a car to your left has lagged direction and (b) magnitude of the ball’s average acceler-
50.0 m behind you at the same speed of 25.0 mi>h. ation due to the floor.
Motion in Two
3 Dimensions †
3.1 Components of motion (68)
■ vector components
■ kinematic equations for
vector components
3.2 Vector addition

and subtraction (72)
3.3 Projectile motion (80)

■ horizontal projections
■ projections at
arbitrary angles
*3.4 Relative velocity (88)

■ Relative velocity in PHYSICS FACTS
Y
one dimension
✦ Word origins:
■ Relative velocity in – kinematics: from the Greek ou can get there from here!
two dimensions kinema, meaning “motion.”
– velocity: from the Latin velocitas,
It’s a matter of knowing
meaning “swiftness.” which way to head at the cross-
– acceleration: from the Latin
accelerare, meaning “hasten.” roads (chapter-opening photo). But
✦ Projectiles:
did you ever wonder why so many
– “Big Bertha,” a gun used by the
Germans in World War I, with a streets and roads meet at right
barrel length of 6.7 m (22 ft),
could project an 820-kg angles? There’s a good reason. Liv-
(1800-lb) shell 15 km (9.3 mi).
ing on the Earth’s surface, we are
– The “Paris Gun,” also used by
the Germans in World War I, used to describing locations mainly
with a barrel length of 34 m
(112 ft), could project a 120-kg in two dimensions, and one of the
(264-lb) shell 131 km (81 mi).
easiest ways to do this is by refer-
Designed to bombard Paris,
France, the shell reached a max- ring to a pair of mutually perpen-
imum height of 40 km (25 mi)
during its 170-s trajectory. dicular axes. When you want to tell
– A bullet fired from a high-
someone how to get to a particular
powered rifle has a muzzle
speed on the order of 2900 place in the city, you might say, “Go
km/h (1800 mi/h).
†
The mathematics needed in this chapter four blocks on Main St., turn right
involves trigonometric functions. You may
want to review these in Appendix I. onto Oak St. and go three more
68 3 MOTION IN TWO DIMENSIONS
blocks.” In the country, it might be “Go south for five miles and then another half
mile east.” In each case, one needs to know how far to go in each of two directions
that are 90° apart.
The same approach can be used to describe motion—and the motion doesn’t
have to be in a straight line. As will be learned shortly, a generalized version of vec-
tors introduced in Section 2.2 can be used to describe motion in curved paths as
well. Such analysis of curvilinear motion will eventually allow you to analyze the
behavior of batted balls, planets circling the Sun, and even the motions of elec-
trons in atoms.
Two-dimensional curvilinear motion can be analyzed by using rectangular
components of motion. Essentially, the curved motion is broken down or resolved
into rectangular (x and y) components so the motion can be considered linearly in
both dimensions. The kinematic equations introduced in Chapter 2 can be applied
to these components. For an object moving in a curved path, for example, the
x- and y-coordinates of its motion will give the object’s position at any time.
3.1 Components of Motion

➥ How is motion in two dimensions described?

➥ What are the magnitudes of the components of velocity (v) in two dimensions?
➥ What is the major restriction for using the kinematic equations for components of
motion?
In Section 2.1, an object moving in a straight line was considered to be moving

along one of the Cartesian axes (x or y). But what if the motion is not along an
axis? For example, consider the situation illustrated in 䉴 Fig. 3.1. Here, three balls
are moving uniformly across a tabletop. The ball rolling in a straight line along the
side of the table, designated as the x-direction, is moving in one dimension. That
is, its motion can be described with a single coordinate, x. Similarly, the motion of
the ball rolling along the end of the table in the y-direction can be described by a
single y-coordinate. However, for this coordinate choice, both x- and y-coordinates
are needed to describe the motion of the ball rolling diagonally across the table;
that is, the motion is described in two dimensions.
You might observe that if the diagonally moving ball were the only object to
consider, the x-axis could be chosen to be in the direction of that ball’s motion, and
the motion would thereby be reduced to one dimension. This observation is true,
but once the coordinate axes are fixed, motions not along the axes must be
described with two coordinates (x, y), or in two dimensions. Also, keep in mind
that not all motions in a plane (two dimensions) are in straight lines. Think about
the path of a ball you toss to another person. The path is curved for such projectile
motion. (This motion will be considered in Section 3.3.) In general, both coordi-
nates are needed.
In considering the motion of the ball moving diagonally across the table in
Fig. 3.1a, it can be thought of as moving in the x- and y-directions simultaneously.
That is, it has a velocity in the x-direction (vx) and a velocity in the y-direction (vy)
at the same time. The combined velocity components describe the actual motion
of the ball. If the ball has a constant velocity v in a direction at an angle u relative to
the x-axis, then the velocities in the x- and y-directions are obtained by resolving,
or breaking down, the velocity vector into components of motion in these direc-
3.1 COMPONENTS OF MOTION 69
䉳 F I G U R E 3 . 1 Components of
y motion (a) The velocity (and dis-
v placement) for uniform, straight-
vy vy line motion—that of the dark purple
ball—may have x- and
vx y-components (vx and vy as shown
vy vy v in the pencil drawing), because of
the chosen orientation of the coordi-
nate axes. Note that the velocity and
vx
displacement of the ball in the x-
vy vy v direction are exactly the same as
those that a ball rolling along the x-
vx axis with a uniform velocity of vx
vy vy v would have. A comparable relation-
ship holds true for the ball’s motion
vx in the y-direction. Since the motion
is uniform, the ratio vy>vx (and
therefore u) is constant. (b) The
x
vx vx vx vx coordinates (x, y) of the ball’s posi-
tion and the distance d the ball has
(a) traveled from the origin can be
y found at any time t.
(x, y)
2
2 +y
y = vyt √x
d=

x
x = vxt
(b)
tions (see the pencil drawing in Fig. 3.1a). As this drawing shows, the vx and vy
components have magnitudes of
vx = v cos u (3.1a)
and
vy = v sin u (3.1b)
(Notice that v = 2v2x + v2y , so v is a combination of the velocities in the x- and

y-directions.)
You are familiar with the use of two-dimensional length components in finding
the x- and y-coordinates in a Cartesian system. For the ball rolling on the table, its
position (x, y), or the distance traveled from the origin in each of the component
directions at time t, is given by (Eq. 2.11 with a = 0)
x = xo + vx t (magnitudes of displacement components

under condition of constant velocity and (3.2a)
zero acceleration)
y = yo + vy t (3.2b)
respectively. (Here, the xo and yo are the ball’s coordinates at t = 0, which may
be other than zero.) The ball’s straight-line distance from the origin at any given
time is then d = 2x 2 + y2 (Fig. 3.1b).
Note that tan u = vy>vx (see Fig. 3.1a.). So the direction of the motion relative to
the x-axis is given by u = tan-11vy >vx2. Also, u = tan-11y>x2.
In this introduction to components of motion, the velocity vector has been taken
to be in the first quadrant 10 6 u 6 90°2, where both the x- and y-components are
positive. But, as will be shown in more detail in the next section, vectors may be in
any quadrant, and one or both of their components can be negative. Can you tell in
which quadrants the vx and vy components would both be negative?
EXAMPLE 3.1 On a Roll: Using Components of Motion

If the diagonally moving ball in Fig. 3.1a has a constant veloc- given by Eq. 3.2, we first need to compute the velocity com-
ity of 0.50 m>s at an angle of 37° relative to the x-axis, find ponents vx and vy (Eq. 3.1):
how far it travels in 3.0 s by using x- and y-components of its
vx = v cos 37° = 10.50 m>s210.802 = 0.40 m>s
motion.
vy = v sin 37° = 10.50 m>s210.602 = 0.30 m>s
THINKING IT THROUGH. Given the magnitude and direction
(angle) of the velocity of the ball, the x- and y-components of Then, taking xo = 0 and yo = 0, the component distances are
the velocity can be found. Then the distance in each direction x = vx t = 10.40 m>s213.0 s2 = 1.2 m
can be computed. Since the x- and y-axes are at right angles to and
each other, the Pythagorean theorem gives the distance of the y = vy t = 10.30 m>s213.0 s2 = 0.90 m
straight-line path of the ball, as shown in Fig. 3.1b. (Note the
procedure: Separate the motion into components, calculate and the distance of the path is
what is needed in each direction, and recombine if necessary.)
d = 2x 2 + y2 = 211.2 m22 + 10.90 m22 = 1.5 m
SOLUTION. Listing the data,
FOLLOW-UP EXERCISE. Suppose that a ball is rolling diago-
Given: v = 0.50 m>s Find: d (distance traveled) nally across a table with the same speed as in this Example,
but from the lower right corner, which is taken as the origin of
u = 37° the coordinate system, toward the upper left corner at an
t = 3.0 s angle of 37° relative to the -x-axis. What would be the veloc-
ity components in this case? (Would the distance change?)
The distance traveled by the ball in terms of its x- and (Answers to all Follow-Up Exercises are given in Appendix VI at
y-components is given by d = 2x 2 + y2. To find x and y as the back of the book.)
Note that for this simple case, the distance can also be obtained directly from
d = vt = 10.50 m>s213.0 s2 = 1.5 m. However, this Example was solved in a more
general way to illustrate the use of components of motion. The direct solution would
have been evident if the equations had been combined algebraically before calcula-
tion, that is, as
x = vx t = 1v cos u2t
and
y = vy t = 1v sin u2t
from which it follows that
d = 2x 2 + y2 = 21v cos u22 t2 + 1v sin u22 t2 = 2v2 t2 (cos2 u + sin2 u) = vt
Before embarking on the first solution strategy that occurs to you, pause for a moment to
see whether there might be an easier or more direct way of approaching the problem.
KINEMATIC EQUATIONS
FOR COMPONENTS OF MOTION
Example 3.1 involved two-dimensional motion in a plane. With a constant veloc-
ity (constant components vx and vy), the motion is in a straight line. The motion
may also be accelerated. For motion in a plane with a constant acceleration that
has components ax and ay, the displacement and velocity components are given
3.1 COMPONENTS OF MOTION 71
by the kinematic equations of Section 2.4 written separately for the x- and y-
directions:
x = xo + vxo t + 12 ax t2
¯˚˚˘˚˚˙
(3.3a)
y = yo + vyo t + 12 a y t2 (3.3b)
(constant acceleration only)
vx = vxo + a x t (3.3c)
vy = vyo + a y t (3.3d)
If an object is initially moving with a constant velocity and suddenly experiences

an acceleration in the direction of the velocity or opposite to it, it will continue in a
straight-line path, either speeding up or slowing down, respectively.
If, however, the acceleration is at some angle other than 0° or 180° to the veloc-
ity vector, the motion will be along a curved path. For the motion of an object to be
curvilinear—that is, to vary from a straight-line path—an acceleration not parallel
to the velocity is required. For such a curved path, the ratio of the velocity compo-
nents varies with time. That is, the direction of the motion, u = tan-11vy >vx2,
varies with time, because one or both of the velocity components do.
Consider a ball initially moving along the x-axis, as illustrated in 䉲 Fig. 3.2.
Assume that, starting at a time to = 0, the ball receives a constant acceleration ay in
the y-direction. The magnitude of the x-component of the ball’s displacement is
given by x = vx t, where the 12 ax t2 term of Eq. 3.3a drops out because there is no
acceleration in the x-direction (ax = 0). Prior to
to , the motion was in a straight line along the
x-axis. But at any time after to , the y-coordinate y
is not zero, but is given by y = 12 ay t2 (Eq. 3.3b
with yo = 0 and vyo = 0). The result is a curved
path for the ball. v3
Note that the length (magnitude) of the vy3 = ayt3
velocity component vy changes with time, while
that of the vx component remains constant. The
total velocity vector at any time is tangent to the 3
curved path of the ball. It is at an angle u relative y3 = 12 ayt 32
-1 vx
to the positive x-axis, given by u = tan 1vy >vx2,
which now changes with time, as can be seen in
Fig. 3.2 and in Example 3.2.
vy2 = ayt2 v2
2
y2 = 12 ayt 22
vx
vy1 = ayt1 v1
1
y1 = 12 ayt12
vx
vx
x
y=0 vx x1 = vxt1 x2 = vxt2 x3 = vxt3
䉴 F I G U R E 3 . 2 Curvilinear to
motion An acceleration not parallel Straight-line Curvilinear
to the instantaneous velocity pro- motion motion
duces a curved path. Here, an acceler- vy = 0 ay
ation ay is applied at to = 0 to a ball vx
initially moving with a constant
velocity vx . The result is a curved path at to = 0
with the velocity components as
shown. Notice how vy increases with
time, while vx remains constant.
EXAMPLE 3.2 A Curving Path: Vector Components

Suppose that the ball in Fig. 3.2 has an initial velocity of T H I N K I N G I T T H R O U G H . Keep in mind that the motions in
1.50 m>s along the x-axis. Starting at to = 0, the ball receives the x- and y-directions can be analyzed independently. For
an acceleration of 2.80 m>s2 in the y-direction. (a) What is the part (a), simply compute the x- and y-positions at the given
position of the ball 3.00 s after to? (b) What is the velocity of time, taking into account the acceleration in the y-direction.
the ball at that time? For part (b), find the component velocities, and vectorially
combine them to get the total velocity.
SOLUTION. Referring to Fig. 3.2,
Given: vxo = vx = 1.50 m>s Find: (a) (x, y) (position coordinates)

v yo = 0 (b) v (velocity, magnitude and direction)
ax = 0
a y = 2.80 m>s2
t = 3.00 s
(a) At 3.00 s after to = 0, Eqs. 3.3a and 3.3b tell us that the ball (This component is constant, since there is no acceleration in
has traveled the following distances from the origin the x-direction.) Similarly, the y-component of the velocity is
1xo = yo = 02 in the x- and y-directions: given by Eq. 3.3d:
x = vxo t + 12 a x t2 = 11.50 m>s213.00 s2 + 0 = 4.50 m vy = vyo + ay t = 0 + 12.80 m>s2213.00 s2 = 8.40 m>s
y = vyo t + 12 a y t2 = 0 + 12 12.80 m>s2213.00 s22 = 12.6 m The velocity therefore has a magnitude of
Thus, the position of the ball is 1x, y2 = 14.50 m, 12.6 m2. v = 2v2x + v2y = 211.50 m>s22 + 18.40 m>s22 = 8.53 m>s
If you had computed the distance d = 2x 2 + y2, what
would have been obtained? [This quantity is the magnitude and its direction relative to the + x-axis is
of the displacement, or straight-line distance, from the origin to vy 8.40 m>s
the 1x, y2 = 14.50 m, 12.6 m2 position.] u = tan-1 a b = tan-1 a b = 79.9°
vx 1.50 m>s
(b) The x-component of the velocity is given by Eq. 3.3c:
vx = vxo + a x t = 1.50 m>s + 0 = 1.50 m>s
FOLLOW-UP EXERCISE. Suppose that the ball in this Example also received an acceleration of 1.00 m>s 2 in the +x-direction start-
ing at to . What would be the position of the ball 3.00 s after to in this case?
When using the kinematic equations, it is important to note that motion in the x- and
y-directions can be analyzed independently—the factor connecting them being time t.
That is, you can find (x, y) and/or (vx, vy) at a given time t. Also, keep in mind that the
initial positions are often set xo = 0 and yo = 0, which means that the object is located at
the origin at to = 0. If the object is actually elsewhere at to = 0, then the values of xo
and/or yo would have to be used in the appropriate equations. (See Eqs. 3.3a and b.)
DID YOU LEARN?

➥ Two-dimensional motion is described by considering an object to be moving in the
x- and y- directions simultaneously.
➥ The magnitudes of velocity components are vx = v cos u and vy = v sin u,
where u is the angle of the x- and y-components.
➥ The kinetic equations for the components of motion are for constant
acceleration only.
3.2 Vector Addition and Subtraction

➥ What is a resultant vector?

➥ Given the rectangular vector components, Cx = C cos u and Cy = C sin u, how is the
angle determined?
➥ What is a unit vector?
3.2 VECTOR ADDITION AND SUBTRACTION 73
Many physical quantities, including those describing motion, have a direction

associated with them—that is, they are vectors. You have already worked with a
few such quantities related to motion (displacement, velocity, and acceleration)
and will encounter more during the course of study. A very important technique
in the analysis of many physical situations is the addition (and subtraction) of vec-
tors. By adding or combining such quantities (vector addition), the resultant, or
net, vector is obtained. This resultant vector is the vector sum.
You have already been adding vectors. In Section 2.2, displacements in one
dimension were added to get the net displacement. In this chapter, vector compo-
nents of motion in two dimensions will be added to get net effects. Notice that in
Example 3.2, the velocity components vx and vy were combined to get the resultant
velocity.
In this section, vector addition and subtraction in general, along with common
vector notation, will be considered. As will be learned, these operations are not the
same as scalar or numerical addition and subtraction, with which you are already
familiar. Vectors have magnitudes and directions, so different rules apply.
In general, there are geometrical (graphical) methods and analytical (computa-
tional) methods of vector addition. The geometrical methods are useful in helping
you visualize the concepts of vector addition, particularly with a quick sketch.
Analytical methods are more commonly used, however, because they are faster
and more precise.
In Section 3.1, the concern was chiefly about vector components. The notation
for the magnitudes of components was, for example, vx and vy. To represent vec-
B B
tors, the notation A and B—a boldface symbol with an overarrow—will be used.
VECTOR ADDITION: GEOMETRIC METHODS

B B B B
Triangle Method To add two vectors—say, to add B to A (that is, to find A + B)
B
by the triangle method—you first draw A on a sheet of graph paper to some scale
B
(䉲 Fig. 3.3a). For example, if A represents a displacement in meters, a convenient
Draw first vector (A) from origin. Draw second vector (B) Draw vector from tail of A to tip
from tip of first vector. of B. This is the resultant (R).
(a)
䉱 F I G U R E 3 . 3 Triangle
B
method
B
of vector addi-
y y tion (a) The vectors A and B are placed Btip to tail.
The vector
B
that extends from the tail of A to the
R=A+B R=A+B tip of B, forming the third
B
side of
B B
the triangle, is
the resultant or sum R = A + B. (b) When the B
R B vectors are drawn to scale, the magnitudeBof R
can be found by measuring the length of R and
B using the scale conversion, and the direction
R angle uR can be measured with a protractor. Ana-
A B
R lytical methods can also be used. For a nonright
A R A triangle, as in part (b), the laws of sines and
x x cosines
B
can be used to determine the magnitude
Scale: 1 cm = 1 m Scale: 1 cm = 1 m of R and uR (AppendixB
I). (c) If the vector triangle
(b) (c) is a right triangle, R is easily obtained via the
Pythagorean theorem, and the direction angle is
given by an inverse trigonometric function.
scale is 1 cm : 1 m, or 1 cm of vector length on the graph corresponds to 1 m of dis-

B
placement. As shown in Fig. 3.3b, the direction of the A vector is specified as being
at an angle uA relative to a coordinate axis, usually the x-axis.
B B
Next, draw B with its tail starting at the tip of A. (Thus, this method is also
B B
called the tip-to-tail method.) The vector from the tail of A to the tip of B is then the
B B B B
vector sum R, or the resultant of the two vectors: R = A + B.
B
If the vectors are drawn to scale, the magnitude of R can be found by measur-
ing its length and using the scale conversion. In such a graphical approach, the
direction angle uR is measured with a protractor. If the magnitudes and directions
B B B
(angles u) of A and B are known, the magnitude and direction of R can be found
y
analytically by using trigonometric methods. For the nonright triangle in Fig. 3.3b,
the laws of sines and cosines can be used. (See Appendix I.)
This tip-to-tail method can be extended to any number of vectors. The vector
(A + B) from the tail of the first vector to the tip of the last vector is the resultant or vector
sum. For more than two vectors, it is called the polygon method.
B
A The resultant of the vector right triangle in Fig. 3.3c would be much easier to
x find using the Pythagorean theorem for the magnitude and an inverse trigono-
B
metric function to find the direction angle. Notice that R is made up of x- and y-
–B B B
vectors A and B. Such x- and y-components are the basis of the convenient
A–B analytical component method, which will be discussed shortly.
䉱 F I G U R E 3 . 4 Vector
subtraction Vector subtraction is a Vector Subtraction Vector subtraction is a special case of vector addition:
special case ofB vector Baddition; that B B B B
B B
is, A - B = A + 1 - B2, where
B
-B A - B = A + 1- B2
B
has the same magnitude as B, but is B B B B
in the oppositeBdirection.
B
(See the That is, to subtract B from A, a negative B is added to A. In Section 2.2, you
sketch.) Thus,
B B
A + B is not the learned that a minus sign simply means that the direction of a vector is opposite
same as B - A, in either length or that of one with a plus sign (for example, +x and -x). The same is true with vec-
direction. Can you show B
B B B B tors represented by boldface notation. The vector -B has the same magnitude as
that B - A = - 1- A - B2 B
geometrically? the vector B, but is in the opposite direction (䉳 Fig. 3.4). The vector diagram in
B B
Fig. 3.4 provides a graphical representation of A - B.
VECTOR COMPONENTS AND THE ANALYTICAL

COMPONENT METHOD
Probably the most widely used analytical method for adding multiple vectors is
the component method. It will be used again and again throughout the course
of our study, so a basic understanding of the method is essential. Learn this sec-
tion well.
Adding Rectangular Vector Components Rectangular components means that

vector components are at right 90° angles to each other, usually taken in the rec-
tangular coordinate x- and y-directions. You have already had an introduction to
the addition of such components in the discussion of the velocity components of
B B
motion in Section 3.1. For the general case, suppose that A and B, two vectors at
right angles, are added, as illustrated in 䉴 Fig. 3.5a. The right angle makes the math
B
easy. The magnitude of C is given by the Pythagorean theorem:
C = 2A2 + B 2 (3.4a)
B
The orientation of C relative to the x-axis is given by the angle
B
u = tan-1 a b (3.4b)
A
This notation is how a resultant is expressed in magnitude–angle form.

Cx = C cos θ 䉳 F I G U R E 3 . 5 Vector
B
compo-
B
y y Cy = C sin θ nents (a) The vectors A and B along
the x- and
B
y-axes, respectively,
B
add
C = 兹Cx2 ⫹ Cy2 to give C. (b) A vector C may be
θ = tan –1 (Cy /Cx) resolved
B
into rectangular
B
compo-
nents Cx and Cy.
C C
B Cy

x x
A Cx
C=A+B C = Cx + Cy
(a) (b)
Resolving a Vector into Rectangular Components; Unit Vectors Resolving a

vector into rectangular components is essentially the reverse of adding the rectan-
B
gular components of the vector. Given a vector C, Fig. 3.5b illustrates how it may
B B
be resolved into x and y vector components Cx and Cy. Simply complete the vector
triangle with x- and y-components. As the diagram shows, the magnitudes, or
vector lengths, of these components are given by
Cx = C cos u (3.5a)
(vector components)
Cy = C sin u (3.5b)
respectively (similar to vx = v cos u and vy = v sin u in Example 3.1).* The angle

B
of direction of C can also be expressed in terms of the components, since
tan u = Cy >Cx, or
Cy (direction of vector
u = tan-1 ¢ ≤ (3.6)
Cx from magnitudes of components)
Another way of expressing the magnitude and direction of a vector involves the
B
use of unit vectors. For example, as illustrated in 䉴Fig. 3.6, a vector A can be written
B
as A = AaN . The numerical magnitude is represented by A, and aN is a unit vector,
which indicates direction. That is, aN has a magnitude of unity, or one, with no units,
A
and simply indicates a vector’s direction. For example, a velocity along the x-axis can (vector)
A –A A
B
be written v = 14.0 m>s2xN (that is, 4.0 m>s magnitude in the +x-direction). (magnitude)
B
Note in Fig. 3.6 how -A would be represented in this notation. Although the
minus sign is sometimes put in front of the numerical magnitude, this quantity is
an absolute number or value. The minus actually goes with the unit vector: â (unit –â
B
-A = - AaN = A1 -aN 2.† That is, the unit vector is in the -aN direction (opposite aN ). vector)
A velocity of vB
= 1-4.0 m>s2xN has a magnitude of 4.0 m>s in the -x direction, that A=Aâ –A = –A â = A(–â)
is, v = 14.0 m>s21 -xN 2.
B
(a) (b)
This notation can be used to express explicitly the rectangular components of a
vector. For example, the ball’s displacement from the origin in Example 3.2 could 䉱 F I G U R E 3 . 6 Unit vectors
B
be written d = 14.50 m2xN + 112.6 m2yN , where xN and yN are unit vectors in the x- (a) A unit vector aN has a magnitude of
unity, or one, and thereby simply
and y-directions. In some instances, it may be more convenient to express a gen- indicates a vector’s direction. Written
eral vector in this unit vector component form: with the magnitude A, it represents
B B
the vector A, and A B= AaN .
B
C = Cx xN + Cy yN (3.7) (b) For the vector B
- A, the unit vec-
tor is -aN , and -A = - AaN = A1 - aN 2.
*Figure 3.5b illustrates only a vector in the first quadrant, but the equations hold for all quadrants
when vectors are referenced to either the positive or negative x-axis. The directions of the components
are indicated by + and - signs, as will be shown shortly.
B B
†
The notation is sometimes written with an absolute value, A = ƒ A ƒ aN , or -A = - ƒ A ƒ aN , so as to
B
clearly show that the magnitude of A is a positive quantity.
y y
F2 F2 F y2
+ F2 Fy2 +
F1 F1
= = Fy = Fy1 + Fy2
F F
F1 Fx Fy1
Fy1 2 Fx1 Fx2
Fx1 Fx = Fx1 + Fx2

x x
(a) (b)
䉱 F I G U R E 3 . 7 Component addition (a) In adding vectors by the component method, each

vector is first resolved into
B
its x- Band y-component
B B B
vectors.
B
(b) BThe sums
B
of the x- and
y-components of vectors F1 and F2 are Fx = Fx1 + Fx2 and Fy = Fy1 + Fy2 , respectively.
VECTOR ADDITION USING COMPONENTS

The analytical component method of vector addition involves resolving the vec-
tors into rectangular vector components and adding the components for each axis
independently. This method is illustrated graphically in 䉱 Fig. 3.7 for two vectors
B B
F1 and F2 .* The sums of the x- and y-component vectors being added are equal to the cor-
responding vector components of the resultant vector.
The same principle applies if you are given three (or more) vectors to add. You
could find the resultant by applying the graphical tip-to-tail method. However,
this technique involves drawing the vectors to scale and using a protractor to mea-
sure angles, which can be time consuming. But in using the component method,
you do not have to draw the vectors tip to tail. In fact, it is usually more conve-
nient to put all of the tails together at the origin, as shown in 䉴 Fig. 3.8a. Also, the
vectors do not have to be drawn to scale, since the approximate sketch is just a
visual aid in applying the analytical method.
Basically, in the component method, the vectors to be added are resolved into their x-
and y-components, and the respective components added and then recombined to find the
resultant. The resultant of the three vectors in Fig. 3.8a is shown in Fig. 3.8b. By
looking at the x-components, it can be seen that the vector sum of these compo-
nents is in the -x-direction. Similarly, the sum of the y-components is in the
B
+y-direction. (Note that v 2 is in the y-direction and has a zero x-component, just as
a vector in the x-direction would have a zero y-component.)
B B B
The x- and y-components of the resultant are v x = vx1 + vx3 and
B B B B
vy = vy1 + vy2 + vy3. When the numerical values of the vector components are
computed and put into these equations, you will have values for vx 6 0 (negative)
and vy 7 0 (positive) as shown in Fig. 3.8b.
Notice also in Fig. 3.8b that the directional angle u of the resultant is referenced
to the x-axis, as are the individual vectors in Fig. 3.8a. In adding vectors by the com-
ponent method, all vectors will be referenced to the nearest x-axis—that is, the +x-axis or
- x-axis. This policy eliminates angles greater than 90° (as occurs when customar-
ily measuring angles counterclockwise from the + x-axis) and the use of double-
angle formulas, such as cos1u + 90°2. This greatly simplifies calculations. The
B
*The symbol F is commonly used to denote force, a very important vector quantity that will be
B
studied in Chapter 4. Here, F is employed as a general vector, but its use provides familiarity with the
notation used in the next chapter, where knowledge of the addition of forces is essential.
y y
5.0 m/s v2
/s
m
v1 vy1
5
4.
vx3 45°
x (re v
30° vx1 su vy = vy + vy + vy
lta 1 2 3
nt
)
vy3 v3 u
/s x
9.0m vx = vx1 + vx3
(a) Component method (b) Component method

(resolving into components) (adding x- and y-components, shown as
offset dashed arrows, and finding resultant)
䉱 F I G U R E 3 . 8 Component method of vector addition (a) In the analytical component method, all the vectors to
B B B
be added (v1 , v2 and v3) are first placed with their tails at the origin so that they may be easily resolved into rectan-
gular components. (b) The respective summations of all the x-components and all the y-components are then
B
added to give the components of the resultant v.
recommended procedures for adding vectors analytically by the component

method can be summarized as follows:
PROCEDURES FOR ADDING VECTORS

BY THE COMPONENT METHOD
1. Resolve the vectors to be added into their x- and y-components. Use the
acute angles (angles less than 90°) between the vectors and the x-axis to find
the magnitudes, and indicate the directions of the components by plus and
minus signs (䉲 Fig. 3.9).
y y
Ay
A
Ay
C Cy
A Cx C
Bx Ax
x x
B Ax Bx
By B
C = Cx xˆ + Cy yˆ By C = 兹Cx2 + Cy2
C
Bx = –B cos B Ax = A cos A Add components C = tan–1 y
Ay = A sin A Cx = Ax + Bx Cx
By = –B sin B With Cx < 0 and
= A cos θA+ (–B cosθ B)
(minus signs indicate Cy = Ay + By Cy > 0, resultant C
components in negative is in 2nd quadrant.
= A sin θA + (–B sinθB)
x- and y-directions)
(a) (b)
䉱 F I G U R E 3 . 9 Vector addition by the analytical component method (a) Resolve the vectors into their x- and
y-components. (b) Add B
all of
B
the x-components and all of the y-components together vectorially to obtain the x-
and y-components Cx and Cy, respectively, of the resultant. Express the resultant in either component form or
magnitude–angle form. All angles are referenced to the + x- or -x-axis to keep them less than 90°.
2. Add all of the x-components together, and all of the y-components together
vectorially to obtain the x- and y-components of the resultant, or vector sum.
3. Express the resultant vector, using:
B
(a) the unit vector component form—for example, C = Cx xN + Cy yN —or
(b) the magnitude–angle form.
For the latter notation, find the magnitude of the resultant by using the summed x-
and y-components and the Pythagorean theorem:
C = 3C 2x + C 2y
Find the angle of direction (relative to the x-axis) by taking the inverse tangent
1tan-12 of the absolute value (that is, the positive value, ignoring any minus signs)
of the ratio of the magnitudes of y- and x-components:
Cy
u = tan-1 ` `
Cx
Designate the quadrant in which the resultant lies. This information is obtained
from the signs of the summed components or from a sketch of their addition via
the triangle method. (See Fig. 3.9.) The angle u is the angle between the resultant
and the x-axis in that quadrant.
EXAMPLE 3.3 Applying the Analytical Component Method:

Separating and Combining x- and y-Components
Let’s apply the procedural steps of the component method to T H I N K I N G I T T H R O U G H . Follow and learn the steps of the
the addition of the vectors in Fig. 3.8a. The vectors with units procedure. Basically, the vectors are resolved into compo-
of meters per second represent velocities. nents and the respective components are added to get the
components of the resultant, which then may be expressed in
(unit vector) component form or magnitude–angle form.
SOLUTION. The rectangular components of the vectors are shown in Fig. 3.8b. Summing these components and taking the val-
ues from Fig. 3.8a,
v = vx xN + vy yN = 1vx1 + vx2 + vx32xN + 1vy1 + vy2 + vy32yN
B
where
vx = vx1 + vx2 + vx3 = v1 cos 45° + 0 - v3 cos 30°
= 14.5 m>s210.7072 - 19.0 m>s210.8662 = - 4.6 m>s
and
vy = vy1 + vy2 + vy3 = v1 sin 45° + v2 - v3 sin 30°
= 14.5 m>s210.7072 + 15.0 m>s2 - 19.0 m>s210.502 = 3.7 m>s
Expressed in tabular form, the components are as follows:
x-Components y-Components
vx1 + v1 cos 45° = + 3.2 m>s vy1 +v1 sin 45° = + 3.2 m>s
vx2 = 0 m>s vy2 = + 5.0 m>s
vx3 - v3 cos 30° = - 7.8 m>s vy3 -v3 sin 30° = - 4.5 m>s
Sums: vx = - 4.6 m>s vy = + 3.7 m>s
B
The directions of the components are indicated by signs. (The + sign is sometimes omitted as being understood.) Here, v 2 has no
x-component. Note that in general, for the analytical component method, the x-components are cosine functions and the y-components are
sine functions, as long as they are referenced to the nearest part of the x-axis.
In component form, the resultant vector is
v = 1-4.6 m>s2xN + 13.7 m>s2yN
B
In magnitude–angle form, the resultant velocity has a magnitude of

v = 3v 2x + v2y = 31- 4.6 m>s22 + 13.7 m>s22 = 5.9 m>s
Since the x-component is negative and the y-component is positive, the resultant lies in the second quadrant at an angle of
vy 3.7 m>s
u = tan-1 ` ` = tan-1 a b = 39°
vx 4.6 m>s
above the -x-axis because of the negative x component (see Fig. 3.8b).
Suppose in this Example that there were an additional velocity vector v4 = 1+ 4.6 m>s2xN . What would be
B
FOLLOW-UP EXERCISE.
the resultant of all four vectors in this case?
Although our discussion is limited to motion in two dimensions (in a plane),

the component method is easily extended to three dimensions. For a velocity in
three dimensions, the vector has x-, y-, and z-components: v = vx xN + vy yN + vz zN
B
and magnitude v = 2v2x + v2y + v 2z . LEARN BY DRAWING 3.1
make a sketch
EXAMPLE 3.4 Find the Vector: Add Them Up
B
and add them up
Given two displacement vectors, A, with a magnitude of 8.0 m in a direction 45° below (a) A sketch is made for the vec-
B
the + x-axis, and B, which has an x-component of +2.0 m and a y-component of +4.0 m. B B
tors A and B. In a vector draw-
B B B B B
Find a vector C so that A + B + C equals a vector D that has a magnitude of 6.0 m in ing, the vector lengths are
the + y-direction. usually set to some scale—for
T H I N K I N G I T T H R O U G H . Here again, a sketch helps to understand the situation and example, 1 cm : 1 m—but in a
B
gives a general idea of the attributes of C. This would be something like the accompanying quick sketch, the vector lengths
B
B
Learn By Drawing 3.1, Make a Sketch and Add Them Up. Note that in part (a) both A and are estimated. (b) By shifting B
B B
B B
B have + x-components, so C would have to have a -x-component to cancel these compo- to the tip of A and putting in D,
B
B B B
nents. (It is given that the resultant D points only in the +y-direction.) By and D are in the the vector C can be found from
B B B B
B B
+ y-direction, but the Ay-component is larger in the -y-direction, so C would have to have A + B + C = D.
B
a + y-component. With this information, it can be seen that C lies in the second quadrant. +y
A polygon sketch [shown in part (b) of the Learn by Drawing] confirms this observation.
B B
So C has second-quadrant components and it has a relatively large magnitude (from the
lengths of the vectors in the polygon drawing). This information gives an idea of what we
are looking for, making it easer to see if the results from the analytic solution are reasonable. By = +4.0 m
SOLUTION. Bx = +2.0 m
B B
-x +x
Given: A: 8.0 m, 45° below the +x-axis Find: C such that 45°
B B B B
(fourth quadrant) A + B + C = D = 1+6.0 m2yN
B
Bx = 12.0 m2xN 8.0 m
B
By = 14.0 m2yN -y
Setting up the components in tabular form again so they can be easily seen:
A
x-Components y-Components (a)
A x = A cos 45° = 18.0 m210.7072 = + 5.7 m A y = - A sin 45° = - 18.0 m210.7072 +y
Bx = + 2.0 m = - 5.7 m
Cx = ? By = + 4.0 m
Dx = 0 Cy = ?
D
Dy = + 6.0 m
B C ?
B B B B B
To find the components of C, where A + B + C = D, the x- and y-components are
summed separately: -x +x
B B B B
x: Ax + Bx + Cx = Dx
or A
+ 5.7 m + 2.0 m + Cx = 0 and Cx = - 7.7 m
B B B B
y: Ay + By + Cy = Dy
or B
- 5.7 m + 4.0 m + Cy = 6.0 m and Cy = + 7.7 m -y
(continued on next page) (b)
So,
B
C = 1-7.7 m2xN + 17.7 m2yN
The result can also be expressed in magnitude–angle form:
C = 2C 2x + C 2y = 21 -7.7 m22 + 17.7 m22 = 11 m
and
Cy 7.7 m
u = tan-1 ` ` = tan-1 ` ` = 45° (above the - x axis; why?)
Cx -7.7 m
B
FOLLOW-UP EXERCISE. Suppose D pointed in the opposite direction [that is,
B B
D = 1-6.0 m2yN ]. What would C be in this case?
DID YOU LEARN?

➥ A resultant is the vector sum of added (or subtracted) vectors.
Cy C sin u Cy
➥ u is given by the component ratio = = tan u, and u = tan-1 ` `
Cx C cos u Cx
➥ A unit vector has a magnitude of unity, or one, with no units, and simply
indicates a vector’s direction.
3.3 Projectile Motion

➥ Why do two balls, one dropped and one horizontally projected from the same
height, strike the ground at the same time?
➥ For projections with vx and vy components, which is constant and why?
A familiar example of two-dimensional, curvilinear motion is that of an object that

is thrown or projected by some means. The motion of a stone thrown across a
stream or a golf ball driven off a tee are examples of projectile motion. A special
case of projectile motion in one dimension occurs when an object is projected ver-
tically upward (or downward or dropped). This case was treated in Section 2.5 in
terms of free fall. General two-dimensional projectile motion is in free fall too,
because the only acceleration of a projectile is that due to gravity (air resistance
neglected). Vector components can be used to analyze projectile motion by simply
breaking up the motion into its x- and y-components and treating them separately.
HORIZONTAL PROJECTIONS
It is instructive to first analyze the motion of an object projected horizontally, or
parallel to a level surface. Suppose that you throw an object horizontally with an
initial velocity vxo as in 䉴 Fig. 3.10. Projectile motion is analyzed beginning at the
instant of release 1t = 02. Once the object is released, there is no longer a horizon-
tal acceleration 1a x = 02, so throughout the object’s path, the horizontal velocity
remains constant: vx = vxo .
According to the equation x = xo + vx t (Eq. 3.2a), the projected object would
continue to travel in the horizontal direction indefinitely. However, you know that
this is not what happens. As soon as the object is projected, it is in free fall in the ver-
tical direction, with vyo = 0 (vertically it behaves as though it had been dropped)
and a y = - g. In other words, the projected object travels at a uniform velocity in the
horizontal direction, while at the same time undergoing acceleration in the down-
ward direction under the influence of gravity. The result is a curved path, as illus-
trated in Fig. 3.10. (Compare the motions in Fig. 3.10 and Fig. 3.2. Do you see any
similarities?) If there were no horizontal motion, the object would simply drop to
the ground in a straight line. In fact, the time of flight of the horizontally projected
object is exactly the same as if it were a dropped object falling vertically.
3.3 PROJECTILE MOTION 81
y (a) 䉳 F I G U R E 3 . 1 0 Horizontal projec-

t=0 tion (a) The velocity components of
vxo = vx a projectile launched horizontally
x show that the projectile travels to
vy o = 0 the right as it falls downward. Note
vxo the increase in vy . (b) A multiflash
photograph shows the paths of two
v1 golf balls. One was projected hori-
vy
1 zontally at the same time that the
other was dropped straight down.
The horizontal lines are 15 cm apart,
and the interval between flashes
1
vx was 30 s. The vertical motions of the
o
balls are the same. Why? Can you
(b) describe the horizontal motion of
the yellow ball?
v2
vy
2
vx
o
vy v3
3
xmax
Note the components of the velocity vector in Fig. 3.10a. The length of the hori-
zontal component of the velocity vector remains the same, but the length of the
vertical component increases with time. What is the instantaneous velocity at any
point along the path? (Think in terms of vector addition, covered in Section 3.2.)
The photo in Fig. 3.10b shows the actual motions of a horizontally projected golf
ball and one that is simultaneously dropped from rest. The horizontal reference
lines show that the balls fall vertically at the same rate. The only difference is that
the horizontally projected ball also travels to the right as it falls.
EXAMPLE 3.5 Starting at the Top: Horizontal Projection

Suppose that the ball in Fig. 3.10a is projected from a height of (a) As noted previously, the time of flight is the same as the
25.0 m above the ground and is thrown with an initial hori- time it takes for the ball to fall vertically to the ground. To find
zontal velocity of 8.25 m>s. (a) How long is the ball in flight this time, the equation y = yo + vyo t - 12 gt2 can be used, in
before striking the ground? (b) How far from the building which the negative direction of g is expressed explicitly, as
does the ball strike the ground? was done in Section 2.4. With vyo = 0,
T H I N K I N G I T T H R O U G H . In looking at the components of y = - 12 gt2
motion, we see that part (a) involves the time it takes the ball
to fall vertically, analogous to a ball dropped from that height. So,
This time is also the time the ball travels in the horizontal 2y 21- 25.0 m2
direction. The horizontal speed is constant, so the horizontal t = = = 2.26 s
A -g C -9.80 m>s2
distance requested in part (b) can be found.
(b) The ball travels in the x-direction for the same amount of
SOLUTION. Writing the data with the origin chosen as the
time it travels in the y-direction (that is, 2.26 s). Since there is
point from which the ball is thrown and downward taken as
no acceleration in the horizontal direction, the ball travels in
the negative direction:
this direction with a uniform velocity. Thus, with xo = 0 and
Given: y = - 25.0 m Find: (a) t (time of flight) a x = 0,
(b) x (horizontal distance)
vxo = 8.25 m>s x = vxo t = 18.25 m>s212.26 s2 = 18.6 m
ax = 0
vyo = 0 FOLLOW-UP EXERCISE. (a) Choose the axes to be at the base
of the building, and show that the resulting equation is the
ay = - g same as in the Example. (b) What is the velocity (in compo-
(xo = 0 and yo = 0 because of our choice of axes location.) nent form) of the ball just before it strikes the ground?
y
vy3 = 0
ymax vy2 vxo
vxo vxo
vy1
vy4
vxo vxo
vyo
vo v y5
vxo
x
vxo
v y6
Range R = x max v6
䉱 F I G U R E 3 . 1 1 Projection at an angle The velocity components of the ball are shown for various times.
(Directions are indicated by signs, with the + sign being omitted as conventionally understood.) Note that
vy = 0 at the top of the arc, or at ymax . The range R is the maximum horizontal distance, or xmax . (Notice
that vo = v6 in magnitude. Why?)
PROJECTIONS AT ARBITRARY ANGLES

The general case of projectile motion involves an object projected at an arbitrary
angle u relative to the horizontal—for example, a golf ball hit by a club (䉱 Fig. 3.11).
During projectile motion, the object travels up and down while traveling horizon-
tally with a constant velocity. (Does the ball have acceleration? Yes. At each point
of the motion, gravity acts, and aB = - gyN .)
This motion is also analyzed by using its components. As before, upward is
taken as the positive direction and downward as the negative direction. The initial
velocity vo is first resolved into rectangular components:
vxo = vo cos u (3.8a)

vyo = vo sin u (initial velocity components) (3.8b)
There is no horizontal acceleration and the acceleration due to gravity acts in the
negative y-direction. Thus, the x-component of the velocity is constant and the y-
component varies with time (see Eq. 3.3d):
vx = vxo = vo cos u (3.9a)

(projectile motion velocity components)
vy = vyo - gt = vo sin u - gt (3.9b)
The components of the instantaneous velocity at various times are illustrated in

Fig. 3.11. The instantaneous velocity is the sum of these components and is tan-
gent to the curved path of the ball at any point. Notice that the ball strikes the
ground at the same speed (but with -vyo) and at the same angle below the hori-
zontal as it was launched.
Similarly, the displacement components are given by 1xo = yo = 02:
x = vxo t = 1vo cos u2t (projectile motion (3.10a)

1 2 1 2
y = vyo t - 2 gt = 1vo sin u2t - 2 gt
locations) (3.10b)
The curve described by these equations, or the path of motion (trajectory) of the
projectile, is called a parabola. The path of projectile motion is often referred to as
a parabolic arc. Such arcs are commonly observed (䉳 Fig. 3.12).
Note that, as in the case of horizontal projection, time is the common feature shared
䉱 F I G U R E 3 . 1 2 Parabolic
arcs Sparks of hot metal projectiles
by the components of motion. Aspects of projectile motion that may be of interest in
from welding describe parabolic various situations include the time of flight, the maximum height reached, and the
arcs. range (R), which is the maximum horizontal distance traveled.
EXAMPLE 3.6 Teeing Off: Projection at an Angle

Suppose a golf ball is hit off the tee with an initial velocity of Solving for tu ,
30.0 m>s at an angle of 35° to the horizontal, as in Fig. 3.11. vyo 17.2 m>s
(a) What is the maximum height reached by the ball? (b) What tu = = = 1.76 s
is its range? g 9.80 m>s 2
T H I N K I N G I T T H R O U G H . The maximum height involves the (Note that tu represents the amount of time the ball moves
y-component; the procedure for finding this is like that for upward.)
finding the maximum height of a ball projected vertically The maximum height ymax is then obtained by substituting
upward. The ball travels in the x-direction for the same tu into Eq. 3.10b:
amount of time it would take for the ball to go up and down. ymax = vyo tu - 12 gt2u
SOLUTION. = 117.2 m>s211.76 s2 - 12 19.80 m>s2 2(1.76 s22 = 15.1 m
Given: vo = 30.0 m>s Find: (a) ymax The maximum height could also be obtained directly from
u = 35° (b) R = xmax Eq. 2.11⬘, v2y = v2yo - 2gy, with y = ymax and vy = 0. How-
ay = - g ever, the method of solution used here illustrates how the
(xo and yo = 0 and final y = 0) time of flight is obtained.
Let us compute vxo and vyo explicitly so simplified kinematic (b) As in the case of vertical projection, the time in going up is
equations can be used: equal to the time in coming down, so the total time of flight is
t = 2tu (to return to the elevation from which the object was
vxo = vo cos 35° = 130.0 m>s210.8192 = 24.6 m>s
projected, y = yo = 0, as can be seen from
vyo = vo sin 35° = 130.0 m>s210.5742 = 17.2 m>s y - yo = vyo t - 12 gt2 = 0, and t = 2vyo >g = 2tu.)
The range R is equal to the horizontal distance traveled
(a) Just as for an object thrown vertically upward, vy = 0 at
(xmax), which is easily found by substituting the total time of
the maximum height (ymax). Thus, the time to reach the maxi-
flight t = 2tu = 211.76 s2 = 3.52 s into Eq. 3.10a:
mum height (tu) can be found by using Eq. 3.3b, with vy set
equal to zero: R = xmax = vx t = vxo12tu2 = 124.6 m>s213.52 s2 = 86.6 m
vy = 0 = vyo - gtu
F O L L O W - U P E X E R C I S E . How would the values of maximum height (ymax) and the range (xmax) compare with those found in this
Example if the golf ball had been similarly teed off on the surface of the Moon? [Hint: gM = g>6; that is, acceleration due to gravity
on the Moon is one-sixth of that on the Earth.] Do not do any numerical calculations. Find the answers by “sight reading” the
equations.
The range of a projectile is an important consideration in various applications.

This factor is particularly important in sports in which a maximum range is
desired, such as golf and javelin throwing.
In general, what is the range of a projectile launched with velocity vo at an angle
u? In order to answer this question, consider the equation used in Example 3.6 to
calculate the range, R = vx t. First let’s look at the expressions for vx and t. Since
there is no acceleration in the horizontal direction,
vx = vxo = vo cos u
and the total time t (as shown in Example 3.6) is

2vyo 2vo sin u
t = =
g g
and R is given by,

2vo sin u 2v 2o sin u cos u
R = vx t = 1vo cos u2a b =
g g
Using the trigonometric identity sin 2u = 2 cos u sin u (see Appendix I),
v2o sin 2u (projectile range xmax ,
R = (3.11)
g only for yinitial = yfinal)
Note that the range depends on the magnitude of the initial velocity (or speed), vo ,
and that the angle of projection, u, and g are assumed to be constant. Keep in mind
that this equation applies only to the special, but common, case of yinitial = yfinal , that
is, when the landing point is at the same height as the launch point.
EXAMPLE 3.7 A Throw from the Bridge

A young girl standing on a bridge throws a stone with an ini- y
tial velocity of 12 m>s at a downward angle of 45° to the hori-
xo = 0, yo = 0 (launch point)
zontal, in an attempt to hit a block of wood floating in the
river below (䉴 Fig. 3.13). If the stone is thrown from a height of
20 m and it just reaches the water when the block is 13 m from vo 45° x
the bridge, does the stone hit the block? (Assume that the
block does not move appreciably and that it is in the plane of
the throw.)
T H I N K I N G I T T H R O U G H . The question is, what is the range of
the stone? If this range is the same as the distance between the
block and the bridge, then the stone hits the block. To find the
range of the stone, we need to find the time of descent (from
the y-component of motion) and then use this time to find the
distance xmax . (Time is the connecting factor.)
y = –20 m ??
x
xblock = 13 m
䉱 F I G U R E 3 . 1 3 A throw from the bridge—hit or miss? See

Example text for description.
SOLUTION.
Given: vo = 12 m>s Find: Range or xmax of stone from bridge. (Is it the
u = 45° vxo = vo cos 45° = 8.5 m>s same as the block’s distance from the bridge?)
y = - 20 m vyo = - vo sin 45° = - 8.5 m>s
xblock = 13 m
1xo = yo = 02
To find the time for upward travel, vy = vyo - gt was used in The stone’s horizontal distance from the bridge at this time is
Example 3.6, where vy = 0 at the top of the arc. However, in
xmax = vxo t = 18.5 m>s211.4 s2 = 12 m
this case, vy is not zero when the stone reaches the river, so to
use this equation, vy is needed. This may be found from the So the girl’s throw falls short by a meter (the block is at 13 m).
kinematic equation Eq. 2.11’, Note that Eq. 3.10b, y = yo + vyo t - 12 gt2, could have been
used to find the time, but this calculation would have
v2y = v2yo - 2gy
involved solving a quadratic equation.
as
F O L L O W - U P E X E R C I S E . (a) Why was it assumed that the
vy = 21- 8.5 m>s22 - 219.8 m>s221 - 20 m2 = - 22 m>s block was in the plane of the throw? (b) Why wasn’t Eq. 3.11
used in this Example to find the range? Show that Eq. 3.11
(negative root because vy is downward).
works in Example 3.6, but not in Example 3.7, by computing
Then solving vy = vyo - gt for t,
the range in each case and comparing your results with the
vyo - vy - 8.5 m>s - 1-22 m>s2 answers found in the Examples.
t = = = 1.4 s
g 9.8 m>s 2
CONCEPTUAL EXAMPLE 3.8 Which Has the Greater Speed?

Consider two balls, both thrown with the same initial speed REASONING AND ANSWER. At first, you might think the
vo , but one at an angle of 45° above the horizontal and the answer is (b), because this ball is projected downward. But
other at an angle of 45° below the horizontal (䉲 Fig. 3.14). the ball projected upward falls from a greater maximum
Determine whether, upon reaching the ground, (a) the ball height, so perhaps the answer is (a). To solve this dilemma,
projected upward will have the greater speed, (b) the ball pro- look at the horizontal line in Fig. 3.14 between the two veloc-
jected downward will have the greater speed, or (c) both balls ity vectors that extends beyond the upper trajectory. From
will have the same speed. Clearly establish the reasoning and this diagram, it can be seen that the trajectories for both balls
physical principle(s) used in determining your answer before check- are the same below this line. Moreover, the downward veloc-
ing it. That is, why did you select your answer? ity of the upper ball on reaching this line is vo at an angle of
45° below the horizontal. (See Fig. 3.11.) Therefore, relative to
the horizontal line and below, the conditions are identical,
vo with the same y-component and the same constant
2 x-component. So the answer is (c).
45°
45° F O L L O W - U P E X E R C I S E . Suppose the ball thrown downward
vo was thrown at an angle of - 40° and the upward ball at 45°.
Which ball would hit the ground with the greater vertical
velocity?
䉳 F I G U R E 3 . 1 4 Which has the

greater speed? See Example text
for description.
The range of a projectile projected downward, as in Fig. 3.14, is found as illustrated in

Example 3.7. But what about the range of a projectile projected upward? This case might
be thought of as an “extended range” problem. One way to solve it is to divide the
trajectory into two parts—(1) the arc above the horizontal line and (2) the downward
part below the horizontal line—such that xmax = x1 + x2 . You know how to find x1
(Example 3.6) and x2 (Example 3.7). Another way to solve the problem is to use
y = yo + vyo t - 12 gt2, where y is the final position of the projectile, and solve for t, the
total time of flight. You would then use that value in the equation x = vxo t.
v2o sin 2u
Equation 3.11, R = , allows the range to be computed for a particular
g
projection angle and initial velocity on a level surface. However, we are some-
times interested in the maximum range for a given initial velocity—for example,
the maximum range of an artillery piece that fires a projectile with a particular
muzzle velocity. Is there an optimum angle that gives the maximum range? Under
ideal conditions, the answer is yes.
䉴 F I G U R E 3 . 1 5 Range For a pro- y

jectile with a given initial speed, the
maximum range is ideally attained 75°
with a projection of 45° (no air resis-
tance). For projection angles above
and below 45°, the range is shorter, 60°
and it is equal for angles equally dif-
ferent from 45° (for example, 30° 45°
and 60°).
30°
15°
x
For a particular vo , the range is a maximum 1Rmax2 when sin 2u = 1, since this
value of u yields the maximum value of the sine function (which varies from 0
to 1). Thus,
v2o
Rmax = (yinitial = yfinal) (3.12)
g
Because this maximum range is obtained when sin 2u = 1 and because sin 90° = 1,
2u = 90° or u = 45°
for the maximum range for a given initial speed when the projectile returns to the elevation
from which it was projected. At a greater or smaller angle, for a projectile with the
same initial speed, the range will be less, as illustrated in 䉱 Fig. 3.15. Also, the
range is the same for angles equally above and below 45°, such as 30° and 60°.
Thus, to get the maximum range, a projectile ideally should be projected at an
angle of 45°. However, up to now, air resistance has been neglected. In actual situa-
tions, such as when a baseball is thrown or hit, this factor may have a significant
effect. Air resistance reduces the speed of the projectile, thereby reducing the
range. As a result, when air resistance is a factor, the angle of projection for maxi-
mum range is less than 45°, which gives a greater initial horizontal velocity
(䉲 Fig. 3.16). Other factors, such as spin and wind, may also affect the range of a
projectile. For example, backspin on a driven golf ball provides lift and the projec-
tion angle for the maximum range may be considerably less than 45°.
45° With no air resistance
< 45° With air resistance

45° With air resistance
x
(a) (b)
䉱 F I G U R E 3 . 1 6 Air resistance and range (a) When air resistance is a factor, the angle of projection for maximum range is less than
45°. (b) Javelin throw. Because of air resistance, the javelin is thrown at an angle less than 45° in order to achieve maximum range.
Keep in mind that for the maximum range to occur at a projection angle of 45°,
the components of initial velocity must be equal—that is, tan-11vyo >vxo 2 = 45°
and tan 45° = 1, so that vyo = vxo. However, this condition may not always be
physically possible, as Conceptual Example 3.9 shows.
CONCEPTUAL EXAMPLE 3.9 The Longest Jump: Theory and Practice

In a long-jump event, does the jumper normally have a mize both velocity components. The initial vertical velocity
launch angle of (a) less than 45°, (b) exactly 45°, or (c) greater component vyo depends on the upward push of the jumper’s
than 45°? Clearly establish the reasoning and physical principle(s) legs, whereas the initial horizontal velocity component vxo
used in determining your answer before checking it. That is, why depends mostly on the running speed toward the jump point.
did you select your answer? In general, a greater velocity can be achieved by running
than by jumping, so vxo 7 vyo. Then, since u = tan-11vyo >vxo2,
REASONING AND ANSWER. Air resistance is not a major fac-
then u 6 45°, where vyo >vxo 6 1 in this case. Hence, the
tor here (although wind speed is taken into account for record
answer is (a)—it certainly could not be (c). A typical launch
setting in track-and-field events). Therefore, it would seem
angle for a long jump is 20° to 25°. (If a jumper increased the
that in order to achieve maximum range, the jumper would
launch angle to be closer to the ideal 45°, then the running
take off at an angle of 45°. But there is another physical con-
speed would have to decrease, resulting in a decrease in range.)
sideration. Let’s look more closely at the jumper’s initial
velocity components (䉲 Fig. 3.17a). F O L L O W - U P E X E R C I S E . When driving in and jumping to score,
To maximize a long jump, the jumper runs as fast as possi- basketball players seem to be suspended momentarily, or to
ble and then pushes upward as strongly as possible to maxi- “hang” in the air (Fig. 3.17b). Explain the physics of this effect.
䉳 F I G U R E 3 . 1 7 Athletes in action (a) To

maximize a long jump, a jumper runs as fast
as possible and then pushes upward as
strongly as he can to maximize the velocity
components (vx and vy ). (b) When driving in
toward the basket and jumping to score, bas-
ketball players seem to be suspended momen-
tarily, or “hang” in the air.
(a) (b)
EXAMPLE 3.10 A “Slap Shot”: Is It Good?

A hockey player hits a “slap shot” in practice (with no goalie T H I N K I N G I T T H R O U G H . First let’s make a sketch of the situa-
present) when he is 15.0 m directly in front of the net. The net tion using x–y coordinates, assuming that the puck is at the
is 1.20 m high, and the puck is initially hit at an angle of 5.00° origin at the time it is hit and showing the net and its height
above the ice with a speed of 35.0 m>s. (a) Determine whether as in 䉲 Fig. 3.18. Note that the launch angle is exaggerated. An
the puck makes it into the net. (b) If it does, determine angle of 5.00° is quite small, but then again, the top of the net
whether the puck is rising or falling vertically as it crosses the is not overly high (1.20 m).
front plane of the net.
y
1.20 m
5.00° = (exaggerated for clarity)
x
(Ice)
x = 0 (launch point)
x = 15.0 m 䉳 F I G U R E 3 . 1 8 Slap shot Is it a goal? See
y=0 Example text for description.
To determine whether the shot is of goal quality, we need to know whether the puck’s trajectory takes it above the net or into the
net. That is, what is the puck’s height (y) when its horizontal distance is x = 15.0 m? Whether the puck is rising or falling at this
horizontal distance depends on when the puck reaches its maximum height. The appropriate equation(s) should provide this infor-
mation; but keep mind that time is the connecting factor between the x- and y-components.
SOLUTION. Listing the data as usual,

Given: x = 15.0 m, xo = 0 Find: (a) Whether the puck goes into the net
ynet = 1.20 m, yo = 0 (b) If so, is it rising or falling?
u = 5.00°
vo = 35.0 m>s
vxo = vo cos 5.00° = 34.9 m>s
vyo = vo sin 5.00° = 3.05 m>s
The vertical location of the puck at any time t is given by y = vyo t - 12 gt2, so we need to know how long the puck takes to travel
the 15.0 m to the net. The connecting factor of the components is time, so this time can be found from the x motion:
x 15.0 m
x = vxo t or t = = = 0.430 s
vxo 34.9 m>s
So on reaching the front of the net, the puck is at a height of

y = vyo t - 12 gt2 = (3.05 m>s210.430 s2 - 12 19.80 m>s2210.430 s22
= 1.31 m - 0.906 m = 0.40 m
Goal!
The time (tu) for the puck to reach its maximum height is given by vy = vyo - gtu , where vy = 0 and
vyo 3.05 m>s
tu = = = 0.311 s
g 9.80 m>s2
and with the puck reaching the net in 0.430 s, it is descending.

FOLLOW-UP EXERCISE. At what distance from the net did the puck start to descend?
DID YOU LEARN?

➥ The vertical components of a dropped ball and a ball horizontally projected from
the same height are the same, and the balls fall at the same rate hitting the ground
at the same time.
➥ The vx component is constant in projectile motion because there is no acceleration
in that direction, whereas the vy component changes with time because of the
acceleration due to gravity.
*3.4 Relative Velocity

➥ What is relative in relative velocity?

ac = vab + vbc, what do the subscripts
B B B
➥ When adding velocity vectors, such as v
stand for and how is it known that the equation is set up correctly?
Velocity is not absolute, but is dependent on the observer. That is, its description is
relative to the observer’s state of motion. If an object is observed moving with a
certain velocity, then that velocity must be relative to something else. For example,
a bowling ball moves down the alley with a certain velocity, and its velocity is rela-
tive to the alley. The motions of objects are often described as being relative to the
Earth or ground, which is commonly thought of as a stationary frame of reference.
In other instances it may be convenient to use a moving frame of reference.
Measurements must be made with respect to some reference. This reference is
usually taken to be the origin of a coordinate system. The point you designate as
the origin of a set of coordinate axes is arbitrary and entirely a matter of choice.
For example, you may “attach” the coordinate system to the road or the ground
and then measure the displacement or velocity of a car relative to these axes. For a
“moving” frame of reference, the coordinate axes may be attached to a car moving
along a highway. In analyzing motion from another reference frame, you do not
*3.4 RELATIVE VELOCITY 89
change the physical situation or what is taking place, only the point of view from
which you describe it. Hence, motion is relative (to some reference frame), and is
referred to as relative velocity. Since velocity is a vector, vector addition and sub-
traction are helpful in determining relative velocities.
RELATIVE VELOCITIES IN ONE DIMENSION

When the velocities are linear (along a straight line) in the same or opposite direc-
tions and all have the same reference (such as the ground), the relative velocities
can be found by using vector subtraction. As an illustration, consider cars moving
with constant velocities along a straight, level highway, as in 䉲 Fig. 3.19. The veloc-
ities of the cars shown in the figure are relative to the Earth, or the ground, as indi-
cated by the reference set of coordinate axes in Fig. 3.19a, with motions along the
x-axis. They are also relative to the stationary observers standing by the highway
and sitting in the parked car A. That is, these observers see the cars as moving
B B
with velocities v B = + 90 km>h and vC = - 60 km>h. The relative velocity of two
objects is given by the velocity (vector) difference between them. For example, the
velocity of car B relative to car A is given by
B
v B B
N - 0 = 1+90 km>h 2xN
BA = vB - vA = 1+ 90 km>h 2x
Thus, a person sitting in car A would see car B move away (in the positive x-direction)
with a speed of 90 km>h. For this linear case, the directions of the velocities are indi-
cated by plus and minus signs (in addition to the minus sign in the equation).
Similarly, the velocity of car C relative to an observer in car A is
B
v B B
N - 0 = 1-60 km>h2xN
CA = vC - vA = 1- 60 km>h2x
The person in car A would see car C approaching (in the negative x-direction) with
a speed of 60 km>h.
䉳 F I G U R E 3 . 1 9 Relative velocity The

observed velocity of a car depends on, or is rela-
tive to, the frame of reference. The velocities
C shown in (a) are relative to the ground or to the
parked car. In (b), the frame of reference is with
vCA = 60 km/h respect to car B, and the velocities are those that
y a driver of car B would observe. (See text for
B description.) (c) These aircraft, performing air-
to-air refueling, are normally described as trav-
vBA = 90 km/h eling at hundreds of kilometers per hour. To
A what frame of reference do these velocities refer?
x What is their velocity relative to each other?
0 vA = 0
(a)
y‘ C
vCB = 150 km/h
B
x‘
0 vB = 0
A
vAB = 90 km/h
(b) (c)
But suppose that you want to know the velocities of the other cars relative to
car B (that is, from the point of view of an observer in car B) or relative to a set of
coordinate axes with the origin fixed on car B (Fig. 3.19b). Relative to these axes,
car B is not moving; it acts as the fixed reference point. The other cars are moving
relative to car B. The velocity of car C relative to car B is
B
v B B
N - 1+ 90 km>h2xN = 1 -150 km>h2xN
CB = vC - vB = 1- 60 km>h2x
Similarly, car A has a velocity relative to car B of
B
v B B
N = 1-90 km>h2xN
AB = vA - vB = 0 - 1+90 km>h2x
Notice that relative to B, the other cars are both moving in the negative x-direction.
That is, car C is approaching car B with a velocity of 150 km>h in the negative x-
direction, and car A appears to be receding from car B with a velocity of 90 km>h
in the negative x-direction. (Imagine yourself in car B, and take that position as
stationary. Car C would appear to be coming toward you at a high speed, and
car A would be getting farther and farther away, as though it were moving back-
ward relative to you.) Note that in general,
B B
v AB = - vBA
(Prove this for yourself.)
What about the velocities of cars A and B relative to car C? From the point of
view (or reference point) of car C, both cars A and B would appear to be approach-
ing or moving in the positive x-direction. For the velocity of car B relative to car C,
B
v B B
N - 1- 60 km>h2xN = 1 +150 km>h2xN
BC = vB - vC = 190 km>h2x
Can you show that v B
AC = 1+ 60 km>h2x N ? Also note the situation in Fig. 3.19c.
In some instances, velocities do not all have the same reference point. In such
cases, relative velocities can be found by means of vector addition. To solve prob-
lems of this kind, it is essential to identify the velocity references with care.
Let’s look first at a one-dimensional (linear) example. Suppose that a straight
moving walkway in a major airport moves with a velocity of v B
N,
wg = 1 +1.0 m>s2x
where the subscripts indicate the velocity of the walkway (w) relative to the
ground (g). A passenger (p) on the walkway (w) trying to make a flight connection
walks with a velocity of v B
pw = 1+ 2.0 m>s2x N relative to the walkway. What is the
passenger’s velocity relative to an observer standing next to the walkway (that is,
relative to the ground)?
B
This velocity, v pg , is given by
B
vpg = v B B
pw + vwg = 12.0 m>s2x N + 11.0 m>s2xN = 13.0 m>s2xN
Thus, the stationary observer sees the passenger as traveling with speed of 3.0 m>s
down the walkway. (Make a sketch, and show how the vectors add.) An explana-
tion of the indicator line on the w symbols follows.
Notice the pattern of the subscripts in this example. On the right side of the equation, the
two inner subscripts out of the four total subscripts are the same (w). Basically, the walk-
way (w) is used as an intermediate reference frame. The outer subscripts (p and g) are
sequentially the same as those for the relative velocity on the left side of the equation.
When adding relative velocities, always check to make sure that the subscripts have this
relationship—it indicates that you have set up the equation correctly.
What if a passenger got on the walkway going in the opposite direction and
walked with the same speed as that of the walkway? Now it is essential to indicate
the direction in which the passenger is walking by means of a minus sign:
B
v pw = 1- 1.0 m>s2xN . In this case, relative to the stationary observer,
B
v B B
pg = vpw + vwg = 1- 1.0 m>s2x N + 11.0 m>s2xN = 0
so the passenger is stationary with respect to the ground, and the walkway acts as
a treadmill. (Good physical exercise.)
RELATIVE VELOCITIES IN TWO DIMENSIONS

Of course, velocities are not always in the same or opposite directions. However,
with the knowledge of how to use rectangular components to add or subtract vec-
tors, problems involving relative velocities in two dimensions can be solved, as
Examples 3.11 and 3.12 show.
EXAMPLE 3.11 Across and Down the River: Relative Velocity and Components of Motion
The current of a 500-m-wide straight river has a flow rate of T H I N K I N G I T T H R O U G H . Careful designation of the given
2.55 km>h. A motorboat that travels with a constant speed of quantities is very important—the velocity of what, relative to
8.00 km>h in still water crosses the river (䉲 Fig. 3.20). (a) If the what? Once this is done, part (a) should be straightforward.
boat’s bow points directly across the river toward the oppo- (See the previous Problem-Solving Hint.) For part (b), kine-
site shore, what is the velocity of the boat relative to the sta- matics is used, where the time it takes the boat to cross the
tionary observer sitting at the corner of the bridge? (b) How river is the key.
far downstream will the boat’s landing point be from the
point directly opposite its starting point?
x
vrs
vrs
v br v bs
ymax
500 m
vrs

v br v bs
䉳 F I G U R E 3 . 2 0 Relative velocity
and components of motion
x As the boat moves across the river,
x=0
y=0 it is carried downstream by the
current.
B
SOLUTION. As indicated in Fig. 3.20, the river’s flow velocity (vrs, river to shore) is taken to be in the x-direction and the boat’s
B
velocity (vbr, boat to river) to be in the y-direction. Note that the river’s flow velocity is relative to the shore and that the boat’s
velocity is relative to the river, as indicated by the subscripts. Listing the data,
B
Given: ymax = 500 m (river width) Find: (a) vbs (velocity of boat relative to shore)
(b) x (distance downstream)
vrs = 12.55 km>h 2xN
B
(velocity of river relative to shore)
= 10.709 m>s2xN
vbr = 18.00 km>h2yN
B
= (2.22 m>s2yN (velocity of boat relative to river)
Notice that as the boat moves toward the opposite shore, it velocity. Since the velocity components are constant, the boat
is also carried downstream by the current. These velocity travels in a straight line diagonally across the river (much like
components would be clearly apparent to the jogger crossing the ball rolling across the table in Example 3.1).
B
the bridge and to the person sauntering downstream in (a) The velocity of the boat relative to the shore 1vbs2 is given
Fig. 3.20. If both observers stay even with the boat, the veloc- by vector addition. In this case,
ity of each will match one of the components of the boat’s B B B
vbs = vbr + vrs
Since the velocities are not in the same direction and not (b) To find the distance x that the current carries the boat
along one axis, their magnitudes cannot be added directly. Noti- downstream, we use components. Note that in the y-direction,
ce in Fig. 3.20 that the vectors form a right triangle, so the Pytha- ymax = vbr t, and
gorean theorem can be applied to find the magnitude of vbs:
ymax 500 m
vbs = 2v2br + v 2rs = 212.22 m>s22 + 10.709 m>s22 t = = = 225 s
vbr 2.22 m>s
= 2.33 m>s
which is the time it takes the boat to cross the river.
The direction of this velocity is defined by During this time, the boat is carried downstream by the
vrs 0.709 m>s current a distance of
u = tan-1 a b = tan-1 a b = 17.7°
vbr 2.22 m>s x = vrs t = 10.709 m>s21225 s2 = 160 m
FOLLOW-UP EXERCISE. What is the distance traveled by the boat in crossing the river?
EXAMPLE 3.12 Flying into the Wind: Relative Velocity

An airplane with an airspeed of 200 km>h (its speed in still air) vag
flies in a direction such that with a west wind of 50.0 km>h, it
travels in a straight line northward. (Wind direction is specified West wind
by the direction from which the wind blows, so a west wind
blows from west to east.) To maintain its course due north, the
plane must fly at an angle, as illustrated in 䉴Fig. 3.21. What is
the speed of the plane along its northward path?
vpa
THINKING IT THROUGH. Here again, the velocity designa-
tions are important, but as Fig. 3.21 shows, the velocity vec- N
䉳 F I G U R E 3 . 2 1 Flying into
tors form a right triangle, and the magnitude of the unknown vpg
velocity can be found by using the Pythagorean theorem. W E the wind To fly directly
north, the plane’s heading
S (u-direction) must be west of
north.
SOLUTION. As always, it is important to identify the reference frame to which the given velocities are relative.
B
Given: vpa = 200 km>h at angle u (velocity of plane with respect to still Find: vpg (ground speed of plane)
air = air speed)
B
vag = 50.0 km>h east (velocity of air with respect to the Earth, or
ground = wind speed)
B
Plane flies due north with velocity vpg
The speed of the plane with respect to the Earth, or the ground, vpg , is called the plane’s ground speed, and vpa is its airspeed. Vec-
torially, the respective velocities are related by
B B B
vpg = vpa + vag
If no wind were blowing 1vag = 02, the ground speed and airspeed would be equal. However, a headwind (a wind blowing
directly toward the plane) would cause a slower ground speed, and a tailwind would cause a faster ground speed. The situation
is analogous to that of a boat going upstream versus downstream.
B
Here, vpg is the resultant of the other two vectors, which can be added by the triangle method. Using the Pythagorean theo-
rem to find vpg , noting that vpa is the hypotenuse of the triangle:
vpg = 2v 2pa - v 2ag = 21200 km>h22 - 150.0 km>h22 = 194 km>h
(Note that it was convenient to use the units of kilometers per hour, since the calculation did not involve any other units.)
FOLLOW-UP EXERCISE. What must be the plane’s heading (u-direction) in this Example for the plane to fly directly north?
DID YOU LEARN?

➥ Relative velocity describes the relative motion between different frames of
reference.
➥ The subscripts in an equation to compute relative velocity stand for the different
velocities, and for the overall motion, the innermost two subscripts must be the same.
PULLING IT TOGETHER Soccer, Kinematics, and Vectors

A soccer player kicks a ball at an angle of 50° above the hori-
zontal as shown in 䉴 Fig. 3.22. The player is 10.0 m from the
base of a rectangular building that is 5.0 m high (with a flat
5.0 m
roof). The roof is 10.0 m wide. The ball is kicked with an ini-
tial speed of 15.0 m>s. 50°
(a) Show that the ball will clear the front wall of the build-
10.0 m 10.0 m
ing. (b) Does the maximum height of the ball occur while it is
over the roof, before it passes over the roof, or after it passes
䉱 F I G U R E 3 . 2 2 Where does the ball go? See Exampe text
over the roof? (c) Determine whether the ball lands on the for description.
roof or beyond the building. (d) Wherever the ball lands,
determine its location relative to the back wall of the building. horizontal velocity. Hence, determining the initial velocity
(e) What is the ball’s velocity, in unit vector notation, just components is crucial. (b) Maximum height occurs when the
before it lands? vertical velocity is zero. This enables the determination of
the time and hence the horizontal distance. (c) This involves
T H I N K I N G I T T H R O U G H . This example demonstrates the use finding the time when the ball is 5.0 m above the ground
of kinematic equation in 2-dimensions, components, and unit level. (d) Determining where the ball lands means finding its
vectors. (a) This involves determining the ball’s height above x-coordinate. Part (c) gives the time. (e) The horizontal
the ground at the time it is in the vertical plane of the front velocity does not change, so all that is needed is the vertical
wall. The key here is to determine the time from the constant velocity at the time of landing.
SOLUTION.
Given: 10.0 m (distance of player from building) Find: (a) ball’s height when it has
vo = 15.0 m>s at 50° (ball initial velocity) traveled horizontally 10.0 m
h = 5.0 m (wall height) (b) where ymax occurs
L = 100 m (roof width) (c) if ball lands on roof
(d) landing location
B
(e) v (landing velocity)
(a) Locating the kick at the origin of the x–y coordinate sys- Since x is less than the 20 m horizontal distance to the back
tem, the ball’s coordinates vary with time according to wall but greater than the 10 m horizontal distance to the near
wall, it reaches its maximum height while over the roof.
x = vxot
(c) To determine whether the ball lands on the roof, the time
y = vyo t - 12 gt2 for it to reach the roof height 1y = 5.0 m2 is needed. There will
be two such times (why?). The general equation for the
where vxo = vo cos 50° = 9.64 m>s and vyo = vo sin 50° = ball’s height above the ground is y = vyo t - 12 gt2, which, after
11.5 m>s. To find the time to reach the vertical plane of the substituting, becomes 5.00 = 11.5t - 4.90t2, expressed in
near wall, set x = 10.0 m in the x equation and solve for t: meters and seconds. (Units are omitted for convenience.) In
standard quadratic form this is 4.90t2 - 11.5t + 5.00 = 0.
x 10.0 m
t = = = 1.04 s The quadratic formula yields the two roots:
vxo 9.64 m>s
11.5 ⫾ 41 -11.522 - 414.90215.002 11.5 ⫾ 5.85
The ball’s height at this time is found by substituting this time t = =
214.902 9.80
into the y equation:
and t = 0.58 s or 1.77 s
y = 111.5 m>s211.04 s2 - 12 19.80 m>s2211.04 s22 = 6.35 m
Since it takes the ball 1.04 s to reach the front wall’s plane, the
Since this is greater than the wall height of 5.0 m, the ball does 0.58 s is the time when the ball is 5.0 m high as it rises on its
clear the wall. way to the building. The 1.77 s refers to the time at which the
(b) To find the time 1tmax2 for the ball to get to its maximum ball is 5.0 m high on the way down after clearing the front
height 1ymax2, set the vertical velocity component to zero and wall. This latter time needs to be compared to the time to
solve. The general equation for the vertical velocity is reach the vertical plane of the back wall, where x = 20.0 m. To
vy = vyo - gt, hence 0 = vyo - gtmax , and get to the plane of the back wall requires a time of
vyo x 20.0 m
11.5 m>s t = = = 2.08 s
tmax = = = 1.17 s vxo 9.64 m>s
g 9.80 m>s2
Since this time is longer than 1.77 s, the ball lands on the roof.
Use this time to determine the ball’s horizontal location (x) (d) To determine where on the roof the ball lands, find x for
when it is at its maximum height: the 1.77 s time in part (c).
x = vxo tmax = 19.64 m>s211.17 s2 = 11.3 m x = vxo t = 19.64 m>s211.77 s2 = 17.1 m
Since the back wall is at x = 20.0 m, the ball lands 20.0 - 17.1 As expected, the vertical component is negative and smaller
or 2.9 m before the rear edge of the roof. in magnitude than the initial vertical component (why?).
(e) The velocity just before landing on the roof requires both Finally, the velocity just before hitting the rooftop in unit vec-
x- and y-components. The x-component is constant, hence tor notation is
vx = + 9.64 m>s. The y-component is determined from the vB = vx xN + vy yN = 19.64xN - 5.85yN 2 m>s
1.77 s elapsed time and the vertical velocity equation,
vy = vyo - gt = 11.5 m>s - 19.80 m>s2211.77 s2 = - 5.85 m>s
■ Motion in two dimensions is analyzed by considering the Vector Representation:

motion of linear components. The connecting factor between
components is time. C = 2C 2x + C 2y
Cy ∂ (magnitude–angle form) (3.4a)
Components of Initial Velocity:
u = tan-1 ` `
vxo = vo cos u (3.1a) Cx
B
vyo = vo sin u (3.1b) C = CxxN + CyyN (component form) (3.7)
Components of Displacement (constant acceleration only): ■ Projectile motion is analyzed by considering horizontal
x = xo + vxo t + 12 ax t2 (3.3a) and vertical components separately—constant velocity in
the horizontal direction and an acceleration due to gravity,
y = yo + vyo t + 12 ay t2 (3.3b) g, in the downward vertical direction. (The foregoing equa-
Components of Velocity (constant acceleration only): tions for constant acceleration then have an acceleration of
a = - g instead of a.)
vx = vxo + ax t (3.3c)
vy = vyo + ay t (3.3d)
■ Range (R) is the maximum horizontal distance traveled.
v 2o sin 2u (projectile range xmax
R = (3.11)
g only for yinitial = yfinal)
y
vy
v v2o
vy
Rmax = (yinitial = yfinal) (3.12)
g
vx
vy vy v y
vy3 = 0
ymax vy2 vxo
vx
vy vy v vxo vxo
vy1
-vy4
vxo
vx vxo
vyo
vy vy v -vy5
vo
vxo
x
vxo
vx
-vyo
Range R = x max v6
x
vx vx vx vx
■ Relative velocity is expressed relative to a particular refer-

ence frame.
■ Of the various methods of vector addition, the component
method is most useful. A resultant vector can be expressed in
magnitude–angle form or in unit vector component form.

3.1 COMPONENTS OF MOTION 2. The equation x = xo + vxot + 12 a xt2 applies (a) to all
1. On Cartesian axes, the x-component of a vector is gener- kinematic problems, (b) only if vyo is zero, (c) to constant
ally associated with a (a) cosine, (b) sine, (c) tangent, accelerations, (d) to negative times.
(d) none of the foregoing.
3. For an object in curvilinear motion, (a) the object’s veloc- eration combined with (a) an equal horizontal accelera-
ity components are constant, (b) the y-velocity component tion, (b) a uniform horizontal velocity, (c) a constant
is necessarily greater than the x-velocity component, upward velocity, (d) an acceleration that is always per-
(c) there is an acceleration nonparallel to the object’s path, pendicular to the path of motion.
(d) the velocity and acceleration vectors must be at right
9. A football is thrown on a long pass. Compared to the
angles (90°).
ball’s initial horizontal velocity component, the velocity
4. Which one of the following cannot be a true statement at the highest point is (a) greater, (b) less, (c) the same.
about an object: (a) It has zero velocity and a nonzero
acceleration; (b) it has velocity in the x-direction and 10. A football is thrown on a long pass. Compared to the
acceleration in the y-direction; (c) it has velocity in the y- ball’s initial vertical velocity, the vertical component of
direction and acceleration in the y-direction; of (d) it has its velocity at the highest point is (a) greater, (b) less,
constant velocity and changing acceleration? (c) the same.
3.2 VECTOR ADDITION *3.4 RELATIVE VELOCITY

AND SUBTRACTION 11. You are traveling in a car on a straight, level road going
5. Two linear vectors of magnitudes 3 and 4 are added. The 70 km>h. A car coming toward you appears to be travel-
magnitude of the resultant vector is (a) 1, (b) 7, ing 130 km>h. How fast is the other car going:
(c) between 1 and 7. (a) 130 km>h, (b) 60 km>h, (c) 70 km>h, or (d) 80 km>h?
B B B B
6. The resultant of A - B is the same as (a) B - A, 12. Two cars approach each other on a straight, level high-
B B B B B B
(b) - A + B, (c) - 1A + B2, (d) -1B - A2. way. Car A travels at 60 km>h and car B at 80 km>h. The
7. A unit vector has (a) magnitude, (b) direction, (c) neither driver of car B sees car A approaching at a speed of
of these, (d) both of these. (a) 60 km>h, (b) 80 km>h, (c) 20 km>h, (d) greater than
100 km>h.
13. For the situation in Exercise 12, at what speed does the
3.3 PROJECTILE MOTION
driver of car A see car B approaching: (a) 60 km>h,
8. If air resistance is neglected, the motion of an object pro- (b) 80 km>h, (c) 20 km>h, or (d) greater than 100 km>h?
jected at an angle consists of a uniform downward accel-
3.1 COMPONENTS OF MOTION y

1. Can the x-component of a vector be greater than the
magnitude of the vector? How about the y-component?
Explain. A
2. Is it possible for an object’s velocity to be perpendicular
to the object’s acceleration? If so, describe the motion. F
3. Describe the motion of an object that is initially travel- C
B
ing with a constant velocity and then receives an accel-
eration of constant magnitude (a) in a direction
parallel to the initial velocity, (b) in a direction perpen- E
dicular to the initial velocity, and (c) that is always per- x
0
pendicular to the instantaneous velocity or direction of
motion. 䉳 FIGURE 3.23
D G Different vectors?
See Conceptual
3.2 VECTOR ADDITION Question 9.
AND SUBTRACTION
4. What are the conditions for two vectors to add to zero? 3.3 PROJECTILE MOTION
5. (a) Can a vector be less than one of its components?
(b) How about equal to one of its components? 10. A golf ball is hit on a level fairway. When it lands, its
velocity vector has rotated through an angle of 90°. What
6. Can a nonzero vector have a zero x-component? Explain.
was the launch angle of the golf ball? [Hint: See
7. Is it possible to add a vector quantity to a scalar Fig. 3.11.]
quantity?
B B B B 11. Figure 3.10b shows a multiflash photograph of one ball
8. Can A + B equal zero, when A and B have nonzero dropping from rest and, at the same time, another ball
magnitudes? Explain. projected horizontally from the same height. The two
9. Are any of the vectors in 䉴 Fig. 3.23 equal? balls hit the ground at the same time. Explain.
12. In 䉲 Fig. 3.24, a spring-loaded “cannon” on a wheeled car sighting is used to shoot a target uphill, should one aim
fires a metal ball vertically. The car is given a push and above, below, or right at the bull’s-eye? (b) How about
set in motion horizontally with constant velocity. A pin is shooting downhill?
pulled with a string to launch the ball, which travels
upward and then falls back into the moving cannon
*3.4 RELATIVE VELOCITY
every time. Why does the ball always fall back into the
cannon? Explain. 14. Sitting in a parked bus, you suddenly look up at a bus
? moving alongside and it appears that you are moving.
vy Why is this? How about with both buses moving in
opposite directions?
15. A student walks on a treadmill moving at 4.0 m>s and
remains at the same place in the gym. (a) What is the stu-
dent’s velocity relative to the gym floor? (b) What is the
student’s speed relative to the treadmill?
16. You are running in the rain along a straight sidewalk to
your dorm. If the rain is falling vertically downward rel-
vx ative to the ground, how should you hold your umbrella
so as to minimize the rain landing on you? Explain.
17. When driving to the basket for a layup, a basketball
䉱 F I G U R E 3 . 2 4 A ballistics car See Conceptual player usually tosses the ball gently upward relative to
Question 12 and Exercise 52. herself. Explain why.
18. When you are riding in a fast-moving car, in what direc-
13. A rifle is sighted-in so that a bullet hits the bull’s-eye of a tion would you throw an object up so it will return to
target 1000 m away on the same level. (a) If the same your hand? Explain.
EXERCISES
Many exercise sections will include “paired” exercises. These exercise pairs, identified with red numbers, are
3.1 COMPONENTS OF MOTION 6. IE ● ● A student walks 100 m west and 50 m south. (a) To
get back to the starting point, the student must walk in a
1. ● An airplane climbs at an angle of 15° with a horizontal
general direction of (1) south of west, (2) north of east,
component of speed of 200 km>h. (a) What is the plane’s
(3) south of east, (4) north of west. (b) What displace-
actual speed? (b) What is the magnitude of the vertical
ment will bring the student back to the starting point?
component of its velocity?
2. IE ● A golf ball is hit with an initial speed of 35 m>s at 7. ●● A student strolls diagonally across a level rectangular
an angle less than 45° above the horizontal. (a) The hori- campus plaza, covering the 50-m distance in 1.0 min
zontal velocity component is (1) greater than, (2) equal (䉲 Fig. 3.25). (a) If the diagonal route makes a 37° angle
to, (3) less than the vertical velocity component. Why? with the long side of the plaza, what would be the dis-
(b) If the ball is hit at an angle of 37°, what are the initial tance traveled if the student had walked halfway around
horizontal and vertical velocity components? the outside of the plaza instead of along the diagonal
route? (b) If the student had walked the outside route in
3. IE ● The x- and y-components of an acceleration vector 1.0 min at a constant speed, how much time would she
are 3.0 m>s2 and 4.0 m>s2, respectively. (a) The magni- have spent on each side?
tude of the acceleration vector is (1) less than 3.0 m>s2,
(2) between 3.0 m>s2 and 4.0 m>s2, (3) between 4.0 m>s2
and 7.0 m>s2, (4) equal to 7.0 m>s2. (b) What are the mag-
nitude and direction of the acceleration vector?
4. ● If the magnitude of a velocity vector is 7.0 m>s and the
x-component is 3.0 m>s, what is the y-component?

m
50
5. ●● The x-component of a velocity vector that has an

37°
angle of 37° to the + x-axis has a magnitude of 4.8 m>s.
(a) What is the magnitude of the velocity? (b) What is the
magnitude of the y-component of the velocity? 䉱 F I G U R E 3 . 2 5 Which way? See Exercise 7.
EXERCISES 97
8. ●● A ball rolls at a constant velocity of 1.50 m>s at 21. ●● If the vector is added to vector , the result is . If is
an angle of 45° below the + x-axis in the fourth quadrant. subtracted from , the result is . What is the magnitude of
B
If we take the ball to be at the origin at t = 0 what are its A?
coordinates (x, y) 1.65 s later?
22. ●● Two boys are pulling a box across a horizontal floor
9. ●● A ball rolling on a table has a velocity with rectangu- as shown in 䉲 Fig 3.26. If F1 = 50.0 N and F2 = 100 N,
lar components vx = 0.60 m>s and vy = 0.80 m>s. What find the resultant (or sum) force by (a) the graphical
is the displacement of the ball in an interval of 2.5 s? method and (b) the component method.
10. ●● A hot air balloon rises vertically with a speed of
1.5 m>s. At the same time, there is a horizontal 10 km>h N
wind blowing. In which direction is the balloon moving? F2
11. IE ● ● During part of its trajectory (which lasts exactly
1 min) a missile travels at a constant speed of 2000 mi>h
while maintaining a constant orientation angle of 20° 60°
from the vertical. (a) During this phase, what is true F1
about its velocity components: (1) vy 7 vx, (2) vy = vx, or 30°
(3) vy 6 vx? [Hint: Make a sketch and be careful of the W E
angle.] (b) Determine the two velocity components ana- (overhead view)
lytically to confirm your choice in part (a) and also calcu-
late how far the missile will rise during this time.
䉳 FIGURE 3.26
12. ●● At the instant a ball rolls off a rooftop it has a hori-
Adding force vectors
zontal velocity component of +10.0 m>s and a vertical See Exercises 22 and 43.
S
component (downward) of 15.0 m>s. (a) Determine the
angle of the roof. (b) What is the ball’s speed as it leaves
the roof? 23. ●● For each of the given vectors, give a vector that,
when added to it, yields a null vector (a vector with a
13. ●● A particle moves at a speed of 3.0 m>s in the magnitude of zero). Express the vector in the form other
+ x-direction. Upon reaching the origin, the particle than that in which it is given (component or magnitude–
receives a continuous constant acceleration of 0.75 m>s2 B
angle): (a) A = 4.5 cm, 40° above the + x-axis;
in the - y-direction. What is the position of the particle B B
(b) B = 12.0 cm2xN - 14.0 cm2yN ; (c) C = 8.0 cm at an
4.0 s later? angle of 60° above the - x-axis.
14. ●● At a constant speed of 60 km>h, an automobile trav-
24. IE ● ● (a) If each of the two components (x and y) of a
els 700 m along a straight highway that is inclined 4.0° to
vector are doubled, (1) the vector’s magnitude doubles,
the horizontal. An observer notes only the vertical
but the direction remains unchanged; (2) the vector’s
motion of the car. What is the car’s (a) vertical velocity
magnitude remains unchanged, but the direction angle
magnitude and (b) vertical travel distance?
doubles; or (3) both the vector’s magnitude and direc-
15. ● ● ● A baseball player hits a home run into the right field tion angle double. (b) If the x- and y-components of a
upper deck. The ball lands in a row that is 135 m hori- vector of 10 m at 45° are tripled, what is the new vector?
zontally from home plate and 25.0 m above the playing B
field. An avid fan measures its time of flight to be 4.10 s. 25. ●● Two vectors are given by A = 4.0 xN - 2.0 yN and
B B B B B
(a) Determine the ball’s average velocity components. B = 1.0 xN + 5.0 yN . What is (a) A + B, (b) B - A, and
B B B B
(b) Determine the magnitude and angle of its average (c) a vector C such that A + B + C = 0?
velocity. (c) Explain why you cannot determine its aver- 26. ●● Two brothers are pulling their other brother on a sled
age speed from the data given. (䉲 Fig 3.27). (a) Find the resultant (or sum) of the vectors
B B B
F1 and F2 (b) If F1 in the figure were at an angle of 27°
3.2. VECTOR ADDITION instead of 37° with the +x-axis, what would be the resul-
B B
AND SUBTRACTION tant (or sum) of F1 and F2?
16. ● Using the triangle method, show graphically that

B B B B B B B
(a) A + B = B + A and (b) if A - B = C, then
B B B
A = B + C
17. IE ● (a) Is vector addition associative? That is, does y
B B B B B B
1A + B2 + C = A + 1B + C2? (b) Justify your answer
graphically. 12
.0 N
B B
F2 N .0
18. The vectors A and B are perpendicular to one another
●
B B B B
12 F1
and in the same plane. Prove that AxBx + AyBy = 0. 37° 37° x
B
19. ● (a) What is the sum of A = 3.0 xN + 5.0 yN and
B
B = 1.0 xN - 3.0 yN ? (b) What are the magnitude and
B B
direction of A + B?
20. ● For the two vectors xB1 = 120 m2xN and xB2 = 115 m2xN , 䉳 FIGURE 3.27
compute and show graphically (a) xB1 + xB2 , (b) xB1 - xB2 , Vector addition
and (c) xB2 - xB1 . See Exercise 26.
B
27. ●● Given two vectors, A which has a length of 10.0 and 35. ●● A student works three problems involving the addi-
B B B
makes an angle of 45° below the - x-axis, and B which tion of two different vectors F1 and F2 . He states that the
has an x-component of + 2.0 and a y-component of +4.0, magnitudes of the three resultants are given by (a) F1 + F2 ,
(a) sketch the vectors on x–y axes, with all their “tails” (b) F1 - F2 , and (c) 2F 21 + F 22 . Are these results
B B
starting at the origin, and (b) calculate A + B. possible? If so, describe the vectors in each case.
28. ●● The velocity of object 1 in component form is 36. ●● A block weighing 50 N rests on an inclined plane. Its
v1 = 1 +2.0 m>s2xN + 1- 4.0 m>s2yN . Object 2 has twice the
B
weight is a force directed vertically downward, as illus-
speed of object 1 but moves in the opposite direction. trated in 䉲 Fig. 3.30. Find the components of the force par-
(a) Determine the velocity of object 2 in component nota- allel to the surface of the plane and perpendicular to it.
tion. (b) What is the speed of object 2?
29. ●● For the vectors shown in 䉲 Fig. 3.28, determine
B B B
A + B + C.
w (50 N)
B (10 m/s)
C (15 m/s)
䉳 FIGURE 3.30
Block on an inclined
= 37° plane See Exercise 36.
30° 60°
x 37. ●● Two displacements, one with a magnitude of 15.0 m
A (5.0 m/s)
and a second with a magnitude of 20.0 m, can have any
angle you want. (a) How would you create the sum of
䉱 F I G U R E 3 . 2 8 Adding vectors these two vectors so it has the largest magnitude possi-
See Exercises 29 and 30. ble? What is that magnitude? (b) How would you orient
them so the magnitude of the sum was at its minimum?
30. ●● For the velocity vectors shown in Fig. 3.28, determine
B B B What value would that be? (c) Generalize the result to
A - B - C.
any two vectors.
B B
31. ●● Given two vectors A and B with magnitudes A and
B B 38. ●●● A person walks from point A to point B as shown in
B, respectively, you can subtract B from A to get a third
B B B B 䉲 Fig.3.31. What is the person’s displacement relative to
vector C = A - B. If the magnitude of C is equal to
B point A?
C = A + B, what is the relative orientation of vectors A
B
and B?
y
32. ●● In two successive chess moves, a player first moves 40 m
his queen two squares forward, then moves the queen 45°
three steps to the left (from the player’s view). Assume 20 m
each square is 3.0 cm on a side. (a) Using forward
(toward the player’s opponent) as the positive y-axis and B 30 m
right as the positive x-axis, write the queen’s net dis-
placement in component form. (b) At what net angle was
the queen moved relative to the leftward direction?
20 m
33. ●● Referring to the parallelogram in 䉲 Fig. 3.29, express 䉳 FIGURE 3.31
30° x
B B B B B B B
C, C - B, and 1E - D + C2 in terms of A and B.
B Adding displace-
A
ment vectors
See Exercise 38.
39. IE ● ● ● A meteorologist tracks the movement of a thun-
C D derstorm with Doppler radar. At 8:00 PM, the storm was
60 mi northeast of her station. At 10:00 PM, the storm is at
75 mi north. (a) The general direction of the thunder-
A storm’s velocity is (1) south of east, (2) north of west,
(3) north of east, (4) south of west. (b) What is the aver-
age velocity of the storm?
B E 䉳 FIGURE 3.29
40. IE ● ● ● A flight controller determines that an airplane is
Vector combos
See Exercise 33. 20.0 mi south of him. Half an hour later, the same plane
is 35.0 mi northwest of him. (a) The general direction of
B
34. ●● Two force vectors, F1 = 13.0 N2xN - 14.0 N2yN and the airplane’s velocity is (1) east of south, (2) north of
B
F2 = 1 - 6.0 N2xN + 14.5 N2yN , are applied to a particle. west, (3) north of east, (4) west of south. (b) If the plane is
B
What third force F3 would make the net, or resultant, flying with constant velocity, what is its velocity during
force on the particle zero? this time?
EXERCISES 99
41. IE ● ● ● 䉲Fig. 3.32 depicts a decorative window (the thick 49. ●● A pitcher throws a fastball horizontally at a speed of
inner square) weighing 100 N suspended in a patio open- 140 km>h toward home plate, 18.4 m away. (a) If the batter’s
ing (the thin outer square). The upper two corner cables combined reaction and swing times total 0.350 s, how long
are each at 45° and the left one exerts a force (F1) of 100 N can the batter watch the ball after it has left the pitcher’s hand
on the window. (a) How does the magnitude of the force before swinging? (b) In traveling to the plate, how far does
exerted by the upper right cable (F2) compare to that the ball drop from its original horizontal line?
exerted by the upper left cable: (1) F2 7 F1 , (2) F2 = F1, or 50. IE ● ● Ball A rolls at a constant speed of 0.25 m>s on a
F2 6 F1? (b) Use your result from part (a) to help deter- table 0.95 m above the floor, and ball B rolls on the floor
mine the force exerted by the bottom cable (F3). directly under the first ball with the same speed and
direction. (a) When ball A rolls off the table and hits the
floor, (1) ball B is ahead of ball A, (2) ball B collides with
F1 F2 ball A, (3) ball A is ahead of ball B. Why? (b) When ball A
hits the floor, how far from the point directly below the
edge of the table will each ball be?
51. ● ● The pilot of a cargo plane flying 300 km>h at an altitude
Suspended
window of 1.5 km wants to drop a load of supplies to campers at a
particular location on level ground. Having the designated
point in sight, the pilot prepares to drop the supplies.
(a) What should the angle be between the horizontal and the
F3 pilot’s line of sight when the package is released? (b) What is
the location of the plane when the supplies hit the ground?
52. ● ● A wheeled car with a spring-loaded cannon fires a
䉱 F I G U R E 3 . 3 2 A suspended
patio window See Exercise 41. metal ball vertically (Fig. 3.24). If the vertical initial speed
of the ball is 5.0 m>s as the cannon moves horizontally at
a speed of 0.75 m>s, (a) how far from the launch point
42. ● ● ● A golfer lines up for her first putt at a hole that is
does the ball fall back into the cannon, and (b) what
10.5 m exactly northwest of her ball’s location. She hits
would happen if the cannon were accelerating?
the ball 10.5 m and straight, but at the wrong angle, 40°
53. ● ● A convertible travels down a straight, level road at a
from due north. In order for the golfer to have a “two-
putt green,” determine (a) the angle of the second putt slow speed of 13 km>h. A person in the car throws a ball
and (b) the magnitude of the second putt’s displacement. with a speed of 3.6 m>s forward at an angle of 30° to the
(c) Determine why you cannot determine the length of horizontal. Where is the car when the ball lands?
travel of the second putt. 54. ● ● A good-guy stuntman is being chased by bad guys on
a building’s level roof. He comes to the edge and is to

43. ● ● ● Two students are pulling a box as shown in
jump to the level roof of a lower building 4.0 m below and
Fig. 3.26, where F1 = 100 N and F2 = 150 N. What third
5.0 m away. What is the minimum launch speed the stunt-
force would cause the box to be stationary when all three
man needs to complete the jump? (Landing on the edge is
forces are applied?
assumed complete.)
55. ● ● An astronaut on the Moon fires a projectile from a
3.3 PROJECTILE MOTION launcher on a level surface so as to get the maximum
(Assume angles to be exact for significant figure purposes.) range. If the launcher gives the projectile a muzzle veloc-
ity of 25 m>s, what is the range of the projectile? [Hint:
44. ● A ball with a horizontal speed of 1.0 m>s rolls off a
The acceleration due to gravity on the Moon is only one-
bench 2.0 m high. (a) How long will the ball take to reach
sixth of that on the Earth.]
the floor? (b) How far from a point on the floor directly
56. ● ● In 2004 two Martian probes successfully landed on the
below the edge of the bench will the ball land?
Red Planet. The final phase of the landing involved bounc-
45. ● An electron is ejected horizontally at a speed of
ing the probes until they came to rest (they were surrounded
1.5 * 106 m>s from the electron gun of a computer mon-
by protective inflated “balloons”). During one of the
itor. If the viewing screen is 35 cm from the end of the
bounces, the telemetry (electronic data sent back to Earth)
gun, how far will the electron travel in the vertical direc-
indicated that the probe took off at 25.0 m>s at an angle of
tion before hitting the screen? Based on your answer, do
20° and landed 110 m away (and then bounced again).
you think designers need to worry about this gravita-
Assuming the landing region was level, determine the accel-
tional effect?
eration due to gravity near the Martian surface.
46. ● A ball rolls horizontally with a speed of 7.6 m>s off the 57. ● ● In laboratory situations, a projectile’s range can be
edge of a tall platform. If the ball lands 8.7 m from the used to determine its speed. To see how this is done, sup-
point on the ground directly below the edge of the plat- pose a ball rolls off a horizontal table and lands 1.5 m out
form, what is the height of the platform? from the edge of the table. If the tabletop is 90 cm above
47. ● A ball is projected horizontally with an initial speed of the floor, determine (a) the time the ball is in the air, and
5.0 m>s. Find its (a) position and (b) velocity at t = 2.5 s. (b) the ball’s speed as it left the table top.
48. ● An artillery crew wants to shell a position on level 58. ● ● A stone thrown off a bridge 20 m above a river has an
ground 35 km away. If the gun has a muzzle velocity of initial velocity of 12 m>s at an angle of 45° above the hori-
770 m>s, to what angle of elevation should the gun be zontal (䉲 Fig. 3.33). (a) What is the range of the stone?
raised? (b) At what velocity does the stone strike the water?
y which is released at the same time that the gun is fired.

45°
x This gun won’t miss as long as the initial speed of the
bullet is sufficient to reach the falling target before the
target hits the floor. Verify this statement, using the fig-
ure. [Hint: Note that yo = x tan u.]
63. IE ● ● ● A shot-putter launches the shot from a vertical

distance of 2.0 m off the ground (from just above her ear)
20 m at a speed of 12.0 m>s. The initial velocity is at an angle
of 20°above the horizontal. Assume the ground is flat.
(a) Compared to a projectile launched at the same angle
R and speed at ground level, would the shot be in the air
(1) a longer time, (2) a shorter time, or (3) the same
amount of time? (b) Justify your answer explicitly; deter-
mine the shot’s range and velocity just before impact in
䉱 F I G U R E 3 . 3 3 A view from the bridge unit vector (component) notation.
See Exercise 58. 64. ● ● ● A ditch 2.5 m wide crosses a trail bike path (䉲Fig 3.36).
An upward incline of 15° has been built on the approach so

59. If the maximum height reached by a projectile
●●
that the top of the incline is level with the top of the ditch.
launched on level ground is equal to half the projectile’s What is the minimum speed a trail bike must be moving to
range, what is the launch angle? clear the ditch? (Add 1.4 m to the range for the back of the
60. ● ● William Tell is said to have shot an apple off his son’s bike to clear the ditch safely.)
head with an arrow. If the arrow was shot with an initial
speed of 55 m>s and the boy was 15 m away, at what
launch angle did Bill aim the arrow? (Assume that the
arrow and apple are initially at the same height above
the ground.) 15°
61. ● ● ● This time, William Tell is shooting at an apple that
hangs on a tree (䉲 Fig. 3.34). The apple is a horizontal dis-
tance of 20.0 m away and at a height of 4.00 m above the 2.5 m
ground. If the arrow is released from a height of 1.00 m
above the ground and hits the apple 0.500 s later, what is
䉱 F I G U R E 3 . 3 6 Clear the ditch See Exercise 64.
the arrow’s initial velocity?
Apple Apple tree 65. ● ● ● A ball rolls down a roof that makes an angle of 30°
to the horizontal (䉲 Fig. 3.37). It rolls off the edge with a

vo speed of 5.00 m>s. The distance to the ground from that
point is two stories or 7.00 m. (a) How long is the ball in
4.00 m
the air? (b) How far from the base of the house does it
1.00 m
land? (c) What is its speed just before landing?
20.0 m
䉱 F I G U R E 3 . 3 4 Hit the apple See Exercise 61. 30°

(Not drawn to scale.)
62. ● ● ● The apparatus for a popular lecture demonstration
is shown in 䉲 Fig. 3.35. A gun is aimed directly at a can,
Can held
y by magnet
dropped at
t=0
t
fs igh
eo
Lin yo
vo 䉱 F I G U R E 3 . 3 7 There she rolls See Exercise 65.

y (Not drawn to scale.)
x
x
66. ●●● A quarterback passes a football—at a velocity of
Gun with switch,
50 ft>s at an angle of 40° to the horizontal—toward an
fired at t = 0
intended receiver 30 yd downfield. The pass is released
䉱 F I G U R E 3 . 3 5 A sure shot See Exercise 62. 5.0 ft above the ground. Assume that the receiver is sta-
(Not drawn to scale.) tionary and that he will catch the ball if it comes to him.
EXERCISES 101
Will the pass be completed? If not, will the throw be long

or short?
67. ● ● ● A 2.05-m-tall basketball player takes a shot when he
is 6.02 m from the basket (at the three-point line). If the

launch angle is 25° and the ball was launched at the level
of the player’s head, what must be the release speed of
the ball for the player to make the shot? The basket is
3.05 m above the floor.
Current 150 m
*3.4 RELATIVE VELOCITY
68. ● While you are traveling in a car on a straight, level
interstate highway at 90 km>h, another car passes you in
the same direction; its speedometer reads 120 km>h.
(a) What is your velocity relative to the other driver?
(b) What is the other car’s velocity relative to you?
69. ● A shopper is in a hurry to catch a bargain in a depart-
ment store. She walks up the escalator, rather than let- 䉱 F I G U R E 3 . 3 8 Over and back See Exercise 76.
ting it carry her, at a speed of 1.0 m>s relative to the (Not drawn to scale.)
escalator. If the escalator is 10 m long and moves at a
speed of 0.50 m>s, how long does it take for the shopper 77. ●● A pouring rain comes straight down with a raindrop
to get to the next floor? speed of 6.0 m>s. A woman with an umbrella walks east-
ward at a brisk clip of 1.5 m>s to get home. At what
70. ● A motorboat’s speed in still water is 2.0 m>s. The dri- angle should she tilt her umbrella to get the maximum
ver wants to go directly across a river with a current protection from the rain?
speed of 1.5 m>s. At what angle upstream should the
78. IE ● ● It is raining, and there is no wind. When you are
boat be steered?
sitting in a stationary car, the rain falls straight down rel-
71. ● A person riding in the back of a pickup truck traveling ative to the car and the ground. But when you’re dri-
at 70 km>h on a straight, level road throws a ball with a ving, the rain appears to hit the windshield at an angle.
speed of 15 km>h relative to the truck in the direction (a) As the velocity of the car increases, this angle (1) also
opposite to the truck’s motion. What is the velocity of increases, (2) remains the same, (3) decreases. Why? (b) If
the ball (a) relative to a stationary observer by the side of the raindrops fall straight down at a speed of 10 m>s, but
the road, and (b) relative to the driver of a car moving in appear to make an angle of 25° to the vertical, what is the
the same direction as the truck at a speed of 90 km>h? speed of the car?
79. ● ● If the flow rate of the current in a straight river is
72. ● In Exercise 71, what are the relative velocities if the
ball is thrown in the direction of the truck? greater than the speed of a boat in the water, the boat
cannot make a trip directly across the river. Prove this
73. ●● In a 500-m stretch of a river, the speed of the current statement.
is a steady 5.0 m>s. How long does a boat take to finish a 80. IE ● ● You are in a fast powerboat that is capable of a
round trip (upstream and downstream) if the speed of sustained steady speed of 20.0 m>s in still water. On a
the boat is 7.5 m>s relative to still water? swift, straight section of a river you travel parallel to the
74. ●● A moving walkway in an airport is 75 m long and bank of the river. You note that you take 15.0 s to go
moves at a speed of 0.30 m>s. A passenger, after travel- between two trees on the riverbank that are 400 m apart.
ing 25 m while standing on the walkway, starts to walk (a) (1) Are you traveling with the current, (2) are you
at a speed of 0.50 m>s relative to the surface of the walk- traveling against the current, or (3) is there no current?
way. How long does she take to travel the total distance (b) If there is a current [reasoned in part (a)], determine
of the walkway? its speed.
81. ● ● An observer by the side of a straight, level, north-
75. IE ● ● A swimmer swims north at 0.15 m>s relative to south road watches a car (A) moving south at a rate of
still water across a river that flows at a rate of 0.20 m>s 75 km>h. A driver in another car (B) going north at
from west to east. (a) The general direction of the swim- 50 km>h also observes car A. (a) What is car A’s velocity
mer’s velocity, relative to the riverbank, is (1) north of as observed from car B? (Take north to be positive.) (b) If
east, (2) south of west, (3) north of west, (4) south of east. the roadside observer sees car A brake to a stop in 6.0 s,
(b) Calculate the swimmer’s velocity relative to the what constant acceleration would be measured?
riverbank. (c) What constant acceleration would the driver in car B
76. ●● A boat that travels at a speed of 6.75 m>s in still water measure for the braking car A?
is to go directly across a river and back (䉴Fig. 3.38). 82. ● ● ● An airplane flies due north with an air speed of
The current flows at 0.50 m>s. (a) At what angle(s) must 250 km>h. A steady wind at 75 km>h blows eastward.
the boat be steered? (b) How long does it take to make the (Air speed is the speed relative to the air.) (a) What is the
round trip? (Assume that the boat’s speed is constant at plane’s ground speed (vpg)? (b) If the pilot wants to fly
all times, and neglect turnaround time.) due north, what should his heading be?
83. ● ● ● A shopper in a mall is on an escalator that is moving 84. ● ● ● An airplane is flying at 150 mi>h (its speed in still
downward at an angle of 41.8° below the horizontal at a air) in a direction such that with a wind of 60.0 mi>h
constant speed of 0.75 m>s. At the same time a little boy blowing from east to west, the airplane travels in a
drops a toy parachute from a floor above the escalator straight line southward. (a) What must be the plane’s
and it descends at a steady vertical speed of 0.50 m>s. heading (direction) for it to fly directly south? (b) If the
Determine the speed of the parachute toy as observed plane has to go 200 mi in the southward direction, how
from the moving escalator. long does it take?
85. A hockey puck slides along a horizontal ice surface at 88. A sailboat is traveling due north at 2.40 m>s on a calm
20.0 m>s, hits a flat vertical wall, and bounces off. Its ini- lake with no noticeable water currents. From the crow’s
tial velocity vector makes an angle of 35° with the wall nest at the top of its 10.0-m-high mast, one of the passen-
and it comes off at an angle of 25° moving at 10.0 m>s. gers drops her digital camera. (a) Make a sketch of the
Choose the + x-axis to be along the wall in the direction camera’s trajectory from the point of view of the passen-
of motion and the y-axis to be perpendicular (into) to the gers on the deck below and from the point of view of
wall. (a) Write each velocity in unit vector notation. passengers on a nearby boat at rest relative to the lake.
(b) Determine the change in velocity in unit vector nota- (b) Determine the camera’s initial velocity relative to the
tion. (c) Determine the magnitude and direction, relative ship and relative to the lake surface. (c) How long does
to the wall, of this velocity change. the camera take to hit the deck? (d) What is its total
travel distance as determined by the boat passengers?
86. A football is kicked off the flat ground at 25.0 m>s at an (e) Compare this to the magnitude of the net displace-
angle of 30° relative to the ground. (a) Determine the ment, as determined by the passengers on the nearby
total time it is in the air. (b) Find the angle of its velocity boat and comment on why they are different.
with respect to the ground after it has been in the air for
one-fourth of this time. (c) Repeat for one-half and three- 89. At a merging on-ramp of a busy Los Angeles freeway,
fourths of the total time. (d) For each of these times, car A is moving directly east on the freeway at a steady
determine its speed. Comment on the speed changes as speed of 35.0 m>s. Car B is merging onto the freeway
it follows its parabolic arc. Do they make sense from the on-ramp, which points 10° north of due east,
physically? moving at 30.0 m>s. (See 䉲 Fig. 3.39.) If the two cars col-
lide, it will be at the point marked x in the figure, which
87. A railroad flatbed car is set up for a physics demonstra- is 350 m down the road from the position of car A. Use
tion. It is set to roll horizontally on its straight rails at a the x–y coordinate system to signify E–W versus N–S
constant speed of 12.0 m>s. On it is rigged a small directions. (a) What is the velocity of car B relative to car
launcher capable of launching a small lead ball vertically A? (b) Show that they do not collide at point x. (c) Deter-
upward, relative to the bed of the car, at a speed of mine how far apart the cars are (and which car is ahead)
25.0 m>s. (a) Compare [by making two sketches] the when car B reaches point x.
description of the motion from the point of view of two
different observers: one riding on the car and one at rest
A
on the ground next to the tracks. (b) How long does it
take the ball to return to its launch location? (c) Compare
the ball’s velocity at the top of its motion from the view-
point of each of the two observers and explain any dif-
ferences. (d) What is the launch angle (relative to the B
ground) and speed of the ball according to the ground
observer? (e) How far down the rails has the car moved
when the ball lands back at the launcher? Compare this 䉱 F I G U R E 3 . 3 9 Los Angeles freeway
distance to how far the ball has moved relative to the car. See Exercise 89.
4 Force and Motion †
4.1 The concepts of force

and net force (104)
■ definition of force
■ net (or unbalanced) force
4.2 Inertia and Newton’s

first law of motion (105)
■ law of inertia
4.3 Newton’s second law

of motion (107)
■ force and acceleration
4.4 Newton’s third law

of motion (113) PHYSICS FACTS
■ action and reaction
✦ Isaac Newton was born on Christ-
mas Day, 1642, the same day that
Galileo died. (By our current Gre-
gorian calendar, Newton’s birth
Y ou don’t have to understand
any physics to know that what’s
needed to get the car in the chapter-
4.5 More on Newton’s laws: date is January 4, 1643. England
free-body diagrams and did not begin using the Gregorian
opening photo moving is a push or a
translational equilibrium (116) calendar until 1752.) pull. If the frustrated men (or the tow
■
a Fxn = 0, a Fyn = 0 ✦ Issac Newton
– demonstrated white light to be
truck that may soon be called) can
a mixture of colors and theo- apply enough force, the car will
rized that light was made up of
4.6 Friction (121) particles, which he called cor- move. But what’s keeping the car
■ static friction puscles, rather than waves. We
■ kinetic friction now have the dual nature of
stuck in the snow? A car’s engine can
light, with light both behaving generate plenty of force—so why
as a wave and made up of parti-
cles called photons. doesn’t the driver just put the car
– developed the fundamentals of
calculus. Gottfried Leibniz, a
into reverse and back out? For a car
German mathematician, inde- to move, another force is needed
pendently developed a similar
version of calculus. There was a besides that exerted by the
lifelong, bitter dispute between
engine—friction. Here, the problem
Newton and Leibniz over who
should receive the credit for is most likely that there is not
doing so first.
†
The mathematics needed in this chapter
– built the first reflecting tele-
enough friction between the tires
involves trigonometric functions. You may
want to review them in Appendix I. scope with a power of 40x. and the snowy surface.
104 4 FORCE AND MOTION
Chapters 2 and 3 covered how to analyze motion in terms of kinematics. Now

our attention turns to the study of dynamics—that is, what causes motion and
changes in motion. This leads to the concepts of force and inertia. The study of
force and motion occupied many early scientists. The English scientist Isaac
Newton (1642–1727; 䉳 Fig. 4.1) summarized the various relationships and princi-
ples of those early scientists into three statements, or laws, which not surprisingly
are known as Newton’s laws of motion. These laws sum up the concepts of dynam-
ics. In this chapter, you’ll learn what Newton had to say about force and motion.
4.1 The Concepts of Force and Net Force

䉱 F I G U R E 4 . 1 Isaac Newton ➥ What is meant by a force?

Newton (1642–1727), one of the ➥ What is meant by a net force?
greatest scientific minds of all time,
made fundamental contributions to Let’s first take a closer look at the meaning of force. It is easy to give examples of
mathematics, astronomy, and sev-
eral branches of physics, including
forces, but how would you generally define this concept? An operational defini-
optics and mechanics. He formu- tion of force is based on observed effects. That is, a force is recognized and
lated the laws of motion and univer- described in terms of what it does. From your own experience, you know that
sal gravitation (Section 7.5) and was forces can produce changes in motion. A force can set a stationary object into motion.
one of the inventors of calculus. It can also speed up or slow down a moving object and/or change the direction of
He did some of his most profound
work when he was in his
its motion. In other words, a force can produce a change in velocity (speed and/or
mid-twenties. direction)—that is, an acceleration. Therefore, an observed change in motion, includ-
ing motion starting from rest, is evidence of a force. This concept leads to a com-
mon definition of force:
A force is something that is capable of changing an object’s state of motion, that is,
changing its velocity or producing an acceleration.
The word capable is very significant here. It takes into account the fact that a force
may be acting on an object, but its capability to produce a change in motion may
be balanced, or canceled, by one or more other forces. The net effect is then zero.
Thus, a force may not necessarily produce a change in motion. However, it follows
that if a force acts alone, the object on which it acts will have a change in velocity or
an acceleration.
Since a force can produce an acceleration—a vector quantity—force must itself
be a vector quantity, with both magnitude and direction. When several forces act
on an object, the interest is often in their combined effect—the net force. The net
force, Fnet , is the vector sum gF i , or resultant, of all the forces acting on an object
B B
or system.* Consider the opposite forces illustrated in 䉴 Fig. 4.2a. The net force is
zero when forces of equal magnitude act in opposite directions (Fig. 4.2b, where
signs are used to indicate directions). Such forces are said to be balanced. A
nonzero net force is referred to as an unbalanced force (Fig. 4.2c). In this case, the
situation can be analyzed as though only one force equal to the net force were act-
ing. An unbalanced, or nonzero, net force always produces an acceleration. In
some instances, an applied unbalanced force may also deform an object, that is,
change its size and/or shape (as will be seen in Section 9.1). A deformation
involves a change in motion for some part of an object; hence, there is an
acceleration.
Forces are sometimes divided into two types or classes. The more familiar of these
classes is contact forces. Such forces arise because of physical contact between objects.
*In the notation g Fi the Greek letter sigma means the “sum of” the individual forces, as indicated
B
by the i subscript: g Fi = F1 + F2 + F3 + Á , that is, a vector sum. The i subscript is sometimes omitted
B B B B
as being understood, g F.
B
4.2 INERTIA AND NEWTON’S FIRST LAW OF MOTION 105
䉳 F I G U R E 4 . 2 Net force
(a) Opposite forces are applied to a
F1 F2
crate. (b) If the forces are of equal
magnitude, the vector resultant, or
the net force acting on the crate is
zero. The forces acting on the crate
are said to be balanced. (c) If the
(a) forces are unequal in magnitude,
the resultant is not zero. A nonzero
net force 1Fnet2, or unbalanced force,
then acts on the crate, producing an
F2 acceleration (for example, setting
F1 F2 F1 the crate in motion if it was initially
Fnet = F2 – F1 = 0 at rest).
x
(b) Zero net force (balanced forces)
F2
F1
a
Fnet = F2 – F1 ≠ 0
F1 F2 a
x
Fnet
(c) Nonzero net force (unbalanced forces)
For example, when you push on a door to open it or throw or kick a ball, you exert a
contact force on the door or ball.
The other class of forces is called action-at-a-distance forces. Examples of these
forces include the gravitational force, the electrical force between two charges, and
the magnetic force between two magnets. The Moon is attracted to the Earth and
maintained in orbit by a gravitational force, but there seems to be nothing physi-
cally transmitting that force. (In Chapter 30, the modern view of how such action-
at-a-distance forces are thought to be transmitted is given.)
Now, with a better understanding of the concept of force, let’s see how force
and motion are related through Newton’s laws.
DID YOU LEARN?

➥ A force is something that is capable of changing an object’s velocity, or producing
an acceleration.
➥ A net force is the vector sum of the forces acting on an object.
4.2 Iner tia and Newton’s First Law of Motion

➥ What is inertia?
➥ How is inertia related to mass?
➥ In the absence of an unbalanced force, what can be said about an object’s motion?
The groundwork for Newton’s first law of motion was laid by Galileo. In his
experimental investigations, Galileo dropped objects to observe motion under
the influence of gravity. (See the related Chapter 2 Insight 2.1, Galileo Galilei and
the Leaning Tower of Pisa.) However, the relatively large acceleration due to
gravity causes dropped objects to move quite fast and quite far in a short time.
From the kinematic equations in Section 2.4, it can be seen that 3.0 s after being
dropped, an object in free fall has a speed of about 29 m>s 164 mi>h2 and has
fallen a distance of 44 m (about 48 yd, or almost half the length of a football field).
䉴 F I G U R E 4 . 3 Galileo’s
experiment A ball rolls farther
along the upward incline as the
angle of incline is decreased. On a
smooth, horizontal surface, the ball
rolls a greater distance before com-
ing to rest. How far would the ball
travel on an ideal, perfectly smooth
surface? (The ball would slide,
rather than roll, in this case because Thus, experimental measurements of free-fall distance versus time were particu-
of the absence of friction.) larly difficult to make with the instrumentation available in Galileo’s time.
To slow things down so he could study motion, Galileo used balls rolling on
inclined planes. He allowed a ball to roll down one inclined plane and then up
another with a different degree of incline (䉱 Fig. 4.3). Galileo noted that the ball
rolled to approximately the same height in each case, but it rolled farther in the
horizontal direction when the angle of incline was smaller. When allowed to roll
onto a horizontal surface, the ball traveled a considerable distance and went even
farther when the surface was made smoother. Galileo wondered how far the ball
would travel if the horizontal surface could be made perfectly smooth (friction-
less). Although this situation is impossible to attain experimentally, Galileo rea-
soned that in this ideal case with an infinitely long surface, the ball would
continue to travel indefinitely with straight-line, uniform motion, since there
would be nothing (no net force) to cause its motion to change. (The ball would
actually slide, not roll, in this ideal case of the absence of friction.)
According to Aristotle’s theory of motion, which had been accepted for about
1900 years prior to Galileo’s time, the normal state of a body was to be at rest (with
the exception of celestial bodies, which were thought to be naturally in motion).
Aristotle no doubt observed that objects moving on a surface tend to slow down
and come to rest, so this conclusion would have seemed logical to him. However,
from his experiments, Galileo concluded that bodies in motion exhibit the behav-
ior of maintaining that motion, and if an object were initially at rest, it would
remain so unless something caused it to move.
Galileo called this tendency of an object to maintain its initial state of motion
inertia. That is,
Inertia is the natural tendency of an object to maintain a state of rest or to remain in
uniform motion in a straight line (constant velocity).
For example, if you’ve ever tried to stop a slowly rolling automobile by pushing
on it, you felt its resistance to a change in motion. Physicists describe the property
of inertia in terms of observed behavior. A comparative example of inertia is illus-
trated in 䉳 Fig. 4.4. If the two punching bags have the same density (mass per unit
䉱 F I G U R E 4 . 4 A difference in volume; see Section 1.4), the larger one has more mass and therefore more inertia,
inertia The larger punching bag has as you would quickly notice when punching each bag.
more mass and hence more inertia, Newton related the concept of inertia to mass. Originally, he called mass a
or resistance to a change in motion.
quantity of matter, but later redefined it as follows:
Mass is a quantitative measure of inertia.
That is, a massive object has more inertia, or more resistance to a change in
motion, than does a less massive object. For example, a car has more inertia than a
bicycle.
Newton’s first law of motion, sometimes called the law of inertia, summarizes
these observations:
B
In the absence of an unbalanced applied force (Fnet = 0), a body at rest remains at
rest, and a body in motion remains in motion with a constant velocity (constant speed
and direction).
That is, if the net force acting on an object is zero, then its acceleration is zero. It
B
may be moving with a constant velocity, or be at rest—in both cases, ¢v = 0 or
B
v = constant.
4.3 NEWTON’S SECOND LAW OF MOTION 107
DID YOU LEARN?

➥ Inertia is the natural tendency of an object to remain at rest or in motion with a
constant velocity.
➥ Mass is a measure of inertia—the greater the mass, the greater the inertia.
➥ With no unbalanced force acting, if an object is at rest, it will remain at rest; if in
motion, it has a constant velocity.
4.3 Newton’s Second Law of Motion

➥ In the expression F = ma, what do F and m represent?

➥ What is the difference between mass and weight?
➥ What is the SI unit of weight?
A change in motion, or an acceleration (that is, a change in velocity—speed

and/or direction), is evidence of a net force. All experiments indicate that the
acceleration of an object is directly proportional to, and in the direction of, the
applied net force; that is, in vector notation,
B
aB r Fnet
For example, suppose you separately cued two identical billiard balls. If you hit
the second ball twice as hard as the first (that is, you applied twice as much force),
you would expect the acceleration of the second ball to be twice as great as that of
the first ball (and still in the direction of the force).
However, as Newton recognized, the inertia or mass of the object also plays a
role. For a given net force, the more massive the object, the less its acceleration will
be. For example, if you hit two balls of different masses with the same force, the
less massive ball would experience a greater acceleration. Specifically, the accelera-
tion is inversely proportional to mass:
B
Fnet
aB r
m
or in words,
The acceleration of an object is directly proportional to the net force acting on it and
inversely proportional to its mass.The direction of the acceleration is in the direction
of the applied net force.
䉲 Figure 4.5 presents some illustrations of this principle.
䉲 F I G U R E 4 . 5 Newton’s second
law The relationships among force,
acceleration, and mass shown here
are expressed by Newton’s second
law of motion (assuming no friction
on the cart wheels, which would
slide).
a 2a a/2
F 2F F
m m m m
THIS SIDE UP THIS SIDE UP THIS SIDE UP THIS SIDE UP
(a) (b) (c)

A nonzero net force accelerates the crate: a F/m If the net force is doubled, If the mass is doubled,
the acceleration is doubled. the acceleration is halved.
B
Rewritten as Fnet r maB. Newton’s second law of motion is commonly
expressed in equation form as
B
Fnet = maB (Newton’s second law) (4.1)
SI unit of force: newton (N) or

kilogram-meter per second squared (kg # m>s2)
a
where Fnet = gFi . Equation 4.1 defines the SI unit of force, which is appropriately
B B
1.0 m/s2 called the newton (N).

Fnet By unit analysis, Eq. 4.1 shows that a newton in base units is defined as
m
1.0 kg 1 N = 1 kg # m>s 2. That is, a net force of 1 N gives a mass of 1 kg an acceleration of
1.0 N
1 m>s 2 (䉳 Fig. 4.6). The British system unit of force is the pound (lb). One pound is
equivalent to about 4.5 N (actually, 4.448 N). An average apple weighs about 1 N.
B
Fnet = ma Newton’s second law, Fnet = maB, allows the quantitative analysis of force and
1.0 N = (1.0 kg) (1.0 m/s2) motion. It might be thought of as a cause-and-effect relationship, with the force being
the cause and acceleration being the motional effect. Notice that if the net force acting
䉱 F I G U R E 4 . 6 The newton (N) A
net force of 1.0 N acting on a mass of on an object is zero, the object’s acceleration is zero, and it remains at rest or in uni-
1.0 kg produces an acceleration of form motion, which is consistent with Newton’s first law. For a nonzero net force (an
1.0 m>s2 (on a frictionless surface). unbalanced force), the resulting acceleration is in the same direction as the net force.*
MASS AND WEIGHT

Equation 4.1 can be used to relate mass and weight. Recall from Section 1.2 that
weight is the gravitational force of attraction that a celestial body exerts on an
object. For us, this is mainly the force of the gravitational attraction of the Earth. Its
effects are easily demonstrated: When you drop an object, it falls (accelerates)
toward the Earth. Since there is only one force is acting on the object (air resistance
B B
neglected), its weight (w ) is the net force Fnet, and the acceleration due to gravity
1g2 can be substituted for a in Eq. 4.1. Therefore in terms of magnitudes,
B B
w = mg (4.2)
1Fnet = ma2
Thus the weight of an object with 1.0 kg of mass is w = mg = 11.0 kg219.8 m>s22 =
9.8 N.
College College That is, 1.0 kg of mass has a weight of approximately 9.8 N, or 2.2 lb, near the
Physics Physics Earth’s surface. Although weight and mass are simply related through Eq. 4.2, keep
m 2m in mind that mass is the fundamental property. Mass doesn’t depend on the value of g,
but weight does. As pointed out previously, the acceleration due to gravity on the
Moon is about one-sixth that on the Earth. The weight of an object on the Moon
would thus be one-sixth of its weight on the Earth, but its mass, which reflects the
quantity of matter it contains and its inertia, would be the same in both places.
g F g
2F
Newton’s second law, along with the fact that w r m, explains why all objects in
free fall have the same acceleration (Section 2.5). Consider, for example, two falling
objects, one with twice the mass of the other. The object with twice as much mass
would have twice as much weight, or two times as much gravitational force acting
F =g 2F = g on it. But the more massive object also has twice the inertia, so twice as much force is
m 2m
needed to give it the same acceleration. Expressing this relationship mathemati-
䉱 F I G U R E 4 . 7 Newton’s second cally, for the smaller mass (m), the acceleration is a = Fnet>m = mg>m = g , and for
law and free fall In free fall, all the larger mass (2m), the acceleration is the same: a = Fnet>m = 2mg>12m2 = g
objects fall with the same constant (䉳 Fig. 4.7). Some other effects of g, which you may have experienced, are dis-
acceleration g. An object with twice cussed in Insight 4.1, g’s of Force and Effects on the Human Body.
the mass of another has twice as
much gravitational force acting on
it. But with twice the mass, the *It may appear that Newton’s first law of motion is a special case of Newton’s second law, but not so.
object also has twice the inertia, so The first law defines what is called an inertial reference frame (Section 26.1): a frame in which Newton’s
twice as much force is needed to first law holds. That is, in an inertial frame, an object on which there is no net force does not accelerate.
Since Newton’s first law holds in this frame, the second law of motion 1Fnet = ma2 also holds.
B B
give it the same acceleration.
INSIGHT 4.1 g’s of Force and Effects on the Human Body

The value of g at the Earth’s surface is referred to as the The maximum force on astronauts in a space shuttle on
standard acceleration and is used as a nonstandard unit. For blastoff is about 3g. But jet fighter pilots are subjected to as
example, when a spacecraft lifts off, astronauts are said to much as 9g when pulling out of a downward dive. These
experience an acceleration of “several g’s.” This expression pilots wear “g-suits,” which are designed to prevent blood
means that the astronauts’ acceleration is several times the pooling. The common g-suit is inflated by compressed air and
standard acceleration g. Since g = w>m, it can be seen that g is applies pressure to the pilot’s lower body to prevent blood
the (weight) force per unit mass. Thus, the term g’s of force is from accumulating there. Work is being done on the develop-
used to express force in terms of multiples of the standard ment of a hydrostatic g-suit that contains liquid, which is less
acceleration. restrictive than air. When the number of g’s increases, the liq-
To help understand this nonstandard unit of force, let’s uid, like the blood in the body, flows into the lower part of the
look at some examples. During the takeoff of a jet airliner, suit and applies pressure to the legs.
passengers experience an average horizontal force of about On the Earth, where only 1g is experienced, a partial “g-
0.20g. This means that as the plane accelerates down the run- suit” of sorts is used to prevent blood clots in patients who
way, the seat back exerts a horizontal force on you of about have undergone hip replacement surgery. Each year 400 to
one-fifth of your weight (to accelerate you along with the 800 people die in the first three months after such surgery, pri-
plane), but you experience a feeling of being pushed back into marily because blood clots form in a leg, break off into the
the seat. On takeoff at an angle of 30°, the force increases to bloodstream, and lodge in the lungs—giving rise to a condi-
about 0.70g. tion called pulmonary embolism. In other cases, a blood clot in
When a person is subjected to several g’s vertically, blood the leg may slow the flow of blood to the heart. These compli-
can begin to pool in the lower extremities, which may cause cations arise more often after hip replacement surgery than
blood vessels to distend or capillaries to rupture. Under such after almost any other surgery and occur after the patient has
conditions, the heart has a difficult time pumping blood left the hospital.
throughout the body. At about 4g, the pooling of blood in the Studies have shown that pneumatic (operated by air) com-
lower body deprives the head of sufficient oxygen. Lack of pression of the legs during the hospital stay reduces these
blood circulation to the eyes can cause temporary blindness, risks. A plastic leg cuff inflates every few minutes, forcing
and if the brain is deprived of oxygen, a person becomes dis- blood from the lower leg (Fig. 1). This mechanical massaging
oriented and quickly “blacks out” or loses consciousness. The prevents blood from pooling in the veins and clotting. By
average person can withstand several g’s for only a short using both this technique and anticlotting drug therapy,
period of time. many of the postoperative deaths can be prevented.
F I G U R E 1 Pneumatic massage
The leg cuffs inflate periodically,
forcing the blood from the lower
legs and preventing it from
pooling in the veins, particularly
after hip surgery.
B
Newton’s second law, Fnet = maB, allows us to analyze dynamic situations. In
using this law, keep in mind that Fnet is the magnitude of the net force and m is the
total mass of the system. The boundaries defining a system may be real or imaginary.
For example, a system might consist of all the gas molecules in a particular sealed
vessel. But you might also define a system to be all the gas molecules in an arbi-
trary cubic meter of air. In dynamics, there are often occasions to work with sys-
tems made up of two or more discrete masses—the Earth and Moon, for instance,
or a series of blocks on a tabletop, or a tractor and wagon, as in Example 4.1.
EXAMPLE 4.1 Newton’s Second Law: Finding Acceleration

A tractor pulls a loaded wagon on a
level road with a constant horizontal System
force of 440 N (䉴 Fig. 4.8). If the mass
of the wagon is 200 kg and that of the
m = 75 kg
load is 75 kg, what is the magnitude
of the wagon’s acceleration? (Ignore
frictional forces.) m = 200 kg
F = 440 N
THINKING IT THROUGH. This prob-
lem is a direct application of
Newton’s second law. The two sepa-
rate masses (wagon and contents) 䉱 F I G U R E 4 . 8 Force and acceleration See Example text for description.
make up the system.
SOLUTION. Listing the given data and what is to be found, tractor and the frictional force. Then the acceleration would
be (using directional signs)
Given: F = 440 N Find: a (acceleration)
Fnet F - f 440 N - 140 N
m1 = 200 kg (wagon) a = = = = 1.09 m>s2
m2 = 75 kg (load) m m1 + m2 275 kg
Again, the direction of a is in the direction of Fnet.
In this case, F is the net force, and the acceleration is given by With a constant net force, the acceleration is also constant,
Eq. 4.2, Fnet = ma, where m is the total mass. Solving for the so the kinematic equations of Section 2.4 can be applied. Sup-
magnitude of a, pose the wagon started from rest 1vo = 02. Could you find
Fnet Fnet 440 N how far it traveled in 4.00 s? Using the appropriate kinematic
a = = = = 1.60 m>s2 equation (Eq. 2.11, with xo = 0) for the case with friction,
m m1 + m2 200 kg + 75 kg
x = vo t + 12 at2 = 0 + 12 11.09 m>s2214.00 s22 = 8.72 m
and the direction of a is in the direction of Fnet or the direction
in which the tractor is pulling. Note that m is the total mass of F O L L O W - U P E X E R C I S E . Suppose the applied force on the
the wagon and its contents. In reality, there would be a total wagon is 550 N. With the same frictional force, what would be
opposing force of friction, - f. Suppose there were an effec- the wagon’s velocity 4.00 s after starting from rest? (Answers
tive frictional force of magnitude f = 140 N. In this case, the to all Follow-Up Exercises are given in Appendix VI at the back of
net force would be the vector sum of the force exerted by the the book.)
EXAMPLE 4.2 Newton’s Second Law: Finding Mass

A student weighs 588 N. What is her mass? In countries that use the metric system, the kilogram unit of
mass is used to express “weight” rather than a force unit. It
T H I N K I N G I T T H R O U G H . Newton’s second law allows us to
would be said that this student weighs 60.0 “kilos.”
determine an object’s mass if we know the object’s weight
Recall that 1 kg of mass has a weight of 2.2 lb on the Earth’s
(force), since g is known.
surface. Then in British units, she would weigh
Given: w = 588 N Find: m (mass) 60.0 kg 12.2 lb>kg2 = 132 lb.
SOLUTION. Recall that weight is a (gravitational) force and it F O L L O W - U P E X E R C I S E . (a) A person in Europe is a bit over-
is related to the mass of an object by w = mg (Eq. 4.2), where g weight and would like to lose 5.0 “kilos.” What would be the
is the acceleration due to gravity (9.80 m>s2). Rearranging the equivalent loss in pounds? (b) What is your “weight” in kilos?
equation,
w 588 N
m = = = 60.0 kg
g 9.80 m>s2
As has been learned, a dynamic system may consist of more than one object. In
applications of Newton’s second law, it is often advantageous, and sometimes
necessary, to isolate a given object within a system. This isolation is possible
because the motion of any part of a system is also described by Newton’s second law, as
Example 4.3 shows.
EXAMPLE 4.3 Newton’s Second Law: All or Part of the System?

Two blocks with masses a
m1 = 2.5 kg and m2 = 3.5 kg rest
on a frictionless surface and are T T F
connected by a light string m2 m1
(䉴 Fig. 4.9).* A horizontal force (F)
of 12.0 N is applied to m1, as
shown in the figure. (a) What is
the magnitude of the acceleration
of the masses (that is, of the total
system)? (b) What is the magni-
tude of the force (T) in the string?
[When a rope or string is stretched
taut, it is said to be under tension. a a
For a very light string, the force at
the right end of the string has the T F
same magnitude (T) as the force at T
the left end.]
Isolating the masses
䉱 F I G U R E 4 . 9 An accelerated system See Example text for description.
T H I N K I N G I T T H R O U G H . It is important to remember that (b) Under tension, a force is exerted on an object by a string is
Newton’s second law may be applied to a total system or any directed along the string. Note in the figure that it is assumed
part of it (a subsystem, so to speak). This capability allows for the tension is transmitted undiminished through the string. That
the analysis of a particular component of a system, if desired. is, the tension is the same everywhere in the string. Thus, the
Identification of all of the acting forces is critical, as this magnitude of T acting on m2 is the same as that acting on m1.
Example shows. Then Fnet = ma is applied to each subsystem This is actually true only if the string has zero mass. Only such
or component. idealized light (that is, of negligible mass) strings or ropes will
be considered in this book.
SOLUTION. Carefully listing the data and what is to be found: So there is a force of magnitude T on each of the masses,
Given: m1 = 2.5 kg Find: (a) a (acceleration) because of tension in the connecting string. To find the value
m2 = 3.5 kg (b) T (tension, a force) of T, a part of the system that is affected by this force is consid-
F = 12.0 N ered. Each block may be considered as a separate system to
which Newton’s second law applies. In these subsystems, the
Given an applied force, the acceleration of the masses can be tension comes into play explicitly. Note in the sketch of the
found from Newton’s second law. It is important to keep in isolated m2 in Fig. 4.9 that the only force acting to accelerate
mind that Newton’s second law applies to the total system or this mass is T. From the values of m2 and a, the magnitude of
to any part of it—that is, to the total mass 1m1 + m22 or indi- this force is given directly by
Fnet = T = m2 a = 13.5 kg212.0 m>s22 = 7.0 N
vidually to m1 or m2. However, you must be sure to correctly
identify the appropriate force or forces in each case. The net force
acting on the combined masses, for example, is not the same An isolated sketch of m1 is also shown in Fig. 4.9, and
as the magnitude of the net force acting on m2 considered sep- Newton’s second law can equally well be applied to this block to
arately, as will be seen. find T. The forces must be added vectorially to get the net force
on m1 that produces its acceleration. Recalling that vectors in one
(a) First, taking the system as a whole (that is, considering
dimension can be written with directional signs and magnitudes,
both m1 and m2), the net force acting on this system is F. Note
that in considering the total system, we are concerned only Fnet = F - T = m1 a (direction of F taken to be positive)
about the net external force acting on it. The internal equal Then, solving for T,
and opposite T forces are not a consideration in this case,
since they cancel. Then, using Newton’s second law: T = F - m1 a
Fnet F 12.0 N = 12.0 N - 12.5 kg212.0 m>s22 = 12.0 N - 5.0 N = 7.0 N

a = = = = 2.0 m>s2
m m1 + m2 2.5 kg + 3.5 kg
F O L L O W - U P E X E R C I S E . Suppose that an additional horizon-
The acceleration of both blocks is in the direction of the tal force to the left of 3.0 N is applied to m2 in Fig. 4.9. What
applied force, as indicated in the figure. would be the tension in the connecting string in this case?
*When an object is described as being “light,” its mass can be ignored in analyzing the situation given in the problem. That is, here the mass of
the string is negligible relative to the other masses.
THE SECOND LAW IN COMPONENT FORM

Not only does Newton’s second law hold for any part of a system, but it also
applies to each component of the acceleration. For example, a force may be
expressed in component notation in two dimensions as follows:
B
a F i = ma
B
and
a 1Fx xN + Fy yN 2 = m1a x xN + ay yN 2 = ma x xN + may yN (4.3a)
Hence, to satisfy both x and y directions independently,
a Fx = ma x and a Fy = ma y (4.3b)
and Newton’s second law applies separately to each component of motion. (Also,
©Fz = maz in three dimensions.) The components in the equations are scalar com-
ponents, and will be either positive or negative numbers depending on whether
along the positive or negative x or y axis. Example 4.4 illustrates how the second
law is applied using components.
EXAMPLE 4.4 Newton’s Second Law: Components of Force

A block of mass 0.50 kg travels with a speed of 2.0 m>s in the y
positive x-direction on a flat, frictionless surface. On passing y
through the origin, the block experiences a constant force of
3.0 N at an angle of 60° relative to the x-axis for 1.5 s (䉴 Fig. 4.10). F
F
What is the velocity of the block at the end of this time? 60° y
x
THINKING IT THROUGH. With the force at an angle to the ini- Fx
tial motion, it would appear that the solution is complicated. (Top view)
But note in the insert in Fig. 4.10 that the force can be resolved
into components. The motion can then be analyzed in each vo
component direction. x
SOLUTION. Listing the given data and what is to be found: 60°

B
Given: m = 0.50 kg Find: v (velocity at the
vxo = 2.0 m>s end of 1.5 s) F
vyo = 0 (Top view)
F = 3.0 N, u = 60° 䉱 F I G U R E 4 . 1 0 Off the straight and narrow A force is
t = 1.5 s applied to a moving block when it reaches the origin, and the
block then begins to deviate from its straight-line path. The
First let’s find the magnitudes of the force components:
vector components are shown in the box.
Fx = F cos 60° = 13.0 N210.5002 = 1.5 N
Fy = F sin 60° = 13.0 N210.8662 = 2.6 N
Then, applying Newton’s second law to each direction to find vx = vxo + a x t = 2.0 m>s + 13.0 m>s2211.5 s2 = 6.5 m>s
the components of acceleration,
vy = vyo + a y t = 0 + 15.2 m>s2211.5 s2 = 7.8 m>s
Fx 1.5 N
ax = = = 3.0 m>s2
m 0.50 kg And, at the end of the 1.5 s, the velocity of the block is
Fy 2.6 N v = vx xN + vy yN = 16.5 m>s2xN + 17.8 m>s2yN
B
ay = = = 5.2 m>s2
m 0.50 kg
F O L L O W - U P E X E R C I S E . (a) What is the direction of the veloc-
Next, from the kinematic equation relating velocity and accel- ity at the end of the 1.5 s? (b) If the force were applied at an
eration (Eq. 2.8), the magnitudes of the velocity components angle of 30° (rather than 60°) relative to the x-axis, how would
of the block are given by the results of this Example be different?
DID YOU LEARN?

➥ In F = ma, the symbol F is the net force (Fnet) and m is the total mass of the system.
➥ Mass (m) is an invariant fundamental property. Weight is related to mass (w = mg),
but can vary with variations in g.
➥ Since weight is a force (w = mg), it has units of kg # m>s2, which is taken to be a
newton (N).
4.4 NEWTON’S THIRD LAW OF MOTION 113
4.4 Newton’s Third Law of Motion

➥ Is it possible to have a single force?

➥ For a third law force pair, which force is the action and which is the reaction?
➥ What is a normal force?
Newton formulated a third law that is as far-reaching in its physical significance

as the first two laws. For a simple introduction to the third law, consider the forces ? N = mg
involved in seatbelt safety. When the brakes are suddenly applied when you are
riding in a moving car, because of your inertia you continue to move forward as
the car slows. (The frictional force on the seat of your pants is not enough to stop N
you.) In doing so, you exert forward forces on the seatbelt and shoulder strap. The
belt and strap exert corresponding backward reaction forces on you, causing you mg mg
to slow down with the car. If you hadn’t buckled up, you would keep going
ΣFy = N – mg = may = 0
(Newton’s first law) until another backward force, such as that applied by the so N = mg
dashboard or windshield, slowed you down. (In an abrupt collision stop, hope- (a)
fully the air bags would come into effect; see Chapter 6 Insight 6.1, The Automo- y
bile Air Bag and Martian Air Bags.) Fx
We commonly think of forces as occuring singly. However, Newton recognized x
that it is impossible to have a single force. He observed that in any force applica- N = mg + Fy
F Fy
tion, there is always a mutual interaction; therefore, forces occur in pairs. An θ
example given by Newton was the following: If you press on a stone with a finger, N
then the finger is also pressed by, or receives a force from, the stone.
Newton termed the paired forces action and reaction, and Newton’s third law of
motion is as follows:
mg
For every force (action), there is an equal and opposite force (reaction). x
ΣFy = N – mg – Fy = may = 0
In symbol notation, Newton’s third law may be expressed: (b)
B B
F 12 = - F 21 Fx
B B
That is, F 12 is the force exerted on object 1 by object 2, and -F 21 is the equal and Fy θ F
opposite force exerted on object 2 by object 1. (The minus sign indicates the oppo-
B
site direction.) Which force is considered the action or the reaction is arbitrary; F 21 may N = mg – Fy
B
be the reaction to F 12 or vice versa.
At a glance, Newton’s third law may seem to contradict Newton’s second law: N
If there are always equal and opposite forces, how can there be a nonzero net mg
force? An important thing to remember about the force pair of the third law is that
ΣFy = N – mg + Fy = may = 0
the action–reaction forces do not act on the same object. The second law is concerned
(c)
with a force (or forces) acting on a particular object (or system). The opposing
forces of the third law act on different objects. Hence, these forces cannot cancel y
x
each other nor have a vector sum of zero when the second law is applied to the
individual objects.
To illustrate this distinction, consider the situations shown in 䉴Fig. 4.11. We Fx N
often tend to forget reaction forces. For example, in the left portion of Fig. 4.11a, θ
the obvious force that acts on a block sitting on a table is the Earth’s gravita- mg Fy
tional attraction, which is expressed by the weight mg. But, there has to be another
force acting on the block. For the block not to accelerate, the table must exert an N = Fy = mg cos θ
B θ
upward force N of which the magnitude is equal to the block’s weight. Thus, ΣFy = N – Fy = may = 0
gFy = + N - mg = may = 0.
B B (d)
In reaction to N, the block exerts a downward force on the table, -N, whose
B
magnitude is the same as the block’s weight, mg. However, -N is not the object’s 䉱 F I G U R E 4 . 1 1 Distinctions
B
weight. Weight and -N have two different origins: Weight is the action-at-a- between Newton’s second and third
B
distance gravitational force, and -N is a contact force between the two surfaces. laws Newton’s second law deals
You can easily demonstrate that this upward force on the block is there by plac- with the forces acting on a particu-
lar object (or system). Newton’s
ing the block on your hand and holding it stationary—you exert an upward force third law deals with the force pair
B
on the block and you would feel a reaction force of - N on your hand. If you that acts on different objects. See
applied a greater force, that is, N 7 mg, then the block would accelerate upward. text for description.
The force that a surface exerts on an object is called a normal force and the sym-
bol N is used to denote the force. Normal means perpendicular. The normal force
that a surface exerts on an object is always perpendicular to the surface. In
Fig. 4.11a, the normal force is equal and opposite to the weight of the block (but
not a third law pair. Why?). However, the normal force is not always equal and
opposite to an object’s weight. The normal force is a “reaction” force; it reacts to
the situation. Examples of this given in Figs. 4.11b, c, and d, are described here
with the summation of the vertical components 1gFy2.
(Fig. 4.11b) Applied force at a downward angle.
a Fy : N - mg - Fy = may = 0, and N = mg + Fy 1N 7 mg2

(Fig. 4.11c) Applied force at an upward angle.
a Fy : N - mg + Fy = may = 0, and N = mg - Fy 1N 6 mg2

(Fig. 4.11d) Block on an inclined plane. (Normal force perpendicular to the sur-
face of the plane.)
a Fy : N - Fy = may = 0, and N = Fy = mg cos u

where mg cos u is the weight component perpendicular to the plane.
In the case of Fig. 4.11d, the weight component down the plane, Fx , would accel-
erate the block down the plane in the absence of an equal opposing frictional force
between the block and surface of the plane.
CONCEPTUAL EXAMPLE 4.5 Where Are the Newton’s Third Law Force Pairs?
A woman waiting to cross the street holds a
briefcase in her hand as shown in 䉴 Fig. 4.12a.
Identify all of the third law force pairs involv-
ing the briefcase in this situation. F1 on briefcase
REASONING AND ANSWER. The briefcase is Contact forces

being held motionless, so its acceleration is F1 on hand
zero, and g Fy = 0. Focusing only on the case,
two equal and opposite forces acting on it can
be identified—the downward weight of the F2 on briefcase
case and the upward applied force by the hand. a=g
However, these two forces cannot be a third law
force pair because they act on the same object.
On an overall inspection, you should realize Action-at-a-distance forces
that the reaction force to the upward force of
the hand on the briefcase is a downward force
on Earth
on the hand. Then how about the reaction force
F2
to the weight of the case? Since weight is the
attractive gravitational force on the case by the (a) (b)
Earth, the corresponding force on the Earth by 䉱 F I G U R E 4 . 1 2 Force pairs of Newton’s third law B(a) When B
a person holds
the case makes up the third law force pair. a briefcase, there are two force
B
pairs: Ba contact pair (F1 and - F1) and an action-
at-a-distance (gravity) pair (F2 and -F2). The net force acting on the briefcase is
F O L L O W - U P E X E R C I S E . The woman inadver-
zero: The upward contact force 1F1 on the briefcase2 balances the downward
B
tently drops her briefcase as illustrated in weight force. Note, however, that the upward contact force and downward
Fig. 4.12b. Are there any third law force pairs weight force are not a third law pair. (b) Any third law force pairs? See the
in this situation? Explain. Follow-Up Exercise.
Jet propulsion is yet another example of Newton’s third law in action. In the
case of a rocket, the rocket and exhaust gases exert equal and opposite forces on
each other. As a result, the exhaust gases are accelerated away from the rocket, and
the rocket is accelerated in the opposite direction. When a big rocket “blasts off,”
as in a Space Shuttle launch, it produces a fiery release of exhaust. A common mis-
conception is that the exhaust gases “push against” the launch pad to accelerate
4.4 NEWTON’S THIRD LAW OF MOTION 115
the rocket. If this interpretation were true, there would be no space travel, since
there is nothing to “push against” in space. The correct explanation is one of
action (the expanding gases exert a force on the rocket) and reaction (the rocket
exert a force on the gas).
Another action–reaction pair is given in Insight 4.2, Sailing into the Wind—
Tacking.
INSIGHT 4.2 Sailing into The Wind—Tacking

A sailboat can easily sail in the direction of the wind (which (Fig. 1b). The skipper continually repeats the maneuver, and
fills the sails). However, after sailing some distance in the using this zigzag course, the boat gets back to port (Fig. 2a).
windward direction, the skipper usually wants to return to What about the perpendicular force component? You might
home port—which involves somehow “sailing into the think that this would take the boat way off course. It would,
wind.” This may sound rather impossible, but it isn’t. The and does a little, but most of the perpendicular force is bal-
process is called tacking, and can be explained and under- anced by the force of water on the keel of the boat, which is
stood by using force vectors and Newton’s laws. underneath (Fig. 2b). The water resistance exerts an opposite
A sailboat cannot sail directly upwind, since the wind force force on the keel, which cancels out most of the sideways per-
on the sail would accelerate the boat backward, or opposite pendicular force, producing little, if any, acceleration in that
the desired direction. The wind filling the sail exerts a force Fs direction.
perpendicular to the sail (Fig. 1a). If the boat is steered at an
angle relative to the wind direction, there is a component of
force parallel to the boat’s heading 1F‘ 2. On this course, some
distance upwind is gained, but it would never get the boat
back to port. The perpendicular component 1F⬜2 acts side-
ways and would put the boat way off course.
So, being an old salt, the skipper “tacks” or maneuvers the
boat so that the parallel force component is changed by 90°
Fs (force perpendicular to sail)

F
Wind velocity
(a)
(a)
F
Wind velocity
F
F
(b)
(b)
F I G U R E 1 Let’s go tacking (a) The wind filling the
sail exerts a force perpendicular to the sail (Fs) We can F I G U R E 2 Into the wind (a) As the skipper turns the boat into
resolve this force vector into components. The one par- the wind, the tacking begins. (b) The perpendicular force com-
allel to the motion of the boat 1F‘ 2 has an upwind component in tacking would take the boat off course sideways. But
ponent. (b) By changing the direction of the sail, the the water resistance on the keel under the boat exerts an oppo-
skipper can “tack” the boat upwind. site force and cancels out most of the sideways force.
DEMONSTRATION 1 Tension in a String: Action and Reaction Forces
(a) Two suspended 2-kilogram masses are (b) No, think of it in this manner. The effect of (c) Similarly, the other end of
attached to opposite sides of a scale (calibrated the weight of the mass on the right is the scale can be fixed. (A
in newtons).The total suspended weight is replaced by fixing the end of the string. fixed pulley merely changes
w = mg = 14.00 kg219.80 m>s22 = 39.2 N, yet The other mass stretches the scale spring, the direction of the force,
the scale reads about 20 N. Is something wrong giving a reading of about 20 N [or w = mg = and the scale can be hung
with the scale? 12.00 kg219.80 m>s 22 = 19.6 N]. vertically with the same
effect.) In all cases, the ten-
sion in the string is 19.6 N, as
the scale shows.
DID YOU LEARN?

➥ Forces always occur in pairs, but act on different objects.
B B
➥ If F1 is the action, then -F2 is the reaction, and vice versa.
➥ A normal force is the force a surface exerts on an object and is always
perpendicular to the surface.
4.5 More on Newton’s Laws: Free-Body Diagrams

and Translational Equilibrium
➥ What is the difference between a space diagram and a free-body diagram?

➥ What is the condition for translational equilibrium, and does equilibrium mean that
an object is at rest?
Now that you have been introduced to Newton’s laws and some applications in
analyzing motion, the importance of these laws should be evident. They are so
simply stated, yet so far-reaching. The second law is probably the most often
applied, because of its mathematical relationship. However, the first and third
laws are often used in qualitative analysis, as our continuing study of the different
areas of physics will reveal.
In general, we will be concerned with applications that involve constant forces.
Constant forces result in constant accelerations and allow the use of the kinematic
equations from Section 2.4 in analyzing the motion. When there is a variable force,
Newton’s second law holds for the instantaneous force and acceleration, but the
acceleration will vary with time, requiring advanced mathematics to analyze. So
in general, our study will be limited to constant accelerations and forces. Several
examples of applications of Newton’s second law are presented in this section so
that you can become familiar with its use. This small but powerful equation will
be used again and again throughout the book.
There is still one more item in the problem-solving arsenal that is a great help
with force applications—free-body diagrams. These are explained in the follow-
ing Problem-Solving Strategy.
4.5 MORE ON NEWTON’S LAWS: FREE-BODY DIAGRAMS AND TRANSLATIONAL EQUILIBRIUM 117
PROBLEM-SOLVING STRATEGY: FREE-BODY DIAGRAMS
In illustrations of physical situations, sometimes called space diagrams, force vectors are
drawn at different locations to indicate their points of application. However, presently being
concerned with only linear motions, vectors in free-body diagrams (FBDs) may be shown as LEARN BY DRAWING 4.1
emanating from a common point, which is usually chosen as the origin of the x–y axes. One
of the axes is generally chosen along the direction of the net force acting on an object, since forces on an object on
that is the direction in which the object will accelerate. Also, it is often important to resolve
force vectors into components, and properly chosen x–y axes simplify this task.
an inclined plane and
In a free-body diagram, the vector arrows do not have to be drawn exactly to scale. free-body diagrams
However, the diagram should clearly show whether there is a net force and whether
forces balance each other in a particular direction. When the forces aren’t balanced, by
Newton’s second law, there must be an acceleration.
In summary, the general steps in constructing and using free-body diagrams are as
N T
follows. (Refer to the accompanying Learn by Drawing 4.1, Forces on an Object on an
Inclined Plane and Free-Body Diagrams as you read.)
1. Make a sketch, or space diagram, of the situation (if one is not already available) 1
and identify the forces acting on each body of the system. A space diagram is an g
illustration of the physical situation that identifies the force vectors. m2 g
2. Isolate the body for which the free-body diagram is to be constructed. Draw a set of
Cartesian axes, with the origin at a point through which the forces act and with one
of the axes along the direction of the body’s acceleration. (The acceleration will be in
the direction of the net force, if there is one.)
3. Draw properly oriented force vectors (including angles) on the diagram, emanating
from the origin of the axes. If there is an unbalanced force, assume a direction of
acceleration and indicate it with an acceleration vector. Be sure to include only those
forces that act on the isolated body of interest.
4. Resolve any forces that are not directed along the x- or y-axis into x- or y-components
(use plus and minus signs to indicate direction). Use the free-body diagram and force
components to analyze the situation in terms of Newton’s second law of motion. (Note: 2
If you assume that the acceleration is in one direction, and in the solution it comes out
with the opposite sign, then the acceleration is actually in the opposite direction from
that assumed. For example, if you assume that aB is in the + x-direction, but you get a
negative answer, then aB is in the -x-direction. )
a
N
Free-body diagrams are a particularly useful way of following one of the suggested
problem-solving procedures in Section 1.7: Draw a diagram as an aid in visualizing and T
analyzing the physical situation of the problem. Make it a practice to draw free-body dia-
grams for force problems, as done in the following Examples.
g
EXAMPLE 4.6 Up or Down? Motion on a Frictionless Inclined Plane 3
Two masses are connected by a light string running over a light pulley of negligible
friction, as illustrated in the Learn by Drawing (LBD) 4.1 space diagram. One mass
1m1 = 5.0 kg2 is on a frictionless 20° inclined plane, and the other 1m2 = 1.5 kg2 is freely N
a
suspended. What is the acceleration of the masses?
THINKING IT THROUGH. Apply the preceding Problem-Solving Strategy. T
SOLUTION. Following the usual procedure of listing the data and what is to be found:
B
Given: m1 = 5.0 kg Find: a (acceleration) 4
g
m2 = 1.5 kg
u = 20°
(To help visualize the forces involved, isolate m1 and m2 and draw free-body diagrams for
each mass.) For mass m1, there are three concurrent forces (forces acting through a common
point). These forces are T, its weight m1 g, and N, where T is the tension force of the string on
m1 and N is the normal force of the plane on the block. (See 3 in the LBD 4.1.) The forces are
shown as emanating from their common point of action. (Recall that a vector can be moved
as long as its direction and magnitude are not changed.)
Start by assuming that m1 accelerates up the plane, which is taken to be in the

+x-direction. (It makes no difference whether it is assumed that m1 accelerates up or
down the plane, as will be seen shortly.) Notice that m1 g (the weight) is broken down into
components. The x-component is opposite to the assumed direction of acceleration, and
the y-component acts perpendicularly to the plane and is balanced by the normal force N.
(There is no acceleration in the y-direction, so there is no net force in this direction.)
Then, applying Newton’s second law in component form (Eq. 4.3b) to m1 ,
a Fx1 = T - m1 g sin u = m1 a
a Fy1 = N - m1 g cos u = m1ay = 0 (ay = 0, no net force,
so the forces cancel)
And for m2,
a Fy2 = m2 g - T = m2 a y = m2 a
where the masses of the string and pulley have been neglected. Since they are connected
by a string, the accelerations of m1 and m2 have the same magnitudes, so ax = ay = a.
Then adding the first and last equations to eliminate T,
a m2 g - m1 g sin u = 1m1 + m22a
F 1net force = total mass * acceleration2

30° (Note that this is the equation that would be obtained by applying Newton’s second
N law to the system as a whole, because in the system of both blocks, the T forces are
w internal forces and cancel.)
Then, solving for a:
m2 g - m1 g sin 20°
a =
m1 + m2
11.5 kg219.8 m>s22 - 15.0 kg219.8 m>s2210.3422
=
y 5.0 kg + 1.5 kg
= - 0.32 m>s2
The minus sign indicates that the acceleration is opposite to the assumed direction.
N a That is, m1 actually accelerates down the plane, and m2 accelerates upward. As this
example shows, if you assume the acceleration to be in the wrong direction, the sign on
Fx
x the result will give the correct direction anyway.
30° Fy Could you find the tension force T in the string if asked to do so? How this task
F could be done should be quite evident from the free-body diagram.
w = mg FOLLOW-UP EXERCISE. (a) In this Example, what is the minimum amount of mass for m2
Free-body diagram that would cause m1 not to accelerate up or down the plane? (b) Keeping the masses the
same as in the Example, how should the angle of incline be adjusted so that m1 would not
䉱 F I G U R E 4 . 1 3 Finding force from accelerate up or down the plane?
motional effects See Example 4.7.
EXAMPLE 4.7 Components of Force and Free-Body Diagrams

A force of 10.0 N is applied at an angle of 30° to the horizontal (a) The acceleration of the block can be calculated using New-
on a 1.25-kg block initially at rest on a frictionless surface, as ton’s second law, and the axes are chosen so that a is in the
illustrated in 䉱 Fig. 4.13. (a) What is the magnitude of the block’s + x-direction. As the free-body diagram shows, only a compo-
acceleration? (b) What is the magnitude of the normal force? nent (Fx) of the applied force F acts in this direction. The com-
ponent of F in the direction of motion is Fx = F cos u.
T H I N K I N G I T T H R O U G H . The applied force may be resolved
Applying Newton’s second law in the x-direction to calculate
into components. The horizontal component accelerates the
the acceleration:
block. The vertical component affects the normal force.
(Review Fig. 4.11 if necessary.) And drawing a free-body dia- Fx = F cos 30° = max
gram for the block, as in Fig. 4.13, is helpful. and
SOLUTION. F cos 30° 110.0 N210.8662
ax = = = 6.93 m>s2
Given: F = 10.0 N Find: (a) a (acceleration) m 1.25 kg
m = 1.25 kg (b) N (normal force) (b) The acceleration found in part (a) is the acceleration of the
u = 30° block, since the block accelerates only in the x-direction. Since
vo = 0 a y = 0, the sum of the forces in the y-direction must be zero.
4.5 MORE ON NEWTON’S LAWS: FREE-BODY DIAGRAMS AND TRANSLATIONAL EQUILIBRIUM 119
That is, the downward component of F acting on the block, or

Fy , and its downward weight force, w, must be balanced by N - F sin 30° - mg = 0
the upward normal force N that the surface exerts on the
block. If this were not the case, then there would be a net force and
and an acceleration in the y-direction. N = F sin 30° + mg = 110.0 N210.5002 + 11.25 kg219.80 m>s22
Summing the forces in the y-direction with upward taken = 17.3 N
as positive,
The surface then exerts a force of 17.3 N upward on the block,
a Fy = N - Fy - w = 0 which balances the sum of the downward forces acting on it.
F O L L O W - U P E X E R C I S E . (a) Suppose the applied force on the block is applied for only a short time. What is the magnitude of the
normal force after the applied force is removed? (b) If the block slides off the edge of the table, what would be the net force on the
block just after it leaves the table (with the applied force removed)?
There is no single fixed way to go about solving a problem. However, some general
strategies or procedures are helpful in solving problems involving Newton’s second law.
When using the suggested problem-solving procedures introduced in Section 1.7, you
might include the following steps when solving problems involving force applications:
■ Draw a free-body diagram for each individual body, showing all of the forces acting
on that body.
F3
■ Depending on what is to be found, apply Newton’s second law either to the system
as a whole (in which case internal forces cancel) or to a part of the system. Basically,
you want to obtain an equation (or set of equations) containing the quantity for which you
f
want to solve. Review Example 4.3. (If there are two unknown quantities, application
of Newton’s second law to two parts of the system may give you two equations and
two unknowns. See Example 4.6.)
■ Keep in mind that Newton’s second law may be applied to components of accelera-
tion and that forces may have to be resolved into components to do this. Review
Example 4.7.
Fg
TRANSLATIONAL EQUILIBRIUM
Several forces may act on an object without producing an acceleration. In such a f
case, with aB = 0, from Newton’s second law,
B
a Fi = 0 (4.4)
F1
That is, the vector sum of the forces, or the net force, is zero, so the object either
remains at rest (as in 䉴 Fig. 4.14) or moves with a constant velocity (Newton’s first
law). In such cases, objects are said to be in translational equilibrium. When
remaining at rest, an object is said to be in static translational equilibrium. (a)
It follows that the sums of the rectangular components of the forces for an
object in translational equilibrium are also zero (why?):
a Fxn = 0 (translational equilibrium only) (4.5)
a FyN = 0
For three-dimensional problems, g Fzn = 0 also applies. However, our discus- F3
sion will be restricted to forces in two dimensions.
f
Equations 4.5 give what is often referred to as the condition for translational equi-
librium. Let’s apply this translational equilibrium condition to a case involving static Fg
equilibrium. F1
f
䉴 F I G U R E 4 . 1 4 Many forces, no acceleration (a) At least five
different external forces act on this physics professor. (Here, f is
the force of friction.) Nevertheless, she experiences no accelera-
tion. Why? (b) Adding the force vectors by the polygon method
reveals that the vector sum of the forces is zero. The professor is
in static translational equilibrium. (b)
EXAMPLE 4.8 Keep It Straight: In Static Equilibrium

Keeping a broken leg bone straight while it is healing some- the strings are ideal, determine the magnitude of the tension
times requires traction, which is the procedure in which the in the horizontal cord.
bone is held under stretching tension forces at both ends to
T H I N K I N G I T T H R O U G H . The pulley is in a static equilibrium
keep it aligned. Consider a leg under tractional tension as
and thus has no net force on it. If the forces are summed both
shown in 䉲 Fig. 4.15. The cord is attached to a suspended mass
vertically and horizontally, they independently should add to
of 5.0 kg and runs over a pulley. The attached cord above the
zero. This should allow the tension in the horizontal string to
pulley makes an angle of u = 40° with the vertical. Neglect-
be found.
ing the mass of the lower leg and the pulley and assuming all
T1 䉳 F I G U R E 4 . 1 5 Static transla-
tional equilibrium See Example text
u u for description.
T2
T
m T1
mg
(pulley) (weight)
SOLUTION. and summing the horizontal forces:

Given: Listing the data: Find: T (tension) in the a Fxn = + T2 sin u - T = 0
m = 5.0 kg horizontal cord
u = 40° Solving the latter equation for T, and substituting T2 from the
first:
Draw free-body diagrams for the pulley and suspended
mass (shown in Fig. 4.15). It should be clear that the horizon- T1
tal string must exert a force to the left on the pulley as shown. T = T2 sin u = sin u = mg tan u
cos u
Summing the vertical forces on m, it can be seen that
T1 = mg. Then, summing the vertical forces on the pulley, where T1 = mg. Putting in the numbers,
a Fy = + T2 cos u - T1 = 0 T = mg tan u = 15.0 kg219.8 m>s22 tan 40° = 41 N
F O L L O W - U P E X E R C I S E . Suppose the attending physician requires a tractional force on the bottom of the foot of 55 N. If the sus-
pended mass was kept the same, would you increase or decrease the angle of the upper string? Prove your answer by calculat-
ing the required angle.
EXAMPLE 4.9 On Your Toes: In Static Equilibrium

An 80-kg person stands on one foot with the heel
elevated (䉴 Fig. 4.16a). This gives rise to a tibia Gastrocnemius muscle
force F1 and an Achilles tendon “pull” force F2 as
illustrated in Fig. 4.16b. Typical angles are
u1 = 15° and u2 = 21°, respectively. (a) Find gen- Tibia
F1
eral equations for F1 and F2, and show that u2 must u1 F 2 u2
be greater than u1 to prevent damage to the Achilles tendon
Achilles tendon. (b) Compare the force applied by
the Achilles tendon with the weight of the person.
T H I N K I N G I T T H R O U G H . This is a case of static
translational equilibrium, so the x- and y-compo-
nents can be summed to get equations for F1 and F2.
Bone
mf g
䉴 F I G U R E 4 . 1 6 On your toes (a) A per- N

son stands on one foot with the heel ele-
vated. (b) The foot forces involved for this
position (not to scale). (a) (b)
4.6 FRICTION 121
SOLUTION. Listing what is given and what is to be found,

Given: m = 80 kg Find: (a) general equations for F1 and F2
F1 = tibia force (b) comparison of tendon force F2
F2 = tendon “pull” and the person’s weight
u1 = 15°, u2 = 21°
(The mass of the foot mf is not given.)
(a) It is assumed that the person of mass m is at rest, standing finite force, we must have tan u2 7 tan u1 or u2 7 u1, and
on one foot. Then, summing the force components on the foot 21° 7 15°, so Nature obviously knows her physics.
(Fig. 4.16b), Then, substituting F2 into Eq. 1 to find F1,
1m - mf2 g
F1 = F2 a b = a b
sin u2 sin u2
a Fxn = + F1 sin u1 - F2 sin u2 = 0
J cos u2 a b - 1K
sin u1 tan u2 sin u1
a Fyn = + N - F1 cos u1 + F2 cos u2 - mf g = 0 tan u1
where mf is the mass of the foot. From the Fx equation, 1m - mf2g tan u2 tan u2 1m - mf2g
= =
a
tan u2 cos u1 tan u2 - sin u1
F1 = F2 a b
sin u2 - 1b sin u1
(1) tan u1
sin u1
(Check the trig manipulation on this last step.)
Substituting into the Fy equation,
(b) The person’s weight is w = mg, where m is the mass of the
person’s body. This is to be compared with F2. Then, with
N - F2 a b cos u1 + F2 cos u2 - mf g = 0
sin u2
sin u1 m W mf (total body mass much greater than mass of the
foot), to a good approximation, mf may be assumed negligible
Solving for F2 with N = mg yields, compared to m, that is, w - mf g = mg - mf g L w. So for F2,
mg - mfg w - mf g w
N - mf g F2 = L = 2.5w
F2 = (2)
cos u2 a cos 21° a - 1b
= tan u2 tan 21°
a b - cos u2 cos u2 a
sin u2 tan u2 - 1b
- 1b tan u1 tan 15°
tan u1 tan u1
The Achilles tendon force is thus approximately 2.5 times the
Examining the F2 in Eq. 2, we see that if u2 = u1 , or person’s weight. No wonder folks stretch or tear this tendon,
tan u2 = tan u1 , then F2 is very large. (Why?) So to have a even without jumping.
F O L L O W - U P E X E R C I S E . (a) Compare the tibia force with the weight of the person. (b) Suppose the person jumped upward from
the one-foot toe position (as in taking a running jump shot in basketball). How would this jump affect F1 and F2?
DID YOU LEARN?

➥ A space diagram shows force vectors drawn at the points of application; a free-
body diagram shows the force vectors emanating from a common point, usually
the origin of the x–y axes.
➥ The condition for translational equilibrium is gFi = 0.This does not mean the
B
object is at rest. By Newton’s first law, it could be moving with a constant velocity.
4.6 Friction
➥ What is the difference between static friction and kinetic friction?

➥ What is a coefficient of friction?
➥ Why does air resistance depend on the size and shape of an object?
Friction refers to the ever-present resistance to motion that occurs whenever two
materials, or media, are in contact with each other. This resistance occurs for all types
B
of media—solids, liquids, and gases—and is characterized as the force of friction (f ).
For simplicity, up to now various kinds of friction (including air resistance) have
been generally ignored in examples and exercises. Now knowing how to describe
motion, you are ready to consider situations that are more realistic, in that the
effects of friction are included.
In some situations, an increase in friction is desired—for example, when
putting sand on an icy road or sidewalk to improve traction. This might seem con-
tradictory, since an increase in friction presumably would increase the resistance
to motion. We commonly say that friction opposes motion, and think that the force
of friction is in the opposite direction of motion. However, consider the forces
involved in walking, as illustrated in 䉳 Fig. 4.17. The force of friction does resist
motion (that of the foot), but is in the direction of the (walking) motion. Without
friction, the foot would slip backward, as when walking on a slippery surface.
As another example, consider a worker standing in the center of the bed of a
flatbed truck that is accelerating in the forward direction. If there were no friction
between the worker’s shoes and the truck bed, the truck would slide out from
under him. Obviously, there is a frictional force between the shoes and the bed,
and it is in the forward direction. This is necessary for the worker to accelerate
with the truck.
So there are situations where friction is desired (䉲 Fig. 4.18a), and situations
where reduced friction is needed (Fig. 4.18b). Another situation where reduced
F f friction is promoted is in the lubrication of moving machine parts. This allows the
F f parts to move more freely, thereby lessening wear and reducing the expenditure of
Force exerted Frictional force energy. Automobiles would not run without friction-reducing oils and greases.
on ground exerted on
by foot foot by ground This section is concerned chiefly with friction between solid surfaces. All sur-
faces are microscopically rough, no matter how smooth they appear or feel. It was
originally thought that friction was due primarily to the mechanical interlocking
䉱 F I G U R E 4 . 1 7 Friction andB of surface irregularities, or asperities (high spots). However, research has shown
walking The force of friction, f , is
shown in the direction of the walk- that friction between the contacting surfaces of ordinary solids (metals in particu-
ing motion. The force of friction pre- lar) is due mostly to local adhesion. When surfaces are pressed together, local
vents the foot from slipping welding or bonding occurs in a few small patches where the largest asperities
backward while the other foot is make contact. To overcome this local adhesion, a force great enough to pull apart
brought forward. B
If you walk on a the bonded regions must be applied.
deep-pile rug, F is evident in that
the pile will be bent backward. Friction between solids is generally classified into three types: static, kinetic
(sliding), and rolling. Static friction includes all cases in which the frictional force
is sufficient to prevent relative motion between surfaces. Suppose you want to
move a large desk. You push on it, but the desk doesn’t move. The force of static
friction between the desk’s legs and the floor opposes and equals the horizontal
force you are applying, so there is no motion—a static condition.
Kinetic friction (or sliding) friction, occurs when there is relative (sliding)
motion at the interface of the surfaces in contact. When pushing on the desk, you
can eventually get it sliding, but there is still a great deal of resistance between the
desk’s legs and the floor—kinetic friction.
䉴 F I G U R E 4 . 1 8 Increasing and
decreasing friction (a) To get a fast
start, drag racers need to make sure
that their wheels don’t slip when
the starting light goes on. Just
before the start of the race, they
floor the accelerator to maximize
the friction between their tires and
the track by “burning in” the tires.
This “burn in” is done by spinning
the wheels with the brakes on until
the tires are extremely hot. The rub-
ber becomes so sticky that it almost
welds itself to the surface of the
road. (b) Water serves as a good
lubricant to reduce friction in rides
such as this one. (a) (b)
4.6 FRICTION 123
Rolling friction occurs when one surface rotates as it moves over another surface, but
does not slip or slide at the point or area of contact. Rolling friction, such as that occur-
ring between a train wheel and a rail, is attributed to small, local deformations in the
contact region. This type of friction is difficult to analyze and will not be considered.
FRICTIONAL FORCES AND COEFFICIENTS OF FRICTION

Now let’s look at the forces of friction on stationary and sliding objects. These
forces are called the force of static friction and the force of kinetic (sliding) friction,
respectively. Experimentally, it has been found that the force of friction 1f 2 depends
B
on both the nature of the two surfaces, and, to a good approximation, the normal
force 1N2 that a surface exerts on an object, that is, f r N. For an object on a horizontal
B B B
surface, and with no other vertical forces, this force is equal in magnitude to the
object’s weight. (Why?) However, as was shown in the previous Learn By Drawing
4.1, on an inclined plane the normal force is in response to only a component of the
weight force. 䉲 F I G U R E 4 . 1 9 Force of friction
B
The force of static friction fs between surfaces in contact acts in the direction versus applied force (a) In the static
that opposes the initiation of relative motion between the surfaces. The magnitude region of the graph, as the applied
force F increases, so does fs ; that is,
takes on a range of values given by fs = F and fs 6 ms N (b) When the
applied force F exceeds fsmax = ms N
fs … ms N (static conditions) (4.6) the heavy file cabinet is set into
motion. (c) Once the cabinet is mov-
where ms is a constant of proportionality called the coefficient of static friction. ing, the frictional force decreases,
(“ms” is the Greek letter mu. Note that it is dimensionless. How do you know this since kinetic friction is less than
from the equation?) static friction 1fk 6 fsmax2 Thus, if
The less-than-or-equal-to sign (…) indicates that the force of static friction may the applied force is maintained,
have different values from zero up to some maximum value. To understand this there is a net force, and the cabinet
is accelerated. For the cabinet to
concept, look at 䉲 Fig. 4.19. In Fig. 4.19a, one person pushes on a file cabinet, but it move with constant velocity, the
doesn’t move. With no acceleration, the net force on the cabinet is zero, and applied force must be reduced to
F - fs = 0, or F = fs . Suppose that a second person also pushes, and the file cabi- equal the kinetic friction force:
net still doesn’t budge. Then fs must now be larger, since the applied force has fk = mk N.
Fnet = F – fk
dkaj dkaj dkaj
F F F
dkaj dkaj dkaj
a
dkaj dkaj dkaj
dkaj dkaj dkaj
fs < µ sN fsmax = µ sN fk = µ k N
(a) (b) (c)
f
fsmax (a) (b)
(c) fk = µkN
F = fs
Kinetic friction
F
Static friction
Applied force = static frictional force F = fsmax
been increased. Finally, if the applied force is made large enough to overcome the
static friction, motion occurs (Fig. 4.19c). The greatest, or maximum, force of static
friction is exerted just before the cabinet starts to slide (Fig. 4.19b), and for this
case, Eq. 4.7 gives the maximum value of static friction:
fsmax = ms N (maximum value of static friction) (4.7)
Once an object is sliding, the force of friction changes to kinetic friction 1fk2
B
This force acts in the direction opposite to the direction of the object’s motion and
has a magnitude of
fk = mk N (sliding conditions) (4.8)
where mk is the coefficient of kinetic friction (sometimes called the coefficient of

sliding friction). Note that Eqs. 4.7 and 4.8 are not vector equations, since f and N are
in different directions. Generally, the coefficient of kinetic friction is less than the
coefficient of static friction 1mk 6 ms2, which means that the force of kinetic fric-
tion is less than fsmax . The coefficients of friction between some common materials
are listed in 䉲 Table 4.1.
Note that the force of static friction ( fs) exists in response to an applied force.
The magnitude of fs and its direction depend on the magnitude and direction of
the applied force. Up to its maximum value, the force of static friction is equal in
magnitude and opposite in direction to the applied force (F), since there is no
acceleration 1F - fs = ma = 02. Thus, if the person in Fig. 4.19a were to push on
the cabinet in the opposite direction, fs would also change direction to oppose the
new push. If there were no applied force F, then fs would be zero. When the mag-
nitude of F exceeds that of fsmax , the cabinet begins moving (accelerates), and
kinetic friction comes into play, with fk = mk N. If the magnitude of F is reduced to
that of fk , the cabinet will slide with a constant velocity; if the magnitude of F is
maintained greater than that of fk , the cabinet will continue to accelerate.
It has been experimentally determined that the coefficients of friction (and
therefore the forces of friction) are nearly independent of the contact area between
TABLE 4.1 Approximate Values for Coefficients of Static and Kinetic

Friction between Certain Surfaces
Friction between Materials Ms Mk
Aluminum on aluminum 1.90 1.40

Glass on glass 0.94 0.35
Rubber on concrete
dry 1.20 0.85
wet 0.80 0.60
Steel on aluminum 0.61 0.47
Steel on steel
dry 0.75 0.48
lubricated 0.12 0.07
Teflon on steel 0.04 0.04
Teflon on Teflon 0.04 0.04
Waxed wood on snow 0.05 0.03
Wood on wood 0.58 0.40
Lubricated ball bearings 60.01 60.01
Synovial joints (at the ends of most long bones—for 0.01 0.01
example, elbows and hips)
4.6 FRICTION 125
metal surfaces. This means that the force of friction between a brick-shaped metal
block and a metal surface is the same regardless of whether the block is lying on a
larger side or a smaller side.
Finally, keep in mind that although the equation f = mN holds in general for
frictional forces, it may not remain linear. That is, m is not always constant. For
example, the coefficient of kinetic friction varies somewhat with the relative speed
of the surfaces. However, for speeds up to several meters per second, the coeffi-
cients are relatively constant. For simplicity, our discussion will neglect any varia-
tions due to speed (or area), and the forces of static and kinetic friction will be
assumed to depend only on the load (N) and the nature of the two surfaces as
expressed by the given coefficients of friction.
EXAMPLE 4.10 Pulling a Crate: Static and Kinetic Forces of Friction

(a) In 䉲Fig. 4.20, if the coefficient of static friction between the the worker needs to apply. The weight of the crate and the
40.0-kg crate and the floor is 0.650, what is the magnitude of normal force are equal in magnitude in this case (see the free-
the minimum horizontal force the worker must pull to get the body diagram in Fig. 4.20), so the magnitude of the maximum
crate moving? (b) If the worker maintains that force once the force of static friction is
crate starts to move and the coefficient of kinetic friction fsmax = msN = ms1mg2
between the surfaces is 0.500, what is the magnitude of the
acceleration of the crate? = 10.6502140.0 kg219.80 m>s22 = 255 N
So the crate will begin to move when the applied force F
THINKING IT THROUGH. This situation involves applications
exceeds 255 N.
of the forces of friction. In (a), the maximum force of static fric-
tion must be calculated. In (b), if the worker maintains an (b) Now with the crate in motion, the kinetic friction fk acts on
applied force of this magnitude after the crate is in motion, the crate. However, this force is smaller than the applied force
there will be an acceleration, since fk 6 fsmax F = fsmax = 255 N, because mk 6 ms . Hence, there is a net
force on the crate and the acceleration of the crate can be
SOLUTION. As usual, listing the given data and what is to be found by using Newton’s second law in the x-direction:
found:
a Fx = + F - fk = F - mk N = max
Given: m = 40.0 kg Find: (a) F (minimum force neces-
ms = 0.650 sary to move crate) Solving for ax ,
mk = 0.500 (b) a (acceleration) F - mk N F - mk1mg2
ax = =
(a) The crate will not move until the magnitude of the m m
applied force F slightly exceeds that of the maximum static 255 N - 10.5002140.0 kg219.80 m>s22
frictional force fsmax . So fsmax must be found to see what force = = 1.48 m>s2
40.0 kg
䉴 FIGURE 4.20
Forces of static and
kinetic friction See
Example text for
description.
fs F
F
w = mg N
g
Free-body diagram
fs
FOLLOW-UP EXERCISE. On the average, by what factor does ms exceed mk for nonlubricated, metal-on-metal surfaces? (See Table 4.1.)
Let’s look at another worker with the same crate, but this time with the worker
applying the force at an angle (䉲 Fig. 4.21).
N
F
F sin 30°
F 30°
x
30° fs F cos 30°
N w = mg
mg
fs Free-body diagram
䉱 F I G U R E 4 . 2 1 Pulling at an angle: a closer look at the normal force See Example 4.11.
EXAMPLE 4.11 Pulling at an Angle: A Closer Look at the Normal Force

A worker pulling a crate applies a force at an angle of 30° to nent of the applied force. (See the free-body diagram in
the horizontal, as shown in Fig. 4.21. What is the magnitude Fig. 4.21.) Then by Newton’s second law, since ay = 0.
of the minimum force he must apply to move the crate?
(Before looking at the solution, would you expect that the a Fy = + N + F sin 30° - mg = 0
force needed in this case would be greater or less than that in and
Example 4.10?)
N = mg - F sin 30°
THINKING IT THROUGH. Since the applied force is at an angle
In effect, the applied force here partially supports the weight
to the horizontal surface, the vertical component will affect
of the crate. Substituting this expression for N into the first
the normal force. (See Fig. 4.11.) This change in the normal
equation gives
force will, in turn, affect the maximum force of static friction.
F cos 30° = ms1mg - F sin 30°2
SOLUTION. The data are the same as in Example 4.10, except
that the force is applied at an angle. Solving for F,
Given: u = 30° Find: F (minimum force necessary
mg
1cos 30°>ms2 + sin 30°
to move the crate) F =
In this case, the crate will begin to move when the horizontal
component of the applied force, F cos 30°, slightly exceeds the 140.0 kg219.80 m>s 22
10.866>0.6502 + 0.500
maximum static friction force. So for the maximum friction: = = 214 N
F cos 30° = fsmax = ms N

Thus, less applied force is needed in this case, reflecting the
However, the magnitude of the normal force is not equal to fact that the frictional force is less, because of the reduced nor-
the weight of the crate here, because of the upward compo- mal force.
F O L L O W - U P E X E R C I S E . Note that in this Example, applying the force at an angle produces two effects. As the angle between the
applied force and the horizontal increases, the horizontal component of the applied force is reduced. However, the normal force
also gets smaller, resulting in a lower fsmax . Does one effect always outweigh the other? That is, does the applied force F neces-
sary to move the crate always decrease with increasing angle? [Hint: Investigate F for different angles. For example, compute F
for 20° and 50°. You already have a value for 30°. What do the results tell you?]
EXAMPLE 4.12 No Slip, No Slide: Static Friction

A crate sits in the middle of the bed on a flatbed truck that is trav- There is a net force in the -x-direction, and hence there should
eling at 80 km>h on a straight, level road. The coefficient of static be an acceleration in that direction. What does this mean? It
friction between the crate and the truck bed is 0.40. When the means that relative to the ground, the crate is decelerating at the
truck comes uniformly to a stop, the crate does not slide, but same rate as the truck, which is necessary for the crate not to
remains stationary on the truck. What is the minimum stopping slide—the crate and the truck slow down uniformly together.
distance for the truck so the crate does not slide on the truck bed? The force creating this acceleration for the crate is the static
force of friction. The acceleration is found using Newton’s
T H I N K I N G I T T H R O U G H . There are three forces on the crate,
second law, and then is used in one of the kinematic equa-
as shown in the free-body diagram in 䉴 Fig. 4.22 (assuming
tions to find the distance.
that the truck is initially traveling in the +x-direction). But wait.
4.6 FRICTION 127
SOLUTION. y
Given: vxo = 80 km>h = 22 m>s Find: x (minimum stopping distance)
ms = 0.40
Applying Newton’s second law to the crate using the maximum fs to find the minimum
stopping distance,
N
a Fx = - fsmax = - ms N = - ms mg = ma x
Solving for ax, fs
ax = - ms g = - 10.40219.8 m>s 2 = - 3.9 m>s
2 2 x
which is the maximum deceleration of the truck so the crate does not slide. w = mg
Hence, the minimum stopping distance (x) for the truck is based on this acceleration
and given by Eq. 2.12, where vx = 0 and xo is taken to be zero. So,
v2x = 0 = v2xo + 21a x2x
Solving for x,
v 2xo 122 m>s22 䉱 F I G U R E 4 . 2 2 Free-body diagram
See Example text for description.
-2 A -3.9 m>s2 B
x = = = 62 m
- 2a x
Is the answer reasonable? This distance is about two-thirds the length of a football field.
F O L L O W - U P E X E R C I S E . Draw a free-body diagram and describe what happens in terms of accelerations and coefficients of friction
if the crate starts to slide forward on the truck bed when the truck is braking to a stop (in other words, if ax exceeds - 3.9 m>s2).
AIR RESISTANCE
Air resistance refers to the resistance force acting on an object as it moves through
air. In other words, air resistance is a type of frictional force. In analyses of falling
objects, you can usually ignore the effect of air resistance and still get good
approximations for those falling relatively short distances. However, for longer
distances, air resistance cannot be ignored.
Air resistance occurs when a moving object collides with air molecules. Therefore,
air resistance depends on the object’s shape and size (which determine the area of the
object that is exposed to collisions) as well as its speed. The larger the object and the
faster it moves, the more collisions there will be with air molecules. (Air density is
also a factor, but this quantity can be assumed to be constant near the Earth’s sur-
face.) To reduce air resistance (and fuel consumption), automobiles are made more
“streamlined,” and airfoils are used on trucks and campers (䉴 Fig. 4.23).
Consider a falling object. Since air resistance depends on speed, as a falling
object accelerates under the influence of gravity, the retarding force of air resis- 䉱 F I G U R E 4 . 2 3 Airfoil The air-
tance increases (䉲 Fig. 4.24a). Air resistance for human-sized objects as a general foil at the top of the truck’s cab
rule is proportional to the square of the speed, v2, so the resistance builds up rather makes the truck more streamlined
and therefore reduces air resistance.
䉳 F I G U R E 4 . 2 4 Air resistance and terminal velocity (a) As the speed

f1
of a falling object increases, so does the frictional force of air resistance.
(b) When this force of friction equals the weight of the object, the net
v1 force is zero, and the object falls with a constant (terminal) velocity.
mg (c) A plot of speed versus time, showing these relationships.
Speed
f
f2
vt
v2 vt
mg mg
Time
(a) As v increases, so does f. (b) When f = mg, the object (c)
falls with a constant
(terminal) velocity.
䉴 F I G U R E 4 . 2 5 Terminal velocity
Skydivers assume a spread-eagle
position to maximize air resistance.
This causes them to reach terminal
velocity more quickly and prolongs
the time of fall. Shown here is a for-
mation of sky divers viewed from
below.
rapidly. Thus when the speed doubles, the air resistance increases by a factor of 4.
Eventually, the magnitude of the retarding force equals that of the object’s weight
force (Fig. 4.24b), so the net force on it is zero. The object then falls with a maximum
constant velocity, which is called the terminal velocity, with magnitude vt.
This can be easily seen from Newton’s second law. For the falling object,
Fnet = ma
or
mg - f = ma
where f is the air resistance (friction) and downward has been taken as positive for
convenience. Solving for a,
f
a = g -
m
where a is the magnitude of the instantaneous downward acceleration.
Notice that the acceleration for a falling object when air resistance is included
is less than g; that is, a 6 g. As the object continues to fall, its speed increases, and
the force of air resistance, f, increases (since it is speed dependent) until a = 0 when
f = mg and f - mg = 0. The object then falls at its constant terminal velocity.
For a skydiver with an unopened parachute, terminal velocity is about
200 km>h (about 125 mi>h). To reduce the terminal velocity so that it can
be reached sooner and the time of fall extended, a skydiver will try to increase
exposed body area to a maximum by assuming a spread-eagle position (䉱Fig. 4.25).
This position takes advantage of the dependence of air resistance on the size and
shape of the falling object. Once the parachute is open (giving a larger exposed area
and a shape that catches the air), the additional air resistance slows the diver down
to about 40 km>h (25 mi>h), which is more preferable for landing.
CONCEPTUAL EXAMPLE 4.13 Race You Down: Air Resistance and Terminal Velocity
From a high altitude, a balloonist simultaneously drops two establish the reasoning and physical principle(s) used in determin-
balls of identical size, but appreciably different in mass. ing your answer before checking it next. That is, why did you
Assuming that both balls reach terminal velocity during the select your answer?
fall, which of the following is true? (a) The heavier ball
reaches terminal velocity first; (b) the balls reach terminal REASONING AND ANSWER. Terminal velocity is reached when
velocity at the same time; (c) the heavier ball hits the ground the weight of a ball is balanced by the frictional air resistance.
first; (d) the balls hit the ground at the same time. Clearly Both balls initially experience the same acceleration, g, and their
4.6 FRICTION 129
speeds and the retarding forces of air resistance increase at the velocity 1a = 02 first, the heavier ball continues to accelerate
same rate. The weight of the lighter ball will be balanced first, so and pulls ahead of the lighter ball. Hence, the heavier ball hits
(a) and (b) are incorrect with the lighter ball reaching terminal the ground first, and the answer is (c), and (d) is incorrect.
FOLLOW-UP EXERCISE. Suppose the heavier ball were much larger in size than the lighter ball. How might this difference affect
the outcome?
You see an example of terminal velocity quite often. Why do clouds stay seemingly
suspended in the sky? Certainly the water droplets or ice crystals (high clouds)
should fall—and they do. However, they are so small that their terminal velocity is
reached quickly, and the very slow rate of their descent goes unnoticed. In addition,
there may be some helpful updrafts that keep the water droplets and ice crystals
from reaching the ground.
An extraterrestrial use of “air” resistance is called aerobraking. This spaceflight
technique uses a planetary atmosphere to slow down a spacecraft. As the craft
passes through the top layer of the planetary atmosphere, the atmospheric “drag”
slows and lowers the craft’s speed so as to put it in the desired orbit. Many passes
may be needed, with the spacecraft passing in and out of the atmosphere to
achieve the proper final orbit.
Aerobraking is a worthwhile technique because it eliminates the need for a heavy
load of chemical propellants that would otherwise be needed to place the spacecraft
in orbit. This allows a greater payload of scientific instruments for investigations.
DID YOU LEARN?

➥ Static friction prevents motion and has a maximum value of fsmax = ms N. Kinetic
friction acts on a sliding body.
➥ The coefficient of friction is a constant of proportionality between frictional and
normal forces.
➥ Air resistance depends on the area of an object exposed to air molecule collisions.
PULLING IT TOGETHER Newton Helps Superman: Kinematics and Forces

Traveling at 90.0 mi>h, a driver applies the brakes to his fast- +y
moving car and skids out of control on a wet, concrete hori- N
zontal road. The 2000-kg car is headed directly toward a
student waiting to catch a bus to campus who is standing fk
+x
58.0 m down the road. Fortunately, Superman is flying over-
30°
head and surveys the situation. Remembering from his w
physics class that the coefficient of kinetic friction between F
rubber and rough wet concrete is 0.800, he quickly determines Car’s Free-Body Diagram
that friction alone will not stop the car in time. v
So, he flies down and exerts a constant force of
F = 13 000 N 12925 lb2 on the car’s hood at a downward
angle of 30° (䉴 Fig. 4.26). 30°
(a) Show that Superman was correct in his determination that
the force of friction (assumed constant) alone would not stop the
car in time. (b) Show that Superman’s applied force saves the day.
(c) How close did the car come to the student before stopping? 䉱 F I G U R E 4 . 2 6 Superman to the rescue. Superman
applies a force to the skidding car. Will it be enough to save
T H I N K I N G I T T H R O U G H . Involved in this example are the student? See Example text for description.
Newton’s laws, friction, free-body diagrams, the summing of
forces in two dimensions, and constant acceleration kinematics. to friction alone can be obtained. (b) This involves using the net
(a) Stopping the car before a collision requires a minimum decel- force (which includes Superman’s applied force) to find the
eration that can be calculated from the initial speed and distance. deceleration using Newton’s second law and ensuring that it
Comparing this to the deceleration provided by friction will tell matches or exceeds the required minimum deceleration. See
whether friction alone can do the job—if it can, the frictional the free-body diagram of the forces in Fig. 4.26. (c) Using the
acceleration will be greater than the required minimum accelera- deceleration and initial speed, the stopping distance can be
tion. The force of friction can be determined from the coefficient computed from the appropriate kinematic equation.
of friction and the normal force. From this, the deceleration due (continued on next page)
SOLUTION.
Given: initial car speed, vo = 90.0 mi>h a b = 40.2 m>s

0.447 m>s
Find: (a) Show that friction alone isn’t enough to
mi>h
stop the car to avoid a tragedy.
m = 2000 kg
(b) Show that Superman’s applied force is
mk = 0.800
enough to stop the car in time.
Superman’s force and angle: F = 13 000 N, at an angle of u = 30°
(c) How close the car comes to the student.
d = 58.0 m
(a) Assuming a constant frictional force, the minimum decel- See the free-body diagram in Fig. 4.26.) Summing the forces in
eration can be determined from kinematics, using the vertical direction, which add up to zero (why?):
v2 = v2o + 2a1x - xo2. Setting the final velocity to zero gives Fy = N - w - F sin 30° = 0
the minimum deceleration a min as follows:
Solving for N, which is needed to determine the force of friction:
0 = v 2o + 2amin1x - xo2
N = w + F sin 30°
and = mg + F sin 30°
vo2 1 -40.2 m>s22 = 12000 kg219.80 m>s22 + 11.30 * 104 N210.5002
= + 13.9 m>s2
21x - xo2
a min = - = - = 2.61 * 104 N
21- 58.0 m2
Then the force of friction is
fk = mk N = 10.800212.61 * 104 N2 = 2.09 * 104 N
Notice how the signs work out: the displacement is negative and
the acceleration comes out positive, opposite the velocity, exactly
what is needed for slowing down. (That is, x and v are in the To determine the deceleration with Superman in the picture,
negative direction in this case, and a is in the positive direction.) the forces in the horizontal direction are summed for Newton’s
The net force in this friction-only case is the backward- second law:
pointing force of kinetic friction. The normal force on the car is Fx = fk + F cos 30°
= 2.09 * 104 N + 11.30 * 104 N210.8662
the same magnitude as the car’s weight (why?), hence
fk = mk N = mk w = mk mg = 10.800212000 kg219.80 m>s22 = 3.22 * 104 N = ma

= 1.57 * 104 N Solving for the deceleration:
Then, being the only force in the x (horizontal) direction, by Fx 3.22 * 104 N
a = = = + 16.1 m>s2
Newton’s second law, m 2000 kg
Fx = fk = 1.57 * 104 N = ma This exceeds the minimum value of a min = + 13.9 m>s2, so
Superman does save the day.
and
(c) Since the acceleration, a = 16.1 m>s2, and the initial and
Fx 1.57 * 104 N final velocities, vo = - 40.2 m>s and v = 0, are known, the
a = = = + 7.85 m>s2
m 2000 kg stopping distance can be found using the appropriate kine-
matic equation:
Note that the positive sign means that the deceleration is
opposite the car’s velocity, consistent with slowing down. v2 = v 2o + 2ax
Since this is less than the minimum required deceleration, Then,
0 - 1- 40.2 m>s22
amin , Superman’s push is needed.
v2 - v2o
x = = - 50.2 m
2116.1 m>s22
(b) To find the deceleration with Superman’s applied force, =
2a
first the normal force N is needed, which involves the vertical
component of the applied force. (There are four forces acting So the car stops a distance of 58.0 m - 50.2 m = 7.8 m from
on the car: weight, normal, frictional, and Superman forces. the student.
■ A force is something that is capable of ■ Newton’s first law of motion is also called the law of inertia,
F2
changing an object’s state of motion. To F1 where inertia is the natural tendency of an object to main-
produce a change in motion, there tain its state of motion. It states that in the absence of a net
Fnet = F2 – F1 ≠ 0
must be a nonzero net, or unbalanced, a
applied force, a body at rest remains at rest, and a body in
force: motion remains in motion with constant velocity.
B B Fnet
Fnet = a Fi
LEARNING PATH QUESTIONS AND EXERCISES 131
■ Newton’s second law relates the net force acting on an ■ An object is said to be in translational equilibrium when it
object or system to the (total) mass and the resulting acceler- either is at rest or moves with a constant velocity. When
ation. It defines the cause-and-effect relationship between remaining at rest, an object is said to be in static translational
force and acceleration: equilibrium. The condition for translational equilibrium is
B B represented as
a Fi = Fnet = ma
B
(4.1)
B
a Fi = 0 (4.4)
a
or
F
a Fxn = 0 and a Fyn = 0 (4.5)
m
■ Friction is the resistance to motion that occurs between con-
THIS SIDE UP
tacting surfaces. (In general, friction occurs for all types of

media—solids, liquids, and gases.)
A nonzero net force accelerates the crate: a F/m
The equation for weight in terms of mass is a form of New-

ton’s second law:
F f
F f
w = mg (4.2) Force exerted Frictional force
on ground exerted on
by foot foot by ground
The component form of Newton’s second law:
a 1Fx xN + Fy yN 2 = m1ax xN + a y yN 2 = max xN + may yN (4.3a) ■ The frictional force between surfaces is characterized by
coefficients of friction 1m2, one for the static case and one for
and
the kinetic (moving) case. In many cases, f = mN where N
a Fx = max and a Fy = may (4.3b) is the normal force—the force perpendicular to the surface
(that is, the force exerted by the surface on the object). As a
y ratio of forces 1f>N2, m is unitless.
F f
F
60° y fsmax
x
f k = µ kN
Fx F = fs
Kinetic friction
Static friction
■ Newton’s third law states that for every force, there is an
equal and opposite reaction force. The opposing forces of a Force of Static Friction:
third law force pair always act on different objects. fs … ms N (4.6)
fsmax = ms N (maximum value of static friction) (4.7)
F1 on briefcase
Contact forces
Force of Kinetic (Sliding) Friction:
F1 on hand
fk = mk N (4.8)
F2 on briefcase ■ The force of air resistance on a falling object increases with
increasing speed. It eventually attains a constant velocity,
Action-at-a-distance forces
called the terminal velocity.
on Earth
F2

4.1 THE CONCEPTS OF FORCE AND NET 3. If an object is moving at constant velocity, (a) there must
FORCE be a force in the direction of the velocity, (b) there must
AND be no force in the direction of the velocity, (c) there must
4.2 INERTIA AND NEWTON’S FIRST LAW be no net force, (d) there must be a net force in the direc-
tion of the velocity.
OF MOTION
1. Mass is related to an object’s (a) weight, (b) inertia, (c) 4. If the net force on an object is zero, the object could (a) be
density, (d) all of the preceding. at rest, (b) be in motion at a constant velocity, (c) have
zero acceleration, (d) all of the preceding.
2. A force (a) always produces motion, (b) is a scalar quan-
tity, (c) is capable of producing a change in motion, (d) 5. The force required to keep a rocket ship moving at a con-
both a and b. stant velocity in deep space is (a) equal to the weight of
the ship, (b) dependent on how fast the ship is moving, than the magnitude of the force of the car on the truck,
(c) equal to that generated by the rocket’s engines at half (c) the magnitude of the force of the truck on the car is
power, (d) zero. equal to the magnitude of the force of the car on the
truck, (d) none of the preceding.
4.3 NEWTON’S SECOND LAW OF

MOTION 4.5 MORE ON NEWTON’S LAWS: FREE-
BODY DIAGRAMS AND TRANSLATIONAL
6. The newton unit of force is equivalent to (a) kg # m>s, (b)
kg # m>s2, (c) kg # m2>s, (d) none of the pre- EQUILIBRIUM
ceding. 12. The kinematic equations of Chapter 2 can be used (a) only
7. The acceleration of an object is (a) inversely proportional with constant forces, (b) only with constant velocities,
to the acting net force, (b) directly proportional to its (c) with variable accelerations, (d) all of the preceding.
mass, (c) directly proportional to the net force and 13. The condition(s) for translational equilibrium is (are)
inversely proportional to its mass, (d) none of these. (a) gFx = 0, (b) gFy = 0, (c) gFi = 0, (d) all of the pre-
B
8. The weight of an object is directly proportional to (a) its ceding.

mass, (b) its inertia, (c) the acceleration due to gravity,
(d) all of the preceding.
4.6 FRICTION
14. In general, the frictional force (a) is greater for smooth
4.4 NEWTON’S THIRD LAW OF MOTION
than rough surfaces, (b) depends significantly on sliding
9. The action and reaction forces of Newton’s third law (a) speeds, (c) is proportional to the normal force,
are in the same direction, (b) have different magnitudes, (d) depends significantly on the surface area of contact.
(c) act on different objects, (d) can be the same force.
15. The coefficient of kinetic friction, mk , (a) is usually
10. A brick hits a glass window. The brick breaks the glass, greater than the coefficient of static friction, ms , (b) usu-
so (a) the magnitude of the force of the brick on the glass ally equals ms , (c) is usually smaller than ms , (d) equals
is greater than the magnitude of the force of the glass on the applied force that exceeds the maximum static force.
the brick, (b) the magnitude of the force of the brick on
the glass is smaller than the magnitude of the force of the 16. A crate sits in the middle of the bed of a flatbed truck.
glass on the brick, (c) the magnitude of the force of the The driver accelerates the truck gradually from rest to a
brick on the glass is equal to the magnitude of the force normal speed, but then has to make a sudden stop to
of the glass on the brick, (d) none of the preceding. avoid hitting a car. If the crate slides as the truck stops,
the frictional force would be (a) in the forward direction,
11. A freight truck collides head-on with a passenger car,
(b) in the backward direction, (c) zero.
causing a lot more damage to the car than to the truck.
From this condition, we can say that (a) the magnitude 17. Two people, a 100-kg man and a 50-kg woman, jump out
of the force of the truck on the car is greater than the of a plane together and open their identical parachutes
magnitude of the force of the car on the truck, (b) the at the same time. Who will strike the ground first: (a) the
magnitude of the force of the truck on the car is smaller man, (b) the woman, or (c) both together?
4.1 THE CONCEPTS OF FORCE AND NET pushed and a force is applied to accelerate it, which way
FORCE would the bubble move? Which way would the bubble
AND move if the force is then removed and the level slows
4.2 INERTIA AND NEWTON’S FIRST LAW down, due to friction? (b) Such a level is sometimes used
as an “accelerometer” to indicate the direction of the
OF MOTION
acceleration. Explain the principle involved. [Hint: Think
1. (a) If an object is at rest, there must be no forces acting on about pushing a pan of water.]
it. Is this statement correct? Explain. (b) If the net force
on an object is zero, can you conclude that the object is at a
rest? Explain.
?
2. When on a jet airliner that is taking off, you feel that you
are being “pushed” back into the seat. Use Newton’s
first law to explain why.
3. An object weighs 300 N on Earth and 50 N on the Moon. m
Does the object also have less inertia on the Moon?
4. Consider an air-bubble level that is sitting on a horizon-
tal surface (䉴 Fig. 4.27). Initially, the air bubble is in the 䉱 F I G U R E 4 . 2 7 An air-bubble
middle of the horizontal glass tube. (a) If the level is level/accelerometer See Conceptual Question 4.
5. As a follow-up to Conceptual Question 4, consider a 12. In football, good wide receivers usually have “soft”
child holding a helium balloon in a closed car at rest. hands for catching balls (䉲 Fig. 4.30). How would you
What would the child observe when the car (a) acceler- interpret this description on the basis of Newton’s sec-
ates from rest and (b) brakes to a stop? (The balloon does ond law?
not touch the roof of the car.)
6. The following is an old trick (䉲 Fig. 4.28). If a tablecloth is
yanked out very quickly, the dishes on it will barely
move. Why?
䉱 F I G U R E 4 . 2 8 Magic or physics? See Conceptual

Question 6.
7. Another old one: Referring to 䉱 F I G U R E 4 . 3 0 Soft hands See Conceptual Question 12.
䉴 Fig. 4.29, (a) how would you
pull to get the upper string to
break? (b) How would you 4.4 NEWTON’S THIRD LAW OF MOTION
pull to get the lower string to
break? 13. Here is a story of a horse and a farmer: One day, the
farmer attaches a heavy cart to the horse and demands
that the horse pull the cart. “Well,” says the horse, “I can-
not pull the cart, because, according to Newton’s third
law, if I apply a force to the cart, the cart will apply an
equal and opposite force on me. The net result will be
that I cannot pull the cart, since all the forces will cancel.
Therefore, it is impossible for me to pull this cart.” The
䉱 F I G U R E 4 . 2 9 Give
farmer was very upset! What could he say to persuade
it a pull. See Conceptual
Question 7. the horse to move?
14. Is something wrong with the following statement? When
8. A student weighing 600 N crouches on a scale and sud- a baseball is hit with a bat, there are equal and opposite
denly springs vertically upward. Will the scale read forces on the bat and baseball. The forces then cancel,
more or less than 600 N just before the student leaves the and there is no motion.
scale?
4.5 MORE ON NEWTON’S LAWS: FREE-

4.3 NEWTON’S SECOND LAW OF BODY DIAGRAMS AND TRANSLATIONAL
MOTION EQUILIBRIUM
9. An astronaut has a mass of 70 kg when measured on 15. Draw the free-body diagram for a person sitting in the
Earth. What is her weight in deep space, far from any seat of an aircraft (a) that is accelerating down the run-
celestial body? What is her mass there? way for takeoff, and (b) after takeoff at a 20° angle to the
10. In general, this chapter has considered forces that are ground.
applied to objects of constant mass. What would be the 16. A person pushes perpendicularly on a block of wood
situation if mass were added to or lost from a system that has been placed against a wall. Draw a free-body
while a constant force was being applied to the system? diagram of the block and identify the reaction forces to
Give examples of situations in which this set of events all the forces on the block.
might happen. 17. A person on a bathroom scale (not the digital type) stands
11. The engines of most rockets produce a constant thrust on the scale with his arms at his side. He then quickly
(forward force). However, when a rocket is fired, its raises his arms over his head, and notices that the scale
acceleration increases with time as the engine continues reading increases as he brings his arms upward. Similarly,
to operate. Is this situation a violation of Newton’s sec- there is a decrease as he brings his arms downward. Why
ond law? Explain. does the scale reading change? (Try this yourself.)
4.6 FRICTION 21. (a) We commonly say that friction opposes motion. Yet
when we walk, the frictional force is in the direction of
18. Identify the direction of the friction force in the follow-
our motion (Fig. 4.17). Is there an inconsistency in terms
ing cases: (a) a book sitting on a table; (b) a box sliding
of Newton’s second law? Explain. (b) What effects
on a horizontal surface; (c) a car making a turn on a flat
would wind have on air resistance? [Hint: The wind can
road; (d) the initial motion of a machine part delivered
blow in different directions.]
on a conveyor belt in an assembly line.
19. The purpose of a car’s antilock brakes is to prevent the 22. Why are drag-racing tires wide and smooth, whereas
wheels from locking up so as to keep the car rolling passenger-car tires are narrower and have tread (䉲 Fig.
rather than sliding. Why would rolling decrease the 4.32)? Are there frictional and/or safety considerations?
stopping distance as compared with sliding? Does this difference between the tires contradict the fact
20. Shown in 䉲 Fig. 4.31 are the front and rear wings of an that friction is independent of surface area?
Indy racing car. These wings generate down force, which
is the vertical downward force produced by the air mov-
ing over the car. Why is such a down force desired? An
Indy car can create a down force equal to twice its
weight. Why not simply make the cars heavier?
䉱 F I G U R E 4 . 3 2 Racing tires versus passenger-car tires:

safety See Conceptual Question 22.
23. How could you approximately determine the coefficient

of kinetic friction between your shoes and a fairly
䉱 F I G U R E 4 . 3 1 Down force See Conceptual Question 20. smooth floor? [Hint: See Exercise 70.]
EXERCISES*
Throughout the text, many exercise sections will include “paired” exercises. These exercise pairs, identified with
red numbers, are intended to assist you in problem solving and learning. In a pair, the first exercise (even num-
bered) is worked out in the Study Guide so that you can consult it should you need assistance in solving it. The sec-
ond exercise (odd numbered) is similar in nature, and its answer is given in Appendix VII at the back of the book.
4.1 THE CONCEPTS OF FORCE AND NET 4. ●A net force of 4.0 N gives an object an acceleration of
FORCE 10 m>s2. What is the mass of the object?
AND 5. ● Consider a 2.0-kg ball and a 6.0-kg ball in free fall.
4.2 INERTIA AND NEWTON’S FIRST LAW (a) What is the net force acting on each? (b) What is the
OF MOTION acceleration of each?
6. IE ● ● A hockey puck with a weight of 0.50 lb is sliding
1. ● Which has more inertia, 20 cm3 of water or 10 cm3 of
freely across a section of very smooth (frictionless) hori-
aluminum, and how many times more? (See Table 9.2.)
zontal ice. (a) When it is sliding freely, how does the
2. ● Two forces act on a 5.0-kg object sitting on a friction- upward force of the ice on the puck (the normal force)
less horizontal surface. One force is 30 N in the compare with the upward force when the puck is sitting
+ x-direction, and the other is 35 N in the - x-direction. permanently at rest: (1) The upward force is greater
What is the acceleration of the object? when the puck is sliding; (2) the upward force is less
3. ● In Exercise 2, if the 35-N force acted downward at an when it is sliding; (3) the upward force is the same in
angle of 40° relative to the horizontal, what would be the both situations? (b) Calculate the upward force on the
acceleration in this case? puck in both situations.
*Unless otherwise stated, all objects are located near the Earth’s surface, where g = 9.80 m>s2.
EXERCISES 135
7. ●● A 5.0-kg block at rest on a frictionless surface is acted 14. ●A loaded Boeing 747 jumbo jet has a mass of
on by forces F1 = 5.5 N and F2 = 3.5 N as illustrated in 2.0 * 105 kg. What net force is required to give the plane
䉲 Fig. 4.33. What additional force will keep the block an acceleration of 3.5 m>s 2 down the runway for take-
at rest? offs?
15. IE ● A 6.0-kg object is brought to the Moon, where the
acceleration due to gravity is only one-sixth of that on
y
the Earth. (a) The mass of the object on the Moon is (1)
zero, (2) 1.0 kg, (3) 6.0 kg, (4) 36 kg. Why? (b) What is the
weight of the object on the Moon?
F1 F2
16. ● ● A gun is fired and a 50-g bullet is accelerated to a
30° 37° muzzle speed of 100 m>s. If the length of the gun barrel
x
is 0.90 m, what is the magnitude of the accelerating
force? (Assume the acceleration to be constant.)
17. IE ● ● ● 䉲 Fig. 4.34 shows a product label. (a) This label is
䉱 F I G U R E 4 . 3 3 Two applied forces See Exercise 7. correct (1) on the Earth; (2) on the Moon, where the accel-
eration due to gravity is only one-sixth of that on the
8. IE ● ● (a) You are told that an object has zero accelera- Earth; (3) in deep space, where there is little gravity; (4) all
tion. Which of the following is true: (1) The object is at of the preceding. (b) What mass of lasagne would a label
rest; (2) the object is moving with constant velocity; (3) show for an amount that weighs 2 lb on the Moon?
either (1) or (2) is possible; or (4) neither 1 nor 2 is possi-
ble. (b) Two forces on the object are F1 = 3.6 N at 74°
below the + x-axis and F2 = 3.6 N at 34° above the
- x-axis. Is there a third force on the object? Why or why
not? If there is a third force, what is it?
9. IE ● ● A fish weighing 25 lb is caught and hauled onto
the boat. (a) Compare the tension in the fishing line
when the fish is brought up vertically at a constant speed
to the tension when the fish is held vertically at rest for
the picture-taking ceremony on the wharf. In which case
is the tension largest: (1) When the fish is moving up; (2)
when the fish is being held steady; or (3) the tension is
the same in both situations? (b) Calculate the tension in
the fishing line.
10. ● ● ● A 1.5-kg object moves up the y-axis at a constant
speed. When it reaches the origin, the forces F1 = 5.0 N
at 37° above the + x-axis, F2 = 2.5 N in the +x-direction, 䉱 F I G U R E 4 . 3 4 Correct label? See Exercise 17.
F3 = 3.5 N at 45° below the - x-axis, and F4 = 1.5 N in
the - y-direction are applied to it. (a) Will the object con- 18. ●● In a college homecoming competition, eighteen stu-
tinue to move along the y-axis? (b) If not, what simulta- dents lift a sports car. While holding the car off the
neously applied force will keep it moving along the ground, each student exerts an upward force of 400 N.
y-axis at a constant speed? (a) What is the mass of the car in kilograms? (b) What is
its weight in pounds?
11. IE ● ● ● Three horizontal forces (the only horizontal ones)
act on a box sitting on a floor. One (call it F1) acts due 19. IE ● ● (a) A horizontal force acts on an object on a fric-
east and has a magnitude of 150 lb. A second force (call it tionless horizontal surface. If the force is halved and the
F2) has an easterly component of 30.0 lb and a southerly mass of the object is doubled, the acceleration will be (1)
component of 40.0 lb. The box remains at rest. (Neglect four times, (2) two times, (3) one-half, (4) one-fourth as
friction.) (a) Sketch the two known forces on the box. In great. (b) If the acceleration of the object is 1.0 m>s2, and
which quadrant is the unknown third force: (1) the first the force on it is doubled and its mass is halved, what is
quadrant; (2) the second quadrant; (3) the third quad- the new acceleration?
rant; or (4) the fourth quadrant? (b) Find the unknown 20. ● ● A force of 50 N acts on a mass m1, giving it an acceler-
third force in newtons and compare your answer to the ation of 4.0 m>s2. The same force acts on a mass m2 and
sketched estimate. produces an acceleration of 12 m>s 2. What acceleration
will this force produce if the total system is m1 + m2 ?
21. ● ● A student weighing 800 N crouches on a scale and
4.3 NEWTON’S SECOND LAW OF
suddenly springs vertically upward. His roommate
MOTION
notices that the scale reads 900 N momentarily just as he
12. ●A 6.0-N net force is applied to a 1.5-kg mass. What is leaves the scale. With what acceleration does he leave
the object’s acceleration? the scale?
13. ● A force acts on a 1.5-kg, mass, giving it an acceleration 22. ● ● The engine of a 1.0-kg toy plane exerts a 15-N for-
of 3.0 m>s2. (a) If the same force acts on a 2.5-kg mass, ward force. If the air exerts an 8.0-N resistive force on the
what acceleration would be produced? (b) What is the plane, what is the magnitude of the acceleration of the
magnitude of the force? plane?
23. ●● When a horizontal force of 300 N is applied to a 75.0- 28. ●● A force of 10 N acts on two blocks on a frictionless
kg box, the box slides on a level floor, opposed by a force surface (䉲 Fig. 4.37). (a) What is the acceleration of the
of kinetic friction of 120 N. What is the magnitude of the system? (b) What force does block A exert on block B?
acceleration of the box? (c) What force does block B exert on block A?
24. IE ● ● A rocket is far away from all planets and stars, so 䉳 FIGURE 4.37
gravity is not a consideration. It is using its rocket Forces: inside and
engines to accelerate upward with an acceleration out See Example 28.
a = 9.80 m>s2. On the floor of the main deck is a crate
(object with brick pattern) with a mass of 75.0 kg F ⫽ 10 N A
(3.0 kg) B
(䉲 Fig. 4.35). (a) How many forces are acting on the crate:
(2.0 kg)
(1) zero; (2) one; (3) two; (4) three? (b) Determine the nor-
mal force on the crate and compare it to the normal force
the crate would experience if it were at rest on the sur-
face of the Earth.
29. ●● A 2.0-kg object has an acceleration of 1.5 m>s 2 at 30°
above the -x-axis. Write the force vector producing this
acceleration in component form.
30. ● ● ● In a pole-sliding game among friends, a 90-kg man
a
makes a total vertical drop of 7.0 m while gripping the
pole which exerts and upward force (call it Fp) on him.
Starting from rest and sliding with a constant accelera-
tion, his slide takes 2.5 s. (a) Draw the man’s free body
diagram being sure to label all the forces. (b) What is the
magnitude of the upward force exerted on the man by
the pole? (c) A friend whose mass is only 75 kg, slides
down the same distance, but the pole force is only 80%
of the force on his buddy. How long did the second per-
son’s slide take?
4.4 NEWTON’S THIRD LAW OF MOTION

䉳 F I G U R E 4 . 3 5 Away we go
See Exercise 24. 31. IE ● A book is sitting on a horizontal surface. (a) There is
(are) (1) one, (2) two, or (3) three force(s) acting on the book.
25. ●● An object (mass 10.0 kg) slides upward on a slippery (b) Identify the reaction force to each force on the book.
vertical wall. A force F of 60 N acts at an angle of 60° as
32. ●● In an Olympic figure-skating event, a 65-kg male
shown in 䉲 Fig. 4.36. (a) Determine the normal force
skater pushes a 45-kg female skater, causing her to acceler-
exerted on the object by the wall. (b) Determine the
ate at a rate of 2.0 m>s2. At what rate will the male skater
object’s acceleration.
accelerate? What is the direction of his acceleration?
33. IE ● ● A sprinter of mass 65.0 kg starts his race by push-
ing horizontally backward on the starting blocks with a
force of 200 N. (a) What force causes him to accelerate
out of the blocks: (1) his push on the blocks; (2) the
downward force of gravity; or (3) the force the blocks
60° exert forward on him? (b) Determine his initial accelera-
䉳 F I G U R E 4 . 3 6 Up a wall tion as he leaves the blocks.
F See Exercise 25. (Drawing not 34. ●● Jane and John, with masses of 50 kg and 60 kg,
to scale.) respectively, stand on a frictionless surface 10 m apart.
John pulls on a rope that connects him to Jane, giving
26. ●● In an emergency stop to avoid an accident, a Jane an acceleration of 0.92 m>s2 toward him. (a) What is
shoulder-strap seatbelt holds a 60-kg passenger in place. John’s acceleration? (b) If the pulling force is applied
If the car was initially traveling at 90 km>h and came to a constantly, where will Jane and John meet?
stop in 5.5 s along a straight, level road, what was the 35. IE ● ● ● During a daring rescue, a helicopter rescue squad
average force applied to the passenger by the seatbelt? initially accelerates a little girl (mass 25.0 kg) vertically
27. IE ● ● A student is assigned the task of measuring the off the roof of a burning building. They do this by drop-
startup acceleration of a large RV (recreational vehicle) ping a rope down to her, which she holds on to as they
using an iron ball suspended from the ceiling by a long pull her up. Neglect the mass of the rope. (a) What force
string. In accelerating from rest, the ball no longer hangs causes the girl to accelerate vertically upward: (1) her
vertically, but at an angle to the vertical. (a) Is the angle weight; (2) the pull of the helicopter on the rope; (3) the
of the ball forward or backward from the vertical? (b) If pull of the girl on the rope; or (4) the pull of the rope on
the string makes an angle of 3.0 degrees from the verti- the girl? (b) Determine the pull of the rope (the tension) if
cal, what is the initial acceleration of the RV? she initially accelerates upward at 0.750 m>s2.
EXERCISES 137
4.5 MORE ON NEWTON’S LAWS: FREE- 44. IE ● ● (a) An Olympic skier coasts down a slope with an
BODY DIAGRAMS AND TRANSLATIONAL angle of inclination of 37°. Neglecting friction, there is (are)
EQUILIBRIUM (1), one, (2) two, (3) three force(s) acting on the skier.
(b) What is the acceleration of the skier? (c) If the skier has
36. ●● A 75.0-kg person is standing on a scale in an elevator. a speed of 5.0 m>s at the top of the slope, what is his speed
What is the reading of the scale in newtons if the eleva- when he reaches the bottom of the 35-m-long slope?
tor is (a) at rest, (b) moving up at a constant velocity of
45. ● ● A car coasts (engine off) up a 30° grade. If the speed
2.00 m>s, and (c) accelerating up at 2.00 m>s2 ?
of the car is 25 m>s at the bottom of the grade, what is
37. ● ● In Exercise 36, what if the elevator is accelerating
the distance traveled by the car before it comes to rest?
down at 2.00 m>s2 ? 46. ● ● Assuming ideal frictionless conditions for the appara-
38. IE ● (a) When an object is on an inclined plane, the nor- tus shown in 䉲Fig. 4.40, what is the acceleration of the sys-
mal force exerted by the inclined plane on the object is tem if (a) m1 = 0.25 kg, m2 = 0.50 kg, and m3 = 0.25 kg,
(1) less than, (2) equal to, (3) more than the weight of the and (b) m1 = 0.35 kg, m2 = 0.15 kg, and m3 = 0.50 kg?
object. Why? (b) For a 10-kg object on a 30° inclined
plane, what are the object’s weight and the normal force m3
exerted on the object by the inclined place?
39. IE ● ● The weight of a 500-kg object is 4900 N. (a) When
the object is on a moving elevator, its measured weight
could be (1) zero, (2) between zero and 4900 N, (3) more
m1 m2 䉳 FIGURE 4.40
than 4900 N, (4) all of the preceding. Why? (b) Describe
the motion if the object’s measured weight is only Which way will they
4000 N in a moving elevator. accelerate? See
Exercises 46, 80,
40. ● ● A boy pulls a box of mass 30 kg with a force of 25 N
and 81.
in the direction shown in 䉲 Fig. 4.38. (a) Ignoring friction,
what is the acceleration of the box? (b) What is the nor- 47. IE ● ● A rope is fixed at both ends on two trees and a bag is
mal force exerted on the box by the ground? hung in the middle of the rope (causing the rope to sag ver-
tically). (a) The tension in the rope depends on (1) only the
tree separation, (2) only the sag, (3) both the tree separation
and sag, (4) neither the tree separation nor the sag. (b) If the
tree separation is 10 m, the mass of the bag is 5.0 kg, and
25 N the sag is 0.20 m, what is the tension in the line?
30° 48. ● ● A 55-kg gymnast hangs vertically from a pair of parallel
rings. (a) If the ropes supporting the rings are attached to
the ceiling directly above, what is the tension in each rope?
(b) If the ropes are supported so that they make an angle of
䉱 F I G U R E 4 . 3 8 Pulling a box See Exercise 40. 45° with the ceiling, what is the tension in each rope?
49. ● ● A physicist’s car has a small lead weight suspended
from a string attached to the interior ceiling. Starting
41. ●● A girl pushes a 25-kg lawn mower as shown in
from rest, after a fraction of a second the car accelerates
䉲 Fig.4.39. If F = 30 N and u = 37° (a) what is the accel-
at a steady rate for about 10 s. During that time, the
eration of the mower, and (b) what is the normal force
string (with the weight on the end of it) makes a back-
exerted on the mower by the lawn? Ignore friction.
ward (opposite the acceleration) angle of 15.0° from the
vertical. Determine the car’s (and the weight’s) accelera-
tion during the 10-s interval.
50. ● ● A 10-kg mass is suspended as shown in 䉲 Fig. 4.41.
What is the tension in the cord between points A and B?
51. ● ● Referring to Fig. 4.41, what are the tensions in all
F the cords?
䉱 F I G U R E 4 . 3 9 Mowing the lawn See Exercise 41. 45 TAB 45

A B
42. ●●A 3000-kg truck tows a 1500-kg car by a chain. If the 30 30
net forward force on the truck by the ground is 3200 N,
(a) what is the acceleration of the car, and (b) what is the
tension in the connecting chain?
43. ● ● A block of mass 25.0 kg slides down a frictionless
10 kg
surface inclined at 30°. To ensure that the block does not
accelerate, what is the smallest force that you must exert
on it and what is its direction? 䉱 F I G U R E 4 . 4 1 Under tension See Exercises 50 and 51.
52. ●● At the end of most landing runways in airports, an for the magnitude of the string tension T compared to
extension of the runway is constructed using a special other forces: (1) T 7 w2 and T 6 F; (2) T 7 w2 and
substance called formcrete. Formcrete can support the T 7 F; (3) T 6 w2 and T 6 F; or (4) T = w2 and T 6 F?
weight of cars, but crumbles under the weight of air- (b) Apply Newton’s laws to find the required pull, F.
planes to slow them down if they run off the end of a (c) Find the tension in the string, T.
runway. If a plane of mass 2.00 * 105 kg is to stop from a 59. ● ● ● Two blocks on a level, frictionless table are in con-
speed of 25.0 m>s on a 100-m-long stretch of formcrete, tact. The mass of the left block is 5.00 kg and the mass of
what is the average force exerted on the plane by the the right block is 10.0 kg, and they accelerate to the left at
formcrete? 1.50 m>s2. A person on the left exerts a force (F1) of
53. ● ● A rifle weighs 50.0 N and its barrel is 0.750 m long. It 75.0 N to the right. Another person exerts an unknown
shoots a 25.0-g bullet, which leaves the barrel at a speed force (F2) to the left. (a) Determine the force F2. (b) Calcu-
(muzzle velocity) of 300 m>s after being uniformly accel- late the force of contact N between the two blocks (that
erated. What is the magnitude of the force exerted on the is, the normal force at their vertical touching surfaces).
rifle by the bullet? 60. ● ● ● In the frictionless apparatus shown in 䉲 Fig. 4.43,
54. ● ● A horizontal force of 40 N acting on a block on a fric- m1 = 2.0 kg. What is m2 if both masses are at rest? How
tionless, level surface produces an acceleration of about if both masses are moving at constant velocity?
2.5 m>s2. A second block, with a mass of 4.0 kg, is
dropped onto the first. What is the magnitude of the
acceleration of the combination of blocks if the same
force continues to act? (Assume that the second block
does not slide on the first block.) m1
55. ● ● The Atwood machine consists of two masses sus- m2
pended from a fixed pulley, as shown in 䉲 Fig. 4.42. It is
named after the British scientist George Atwood
(1746–1807), who used it to study motion and to mea-
sure the value of g. If m1 = 0.55 kg and m2 = 0.80 kg, 37°
(a) what is the acceleration of the system, and (b) what is
the magnitude of the tension in the string? 䉱 F I G U R E 4 . 4 3 Inclined Atwood machine See Exer-
cises 60, 61, and 79.
61. ● ● ● In the ideal setup shown in Fig. 4.43, m 1 = 3.0 kg
and m2 = 2.5 kg. (a) What is the acceleration of the

masses? (b) What is the tension in the string?
4.6 FRICTION
T
62. IE ● A 20-kg box sits on a rough horizontal surface. When
T a horizontal force of 120 N is applied, the object accelerates
m1
a at 1.0 m>s 2. (a) If the applied force is doubled, the accelera-
a tion will (1) increase, but less than double; (2) also double;
m1g m2 (3) increase, but more than double. Why? (b) Calculate the
acceleration to prove your answer to part (a).
䉳 F I G U R E 4 . 4 2 Atwood 63. ● The coefficients of static and kinetic friction between a
machine See Exercises 55, 56,
50.0-kg box and a horizontal surface are 0.500 and 0.400
m2 g and 57.
respectively. (a) What is the acceleration of the object if a
56. ●● An Atwood machine (see Fig. 4.42) has suspended 250-N horizontal force is applied to the box? (b) What is
masses of 0.25 kg and 0.20 kg. Under ideal conditions, the acceleration if the applied force is 235 N?
what will be the acceleration of the smaller mass?
64. ● In moving a 35.0-kg desk from one side of a classroom
57. ● ● ● One mass, m1 = 0.215 kg, of an ideal Atwood to the other, a professor finds that a horizontal force of
machine (see Fig. 4.42) rests on the floor 1.10 m below 275 N is necessary to set the desk in motion, and a force
the other mass, m2 = 0.255 kg, (a) If the masses are of 195 N is necessary to keep it in motion at a constant
released from rest, how long does it take m2 to reach the speed. What are the coefficients of (a) static and
floor? (b) How high will mass m1 ascend from the floor? (b) kinetic friction between the desk and the floor?
(Hint: When m2 hits the floor, m1 continues to move
upward.) 65. ● A 40-kg crate is at rest on a level surface. If the coeffi-
cient of static friction between the crate and the surface is
58. IE ● ● ● Two blocks are connected by a light string and
0.69, what horizontal force is required to get the crate
accelerated upward by a pulling force F. The mass of
moving?
the upper block is 50.0 kg and that of the lower block is
100 kg. The upward acceleration of the system as a 66. ●● A packing crate is placed on a 20° inclined plane. If
whole is 1.50 m>s2. Neglect the mass of the string. the coefficient of static friction between the crate and the
(a) Draw the free-body diagram of each block. Use the plane is 0.65, will the crate slide down the plane if
diagrams to determine which of the following is true released from rest? Justify your answer.
EXERCISES 139
67. ●● A 1500-kg automobile travels at 90 km>h along a 75. ●● A block that has a mass of 2.0 kg and is 10 cm wide
straight concrete highway. Faced with an emergency on each side just begins to slide down an inclined plane
situation, the driver jams on the brakes, and the car skids with a 30° angle of incline (䉲 Fig. 4.45). Another block of
to a stop. What is the car’s stopping distance for (a) dry the same height and same material has base dimensions
pavement and (b) wet pavement? of 20 cm * 10 cm and thus a mass of 4.0 kg. (a) At what
68. ● ● A hockey player hits a puck with his stick, giving the
critical angle will the more massive block start to slide
puck an initial speed of 5.0 m>s. If the puck slows uni- down the plane? Why? (b) Estimate the coefficient of sta-
formly and comes to rest in a distance of 20 m, what is the tic friction between the block and the plane.
coefficient of kinetic friction between the ice and the puck? 4.0 kg
2.0 kg
69. ● ● A crate sits on a flat-bed truck that is traveling with a
speed of 50 km>h on a straight, level road. If the coeffi-

cient of static friction between the crate and the truck
bed is 0.30, in how short a distance can the truck stop
with a constant acceleration without the crate sliding?
70. ● ● A block is projected with a speed of 2.5 m>s on a hor-
izontal surface. If the block comes to rest in 1.5 m, what 30˚ θ=?
is the coefficient of kinetic friction between the block and
the surface? 䉱 F I G U R E 4 . 4 5 At what angle will it begin to slide? See
71. ● ● A block is projected with a speed of 3.0 m>s on a hor- Exercise 75.
izontal surface. If the coefficient of kinetic friction
between the block and the surface is 0.60, how far does 76. ●● In the apparatus shown in 䉲 Fig. 4.46, m1 = 10 kg and
the block slide before coming to rest? the coefficients of static and kinetic friction between m1
72. IE ● ● A person has a choice while trying to push a crate and the table are 0.60 and 0.40, respectively. (a) What mass
across a horizontal pad of concrete: push it at a down- of m2 will just barely set the system in motion? (b) After
ward angle of 30°, or pull it at an upward angle of 30°. the system begins to move, what is the acceleration?
(a) Which choice is most likely to require less force on
the part of the person: (1) pushing at a downward angle;
m1
(2) pulling at the same angle, but upward; or (3) pushing
or pulling shouldn’t matter? (b) If the crate has a mass of
50.0 kg and the coefficient of kinetic friction between it
and the concrete is 0.750, calculate the required force to
move it across the concrete at a steady speed for both
situations. m2
73. ● ● Suppose the slope conditions for the skier shown in
䉲 Fig. 4.44 are such that the skier travels at a constant 䉱 F I G U R E 4 . 4 6 Friction and motion See Exercise 76.
velocity. From the photo, could you find the coefficient
77. ●● In loading a fish delivery truck, a person pushes a
of kinetic friction between the snowy surface and the
block of ice up a 20° incline at constant speed. The push
skis? If so, describe how this would be done.
is 150 N in magnitude and parallel to the incline. The
block has a mass of 35.0 kg. (a) Is the incline frictionless?
(b) If not, what is the force of kinetic friction on the block
of ice?
78. ● ● ● An object (mass 3.0 kg) slides upward on a vertical
wall at constant velocity when a force F of 60 N acts on it

at an angle of 60° to the horizontal. (a) Draw the free-
body diagram of the object. (b) Using Newton’s laws
find the normal force on the object. (c) Determine the
force of kinetic friction on the object.
79. ● ● ● In the apparatus shown in Fig. 4.43, m 1 = 2.0 kg
and the coefficients of static and kinetic friction between

m1 and the inclined plane are 0.30 and 0.20, respectively.
(a) What is m2 if both masses are at rest? (b) What is m2 if
both masses are moving at constant velocity?
80. ● ● ● For the apparatus shown in Fig. 4.40, what is the
minimum value of the coefficient of static friction between

䉱 F I G U R E 4 . 4 4 A down slope run See Exercise 73. the block (m3) and the table that would keep the system at
rest if m1 = 0.25 kg, m2 = 0.50 kg, and m3 = 0.75 kg?
74. ●● A 5.0-kg wooden block is placed on an adjustable 81. ● ● ● If the coefficient of kinetic friction between the block
wooden inclined plane. (a) What is the angle of incline and the table in Fig. 4.40 is 0.560, and m1 = 0.150 kg and
above which the block will start to slide down the plane? m2 = 0.250 kg, (a) what should m3 be if the system is to
(b) At what angle of incline will the block then slide move with a constant speed? (b) If m3 = 0.100 kg, what is
down the plane at a constant speed? the magnitude of the acceleration of the system?
82. IE One block (A, mass 2.00 kg) rests atop another (B, from the horizontal and the boat is being momentarily
mass 5.00 kg) on a horizontal surface. The surface is a held at rest. Compare this to the tension when the boat is
powered walkway accelerating to the right at 2.50 m>s2. raised and held at rest so the angle becomes 30°.
B does not slip on the walkway surface, nor does A slip on 85. You are in charge of an accident reconstruction case for
B’s top surface. (a) Sketch the free-body diagram of each the local police department. In order to determine car
block. Use these to determine the force responsible for A’s speeds, skid mark lengths are measured. To determine
acceleration. Is it (1) the pull of the walkway, (2) the nor- the coefficient of kinetic friction, you get into an identical
mal force on A by the top surface of B, (3) the force of static car, and at a speed of 65.2 mi>h, you lock its brakes and
friction on the bottom surface of B, or (4) the force of static skid 51.5 m to rest. (a) Determine the car’s deceleration.
friction acting on A due to the top surface of B? (b) Deter- (b) What is the coefficient of kinetic friction between the
mine the forces of static friction on each block. tires and road surface? (c) The car in the accident actu-
83. Two blocks (A and B) remain stuck together as they are ally skidded 57.3 m. What was its initial speed?
pulled to the right by a force F = 200 N (䉲 Fig. 4.47). B is 86. IE Compare two different situations in which a ball and
on a rough horizontal tabletop (coefficient of kinetic fric- hard surface exert forces on one another. First, a putty
tion of 0.800). (a) What is the acceleration of the system? ball is placed gently on the floor and left at rest. Then it
(b) What is the force of friction between the two objects? is dropped from a height of 2.00 m and comes to rest
without a bounce, leaving a 1.15-cm-deep dent in the
mA = 5.00 kg putty. (a) In which case does the ball exert more force on
the floor? In which case is it most likely to dent the floor?
F = 200 N
A Explain. (b) Calculate the force exerted by the ball on the
floor (in terms of its weight w) in the first case. (c) Deter-
mB = 10.0 kg B mine the average acceleration of the ball and the average
force exerted by the ball on the floor (in terms of the
ball’s weight w) in the second case.
䉱 F I G U R E 4 . 4 7 Take it away See Exercise 83. 87. A hockey puck impacts a goalie’s plastic mask horizon-
tally at 122 mi>h and rebounds horizontally off the mask
84. IE To haul a boat out of the water for the winter, a at 47 mi>h. If the puck has a mass of 170 g and it is in
worker at the storage facility uses a wide strap with contact with the mask for 25 ms, (a) what is the average
cables operating at the same angle (measured from the force (including direction) that the puck exerts on the
horizontal) on either side of the boat (䉲 Fig. 4.48). (a) As mask? (b) Assuming that this average force accelerates
the boat comes up vertically and u decreases, the tension the goalie (neglect friction with the ice), with what speed
in the cables (1) increases, (2) decreases, (3) stays the will the goalie move, assuming she was at rest initially
same. (b) Determine the tension in each cable if the boat and has a total mass of 85 kg?
has a mass of 500 kg and the angle of each cable is 45° 88. A 2.50-kg block is placed on a rough surface inclined at
30°. The block is propelled and launched at a speed of
1.60 m>s down the incline and comes to rest after sliding
T2 T1
1.10 m. (a) Draw the free-body diagram of the block
while it is sliding. Also indicate your coordinate system
axes. (b) Starting with Newton’s second law applied
along both axes of your coordinate system, use your
u u free-body diagram to generate two equations. (c) Solve
these equations for the coefficient of kinetic friction
between the block and the incline surface. [Hint: You will
䉱 F I G U R E 4 . 4 8 Hoist it up See Exercise 84. need to first determine the block’s acceleration.]
5 Work and Energy †
5.1 Work done by a

constant force (142)
5.2 Work done by a

variable force (147)
5.3 The work–energy theorem:

kinetic energy (150)
■ energy of motion
5.4 Potential energy (154)

■ energy of position
Conservation of energy (157) PHYSICS FACTS
A
5.5
total energy
■ ✦ Kinetic comes from the Greek
description of pole vaulting,
■ mechanical energy kinein, meaning “to move.” as shown in the chapter-
✦ Energy comes from the Greek
energeia, meaning “activity.” opening photo, might be as fol-
5.6 Power (166)
✦ The United States has 5% of the lows: The athlete runs with a pole,
world’s population, yet consumes
■ efficiency: work out/energy in about 26% of its energy supply. plants it into the ground, and tries
✦ Recycling aluminum takes 95% to vault his body over a bar set at
less energy than making alu-
minum from raw materials. a certain height. However, a physi-
✦ The human body uses muscles to cist might give a different descrip-
propel itself, turning stored energy
into motion. There are 630 active tion: The athlete has chemical
muscles in your body and they act
in groups.
potential energy stored in his
✦ The human body operates within body. He uses this potential
the limits imposed by the law of
conservation of total energy, need- energy to do work in running
ing dietary energy equal to the
down the path to gain speed, or
energy expended in the overall
work of daily activities, internal kinetic energy. When he plants
activities, and system heat losses.
✦ Energy is neither created nor
the pole, most of his kinetic
†
destroyed. The amount of energy energy goes into elastic potential
The mathematics in this chapter involves in the universe is constant or
trigonometric functions. You may want to review conserved. energy of the bent pole.
these in Appendix I.
142 5 WORK AND ENERGY
This potential energy is used to lift the vaulter in doing work against gravity, and is
partially converted into gravitational potential energy. At the top, there is just
enough kinetic energy left to carry the vaulter over the bar. On the way down, the
gravitational potential energy is converted back to kinetic energy, which is
absorbed by the mat in doing work to stop the fall. The pole vaulter participates in
a game of work–energy, a game of give and take.
This chapter centers on two concepts that are important in both science and
everyday life—work and energy. We commonly think of work as being associated
with doing or accomplishing something. Because work makes us physically (and
sometimes mentally) tired, machines have been invented to decrease the amount
of effort expended personally. Thinking about energy tends to bring to mind the
cost of fuel for transportation and heating, or perhaps the food that supplies the
energy our bodies need to sustain life processes and to do work.
Although these notions do not really define work and energy, they point in the
right direction. As you may have surmised, work and energy are closely related. In
physics, as in everyday life, when something possesses energy, it has the ability to
do work. For example, water rushing through the sluices of a dam has energy of
motion, and this energy allows the water to do the work of driving a turbine or
dynamo to generate electricity. Conversely, no work can be performed without
energy.
Energy exists in various forms: mechanical energy, chemical energy, electrical
energy, heat energy, nuclear energy, and so on. A transformation from one form to
another may take place, but the total amount of energy is conserved, meaning
there is always the same amount. This point makes the concept of energy very
useful. When a physically measurable quantity is conserved, it not only gives us an
insight that leads to a better understanding of nature, but also usually provides
another approach to practical problems. (You will be introduced to other con-
served quantities and conservation laws during the course of our study of physics.)
5.1 Work Done by a Constant Force

➥ What is the work done by a constant force?

➥ How does negative work arise?
➥ What is meant by total, or net, work?
The word work is commonly used in a variety of ways: We go to work; work on

projects; work at our desks or on computers; work on problems. In physics, how-
ever, work has a very specific meaning. Mechanically, work involves force and dis-
placement, and the word work is used to describe quantitatively what is
accomplished when a force acts on an object as it moves through a distance. In the
simplest case of a constant force acting on an object, work that the force does is
defined as follows:
The work done by a constant force acting on an object is equal to the product of the
magnitudes of the displacement and the force, or component of the force, parallel to
that displacement.
5.1 WORK DONE BY A CONSTANT FORCE 143
F
F
F F cos u
=
u
F
F F sin u
u
d
d0
d
(a) (b) (c)
䉱 F I G U R E 5 . 1 Work done by a constant force—the product of the magnitudes of the

parallel component of force and the displacement (a) If there is no displacement, no work
is done: W = 0. (b) For a constant force in the same direction as the displacement, W = Fd.
(c) For a constant force at an angle to the displacement, W = 1F cos u2d.
Work then involves a force acting on an object and moving it through a distance. A
force may be applied, as in 䉱 Fig. 5.1a, but if there is no motion (no displacement), then
no work is done. However when there is motion, a constant force F acting in the same
direction as the displacement d does work (Fig. 5.1b). The work (W) done in this
case is defined as the product of their magnitudes:
W = Fd (5.1)
and work is a scalar quantity. (As you might expect, when work is done as in Fig.
5.1b, energy is expended. The relationship between work and energy is discussed
in Section 5.3.)
In general, work is done on an object by a force, or force component, parallel to
the line of motion or displacement of the object (Fig. 5.1c). That is, if the force acts
at an angle u to the object’s displacement, then F‘ = F cos u is the component of
the force parallel to the displacement. Thus, a more general equation for work
done by a constant force is*
W = F‘ d = 1F cos u2d (work done by a constant force) (5.2)
Notice that u is the angle between the force and the displacement vectors. As a
reminder of this factor, cos u may be written between the magnitudes of the force
and displacement, W = F1cos u2d. If u = 0° (that is, force and displacement are in
the same direction, as in Fig. 5.1b), then W = F1cos 0°2d = Fd, so Eq. 5.2 reduces
to Eq. 5.1. The perpendicular component of the force, F⬜ = F sin u, does no work,
since there is no displacement in this direction.
The units of work can be determined from the equation W = Fd. With force in
newtons and displacement in meters, work has the SI unit of newton-meter
1N # m2. This unit is called a joule (J):†
Fd = W
1N#m = 1J
For example, the work done by a force of 25 N on an object as the object moves
through a parallel displacement of 2.0 m is W = Fd = 125 N212.0 m2 = 50 N # m,
or 50 J.
*The product of two vectors (force and displacement) is a special type of vector multiplication and
yields a scalar quantity equal to (F cos u)d. Thus, work is a scalar—it does not have direction. It can,
however, be positive, zero, or negative, depending on the angle.
†
The joule (J), pronounced “jool,” was named in honor of James Prescott Joule (1818–1889), a British
scientist who investigated work and energy.
From the previous displayed equation, it can also be seen that in the British sys-
tem, work would have the unit pound-foot. However, this name is commonly
written in reverse. The British standard unit of work is the foot-pound (ft # lb).
One ft # lb is equal to 1.36 J.
LEARN BY DRAWING 5.1
Work can be analyzed graphically. Suppose a constant force F in the x-direction
work: area under the acts on an object as it moves a distance x. Then W = Fx and if F versus x is plotted,
a horizontal straight-line graph is obtained such as shown in the accompanying
F-versus-x cur ve Learn by Drawing 5.1, Work: Area under the F-versus-x Curve.
The area under the line is Fx, so this area is equal to the work done by the force
over the given distance. Work done by a nonconstant, or variable, force will be
considered later.*
Remember that work is a scalar quantity and may have a positive or negative
F value. In Fig. 5.1b, the work is positive, because the force acts in the same direction
as the displacement (and cos 0° is positive). The work is also positive in Fig. 5.1c,
because a force component acts in the direction of the displacement (and cos u is
Work positive).
W = Fx However, if the force, or a force component, acts in the opposite direction of the
displacement, the work is negative, since the cosine term is negative. For example,
for u = 180° (force opposite to the displacement), cos 180° = - 1, so the work is
x negative: W = F‘ d = F1cos 180°2d = - Fd. An example is a braking force that
slows down or decelerates an object. See the associated Learn by Drawing 5.2,
Determining the Sign of Work.
EXAMPLE 5.1 Applied Psychology: Mechanical Work

A student holds her 1.5-kg psychology textbook out a second-story dormitory window
until her arm is tired; then she releases it (䉴 Fig. 5.2). (a) How much work is done on the
book by the student in simply holding it out the window? (b) How much work is done by
the force of gravity during the time in which the book falls 3.0 m?
T H I N K I N G I T T H R O U G H . Analyze the situations in terms of the definition of work, keep-
ing in mind that force and displacement are the key factors.
Given: vo = 0 (initially at rest) Find: (a) W (work done by student in holding)
m = 1.5 kg (b) W (work done by gravity in falling)
d 3.0 m
d = 3.0 m
(a) Even though the student gets tired (because work is performed within the body to
maintain muscles in a state of tension), she does no work on the book in merely holding it
stationary. She exerts an upward force on the book (equal in magnitude to its weight), but
the displacement is zero in this case 1d = 02. Thus, W = Fd = F * 0 = 0 J.
(b) While the book is falling, the only force acting on it is the force of gravity (neglecting
air resistance), which is equal in magnitude to the weight of the book: F = w = mg. The
displacement is in the same direction as the force 1u = 0°2 and has a magnitude of
w
d = 3.0 m so the work done by gravity is
W = F1cos 0°2d = 1mg2d = 11.5 kg219.8 m>s 2213.0 m2 = + 44 J
( + because the force and displacement are in the same direction.)

F O L L O W - U P E X E R C I S E . A 0.20-kg ball is thrown upward. How much work is done on
the ball by gravity as the ball rises between heights of 2.0 m and 3.0 m? (Answers to all
Follow-Up Exercises are given in Appendix VI at the back of the book.) 䉱 F I G U R E 5 . 2 Mechanical work
requires motion See Example text
for description.
*Work is the area under the F-versus-x curve even if the curve is not a straight line. Finding the
work in such cases generally requires advanced mathematics.
5.1 WORK DONE BY A CONSTANT FORCE 145
EXAMPLE 5.2 Hard Work

A worker pulls a 40.0-kg crate with a rope, as illustrated in 䉲 Fig. 5.3. The coefficient of
kinetic (sliding) friction between the crate and the floor is 0.550. If he moves the crate LEARN BY DRAWING 5.2
with a constant velocity a distance of 7.00 m, how much work is done?
THINKING IT THROUGH. A good thing to do first in problems such as this is to draw a
determining the sign
free-body diagram. This is shown in the figure. To find the work, the force F must be of work
known. As usual in such cases, this is done by summing the forces.
Given: m = 40.0 kg Find: W (work done in moving the crate

mk = 0.550 7.00 m)
d = 7.00 m
u = 30° (from figure)
v (constant)
␪ = 0°
SOLUTION. Then, summing the forces in the x- and y-directions and setting these
equal to zero (with a constant velocity Fnet = 0):
a Fx = F cos 30° - fk = F cos 30° - mk N = max = 0 ␪ < 90° ␪
a Fy = N + F sin 30° - mg = may = 0
To find F, the second equation may be solved for N, which is then substituted in the first
equation.
N = mg - F sin 30° ␪ = 90°
␪
(Notice that N is not equal to the weight of the crate. Why?) And, substituting N into
the first equation,
F cos 30° - mk1mg - F sin 30°2 = 0
Solving for F and putting in values:
mk mg 10.5502140.0 kg219.80 m>s 22
␪
1cos 30° + mk sin 30°2 10.8662 + 10.550210.50024
F = = = 189 N
␪ > 90°
Then,
W = F1cos 30°2d = 1189 N210.866217.00 m2 = 1.15 * 103 J
FOLLOW-UP EXERCISE. It takes about 3.80 * 104 J of work to lose 1.00 g of body fat.
␪ = 180°
What distance would the worker have to pull the crate to lose 1 g of fat? (Assume all the
work goes into fat reduction.) Make an estimate before solving and see how close you
come.
N
F
F sin 30
30 x
30
fk F cos 30
N w mg
mg
fk Free-body diagram
䉱 F I G U R E 5 . 3 Doing some work See Example 5.2.

It is commonly said that a force does work on an object. For example, the force
of gravity does work on a falling object, such as the book in Example 5.1. Also,
when you lift an object, you do work on the object. This is sometimes described as
doing work against gravity, because the force of gravity acts in the direction oppo-
site that of the applied lift force and opposes it. For example, an average-sized
apple has a weight of about 1 N. So, when lifting such an apple a distance of 1 m
with a force equal to its weight, 1 J of work is done against gravity
3W = Fd = 11 N211 m2 = 1 J4. This gives an idea of how much work 1 J represents.
In both Examples 5.1 and 5.2, work was done by a single constant force. If more
than one force acts on an object, the work done by each can be calculated sepa-
rately. That is:
The total, or net, work is defined as the work done by all the forces acting on an object,
or the scalar sum of the work done by each force.
This concept is illustrated in Example 5.3.
EXAMPLE 5.3 Total or Net Work

A 0.75-kg block slides with a uniform velocity down a 20° T H I N K I N G I T T H R O U G H . (a) The length of the plane can be
inclined plane (䉲 Fig. 5.4). (a) How much work is done by the found using trigonometry, so this part boils down to finding the
force of friction on the block as it slides the total length of the force of friction. (b) The net work is the sum of all the work done
plane? (b) What is the net work done on the block? (c) Discuss by the individual forces. (Note: Since the block has a uniform, or
the net work done if the angle of incline is adjusted so that the constant, velocity, the net force on it is zero. This observation
block accelerates down the plane. should tell you the answer, but it will be shown explicitly in the
solution.) (c) If there is acceleration, Newton’s second law
applies, which involves a net force, so there may be net work.
䉴 F I G U R E 5 . 4 Total or
net work See Example text y
for description.
d v N
fk x
N fk
mg sin u
v
u mg cos u
mg mg
u = 20°
Free-body diagram
L = 1.2 m
SOLUTION. Listing the data given, and specifically what is to be found:

Given: m = 0.75 kg Find: (a) Wf (work done on the block by friction)
u = 20° (b) Wnet (net work on the block)
L = 1.2 m (from Fig. 5.4) (c) W (discuss net work with block accelerating)
(a) Note from the Fig. 5.4 free-body diagram that only two The angle 180° indicates that the force and displacement are
forces do work, because there are only two forces parallel to the in opposite directions. (It is common in such cases to write
motion: fk , the force of kinetic friction, and mg sin u, the compo- Wf = - fk d directly, since kinetic friction typically opposes
nent of the block’s weight acting down the plane. The normal motion.) The distance d the block slides down the plane can
force N and mg cos u, the component of the block’s weight, act be found by using trigonometry. Note that cos u = L>d ,so
perpendicular to the plane and do no work. (Why?)
First finding the work done by the frictional force: L
d =
cos u
Wf = fk1cos 180°2d = - fk d = - mkNd
5.2 WORK DONE BY A VARIABLE FORCE 147
We know that N = mg cos u, but what is mk? It would appear where the calculation is the same as in part (a) except for the
that some information is lacking. When this situation occurs, sign. Then, the net work is
Wnet = Wg + Wf = + 3.2 J + 1 - 3.2 J2 = 0
look for another approach to solve the problem. As noted ear-
lier, there are only two forces parallel to the motion, and they
are opposite, so with a constant velocity their magnitudes are (constant velocity, zero net force, zero net work). Remember
equal, fk = mg sin u. Thus, that work is a scalar quantity, so scalar addition is used to find
net work.
Wf = - fk d = - 1mg sin u2 a b = - mg L tan 20°
L
cos u (c) If the block accelerates down the plane, then from New-
= - 10.75 kg219.8 m>s 2211.2 m210.3642 = - 3.2 J ton’s second law, Fnet = mg sin u - fk = ma. The component
of the gravitational force 1mg sin u2 is greater than the oppos-
(b) To find the net work, the work done by gravity needs to be ing frictional force ( fk), so net work is done on the block,
calculated and then added to the result in part (a). Since F‘ for because now ƒ Wg ƒ 7 ƒ Wf ƒ . You may be wondering what the
gravity is just mg sin u, effect of nonzero net work is. As will be shown shortly,
nonzero net work causes a change in the amount of kinetic
Wg = F‘ d = 1mg sin u2a b = mgL tan 20° = + 3.2 J
L
energy an object has.
cos u
F O L L O W - U P E X E R C I S E . In part (c) of this Example, is it possible for the frictional work to be greater in magnitude than the gravi-
tational work? What would this condition mean in terms of the block’s speed?
Note that in part (a) of Example 5.3, the equation for Wf was simplified by using alge-
braic expressions for N and d instead of by computing these quantities initially. It is a
good rule of thumb not to plug numbers into an equation until you have to. Simplifying
an equation through cancellation is easier with symbols and saves computation time.
DID YOU LEARN?

➥ The product of the magnitudes of the displacement and the force, or component of
force, parallel to the displacement gives the work done by a constant force.
➥ If a force or a force component acts in the opposite direction of the displacement,
the work done by the force is negative.
➥ The work done by all the forces acting on an object, or the scalar sum of all the
work, gives the total, or net, work.
5.2 Work Done by a Variable Force

➥ What is meant by “a spring force is a function of position”?

➥ If a spring, or force, constant of one spring is greater than that of another, what does
this imply?
The discussion in the preceding section was limited to work done by constant
forces. In general, however, forces are variable; that is, they change in magnitude
and>or angle with time and>or position.
An example of a variable force that does work is illustrated in 䉲 Fig. 5.5, which
depicts a spring being stretched by an applied force Fa. As the spring is stretched
(or compressed) farther and farther, its restoring force (the spring force that
opposes the stretching or compression) becomes greater, and an increased applied
force is required. For most springs, the spring force ( Fs) is directly proportional to
the change in length of the spring from its unstretched length. In equation form,
this relationship is expressed
Fs = - k¢x = - k1x - xo2
or, if xo = 0,
Fs = - kx (ideal spring force) (5.3)

䉴 F I G U R E 5 . 5 Spring force
(a) An applied force Fa stretches the Unstretched
spring, and the spring exerts an
equal and opposite force Fs on the
hand. (b) The magnitude of the
force depends on the change ¢x in
the spring’s length. This change is
measured from to the end of the
unstretched spring at xo. Fs Fa
Spring Applied
force force
xo
(a) Fs k∆x k (x xo)
Fs kx with xo 0
Fs Fa
xo x
∆x x xo
(b)
where x now represents the distance the spring is stretched (or compressed) from
its unstretched length. As can be seen, the force varies with x. This is described by
saying that the force is a function of position.
The k in this equation is a constant of proportionality and is commonly called
the spring constant, or force constant. The greater the value of k, the stiffer or
stronger the spring. As you should be able to prove to yourself, the SI unit of k is
newtons per meter (N>m). The minus sign in Eq. 5.3 indicates that the spring force
acts in the direction opposite to the displacement when the spring is either
stretched or compressed. Equation 5.3 is a form of what is known as Hooke’s law,
named after Robert Hooke, a contemporary of Newton.
The relationship expressed by the spring force equation holds only for ideal
springs. Real springs approximate this linear relationship between force and dis-
placement within certain limits. If a spring is stretched beyond a certain point,
called its elastic limit, the spring will be permanently deformed, and the linear rela-
tionship will no longer apply.
F Computing the work done by variable forces generally requires calculus. But it
is fortunate that the spring force is a special case that can be computed graphically.
F A plot of F (the applied force) versus x is shown in 䉳 Fig. 5.6. The graph has a
F = kx straight-line slope of k, with F = kx, where F is the applied force doing work in
= k stretching the spring.
pe
Slo As described earlier, work is the area under an F-versus-x curve, and here it is
Area = W
0 in the form of a triangle, as indicated by the shaded area in the figure. Then, com-
x puting this area,
䉱 F I G U R E 5 . 6 Work done by a uni- area = W = 12 1altitude * base2
formly variable spring force A graph
of F versus x, where F is the applied or
W = 12 Fx = 12 1kx2x = 12 kx 2
force doing work in stretching a
spring, is a straight line with a slope
of k. The work is equal to the area where F = kx. Thus,
under the line, which is that of a trian-
gle with area = 12 1altitude * base2.
Then W = 12 Fx = 12 1kx2x = 12 kx 2. (work done in stretching or compressing
W = 12 kx 2 (5.4)
a spring from xo = 0)
5.2 WORK DONE BY A VARIABLE FORCE 149
EXAMPLE 5.4 Determining the Spring Constant

A 0.15-kg mass is attached to a vertical spring and hangs at rest a distance of 4.6 cm
below its original position (䉴 Fig. 5.7). An additional 0.50-kg mass is then suspended from
the first mass and the system is allowed to descend to a new equilibrium position. What
is the total extension of the spring? (Neglect the mass of the spring.)
T H I N K I N G I T T H R O U G H . The spring constant k appears in Eq. 5.3. Therefore, to find the
value of k for a particular instance, the spring force and distance the spring is stretched
(or compressed) must be known. xo 0
SOLUTION. The data given are as follows: x1
Given: m1 = 0.15 kg Find: x (total stretch distance)
Fs
x1 = 4.6 cm = 0.046 m
m1
m2 = 0.50 kg
The total stretch distance is given by x = F>k, where F is the applied force, which in this F m1g x
case is the weight of the mass suspended on the spring. (The minus sign in Eq. 5.3 is
(a)
ignored here for convenience.) However, the spring constant k is not given. But, k may be
found from the data pertaining to the suspension of m1 and resulting displacement x1.
Fs
(This method is commonly used to determine spring constants.) As seen in Fig. 5.7a, the
magnitudes of the weight force and the restoring spring force are equal, since a = 0, their
magnitudes may be equated: m1
Fs = kx1 = m1 g
m2
Solving for k,
m1 g 10.15 kg219.8 m>s 22
k = = = 32 N>m
x1 0.046 m
F (m1 m2)g
Then, knowing k, the total extension of the spring can be found from the balanced-force
situation shown in Fig. 5.7b: (b)
Fs = 1m1 + m22g = kx 䉱 F I G U R E 5 . 7 Determining

the spring constant and the work
Thus, done in stretching a spring See
1m1 + m22g 10.15 kg + 0.50 kg219.8 m>s22 Example text for description.
x = = = 0.20 m 1or 20 cm2
k 32 N>m
F O L L O W - U P E X E R C I S E . How much work is done by gravity in stretching the spring

through both displacements in Example 5.4?
The reference position xo used to determine the change in length of a spring is arbitrary
but is usually chosen as xo = 0 for convenience. The important quantity in computing work
is the difference in position, ¢x, or the net change in the length of the spring from its unstretched
length. As shown in 䉲 Fig. 5.8 for a mass suspended on a spring, xo can be referenced to
the unloaded length of the spring or to the loaded position, which may be taken as the
zero position for convenience. In Example 5.4, xo was referenced to the end of the
unloaded spring.
When the net force on the suspended mass is zero, the mass is said to be at its equilibrium
position (as in Fig. 5.7a with m1 suspended). This position, rather than the unloaded length,
may be taken as a zero reference (xo = 0; see Fig. 5.8b). The equilibrium position is a conve-
nient reference point for cases in which the mass oscillates up and down on the spring.
Also, since the displacement is in the vertical direction, the x’s are often replaced by y’s.
DID YOU LEARN?

➥ The spring force depends on the length of the spring from its unstretched position,
for either an extension or a compression.
➥ A spring with a greater spring constant would apply a greater spring force, or is a
stiffer spring.
䉴 F I G U R E 5 . 8 Displacement ref-
erence The reference position xo is
arbitrary and is usually chosen for
convenience. It may be (a) at the end
of the spring at its unloaded posi-
tion or (b) at the equilibrium posi-
tion when a mass is suspended on xo
the spring. The latter is particularly
convenient in cases in which the
mass oscillates up and down on the ∆x +x
spring.
Fs Fs
x
m m Equilibrium
xo = 0
position
mg mg –x
(a) (b)
5.3 The Work–Energy Theorem: Kinetic Energy

➥ Why is kinetic energy called “the energy of motion”?

➥ How does the work–energy theorem relate work and energy?
➥ How is a change in kinetic energy computed?
Now that we have an operational definition of work, let’s take a look at how work
is related to energy. Energy is one of the most important concepts in science. It is
described as something that objects or systems possess. Basically, work is some-
thing that is done on objects, whereas energy is something that objects have, which
is the ability to do work.
One form of energy that is closely associated with work is kinetic energy.
(Another basic form of energy, potential energy, will be discussed in Section 5.4.)
Consider an object at rest on a frictionless surface. Let a horizontal force act on the
object and set it in motion. Work is done on the object, but where does the work
“go,” so to speak? It goes into setting the object into motion, or changing its kinetic
conditions. Because of its motion, we say the object has gained energy—kinetic
energy, which gives it the capability to do work.
For a constant force doing work on a moving object parallel to the direction of
motion, as illustrated in 䉲 Fig. 5.9, the force does an amount of work W = Fx. But
what are the kinematic effects? The force gives the object a constant acceleration,
and from Eq. 2.12, v2 = v 2o + 2ax (with xo = 0),
v 2 - v2o
a =
2x
W = K – Ko = ∆ K
Ko = 1 mvo2 K = 1 mv2
2 2
vo v
䉴 F I G U R E 5 . 9 The relation- F m F m
ship of work and kinetic energy
The work done on a block by a (Frictionless)
constant force in moving it
along a horizontal frictionless
surface is equal to the change x
in the block’s kinetic energy:
W = ¢K. W = Fx
5.3 THE WORK–ENERGY THEOREM: KINETIC ENERGY 151
where vo may or may not be zero. Writing the magnitude of the force in the form
of Newton’s second law and substituting in the expression for a from the previous
equation gives
v2 - v 2o
F = ma = ma b
2x
Using this expression in the equation for work,
v 2 - v2o
W = Fx = ma bx
2x
= 12 mv2 - 12 mv 2o
The term 12 mv 2 is defined as the kinetic energy (K) of the moving object:
K = 12 mv 2 (kinetic energy) (5.5)
SI unit of energy: joule (J)

Kinetic energy is often called the energy of motion. Note that it is directly propor-
tional to the square of the (instantaneous) speed of a moving object, and therefore
cannot be negative.
Then, in terms of kinetic energy, the previous expression for work may be writ-
ten as
W = 12 mv 2 - 12 mv 2o = K - Ko = ¢K
or
W = ¢K (5.6)
where it is understood that W is the net work if more than one force acts on the object,
as shown in Example 5.3. This equation is called the work–energy theorem, and it
relates the work done on an object to the change in the object’s kinetic energy. That
is, the net work done on a body by all the forces acting on it is equal to the change in kinetic
energy of the body. Both work and energy have units of joules, and both are scalar
quantities. The work–energy theorem is true in general for variable forces and not
just for the special case considered in deriving Eq. 5.6.
To illustrate that net work is equal to the change in kinetic energy, recall that in
Example 5.1 the force of gravity did +44 J of work on a book that fell from rest
through a distance of y = 3.0 m. At that position and instant, the falling book had
44 J of kinetic energy. Since vo = 0 in this case, 12 mv 2 = mgy. Substituting this
expression into the equation for the work done on the falling book by gravity,
mv2
W = Fd = mgy = = K = ¢K
2
where Ko = 0. Thus the kinetic energy gained by the book is equal to the net work
done on it: 44 J in this case. (As an exercise, confirm this fact by calculating the
speed of the book and computing its kinetic energy.)
The work–energy theorem tells us that when work is done on an object, there is
a change in or a transfer of energy. In general, then, it might be said that work is a
measure of the transfer of kinetic energy to the object. For example, a force doing work
on an object that causes the object to speed up gives rise to an increase in the
object’s kinetic energy. Conversely, (negative) work done by the force of kinetic
friction may cause a moving object to slow down and decrease its kinetic energy.
So for an object to have a change in its kinetic energy, net work must be done on
the object, as Eq. 5.6 indicates.
When an object is in motion, it possesses kinetic energy and thus has the capabil- 䉱 F I G U R E 5 . 1 0 Kinetic energy
and work A moving object, such as a
ity to do work. For example, a moving automobile has kinetic energy and can do wrecking ball, processes kinetic
work in crumpling a fender in a fenderbender—not useful work in that case, but still energy and can do work. A massive
work. Another example of work done by kinetic energy is shown in 䉴 Fig. 5.10. ball is used in demolishing buildings.
EXAMPLE 5.5 A Game of Shuffleboard: The Work–Energy Theorem

A shuffleboard player (䉲 Fig. 5.11) pushes a 0.25-kg puck that T H I N K I N G I T T H R O U G H . Apply the work–energy theorem. If
is initially at rest such that a constant horizontal force of 6.0 N the amount of work done can be found, then this gives the
acts on it through a distance of 0.50 m. (Neglect friction.) change in kinetic energy.
(a) What are the kinetic energy and the speed of the puck
when the force is removed? (b) How much work would be
required to bring the puck to rest?
vo 0
F v
䉱 F I G U R E 5 . 1 1 Work and kinetic energy See Example text for description.
SOLUTION. Listing the given data as usual, The speed can be found from the kinetic energy.
Since K = 12 mv2,
Given: m = 0.25 kg Find: (a) K (kinetic energy)
F = 6.0 N v (speed) 2K 213.0 J2
v = = = 4.9 m>s
d = 0.50 m (b) W (work done in A m B 0.25 kg
vo = 0 stopping puck)
(b) As you might guess, the work required to bring the puck
(a) Since the speed is not known, the kinetic energy to rest is equal to the puck’s kinetic energy (that is, the
1K = 12 mv22 cannot be computed directly. However, kinetic amount of energy that the puck must lose to come to a stop).
energy is related to work by the work–energy theorem. The To confirm this equality, the previous calculation is essentially
work done on the puck by the player’s applied force F is performed in reverse, with vo = 4.9 m>s and v = 0:
W = Fd = 16.0 N210.50 m2 = 3.0 J W = K - Ko = 0 - Ko = - 12 mv2o = - 12 10.25kg214.9 m>s22
= - 3.0 J
Then, by the work–energy theorem,
The minus sign indicates that the puck loses energy as it
W = ¢K = K - Ko = 3.0 J slows down. The work is done against the motion of the puck;
But Ko = 12 mv2o = 0, because vo = 0, so that is, the opposing force is in a direction opposite that of the
motion. (In the real-life situation, the opposing force could be
K = 3.0 J friction.)
F O L L O W - U P E X E R C I S E . Suppose the puck in this Example had twice the final speed when released. Would it then take twice as
much work to stop the puck? Justify your answer numerically.
Notice how work–energy considerations were used to find speed in Example 5.5. This
operation can be done in another way as well. First, the acceleration could be found from
a = F>m, and then the kinematic equation v2 = v2o + 2ax could be used to find v (where
x = d = 0.50 m). The point is that many problems can be solved in different ways, and
finding the fastest and most efficient way is often the key to success. As our discussion of
energy progresses, it will be seen how useful and powerful the notions of work and energy
are, both as theoretical concepts and as practical tools for solving many kinds of problems.
5.3 THE WORK–ENERGY THEOREM: KINETIC ENERGY 153
CONCEPTUAL EXAMPLE 5.6 Kinetic Energy: Mass versus Speed

In a football game, a 140-kg guard runs at a speed of 4.0 m>s, However, doubling the speed increases the kinetic
and a 70-kg free safety moves at 8.0 m>s. Which of the follow- energy, not by a factor of 2 but by a factor of 22, or 4. Thus,
ing is a correct statement? (a) The players have the same the safety, with half the mass but twice the speed, would
kinetic energy. (b) The safety has twice as much kinetic energy have 12 * 4 = 2 times as much kinetic energy as the guard,
as the guard. (c) The guard has twice as much kinetic energy and so the answer is (b).
as the safety. (d) The safety has four times as much kinetic Note that to answer this question, it was not necessary to
energy as the guard. calculate the kinetic energy of each player. But this can be
done to verify the answer:
REASONING AND ANSWER. The kinetic energy of a body
depends on both its mass and speed. You might think that, Ksafety = 12 ms v 2s = 12 170 kg218.0 m>s22 = 2.2 * 103 J
Kguard = 12 mg v2g = 12 1140 kg214.0 m>s22 = 1.1 * 103 J
with half the mass but twice the speed, the safety would have
the same kinetic energy as the guard, but this is not the case.
As observed from the relationship K = 12 mv2, kinetic energy
which explicitly shows the answer to be correct.
is directly proportional to the mass, but is proportional to the
square of the speed. Thus, having half the mass decreases the F O L L O W - U P E X E R C I S E . Suppose that the safety’s speed
kinetic energy by a factor of 2. So if the two athletes had equal were only 50% greater than the guard’s, or 6.0 m>s. Which
speeds, the safety would have half as much kinetic energy as athlete would then have the greater kinetic energy, and how
the guard. much greater?
Note that the work–energy theorem relates the work done to the change in the kinetic
energy. Often, vo = 0 and Ko = 0, so W = ¢K = K. But take care! You cannot simply use
the square of the change or difference in speed, 1v - vo22 = 1¢v22, to calculate ¢K, as
you might at first think. In terms of speed,
W = ¢K = K - Ko = 12 mv2 - 12 mv2o = 12 m1v2 - v2o2

Note that 1v 2 - v 2o2 is not the same as 1v - vo22 = 1¢v22 because 1v - vo22
= v2 - 2vvo + v 2o . Hence, the change in kinetic energy is not equal to 12 m1v - vo22
= 12 m1¢v22 Z ¢K .
This observation means that to calculate work, or the change in kinetic energy, you
must compute the kinetic energy of an object at one point or time (using the instanta-
neous speed to get the instantaneous kinetic energy) and also at another location or
time. Then subtract the quantities to find the change in kinetic energy, or the work.
Alternatively, you can find the difference of the squares of the speeds 1v 2 - v2o2 first in
computing the change, but remember never to use the square of the difference of the
speeds. To see this hint in action, look at Conceptual Example 5.7.
CONCEPTUAL EXAMPLE 5.7 An Accelerating Car: Speed and Kinetic Energy

A car traveling at 5.0 m>s speeds up to 10 m>s, with an So the greater the speed of an object, the greater its kinetic
increase in kinetic energy that requires work W1. Then the energy. The difference in kinetic energy in changing speeds (or
car’s speed increases from 10 m>s to 15 m>s, requiring addi- the work required to change speed) would then be greater for
tional work W2. Which of the following relationships accu- higher speeds for the same ¢v. Therefore, (c) is the answer.
rately compares the two amounts of work: (a) W1 7 W2 , The main point is that the ¢v values are the same, but
(b) W1 = W2 , or (c) W2 7 W1? more work is required to increase the kinetic energy of an
object at higher speeds.
REASONING AND ANSWER. As noted previously, the
work–energy theorem relates the work done on the car to the F O L L O W - U P E X E R C I S E . Suppose the car speeds up a third
change in its kinetic energy. Since the speeds have the same time, from 15 m>s to 20 m>s, a change requiring work W3.
increment in each case 1¢v = 5.0 m>s2, it might appear that How does the work done in this increment compare with W2?
(b) would be the answer. However, keep in mind that the Justify your answer numerically. [Hint: Use a ratio.]
work is equal to the change in kinetic energy and involves
v22 - v21 , not 1¢v22 = 1v2 - v122.
DID YOU LEARN?

➥ By definition, K = 12 mv2, so there must be motion (v Z 0) to have kinetic energy.
➥ The work–energy theorem relates the net work done on an object to its change in
kinetic energy, W = ¢K.
➥ K must be computed at different times, K1 = 12 mv 21 and K2 = 12 mv 22, to find the
change in kinetic energy, ¢K = K2 - K1.
5.4 Potential Energy

➥ How may an object’s potential energy be changed?

➥ How do different reference points affect the difference in the gravitational potential
energy of two positions?
An object in motion has kinetic energy. However, whether an object is in motion or

not, it may have another form of energy—potential energy. As the name implies,
an object having potential energy has the potential to do work. You can probably
think of many examples: a compressed spring, a drawn bow, and water held back
by a dam. In all such cases, the potential to do work derives from the position or
configuration of bodies. A spring has energy because it is compressed, a bow
because it is drawn, and the water because it has been lifted above the surface of
the Earth (䉳 Fig. 5.12). Consequently, potential energy (U), is often called the energy
of position (and>or configuration).
(a) Unlike kinetic energy, which is associated with motion, potential energy is a
form of mechanical energy associated with the position of an object within a sys-
tem (or configuration). Potential energy is a property of the system, rather than
the object. If the configuration of a system of objects changes, so does the potential
energy of a particular object within that system.
In a sense, potential energy can be thought of as stored work. You have already
seen an example of potential energy in Section 5.2 when work was done in stretch-
ing a spring from its equilibrium position. Recall that the work done in such a case
is W = 12 kx 2 (with xo = 0). Note that the amount of work done depends on the
amount of stretching (x). Because work is done, there is a change in the spring’s
potential energy (¢U), which is equal to the work done by the applied force in
stretching (or compressing) the spring:
W = ¢U = U - Uo = 12 kx 2 - 12 kx 2o
Thus, with xo = 0 and Uo = 0, as they are commonly taken for convenience, the
potential energy of a spring is
(b)
U = 12 kx 2 (potential energy of a spring) (5.7)
䉱 F I G U R E 5 . 1 2 Potential energy
Potential energy has many forms. SI unit of energy: joule (J)
(a) Work must be done to bend the
bow, giving it potential energy. That [Note: Since the potential energy varies as x2, the previous Problem-Solving Hint also
energy is converted into kinetic applies, and when xo Z 0, then x 2 - x 2o Z 1x - xo22. That is, the potential energy of
energy when the arrow is released. a spring must be calculated at different positions and then subtracted to find ¢U.]
(b) Gravitational potential energy is
converted into kinetic energy when Perhaps the most well known type of potential energy is gravitational potential
something falls. (Where did the energy. In this case, position refers to the height of an object above some reference
gravitational potential energy of the point, such as the floor or the ground. Suppose that an object of mass m is lifted a dis-
water and the diver come from?) tance ¢y (䉴Fig. 5.13). Work is done against the force of gravity, and an applied force at
least equal to the object’s weight is necessary to lift the object: F = w = mg. The
work done in lifting is then equal to the change in potential energy. Expressing this
relationship in equation form, since there is no overall change in kinetic energy,
work done by external force = change in gravitational potential energy
or
W = F¢y = mg1y - yo2 = mgy - mgyo = ¢U = U - Uo
5.4 POTENTIAL ENERGY 155
F 䉳 F I G U R E 5 . 1 3 Gravitational
potential energy The work done in
lifting an object is equal to the
change in gravitational potential
energy: W = F¢y = mg1y - yo2.
m
y U = mgy
mg
∆y = y – y o W = U – Uo
= ∆U = mg ∆y
m
yo Uo = mgyo
where y is used as the vertical coordinate. With the common choice of yo = 0, such
that Uo = 0, the gravitational potential energy is
U = mgy (5.8)
SI unit of energy: joule (J)
(Eq. 5.8 represents the gravitational potential energy on or near the Earth’s sur-
face, where g is considered to be constant. A more general form of gravitational
potential energy will be given in Chapter 7.5.)
EXAMPLE 5.8 More Energy Needed

To walk 1000 m on level ground, a 60-kg person requires an expenditure of about
1.0 * 105 J of energy. What is the total energy required if the walk is extended another
1000 m along a 5.0° incline as shown in 䉴 Fig. 5.14? (Neglect any frictional changes.)
THINKING IT THROUGH. To walk an additional 1000 m would require the 1.0 * 105 J plus
the additional energy for doing work against gravity in walking up the incline. From the 1000 m
figure, the increase in height can be seen to be h = d sin u, where d is 1000 m.
S O L U T I O N . Listing the given data:
Given: m = 60 kg Find: E (total expended energy)
Eo = 1.0 * 105 J 1for 1000 m2
u = 5.0°
d = 1000 m (for each part of the work)
1000 m
The additional expended energy in going up the incline is equal to gravitational potential
energy gained. So,
¢U = mgh = 160 kg219.8 m>s2211000 m2sin 5.0° = 5.1 * 104 J
Then, the total energy expended for the 2000-m walk is u h
䉱 F I G U R E 5 . 1 4 Adding poten-
Total E = 2Eo + ¢U = 211.0 * 105 J2 + 0.51 * 105 J = 2.5 * 105 J tial energy See Example text for
(continued on next page) description.
Notice that the value of ¢U was expressed as a multiple of 105 F O L L O W - U P E X E R C I S E . If the angle of incline were doubled
in the last equation so it could be added to the Eo term, and and the walk just up the incline is repeated, will the addi-
the result was rounded to two significant figures per the rules tional energy expended by the person in doing work against
given in Chapter 1.6. gravity be doubled? Justify your answer.
EXAMPLE 5.9 A Thrown Ball: Kinetic Energy and Gravitational Potential Energy
A 0.50-kg ball is thrown vertically upward with an initial velocity of 10 m>s (䉴 Fig. 5.15). v=0
(a) What is the change in the ball’s kinetic energy between the starting point and the y = ymax
ball’s maximum height? (b) What is the change in the ball’s potential energy between
the starting point and the ball’s maximum height? (Neglect air resistance.)
T H I N K I N G I T T H R O U G H . Kinetic energy is lost and gravitational potential energy is
gained as the ball travels upward. ∆U = mgymax
SOLUTION. Studying Fig. 5.15 and listing the given data,
Given: m = 0.50 kg Find: (a) ¢K (the change in kinetic energy between yo and ymax) vo
vo = 10 m>s (b) ¢U (the change in potential energy between yo +y
a = g and ymax) y=0
g –y
(a) To find the change in kinetic energy, the kinetic energy is computed at each point.
The initial velocity is vo and at the maximum height v = 0, so K = 0. Thus,
¢K = K - Ko = 0 - Ko = - 12 mv2o = - 12 10.50 kg2110 m>s22 = - 25 J
That is, the ball loses 25 J of kinetic energy as negative work is done on it by the force of
gravity. (The gravitational force and the ball’s displacement are in opposite directions.)
(b) To find the change in potential energy, we need to know the ball’s height above its starting
point when v = 0. Using Eq. 2.11¿, v 2 = v2o - 2 gy (with yo = 0 and v = 0), to find ymax ,
䉱 F I G U R E 5 . 1 5 Kinetic and potential
v2o 110 m>s22 energies (The ball is displaced side-
ymax = = = 5.1 m
2g 219.8 m>s2) ways for clarity.) See Example text for
description.
Then, with yo = 0 and Uo = 0
¢U = U = mgymax = 10.50 kg219.8 m>s 2215.1 m2 = + 25 J
The potential energy increases by 25 J, as might be expected. This is an example of the
conservation of energy, as will be discussed shortly.
F O L L O W - U P E X E R C I S E . In this Example, what are the overall changes in the ball’s kinetic and potential energies when the ball
returns to the starting point?
ZERO REFERENCE POINT

An important point is illustrated in Example 5.9, namely, the choice of a zero refer-
ence point. Potential energy is the energy of position, and the potential energy at a
particular position (U) is meaningful only when referenced to the potential energy
at some other position (Uo). The reference position or point is arbitrary, as is the
origin of a set of coordinate axes for analyzing a system. Reference points are usu-
ally chosen with convenience in mind—for example, yo = 0. The value of the
potential energy at a particular position depends on the reference point used.
However, the difference, or change, in potential energy associated with two positions is
the same regardless of the reference position.
If, in Example 5.9, ground level had been taken as the zero reference point, then
Uo at the release point would not have been zero. However, U at the maximum
height would have been greater, and ¢U = U - Uo would have been the same.
This concept is illustrated in 䉴 Fig. 5.16. Note in Fig. 5.16a that the potential energy
can be negative. When an object has a negative potential energy, it is said to be in a
potential energy well, which is analogous to being in an actual well. Work is
needed to raise the object to a higher position in the well or to get it out of the well.
5.5 CONSERVATION OF ENERGY 157
U2 = mgy U2 = 2mgy
m m
y y
m Uo = mgyo = 0 U1 = mgy m
yo = 0 ∆U 2y
Potential
–y y
energy well
m m
yo = 0
U1 = –mgy Uo = mgyo = 0
(a) ∆U = U2 – U1 = mgy – (– mgy) = 2mgy (b) ∆U = U2 – Uo = 2mgy – 0 = 2mgy
䉱 F I G U R E 5 . 1 6 Reference point and change in potential energy (a) The choice of a reference point (zero height)
is arbitrary and may give rise to a negative potential energy. An object is said to be in a potential energy well in this
case. (b) The well may be avoided by selecting a new zero reference. Note that the difference, or change, in poten-
tial energy (¢U) associated with the two positions is the same, regardless of the reference point. There is no physi-
cal difference, even though there are two coordinate systems and two different zero reference points.
It is also said that gravitational potential energy is independent of path. This

means that only the change in height ¢h (or ¢y) is the consideration, not the path
that leads to the change in height. An object could travel many paths leading to
the same ¢h.
DID YOU LEARN?

➥ A change in position (or configuration) may result in a change in potential energy
for an object.
➥ The difference, or change, in potential energy associated with two positions is the
same irrespective of reference points.
5.5 Conser vation of Energy

➥ What is meant by “the total energy of the universe is conserved”?

➥ When is the total mechanical energy conserved?
➥ How does work done by nonconservative force affect the mechanical energy of a
system?
Conservation laws are the cornerstones of physics, both theoretically and practi-
cally. Most scientists would probably name conservation of energy as the most
profound and far-reaching of these important laws. Saying that a physical quan-
tity is conserved means it is constant, or has a constant value. Because so many
things continually change in physical processes, conserved quantities are
extremely helpful in our attempts to understand and describe a situation. Keep in
mind, though, that many quantities are conserved only under special conditions.
One of the most important conservation laws is that concerning conservation of
energy. (You have seen this topic in Example 5.9.) A familiar statement is that the
total energy of the universe is conserved. This statement is true, because the whole uni-
verse is taken to be a system. A system is defined as a definite quantity of matter
enclosed by boundaries, either real or imaginary. In effect, the universe is the largest
possible closed, or isolated, system we can imagine. On a smaller scale, a classroom
might be considered a system, and so might an arbitrary cubic meter of air.
INSIGHT 5.1 People Power: Using Body Energy

The human body is energy inefficient. That is, a lot of energy turns a gear connected to a simple magnetic coil generator, simi-
doesn’t go into doing useful work and is wasted. It would be lar to those used in flashlights that are energized by a rhythmic
advantageous to convert some of this energy into useful work. shaking (see Chapter 20 Insight 20.1, Electromagnetic Induction
Attempts are being made to do this through “energy harvest- at Work; Flashlights and Antiterrorism). The body’s mechanical
ing” from the human body. Normal body activities produce energy with this device can generate up to 7 watts of electrical
motion, flexing and stretching, compression, and body heat— energy. A typical cell phone operates on about 1 watt. (The watt
this is energy there for the taking. Harvesting the energy is a is a unit of power, J>s, energy>second; see Section 5.6).
difficult job, but using advances in nanotechnology (Chapter Who knows what the future of technology may hold?
1.3) and materials science, the effort is being made. Reflect on how many advances have occurred in your own
One older example of using body energy is the self-winding lifetime.
wristwatch, which is wound mechanically from the wearer’s
wrist movements. (Today, batteries have all but taken over.) An
ultimate goal in “energy harvesting” is to convert some of the
body’s energy into electricity—even if only a small amount.
How might this be done? Here are a couple of ways:
■ Piezoelectric devices. When mechanically stressed, piezo-
electric substances, like some ceramics, can generate
electrical energy.
■ Thermoelectric materials, which convert heat resulting
from some temperature difference into electrical energy.
These methods have severe limitations and produce only small
amounts of electricity. But with miniaturization and nanotech-
nology, the results could be practical. Researchers have already
F I G U R E 1 Backpack generator The backpack frame straps to
developed boots that use the compression of walking on a com-
the body and the load is suspended from springs. The load
pound to produce enough energy to power a radio. moves up and down on the springs when the wearer walks and
A more recent application is the “backpack generator.” The the toothed bar turns the gear on the electrical generator. With
mounted backpack’s load is suspended by springs (䉴 Fig. 1). good strides, more than 15 watts of power is generated. That’s
The up-and-down hip motion of a person wearing the backpack enough to power a GPS (Global Positioning System) locator and
makes the suspended load bounce up and down. This motion a head-lamp.
Within a closed system, particles can interact with each other, but have absolutely
no interaction with anything outside. In general, then, the amount of energy in a sys-
tem remains constant when no work is done on or by the system, and no energy is
transferred to or from the system (including thermal energy and radiation).
Thus, the law of conservation of total energy may be stated as follows:
The total energy of an isolated system is always conserved.
Within such a system, energy may be converted from one form to another, but the
total amount of all forms of energy is constant, or unchanged. Energy can never be
created or destroyed.
An application of energy in a nonconservative system is discussed in the
accompanying Insight 5.1, People Power: Using Body Energy.
CONCEPTUAL EXAMPLE 5.10 Violation of the Conservation of Energy?

A static, uniform liquid is in one side of a double container U = 1m>22g1h>42 + 1m>22g1h>42 = 21m>22g1h>42 = 1mg2h>4.
as shown in 䉴 Fig. 5.17a. If the valve is open, the level will Whoa. Was half of the energy lost?
fall, because the liquid has (gravitational) potential energy.
This may be computed by assuming all the mass of the liq- REASONING AND ANSWER. No; by the conservation of total
uid to be concentrated at its center of mass, which is at a energy, it must be around somewhere. Where might it have
height h/2. (More on the center of mass in Chapter 6.5.) gone? When the liquid flows from one container to the other,
When the valve is open, the liquid flows into the container because of internal friction and friction against the walls,
on the right, and when static equilibrium is reached, each half of the potential energy is first converted to kinetic
container has liquid to a height of h/2, with centers of mass energy (flow of liquid), then to heat (thermal energy), which
at h/4. This being the case, the potential energy of the liquid is transferred to the surroundings as the liquid comes to
before opening the valve was Uo = 1mg2h>2, and afterward, equilibrium. (This means a constant temperature and no
with half the total mass in each container (Fig. 5.16b), internal fluctuations.)
Closed Opened
valve valve
h
h/2 h/2
(a) (b)
䉱 F I G U R E 5 . 1 7 Is energy lost? See Conceptual Example text for description.
FOLLOW-UP EXERCISE. What would happen in this Example in the absence of friction?
CONSERVATIVE AND NONCONSERVATIVE FORCES

A general distinction can be made among systems by considering two categories
of forces that may act within or on them: conservative and nonconservative forces.
You have already been introduced to a couple of conservative forces: the force due
to gravity and the spring force. A classic nonconservative force, friction, was con-
sidered in Chapter 4.6.
A conservative force is defined as follows:
A force is said to be conservative if the work done by it in moving an object is
independent of the object’s path.
This definition means that the work done by a conservative force depends only on
the initial and final positions of an object.
The concept of conservative and nonconservative forces is sometimes difficult
to comprehend at first. Because this concept is so important in the conservation of
energy, let’s consider some illustrative examples to increase understanding.
First, what does independent of path mean? As an example of path independence,
consider picking an object up from the floor and placing it on a table. This is doing
work against the conservative force of gravity. The work done is equal to the poten-
tial energy gained, mg¢h, where ¢h is the vertical distance between the object’s
position on the floor and its position on the table. This is the important point. You
may have carried the object over to the sink before putting it on the table, or
walked around to the other side of the table. But only the vertical displacement
makes a difference in the work done because that is in the direction of the vertical
force. (Note that it was said in the last section that gravitational potential energy is
independent of path. Now you know why.)
For any horizontal displacement no work is done, since the displacement and
force are at right angles. The magnitude of the work done is equal to the change in
potential energy (under frictionless conditions only), and in fact, the concept of
potential energy is associated only with conservative forces. A change in potential
energy can be defined in terms of the work done by a conservative force.
Conversely, a nonconservative force does depend on path.
A force is said to be nonconservative if the work done by it in moving an object

depends on the object’s path.
Friction is a nonconservative force. A longer path would produce more work done by
friction than a shorter one, and more energy would be lost to heat on the longer
path. So the work done against friction certainly depends on the path. Hence, in a
sense, a conservative force allows you to conserve or store energy as potential
energy, whereas a nonconservative force does not.
Another approach to explain the distinction between conservative and noncon-

servative forces is through an equivalent statement of the previous definition of
conservative force:
A force is conservative if the work done by it in moving an object through a round trip
is zero.
Notice that for the conservative gravitational force, the force and displacement
are sometimes in the same direction (in which case positive work is done by the
force) and sometimes in opposite directions (in which case negative work is done
by the force) during a round trip. Think of the simple case of the book falling to
the floor and being placed back on the table. With positive and negative work, the
total work done by gravity is zero.
However, for a nonconservative force like that of kinetic friction, which opposes
motion or is in the opposite direction to the displacement, the total work done by such
a force in a round trip can never be zero and is always negative (that is, energy is lost).
But don’t get the idea that nonconservative forces only take energy away from a sys-
tem. On the contrary, nonconservative pushes and pulls (forces) that add to the energy
of a system are often supplied, such as when you push a stalled car and get it moving.
CONSERVATION OF TOTAL MECHANICAL ENERGY

The idea of a conservative force allows us to extend the conservation of energy to
the special case of mechanical energy, which greatly helps to better analyze many
physical situations. The sum of the kinetic and potential energies is called the total
mechanical energy:
E = K + U
total kinetic potential (5.9)
mechanical energy energy
energy
For a conservative system (that is, a system in which only conservative forces
do work), the total mechanical energy is constant, or conserved:
E = Eo
Substituting for E and Eo from Eq. 5.9,
K + U = Ko + Uo (5.10a)
or
1
2 mv
2
+ U = 12 mv 2o + Uo (5.10b)
Equation 5.10b is a mathematical statement of the law of the conservation of

mechanical energy:
In a conservative system, the sum of all types of kinetic energy and potential energy is
constant and equals the total mechanical energy of the system at any time.
Note: While the kinetic and potential energies in a conservative system may
change, their sum is always constant. For a conservative system when work is
done and energy is transferred within a system, Eq. 5.10a can be written as
1K - Ko2 + 1U - Uo2 = 0 (5.11a)
or
¢K + ¢U = 0 (for a conservative system) (5.11b)
This expression indicates that these quantities are related in a seesaw fashion: If
there is a decrease in potential energy, then the kinetic energy must increase by an
equal amount to keep the sum of the changes equal to zero. However, in a noncon-
servative system, mechanical energy is usually lost (for example, to the heat of
friction), and thus ¢K + ¢U 6 0. But as pointed out previously, a nonconserva-
tive force may instead add energy to a system (or have no effect at all).
EXAMPLE 5.11 Look Out Below! Conservation of Mechanical Energy

A painter on a scaffold drops a 1.50-kg can of paint from a T H I N K I N G I T T H R O U G H . Total mechanical energy is con-
height of 6.00 m. (a) What is the kinetic energy of the can served, since only the conservative force of gravity acts on the
when the can is at a height of 4.00 m? (b) With what speed system (the can). The initial total mechanical energy can be
will the can hit the ground? (Neglect air resistance.) (c) Show found, and the potential energy decreases as the kinetic
that the expression for speed from energy considerations is energy (as well as speed) increases.
the same as that from kinematics (Chapter 2.5).
SOLUTION. Listing the given data and what is to be found:
Given: m = 1.50 kg Find: (a) K (kinetic energy at y = 4.00 m)

yo = 6.00 m (b) v (speed just before hitting the ground)
y = 4.00 m (c) Compare speeds
vo = 0
(a) First, it is convenient to find the can’s initial total mechanical energy, since this quantity is conserved while the can is falling.
With vo = 0, the can’s total mechanical energy is initially all potential energy. Taking the ground as the zero reference point,
E = Ko + Uo = 0 + mgyo = 11.50 kg219.80 m>s2216.00 m2 = 88.2 J
The relation E = K + U continues to hold while the can is falling, and now E is known. Rearranging the equation, K = E - U
and K can be found at y = 4.00 m
K = E - U = E - mgy = 88.2 J - 11.50 kg219.80 m>s2214.00 m2 = 29.4 J
Alternatively, the change in (in this case, the loss of) potential energy, ¢U, could have been computed. Whatever potential
energy was lost must have been gained as kinetic energy (Eq. 5.11). Then,
¢K + ¢U = 0
1K - Ko2 + 1U - Uo2 = 1K - Ko2 + 1mgy - mgyo2 = 0
With Ko = 0 (because vo = 0),
K = mg1yo - y2 = 11.50 kg219.8 m>s2216.00 m - 4.00 m2 = 29.4 J
(b) Just before the can strikes the ground 1y = 0, U = 02, the total mechanical energy is all kinetic energy,
E = K = 12 mv2
Thus, the speed is,

2E 2188.2 J2
v = = = 10.8 m>s
Am A 1.50 kg
(c) Basically, all of the potential energy of a free-falling object released from some height y is converted into kinetic energy just
before the object hits the ground, so
ƒ ¢K ƒ = ƒ ¢U ƒ
(Why absolute values?) Thus,
1 2
2 mv = mgy
and
v = 22gy = 2219.8 m>s2 216.00 m2 = 10.8 m>s
Note that the mass cancels and is not a consideration. This result is also obtained from a kinematic equation (Eq. 2.12):
v2 = v2o - 2g1y - yo2. With vo = 0, yo = 0, and -y (downward),
v = 22gy
F O L L O W - U P E X E R C I S E . A painter on the ground wishes to mechanical energy to determine the minimum speed that she
toss a paintbrush vertically upward a distance of 5.0 m to her must give to the brush.
partner on the scaffold. Use methods of conservation of
CONCEPTUAL EXAMPLE 5.12 A Matter of Direction? Speed and Conservation of Energy

Three balls of equal mass m are projected with the same speed in REASONING AND ANSWER. All of the balls have the
different directions, as shown in 䉲 Fig. 5.18. If air resistance is same initial kinetic energy, Ko = 12 mv2o . (Recall that
neglected, which ball would you expect to strike the ground with energy is a scalar quantity, and the different direc-
the greatest speed: (a) ball 1, (b) ball 2, (c) ball 3, or (d) all balls tions of projection do not produce any difference in
strike with the same speed? the kinetic energies.) Regardless of their trajectories,
vo all of the balls ultimately descend a distance y rela-
tive to their common starting point, so they all lose
45° vo 2 the same amount of potential energy. (Recall that U
45° is energy of position and is independent of path.)
vo By the law of conservation of mechanical energy,
the amount of potential energy each ball loses is
3 equal to the amount of kinetic energy it gains. Since
all of the balls start with the same amount of kinetic
energy and gain the same amount of kinetic energy,
all three will have equal kinetic energies just before
striking the ground. This means that their speeds
1
must be equal, so the answer is (d).
y Although balls 1 and 2 are projected at 45° angles,
this factor is not relevant. Since the change in poten-
tial energy is independent of path, it is independent of
the projection angle. The vertical distance between
the starting point and the ground is the same (y) for
projectiles at any angle. (Note: Although the strike
speeds are equal, the times the balls take to reach the
ground are different. Refer to Chapter 3 Conceptual
Example 3.11 for another approach.)
F O L L O W - U P E X E R C I S E . Would the balls strike the
ground with different speeds if their masses were
䉱 F I G U R E 5 . 1 8 Speed and energy See Example text for description. different? (Neglect air resistance.)
EXAMPLE 5.13 Conservative Forces: Mechanical Energy

of a Spring
A 0.30-kg block sliding on a horizontal frictionless surface with a speed of 2.5 m>s, as
depicted in 䉲 Fig. 5.19, strikes a light spring that has a spring constant of 3.0 * 103 N>m.
(a) What is the total mechanical energy of the system? (b) What is the kinetic energy K1
of the block when the spring is compressed a distance x1 = 1.0 cm? (Assume that no
energy is lost in the collision.)
䉳 F I G U R E 5 . 1 9 Conservative
v force and the mechanical energy
of a spring See Example text for
k description.
m
T H I N K I N G I T T H R O U G H . (a) Initially, the total mechanical energy is all kinetic energy.

(b) The total energy is the same as in part (a), but it is now divided between kinetic
energy and spring potential energy (assuming the spring is not fully compressed).
SOLUTION.
Given: m = 0.30 kg Find: (a) E (total mechanical energy)
vo = 2.5 m>s (b) K1 (kinetic energy)
k = 3.0 * 103 N>m
x1 = 1.0 cm = 0.010 m
(a) Just before the block makes contact with the spring, the total mechanical energy of
the system is all in the form of kinetic energy,
E = Ko = 12 mv2o = 12 10.30 kg212.5 m>s22 = 0.94 J
Since the system is conservative (that is, no mechanical energy is lost), this quantity is
the total mechanical energy at any time.
(b) When the spring is compressed a distance x1, it has gained potential energy
U1 = 12 kx 21 , and the block has kinetic energy K1, so
E = K1 + U1 = K1 + 12 kx 21
Solving for K1,
K1 = E - 12 kx 21
= 0.94 J - 12 13.0 * 103 N>m210.010 m22
= 0.94 J - 0.15 J = 0.79 J
F O L L O W - U P E X E R C I S E . How far will the spring in this Example be compressed when
the block comes to a stop? (Solve using energy principles.)
See the accompanying Learn by Drawing 5.3, Energy Exchanges: A Falling Ball
for another example of energy exchange.
energy exchanges: a falling ball
v=0
Us = 0
Maximum
compression
v=0
Ug = 0
Ug K Us Ug K Us Ug K Us Ug K Us Ug K Us
Both the physical situation and the graphs of gravitational potential energy (Ug), kinetic energy (K), and spring potential energy
(Us) are drawn to scale. (Air resistance, the mass of the spring, and any energy loss in the collision are assumed to be negligible.)
Why is the spring energy only one-quarter of the total when the spring is halfway compressed?
TOTAL ENERGY AND NONCONSERVATIVE FORCES

In the preceding examples, the force of friction was ignored; however, friction is
probably the most common nonconservative force. In general, both conservative
and nonconservative forces can do work on objects. But when nonconservative
forces do work, the total mechanical energy is not conserved. Mechanical energy
is “lost” through the work done by nonconservative forces, such as friction.
You might think that an energy approach can no longer be used to analyze
problems involving such nonconservative forces, since mechanical energy can be
lost or dissipated (䉳 Fig. 5.20). However, in some instances, the total energy can be
used to find out how much energy was lost to the work done by a nonconservative
force. Suppose an object initially has mechanical energy and that nonconservative
forces do an amount of work Wnc on it. Starting with the work–energy theorem,
W = ¢K = K - Ko
In general, the net work (W) may be done by both conservative forces (Wc) and
nonconservative forces (Wnc), so
䉱 F I G U R E 5 . 2 0 Nonconservative Wc + Wnc = K - Ko (5.12)
force and energy loss Friction is a
nonconservative force—when fric- But from Eq. 5.10a, the work done by conservative forces is equal to - ¢U =
tion is present and does work, - (U - Uo), and Wc = Uo - U, so Eq. 5.12 then becomes
Wnc = K - Ko - 1Uo - U2
mechanical energy is not conserved.
Can you tell from the photo what is
= 1K + U2 - 1Ko + Uo2
happening to the work being done
by the motor on the grinding wheel
after the work is converted into Therefore,
rotational kinetic energy? (Note that
the worker is wisely wearing a face
shield rather than just goggles as the Wnc = E - Eo = ¢E (5.13)
sign in the background suggests.)
Hence, the work done by the nonconservative forces acting on a system is equal to
the change in mechanical energy. Notice that for dissipative forces, Eo 7 E Thus,
the change is negative, indicating a decrease in mechanical energy. This condition
agrees in sign with Wnc which, for friction, would also be negative. Example 5.14
illustrates this concept.
EXAMPLE 5.14 Nonconservative Force: Downhill Racer

A skier with a mass of 80 kg starts from rest at the top of a slope
and skis down from an elevation of 110 m (䉴 Fig. 5.21). The
speed of the skier at the bottom of the slope is 20 m>s. (a) Show
that the system is nonconservative. (b) How much work is
done by the nonconservative force of friction?
T H I N K I N G I T T H R O U G H . (a) If the system is nonconservative,
then Eo Z E, and these quantities can be computed. (b) The work
cannot be determined from force–distance considerations, but
Wnc is equal to the difference in total energies (Eq. 5.13). v = 20 m/s 110 m
SOLUTION.
Given: m = 80 kg Find: (a) Show that E is not

vo = 0 conserved.
v = 20 m>s (b) Wnc (work done by
yo = 110 m friction) 䉱 F I G U R E 5 . 2 1 Work done by a nonconservative force
(a) If the system is conservative, the total mechanical energy And the energy at the bottom of the slope is all kinetic, thus
is constant. Taking Uo = 0 at the bottom of the hill, the initial
energy at the top of the hill is E = K = 12 mv2 = 12 180 kg2120 m>s22 = 1.6 * 104 J
Eo = U = mgyo = 180 kg219.8 m>s 221110 m2 = 8.6 * 104 J Therefore, Eo Z E, so this system is not conservative.
(b) The amount of work done by the nonconservative force of F O L L O W - U P E X E R C I S E . In free fall, air resistance is some-
friction is equal to the change in the mechanical energy, or to times negligible, but for skydivers, air resistance has a very
the amount of mechanical energy lost (Eq. 5.13): practical effect. Typically, a skydiver descends about 450 m
Wnc = E - Eo = 11.6 * 104 J2 - 18.6 * 104 J2 = - 7.0 * 104 J
before reaching a terminal velocity (Chapter 4.6) of 60 m>s.
(a) What is the percentage of energy loss to nonconservative
This quantity is more than 80% of the initial energy. (Where forces during this descent? (b) Show that after terminal veloc-
did this energy actually go?) ity is reached, the rate of energy loss in J>s is given by 160 mg2,
where m is the mass of the skydiver.
EXAMPLE 5.15 Nonconservative Force: One More Time

A 0.75-kg block slides on a frictionless surface with a speed T H I N K I N G I T T H R O U G H . The task of finding the final speed
of 20 m>s. It then slides over a rough area 1.0 m in length implies that equations involving kinetic energy can be used,
and onto another frictionless surface. The coefficient of where the final kinetic energy can be found by using the con-
kinetic friction between the block and the rough surface is servation of total energy. Note that the initial and final energies
0.17. What is the speed of the block after it passes across the are kinetic energies, since there is no change in gravitational
rough surface? potential energy. It is always good to make a sketch of the situ-
ation for clarity and understanding (䉲 Fig. 5.22).
Eo E
vo v
x =1.0 m
䉱 F I G U R E 5 . 2 2 A nonconservative rough spot See Example text for description.
SOLUTION. Listing the data, Then, rearranging the energy equation and writing the
terms out in detail,
Given: m = 0.75 kg Find: v (final speed of block)
x = 1.0 m K = Ko + Wnc
mk = 0.17 or
vo = 2.0 m>s 1
2 mv
2
= 12 mv2o - mk mgx
For this nonconservative system, from Eq. 5.13
Solving for v yields,
Wnc = E - Eo = K - Ko
v = 2v2o - 2mk gx
In the rough area, the block loses energy, because of the work = 212.0 m>s22 - 210.17219.8 m>s2211.0 m2
done by friction (Wnc) and thus
= 0.82 m>s
Wnc = - fk x = - mk Nx = - mk mgx
Note that the mass of the block was not needed. Also, it
[negative because fk and the displacement x are in opposite can be easily shown that the block lost more than 80% of its
directions; that is, [fk(cos 180°)x = - fk x]. energy to friction.
F O L L O W - U P E X E R C I S E . Suppose the coefficient of kinetic friction between the block and the rough surface were 0.25. What
would happen to the block in this case?
Note that in a closed nonconservative system, the total energy (not the total
mechanical energy) is conserved (including nonmechanical forms of energy, such as
thermal energy). But not all of the energy is available for mechanical work. For a
conservative system, you get back what you put in, so to speak. That is, if you do
work on the system, the transferred energy is available to do work. Conservative
systems are idealizations, because all real systems are nonconservative to some
degree. However, working with ideal conservative systems gives an insight into the
conservation of energy.
Total energy is always conserved in a closed system. During the course of
study, you will learn about other forms of energy, such as thermal, electrical, and
INSIGHT 5.2 Hybrid Energy Conversion

As was learned, energy may be transformed from one form to
another. An interesting example is the conversion that takes
place in the new hybrid automobiles. A hybrid car has both a Generator Four-cylinder engine
gasoline (internal combustion) engine and a battery-driven
electric motor, both of which may be used to power the vehicle. Batteries
A moving car has kinetic energy, and when you step on the
brake pedal to slow the car down, kinetic energy is lost. Nor- Electric motor
mally, the brakes of a car accomplish this slowing by friction,
and energy is dissipated as heat (conservation of energy).
However, in the braking of a hybrid car, some of the energy is
converted to electrical energy and stored in the battery of the
electric motor. This is called regenerative braking. That is,
instead of using regular friction brakes to slow the car, the
electric motor is used. In this mode, the motor runs in reverse
and acts as a generator, converting the lost kinetic energy into Fuel tank
electrical energy. (See Chapter 20.2 for generator operation.)
The energy is stored in the battery for later use (䉴 Fig. 1).
F I G U R E 1 Hybrid car A diagram showing the major compo-
Hybrid cars must also have regular friction brakes to be
nents. See text for description.
used when rapid braking is needed. (See the Chapter 20
Insight 20.2, Electromagnetic Induction: Hobbies and Trans-
portation for a more detailed discussion on hybrids.)
nuclear energies. In general, on the microscopic and submicroscopic levels, these

forms of energy can be described in terms of kinetic energy and potential energy.
Also, you will learn that mass is a form of energy and that the law of conservation
of energy must take this form into account in order to be applied to the analysis of
nuclear reactions.
A modern example of energy conversion is given in Insight 5.2, Hybrid Energy
Conversion.
DID YOU LEARN?

➥ The total energy of a closed or isolated system is always conserved.
➥ In a conservative system (one in which only conservative forces act), the total
mechanical energy, E = K + U, is conserved.
➥ When a nonconservative force, such as friction, does work, mechanical energy is not
conserved.
5.6 Power
➥ What does power tell you?

➥ What does greater efficiency mean?
A particular task may require a certain amount of work, but that work might be
done over different lengths of time or at different rates. For example, suppose that
you have to mow a lawn. This task takes a certain amount of work, but you might
do the job in a half hour, or you might take an hour. There’s a practical distinction
to be made here. That is, there is usually not only an interest in the amount of
work done, but also an interest in how fast it is done—that is, the rate at which it is
done. The time rate of doing work is called power.
The average power 1P2 is the work done divided by the time it takes to do the
work, or work per unit of time:
W
P = (5.14)
t
5.6 POWER 167
The work (and power) done by a constant force of magnitude F acting while an
object moves through a parallel displacement of magnitude d is
= Fa b = F v
W Fd d
P = = (5.15)
t t t
SI unit of power: J>s or watt (W)

where it is assumed that the force is in the direction of the displacement. Here, v is the
magnitude of the average velocity. If the velocity is constant, then P = P = Fv. If the
force and displacement are not in the same direction, then we can write
F1cos u2d
P = = F v cos u (5.16)
t
where u is the angle between the force and the displacement.
As can be seen from Eq. 5.15, the SI unit of power is joules per second (J>s), but
this unit is given another name, the watt (W):
1 J>s = 1 watt 1W2
The SI unit of power is named in honor of James Watt (1736–1819), a Scottish engi-
neer who developed one of the first practical steam engines. A common unit of
electrical power is the kilowatt (kW).
The British unit of power is foot-pound per second (ft # lb>s). However, a larger
unit coined by Watt, the horsepower (hp), is more commonly used:*
1 hp = 550 ft # lb>s = 746W
Power tells how fast work is being done or how fast energy is transferred. For
example, motors have power ratings commonly given in horsepower. A 2-hp
motor can do a given amount of work in half the time that a 1-hp motor would
take, or twice the work in the same amount of time. That is, a 2-hp motor is twice
as “powerful” as a 1-hp motor.
EXAMPLE 5.16 A Crane Hoist: Work and Power

A crane hoist like the one shown in 䉲 Fig. 5.23 lifts a load of 1.0 SOLUTION.
metric ton a vertical distance of 25 m in 9.0 s at a constant veloc- Given: m = 1.0 metric ton Find: P (power, work per
ity. How much useful work is done by the hoist each second? = 1.0 * 103 kg second)
y = 25 m
䉳 FIGURE 5.23
t = 9.0 s
Power delivery
See Example text Since the load moves with a constant velocity, P = P. (Why?)
for description. The work is done against gravity, so F = mg, and
W Fd mgy
P = = =
t t t
11.0 * 103 kg219.8 m>s 22125 m2
=
9.0 s
= 2.7 * 104 W 1or 27 kW2
Thus, since a watt (W) is a joule per second ( J>s), the hoist did
2.7 * 104 J of work each second. Note that the velocity has a
T H I N K I N G I T T H R O U G H . The useful work done each second magnitude of v = d>t = 25 m>9.0 s = 2.8 m>s, and the dis-
(that is, per second) is the power output, so this is what needs placement is parallel to the applied force, therefore the power
to be found (Eq. 5.15). could be found using P = Fv.
FOLLOW-UP EXERCISE. If the hoist motor of the crane in this Example is rated at 70 hp, what percentage of this power output
goes into useful work?
*In Watt’s time, steam engines were replacing horses for work in mines and mills. To characterize
the performance of his new engine, which was more efficient than existing ones, Watt used the average
rate at which a horse could do work as a unit—a horsepower.
EXAMPLE 5.17 Cleaning Up: Work and Time

The motors of two vacuum cleaners have net power outputs of 1.00 hp and 0.500 hp,
respectively. (a) How much work in joules can each motor do in 3.00 min? (b) How long
does each motor take to do 97.0 kJ of work?
T H I N K I N G I T T H R O U G H . (a) Since power is work>time 1P = W>t2, the work can be
computed. Note that power is given in horsepower units, which is converted to watts.
(b) This part of the problem is another application of Eq. 5.15.
SOLUTION.
Given: P1 = 1.00 hp = 746 W Find: (a) W (work for each)

P2 = 0.500 hp = 373 W (b) t (time for each)
(a) t = 3.00 min = 180 s
(b) W = 97.0 kJ = 97.0 * 103 J
(a) Since P = W>t, for the 1.00-hp motor:

W1 = P1 t = 1746 W21180 s2 = 1.34 * 105 J
And for the 0.500-hp motor:
W2 = P2 t = 1373 W21180 s2 = 0.67 * 105 J
Note that in the same amount of time, the smaller motor does half as much work as the
larger one, as would be expected.
(b) The times are given by t = W>P, and for the same amount of work,
W 97.0 * 103 J
t1 = = = 130 s
P1 746 W
and
W 97.0 * 103 J
t2 = = = 260 s.
P2 373 W
So, the smaller motor takes twice as long as the larger one to do the same amount of work.
FOLLOW-UP EXERCISE. (a) A 10-hp motor breaks down and is temporarily replaced
with a 5-hp motor. What can you say about the rate of work output? (b) Suppose the sit-
uation were reversed—a 5-hp motor is replaced with a 10-hp motor. What can you say
about the rate of work output for this case?
EFFICIENCY
Machines and motors are commonly used items in our daily lives, and comments
are made about their efficiencies—for example, one machine is more efficient than
another. Efficiency involves work, energy, and>or power. Both simple and com-
plex machines that do work have mechanical parts that move, so some input
energy is always lost because of friction or some other cause (perhaps in the form
of sound). Thus, not all of the input energy goes into doing useful work.
Mechanical efficiency is essentially a measure of what you get out for what you
put in—that is, the useful work output compared with the energy input.
Efficiency, E, is given as a fraction (or percentage):
work output
1* 100%2 = 1* 100%2
Wout
e = (5.17)
energy input Ein
Efficiency is a unitless quantity

For example, if a machine has a 100-J (energy) input and a 40-J (work) output, then
its efficiency is
= 0.40 1* 100%2 = 40%

Wout 40 J
e = =
Ein 100 J
5.6 POWER 169
An efficiency of 0.40, or 40%, means that 60% of the energy input is lost because of
friction or some other cause and doesn’t serve its intended purpose. Note that if
both terms of the ratio in Eq. 5.17 are divided by time t, we obtain Wout>t = Pout
and Ein>t = Pin . So efficiency can be written in terms of power, P:
1 * 100%2
Pout
e = (5.18)
Pin
EXAMPLE 5.18 Home Improvement: Mechanical Efficiency and

Work Output
The motor of an electric drill with an efficiency of 80% has a power input of 600 W.
How much useful work is done by the drill in 30 s?
THINKING IT THROUGH. Given the efficiency and power input, the power output Pout
can readily be found from Eq. 5.18. This quantity is related to the work output
1Pout = Wout>t2, from which Wout may be found.
SOLUTION.
Given: e = 80% = 0.80 Find: Wout (work output)

Pin = 600 W
t = 30 s
First, rearranging Eq. 5.18 to find the power output:
Pout = ePin = 10.8021600 W2 = 4.8 * 102 W
Then, substituting this value into the equation relating power output and work output,
Wout = Pout t = 14.8 * 102 W2130 s2 = 1.4 * 104 J
F O L L O W - U P E X E R C I S E . (a) Is it possible to have a mechanical efficiency of 100%?

(b) What would an efficiency of greater than 100% imply?
䉲 Table 5.1 lists the typical efficiencies of some machines. You may be surprised
by the relatively low efficiency of the automobile. Much of the energy input (from
gasoline combustion) is lost as exhaust heat and through the cooling system (more
than 60%), and friction accounts for a good deal more. About 20% of the input
energy is converted to useful work that goes into propelling the vehicle. Air condi-
tioning, power steering, radio, and MP3 and CD players are nice, but they also use
energy and contribute to the car’s decrease in efficiency.
DID YOU LEARN?

➥ How fast work is done or how fast energy is transferred is expressed by power.
➥ A machine with more useful work output for a given energy input has a greater
efficiency.
TABLE 5.1 Typical Efficiencies of Some Machines

Machine Efficiency (approximate %)
Compressor 85
Electric motor 70–95
Automobile (hybrid cars 20
with an efficiency of 25%)
Human muscle* 20–25
Steam locomotive 5–10
*Technically not a machine, but used to perform work.

PULLING IT TOGETHER Springs, Energy, and Friction

A spring with a spring constant of 2000 N>m, is in contact
with a 1.00-kg block on a table (䉴 Fig. 5.24). The spring with
the block is compressed 10.0 cm and then released, and the
block accelerates on a smooth (frictionless) table surface.
10.0 cm Elastic
Once the spring reaches its fully relaxed position, the block
wall
continues on without the spring, but the table surface is now
rough, having a coefficient of kinetic friction of 0.500. The
table is against an elastic wall 50.0 cm from where the block
leaves the spring.
(a) Assume the block rebounds off the wall with no loss of
speed. Using work-energy concepts, show how to determine Rough section of surface
whether the block makes it back to the spring or whether it 50.0 cm
stops before doing so. (b) If the block does make it back to the
spring, how far is the spring compressed? If it doesn’t, where 䉱 F I G U R E 5 . 2 4 Does it make it back? The block is propelled
does the block come to rest? by a compressed spring over a rough surface and rebounds
from an elastic wall. Does the block make it back to the spring?
T H I N K I N G I T T H R O U G H . Energy, a nonconservative force
(friction), and Newton’s laws are involved in this example.
(a) The spring’s potential energy will be converted to the spring. Otherwise, its remaining energy will be used to
block’s kinetic energy. Whether the block makes it back to the recompress the spring. (b) Depending on the result in part (a),
spring or not depends on the balance of the total initial either the leftover kinetic energy will be stored in the spring,
mechanical energy and the work done by friction during the enabling the recompression to be determined, or, if the initial
distance down the table and back (100 cm). If the latter is kinetic energy is all lost due to frictional (nonconservative)
larger than the former, the block will stop short of the spring, work, then the distance needed to do that and therefore the
that is, its kinetic energy will drop to zero before it hits the final resting location of the block can be determined.
SOLUTION.
Given: k = 2.00 * 103 N>m (spring constant) Find: (a) whether the block makes it back to the spring
m = 1.00 kg (block mass) (b) the final compression of the spring, or the
d = 50.0 cm = 0.500 m (distance to wall) final location of the block
xo = 10.0 cm = 0.100 m (initial spring
compression)
mk = 0.500 (coefficient of kinetic friction)
(a) Because the first part of the table surface is smooth, the This is less than the total mechanical energy, so there is some
spring’s potential energy 1Us2 is completely converted into mechanical energy “left over” upon the block’s return to the
the block’s kinetic energy 1Kb2. Thus the total initial mechani- spring. Thus the spring will be (partially) recompressed.
cal energy is (Why partially?)
Kb = Us = 12 kx 2o = 12 12.00 * 103 N>m210.100 m22 = 10.0 J

(b) When the block returns to the spring, it has
K = Kb - Wf = 10.0 J - 4.90 J = 5.10 J of kinetic energy
left. This energy will go into recompressing the spring a dis-
Now how much nonconservative work the force of kinetic
friction 1fk2 does must be determined. First the force of fric-
tance x according to the energy conversion:
tion is found by setting equal the magnitudes of the normal K = U
force (N) and the block’s weight (mg): 5.10 J = 12 kx 2
fk = mk N = mk mg = 10.500211.00 kg219.80 m>s 22 = 4.90 N

Solving for x,
215.10 J2
x = = 0.0714 m = 7.14 cm
Therefore, the maximum magnitude of work that could be C 2.00 * 103 N>m
done by this force over the 100 cm of rough surface is
Note that this distance is less than the original compression
Wf = fk12d2 = 14.90 N211.00 m2 = 4.90 J (10.0 cm), as expected.
■ Work done by a constant force is the product of the magni- F
tude of the displacement and the component of the force

parallel to the displacement:
W = 1F cos u2d (5.2) m
y U = mgy
F = F cos u
=
u
F
mg
∆y = y – y o W = U – Uo
F = F sin u
u = ∆U = mg ∆y
d
■ Calculating work done by a variable force requires
m
advanced mathematics. An example of a variable force is yo Uo = mgyo
the spring force, given by Hooke’s law:
Fs = - kx (5.3)
Unstretched
■ Conservation of energy: The total energy of the universe or

Fs Fa
Spring Applied
of an isolated system is always conserved.
force force Conservation of mechanical energy: The total mechanical
xo energy (kinetic plus potential) is constant in a conservative
(a) Fs = –k∆x = –k (x – xo)
Fs = –kx with xo = 0 system:
1
2 mv
2
+ U = 12 mv2o + Uo (5.10b)
Fs Fa
v0
xo x y ymax
∆x = x – xo
(b)
The work done by a spring force is given by ∆U mgymax
1 2
W = 2 kx (5.4) vo
y
■ Kinetic energy is the energy of motion and is given by y0
g y
K = 12 mv2 (5.5)
■ By the work–energy theorem, the net work done on an object
is equal to the change in the kinetic energy of the object:
W = K - Ko = ¢K (5.6)
■ In systems with nonconservative forces, where mechanical
W = K – Ko = ∆K
energy is lost, the work done by a nonconservative force is
Ko = 1 mvo2 K = 1 mv2
2 2 given by
vo v
F m F m
Wnc = E - Eo = ¢E (5.13)
(Frictionless)
W = Fx
■ Potential energy is the energy of position and>or configura-

tion. The elastic potential energy of a spring is given by
U = 12 kx 2 1with xo = 02 (5.7)
The most common type of potential energy is gravitational v = 20 m/s 110 m
potential energy, associated with the gravitational attrac-
tion near the Earth’s surface.
U = mgy 1with yo = 02 (5.8)
■ Power is the time rate of doing work (or expending energy). ■ Efficiency relates work output to energy (work) input as a
Average power is given by fraction or percent:
1* 100%2
W Fd Wout
P = = = Fv (5.15) e = (5.17)
t t Ein
(constant force in direction of d and v)
1* 100%2
Pout
e = (5.18)
F1cos u2d Pin
P = = F v cos u (5.16)
t
1constant force acts at an angle u between d and v2

5.1 WORK DONE BY A CONSTANT FORCE 12. The change in gravitational potential energy can be
1. The units of work are (a) N # m, (b) kg # m >s , (c) J, (d) all
2 2 found by calculating mg¢h and subtracting the reference
point potential energy: (a) true, (b) false.
of the preceding.
2. For a particular force and displacement, the most work 13. The reference point for gravitational potential energy
is done when the angle between them is (a) 30°, (b) 60°, may be (a) zero, (b) negative, (c) positive, (d) all of the
(c) 90°, (d) 180°. preceding.
3. A pitcher throws a fastball. When the catcher catches it,
(a) positive work is done, (b) negative work is done,
(c) the net work is zero. 5.5 CONSERVATION OF ENERGY
4. Work done in free fall (a) is only positive, (b) is only neg- 14. Energy cannot be (a) transferred, (b) conserved, (c) cre-
ative, or (c) can be either positive or negative. ated, (d) in different forms.
5. Which one of the following has units of work: (a) N,
15. If a nonconservative force acts on an object, and does
(b) N>s, (c) J # s, or (d) N # m?
work, then (a) the object’s kinetic energy is conserved,
(b) the object’s potential energy is conserved, (c) the
5.2 WORK DONE BY A VARIABLE FORCE mechanical energy is conserved, (d) the mechanical
6. The work done by a variable force of the form F = kx is energy is not conserved.
equal to (a) kx2, (b) kx, (c) 12 kx 2, (d) none of the preceding. 16. The speed of a pendulum is greatest (a) when the pendu-
lum’s kinetic energy is a minimum, (b) when the pendu-
5.3 THE WORK—ENERGY THEOREM: lum’s acceleration is a maximum, (c) when the
KINETIC ENERGY pendulum’s potential energy is a minimum, (d) none of
the preceding.
7. Which of the following is a scalar quantity: (a) work,
(b) force, (c) kinetic energy, or (d) both a and c? 17. Two springs are identical except for their force constants,
8. If the angle between the net force and the displacement k2 7 k1. If the same force is used to stretch the springs,
of an object is greater than 90°, (a) kinetic energy (a) spring 1 will be stretched farther than spring 2,
increases, (b) kinetic energy decreases, (c) kinetic energy (b) spring 2 will be stretched farther than spring 1,
remains the same, (d) the object stops. (c) both will be stretched the same distance.
9. Two identical cars, A and B, traveling at 55 mi>h collide 18. If the two springs in Exercise 17 are compressed the
head-on. A third identical car, C, crashes into a brick wall same distance, on which spring is more work done:
going 55 mi>h. Which car has the least damage: (a) car A, (a) spring 1, (b) spring 2, or (c) equal work on both?
(b) car B, (c) car C, or (d) all the same?
19. Two identical stones are thrown from the top of a tall
10. Which of the following objects has the least kinetic
building. Stone 1 is thrown vertically downward with
energy: (a) an object of mass 4m and speed v, (b) an
an initial speed v, and stone 2 is thrown vertically
object of mass 3m and speed 2v, (c) an object of mass 2m
upward with the same initial speed. Neglecting air
and speed 3v, or (d) an object of mass m and speed 4v?
resistance, which stone hits the ground with a greater
speed: (a) stone 1, (b) stone 2, or (c) both have the same
5.4 POTENTIAL ENERGY speed?
11. A change in gravitational potential energy (a) is always 20. In Exercise 19, if air resistance is taken into account,
positive, (b) depends on the reference point, (c) depends on which stone hits the ground with a greater speed:
the path, (d) depends only on the initial and final positions. (a) stone 1, (b) stone 2, or (c) both have the same speed?
5.6 POWER 1.0-hp motor can (a) do twice as much work in half the
time, (b) half the work in the same time, (c) one quarter
21. Which of the following is not a unit of power: (a) J>s,
of the work in three quarters of the time, (d) none of the
(b) W # s, (c) W, or (d) hp?
preceding.
22. Consider a 2.0-hp motor and a 1.0-hp motor. Compared
to the 2.0-hp motor, for a given amount of work, the
1. (a) As a weightlifter lifts a barbell from the floor in the mass by half or reducing the speed by half. Which
“clean” procedure (䉲Fig. 5.25a), has he done work? Why option should you pick, and why?
or why not? (b) In raising the barbell above his head in 9. A certain amount of work W is required to accelerate a
the “jerk” procedure, is he doing work? Explain. (c) In car from rest to a speed v. How much work is required to
holding the barbell above his head (Fig. 5.25b), is he accelerate the car from rest to a speed of v>2?
doing more work, less work, or the same amount of 10. A certain amount of work W is required to accelerate a
work as in lifting the barbell? Explain. (d) If the car from rest to a speed v. If instead an amount of work
weightlifter drops the barbell, is work done on the bar- equal to 2W is done on the car, what is the car’s speed?
bell? Explain what happens in this situation.
11. Car B is traveling twice as fast as car A, but car A has
four times the mass of car B. Which car has the greater
kinetic energy?
5.4 POTENTIAL ENERGY

12. If a spring changes its position from xo to x, what is the
change in potential energy then proportional to?
(Express the answer in terms of xo and x.)
13. A lab notebook sits on a table 0.75 m above the floor.
Your lab partner tells you the book has zero potential
energy, and another student says it has 8.0 J of potential
energy. Who is correct?
(a) (b)
14. An object is said to have a negative potential energy.
䉱 F I G U R E 5 . 2 5 Man at work? See Conceptual Question 1. Because you prefer not to work with negative numbers,
how could you make the object to have a positive poten-
2. You are carrying a backpack across campus. What is the tial energy without moving it?
work done by your vertical carrying force on the back-
pack? Explain.
5.5 CONSERVATION OF ENERGY
3. A jet plane flies in a vertical circular loop. In what
regions of the loop is the work done by the force of grav- 15. For a classroom demonstration, a bowling ball sus-
ity on the plane positive and>or negative? Is the work pended from a ceiling is displaced from the vertical posi-
constant? If not, are there maximum and minimum tion to one side and released from rest just in front of the
instantaneous values? Explain. nose of a student (䉲 Fig. 5.26). If the student doesn’t
4. When walking up stairs, it is easier to do so in a zigzag move, why won’t the bowling ball hit his nose?
path rather than going straight up. Why is this? (Hint:
Think of an inclined plane.)
5. Can an object possess work?
5.2 WORK DONE BY A VARIABLE FORCE

6. Does it take twice the work to stretch a spring 2 cm from
its equilibrium position as it does to stretch it 1 cm from
its equilibrium position? 䉱 F I G U R E 5 . 2 6 In your face? See Conceptual Question 15.
7. If a spring is compressed 2.0 cm from its equilibrium 16. When you throw an object into the air, is its initial speed
position and then compressed an additional 2.0 cm, how the same as its speed just before it returns to your hand?
much more work is done in the second compression Explain by applying the concept of conservation of
than in the first? Explain. mechanical energy.
17. A student throws a ball vertically upward so it just
reaches the height of a window on the second floor of a
5.3 THE WORK—ENERGY THEOREM:
dormitory. At the same time that the ball is thrown
KINETIC ENERGY
upward, a student at the window drops an identical ball.
8. You want to decrease the kinetic energy of an object as Are the mechanical energies of the balls the same at half
much as you can. You can do so by either reducing the the height of the window? Explain.
5.6 POWER 20. Two students who weigh the same start at the same
ground floor location at the same time to go to the same
18. If you check your electricity bill, you will note that you
classroom on the third floor by different routes. If they
are paying the power company for so many kilowatt-
arrive at different times, which student will have
hours (kWh). Are you really paying for power? Explain.
expended more power? Explain.
19. (a) Does efficiency describe how fast work is done?
Explain. (b) Does a more powerful machine always per-
form more work than a less powerful one? Explain.
EXERCISES
1. ● If a person does 50 J of work in moving a 30-kg box the girl is 35 kg and the coefficient of kinetic friction
over a 10-m distance on a horizontal surface, what is the between the sled runners and the snow is 0.20, how
minimum force required? much work does the father do?
2. ● A 5.0-kg box slides a 10-m distance on ice. If the coeffi- 7. ● ● A father pushes horizontally on his daughter’s sled
cient of kinetic friction is 0.20, what is the work done by to move it up a snowy incline, as illustrated in Fig. 5.27b.
the friction force? If the sled moves up the hill with a constant velocity,
3. ● A passenger at an airport pulls a rolling suitcase by its how much work is done by the father in moving it from
handle. If the force used is 10 N and the handle makes an the bottom to the top of the hill? (Some necessary data
angle of 25° to the horizontal, what is the work done by are given in Exercise 6.)
the pulling force while the passenger walks 200 m? 8. ● ● A block on a level frictionless surface has two hori-
4. ● ● A 3.00-kg block slides down a frictionless plane inclined zontal forces applied, as shown in 䉲 Fig. 5.28. (a) What
20° to the horizontal. If the length of the plane’s surface is force F2 would cause the block to move in a straight line
1.50 m, how much work is done, and by what force? to the right? (b) If the block moves 50 cm, how much
5. ● ● Suppose the coefficient of kinetic friction between
work is done by each force? (c) What is the total work
the block and the plane in Exercise 4 is 0.275. What done by the two forces?
would be the net work done in this case?
F2 = ?
6. ● ● A father pulls his young daughter on a sled with a
constant velocity on a level surface a distance of 10 m, as

illustrated in 䉲 Fig. 5.27a. If the total mass of the sled and 60°
30°
F1 = 90 N
v
F 䉱 F I G U R E 5 . 2 8 Make it go straight See Exercise 8.
30°
9. ●●A 0.50-kg shuffleboard puck slides a distance of
3.0 m on the board. If the coefficient of kinetic friction
10 m between the puck and the board is 0.15, what work is
(a) done by the force of friction?
10. ● ● A crate is dragged 3.0 m along a rough floor with a
constant velocity by a worker applying a force of 500 N
v
to a rope at an angle of 30° to the horizontal. (a) How
many forces are acting on the crate? (b) How much work
F does each of these forces do? (c) What is the total work
done on the crate?
11. IE ● ● A hot-air balloon ascends at a constant rate.
3.6 m (a) The weight of the balloon does (1) positive work,
15° (2) negative work, (3) no work. Why? (b) A hot-air
balloon with a mass of 500 kg ascends at a constant rate
(b) of 1.50 m>s for 20.0 s. How much work is done by the
䉱 F I G U R E 5 . 2 7 Fun and work See Exercises 6 and 7. upward buoyant force? (Neglect air resistance.)
EXERCISES 175
12. IE ● ● A hockey puck with a mass of 200 g and an initial F

speed of 25.0 m>s slides freely to rest in the space of 8.0
100 m on a sheet of horizontal ice. How many forces do 6.0
nonzero work on it as it slows: (a) (1) none, (2) one,
(3) two, or (4) three? Explain. (b) Determine the work 4.0
done by all the individual forces on the puck as it slows.
2.0
Force (N)
Distance (m)
13. IE ● ● An eraser with a mass of 100 g sits on a book at
rest. The eraser is initially 10.0 cm from any edge of the 0 x
0 1.0 2.0 3.0 4.0 5.0
book. The book is suddenly yanked very hard and slides –2.0
out from under the eraser. In doing so, it partially drags
the eraser with it, although not enough to stay on the –4.0
book. The coefficient of kinetic friction between the book –6.0
and the eraser is 0.150. (a) The sign of the work done by
the force of kinetic friction of the book on the eraser is –8.0
(1) positive, (2) negative, or (3) zero work is done by
kinetic friction. Explain. (b) How much work is done by 䉱 F I G U R E 5 . 2 9 How much work is done? See Exercise 22.
the book’s frictional force on the eraser by the time it
falls off the edge of the book?
23. IE ● ● A spring with a force constant of 50 N>m is to be
14. ● ● ● A 500-kg, light-weight helicopter ascends from the stretched from 0 to 20 cm. (a) The work required to
ground with an acceleration of 2.00 m>s2. Over a 5.00-s stretch the spring from 10 cm to 20 cm is (1) more than,
interval, what is (a) the work done by the lifting force, (2) the same as, (3) less than that required to stretch it
(b) the work done by the gravitational force, and (c) the from 0 to 10 cm. (b) Compare the two work values to
net work done on the helicopter? prove your answer to part (a).
15. ● ● ● A man pushes horizontally on a desk that rests on a
rough wooden floor. The coefficient of static friction 24. IE ● ● In gravity-free interstellar space, a spaceship fires
between the desk and floor is 0.750 and the coefficient of its rockets to speed up. The rockets are programmed to
kinetic friction is 0.600. The desk’s mass is 100 kg. He increase thrust from zero to 1.00 * 104 N with a linear
pushes just hard enough to get the desk moving and increase over the course of 18.0 km. Then the thrust
continues pushing with that force for 5.00 s. How much decreases linearly back to zero over the next 18.0 km.
work does he do on the desk? Assuming the rocket was stationary to start, (a) during
which segment will more work (magnitude) be done:
16. IE ● ● ● A student could either pull or push, at an angle
(1) the first 60 s, (2) the second 60 s, or (3) the work done
of 30° from the horizontal, a 50-kg crate on a horizontal
is the same in both segments? Explain your reasoning.
surface, where the coefficient of kinetic friction between
(b) Determine quantitatively how much work is done in
the crate and surface is 0.20. The crate is to be moved a
each segment.
horizontal distance of 15 m. (a) Compared with pushing,
pulling requires the student to do (1) less, (2) the same, 25. ●● A particular spring has a force constant of
or (3) more work. (b) Calculate the minimum work 2.5 * 103 N>m. (a) How much work is done in stretching
required for both pulling and pushing. the relaxed spring by 6.0 cm? (b) How much more work
is done in stretching the spring an additional 2.0 cm?
26. ●● For the spring in Exercise 25, how much mass would
5.2 WORK DONE BY A VARIABLE FORCE have to be suspended from the vertical spring to stretch
17. ● To measure the spring constant of a certain spring, a it (a) the first 6.0 cm and (b) the additional 2.0 cm?
student applies a 4.0-N force, and the spring stretches by
27. ● ● ● In stretching a spring in an experiment, a student
5.0 cm. What is the spring constant?
inadvertently stretches it past its elastic limit; the force-
18. ● A spring has a spring constant of 30 N>m. How much
versus-stretch graph is shown in 䉲 Fig. 5.30. Basically,
work is required to stretch the spring 2.0 cm from its after it reaches its limit, the spring begins to behave as if
equilibrium position? it were considerably stiffer. How much work was done
19. ● If it takes 400 J of work to stretch a spring 8.00 cm, on the spring? Assume that on the force axis, the tick
what is the spring constant? marks are every 10 N, and on the x-axis, they are every
20. ● If a 10-N force is used to compress a spring with a 10 cm or 0.10 m.
spring constant of 4.0 * 102 N>m, what is the resulting
spring compression?
21. IE ● A certain amount of work is required to stretch a
Force (N)
spring from its equilibrium position. (a) If twice the

work is performed on the spring, the spring will stretch
more by a factor of (1) 12 (2) 2, (3) 1> 12 (4) 12 . Why?
(b) If 100 J of work is done to pull a spring 1.0 cm, what
work is required to stretch it 3.0 cm?
22. ● ● Compute the work done by the variable force in the Distance (m)
graph of F versus x in 䉴 Fig. 5.29. [Hint: The area of a tri-
angle is A = 12 altitude * base] 䉱 F I G U R E 5 . 3 0 Past the limit See Exercise 27.
28. ●●● A spring (spring 1) with a spring constant of 38. IE ● You are told that the gravitational potential energy
500 N>m is attached to a wall and connected to another of a 2.0-kg object has decreased by 10 J. (a) With this
weaker spring (spring 2) with a spring constant of information, you can determine (1) the object’s initial
250 N>m on a horizontal surface. Then an external force height, (2) the object’s final height, (3) both the initial
of 100 N is applied to the end of the weaker spring 1#22. and the final height, (4) only the difference between the
How much potential energy is stored in each spring? two heights. Why? (b) What can you say has physically
happened to the object?
5.3 THE WORK—ENERGY THEOREM: 39. ●● Six identical books, 4.0 cm thick and each with a
KINETIC ENERGY mass of 0.80 kg, lie individually on a flat table. How
much work would be needed to stack the books one on
29. IE ● A 0.20-kg object with a horizontal speed of 10 m>s top of the other?
hits a wall and bounces directly back with only half the
original speed. (a) What percentage of the object’s initial 40. IE ● ● The floor of the basement of a house is 3.0 m
kinetic energy is lost: (1) 25%, (2) 50%, or (3) 75%? below ground level, and the floor of the attic is 4.5 m
(b) How much kinetic energy is lost in the ball’s collision above ground level. (a) If an object in the attic were
with the wall? brought to the basement, the change in potential energy
30. ● A 1200-kg automobile travels at 90 km>h. (a) What is will be greatest relative to which floor: (1) attic,
its kinetic energy? (b) What net work would be required (2) ground, (3) basement, or (4) all the same? Why?
to bring it to a stop? (b) What are the respective potential energies of 1.5-kg
objects in the basement and attic, relative to ground
31. ● A constant net force of 75 N acts on an object initially
level? (c) What is the change in potential energy if the
at rest as it moves through a parallel distance of 0.60 m.
object in the attic is brought to the basement?
(a) What is the final kinetic energy of the object? (b) If the
object has a mass of 0.20 kg, what is its final speed? 41. ●● A 0.50-kg mass is placed on the end of a vertical
32. IE ● ● A 2.00-kg mass is attached to a vertical spring with spring that has a spring constant of 75 N>m and eased
a spring constant of 250 N>m. A student pushes on the down into its equilibrium position. (a) Determine the
mass vertically upward with her hand while slowly low- change in spring (elastic) potential energy of the system.
ering it to its equilibrium position. (a) How many forces (b) Determine the system’s change in gravitational
do nonzero work on the object: (1) one, (2) two, or (3) potential energy.
three? Explain your reasoning. (b) Calculate the work
42. ●● A horizontal spring, resting on a frictionless tabletop,
done on the object by each of the forces acting on it as it
is stretched 15 cm from its unstretched configuration and
is lowered into position.
a 1.00-kg mass is attached to it. The system is released
33. ● ● The stopping distance of a vehicle is an important from rest. A fraction of a second later, the spring finds
safety factor. Assuming a constant braking force, use the itself compressed 3.0 cm from its unstretched configura-
work–energy theorem to show that a vehicle’s stopping tion. How does its final potential energy compare to its
distance is proportional to the square of its initial speed. initial potential energy? (Give your answer as a ratio,
If an automobile traveling at 45 km>h is brought to a final to initial.)
stop in 50 m, what would be the stopping distance for an
initial speed of 90 km>h? 43. ● ● ● A student has six textbooks, each with a thickness of
34. IE ● ● A large car of mass 2m travels at speed v. A small 4.0 cm and a weight of 30 N. What is the minimum work
car of mass m travels with a speed 2v. Both skid to a stop the student would have to do to place all the books in a
with the same coefficient of friction. (a) The small car single vertical stack, starting with all the books on the
will have (1) a longer, (2) the same, (3) a shorter stopping surface of the table?
distance. (b) Calculate the ratio of the stopping distance 44. ● ● ● A 1.50-kg mass is placed on the end of a spring that
of the small car to that of the large car. (Use the has a spring constant of 175 N>m. The mass–spring sys-
work–energy theorem, not Newton’s laws.) tem rests on a frictionless incline that is at an angle of 30°
35. ● ● ● An out-of-control truck with a mass of 5000 kg is from the horizontal (䉲 Fig. 5.31). The system is eased into
traveling at 35.0 m>s (about 80 mi>h) when it starts its equilibrium position, where it stays. (a) Determine
descending a steep (15°) incline. The incline is icy, so the the change in elastic potential energy of the system.
coefficient of friction is only 0.30. Use the work–energy (b) Determine the system’s change in gravitational
theorem to determine how far the truck will skid potential energy.
(assuming it locks its brakes and skids the whole way)
before it comes to rest.
36. ● ● ● If the work required to speed up a car from 10 km>h
to 20 km>h is 5.0 * 103 J, what would be the work
required to increase the car’s speed from 20 km>h to
30 km>h? M
5.4 POTENTIAL ENERGY

30°
37. ● How much more gravitational potential energy does a
1.0-kg hammer have when it is on a shelf 1.2 m high than 䉱 F I G U R E 5 . 3 1 Changes in potential energy
when it is on a shelf 0.90 m high? See Exercise 44.
EXERCISES 177
5.5 CONSERVATION OF ENERGY ?

45. ● A 0.300-kg ball is thrown vertically upward with an ini- C
tial speed of 10.0 m>s. If the initial potential energy is
taken as zero, find the ball’s kinetic, potential, and 5.0 m/s
8.0 m
mechanical energies (a) at its initial position, (b) at 2.50 m
A
above the initial position, and (c) at its maximum height.
5.0 m
46. ● What is the maximum height reached by the ball in
Exercise 45?
B
47. ●● Referring to Fig. 3.13, find the speed with which the
stone strikes the water using energy considerations.
䉱 F I G U R E 5 . 3 3 Energy conversion(s) See Exercise 53.
48. IE ● ● A girl swings back and forth on a swing with
ropes that are 4.00 m long. The maximum height she
reaches is 2.00 m above the ground. At the lowest point
54. ●● A simple pendulum has a length of 0.75 m and a bob
of the swing, she is 0.500 m above the ground. (a) The
with a mass of 0.15 kg. The bob is released from an angle
girl attains the maximum speed (1) at the top, (2) in the
of 25° relative to a vertical reference line (䉲 Fig. 5.34).
middle, (3) at the bottom of the swing. Why? (b) What is
(a) Show that the vertical height of the bob when it is
the girl’s maximum speed?
released is h = L11 - cos 25°2. (b) What is the kinetic
49. ●● A 1.00-kg block (M) is on a flat frictionless surface energy of the bob when the string is at an angle of 9.0°?
(䉲 Fig. 5.32). This block is attached to a spring initially at (c) What is the speed of the bob at the bottom of the
its relaxed length (spring constant is 50.0 N>m). A light swing? (Neglect friction and the mass of the string.)
string is attached to the block and runs over a frictionless
pulley to a 450-g dangling mass (m). If the dangling mass
is released from rest, how far does it fall before stopping?
θ = 25° L
L
䉱 F I G U R E 5 . 3 2 How far does it go? See Exercise 49. h

y=0 m
50. IE ● ● A 500-g (small) mass on the end of a 1.50-m-long
string is pulled aside 15° from the vertical and shoved
䉱 F I G U R E 5 . 3 4 A pendulum swings See Exercise 54.
downward (toward the bottom of its motion) with a
speed of 2.00 m>s. (a) Is the angle on the other side
(1) greater than, (2) less than, or (3) the same as the angle
55. ●● Suppose the simple pendulum in Exercise 54 were
on the initial side (15°)? Explain in terms of energy.
released from an angle of 60°. (a) What would be the
(b) Calculate the angle it goes to on the other side,
speed of the bob at the bottom of the swing? (b) To what
neglecting air resistance.
height would the bob swing on the other side? (c) What
51. ●● A 0.20-kg rubber ball is dropped from a height of angle of release would give half the speed of that for the
1.0 m above the floor and it bounces back to a height of 60° release angle at the bottom of the swing?
0.70 m. (a) What is the ball’s speed just before hitting the
56. ●● A 1.5-kg box that is sliding on a frictionless surface
floor? (b) What is the speed of the ball just as it leaves the
with a speed of 12 m>s approaches a horizontal spring.
ground? (c) How much energy was lost and where did
(See Fig. 5.19.) The spring has a spring constant of
it go?
2000 N>m. If one end of the spring is fixed and the other
52. ●● A skier coasts down a very smooth, 10-m-high slope end changes its position, (a) how far will the spring be
similar to the one shown in Fig. 5.21. If the speed of the compressed in stopping the box? (b) How far will the
skier on the top of the slope is 5.0 m>s, what is his speed spring be compressed when the box’s speed is reduced
at the bottom of the slope? to half of its initial speed?
53. ●● A roller coaster travels on a frictionless track as 57. ●● A 0.50-kg mass is suspended on a spring that
shown in 䉴 Fig. 5.33. (a) If the speed of the roller coaster stretches 3.0 cm. (a) What is the spring constant?
at point A is 5.0 m>s, what is its speed at point B? (b) Will (b) What added mass would stretch the spring an addi-
it reach point C? (c) What minimum speed at point A is tional 2.0 cm? (c) What is the change in potential energy
required for the roller coaster to reach point C? when the mass is added?
58. ●● A vertical spring with a force constant of 300 N>m is 65. ● The two 0.50-kg weights of a cuckoo clock descend
compressed 6.0 cm and a 0.25-kg ball placed on top. The 1.5 m in a three-day period. At what rate is their total
spring is released and the ball flies vertically upward. gravitational potential energy decreased?
How high does the ball go? 66. ● ● A pump lifts 200 kg of water per hour a height of
59. ● ● A block with a mass m1 = 6.0 kg sitting on a friction- 5.0 m. What is the minimum necessary power output
less table is connected to a suspended mass m2 = 2.0 kg rating of the water pump in watts and horsepower?
by a light string passing over a frictionless pulley. Using 67. ● ● A race car is driven at a constant velocity of
energy considerations, find the speed at which m2 hits 200 km>h on a straight, level track. The power delivered
the floor after descending 0.75 m. (Note: A similar prob- to the wheels is 150 kW. What is the total resistive force
lem in Example 4.6 was solved using Newton’s laws.) on the car?
60. ● ● ● A hiker plans to swing on a rope across a ravine in 68. ● ● An electric motor with a 2.0-hp output drives a
the mountains, as illustrated in 䉲 Fig. 5.35, and to drop machine with an efficiency of 40%. What is the energy
when she is just above the far edge. (a) At what horizon- output of the machine per second?
tal speed should she be moving when she starts to 69. ● ● Water is lifted out of a well 30.0 m deep by a motor
swing? (b) Below what speed would she be in danger of rated at 1.00 hp. Assuming 90% efficiency, how many
falling into the ravine? Explain. kilograms of water can be lifted in 1 min?
70. ● ● How much power must you exert to horizontally
drag a 25.0-kg table 10.0 m across a brick floor in 30.0 s at

constant velocity, assuming the coefficient of kinetic fric-
tion between the table and floor is 0.550?
71. ● ● ● A 3250-kg aircraft takes 12.5 min to achieve its
cruising altitude of 10.0 km and cruising speed of

L = 4.0 m 850 km>h. If the plane’s engines deliver, on average,
1500 hp during this time, what is the efficiency of the
engines?
vo 72. ● ● ● A sleigh and driver with a total mass of 120 kg are
pulled up a hill with a 15° incline by a horse, as illus-

trated in 䉲 Fig. 5.36. (a) If the overall retarding frictional
force is 950 N and the sled moves up the hill with a con-
1.8 m
stant velocity of 5.0 km>h, what is the power output of
the horse? (Express in horsepower, of course. Note the
magnitude of your answer, and explain.) (b) Suppose
that in a spurt of energy, the horse accelerates the sled
uniformly from 5.0 km>h to 20 km>h in 5.0 s. What is the
䉱 F I G U R E 5 . 3 5 Can she make it? See Exercise 60. horse’s maximum instantaneous power output? Assume
the same force of friction.
61. ● ● ● In Exercise 52, if the skier has a mass of 60 kg and
the force of friction retards his motion by doing 2500 J of

work, what is his speed at the bottom of the slope?
62. ● ● ● A 1.00-kg block (M) is on a frictionless, 20° inclined
plane. The block is attached to a spring 1k = 25 N>m2
that is fixed to a wall at the bottom of the incline. A light
string attached to the block runs over a frictionless pul-
ley to a 40.0-g suspended mass. The suspended mass is
given an initial downward speed of 1.50 m>s. How far
does it drop before coming to rest? (Assume the spring is
unlimited in how far it can stretch.) 15°
f
5.6 POWER
䉱 F I G U R E 5 . 3 6 A one-horse open sleigh See Exercise 72.
63. A girl consumes 8.4 * 106 J (2000 food calories) of
●
energy per day while maintaining a constant weight. 73. ● ● ● A construction hoist exerts an upward force of
What is the average power she produces in a day? 500 N on an object with a mass of 50 kg. If the hoist
64. ● A 1500-kg race car can go from 0 to 90 km>h in 5.0 s. started from rest, determine the power it expended to
What average power is required to do this? lift the object vertically for 10 s under these conditions.
74. Two identical springs (neglect their masses) are used to 14.7 m. (a) Show numerically that total mechanical
“play catch” with a small block of mass 100 g (䉲 Fig. 5.37). energy is not conserved during this part of the ball’s
Spring A is attached to the floor and compressed 10.0 cm motion. (b) Determine the work done on the ball by the
with the mass on the end of it (loosely). Spring A is force of air resistance. (c) Calculate the average air resis-
released from rest and the mass is accelerated upward. It tance force on the ball and the ball’s average acceleration.
impacts the spring attached to the ceiling, compresses it 77. An ideal spring of force constant k is hung vertically
2.00 cm, and stops after traveling a distance of 30.0 cm from the ceiling, and a held object of mass m is attached
from the relaxed position of spring A to the relaxed posi- to the loose end. You carefully and slowly ease that mass
tion of spring B as shown. Determine the spring constant down to its equilibrium position by keeping your hand
of the two springs (same since they are identical). under it until it reaches that position. (a) Show that the
mg
spring’s change in length is given by d = . (b) Show
k
m2g 2
that the work done by the spring is Wsp = - .
2k
m2g 2
2.00 cm (c) Show that the work done by gravity is Wg = .
k
30.0 cm Explain why these two works do not add to zero. Since
the overall change in kinetic energy is zero, you might
think they should, no? (d) Show that the work done by
10.0 cm m2g 2
your hand is Whand = - and that the hand exerted
2k
an average force of half the object’s weight.
78. A winch is capable of hauling a ton of bricks vertically

two stories (6.25 m) in 19.5 s. If the winch’s motor is
䉱 F I G U R E 5 . 3 7 Playing catch See Exercise 74. rated at 5.00 hp, determine its efficiency during raising
the load.
75. A 200-g ball is launched from a height of 20.0 m above a
lake. Its launch angle is 40° and it has an initial kinetic 79. IE A 0.455-kg soccer ball is kicked off level ground at an
energy of 90.0 J. (a) Use energy methods to determine its angle of 40° with an initial speed of 30.0 m>s. Neglecting
maximum height above the lake surface. (b) Use projec- air resistance, (a) at its maximum height off the ground,
tile motion kinematics to repeat part (a). (c) Use energy its kinetic energy will be (1) less than its value at launch,
methods to determine its speed just before impact with but not zero, (2) more than its value at launch, (3) zero.
the water. (d) Repeat part (c) using projectile motion Explain. (b) Determine its kinetic energy when it is at its
kinematics. maximum height above the ground and compare it to
76. A 1.20-kg ball is projected straight upward with an initial the kinetic energy at launch. [Hint: What is its velocity at
speed of 18.5 m>s and reaches a maximum height of the top of its arc? Review projectile motion if necessary.]
Linear Momentum
6 and Collisions
6.1 Linear momentum (181)
■ mass * velocity
6.2 Impulse (186)

■ change in momentum
6.3 Conservation
of linear momentum (189)
■ net force zero,
momentum conserved
6.4 Elastic and

inelastic collisions (195)
■ kinetic energy conserved
■ kinetic energy not conserved PHYSICS FACTS
(linear momentum
conserved in both)
✦ Momentum is the Latin word for
“motion.”
✦ Newton called momentum a
“quantity of motion.” From the
T omorrow, sportscasters may
say that the momentum of the
entire game changed as a result of
Principia: The quantity of motion is
6.5 Center of mass (203)
the measure of the same, arising
the clutch hit shown in the photo-
■ center of gravity
from the velocity and quantity of graph. One team is said to have
matter, conjointly.
✦ Newton called impulse a “motive
gained momentum and went on to
6.6 Jet propulsion
force.” win the game. But regardless of the
✦ A collision is the meeting or inter-
and rockets (208)
action of particles or objects, effect on the team, it’s clear that
resulting in an exchange of energy
and/or momentum.
the momentum of the ball in the
✦ There does not have to be physical chapter-opening photograph must
contact for a collision. A spacecraft
in a gravity-assisted fly-by of a have changed dramatically. The
planet is in a collision (energy and
ball was traveling toward the plate
momentum transfer).
✦ It is a common misconception that at a good rate of speed—with a lot
on rocket blastoff, the fiery engine
exhaust striking and “pushing”
of momentum. But a collision with
against the launch pad propels the a hardwood bat—with plenty of
rocket upward. If this were the
case, how could rocket engines be momentum of its own—changed
used in space, where there is noth-
ing to “push” against?
the ball’s direction in a fraction of a
6.1 LINEAR MOMENTUM 181
second. A fan might say that the batter turned the ball around. After studying
Newton’s second law in Section 4.3, you might say that the force the bat applied to
the ball gave it a large acceleration, reversing its velocity vector. Yet if you summed
the momenta (plural of momentum) of the ball and bat just before the collision
and just afterward, you’d discover that although both the ball and the bat had
momentum changes, the total momentum didn’t change.
If you were bowling and the ball bounced off the pins and rolled back toward
you, you would probably be very surprised. But why? What leads us to expect that
the ball will send the pins flying and continue on its way, rather than rebounding?
You might say that the momentum of the ball carries it forward even after the colli-
sion (and you would be right)—but what does that really mean? In this chapter,
the concept of momentum will be studied and you will learn how it is particularly
useful in analyzing motion and collisions.
6.1 Linear Momentum

➥ What is meant by the total linear momentum of a system?

➥ How is momentum related to Newton’s second law?
The term momentum may bring to mind a football player running down the
field, knocking down players who are trying to stop him. Or you might have
heard someone say that a team lost its momentum (and so lost the game). Such
everyday usages give some insight into the meaning of momentum. They sug-
gest the idea of mass in motion and therefore inertia. We tend to think of heavy
or massive objects in motion as having a great deal of momentum, even if they
move very slowly. However, according to the technical definition of momen-
tum, a light object can have just as much momentum as a heavier one, and
sometimes more.
Newton referred to what modern physicists term linear momentum (p) as “the
quantity of motion Á arising from velocity and the quantity of matter conjointly.”
In other words, the momentum of a body is proportional to both its mass and
velocity. By definition,
the linear momentum of an object is the product of its mass and velocity:
B B
p = mv (6.1)
SI unit of momentum: kilogram-meter per second 1kg # m>s2

It is common to refer to linear momentum as simply momentum. Momentum is a
vector quantity that has the same direction as the velocity, and x- and y-components
with magnitudes of px = mvx and py = mvy , respectively.
Equation 6.1 expresses the momentum of a single object or particle. For a sys-
B
tem of more than one particle, the total linear momentum (P) of the system is the
vector sum of the momenta of the individual particles:
B B
P = p B B
1 + p2 + p3 +
Á = gp
B
i (6.2)
B B
(Note: P signifies the total momentum, while p signifies an individual momentum.)
182 6 LINEAR MOMENTUM AND COLLISIONS
EXAMPLE 6.1 Momentum: Mass and Velocity

A 100-kg football player runs with a velocity of 4.0 m>s straight The magnitude of the momentum of the football player is
pp = mp vp = 1100 kg214.0 m>s2 = 4.0 * 102 kg # m>s
down the field. A 1.0-kg artillery shell leaves the barrel of a gun
with a muzzle velocity of 500 m>s. Which has the greater
momentum (magnitude), the football player or the shell? and that of the shell is
THINKING IT THROUGH. Given the mass and velocity of an ps = ms vs = 11.0 kg21500 m>s2 = 5.0 * 102 kg # m>s
object, the magnitude of its momentum can be calculated
from Eq. 6.1. Thus, the less massive shell has the greater momentum.
Remember, the magnitude of momentum depends on both the
SOLUTION. As usual, first listing the given data and using the mass and the magnitude of the velocity.
subscripts p and s to refer to the player and shell, respectively;
F O L L O W - U P E X E R C I S E . What would the football player’s
Given: mp = 100 kg Find: pp and ps (magnitudes of speed have to be for his momentum to have the same magni-
vp = 4.0 m>s the momenta) tude as the artillery shell’s momentum? Would this speed be
ms = 1.0 kg realistic? (Answers to all Follow-Up Exercises are given in
vs = 500 m>s Appendix VI at the back of the book.)
INTEGRATED EXAMPLE 6.2 Linear Momentum: Some Ballpark Comparisons

Consider the three objects shown in 䉲 Fig. 6.1—a .22-caliber of the glacier counterbalances its huge mass to make its
bullet, a cruise ship, and a glacier. Assuming each to be mov- momentum less than might be expected. Assuming the speed
ing at its normal speed, (a) which would you expect to have difference to be greater than the mass difference for the ship
the greatest linear momentum: (1) the bullet, (2) the ship, or and glacier, the ship would have the larger momentum. Simi-
(3) the glacier? (b) Estimate the masses and speeds and com- larly, because of the fast bullet’s relatively tiny mass, it would
pute order-of-magnitude values of the linear momentum of be expected to have the least momentum. So with this reason-
the objects. ing, the largest momentum goes to the ship and the smallest
(A) CONCEPTUAL REASONING. Certainly the bullet travels the momentum to the bullet, and the answer would be (2).
fastest and the glacier the slowest, with the cruise ship in ( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . With no phys-
between. But momentum, p = mv, is equally dependent on ical data given, you are asked to estimate the masses and
mass as well as speed. The fast bullet has a tiny mass compared velocities (speeds) of the objects so as to be able to compute
with that of the ship and the glacier. The slow glacier has a their momenta [which will verify the reasoning in part (a)].
huge mass that greatly overshadows that of the bullet, and to As is often the case in real-life problems, it may be difficult to
some extent that of the ship. The cruise ship weighs a great estimate the values, so you would try to look up approximate
deal and has considerable mass. Which object has the greater values for the various quantities. For this example, the esti-
momentum also depends on the relative speeds. The glacier mates will be provided. (Note that the units given in refer-
“creeps” along compared with the ship, so the very slow speed ences vary, and it is important to convert units correctly.)
(a) (b) (c)
䉱 F I G U R E 6 . 1 Three moving objects: a comparison of momenta and kinetic energies (a) A .22-caliber bullet shat-
tering a ballpoint pen; (b) a cruise ship; (c) a glacier, Glacier Bay, Alaska. See Example text for description.
Given: Estimates of weight (mass) and speed for the bullet, Glacier: The glacier might be 1 km wide, 10 km long, and
cruise ship, and glacier. 250 m deep, and move at a rate of about 1.0 m per day. (There
is much variation among glaciers. Therefore, these figures
Find: The approximate magnitudes of the momenta for the
must involve more assumptions and rougher estimates than
bullet (pb), cruise ship (ps), and glacier (pg).
those for the bullet or ship. For example, a uniform, rectangu-
Bullet: A typical .22-caliber bullet has a weight of about 30 lar cross-sectional area is assumed for the glacier. The depth is
grains and a muzzle velocity of about 1300 ft>s. (A grain, particularly difficult to estimate from a photograph; a mini-
abbreviated gr, is an old British unit. It was once commonly mum value is given by the fact that glaciers must be at least
used for pharmaceuticals, such as 5-gr aspirin tablets; 50–60 m thick before they can “flow.” Observed speeds range
1 lb = 7000 gr.) from a few centimeters to as much as 40 m a day for valley
glaciers such as the one shown in Fig. 6.1c. The value chosen
Ship: A ship like the one shown in Fig. 6.1b would have a here is considered a typical one.)
weight of about 70 000 tons and a speed of about 20 knots. (A
knot is another old unit, still commonly used in nautical con-
texts; 1 knot = 1.15 mi>h.)
Then, converting the data to metric units and giving orders of magnitude yield the following:
Bullet:
1 kg
ba b = 0.0019 kg L 10-3 kg
1 lb
mb = 30 gra
7000 gr 2.2 lb
vb = 11.3 * 103 ft>s2a b = 4.0 * 102 m>s L 102 m>s

0.305 m>s
ft>s
Ship:
2.0 * 103 lb 1 kg
ms = 7.0 * 104 ton a ba b = 6.4 * 107 kg L 108 kg
ton 2.2 lb
ba b = 10 m>s = 101 m>s

1.15 mi>h 0.447 m>s
vs = 20 knotsa
knot mi>h
Glacier:
width w L 103 m, length l L 104 m, depth d L 102 m
1 day
vg = 11.0 m>day2a b = 1.2 * 10-5 m>s L 10-5 m>s
86 400 s
We have all the speeds and masses except for mg, the mass of the glacier. To compute this value, the density of ice is needed,
since m = rV (Eq. 1.1). The density of ice is less than that of water (ice floats in water), but the two are not very different, so the
density of water, 1.0 * 103 kg>m3, will be used to simplify the calculations.
Thus, the mass of the glacier is approximated as
mg = rV = r1l * v * d2
L 1103 kg>m321104 m21103 m21102 m2 = 1012 kg
Then, calculating the magnitudes of the momenta of the objects,
Bullet: pb = mb vb L 110-3 kg21102 m>s2 = 10-1 kg # m>s
Ship: ps = ms vs L 1108 kg21101 m>s) = 109 kg # m>s
Glacier: pg = mg vg L 11012 kg2110-5 m>s2 = 107 kg # m>s
So the ship does have the largest momentum, and the bullet has the smallest according to the estimates.
F O L L O W - U P E X E R C I S E . Which of the objects in this Example has (1) the greatest kinetic energy and (2) the least kinetic energy?
Justify your choices using order-of-magnitude calculations. (Notice here that the dependence is on the square of the speed, K = 12 mv 2)
EXAMPLE 6.3 Total Momentum: A Vector Sum

What is the total momentum for each of the systems of parti- T H I N K I N G I T T H R O U G H . The total momentum is the vector
cles illustrated in 䉲 Fig. 6.2a and b? sum of the individual momenta (Eq. 6.2). This quantity can be
computed using the components of each vector.
SOLUTION.
(a) Total momentum 1P2 for Fig. 6.2a

B
Given: Magnitudes and directions Find:
of momenta from Fig. 6.2 (b) Total momentum 1P2 for Fig. 6.2b
B
(a) The total momentum of a system is the vector sum of the momenta of the individual particles, so
P = p1 + p2 = 12.0 kg # m>s2xN + 13.0 kg # m>s2xN = 15.0 kg # m>s2xN 1+ x-direction2

B B B
(b) Computing the total momenta in the x- and y-directions gives
Px = p1 + p2 = 15.0 kg # m>s2xN + 1 -8.0 kg # m>s2xN

B B B
= - 13.0 kg # m>s2xN 1- x-direction2
Py = p3 = 14.0 kg # m>s2yN 1+ y-direction2

B B
Then
P = Px + Py = 1-3.0 kg # m>s2xN + 14.0 kg # m>s2yN

B B B
or
P = 5.0 kg # m>s at 53° relative to the negative x-axis
y y
P = 5.0 kgm/s
x x
p2 = 3.0 kgm/s p1 = 2.0 kgm/s
Individual momenta Total momentum of system

(a) P = p1 + p2
y y
P = 5.0 kgm/s
p3 = 4.0 kgm/s
Py = 4.0 kgm/s
53°
x x
p2 = 8.0 kgm/s p1 = 5.0 kgm/s Px = 3.0 kgm/s

(b) P = p1 + p2 + p3
䉱 F I G U R E 6 . 2 Total momentum The total momentum of a system of particles is the

vector sum of the particles’ individual momenta. See Example text for description.
B B B B
FOLLOW-UP EXERCISE. In this Example, if p1 and p2 in part (a) were added to p2 and p3 in part (b), what would be the total
momentum?
y
p2
p2
u
– +
Δp x
u
p2 p1 p2
p1
u
Δp = p2 – p1 = Δpx + Δpy
Δp u Δpx = p2x – p1x
p1 = (p2 cos u)xn – (– p1 cos u)xn
Δp = p2 – p1 = (+2p cos u)xn
= (mv)xn – (–mv)xn = (+2mv)xn
Δpy = p2y – p1y
p1
= (p2 sin u)yn – (p1 sin u)yn = 0
(a) (b)
䉱 F I G U R E 6 . 3 Change in momentum The change in momentum is given by the difference

in the momentum vectors. (a) Here, the vector sum is zero, but the vector difference, or
change in momentum, is not. (The particles are displaced for convenience.) (b) The change
in momentum is found by computing the change in the components.
In Example 6.3a, each of the momenta were along one of the coordinate axes
and thus were added straightforwardly. If the motion of one (or more) of the parti-
cles is not along an axis, its momentum vector may be broken up, or resolved, into
rectangular components, and then individual components can be added to find
the components of the total momentum, just as you learned to do with force com-
ponents in Section 4.3.
Since momentum is a vector, a change in momentum can result from a change
in magnitude and>or direction. Examples of changes in the momenta of particles
because of changes of direction on collision are illustrated in 䉱 Fig. 6.3. In the fig-
ure, the magnitude of a particle’s momentum is taken to be the same both before
and after collision (as indicated by the arrows of equal length). Figure 6.3a illus-
trates a direct rebound—a 180° change in direction. Note that the change in
momentum 1¢p B
2 is the vector difference and that directional signs for the vectors
are important. Figure 6.3b shows a glancing collision, for which the change in
momentum is given by analyzing changes in the x- and y-components.
FORCE AND MOMENTUM

As you know from Section 4.3, if an object has a change in velocity, a net force
must be acting on it. Similarly, since momentum is directly related to velocity, a
change in momentum also requires a net force. In fact, Newton originally
expressed his second law of motion in terms of momentum rather than accelera-
B
tion. The force–momentum relationship may be seen by starting with Fnet = maB
and using aB = 1v - vo2> ¢t, where the mass is assumed to be constant. Thus,
o2
B B B B
B m1v - v B
mv B
- mv o p - p o ¢p
Fnet = maB = = = =
¢t ¢t ¢t ¢t
or
B
B ¢p (Newton’s second law of motion
Fnet = (6.3)
¢t in terms of momentum)
B
where Fnet is the average net force on the object if the acceleration is not constant (or
the instantaneous net force if ¢t goes to zero).
Expressed in this form, Newton’s second law states that the net external force acting
on an object is equal to the time rate of change of the object’s momentum. It is easily seen
>¢t are
B B
from the development of Eq. 6.3 that the equations Fnet = maB and Fnet = ¢p B
䉴 F I G U R E 6 . 4 Change in the y
momentum of a projectile The total
momentum vector of a projectile is p2 = mv2
tangential to the projectile’s path (as
is its velocity); this vector changes in p1= mv1 2
magnitude and direction, because of
the action of an external force (grav- 1 3 p3 = mv3
ity). The x-component of the
momentum is constant. (Why?)
po = mvo
u
x
equivalent if the mass is constant. In some situations, however, the mass may vary.
This factor will not be a consideration here in the discussion of particle collisions,
but a special case will be given later in the chapter. The more general form of New-
ton’s second law, Eq. 6.3, is true even if the mass varies.
B
Just as the equation Fnet = maB indicates that an acceleration is evidence of a net
>¢t indicates that a change in momentum is evidence of a
B B
force, the equation Fnet = ¢p
net force. For example, as illustrated in 䉱 Fig. 6.4, the momentum of a projectile is
tangential to the projectile’s parabolic path and changes in both magnitude and
direction. The change in momentum indicates that there is a net force acting on the
projectile, which of course is the force of gravity. Changes in momentum were
illustrated in Fig. 6.3. Can you identify the forces in these two cases? Think in
terms of Newton’s third law.
DID YOU LEARN?

➥ The vector sum of the momenta of all the individual particles or objects is the total
linear momentum of a system.
> ¢t
B
➥ The time rate of change of momentum is equal to the net force, Fnet = ¢p B
B B
(equivalent to Fnet = ma).
6.2 Impulse
➥ How is the impulse–momentum theorem analogous to the work–energy theorem?

(a)
➥ Can kinetic energy be expressed in terms of momentum?
F ➥ Do objects have to come into contact to have a “collision”?
When two objects—such as a hammer and a nail, a golf club and a golf ball, or
even two cars—collide, they can exert a large force on one another for a short
period of time, or an impulse. (䉳 Fig. 6.5a). The force is not constant in this situation.
However, Newton’s second law in momentum form is still useful for analyzing
such situations by using average values. Written in this form, the law states that
>¢t
B B
the average force is equal to the time rate of change of momentum: Favg = ¢p
t (Eq. 6.3). Rewriting the equation to express the change in momentum (with only
Δt
one force acting on the object),
to tf
B B B B
(b) Favg ¢t = ¢p = p - p o (6.4)
B B
䉱 F I G U R E 6 . 5 Collision impulse The term Favg ¢t is known as the impulse (I ) of the force:
(a) A collision impulse causes the
football to be deformed. (b) The B B B B B
impulse is the area under the curve I = Favg ¢t = ¢p = mv - mv o (6.5)
of an F-versus-t graph. Note that the
impulse force on the ball is not con-
stant, but rises to a maximum. SI unit of impulse and momentum: newton-second 1N # s2
6.2 IMPULSE 187
TABLE 6.1 Some Typical Contact Times (¢t)

≤t (milliseconds)
Golf ball (hit by a driver) 1.0

Baseball (hit off tee) 1.3
Tennis (forehand) 5.0
Football (kick) 8.0
Soccer (header) 23.0 F
Fmax
Thus, the impulse exerted on an object is equal to the change in the object’s momentum.
This statement is referred to as the impulse–momentum theorem. Impulse
has units of newton-second 1N # s2, which are also units of momentum Favg
11 N # s = 1 kg # m>s2 # s = 1 kg # m>s2.
In Section 5.3, it was learned that by the work–energy theorem
1Wnet = Fnet ¢x = ¢K2, the area under an Fnet-versus-x curve is equal to the net t
Δt
work, or change in kinetic energy. Similarly, the area under an Fnet-versus-t curve
is equal to the impulse, or the change in momentum (Fig. 6.5b). Forces between
interacting objects usually varies with time and are therefore not constant forces. 䉱 F I G U R E 6 . 6 Average impulse
However, in general, it is convenient to talk about the equivalent constant average force The area under the average
B force curve (Favg ¢t, within the
force Favg acting over a time interval ¢t to give the same impulse (same area under
dashed red lines) is the same as the
the force-versus-time curve), as shown in 䉴 Fig. 6.6. Some typical contact times in area under the F-versus-t curve,
sports are given in 䉱 Table 6.1. which is usually difficult to evaluate.
EXAMPLE 6.4 Teeing Off: The Impulse–Momentum Theorem

A golfer drives a 0.046-kg ball from an elevated tee, giving the ball an initial horizontal
speed of 40 m>s (about 90 mi>h). What is the magnitude of the average force exerted by
the club on the ball during this time?
T H I N K I N G I T T H R O U G H . The average force on the ball is equal to the time rate of
change of its momentum, and this can be computed (Eq. 6.5). The ¢t of the collision is
obtained from Table 6.1.
SOLUTION.
Given: m = 0.046 kg Find: Favg (average force)

v = 40 m>s
vo = 0
¢t = 1.0 ms = 1.0 * 10-3 s (Table 6.1)
The mass and the initial and final velocities are given, so the change in momentum can
be easily found. Then the magnitude of the average force can be computed from the
impulse–momentum theorem:
Favg ¢t = p - po = mv - mvo
and
10.046 kg2140 m>s2 - 0
= 1.8 * 103 N 1or about 410 lb2
mv - mvo
Favg = =
¢t 1.0 * 10-3 s
[This is a very large force compared with the weight of the ball, w = mg =
10.046 kg219.8 m>s 22 = 0.45 N (or about 0.22 lb).] The force is in the direction of the
acceleration and is the average force. The instantaneous force is even greater than this
value near the midpoint of the time interval of the collision ( ¢t in Fig. 6.6).
F O L L O W - U P E X E R C I S E . Suppose the golfer in this Example drives the ball with the same
average force, but “follows through” on the swing so as to increase the contact time to
1.5 ms. What effect would this change have on the initial horizontal speed of the drive?
Example 6.4 illustrates the large forces that colliding objects can exert on one
another during short contact times. In some cases, the contact time may be short-
ened to maximize the impulse—for example, in a karate chop. However, in other
instances, the ¢t may be manipulated to reduce the force. Suppose there is a fixed
Favg Δt = mvo
change in momentum in a given situation. Then, with ¢p = Favg ¢t, if ¢t could be
made longer, the average impulse Favg would be reduced.
You have probably tried to minimize the impulse on occasion. For example, in
catching a hard, fast-moving ball, you quickly learn not to catch it with your arms
rigid, but rather to move your hands with the ball. This movement increases the con-
tact time and reduces the impulse and the “sting” (䉳 Fig. 6.7).
When jumping from a height onto a hard surface, you should not land stiff-legged.
The abrupt stop (small ¢t) would apply a large impulse to your leg bones and joints
(a) and could cause injury. If you bend your knees as you land, the impulse is vertically
upward, opposite your velocity (Favg ¢t = ¢p = - mvo with the final velocity being
zero). Thus, increasing the time interval ¢t makes the impulse smaller. Another
example in which the contact time is increased to decrease the impulse is given in
Insight 6.1, The Automobile Air Bag and Martian Air Bags.
EXAMPLE 6.5 Impulse and Body Injury

Favg Δt = mvo
A 70.0-kg worker jumps stiff-legged from a height of 1.00 m onto a concrete floor.
(a) What is the magnitude of the impulse he feels on landing, assuming a sudden stop in
8.00 ms? (b) What is the average force?
T H I N K I N G I T T H R O U G H . The impulse is Favg ¢t which cannot be calculated directly
from the given data. But impulse is equal to the change in momentum,
Favg ¢t = ¢p = mv - mvo . So the impulse can be calculated from the difference in
(b) momenta if vo is found.
SOLUTION.
䉱 F I G U R E 6 . 7 Adjust the impulse Given: m = 70.0 kg Find: (a) I (impulse on worker)
(a) The change in momentum in
catching the ball is a constant mvo. h = 1.00 m (b) Favg (average force)
If the ball is stopped quickly (small ¢t = 8.00 ms = 8.00 * 10-3 s
¢t), the impulse force is large (big (a) There are two different parts here: (1) the worker descending after jumping and (2) the
Favg) and stings the catcher’s bare sudden stop after hitting the floor. So we must be careful with notation and distinguish
hands. (b) Increasing the contact between the two parts (a and b) with the subscripts of 1 and 2, respectively. (b) Knowing
time (large ¢t) by moving the hands ¢t, the average force can be calculated.
with the ball reduces the impulse
force and makes catching more (a) Here, the initial velocity of the worker is v1o = 0, and the final velocity just before
enjoyable. hitting the floor may be found using v2 = v 2o - 2gh (Eq. 2.12’), with the result of
v1f = - 2 2gh
where the minus sign indicates direction (downward).
(b) The v1f in (a) is then the initial velocity with which the stiff-legged worker hits the floor,
that is, v2o = v1f = - 2 2gh, and the final velocity at the second phase is v2f = 0. Then,
I = Favg ¢t = ¢p = mv2f - mv2o = 0 - m1 - 2 2gh2 = + m 2 2gh
= 170.0 kg22 219.80 m>s2211.00 m2 = 310 kg # m>s
where the impulse is in the upward direction.
With a ¢t of 8.00 * 10-3 s for the sudden stop on impact, this would give a force of
¢p 310 kg # m>s
Favg = = = 3.88 * 104 N (about 8 730 lb of force!)
¢t 8.00 * 10-3 s
and the force is upward on the stiff legs.
F O L L O W - U P E X E R C I S E . Suppose the worker bent his knees and increased the contact
time to 0.60 s on landing. What would be the average impulse force on him in this case?
In some instances, the impulse force may be relatively constant and the contact
time 1¢t2 deliberately increased to produce a greater impulse, and thus a greater
change in momentum 1Favg ¢t = ¢p2. This is the principle of “following through”
in sports, for example, when hitting a ball with a bat or racquet, or driving a golf ball.
6.3 CONSERVATION OF LINEAR MOMENTUM 189
䉴 F I G U R E 6 . 8 Increasing the
contact time (a) A golfer follows
through on a drive. One reason he
does so is to increase the contact time
so that the ball receives greater
impulse and momentum. (b) The
follow-through on a long putt
increases the contact time for greater
momentum, but the main reason
here is for directional control.
(a)
In the latter case (䉴 Fig. 6.8a), assuming that the golfer supplies the same average
force with each swing, the longer the contact time, the greater the impulse or
change in momentum the ball receives. That is, with Favg ¢t = mv (since vo = 0), (b)
the greater the value of ¢t, the greater the final speed of the ball. (This principle
was illustrated in the Follow-Up Exercise in Example 6.4.) In some instances, a
long follow-through may primarily be used to improve control of the ball’s direc-
tion (Fig. 6.8b).
The word impulse implies that the impulse force acts only briefly (like an
“impulsive” person), and this is true in many instances. However, the definition
of impulse places no limit on the time interval of a collision over which the force
may act. Technically, a comet at its closest approach to the Sun is involved in a col-
lision, because in physics, collision forces do not have to be contact forces. Basi-
cally, a collision is any interaction between objects in which there is an exchange
of momentum and>or energy.
As you might expect from the work–energy theorem and the impulse–momentum
theorem, momentum and kinetic energy are directly related. A little algebraic
manipulation of the equation for kinetic energy (Eq. 5.5) allows us to express
kinetic energy (K) in terms of the magnitude of momentum (p):
1mv22 p2
K = 12 mv 2 = = (6.6)
2m 2m
Thus, kinetic energy and momentum are intimately related, but they are different
quantities.
DID YOU LEARN?

➥ Impulse is equal to the change in momentum. Work is equal to the change in
kinetic energy.
➥ An interaction between objects in which there is an exchange of momentum
and/or energy is a collision. Direct contact is not needed, for example, a comet
making a pass around the Sun is in a collision.
➥ In terms of momentum, the kinetic energy is K = p2>2m.
6.3 Conser vation of Linear Momentum

➥ Why is the total momentum of a system conserved if the net force on the system is
zero?
➥ Why is the net internal force of a system always zero?
Like total mechanical energy, the total momentum of a system is a conserved quan-
tity under certain conditions. This fact allows us to analyze a wide range of situa-
tions and solve many problems readily. Conservation of momentum is one of the
most important principles in physics. In particular, it is used to analyze collisions of
objects ranging from subatomic particles to automobiles in traffic accidents.
INSIGHT 6.1 The Automobile Air Bag and Martian Air Bags
A dark, rainy night—a car goes out of control and hits a big explosion that generates gas to inflate the bag at an explosive
tree head-on! But the driver walks away with only minor rate. The complete process from sensing to full inflation takes
injuries, because he had his seatbelt buckled and his car’s air only on the order of 25 thousandths of a second (0.025 s).
bags deployed. Air bags, along with seatbelts, are safety Air bags have saved many lives. However, in some cases,
devices designed to prevent (or lessen) injuries to passengers the deployment of air bags has caused problems. An air bag is
in automobile collisions. not a soft, fluffy pillow. When activated, it is ejected out of its
When a car collides with something basically immovable, compartment at speeds up to 320 km>h (200 mi>h) and can hit
such as a tree or a bridge abutment, or has a head-on collision a person with enough force to cause severe injury and even
with another vehicle, the car stops almost instantaneously. If death. Adults are advised to sit at least 13 cm (6 in.) from the
the front-seat passengers have not buckled up (and there are air bag compartment and to buckle up—always. Children
no air bags), they keep moving until acted on by an external should sit in the rear seat, out of the reach of air bags.*
force (by Newton’s first law). For the driver, this force is sup-
plied by the steering wheel and column, and for the passen- MARTIAN AIR BAGS
ger, by the dashboard and>or windshield.
Air bags on Mars? They were there in 1997 when a robotic rover
Even when everyone has buckled up, there can be injuries.
from the spacecraft Pathfinder landed on Mars. And in 2004,
Seatbelts absorb energy by stretching, and they widen the
area over which the force is exerted. However, if a car is
going fast enough and hits something truly immovable, there
may be too much energy for the belts to absorb. This is where
the air bag comes in (Fig. 1).
The bag inflates automatically on hard impact, cushioning
the driver (and front-seat passenger if both sides are
equipped with air bags). In terms of impulse, the air bag
increases the stopping contact time—the fraction of a second
it takes your head to sink into the inflated bag is many times
longer than the instant in which you would have stopped oth-
erwise by hitting a solid surface such as the windshield. A
longer contact time means a reduced average impact force
and thus less likelihood of an injury. (Because the bag is large,
the total impact force is also spread over a greater area of the
body, so the force on any one part of the body is also less.)
How does an air bag inflate during the little time that elapses
between a front-end impact and the instant the driver would 䉱 F I G U R E 1 Impulse and safety An automobile air bag
hit the steering column? An air bag is equipped with sensors increases the contact time that a person in a crash would
that detect the sharp deceleration associated with a head-on experience with the dashboard or windshield, thereby
collision the instant it begins. If the deceleration exceeds the decreasing the impulse force that could cause injury.
sensors’ threshold settings, a control unit sends an electric *Guidelines from the National Highway Traffic Safety Adminis-
current to an igniter in the air bag, which sets off a chemical tration (www.nhtsa.gov).
For the linear momentum of a single object to be conserved (that is, to remain con-
stant with time), one condition must hold that is apparent from the momentum form
of Newton’s second law (Eq. 6.3). If the net force acting on a particle is zero, that is,
B
B ¢p
Fnet = = 0
¢t
then
B B B
¢p = 0 = p - p o
B B
where p o is the initial momentum and p is the momentum at some later time.
Since these two values are equal, the momentum is conserved, and
B B B B
p = p o or mv = mvo
final momentum = initial momentum
Note that this conservation is consistent with Newton’s first law: An object
remains at rest 1p
B B
= 02, or in motion with a uniform velocity (constant p Z 0),
unless acted on by a net external force.
(a) (b)
more air bags arrived with the Mars Exploration Rover Mission
and the touchdown of two Rovers. Spacecraft landings are usu-
ally softened by retrorockets fired intermittently toward the
planet surface. However, firing retrorockets very near the Mart-
ian surface would have left trace amounts of foreign combus-
tion chemicals on the surface. Since one objective of the Mars
missions was to analyze the chemical composition of Martian
rocks and soil, another method of landing had to be developed.
The solution? Probably the most expensive air bag system
ever created, costing approximately $5 million to develop and
install. The Rovers were surrounded by 4.6-m-(15-ft)-diameter
“beach balls” for an air bag landing (Fig. 2a).
On entering the Martian atmosphere, the spacecraft was trav- (c)
eling at about 27 000 km>h (17 000 mi>h). A high-altitude
rocket system and parachute slowed it down to about F I G U R E 2 More bounce to the ounce (a) “Beach ball” air
80 – 100 km>h (50–60 mi>h). At an altitude of about 200 m bags were used to protect Pathfinder and Mars Rovers.
(660 ft), gas generators inflated the air bags, which allowed the (b) Artist’s conception of bouncing air bags of a Mars Rover.
bag-covered Rovers to bounce and roll a bit on landing (Fig. 2b). (c) A Rover coming out safely.
The air bags then deflated, and out rolled the Rovers (Fig. 2c).
The conservation of momentum can be extended to a system of particles if

Newton’s second law is written in terms of the net force acting on the system and
of the momenta of the particles: Fnet = g Fi and P = gp i = gmivi.
B B B B B
Because Fnet = ¢P>¢t, and if there is no net external force acting on the system, then
B B
B B B B
Fnet = 0, and ¢F = 0; so P = Po , and the total momentum is conserved. This general-
ized condition is referred to as the law of conservation of linear momentum:
B B
P = Po (6.7)
Thus, the total linear momentum of a system, P = gp

B B
i , is conserved if the net
external force acting on the system is zero.
There are various ways to achieve this condition. For example, recall from
Section 5.5 that a closed, or isolated, system is one on which no net external force
acts, so the total linear momentum of an isolated system is conserved.
Within a system, internal forces may act—for example, when particles collide.
These are force pairs of Newton’s third law, and there is a good reason that such forces
are not explicitly referred to in the condition for the conservation of momentum.
m2 =
m1 =
1.0 kg 2.0 kg
v2
v1
m2 =
m1 =
1.0 kg 2.0 kg
− x=0 +
䉱 F I G U R E 6 . 9 An internal force and the conservation of

momentum The spring force is an internal force, so the
momentum of the system is conserved. See Example 6.6.
By Newton’s third law, these internal forces are equal and opposite and vectorially
cancel each other. Thus, the net internal force of a system is always zero.
An important point to understand, however, is that the momenta of individual
particles or objects within a system may change. But in the absence of a net exter-
B
nal force, the vector sum of all the momenta (the total system momentum P)
remains the same. If the objects are initially at rest (that is, the total momentum is
zero) and then are set in motion as the result of internal forces, the total momen-
tum must still add to zero. This principle is illustrated in 䉱 Fig. 6.9 and analyzed in
Example 6.6. Objects in an isolated system may transfer momentum among them-
selves, but the total momentum after the changes must add up to the initial value,
assuming the net external force on the system is zero.
The conservation of momentum is often a powerful and convenient tool for
analyzing situations involving motion and collisions. Its application is illustrated
in the following Examples. (Notice that conservation of momentum, in many
cases, bypasses the need to know the forces involved.)
EXAMPLE 6.6 Before and After: Conservation of Momentum

Two masses, m1 = 1.0 kg and m2 = 2.0 kg, are held on either of the system is conserved. It should be apparent that the ini-
tial total momentum of the system 1Po2 is zero, and therefore
B
side of a light compressed spring by a light string joining
them, as shown in Fig. 6.9. The string is burned (negligible the final momentum must also be zero. Thus,
external force), and the masses move apart on the frictionless B B B B B
Po = P = 0 and P = p1 + p2 = 0
surface, with m1 having a velocity of 1.8 m>s to the left. What
is the velocity of m2 ? (The momentum of the “light” spring does not come into the
equations, because its mass is negligible.) Then,
T H I N K I N G I T T H R O U G H . With no net external force (the
B B
weights are each canceled by a normal force), the total p2 = - p1
momentum of the system is conserved. It is initially zero, so which means that the momenta of m1 and m2 are equal and
after the string is burned, the momentum of m2 must be equal opposite. Using directional signs (with + indicating the direc-
to and opposite that of m1. (Vector addition gives zero total tion to the right in the figure),
momentum. Also, note that the term light indicates that the
masses of the spring and string can be ignored.) m2 v2 = - m1 v1
SOLUTION. Listing the data: and

1.0 kg
v2 = - a bv = - a b1- 1.8 m>s2 = + 0.90 m>s
m1
Given: m1 = 1.0 kg Find: v2 (velocity—
m2 = 2.0 kg speed and m2 1 2.0 kg
v1 = - 1.8 m>s (left) direction)
Thus, the velocity of m2 is 0.90 m>s in the positive x-direction,
Here, the system consists of the two masses and the spring. or to the right in the figure. This value is half that of v1 as you
Since the spring force is internal to the system, the momentum might have expected, since m2 has twice the mass of m1.
F O L L O W - U P E X E R C I S E . (a) Suppose that the large block in with negligible friction. The first girl tosses a 2.5-kg ball to
Fig. 6.9 were attached to the Earth’s surface so that the block the second. If the speed of the ball is 10 m>s, what is the
could not move when the string was burned. Would speed of each girl after the ball is caught, and what is the
momentum be conserved in this case? Explain. (b) Two girls, momentum of the ball before it is tossed, while it is in the
each having a mass of 50 kg, stand at rest on skateboards air, and after it is caught?
INTEGRATED EXAMPLE 6.7 Conservation of Linear Momentum: Fragments and Components

A 30-g bullet with a speed of 400 m>s strikes a glancing blow y vb
to a target brick of mass 1.0 kg. The brick breaks into two frag-
ments. The bullet deflects at an angle of 30° above the + x-axis
and has a reduced speed of 100 m>s. One piece of the brick vb y
(with mass 0.75 kg) goes off to the right, or in the initial direc- v1
30° vbx
tion of the bullet, with a speed of 5.0 m>s. (a) Taking the
x
+ x-axis to the right, will the other piece of the brick move in u2 v2x
the (1) second quadrant, (2) third quadrant, or (3) fourth v2y
quadrant? (b) Determine the speed and direction of the other v2
piece of the brick immediately after collision (where gravity
can be neglected).
(A) CONCEPTUAL REASONING. The conservation of linear Before After
momentum can be applied because there is no net external vb
force on the system— the bullet and brick. Initially, all of the vbo 30°
momentum is in the forward +x-direction, (䉴 Fig. 6.10). After-
ward, one piece of the brick flies off in the +x-direction, and θ 2 v1
v2
the bullet at an angle of 30° to the x-axis. The bullet’s momen-
tum has a positive y-component, so the other piece of the
brick must have a negative y-component because there was 䉱 F I G U R E 6 . 1 0 A glancing collision Momentum is con-
no initial momentum in the y-direction. Hence, with the total served in an isolated system. The motion in two dimensions
momentum in the + x-direction (before and after), the answer may be analyzed in terms of the components of momentum,
is (3) fourth quadrant. which are also conserved.
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . There is one object with momentum before collision (the bullet), and three with
momenta afterward (the bullet and the two fragments). By the conservation of linear momentum, the total (vector) momentum
after collision equals that before collision. As is often the case, a sketch of the situation is helpful, with the vectors resolved in
component form (Fig. 6.10). Applying the conservation of linear momentum should allow the velocity (speed and direction) of
the second fragment to be determined.
Given: mb = 30 g = 0.030 kg Find: v2 (speed of the smaller brick fragment)
vbo = 400 m>s (initial bullet speed) u2 (direction of the fragment relative to
vb = 100 m>s (final bullet speed) the original direction of the bullet)
ub = 30° (final bullet angle)
M = 1.0 kg (brick mass)
m1 = 0.75 kg and u1 = 0° (mass and angle of
the large fragment)
v1 = 5.0 m>s
m2 = 0.25 kg (mass of small fragment)
With no external forces (gravity neglected), the total linear momentum is conserved. Therefore, both the x- and y-components of
the total momentum can be equated before and after (see Fig. 6.10):
before after
x: mb vbo = mb vb cos ub + m1v1 + m2 v2 cos u2
y: 0 = mb vb sin ub - m2 v2 sin u2
The x-equation can be rearranged to solve for the magnitude of the x-velocity of the smaller fragment:
mb vbo - mb vb cos ub - m1 v1
v2 cos u2 =
m2
10.030 kg21400 m>s2 - 10.030 kg21100 m>s210.8862 - 10.75 kg215.0 m>s2
=
0.25 kg
= 22 m>s (continued on next page)
Similarly, the y-equation can be solved for the magnitude of the y-velocity component of the smaller fragment:
mb vb sin ub 10.030 kg21100 m>s210.502
v2 sin u2 = = = 6.0 m>s
m2 0.25 kg
Forming a ratio,
v2 sin u2 6.0 m>s
= = 0.27 = tan u2
v2 cos u2 22 m>s
sin u2
(where the v2 terms cancel, and = tan u2). Then,
cos u2
u2 = tan-110.272 = 15°
and from the x-equation,
22 m>s 22 m>s
v2 = = = 23 m>s
cos 15° 0.97
FOLLOW-UP EXERCISE. Is the kinetic energy conserved for the collision in this Example? If not, where did the energy go?
EXAMPLE 6.8 Physics on Ice

A physicist is lowered from a helicopter to the middle of a 2mv, where v is relative to the ice and m is the mass of one
smooth, level, frozen lake, the surface of which has negligible mitten.
friction, and challenged to make her way off the ice. Walking When thrown separately, the first mitten would have a
is out of the question. (Why?) As she stands there pondering momentum of mv. The physicist and the second mitten would
her predicament, she decides to use the conservation of then be in motion, and throwing the second mitten would
momentum by throwing her heavy, identical mittens, which add some more momentum to the physicist and increase her
will provide her with the momentum to get herself to shore. speed, but would the speed now be greater than that if both
To get to the shore more quickly, which should this sly physi- mittens were thrown simultaneously?
cist do: throw both mittens at once or throw them separately, Let’s analyze the conditions of the second throw. After
one after the other with the same speed? throwing the first mitten, the physicist “system” would have
less mass. With less mass, the second throw would produce a
T H I N K I N G I T T H R O U G H . The initial momentum of the system greater acceleration and speed things up. But on the other
(physicist and mittens) is zero. With no net external force, by hand, after the first throw, the second mitten is moving with
the conservation of momentum, the total momentum remains the person, and when thrown in the opposite direction, the
zero. So if the physicist throws the mittens in one direction, mitten would have a velocity less than v relative to the ice (or
she will go in the opposite direction (because momenta vec- to a stationary observer). So which effect would be greater?
tors in opposite directions add to zero). Then which way of What do you think? Sometimes situations are difficult to ana-
throwing gives greater speed? If both the mittens were lyze intuitively, and you must apply scientific principles to
thrown together, the magnitude of their momentum would be figure them out.
SOLUTION.
Given: m = mass of single mitten Find: Which method of mitten

M = mass of physicist throwing gives the physicist
- v = velocity of thrown mitten(s), in the the greater speed
negative direction
Vp = velocity of physicist in the positive
direction
When the mittens are thrown together, by the conservation of momentum,
2mv
0 = 2m1- v2 + MVp and Vp = (thrown together) (1)
M
When they are thrown separately,
First throw: 0 = m1 -v2 + 1M + m2Vp

mv
and Vp = (thrown separately) (2)
1 1 M + m
Second throw: 1M + m2Vp = m1Vp - v2 + MVp
1 1 2
6.4 ELASTIC AND INELASTIC COLLISIONS 195
Note that in the last m term of the second throw, the quantities in the parentheses represent the velocity of the mitten relative to
the ice. With an initial velocity of + Vp when the first mitten is thrown in the negative direction, then Vp - v. (Recall relative
1 1
velocities from Section 3.4.)
Solving for Vp :
2
Vp = Vp + a bv = + a bv = a bv
m mv m m m
+ (3)
2 1 M M + m M M + m M
where Eq. 2 was substituted for Vp after the first throw.

1
Now, when the mittens are thrown together (Eq. 1),
Vp = a bv
2m
M
so the question is whether the result of Eq. 3 is greater or less than that of Eq. 1. Notice that with a greater denominator for the
m>1M + m2 term in Eq. 3 it is less than the m/M term. So,
a b 6
m m 2m
+
M + m M M
and therefore, Vp 7 Vp , or (thrown together) 7 (thrown separately).
2
FOLLOW-UP EXERCISE. Suppose the second throw were in the direction of the physicist’s velocity from the first throw. Would
this throw bring her to a stop?
As mentioned previously, the conservation of momentum is used to analyze

the collisions of objects ranging from subatomic particles to automobiles in traffic
accidents. In many instances, however, external forces may be acting on the
objects, which means that the momentum is not conserved.
But, as will be learned in the next section, the conservation of momentum often
allows a good approximation over the short time of a collision, during which the
internal forces (which conserve system momentum) are much greater than the
external forces. For example, external forces such as gravity and friction also act
on colliding objects, but are often relatively small compared with the internal
forces of the collision. (This concept was implied in Example 6.7.) Therefore, if the
objects interact for only a brief time, the effects of the external forces may be negli-
gible compared with those of the large internal forces during that time and the
conservation of linear momentum may be used.
DID YOU LEARN?

➥ If Fnet = ¢P> ¢t = 0, then ¢P = 0 and Pinitial = Pfinal , (conservation of linear
B B B B B
momentum).
➥ By Newton’s third law, the internal forces on particles are equal and opposite and
cancel.
6.4 Elastic and Inelastic Collisions

➥ What are the conditions for elastic and inelastic collisions in an isolated system?
➥ How much energy is lost in an inelastic collision?
In general, a collision may be defined as a meeting or interaction of particles or

objects that causes an exchange of energy and>or momentum. Taking a closer look
at collisions in terms of the conservation of momentum is simpler for an isolated
system, such as a system of particles (or balls) involved in head-on collisions. For
simplicity, only collisions in one dimension will be considered, which can be ana-
lyzed in terms of the conservation of energy. On the basis of what happens to the
total kinetic energy, two types of collisions are defined: elastic and inelastic.
(a) (b)
䉱 F I G U R E 6 . 1 1 Collisions (a) Approximate elastic collisions. (b) An inelastic collision.
In an elastic collision, the total kinetic energy is conserved. That is, the total
kinetic energy of all the objects of the system after the collision is the same as the
total kinetic energy before the collision (䉱 Fig. 6.11a). Kinetic energy may be traded
between objects of a system, but the total kinetic energy in the system remains
constant. That is,
total K after = total K before
(condition for
(6.8)
Kf = Ki an elastic collision)
During such a collision, some or all of the initial kinetic energy is temporarily
converted to potential energy as the objects are deformed. But after the maxi-
mum deformations occur, the objects elastically “spring” back to their original
shapes, and the system regains all of its original kinetic energy. For example,
two steel balls or two billiard balls may have a nearly elastic collision, with
each ball having the same shape afterward as before; that is, there is no perma-
nent deformation.
In an inelastic collision, total kinetic energy is not conserved (Fig. 6.11b). For
example, one or more of the colliding objects may not regain the original shapes,
and>or sound or frictional heat may be generated and some kinetic energy is lost.
Then,
total K after 6 total K before
(condition for
(6.9)
Kf 6 Ki an inelastic collision)
For example, a hollow aluminum ball that collides with a solid steel ball may be
dented. Permanent deformation of the ball takes work, and that work is done at
the expense of the original kinetic energy of the system. Everyday collisions are
inelastic.
For isolated systems, momentum is conserved in both elastic and inelastic
collisions. For an inelastic collision, only an amount of kinetic energy consistent with the
conservation of momentum may be lost. It may seem strange that kinetic energy can
be lost and momentum still conserved, but this fact provides insight into the dif-
ference between scalar and vector quantities and the differences in their conserva-
tion requirements.
MOMENTUM AND ENERGY IN INELASTIC COLLISIONS

To see how momentum can remain constant while the kinetic energy changes
(decreases) in inelastic collisions, consider the examples illustrated in 䉴 Fig. 6.12.
In Fig. 6.12a, two balls of equal mass 1m1 = m22 approach each other with equal
Before Collision After 䉳 F I G U R E 6 . 1 2 Inelastic

collisions In inelastic collisions,
momentum is conserved, but
v1 = v2 = 0 kinetic energy is not. Collisions like
the ones shown here, in which the
v 1o v2o
m1 m2 m1 m2 objects stick together, are called
completely or totally inelastic
collisions. The maximum amount
Po = p1o+ p2o = m1v1o – m2v2o = 0 P= 0 of kinetic energy lost is consistent
with the law of conservation of
Ki ≠ 0 Kf = 0 momentum.
(a)
v1o v
m1 m2 m1 m2
Po = p1o = (m1v1o)xn P = p1 + p2 = (m1 + m2)vxn = Po
Ki ≠ 0 Kf ≠ 0, Kf < Ki
(b)
and opposite velocities 1v1o = - v2o2. Hence, the total momentum before the colli-
sion is (vectorially) zero, but the (scalar) total kinetic energy is not zero. After the
collision, the balls are stuck together and stationary, so the total momentum is
unchanged—still zero.
Momentum is conserved because the forces of collision are internal to the sys-
tem of the two balls—there is no net external force on the system. The total kinetic
energy, however, has decreased to zero. In this case, some of the kinetic energy
went into the work done in permanently deforming the balls. Some energy may
also have gone into doing work against friction (producing heat) or may have
been lost in some other way (for example, in producing sound).
It should be noted that the balls need not stick together after collision. In a less
inelastic collision, the balls may recoil in opposite directions at reduced, but equal,
speeds. The momentum would still be conserved (still equal to zero—why?), but
the kinetic energy would again not be conserved. Under all conditions, the
amount of kinetic energy that can be lost must be consistent with the conservation
of momentum.
In Fig. 6.12b, one ball is initially at rest as the other approaches. The balls stick
together after collision, but are still in motion. Both cases in Fig. 6.12 are examples
of a completely inelastic collision, in which the objects stick together, and hence
both objects have the same velocity after colliding. The coupling of colliding rail-
road cars is a practical example of a completely (or totally) inelastic collision.
Assume that the balls in Fig. 6.12b have different masses. Since the momentum
is conserved even in inelastic collisions,
before after
m1 v1o = 1m1 + m22v
and
v = a bv
m1 (m2 initially at rest,
(6.10)
m1 + m2 1o completely inelastic collision only)
Thus, v is less than v1o , since m1>1m1 + m22 must be less than 1. Now consider
how much kinetic energy has been lost. Initially, Ki = 12 m1 v2o , and after collision
the final kinetic energy is:
Kf = 12 1m1 + m22v 2
Substituting for v from Eq. 6.10 and simplifying the result,

1 2 2
m1v1o 2
2 m1 v 1o
2 1m1 + m22 ¢ ≤
1
Kf = =
m1 + m2 m1 + m2
= a b 1 m v2 = a bK
m1 m1
m1 + m2 2 1 1o m1 + m2 i
and
Kf m1 (m2 initially at rest,
= (6.11)
Ki m1 + m2 completely inelastic collision only)
Equation 6.11 gives the fractional amount of the initial kinetic energy that
remains with the system after a completely inelastic collision. For example, if the
masses of the balls are equal 1m1 = m22, then m1>1m1 + m22 = 12 , and Kf>Ki = 12 ,
or Kf = Ki>2. That is, half of the initial kinetic energy is lost.
Note that not all of the kinetic energy can be lost in this case, no matter what
the masses of the balls are. The total momentum after collision cannot be zero,
since it was not zero initially. Thus, after the collision, the balls must be moving
and must have some kinetic energy 1Kf Z 02. In a completely inelastic collision, the
maximum amount of kinetic energy lost must be consistent with the conservation of
momentum.
EXAMPLE 6.9 Stuck Together: Completely Inelastic Collision

A 1.0-kg ball with a speed of 4.5 m>s strikes a 2.0-kg station- T H I N K I N G I T T H R O U G H . Recall the definition of a completely
ary ball. If the collision is completely inelastic, (a) what are the inelastic collision. The balls stick together after collision; kinetic
speeds of the balls after the collision? (b) What percentage of energy is not conserved, but total momentum is.
the initial kinetic energy do the balls have after the collision?
(c) What is the total momentum after the collision?
SOLUTION. Listing the data:
Given: m1 = 1.0 kg Find: (a) v (speed after collision)

1* 100%2
m2 = 2.0 kg Kf
(b)
vo = 4.5 m>s Ki
B
(c) Pf (total momentum after collision)
(a) The momentum is conserved and
1m1 + m22v = m1vo
B B
Pf = Po or
The balls stick together and have the same speed after collision. This speed is then
1.0 kg
v = a bvo = a b14.5 m>s2 = 1.5 m>s
m1
m1 + m2 1.0 kg + 2.0 kg
(b) The fractional part of the initial kinetic energy that the balls have after the completely inelastic collision is given by Eq. 6.11.
Notice that this fraction, as given by the masses, is the same as that for the speeds (Eq. 6.11) in this special case. By inspection,
1.0 kg
= = 0.33 1* 100%2 = 33%
Kf m1 1
= =
Ki m1 + m2 1.0 kg + 2.0 kg 3
Let’s show this relationship explicitly:
2 1m1 + m22v 2 11.0 kg + 2.0 kg211.5 m>s2

1 2 1 2
= 0.33 1 = 33%2
Kf
2 11.0 kg214.5 m>s2

= 1
= 1
Ki 2 2
2 m1 v 0
Keep in mind that Eq. 6.11 applies only to completely inelastic collisions in which m2 is initially at rest. For other types of collisions,
the initial and final values of the kinetic energy must be computed explicitly.
(c) The total momentum is conserved in all collisions (in the absence of external forces), so the total momentum after collision is
the same as before collision. That value is the momentum of the incident ball, with a magnitude of
Pf = p1o = m1 vo = 11.0 kg214.5 m>s2 = 4.5 kg # m>s
and the same direction as that of the incoming ball. Also, as a double check,
Pf = 1m1 + m22v = 4.5 kg # m>s
F O L L O W - U P E X E R C I S E . A small hard-metal ball of mass m collides with a larger, stationary, soft-metal ball of mass M. A
minimum amount of work W is required to make a dent in the larger ball. If the smaller ball initially has kinetic energy K = W,
will the larger ball be dented in a completely inelastic collision between the two balls?
MOMENTUM AND ENERGY IN ELASTIC COLLISIONS

For elastic collisions, there are two conservation criteria: conservation of momentum
(which holds for both elastic and inelastic collisions) and conservation of kinetic
energy (for elastic collisions only). That is, for the elastic collision of two objects:
before after
B B B
B B
Conservation of momentum P: m1 v1o + m2 v2o = m1 v1 + m2 v 2 (6.12)
1 1
Conservation of kinetic energy K: 2
2 m1 v 1o + 2
2 m2 v 2o = 12 m1 v21 + 1 2
2 m2 v 2 (6.13)
䉲 Figure 6.13 illustrates two objects traveling prior to a one-dimensional, head-
on collision with v1o 7 v2o (both in the positive x-direction). For this two-object
situation,
before after
(1)
Total momentum: m1 v1o + m2 v2o = m1 v1 + m2 v2
(where signs are used to indicate directions and the v’s indicate magnitudes).
1
Kinetic energy: 2
2 m1 v 1o + 12 m2 v22o = 12 m1 v 21 + 12 m2 v 22 (2)
If the masses and the initial velocities of the objects are known (which they usu-
ally are), then there are two unknown quantities, the final velocities after collision.
To find them, equations (1) and (2) are solved simultaneously. First the equation
for momentum conservation is written as follows:
m11v1o - v12 = - m21v2o - v22 (3)
Then, canceling the 1>2 terms in (2), rearranging, and factoring 3a2 - b 2 =
1a - b21a + b24:
m11v1o - v121v1o + v12 = - m21v2o - v221v2o + v22 (4)
Dividing equation (4) by (3) and rearranging yields
v1o - v2o = - 1v1 - v22 (5)
This equation shows that the magnitudes of the relative velocities before and after
collision are equal. That is, the relative speed of approach of object m1 to object m2
before collision is the same as their relative speed of separation after collision. (See
Section 3.4.) Notice that this relation is independent of the values of the masses of
the objects, and holds for any mass combination as long as the collision is elastic
and one-dimensional.
䉳 F I G U R E 6 . 1 3 Elastic collision
v1o v2o coming up Two objects traveling
m1 m2
prior to collision with v1o 7 v2o. See
text for description.
Then, combining equation (5) with (3) to eliminate v2 and get v1 in terms of the
two initial velocities,
v1 = a bv1o + a bv
m1 - m2 2m2
(6.14)
m1 + m2 m1 + m2 2o
Similarly, eliminating v1 to find v2,
v2 = a bv1o - a bv
2m1 m1 - m2
(6.15)
m1 + m2 m1 + m2 2o
ONE OBJECT INITIALLY AT REST

For this common, special case, say with v2o = 0, there are only the first terms in
Eqs. 6.14 and 6.15. In addition, if m1 = m2 , then v1 = 0 and v2 = v1o. That is, the
objects completely exchange momentum and kinetic energy. The incoming object is
stopped on collision, and the originally stationary object moves off with the same
velocity as the incoming ball, obviously conserving the system’s momentum and
kinetic energy. (A real-world example that comes close to these conditions is the
head-on collision of billiard balls.)
You can also get some approximates for special cases from the equations for one
object initially at rest (taken to be m2):
For m1 W m2 1massive incoming ball2: v1 L v1o and v2 L 2v1o
That is, the massive incoming object is slowed down only slightly and the light
(less massive) object is knocked away with a velocity almost twice that of the ini-
tial velocity of the massive object. (Think of a bowling ball hitting a pin.)
For m1 V m2 1light incoming ball2: v1 L - v1o and v2 L 0
That is, if a light (small mass) object elastically collides with a massive stationary
one, the massive object remains almost stationary and the light object recoils back-
ward with approximately the same speed that it had before collision.
EXAMPLE 6.10 Elastic Collision: Conservation of Momentum and Kinetic Energy

A 0.30-kg ball with a speed of 2.0 m>s in the positive T H I N K I N G I T T H R O U G H . The incoming ball is less massive
x-direction has a head-on elastic collision with a stationary than the stationary one, so it might expected that the objects
0.70-kg ball. What are the velocities of the balls after separate in opposite directions after collision, with the less
collision? massive ball recoiling from the more massive one. Equations
6.14 and 6.15 can be used to find the velocities with v2o = 0.
SOLUTION. Using the previous notation in listing the data,
Given: m1 = 0.30 kg and v1o = 2.0 m>s Find: v1 and v2

m2 = 0.70 kg and v2o = 0
Directly from Eqs. 6.13 and Eq. 6.14, the velocities after collision are
0.30 kg - 0.70 kg
v1 = a bv1o = a b12.0 m>s2 = - 0.80 m>s
m1 - m2
m1 + m2 0.30 kg + 0.70 kg
210.30 kg2
v2 = a bv1o = c d12.0 m>s2 = 1.2 m>s
2m1
m1 + m2 0.30 kg + 0.70 kg
FOLLOW-UP EXERCISE. What would be the separation distance of the two objects 2.5 s after collision?
TWO COLLIDING OBJECTS, BOTH INITIALLY MOVING

Now let’s look at some examples where both terms in Eqs. 6.14 and 6.15 are
needed.
EXAMPLE 6.11 Collisions: Overtaking and Coming Together

The precollision conditions for two elastic collisions are shown in
䉴 Fig. 6.14. What are the final velocities in each case? v1o 10 m/s v2o 5.0 m/s
m1 m2
T H I N K I N G I T T H R O U G H . These collisions are direct applications of
Eqs. 6.14 and 6.15. Notice that in (a) the 4.0-kg object will overtake m2 = 1.0 kg
and collide with the 1.0-kg object. m1 = 4.0 kg
SOLUTION. Listing the data from the figure with the +x-direction (a)
taken to the right:
Given: (a) m1 = 4.0 kg v1o = 10 m>s Find: v1 and v2
m2 = 1.0 kg v2o = 5.0 m>s (velocities
v1o 6.0 m/s
after m1 m2
(b) m1 = 2.0 kg v1o = 6.0 m>s
collision) v2o 6.0 m/s
m2 = 4.0 kg v2o = - 6.0 m>s
m1 2.0 kg
Then, substituting into the collision equations,
(a) Eq. 6.14: m2 4.0 kg
(b)
v1 = a bv + a bv
m1 - m2 2m2
m1 + m2 1o m1 + m2 2o
䉱 F I G U R E 6 . 1 4 Collisions (a) Overtaking and
4.0 kg - 1.0 kg 231.0 kg4
= a b 10 m>s + a b5.0 m>s (b) coming together. See Example text for description.
4.0 kg + 1.0 kg 4.0 kg + 1.0 kg
= 35 110 m>s2 + 25 15.0 m>s2 = 8.0 m>s
Similarly, Eq. 6.15 gives:

v2 = 13 m>s
So the more massive object overtakes and collides with the less massive object, transferring momentum (increasing velocity).
(b) Applying the collision equations for this situation, (Eq. 6.14):
2.0 kg - 4.0 kg 234.0 kg4
v1 = a b6.0 m>s + a b1- 6.0 m>s2
2.0 kg + 4.0 kg 2.0 kg + 4.0 kg
= - A 13 B 6.0 m>s + A 43 B 1- 6.0 m>s2 = - 10 m>s

Similarly, Eq. 6.15 gives
v2 = 2.0 m>s
Here, the less massive object goes in the opposite (negative) direction after collision, with a greater momentum obtained from the
more massive object.
F O L L O W - U P E X E R C I S E . Show that in parts (a) and (b) of this Example, the amount of momentum gained by one object is the
same as that lost by the other.
INTEGRATED EXAMPLE 6.12 Equal and Opposite

Two balls of equal mass with equal but opposite velocities (why?), but not kinetic energy, so (2) is not applicable for an
approach each other for a head-on elastic collision. (a) After elastic collision. If they both moved off in the same direction
collision, the balls will (1) move off stuck together, (2) both be after collision, (3), the momentum would not be conserved
at rest, (3) move off in the same direction, or (4) recoil in oppo- (zero before, nonzero after). The answer must be (4). This is
site directions. (b) Prove your answer explicitly. the only option by which momentum and kinetic energy
(A) CONCEPTUAL REASONING. Make a sketch of the situation. could be conserved. To maintain the zero momentum before
Then, looking at the choices, (1) is eliminated because if they collision, the objects would have to recoil in opposite direc-
stuck together it would be an inelastic collision. If they come tions with the same speeds as before collision.
to rest after collision, momentum would be conserved (continued on next page)
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . To explicitly Then, substituting into Eqs. 6.14 and 6.15, without writing the
show that (4) is correct, Eqs. 6.13 and 6.14 may be used. Since equations out [see part (a) in Example 6.11],
no numerical values are given, we work with symbols.
v1 = a bv + a b1- v2o2 = - v2o
0 2m
Given: m1 = m2 = m (taking m1 to be Find: v1 and v2 2m 1o 2m
initially traveling and
in the + x
v2 = a bv + a b1- v2o2 = v1o
direction:) 2m 0
v1o and - v2o (with equal speeds) 2m 1o 2m
From the results, it can be seen that after collision the balls
recoil in opposite directions.
FOLLOW-UP EXERCISE. Show that momentum and kinetic energy are conserved in this Example.
CONCEPTUAL EXAMPLE 6.13 Two In, One Out?

A novelty collision device, as shown in 䉴 Fig. 6.15, consists of five
identical metal balls. When one ball swings in, after multiple colli-
sions, one ball swings out at the other end of the row of balls. When
two balls swing in, two swing out; when three swing in, three
swing out, and so on—always the same number out as in.
Suppose that two balls, each of mass m, swing in at velocity v
and collide with the next ball. Why doesn’t one ball swing out at
the other end with a velocity 2v?
REASONING AND ANSWER. The collisions along the horizontal row
of balls are approximately elastic. The case of two balls swinging in
and one ball swinging out with twice the velocity wouldn’t violate
the conservation of momentum: 12m2v = m12v2. However, another
condition applies if we assume elastic collisions—the conservation
of kinetic energy. Let’s check to see if this condition is upheld for
this case:
before after
Ki = Kf
1
2 12m2v 2 ? 1
= 2 m12v22
mv Z 2mv2
2
Hence, the kinetic energy would not be conserved if this happened,

and the equation is telling us that this situation violates established
physical principles and does not occur. Note that there’s a big vio-
lation—more energy out than in.
FOLLOW-UP EXERCISE. Suppose the first ball of mass m were
replaced with a ball of mass 2m. When this ball is pulled back and
allowed to swing in, how many balls will swing out? [Hint: Think
about the analogous situation for the first two balls as in Fig. 6.14a,
and remember that the balls in the row are actually colliding. It 䉱 F I G U R E 6 . 1 5 One in, one out See Example
may help to think of them as being separated.] text for description.
DID YOU LEARN?

➥ Kinetic energy and momentum are conserved in an elastic collision. Momentum is
conserved in an inelastic collision, but not kinetic energy.
➥ In an inelastic collision, only an amount of kinetic energy consistent with the
conservation of momentum may be lost.
6.5 CENTER OF MASS 203
6.5 Center of Mass

➥ The center of mass concept applies to what type(s) of motion?

➥ How are the center of mass and the center of gravity related?
The conservation of total momentum gives a method of analyzing a “system of

particles.” Such a system may be virtually anything—for example, a volume
of gas, water in a container, or a baseball. Another important concept, the center of
mass, allows us to analyze the overall motion of a system of particles. It involves
representing the whole system as a single particle or point mass. This concept will
be introduced here and applied in more detail in the upcoming chapters.
It has been seen that if no net external force acts on a particle, the particle’s lin-
ear momentum is constant. Similarly, if no net external force acts on a system of
particles, the linear momentum of the system is constant. This similarity implies
that a system of particles might be represented by an equivalent single particle.
Moving rigid objects, such as balls, automobiles, and so forth, are essentially sys-
tems of particles and can be effectively represented by equivalent single particles
when analyzing motion. Such representation is done through the concept of the
center of mass (CM):
The center of mass is the point at which all of the mass of an object or system may be
considered to be concentrated, for the purposes of describing its linear or transla-
tional motion only.
Even if a rigid object is rotating, an important result (beyond the scope of this
text to derive) is that the center of mass still moves as though it were a particle
(䉲 Fig. 6.16). The center of mass is sometimes described as the balance point of a
solid object. For example, if you balance a meterstick on your finger, the center of
mass of the stick is located directly above your finger, and all of the mass (or
weight) seems to be concentrated there.
An expression similar to Newton’s second law for a single particle applies to a
system when the center of mass is used:
B B
Fnet = MACM (6.16)
B
Here, Fnet is the net external force on the system, M is the total mass of the
system or the sum of the masses of the particles of the system
B
(M = m1 + m2 + m3 + Á + mn , where the system has n particles), and ACM
is the acceleration of the center of mass of the system. In words, Eq. 6.16 says
that the center of mass of a system of particles moves as though all the mass of
the system were concentrated there and acted on by the resultant of the external
forces. Note that the movement of the individual parts of the system is not pre-
dicted by Eq. 6.16.
It follows that if the net external force on a system is zero, the total linear momen-
tum of the center of mass is conserved (that is, it stays constant), because
¢1MVCM2
B B B
b =
B B ¢VCM ¢PCM
Fnet = MACM = Ma = = 0 (6.17)
¢t ¢t ¢t
䉳 F I G U R E 6 . 1 6 Center of mass
The center of mass of this sliding
wrench moves in a straight line as
though it were a particle. Note the
white dot on the wrench that marks
the center of mass.
䉴 F I G U R E 6 . 1 7 System of parti- x1
cles in one dimension Where is the
system’s center of mass? See Exam- m3 m1 m2 x
ple 6.14. 0
x3
x2
Then, ¢P>¢t = 0, which means that there is no change in P during a time ¢t, or
B B
B B
the total momentum of the system, P = MVCM is constant (but not necessarily
B
zero). Since M is constant (why?), VCM is a constant in this case. Thus, the center of
mass either moves with a constant velocity or remains at rest.
Although you may more readily visualize the center of mass of a solid object,
the concept of the center of mass applies to any system of particles or objects, even
a quantity of gas. For a system of n particles arranged in one dimension, along the
x-axis (䉱 Fig. 6.17), the location of the center of mass is given by
B m1xB1 + m2xB2 + m3xB3 + Á + mnxBn
X CM = (6.18)
m1 + m2 + m3 + Á + mn
That is, XCM is the x-coordinate of the center of mass (CM) of a system of particles.
In shorthand notation (using signs to indicate vector directions in one dimension),
this relationship is expressed as
gmi xi
XCM = (6.19)
M
where g is the summation of the products mixi for n particles 1i = 1, 2, 3, Á , n2. If

g mi xi = 0, then XCM = 0, and the center of mass of the one-dimensional system is
located at the origin.
Other coordinates of the center of mass for systems of particles are similarly
defined. For a two-dimensional distribution of masses, the coordinates of the cen-
ter of mass are (XCM , YCM).
EXAMPLE 6.14 Finding the Center of Mass: A Summation Process

Three masses—2.0 kg, 3.0 kg, and 6.0 kg—are located at positions (3.0, 0), (6.0, 0), and 1-4.0, 02, respectively, in meters from the ori-
gin (Fig. 6.17). Where is the center of mass of this system?
T H I N K I N G I T T H R O U G H . Since all yi = 0, obviously YCM = 0, and the CM lies somewhere on the x-axis. The masses and the
positions are given, so we can use Eq. 6.19 to calculate XCM directly. However, keep in mind that the positions are located by vec-
tor displacements from the origin and are indicated in one dimension by the appropriate signs 1+ or -2.
Given: m1 = 2.0 kg Find: XCM (CM coordinate)
m2 = 3.0 kg
m3 = 6.0 kg
x1 = 3.0 m
x2 = 6.0 m
x3 = - 4.0 m
Then, performing the summation as indicated in Eq. 6.19 yields,
g m i xi
XCM =
M
12.0 kg213.0 m2 + 13.0 kg216.0 m2 + 16.0 kg21 -4.0 m2
= = 0
2.0 kg + 3.0 kg + 6.0 kg
The center of mass is at the origin.
FOLLOW-UP EXERCISE. At what position should a fourth mass of 8.0 kg be added so the CM is at x = + 1.0 m?
EXAMPLE 6.15 A Dumbbell: Center of Mass Revisited

A dumbbell (䉲 Fig. 6.18) has a connecting bar of negligible (a) XCM is given by a two-term sum.
mass. Find the location of the center of mass (a) if m1 and m2 m 1 x1 + m 2 x2
are each 5.0 kg and (b) if m1 is 5.0 kg and m2 is 10.0 kg. XCM =
m1 + m2
y (m) 15.0 kg210.20 m2 + 15.0 kg210.90 m2
= = 0.55 m
5.0 kg + 5.0 kg
0.20 m Similarly, YCM = 0.10 m, as you can prove for yourself. (You
m1 m2 might have seen this right away, since each center of mass is
0.10 m
at this height.) The center of mass of the dumbbell is then
located at 1XCM , YCM2 = 10.55 m, 0.10 m2, or midway
x (m)
0.20 0.40 0.60 0.80 1.00
between the end masses.
䉱 F I G U R E 6 . 1 8 Location of the center of mass See Exam- (b) With m2 = 10.0 kg,
ple text for description.
m 1 x1 + m 2 x2
XCM =
THINKING IT THROUGH. This Example shows how the loca- m1 + m2
15.0 kg210.20 m2 + 110.0 kg210.90 m2
tion of the center of mass depends on the distribution of
mass. In part (b), you might expect the center of mass to be = = 0.67 m
located closer to the more massive end of the dumbbell. 5.0 kg + 10.0 kg
SOLUTION. Listing the data, with the coordinates to be used which is two-thirds of the way between the masses. (Note
in Eq. 6.19, that the distance of the CM from the center of m1 is
¢x = 0.67 m - 0.20 m = 0.47 m. With the distance
Given: x1 = 0.20 m Find: (a) (XCM, YCM) (CM L = 0.70 m between the centers of the masses,
x2 = 0.90 m coordinates), ¢x>L = 0.47 m>0.70 m = 0.67, or 23 .) You might expect the
y1 = y2 = 0.10 m with m1 = m2 balance point of the dumbbell in this case to be closer to m2
(a) m1 = m2 = 5.0 kg (b) (XCM, YCM), and it is. The y-coordinate of the center of mass is again
with m1 Z m2 YCM = 0.10 m.
(b) m1 = 5.0 kg
m2 = 10.0 kg F O L L O W - U P E X E R C I S E . In part (b) of this Example, take the
origin of the coordinate axes to be at the point where m1 touches
Note that each mass is considered to be a particle located at the the x-axis. What are the coordinates of the CM in this case, and
center of the sphere (its center of mass). how does its location compare with that found in the Example?
In Example 6.15, when the value of one of the masses changed, the x-coordinate
of the center of mass changed. However, the centers of the end masses were still at
the same height, and YCM remained the same. To increase YCM , one or both of the
end masses would have to be in a higher position.
Now let’s see how the concept of the center of mass can be applied to a realistic
situation.
INTEGRATED EXAMPLE 6.16 Internal Motion: Where’s the Center of Mass and the Man?
A 75.0-kg man stands in the far end of a 50.0-kg boat 100 m
from the shore, as illustrated in 䉴 Fig. 6.19. If he walks to the
other end of the 6.00-m-long boat, (a) does the CM (1) move to
the right, (2) move to the left, or (3) remain stationary?
Neglect friction and assume the CM of the boat is at its mid-
point. (b) After walking to the other end of the boat, how far is
he from the shore?
( A ) C O N C E P T U A L R E A S O N I N G . With no net external force, the
6.00 m
acceleration of the center of mass of the man–boat system is
zero (Eq. 6.18), and so is the total momentum by Eq. 6.17 100 m
1P = MVCM = 02. Hence, the velocity of the center of mass of
B B
the system is zero, or the center of mass is stationary and 䉱 F I G U R E 6 . 1 9 Walking toward shore See Example text
remains so to conserve system momentum; that is, for description.
XCMi 1initial2 = XCMf 1final2, so the answer is (3). (continued on next page)
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . The answer is before and after the man walks. Since the CM does not move,
not 100 m - 6.00 m = 94.0 m , because the boat moves as the XCMi = XCMf . Using this fact and finding the value of XCMi ,
man walks. Why? The positions of the masses of the man and this value can be used in the calculation of XCMf , which will
the boat determine the location of the CM of the system, both contain the unknown we are looking for.
Taking the shore as the origin 1x = 02,

Given: mm = 75.0 kg Find: xmf (distance of man from shore)
xmi = 100 m
mb = 50.0 kg
xbi = 94.0 m + 3.00 m = 97.0 m
(CM position of the boat)
Note that if we take the man’s final position to be a distance mm xmf + mb xbf
xmf from the shore, then the final position of the boat’s center XCMf =
mm + mb
of mass will be xbf = xmf + 3.00 m, since the man will be at 175.0 kg2xmf + 150.0 kg21xmf + 3.00 m2
the front of the boat, 3.00 m from its CM, but on the other side. = = 98.8 m
Then initially, 75.0 kg + 50.0 kg
mm xmi + mb xbi Here, XCMf = 98.8 m = XCMi , since the CM does not move.
XCMi = Then, solving for xmf ,
mm + mb
175.0 kg21100 m2 + 150.0 kg2197.0 m2 1125 kg2198.8 m2 = 1125 kg2xmf + 150.0 kg213.00 m2
= = 98.8 m and
75.0 kg + 50.0 kg
Finally, the CM must be at the same location, since VCM = 0. xmf = 97.6 m
Then from Eq. 6.19, from the shore.
F O L L O W - U P E X E R C I S E . Suppose the man then walks back to his original position at the opposite end of the boat. Would he
then be 100 m from shore again?
CENTER OF GRAVITY
As you know, mass and weight are related. Closely associated with the concept of
the center of mass is the concept of the center of gravity (CG), the point where all
of the weight of an object may be considered to be concentrated when the object is
represented as a particle. If the acceleration due to gravity is constant in both mag-
nitude and direction over the extent of the object, Eq. 6.20 can be rewritten as (with
all gi = g),
MgXCM = gmi gxi (6.20)
Then, the object’s weight Mg acts as though its mass were concentrated at XCM,
and the center of mass and the center of gravity coincide. As you may have
noticed, the location of the center of gravity was implied in some previous figures
in Chapter 4, where the vector arrows for weight 1w = mg2 were drawn from a
point at or near the center of an object.
For practical purposes, the center of gravity is usually considered to coincide
with the center of mass. That is, the acceleration due to gravity is constant for all
parts of the object. (Note the constant g in Eq. 6.20.) There would be a difference in
the locations of the two points if an object were so large that the acceleration due
to gravity was different at different parts of the object.
In some cases, the center of mass or the center of gravity of an object may be
located by symmetry. For example, for a spherical object that is homogeneous
(that is, the mass is distributed evenly throughout), the center of mass is at the
geometrical center (or center of symmetry). In Example 6.15a, where the end
masses of the dumbbell were equal, it was probably apparent that the center of
mass was midway between them.
First
suspension
point
1
1
Center of mass lies
along this line
(b)
Second
䉱 F I G U R E 6 . 2 0 Location of the center of mass by suspension (a, right) The center of
suspension
mass of a flat, irregularly shaped object can be found by suspending the object from two or point
more points. The CM (and CG) lies on a vertical line under any point of suspension, so the
2
intersection of two such lines marks its location midway through the thickness of the body.
The sheet could be balanced horizontally at this point. Why? (b, above) The process is
illustrated with a cutout map of the United States. Note that a plumb line dropped from
any other point (third photo) does in fact pass through the CM as located in the first two CM
photos. 1
1
The location of the center of mass or center of gravity of an irregularly shaped
object is not so evident and is usually difficult to calculate (even with advanced 2
mathematical methods that are beyond the scope of this book). In some instances,
Center of mass also
the center of mass may be located experimentally. For example, the center of mass lies along this line
of a flat, irregularly shaped object can be determined experimentally by suspend-
(a)
ing it freely from different points (䉱 Fig. 6.20). A moment’s thought should con-
vince you that the center of mass (or center of gravity) always lies vertically below
the point of suspension. Since the center of mass is defined as the point at which
all the mass of a body can be considered to be concentrated, this is analogous to a
particle of mass suspended from a string. Suspending the object from two or more
points and marking the vertical lines on which the center of mass must lie locates
the center of mass as the intersection of the lines.
The center of mass (or center of gravity) of an object may lie outside the body of
the object (䉲 Fig. 6.21). For example, the center of mass of a homogeneous ring is at
the ring’s center. The mass in any section of the ring is compensated for by the mass
in an equivalent section directly across the ring, and by symmetry, the center of mass
is at the center of the ring. For an L-shaped object with uniform legs, equal in mass
and length, the center of mass lies on a line that makes a 45° angle with both legs. Its
location can easily be determined by suspending the L from a point on one of the legs
and noting where a vertical line from that point intersects the diagonal line.
CM
CM
(a) (b)
䉴 F I G U R E 6 . 2 1 The center of mass may be located outside a body The center of mass
(and center of gravity) may lie either inside or outside a body, depending on the distribu-
tion of that object’s mass. (a) For a uniform ring, the center of mass is at the center of the
ring. (b) For an L-shaped object, if the mass distribution is uniform and the legs are of
equal length, the center of mass lies on the diagonal between the legs.
In the high jump, the location of center of gravity (CG) is very important. Jumping
raises the CG. It takes energy to do this, and the higher the jump, the more energy it
takes. Therefore, a high jumper wants to clear the bar while keeping his CG low. A
jumper will try to keep his CG as close to the bar as possible when passing over it. In
the “Fosbury flop” style, made famous by Dick Fosbury in the 1968 Olympics, the
jumper arches his body backward over the bar (䉳 Fig. 6.22). With the legs, head, and
arms below the bar, the CG is lower than in the “layout” style, where the body is
nearly parallel to the ground when going over the bar. With the “flop,” a jumper may
be able to make his CG (which is outside the body) pass underneath the bar while
successfully clearing the bar.
䉱 F I G U R E 6 . 2 2 Center of gravity DID YOU LEARN?

By arching his body backward over ➥ For the center of mass, or the point at which all the mass of an object may be
the bar, the high jumper lowers his considered concentrated, only linear or translational motion applies (including at rest).
center of gravity. See text for ➥ The point at which all the weight of an object may be considered concentrated (the
description. center of gravity) coincides with the center of mass when the acceleration due to
gravity is constant.
6.6 Jet Propulsion and Rockets

➥ How is jet propulsion explained by the conservation of momentum?

➥ What is meant by “reverse thrust”?
The word jet is sometimes used to refer to a stream of liquid or gas emitted at a
high speed—for example, a jet of water from a fountain or a jet of air from an auto-
mobile tire. Jet propulsion is the application of such jets to the production of
motion. This concept usually brings to mind jet planes and rockets, but squid and
octopi propel themselves by squirting jets of water (䉲 Fig. 6.23).
You have probably tried the simple application of blowing up a balloon and
releasing it. Lacking any guidance or rigid exhaust system, the balloon zigzags
around, driven by the escaping air. In terms of Newton’s third law, the air is forced
out by the contraction of the stretched balloon—that is, the balloon exerts a force
on the air. Thus, there must be an equal and opposite reaction force exerted by the
air on the balloon. It is this force that propels the balloon on its erratic path.
Jet propulsion is explained by Newton’s third law, and in the absence of exter-
nal forces, the conservation of momentum also applies. You may understand this
concept better by considering the recoil of a rifle, taking the rifle and the bullet as
an isolated system (䉴 Fig. 6.24).
䉴 F I G U R E 6 . 2 3 Jet propulsion
Squid and octopi propel themselves
by squirting jets of water. Shown
here is a Giant Octopus jetting away.
6.6 JET PROPULSION AND ROCKETS 209
䉳 F I G U R E 6 . 2 4 Conservation of
momentum (a) Before the rifle is
fired, the total momentum of the
(a) P=0 rifle and bullet (as an isolated sys-
– + tem) is zero. (b) During firing, there
are equal and opposite internal
forces, and the instantaneous total
Fr Fb momentum of the rifle–bullet sys-
tem remains zero (neglecting exter-
nal forces, such as those that arise
when a rifle is being held). (c) When
the bullet leaves the barrel, the total
momentum of the system is still
(b) Fb = –Fr zero. [The vector equation is written
in boldface (vector) notation and
vb then in sign–magnitude notation so
vr as to indicate directions.]
pb = mbvb
pr = mrvr (c) P = pb + pr = (mbv b)xn + (–m r vr)xn = 0
Initially, the total momentum of this system is zero. When the rifle is fired (by
remote control to avoid external forces), the expansion of the gases from the
exploding charge accelerates the bullet down the barrel. These gases push back-
ward on the rifle as well, producing a recoil force (the “kick” experienced by a per-
son firing a weapon). Since the initial momentum of the system is zero and the
force of the expanding gas is an internal force, the momenta of the bullet and of
the rifle must be equal and opposite at any instant. After the bullet leaves the bar-
rel, there is no propelling force, so the bullet and the rifle move with constant
velocities (unless acted on by a net external force such as gravity or air resistance).
Similarly, the thrust of a rocket is created by exhausting the gas from burning
fuel out the rear of the rocket. The expanding gas exerts a force on the rocket that
propels the rocket in the forward direction (䉲 Fig. 6.25). The rocket exerts a reaction
force on the gas, so the gas is directed out the exhaust nozzle. If the rocket is at rest
when the engines are turned on and there are no external forces (as in deep space,
where friction is zero and gravitational forces are negligible), then the instanta-
neous momentum of the exhaust gas is equal and opposite to that of the rocket.
The numerous exhaust gas molecules have small masses and high velocities, and
the rocket has a much larger mass and a smaller velocity.
vr
vex
vex
U
S
A
vex
(vr relative to coordinate axes)
(b)
(vex relative to rocket)
(a)
䉱 F I G U R E 6 . 2 5 Jet propulsion and mass reduction (a) A rocket burning fuel is continu-
ously losing mass and thus becomes easier to accelerate. The resulting force on the rocket
(the thrust) depends on the product of the rate of change of its mass with time and the
velocity of the exhaust gases: 1¢m>¢t2v B
ex. Since the mass is decreasing, ¢m> ¢t is negative,
B
and the thrust is opposite vex. (b) The space shuttle uses a multistage rocket. Both of the
two booster rockets and the huge external fuel tank are jettisoned in flight. (c) The first and
second stages of a Saturn V rocket separating after 148 s of burn time. (c)
Unlike a rifle firing a single shot, a rocket continuously loses mass when burn-
ing fuel. (The rocket is more like a machine gun.) Thus, the rocket is a system for
which the mass is not constant. As the mass of the rocket decreases, it accelerates
more easily. Multistage rockets take advantage of this fact. The hull of a burnt-out
stage is jettisoned to give a further in-flight reduction in mass (Fig. 6.25c). The
payload (cargo) is typically a very small part of the initial mass of rockets for
space flights.
Suppose that the purpose of a spaceflight is to land a payload on the Moon. At
some point on the journey, the gravitational attraction of the Moon will become
greater than that of the Earth, and the spacecraft will accelerate toward the Moon.
A soft landing is desirable, so the spacecraft must be slowed down enough to go
into orbit around the Moon or land on it. This slowing down is accomplished by
using the rocket engines to apply a reverse thrust, or braking thrust. The spacecraft
is maneuvered through a 180° angle, or turned around, which is quite easy to do
in space. The rocket engines are then fired, expelling the exhaust gas toward the
Moon and supplying a braking action. That is, the force on the rocket is opposite
its velocity.
You have experienced a reverse thrust effect if you have flown in a commercial
jet. In this instance, however, the craft is not turned around. Instead, after touch-
down, the jet engines are revved up, and a braking action can be felt. Ordinarily,
revving up the engines accelerates the plane forward. The reverse thrust is
accomplished by activating thrust reversers in the engines that deflect the
exhaust gases forward (䉲 Fig. 6.26). The gas experiences an impulse force and a
change in momentum in the forward direction (see Fig. 6.3b), and the engine and
the aircraft have an equal and opposite momentum change, thus experiencing a
braking impulse force.
Question: There are no end-of-chapter exercises on the material covered in this
section, so test your knowledge with this one: Astronauts use handheld maneu-
vering devices (small rockets) to move around on space walks. Describe how
these rockets are used. Is there any danger on an untethered space walk?
DID YOU LEARN?

➥ If the total momentum of a system is initially zero, an internal explosion will propel
parts of the system in opposite directions (Newton’s third law).
➥ A rocket or jet plane uses thrust to accelerate. By turning the propulsion thrust in
the opposite direction (reverse thrust), the rocket or plane is slowed down or
decelerated.
AIR
Thrust Normal operation

reverser
AIR
Fan
Thrust reverser activated
䉱 F I G U R E 6 . 2 6 Reverse thrust Thrust reversers are activated on jet engines during

landing to help slow the plane. The gas experiences an impulse force and a change in
momentum in the forward direction, and the plane experiences an equal and opposite
momentum change and a braking impulse force.
6.6 JET PROPULSION AND ROCKETS 211
PULLING IT TOGETHER Pendulum and Putty Balls

A simple pendulum 1.50 m in length has a 500-g putty ball as and trigonometry. (a) The speed of the 500-g ball can be deter-
an end bob. The pendulum is pulled aside 60° from the verti- mined by applying the principle of mechanical energy conser-
cal and given an initial tangential speed of 1.20 m>s. At the vation, but this cannot be applied to the inelastic collision.
bottom of the arc path, the bob hits and sticks to a 200-g putty However, the principle of conservation of linear momentum
ball sitting on a tee. (a) Determine the speed of the 500-g ball can be applied horizontally at the very bottom of the arc path
just before it hits the 200-g ball and the speed of the combined where the collision takes place and will give the balls’ combined
masses right after they stick together. (b) How much mechani- speed. (b) The mechanical energy loss during the collision is just
cal energy is lost in the collision? (c) What is the maximum the difference between the final and initial total kinetic energy,
angle to which the combined balls swing on the other side? because where the collision takes place, the gravitational poten-
tial energy does not change. (c) After the collision, it is again
T H I N K I N G I T T H R O U G H . This example employs the use of valid to use mechanical energy conservation to determine the
energy, inelastic collisions, conservation of linear momentum, final height and angle of the combined system.
SOLUTION.
Given: vo = 1.20 m>s (initial tangential speed of the ball) Find: (a) v and V (speed of original ball just before collision
L = 1.50 m (pendulum length) and that of the combined balls just after)
u = 60° (initial pendulum angle) (b) ¢K mechanical (kinetic) energy lost
m1 = 500 g = 0.500 kg (mass of initially moving ball) (c) maximum angle after collision
m2 = 200 g = 0.200 kg (mass of the target ball)
(a) The mechanical energy is conserved from just after the ini- Just after the collision, the final kinetic energy of the com-
tial push of the descending ball to just before the collision. Ini- bined masses, K2 , is:
K2 = 12 1m1 + m22V2 = 12 10.700 kg212.87 m>s22 = 2.88 J
tially the total mechanical energy is part kinetic energy and
part gravitational potential energy. But at the bottom of the
arc path, it is all kinetic energy, assuming that the zero point Thus, K1 - K2 = 4.04 J - 2.88 J = 1.16 J of mechanical
for gravitational potential energy is chosen to be at the bot- energy was lost (to heat and sound).
tom of the arc. The 500-g ball is initially at a height of (c) To find the maximum angle on the other side of the arc,
yi = L11 - cos uo2 = 11.50 m211 - cos 60°2 = 0.75 m. (Make the principle of energy conservation is used from just after the
a sketch to see this if it isn’t clear.) Then, by the conservation collision to the location where the combined masses stop, that
of mechanical energy: is, where the ball combination has no kinetic energy (using i
and f to indicate initial and final, respectively):
Ki + Ui = Kf + Uf
Ki + Ui = Kf + Uf
or
1 The initial potential energy and the final kinetic energy are
2
2 m1 v o + m1 gyi = 12 m1 v 21 + 0
zero: Ui = Kf = 0. Then Ki is equal to K2 = 2.88 J from part
where v is the ball’s speed at the bottom of the arc. Solving for (b), and by the conservation of energy,
this speed, K2 = Uf = 1m1 + m22gyf . Solving for the final height yf ,
v1 = 3 2gyi + v2o K2
1m1 + m22g
yf =
= 4 219.80 m>s2210.75 m2 + 11.20 m>s22 = 4.02 m>s 2.88 J
= 0.420 m
10.700 kg219.80 m>s22
=
By the conservation of linear momentum from just before to
just after the collision, m1 vo = 1m1 + m22V, where V repre-
The angle is determined from the trigonometric relationship
sents the balls’ combined speed. Solving for V,
yf = L11 - cos uf2 solved for the final angle, cos uf = 1 - ,
yf
m1 L
V = ¢ ≤v
m1 + m2 1 or
0.420 m
0.500 kg cos uf = 1 - = 0.72
= a b14.02 m>s2 = 2.87 m>s. 1.50 m
0.500 kg + 0.200 kg
and the angle is
(b) From the speeds and masses, the kinetic energies can be uf = cos -110.722 = 43.9°
found. Just before the collision,
which is less than the initial angle. (How would it be possible
K1 = 12 m1 v12 = 12 10.500 kg214.02 m>s22 = 4.04 J to make the final angle to be greater than the initial angle?)
The linear momentum 1p2 of a particle is a vector and is

B
■ ■ Conditions for an inelastic collision:
defined as the product of mass and velocity: B B
Pf = Pi
B B (6.9)
p = mv (6.1)
Kf 6 Ki
The total linear momentum 1P2 of a system is the vector
B
■
sum of the momenta of the individual particles: Before Collision After
P = p1 + p2 + p3 + Á = g pi
B B B B B
(6.2)
v1 = v2 = 0
y y
v1o v2o
m1 m2 m1 m2
P = 5.0 kgm/s
x x
p2 = 3.0 kgm/s p1 = 2.0 kgm/s

■ Final velocity in a head-on, two-body completely inelastic
■ Newton’s second law in terms of momentum (for a particle): collision 1v2o = 02:
B
¢p
v = a bv
B m1
Fnet = (6.3) (6.10)
¢t m1 + m2 1o
■ The impulse–momentum theorem relates the impulse act-
■ Ratio of kinetic energies in a head-on, two-body com-
pletely inelastic collision 1v2o = 02:
ing on an object to its change in momentum:
B B B B
Impulse = Favg ¢t = ¢p = mv - mvo (6.5)
Kf m1
= (6.11)
Ki m1 + m2
■ Final velocities in a head-on, two-body elastic collision
v1 = a bv + a bv
m1 - m2 2m2
(6.14)
m 1 + m 2 1o m1 + m2 2o
v2 = a bv - a bv
2m1 m1 - m2
(6.15)
m 1 + m 2 1o m1 + m2 2o
■ The center of mass is the point at which all of the mass of an
■ Conservation of linear momentum: In the absence of a net object or system may be considered to be concentrated. The
external force, the total linear momentum of a system is center of mass does not necessarily lie within an object. (The
conserved: center of gravity is the point at which all the weight may be
B B considered to be concentrated.)
P = Po (6.7)
■ In an elastic collision, the total kinetic energy of the sys-
tem is conserved. CM
■ Momentum is conserved in both elastic and inelastic colli-
sions. In a completely inelastic collision, objects stick
together after impact.
■ Conditions for an elastic collision: ■ Coordinates of the center of mass (using signs for
B B directions):
Pf = Pi
(6.8) gmi xi
Kf = Ki XCM = (6.19)
M

6.1 LINEAR MOMENTUM 10. Internal forces do not affect the conservation of momen-
tum because (a) they cancel each other, (b) their effects
1. Linear momentum has units of (a) N>m, (b) kg # m>s,
are canceled by external forces, (c) they can never pro-
(c) N>s, (d) all of the preceding.
duce a change in velocity, (d) Newton’s second law is not
2. Linear momentum is (a) always conserved, (b) a scalar applicable to them.
quantity, (c) a vector quantity, (d) unrelated to force.
3. A net force on an object can cause (a) an acceleration,
(b) a change in momentum, (c) a change in velocity, 6.4 ELASTIC AND INELASTIC
(d) all of the preceding. COLLISIONS
4. A change in momentum requires which of the following: 11. Which of the following is not conserved in an inelastic
(a) an unbalanced force, (b) a change in velocity, (c) an collision: (a) momentum, (b) mass, (c) kinetic energy, or
acceleration, or (d) any of these? (d) total energy?
12. A rubber ball of mass m traveling horizontally with a
6.2 IMPULSE speed v hits a wall and bounces back with the same
speed. The change in momentum is (a) mv, (b) - mv,
5. Impulse has units (a) of kg # m>s, (b) of N # s, (c) the same (c) -mv>2, (d) +2mv.
as momentum, (d) all of the preceding.
13. In a head-on elastic collision, a mass m1 strikes a station-
6. Impulse is equal to (a) F ¢x, (b) the change in kinetic
ary mass m2. There is a complete transfer of energy if
energy, (c) the change in momentum, (d) ¢p>¢t.
(a) m1 = m2 , (b) m1 W m2 , (c) m1 V m2 , (d) the masses
7. Impulse (a) is the time rate of change of momentum, stick together.
(b) is the force per unit time, (c) has the same units as
momentum, (d) none of these. 14. The condition for a two-object inelastic collision is
(a) Kf 6 Ki , (b) pi Z pf , (c) m1 7 m2 , (d) v1 6 v2 .
6.3 CONSERVATION OF LINEAR

MOMENTUM 6.5 CENTER OF MASS
8. The conservation of linear momentum is described by 15. The center of mass of an object (a) always lies at the cen-
(a) the momentum–impulse theorem, (b) the ter of the object, (b) is at the location of the most massive
work–energy theorem, (c) Newton’s first law, (d) conser- particle in the object, (c) always lies within the object,
vation of energy. (d) none of the preceding.
9. The linear momentum of an object is conserved if (a) the 16. The center of mass and center of gravity coincide (a) if
force acting on the object is conservative, (b) a single, the acceleration due to gravity is constant, (b) if momen-
unbalanced internal force is acting on the object, (c) the tum is conserved, (c) if momentum is not conserved,
mechanical energy is conserved, (d) none of the preceding. (d) only for irregularly shaped objects.
6.1 LINEAR MOMENTUM 5. A karate student tries not to follow through in order to
break a board, as shown in 䉲 Fig. 6.27. How can the
1. In a football game, does a fast-running running back
abrupt stop of the hand (with no follow-through) gener-
always have more linear momentum than a slow-
ate so much force?
moving, more massive lineman? Explain.
2. Two objects have the same momentum. Do they neces-
sarily have the same kinetic energy? Explain.
3. Two objects have the same kinetic energy. Do they neces-
sarily have the same momentum? Explain.
6.2 IMPULSE
4. “Follow-through” is very important in many sports, 䉳 FIGURE 6.27
such as in serving a tennis ball. Explain how follow- A karate chop See Con-
through can increase the speed of the tennis ball when it ceptual Question 5 and
is served. Exercise 22.
6. Explain the difference for each of the following pairs of 10. Imagine yourself standing in the middle of a frozen lake.
actions in terms of impulse: (a) a golfer’s long drive and a The ice is so smooth that it is frictionless. How could you
short chip shot; (b) a boxer’s jab and a knockout punch; (c) a get to shore? (You couldn’t walk. Why?)
baseball player’s bunting action and a home-run swing. 11. A stationary object receives a direct hit by another object
7. When jumping from a height to the ground, it is advised to moving toward it. Is it possible for both objects to be at
land with the legs bent rather than stiff-legged. Why is this? rest after the collision? Explain.
8. In the Revolutionary War, the Americans had an advan- 12. Does the conservation of momentum follow from New-
tage in using long rifles, instead of the smooth bore mus- ton’s third law?
kets used by the British. The long rifle had a barrel of
48 in. or more, whereas the musket barrel length was on 6.4 ELASTIC AND INELASTIC
the order of 30 in. (䉲 Fig. 6.28). The long rifle had a much
COLLISIONS
greater range than the musket. Explain this greater range
in terms of work and impulse. (The rifling grooves in the 13. Since K = p2>2m, how can kinetic energy be lost in an
barrel of a rifle cause the bullet to spin, which gyroscopi- inelastic collision while the total momentum is still con-
cally improves stability and accuracy. See Section 8.5.) served? Explain.
14. Can all of the kinetic energy be lost in the collision of two
objects? Explain.
15. Automobiles used to have firm steel bumpers for safety.
Today auto bumpers are made out of materials that
crumple or collapse on sufficient impact. Why is this?
16. Two balls of equal mass collide head on in a completely
inelastic collision and come to rest. (a) Is the kinetic energy
conserved? (b) Is the momentum conserved? Explain.
6.5 CENTER OF MASS

17. 䉲 Figure6.30 shows a flamingo standing on one of its
two legs, with its other leg lifted. What can you say
about the location of the flamingo’s center of mass?
䉳 FIGURE 6.30
Delicate balance See
Conceptual Question 17.
䉱 F I G U R E 6 . 2 8 Long rifle See Conceptual Question 8.
6.3 CONSERVATION OF LINEAR

MOMENTUM
9. An airboat of the type used in swampy and marshy
areas is shown in 䉲 Fig. 6.29. Explain the principle of its
propulsion. Using the concept of conservation of linear
momentum, determine what would happen to the boat
if a sail were installed behind the fan.
18. Two identical objects are located a distance d apart. If

one of the objects remains at rest and the other moves
away with a constant velocity, what is the effect on the
CM of the system?
6.6 JET PROPULSION AND ROCKETS

19. Rockets used in the space program are generally
multistage—that is, they have stacked stages, each with
its own engine and propellant. Starting with the bottom,
the stages are jettisoned when they have run out of fuel.
The first stage is usually the largest, the second stage
䉱 F I G U R E 6 . 2 9 Fan propulsion See Conceptual above it is the next largest, and so on. What are the
Question 9. advantages of multistage rockets and stage sizes?
EXERCISES 215
EXERCISES
1. ● If a 60-kg woman is riding in a car traveling at 11. ●● A 0.20-kg billiard ball traveling at a speed of 15 m>s
90 km>h, what is her linear momentum relative to (a) the strikes the side rail of a pool table at an angle of 60°
ground and (b) the car? (䉲 Fig. 6.31). If the ball rebounds at the same speed and
angle, what is the change in its momentum?
2. ● The linear momentum of a runner in a 100-m dash is
7.5 * 102 kg # m>s. If the runner’s speed is 10 m>s, what
is his mass?
3. ● Find the magnitude of the linear momentum of (a) a
7.1-kg bowling ball traveling at 12 m>s and (b) a 1200-kg
automobile traveling at 90 km>h.
60° 60°
4. ● In a football game, a lineman usually has more mass v m
than a running back. (a) Will a lineman always have v
m
greater linear momentum than a running back? Why?
(b) Who has greater linear momentum, a 75-kg running
back running at 8.5 m>s or a 120-kg lineman moving at 䉱 F I G U R E 6 . 3 1 Glancing collision See Exercises 11, 12,
5.0 m>s? and 33.
5. ●● A 0.150-kg baseball traveling with a horizontal speed
of 4.50 m>s is hit by a bat and then moves with a speed 12. ●● Suppose the billiard ball in Fig. 6.31 approaches the
of 34.7 m>s in the opposite direction. What is the change rail at a speed of 15 m>s and an angle of 60°, as shown,
in the ball’s momentum? but rebounds at a speed of 10 m>s and an angle of 50°.
What is the change in momentum in this case? [Hint: Use
6. ●● A 15.0-g rubber bullet hits a wall with a speed of
components.]
150 m>s. If the bullet bounces straight back with a speed
of 120 m>s, what is the change in momentum of the 13. ●● A loaded tractor-trailer with a total mass of 5000 kg
bullet? traveling at 3.0 km>h hits a loading dock and comes to a
stop in 0.64 s. What is the magnitude of the average force
7. IE ● ● Two protons approach each other with different
exerted on the truck by the dock?
speeds. (a) Will the magnitude of the total momentum of
the two-proton system be (1) greater than the magnitude 14. ●● A 2.0-kg mud ball drops from rest at a height of 15 m.
of the momentum of either proton, (2) equal to the differ- If the impact between the ball and the ground lasts
ence between the magnitudes of momenta of the two 0.50 s, what is the average net force exerted by the ball
protons, or (3) equal to the sum of the magnitudes of on the ground?
momenta of the two protons? Why? (b) If the speeds of
15. IE ● ● In football practice, two wide receivers run differ-
the two protons are 340 m>s and 450 m>s, respectively,
ent pass receiving patterns. One with a mass of 80.0 kg
what is the total momentum of the two-proton system?
runs at 45° northeast at a speed of 5.00 m>s. The second
[Hint: Find the mass of a proton in one of the tables
receiver (mass of 90.0 kg) runs straight down the field
inside the backcover.]
(due east) at 6.00 m>s. (a) What is the direction of their
8. ●● How much momentum is acquired by a 75-kg sky- total momentum: (1) exactly northeast, (2) to the north of
diver in free fall in 2.0 minutes after jumping from the northeast, (3) exactly east, or (4) to the east of northeast?
plane? (b) Justify your answer in part (a) by actually computing
their total momentum.
9. ●● A 5.0-g bullet with a speed of 200 m>s is fired hori-
zontally into a 0.75-kg wooden block at rest on a table. If 16. ●● A major league catcher catches a fastball moving at
the block containing the bullet slides a distance of 0.20 m 95.0 mi>h and his hand and glove recoil 10.0 cm in bring-
before coming to rest, (a) what is the coefficient of kinetic ing the ball to rest. If it took 0.00470 s to bring the ball
friction between the block and the table? (b) What frac- (with a mass of 250 g) to rest in the glove, (a) what are
tion of the bullet’s energy is dissipated in the collision? the magnitude and direction of the change in momen-
tum of the ball? (b) Find the average force the ball exerts
10. ●● Two runners of mass 70 kg and 60 kg, respectively,
on the hand and glove.
have a total linear momentum of 350 kg # m>s. The heav-
ier runner is running at 2.0 m>s. Determine the possible 17. ● ● ● At a basketball game, a 120-lb cheerleader is tossed
velocities of the lighter runner. vertically upward with a speed of 4.50 m>s by a male
cheerleader. (a) What is the cheerleader’s change in for a brief time. These directional rockets exert a constant
momentum from the time she is released to just before force of 100.0 N for only 0.200 s. [Neglect the small loss
being caught if she is caught at the height at which she of mass due to burning fuel and assume the impulse is at
was released? (b) Would there be any difference if she right angles to her initial momentum.] (a) What is the
were caught 0.30 m below the point of release? If so, magnitude of the impulse delivered to the astronaut?
what is the change then? (b) What is her new direction (relative to the initial direc-
tion)? (c) What is her new speed?
18. ● ● ● A ball of mass 200 g is released from rest at a height
of 2.00 m above the floor and it rebounds straight up to a 26. ●● A volleyball is traveling toward you. (a) Which
height of 0.900 m. (a) Determine the ball’s change in action will require a greater force on the volleyball, your
momentum due to its contact with the floor. (b) If the catching the ball or your hitting the ball back? Why?
contact time with the floor was 0.0950 s, what was the (b) A 0.45-kg volleyball travels with a horizontal velocity
average force the floor exerted on the ball, and in what of 4.0 m>s over the net. You jump up and hit the ball
direction? back with a horizontal velocity of 7.0 m>s. If the contact
time is 0.040 s, what was the average force on the ball?
27. ●● A boy catches—with bare hands and his arms rigidly
6.2 IMPULSE extended—a 0.16-kg baseball coming directly toward
19. ● When tossed upward and hit horizontally by a batter, him at a speed of 25 m>s. He emits an audible “Ouch!”
a 0.20-kg softball receives an impulse of 3.0 N # s. With because the ball stings his hands. He learns quickly to
what horizontal speed does the ball move away from the move his hands with the ball as he catches it. If the
bat? contact time of the collision is increased from 3.5 ms to
8.5 ms in this way, how do the magnitudes of the aver-
20. ● An automobile with a linear momentum of age impulse forces compare?
3.0 * 104 kg # m>s is brought to a stop in 5.0 s. What is
the magnitude of the average braking force? 28. ●● A one-dimensional impulse force acts on a 3.0-kg
object as diagrammed in 䉲 Fig. 6.32. Find (a) the magni-
21. ● A pool player imparts an impulse of 3.2 N # s to a sta- tude of the impulse given to the object, (b) the magni-
tionary 0.25-kg cue ball with a cue stick. What is the tude of the average force, and (c) the final speed if the
speed of the ball just after impact? object had an initial speed of 6.0 m>s.
22. ●● For the karate chop in Fig. 6.27, assume that the hand
has a mass of 0.35 kg and that the speeds of the hand just F
before and just after hitting the board are 10 m>s and 0, 1000
respectively. What is the average force exerted by the fist 800
Force (N)
on the board if (a) the fist follows through, so the contact

time is 3.0 ms, and (b) the fist stops abruptly, so the con- 600
tact time is only 0.30 ms? 400
23. IE ● ● When bunting, a baseball player uses the bat to 200
change both the speed and direction of the baseball.
t
(a) Will the magnitude of the change in momentum of 0 .02 .04 .06 .08 .10 .12 .14 .16
the baseball before and after the bunt be (1) greater than
Time (s)
the magnitude of the momentum of the baseball either
before or after the bunt, (2) equal to the difference 䉱 F I G U R E 6 . 3 2 Force-versus-time graph See Exercise 28.
between the magnitudes of momenta of the baseball
before and after the bunt, or (3) equal to the sum of the 29. ●● A 0.45-kg piece of putty is dropped from a height of
magnitudes of momenta of the baseball before and after 2.5 m above a flat surface. When it hits the surface, the
the bunt? Why? (b) The baseball has a mass of 0.16 kg; its putty comes to rest in 0.30 s. What is the average force
speeds before and after the bunt are 15 m>s and 10 m>s, exerted on the putty by the surface?
respectively; the bunt lasts 0.025 s. What is the change in
momentum of the baseball? (c) What is the average force 30. ●● A 50-kg driver sits in her car waiting for the traffic
on the ball by the bat? light to change. Another car hits her from behind in a
head-on, rear-end collision and her car suddenly
24. IE ● ● A car with a mass of 1500 kg is rolling on a level receives an acceleration of 16 m>s2. If all of this takes
road at 30.0 m>s. It receives an impulse with a magni- place in 0.25 s, (a) what is the impulse on the driver?
tude of 2000 N # s and its speed is reduced as much as (b) What is the average force exerted on the driver, and
possible by an impulse of this size. (a) Was this impulse what exerts this force?
caused by (1) the driver hitting the accelerator, (2) the
driver putting on the brakes, or (3) the driver turning the 31. ●● An incoming 0.14-kg baseball has a speed of 45 m>s.
steering wheel? (b) What was the car’s speed after the The batter hits the ball, giving it a speed of 60 m>s. If the
impulse was applied? contact time is 0.040 s, what is the average force of the
bat on the ball?
25. ●● An astronaut (mass of 100 kg, with equipment) is
headed back to her space station at a speed of 0.750 m>s 32. ●● At a shooting competition, a contestant fires and a
but at the wrong angle. To correct her direction, she fires 12.0-g bullet leaves the rifle with a muzzle speed of
rockets from her backpack at right angles to her motion 130 m>s. The bullet hits the thick target backing and
EXERCISES 217
stops after traveling 4.00 cm. Assuming a uniform accel- collision is completely inelastic, to what height do the
eration, (a) what is the impulse on the target? (b) What is balls swing?
the average force on the target? 42. ●● A cherry bomb explodes into three pieces of equal
33. ●● If the billiard ball in Fig. 6.31 is in contact with the mass. One piece has an initial velocity of 10 m>s xN .
rail for 0.010 s, what is the magnitude of the average Another piece has an initial velocity of
force exerted on the ball? (See Exercise 11.) 6.0 m>s xN - 3.0 m>s yN . What is the velocity of the third
piece?
34. ●● A 15000-N automobile travels at a speed of 45 km>h
northward along a street, and a 7500-N sports car travels 43. ●● Two ice skaters not paying attention collide in a com-
at a speed of 60 km>h eastward along an intersecting pletely inelastic collision. Prior to the collision, skater 1,
street. (a) If neither driver brakes and the cars collide at with a mass of 60 kg, has a velocity of 5.0 km>h east-
the intersection and lock bumpers, what will the velocity ward, and moves at a right angle to skater 2, who has a
of the cars be immediately after the collision? (b) What mass of 75 kg and a velocity of 7.5 km>h southward.
percentage of the initial kinetic energy will be lost in the What is the velocity of the skaters after collision?
collision? 44. ●● Two balls of equal mass (0.50 kg) approach the origin
35. ● ● ● In a simulated head-on crash test, a car impacts a
along the positive x- and y-axes at the same speed
wall at 25 mi>h (40 km>h) and comes abruptly to rest. A (3.3 m>s). (a) What is the total momentum of the system?
120-lb passenger dummy (with a mass of 55 kg), without (b) Will the balls necessarily collide at the origin? What is
a seatbelt, is stopped by an air bag, which exerts a force the total momentum of the system after both balls have
on the dummy of 2400 lb. How long was the dummy in passed through the origin?
contact with the air bag while coming to a stop? 45. ●● A 1200-kg car moving to the right with a speed of
25 m>s collides with a 1500-kg truck and locks bumpers
36. ● ● ● A baseball player pops a pitch straight up. The ball
with the truck. Calculate the velocity of the combination
(mass 200 g) was traveling horizontally at 35.0 m>s just
after the collision if the truck is initially (a) at rest,
before contact with the bat, and 20.0 m>s just after con-
(b) moving to the right with a speed of 20 m>s, and
tact. Determine the direction and magnitude of the
(c) moving to the left with a speed of 20 m>s.
impulse delivered to the ball by the bat.
46. ●● A 10-g bullet moving horizontally at 400 m>s pene-
trates a 3.0-kg wood block resting on a horizontal sur-
6.3 CONSERVATION OF LINEAR face. If the bullet slows down to 300 m>s after emerging
MOMENTUM from the block, what is the speed of the block immedi-
ately after the bullet emerges (䉲 Fig. 6.33)?
37. ● A 60-kg astronaut floating at rest in space outside a
space capsule throws his 0.50-kg hammer such that it Before After
moves with a speed of 10 m>s relative to the capsule.
What happens to the astronaut? 400 m/s 300 m/s
38. ● In a pairs figure-skating competition, a 65-kg man and
his 45-kg female partner stand facing each other on
skates on the ice. If they push apart and the woman has a
velocity of 1.5 m>s eastward, what is the velocity of her
partner? (Neglect friction.)
39. ●● To get off a frozen, frictionless lake, a 65.0-kg person 䉱 F I G U R E 6 . 3 3 Momentum transfer? See Exercise 46.
takes off a 0.150-kg shoe and throws it horizontally, 47. ●● An explosion of a 10.0-kg bomb releases only two
directly away from the shore with a speed of 2.00 m>s. If separate pieces. The bomb was initially at rest and a
the person is 5.00 m from the shore, how long does he 4.00-kg piece travels westward at 100 m>s immediately
take to reach it? after the explosion. (a) What are the speed and direction of
40. IE ● ● An object initially at rest explodes and splits into the other piece immediately after the explosion? (b) How
three fragments. The first fragment flies off to the west, much kinetic energy was released in this explosion?
and the second fragment flies off to the south. The third 48. ●● A 1600-kg (empty) truck rolls with a speed of 2.5 m>s
fragment will fly off toward a general direction of under a loading bin, and a mass of 3500 kg is deposited
(1) southwest, (2) north of east, (3) either due north or into the truck. What is the truck’s speed immediately
due east. Why? (b) If the object has a mass of 3.0 kg, the after loading?
first fragment has a mass of 0.50 kg and a speed of
2.8 m>s, and the second fragment has a mass of 1.3 kg 49. IE ● ● A new crowd control method utilizes “rubber”
and a speed of 1.5 m>s, what are the speed and direction bullets instead of real ones. Suppose that, in a test, one of
of the third fragment? these “bullets” with a mass of 500 g is traveling at
250 m>s to the right. It hits a stationary target head-on.
41. ●● Consider two string-suspended balls, both with a The target’s mass is 25.0 kg and it rests on a smooth sur-
mass of 0.15 kg. (Similar to the arrangement in Fig. face. The bullet bounces backward (to the left) off the tar-
6.15, but with only two balls.) One ball is pulled back get at 100 m>s. (a) Which way must the target move after
in line with the other so it has a vertical height of 10 the collision: (1) right, (2) left, (3) it could be stationary,
cm, and is then released. (a) What is the speed of the or (4) you can’t tell from the data given? (b) Determine
ball just before hitting the stationary one? (b) If the the recoil speed of the target after the collision.
Before Collision After
y
v1
v1o = 0.95 m/s

1
50°
1 1
2 2 x
40°
v2
䉱 F I G U R E 6 . 3 4 Another glancing collision See Exercise 53.
50. ●● For a movie scene, a 75-kg stuntman drops from a block and the bullet are known. Using the laws of
tree onto a 50-kg sled that is moving on a frozen lake momentum and energy, show that the initial velocity of
with a velocity of 10 m>s toward the shore. (a) What is the projectile is given by vo = 31m + M2>m4 2 2gh.
the speed of the sled after the stuntman is on board?
(b) If the sled hits the bank and stops, but the stuntman
keeps on going, with what speed does he leave the sled?
(Neglect friction.)
51. ●● A 90-kg astronaut is stranded in space at a point
6.0 m from his spaceship, and he needs to get back in
4.0 min to control the spaceship. To get back, he throws a
0.50-kg piece of equipment so that it moves at a speed of
4.0 m>s directly away from the spaceship. (a) Does he get
M
back in time? (b) How fast must he throw the piece of h
+
m
equipment so he gets back in time? m vo
M
52. ●●● A projectile that is fired from a gun has an initial
velocity of 90.0 km>h at an angle of 60.0° above the
䉱 F I G U R E 6 . 3 5 A ballistic pendulum See Exercises 55 and 73.
horizontal. When the projectile is at the top of its tra-
jectory, an internal explosion causes it to separate into
two fragments of equal mass. One of the fragments
6.4 ELASTIC AND INELASTIC
falls straight downward as though it had been released
COLLISIONS
from rest. How far from the gun does the other frag-
ment land? 56. ●● For the apparatus in Fig. 6.15, one ball swinging in at
a speed of 2vo will not cause two balls to swing out with
53. ● ● ● A moving shuffleboard puck has a glancing colli-
speeds vo. (a) Which law of physics precludes this situa-
sion with a stationary puck of the same mass, as shown
tion from happening: the law of conservation of momen-
in 䉱 Fig. 6.34. If friction is negligible, what are the speeds
tum or the law of conservation of mechanical energy?
of the pucks after the collision?
(b) Prove this law mathematically.
54. ● ● ● A small asteroid (mass of 10 g) strikes a glancing blow
57. ●● A proton of mass m moving with a speed of
at a satellite in empty space. The satellite was initially at
3.0 * 106 m>s undergoes a head-on elastic collision with
rest and the asteroid was traveling at 2000 m>s. The satel-
an alpha particle of mass 4m, which is initially at rest. What
lite’s mass is 100 kg. The asteroid is deflected 10° from its
are the velocities of the two particles after the collision?
original direction and its speed decreases to 1000 m>s, but
neither object loses mass. Determine the (a) direction and 58. ●● A 4.0-kg ball with a velocity of 4.0 m>s in the
(b) speed of the satellite after the collision. + x-direction collides head-on elastically with a station-
ary 2.0-kg ball. What are the velocities of the balls after
55. ●●● A ballistic pendulum is a device used to measure the
the collision?
velocity of a projectile—for example, the muzzle velocity
of a rifle bullet. The projectile is shot horizontally into, 59. ●● A dropped rubber ball hits the floor with a speed of
and becomes embedded in, the bob of a pendulum, as 8.0 m>s and rebounds to a height of 0.25 m. What
illustrated in 䉴 Fig. 6.35. The pendulum swings upward fraction of the initial kinetic energy was lost in the
to some height h, which is measured. The masses of the collision?
EXERCISES 219
60. ●● At a county fair, two children ram each other head- 65. ● ● Two balls approach each other as shown in 䉲 Fig. 6.38,
on while riding on the bumper cars. Jill and her car, trav- where m = 2.0 kg , v = 3.0 m>s , M = 4.0 kg , and
eling left at 3.50 m>s, have a total mass of 325 kg. Jack V = 5.0 m>s. If the balls collide and stick together at the
and his car, traveling to the right at 2.00 m>s, have a total origin, (a) what are the components of the velocity v of
mass of 290 kg. Assuming the collision to be elastic, the balls after collision, and (b) what is the angle u?
determine their velocities after the collision.
y
61. ● ● In a high-speed chase, a policeman’s car bumps a
criminal’s car directly from behind to get his attention.
The policeman’s car is moving at 40.0 m>s to the right v′
and has a total mass of 1800 kg. The criminal’s car is ini-
tially moving in the same direction at 38.0 m>s. His car
has a total mass of 1500 kg. Assuming an elastic colli-
sion, determine their two velocities immediately after v θ
the bump. x
m
62. IE ● ● 䉲 Fig. 6.36 shows a bird catching a fish. Assume that
initially the fish jumps up and that the bird coasts hori-
zontally and does not touch the water with its feet or flap
V
its wings. (a) Is this kind of collision (1) elastic, (2) inelas- 䉳 FIGURE 6.38
tic, or (3) completely inelastic? Why? (b) If the mass of the M A completely
bird is 5.0 kg, the mass of the fish is 0.80 kg, and the bird inelastic collision
coasts with a speed of 6.5 m>s before grabbing, what is See Exercise 65.
the speed of the bird after grabbing the fish?
66. IE ● ● A car traveling east and a minivan traveling south
collide in a completely inelastic collision at a perpendic-
ular intersection. (a) Right after the collision, will the car
and minivan move toward a general direction (1) south
of east, (2) north of west, or (3) either due south or due
east? Why? (b) If the initial speed of the 1500-kg car was
90.0 km>h and the initial speed of the 3000-kg minivan
was 60.0 km>h, what is the velocity of the vehicles
immediately after collision?
67. ● ● A 1.0-kg object moving at 2.0 m>s collides elastically
with a stationary 1.0-kg object, similar to the situation
shown in Fig. 6.37. How far will the initially stationary
object travel along a 37° inclined plane? (Neglect friction.)
䉳 FIGURE 6.36 68. ● ● A fellow student states that the total momentum of a
Elastic or inelastic? three-particle system (m1 = 0.25 kg , m2 = 0.20 kg , and
See Exercise 62. m3 = 0.33 kg ) is initially zero. He calculates that after an
inelastic triple collision the particles have velocities of
63. ●● A 1.0-kg object moving at 10 m>s collides with a sta- 4.0 m>s at 0°, 6.0 m at 120°, and 2.5 m>s at 230°, respec-
tionary 2.0-kg object as shown in 䉲 Fig. 6.37. If the collision tively, with angles measured from the + x-axis. Do you
is perfectly inelastic, how far along the inclined plane will agree with his calculations? If not, assuming the first two
the combined system travel? (Neglect friction.) answers to be correct, what should be the momentum of
the third particle so the total momentum is zero?
69. ● ● A freight car with a mass of 25 000 kg rolls down an
inclined track through a vertical distance of 1.5 m. At the
bottom of the incline, on a level track, the car collides
and couples with an identical freight car that was at rest.
What percentage of the initial kinetic energy is lost in the
2.0 kg collision?
1.0 kg
10 m/s 70. ● ● ● In nuclear reactors, subatomic particles called
37° neutrons are slowed down by allowing them to collide
with the atoms of a moderator material, such as carbon
atoms, which are 12 times as massive as neutrons. (a) In
䉱 F I G U R E 6 . 3 7 How far is up? See Exercises 63 and 67. a head-on elastic collision with a carbon atom, what per-
centage of a neutron’s energy is lost? (b) If the neutron
64. ●● In a pool game, a cue ball traveling at 0.75 m>s hits has an initial speed of 1.5 * 107 m>s, what will be its
the stationary eight ball. The eight ball moves off with a speed after collision?
velocity of 0.25 m>s at an angle of 37° relative to the cue 71. ● ● ● In a noninjury chain-reaction accident on a foggy
ball’s initial direction. Assuming that the collision is freeway, car 1 (mass of 2000 kg) moving at 15.0 m>s to
inelastic, at what angle will the cue ball be deflected, and the right elastically collides with car 2, initially at rest.
what will be its speed? The mass of car 2 is 1500 kg. In turn, car 2 then goes on to
lock bumpers (that is, it is a completely inelastic colli- 80. ●● Locate the center of mass of the system shown in
sion) with car 3, which has a mass of 2500 kg and was 䉲 Fig.
6.39 (a) if all of the masses are equal; (b) if
also at rest. Determine the speed of all cars immediately m2 = m4 = 2m1 = 2m3 ; (c) if m1 = 1.0 kg , m2 = 2.0 kg,
after this unfortunate accident. m3 = 3.0 kg, and m4 = 4.0 kg.
72. ● ● ● Pendulum 1 is made of a 1.50-m string with a small
y
Super Ball attached as a bob. It is pulled aside 30° and
released. At the bottom of its arc, it collides with another (0, 4.0 m) (4.0 m, 4.0 m)
pendulum bob of the same length, but the second pen- m2 m3
dulum has a bob made from a Super Ball whose mass is
twice that of the bob of pendulum 1. Determine the
angles to which both pendulums rebound (when they
come to rest) after they collide and bounce back.
73. ● ● ● Show that the fraction of kinetic energy lost in a bal-
listic-pendulum collision (as in Fig. 6.35) is equal to m4

m1 x
M>1m + M2.
(0, 0) (4.0 m, 0)
6.5 CENTER OF MASS 䉱 F I G U R E 6 . 3 9 Where’s the center of mass? See Exercise 80.
74. ● (a) The center of mass of a system consisting of two 81. ●● Two cups are placed on a uniform board that is bal-
0.10-kg particles is located at the origin. If one of the parti- anced on a cylinder (䉲 Fig. 6.40). The board has a mass of
cles is at (0, 0.45 m), where is the other? (b) If the masses 2.00 kg and is 2.00 m long. The mass of cup 1 is 200 g and
are moved so their center of mass is located at (0.25 m, it is placed 1.05 m to the left of the balance point. The
0.15 m), can you tell where the particles are located? mass of cup 2 is 400 g. Where should cup 2 be placed for
balance (relative to the right end of the board)?
75. ● (a) Find the center of mass of the Earth–Moon system.
[Hint: Use data from the tables on the inside cover of the 1 2
book, and consider the distance between the Earth and
Moon to be measured from their centers.] (b) Where is
that center of mass relative to the surface of the Earth?
76. ●● Find the center of mass of a system composed of
three spherical objects with masses of 3.0 kg, 2.0 kg, and 1.05 m
4.0 kg and centers located at 1- 6.0 m, 02, (1.0 m, 0), and
(3.0 m, 0), respectively. 䉱 F I G U R E 6 . 4 0 Don’t let it roll See Exercise 81.
77. ●● Rework Exercise 52, using the concept of the center 82. ●● Two skaters with masses of 65 kg and 45 kg, respec-
of mass, and compute the distance the other fragment tively, stand 8.0 m apart, each holding one end of a piece
landed from the gun. of rope. (a) If they pull themselves along the rope until
they meet, how far does each skater travel? (Neglect fric-
78. IE ● ● A 3.0-kg rod of length 5.0 m has at opposite ends tion.) (b) If only the 45-kg skater pulls along the rope
point masses of 4.0 kg and 6.0 kg. (a) Will the center of until she meets her friend (who just holds onto the rope),
mass of this system be (1) nearer to the 4.0-kg mass, how far does each skater travel?
(2) nearer to the 6.0-kg mass, or (3) at the center of the
83. ● ● ● Three particles, each with a mass of 0.25 kg, are
located at 1 -4.0 m, 02, (2.0 m, 0), and (0, 3.0 m) and are
rod? Why? (b) Where is the center of mass of the system?
acted on by forces F1 = 1-3.0 N2yN , F2 = 15.0 N2yN , and
B B
79. ●● A piece of uniform sheet metal measures 25 cm by
F3 = 14.0 N2xN , respectively. Find the acceleration (mag-
B
25 cm. If a circular piece with a radius of 5.0 cm is cut
from the center of the sheet, where is the sheet’s center of nitude and direction) of the center of mass of the system.
mass now? [Hint: Consider the components of the acceleration.]
84. A 170-g hockey puck sliding on ice perpendicularly 85. You are traveling north and make a 90° right-hand turn
impacts a flat piece of sideboard. Its incoming momen- east on a flat road while driving a car that has a total
tum is 6.10 kg # m>s. It rebounds along its incoming path weight of 3600 lb. Before the turn, the car was traveling
after having suffered a momentum change (magnitude) at 40 mi>h, and after the turn is completed you have
of 8.80 kg # m>s. (a) If the impact with the board took slowed to 30 mi>h. If the turn took 4.25 s to complete,
35.0 ms, determine the average force (including direc- determine the following: (a) the car’s change in kinetic
tion) exerted by the puck on the board. (b) Determine the energy, (b) the car’s change in momentum (including
final momentum of the puck. (c) Was this collision elastic direction), and (c) the average net force exerted on the
or inelastic? Prove your answer mathematically. car during the turn (including direction).
EXERCISES 221
86. IE In the radioactive decay of a nucleus of an atom 88. IE In a laboratory setup, two frictionless carts are placed
called americium-241 (symbol 241Am, mass of on a horizontal surface. Cart A has a mass of 500 g and
4.03 * 10-25 kg), it emits an alpha particle (designated as cart B’s mass is 1000 g. Between them is placed an ideal
a) with a mass of 6.68 * 10-27 kg to the right with a (very light) spring and they are squeezed together care-
kinetic energy of 8.64 * 10-13 J. (This is typical of fully, thereby compressing the spring by 5.50 cm. Both
nuclear energies, small on the everyday scale.) The carts are then released and B’s recoil speed is measured
remaining nucleus is neptunium-237 (237Np) and has a to be 0.55 m>s. (a) Will cart A’s speed be (1) greater than,
mass of 3.96 * 10-25 kg. Assume the initial nucleus was (2) less than, or (3) the same as B’s speed? Explain.
at rest. (a) Will the neptunium nucleus have (1) more, (b) Determine B’s recoil speed to see if your conjecture in
(2) less, or (3) the same amount of kinetic energy (a) was correct. (c) Determine the spring constant of the
compared to the alpha particle? (b) Determine the spring.
kinetic energy of the 237 Np nucleus afterward.
87. A young hockey player with a mass of 30.0 kg is initially
moving at 2.00 m>s to the east. He intercepts and catches
on the stick a puck initially moving at 35.0 m>s at an angle
of u = 60° (䉲Fig. 6.41). Assume that the puck’s mass is
0.180 kg and the player and puck form a single object for a
few seconds. (a) Determine the direction angle and speed
of the puck and skater after the collision. (b) Was this colli-
sion elastic or inelastic? Prove your answer with numbers.
puck
angle and
speed?
䉱 F I G U R E 6 . 4 1 A player–puck collision See

Exercise 87.
Circular Motion
7 and Gravitation
7.1 Angular measure (223)
■ the radian (rad)
7.2 Angular speed

and velocity (226)
■ tangential speed
7.3 Uniform circular motion and

centripetal acceleration (229)
■ no circular motion without
acceleration vector
toward center of circle
7.4 Angular acceleration (236)

■ tangential acceleration PHYSICS FACTS
7.5 Newton’s law of

gravitation (238)
✦ An ultracentrifuge can spin sam-
ples with a force of 15 000 g. Such
force is needed for harvesting pro-
tein precipitates, bacteria, and
P eople often say that rides like
the spirling circular one in the
chapter-opening photograph “defy
other cells.
■ a universal law gravity.” Of course, you know that
✦ Newton coined the word gravity
■ gravitational potential in reality, gravity cannot be defied;
from gravitas, the Latin word for
energy well
“weight” or “heaviness.”
it commands respect. There is
✦ If you want to “lose” weight, go
to the Earth’s equator. Because nothing that will shield you from it
the Earth bulges slightly, the
7.6 Kepler’s laws and and no place in the universe where
acceleration due to gravity is
Earth satellites (247)
■ planetary motion
slightly less there and you would you can go to be entirely free of
weigh less (but your mass would
■ apparent weightlessness still be the same). gravity.
✦ The centripetal force that keeps Circular motion is everywhere,
planets in orbit is supplied by grav-
ity. The centripetal force that keeps from atoms to galaxies, from fla-
atomic electrons in orbit about the
nuclear proton(s) is supplied by
gella of bacteria to Ferris wheels.
the electrical force. The electrical Two terms are frequently used to
force between an electron and a
proton is on the order of 1040 describe such motion. In general,
times greater than the gravita-
we say that an object rotates when
tional force between them
(Section 15.3). the axis of rotation lies within the
body and that it revolves when the
7.1 ANGULAR MEASURE 223
axis is outside the body. Thus, the Earth rotates on its axis and revolves about
the Sun.
Such motion is in two dimensions, and so can be described by rectangular
components as used in Chapter 3. However, it is usually more convenient to
describe circular motion in terms of angular quantities that will be introduced in
this chapter. Being familiar with the description of circular motion will make the
study of rotating rigid bodies in Chapter 8 much easier.
Gravity plays a major role in determining the motions of the planets, since it
supplies the force necessary to maintain their orbits. Newton’s law of gravitation
will be considered in this chapter. This law describes the fundamental force of
gravity, and will be used to analyze planetary motion. The same considerations will
help you understand the motions of Earth satellites, which include one natural
satellite (the Moon) and many artificial ones.
7.1 Angular Measure*

➥ The radian is defined by what two quantities?

➥ How many radians are there in a full circle?
Motion is described as a change of position with time. (Section 2.1). As you might
y
guess, angular speed and angular velocity also involve a time rate of change of posi-
tion, which is expressed by an angular change. Consider a particle traveling in a cir- x = r cos
cular path, as shown in 䉴 Fig. 7.1. At a particular instant, the particle’s position (P) y = r sin
may be designated by the Cartesian coordinates x and y. However, the position
may also be designated by the polar coordinates r and u. The distance r extends
(x, y)
from the origin, and the angle u is commonly measured counterclockwise from the P or
positive x-axis. The transformation equations that relate one set of coordinates to (r, )
the other are r
x = r cos u (7.1a) x
y = r sin u (7.1b)
as can be seen from the x- and y-coordinates of point P in Fig. 7.1.
Note that r is the same for any point on a given circle. As a particle travels in a
circle, the value of r is constant, and only u changes with time. Thus, circular
motion can be described by using one polar coordinate 1u2 that changes with time,
instead of two Cartesian coordinates (x) and (y), both of which change with time. 䉱 F I G U R E 7 . 1 Polar coordinates
Analogous to linear displacement is angular displacement, the magnitude of A point (P) may be described by
which is given by polar coordinates instead of Carte-
sian coordinates—that is, by (r, u)
¢u = u - uo (7.2) instead of (x, y). For a circle, u is the
or simply ¢u = u when uo = 0°. (The direction of the angular displacement will angular distance and r is the radial
distance. The two types of coordi-
be explained in the next section on angular velocity.) A unit commonly used to nates are related by the transforma-
express angular displacement is the degree (°); there are 360° in one complete tion equations x = r cos u and
circle.† y = r sin u.
It is important to be able to relate the angular description of circular motion to the
orbital or tangential description—that is, to relate the angular displacement to the arc
length s. The arc length is the distance traveled along the circular path, and the angle u
is said to subtend (define) the arc length. A quantity that is very convenient for relating
*Here and throughout the text, angles will be considered exact, that is, they do not determine the
number of significant figures.
†
A degree may be divided into the smaller units of minutes 11 degree = 60 minutes2 and seconds
11 minute = 60 seconds2. These divisions have nothing to do with time units.
224 7 CIRCULAR MOTION AND GRAVITATION
䉴 F I G U R E 7 . 2 Radian measure y
Angular displacement may be mea-
s = ru
sured either in degrees or in radians
( u in radians)
(rad). An angle u is subtended by an
arc length s. When s = r, the angle
subtending s is defined to be 1 rad.
r s=r
More generally, u = s>r, where u is u=
in radians. One radian is equal to 1 rad
57.3°. x
angle to arc length is the radian (rad). The angle in radians is given by the ratio of
the arc length (s) and the radius (r)—that is, u (in radians) equals s>r. When s = r,
the angle is equal to one radian, u = s>r = r>r = 1 rad (䉱 Fig. 7.2).
Thus, (with the angle in radians),
TABLE 7.1
Equivalent Degree and s = ru (7.3)
Radian Measures
which is an important relationship between the circular arc length s and the radius
Degrees Radians of the circle r. (Notice that since u = s>r, the angle in radians is the ratio of two
lengths. This means that a radian measure is a pure number—that is, it is dimen-
360° 2p sionless and has no units.)
180° p To get a general relationship between radians and degrees, consider the dis-
90° p/2
tance around a complete circle (360°). For one full circle, with s = 2pr (the circum-
ference), there are a total of u = s>r = 2pr>r = 2p rad in 360°, that is,
60° p/3
2p rad = 360°
57.3° 1
45° p/4 This relationship can be used to obtain convenient conversions of common angles
(䉳 Table 7.1). Also, dividing both sides of this relationship by 2p, the degree value
30° p/6
of 1 rad is obtained:
1 rad = 360°>2p = 57.3°
Notice in Table 7.1 that the angles in radians are expressed in terms of p explicitly,
for convenience.
EXAMPLE 7.1 Finding Arc Length: Using Radian Measure

A spectator standing at the center of a circular running track observes a runner start a N
practice race 256 m due east of her own position (䉴 Fig. 7.3). The runner runs on the
track to the finish line, which is located due north of the observer’s position. What is Finish
W E
the distance of the run?
T H I N K I N G I T T H R O U G H . Note that the subtending angle of the section of circular track S
is u = 90°. The arc length (s) can be found, since the radius r of the circle is known.
SOLUTION. Listing what is given and what is to be found,
Given: r = 256 m Find: s (arc length)
u = 90° = p>2 rad
Simply using Eq. 7.3 to find the arc length,
s = ru = 1256 m2a b = 402 m

p
2 256 m
Start
Note that the unitless rad is omitted, and the equation is dimensionally correct.
F O L L O W - U P E X E R C I S E . What path length would the runner have traveled when going 䉱 F I G U R E 7 . 3 Arc length—
an angular distance of 210° around the track? (Answers to all Follow-Up Exercises are found by means of radians See
given in Appendix VI at the back of the book.) Example text for description.
7.1 ANGULAR MEASURE 225
EXAMPLE 7.2 How Far Away? A Useful Approximation

A sailor sights a distant tanker ship and finds that it subtends Lines of sight
an angle of 1.15° as illustrated in 䉴 Fig. 7.4a. He knows from
the shipping charts that the tanker is 150 m in length. Approx-
imately how far away is the tanker?
s
THINKING IT THROUGH. Note that in the accompanying
Learn by Drawing 7.1, The Small Angle Approximation, for L
small angles, the arc length approximates the y-length of the
triangle (the opposite side from u), or s L y. Hence, if the
length and the angle are known, the radial distance can be
u = 1.15°
found, which is approximately equal to the tanker’s distance
from the sailor. d
r r
T H I N K I N G I T T H R O U G H . To approximate the distance, the
ship’s length is taken to be nearly equal to the arc length sub-
tended by the measured angle. This approximation is good u
for small angles.
SOLUTION. The data are as follows:
Given: u = 1.15°11 rad>57.3°2 Find: r (radial distance)
= 0.0201 rad
L = 150 m
Knowing the arc length and angle, Eq. 7.3 can be used to find r. (a) (b)
s 150 m 䉱 F I G U R E 7 . 4 Angular distance For small angles, the arc

r = L = 7.46 * 103 m = 7.46 km length is approximately a straight line, or the chord length.
u 0.0201
Knowing the length of the tanker, how far away it can be
(Note that the unitless rad is omitted.) found by measuring its angular size. See Example text for
The distance r is an approximation, obtained by assuming description. (Drawing not to scale for clarity.)
that for small angles, the arc length s and the straight-line
chord length L are very nearly the same length (Fig. 7.4b).
How good is this approximation? To check, let’s compute the
actual distance d to the middle of the ship. From the geome-
try, tan 1u>22 = 1L>22>d, so
L 150 m The first calculation is a pretty good approximation—the val-
d = = = 7.47 * 103 m = 7.47 km ues derived by the two methods are nearly equal.
2 tan1u>22 2 tan11.15°>22
F O L L O W - U P E X E R C I S E . As pointed out, the approximation used in this Example is for small angles. You might wonder what is
small. To investigate this question, what would be the percentage error of the approximated distance to the tanker for angles of
10° and 20°?
the small-angle approximation
r s
y y s
u u small:
x
y s
u not small:
x r
s s y y
u (in rad) =
r
u (in rad) =
r r x
y y
sin u =
r
tan u =
x u (in rad) sin u tan u
In computing trigonometric functions such as tan u or sin u, the angle may be expressed
in degrees or radians; for example, sin 30° = sin[1p>62 rad] = sin10.524 rad2 = 0.500.
When finding trig functions with a calculator, note that there is usually a way to change
the angle entry between deg and rad modes. Hand calculators commonly are set in deg
(degree) mode, so if you want to find the value of, say, sin11.22 rad2, first change to rad
mode and enter sin 1.22, and sin11.22 rad2 = 0.939. (Or you could convert rads to
degrees first and use deg mode.) Some calculators may have a third mode, grad. The grad
is a little-used angular unit. A grad is 1>100 of a right (90°) angle; that is, there are 100
grads in a right angle.
DID YOU LEARN?

➥ The radian is given by the ratio of the arc length (s) and the radius (r).That is,
u (in radians) = s>r.
➥ There are 2p radians (rad) in a circle.That is, 2p rad = 360°, and 1 rad = 57.3°.
7.2 Angular Speed and Velocity

In Circular Motion:
➥ How are tangential speed (v) and angular speed (v) related?
➥ What is the relationship between period (T) and frequency (f )?
➥ How is the angular speed related to the period (T) and frequency (f )?
The description of circular motion in angular form is analogous to the description

of linear motion. In fact, you’ll notice that the equations are almost identical math-
ematically, with different symbols being used to indicate that the quantities have
different meanings. The lowercase Greek letter omega with a bar over it is used to
represent average angular speed (V), the magnitude of the angular displacement
divided by the total time to travel the angular distance:
¢u u - uo
v = = o (average angular speed) (7.4)
¢t t - t
It is commonly said that the units of angular speed are radians per second. Tech-
nically, the unit is 1/s or s -1 since the radian is unitless. But it is useful to keep the
rad to indicate that the quantity is angular speed. The instantaneous angular
speed (V) is given by considering a very small time interval—that is, as ¢t
approaches zero.
As in the linear case, if the angular speed is constant, then v = v. Taking uo and
to to be zero in Eq. 7.4,
u
v = or u = vt (instantaneous angular speed) (7.5)
t
SI unit of angular speed: radians per second 1rad>s or s -12

Another common descriptive unit for angular speed is revolutions per minute
(rpm); for example, a CD (compact disc) rotates at a speed of 200–500 rpm (depend-
ing on the location of the track). This nonstandard unit of revolutions per minute is
readily converted to radians per second, since 1 revolution = 2p rad. For example,
1150 rev>min212p rad>rev211 min> 60 s2 = 5.0p rad>s ( = 16 rad>s2.*
The average angular velocity and the instantaneous angular velocity are analo-
gous to their linear counterparts. Angular velocity is associated with angular dis-
placement. Both are vectors and thus have direction; however, this directionality is,
by convention, specified in a special way. In one-dimensional, or linear, motion, a
*It is often convenient to leave the angular speed with p in symbol form, in this case, 5.0p rad>s.
7.2 ANGULAR SPEED AND VELOCITY 227
particle can go only in one direction or the other ( + or - ), so the displacement

and velocity vectors can have only these two directions. In the angular case, a
particle moves one way or the other, but the motion is along its circular path.
Thus, the angular displacement and angular velocity vectors of a particle
in circular motion can have only two directions, which correspond to going
around the circular path with either increasing or decreasing angular dis-
placement from uo—that is, counterclockwise or clockwise. Let’s focus on
B
the angular velocity vector V . (The direction of the angular displacement
will be the same as that of the angular velocity. Why?)
The direction of the angular velocity vector is given by a right-hand rule as
illustrated in 䉴 Fig. 7.5a. When the fingers of your right hand are curled in the
direction of the circular motion, your extended thumb points in the direction
B
of V . Note that since circular motion can be in only one of two circular senses,
clockwise or counterclockwise, then plus and minus signs can be used to dis-
tinguish circular rotation directions. It is customary to take a counterclockwise
rotation as positive ( + ), since positive angular distance (and displacement) is
conventionally measured counterclockwise from the positive x-axis.
Why not just designate the direction of the angular velocity vector to be (–)
either clockwise or counterclockwise? This designation is not used because (+)

clockwise (cw) and counterclockwise (ccw) are directional senses or indica-
tions rather than actual directions. These rotational senses are like right and
left. If you faced another person and each of you were asked whether some-

thing was on the right or left, your answers would disagree. Similarly, if you
held this book up toward a person facing you and rotated it, would it be rotat-
(a) (b)
ing cw or ccw for both of you? Check it out. We can use cw and ccw to indicate
rotational “directions” when they are specified relative to a reference—for 䉱 F I G U R E 7 . 5 Angular velocity
example, the positive x-axis. The direction of the angular velocity
Referring to Fig. 7.5, imagine yourself being first on one side of one of the rotat- vector for an object in rotational
ing disks and then on the other. Then apply the right-hand rule on both sides. You motion is given by the right-hand
rule: When the fingers of the right
should find that the direction of the angular velocity vector is the same for both hand are curled in the direction of
locations (because it is referenced to the right hand). Relative to this vector—for the rotation, the extended thumb
example, looking at the tip—there is no ambiguity in using + and - to indicate points in the direction of the angu-
rotational senses or directions. lar velocity vector. Circular senses
or directions are commonly indi-
cated by (a) plus and (b) minus
RELATIONSHIP BETWEEN TANGENTIAL AND ANGULAR signs.
SPEEDS
A particle moving in a circle has an instantaneous velocity tangential to its circular
path. For a constant angular velocity, the particle’s orbital speed, or tangential
speed (vt), the magnitude of the tangential velocity, is also constant. How the
angular and tangential speeds are related is revealed by starting with Eq. 7.3
1s = ru2 and Eq. 7.5 1u = vt2:
s = ru = r1vt2
The arc length, or distance, is also given by
s = vtt
Combining the equations for s gives the relationship between the tangential speed
(v) and the angular speed 1v2,
(tangential speed relation

vt = rv (7.6)
to angular speed for circular motion)
where v is in radians per second. Equation 7.6 holds in general for instantaneous
tangential and angular speeds for solid- or rigid-body rotation about a fixed axis,
even when v might vary with time.
Note that all the particles of a solid object rotating with constant angular veloc-
ity have the same angular speed, but the tangential speeds are different at differ-
ent distances from the axis of rotation (䉲 Fig. 7.6a).
vt = rv (v in rad/s)
EXAMPLE 7.3 Merry-Go-Rounds: Do Some Go Faster Than Others?
An amusement park merry-go-round at its constant operational speed makes one
complete rotation in 45 s. Two children are on horses, one at 3.0 m from the center of
s the ride and the other farther out, 6.0 m from the center. What are (a) the angular
speed and (b) the tangential speed of each child?
u = vt T H I N K I N G I T T H R O U G H . The angular speed of each child is the same, since both chil-
r dren make a complete rotation in the same time. However, the tangential speeds will be
different, because the radii are different. That is, the child at the greater radius travels in
a larger circle during the rotation time and thus must travel faster.
SOLUTION.
v Given: u = 2p rad (one rotation) Find: (a) v1 and v2 (angular speeds)
t = 45 s (b) v1 and v2 (tangential speeds)
(a) r1 = 3.0 m
r2 = 6.0 m
(a) As noted, v1 = v2 , that is, both riders rotate at the same angular speed. All points
on the merry-go-round travel through 2p rad in the time it takes to make one rotation.
The angular speed can be found from Eq. 7.5 (constant v) as
u 2p rad
v = = = 0.14 rad>s
t 45 s
Hence, v = v1 = v2 = 0.14 rad>s.
(b) The tangential speed is different at different radial locations on the merry-go-round.
All of the “particles” making up the merry-go-round go through one rotation in the same
(b) amount of time. Therefore, the farther a particle is from the center, the longer its circular
path, and the greater its tangential speed, as Eq. 7.6 indicates. (See also Fig. 7.6a.) Thus,
䉱 F I G U R E 7 . 6 Tangential and
angular speeds (a) Tangential and vt1 = r1 v = 13.0 m210.14 rad>s2 = 0.42 m>s
angular speeds are related by
vt = rv, where v is in radians per and
second. Note that all of the particles vt2 = r2 v = 16.0 m210.14 rad>s2 = 0.84 m>s
of an object rotating about a fixed
axis travel in circles. All the particles (Note that the rad has been dropped from the answer. Why?)
have the same angular speed v, but Then, a rider on the outer part of the ride has a greater tangential speed than a rider closer
particles at different distances from to the center. Here, rider 2 has a radius twice that of rider 1 and therefore goes twice as fast.
the axis of rotation have different
tangential speeds. (b) Sparks from a F O L L O W - U P E X E R C I S E . (a) On an old 45-rpm record, the beginning track is 8.0 cm from
grinding wheel provide a graphic the center, and the end track is 5.0 cm from the center. What are the angular speeds and
illustration of instantaneous tangen- the tangential speeds at these distances when the record is spinning at 45 rpm? (b) For
tial velocity. (Why do the paths races on oval tracks, why do inside and outside runners have different starting points
curve slightly?) (called a “staggered” start), such that some runners start “ahead” of others?
PERIOD AND FREQUENCY

Some other quantities commonly used to describe circular motion are period and
frequency. The time it takes an object in circular motion to make one complete rev-
olution (or rotation), or cycle, is called the period (T). For example, the period of
revolution of the Earth about the Sun is one year, and the period of the Earth’s
axial rotation is 24 h.* The standard unit of period is the second (s). The period is
sometimes given in seconds per cycle (s>cycle).
Closely related to the period is the frequency ( f ), which is the number of revolu-
tions, or cycles, made in a given time, generally a second. For example, if a particle trav-
eling uniformly in a circular orbit makes 5.0 revolutions in 2.0 s, the frequency (of
revolution) is f = 5.0 rev>2.0 s = 2.5 rev>s, or 2.5 cycles>s (cps, or cycles per second).
Revolution and cycle are merely descriptive terms used for convenience and are
not units. Without these descriptive terms, it can be seen that the unit of frequency
is inverse seconds (1/s, or s -1), which is called the hertz (Hz) in the SI.†
*The discussion applies to rotations as well as revolutions. Revolutions will be used as a general
term, as is commonly done, for example a CD rotates at so many revolution per minute (rpm).
†
Named for Heinrich Hertz (1857–1894), a German physicist and pioneering investigator of electro-
magnetic waves, which also are characterized by frequency.
7.3 UNIFORM CIRCULAR MOTION AND CENTRIPETAL ACCELERATION 229
The two quantities are inversely related by
1 (relationship of
f = (7.7)
T frequency and period)
SI unit of frequency: hertz 1Hz, 1>s or s -12
where the period is in seconds and the frequency is in hertz, or inverse seconds.
For uniform circular motion, the tangential orbital speed is related to the period T
by v = 2pr>T—that is, the distance traveled in one revolution divided by the time for
one revolution (one period). The frequency can also be related to the angular speed.
Since an angular distance of 2p rad is traveled in one period (by definition of
the period), then
2p (angular speed
v = = 2pf (7.8)
T in terms of period and frequency)
Notice that v and f have the same units, v = 1>s and f = 1>s. This notation can eas-
ily cause confusion, which is why the unitless radian (rad) term is often used in angu-
lar speed (rad>s) and cycles in frequency (cycles>s).
EXAMPLE 7.4 Frequency and Period: An Inverse Relationship

A compact disc (CD) rotates in a player at a constant speed of T H I N K I N G I T T H R O U G H . The relationships for the frequency
200 rpm. What are the CD’s (a) frequency and (b) period of ( f ), the period (T), and the angular frequency (v), are
revolution? expressed in Eqs. 7.7 and 7.8, so these equations can be used.
SOLUTION. The angular speed is not in standard units and so must be converted. Revolutions per minute (rpm) is converted to
radians per second (rad>s).
v = a ba ba b = 20.9 rad>s
Given: 200 rev 1 min 2p rad Find: (a) f (frequency)
min 60 s rev (b) T (period)
[Note that the above unit conversion could be done using one (b) Equation 7.8 could be used to find T, but Eq. 7.7 is a bit
convenient factor: 11rev>min2 = 1p>302 rad>s]* simpler:
(a) Rearranging Eq. 7.8 and solving for f, 1 1
T = = = 0.300 s
v 20.9 rad>s f 3.33 Hz
f = = = 3.33 Hz
2p 2p rad>cycle Thus, the CD takes 0.300 s to make one revolution. (Notice
The units of 2p are rad>cycle or revolution, so the result is in that since Hz = 1>s, the equation is dimensionally correct.)
cycles>second or inverse seconds, which is the hertz.
*11 rev>min2[11 min>60 s212p rad>rev2] = 1p>302 rad>s. That is, 1p>30 rad>s2>rpm.
FOLLOW-UP EXERCISE. If the period of a particular CD is 0.500 s, what is the CD’s angular speed in revolutions per minute?
DID YOU LEARN?

➥ The tangential and angular speeds are directly proportional: vt = rv.
➥ Period and frequency are inversely proportional: T = 1>f, or f = 1/T.
➥ The angular speed (v) is inversely proportional to the period (T) and directly
proportional to the frequency (f ), that is, v = 2p>T = 2pf.
7.3 Uniform Circular Motion and Centripetal Acceleration

➥ What is necessary for uniform circular motion?

➥ How is it known that there is an acceleration for uniform circular motion?
A simple, but important, type of circular motion is uniform circular motion,

which occurs when an object moves at a constant speed in a circular path. An exam-
ple of this motion may be a car going around a circular track at a constant speed.
䉴 F I G U R E 7 . 7 Uniform circular v1 ∆v
motion The speed of an object in
uniform circular motion is constant,
but the object’s velocity changes in v2 = v1 + ∆v
the direction of motion. Thus, there
is an acceleration. v2 v2
v3 = ∆v
v2 + ∆v
v1
Speed is constant:
v1 = v2 = v3 v3
but velocity, due to change
in direction, is not:
v1 ≠ v2 ≠ v3
䉲 F I G U R E 7 . 8 Analysis of cen-
tripetal acceleration (a) The velocity
vector of an object in uniform circu-
lar motion is constantly changing
direction. (b) As ¢t, the time inter- The motion of the Moon around the Earth is approximated by uniform circular
val for ¢u, is taken to be smaller and
smaller and approaches zero, ¢v
B motion. Such motion is curvilinear, and as discussed in Section 3.1 there must be
(the change in the velocity, and an acceleration. But what are its magnitude and direction?
therefore an acceleration) is directed
toward the center of the circle. The
result is a centripetal, or center- CENTRIPETAL ACCELERATION
seeking, acceleration that has a
magnitude of ac = v2>r. The acceleration of uniform circular motion is not in the same direction as the instan-
taneous velocity (which is tangent to the circular path at any point). If it were, the
∆s
v1 object would speed up, and the circular motion wouldn’t be uniform. Recall that
acceleration is the time rate of change of velocity and that velocity has both magnitude
v2 and direction. In uniform circular motion, the direction of the velocity is continuously
changing, which is a clue to the direction of the acceleration. (䉱 Fig. 7.7).
r ∆u
The velocity vectors at the beginning and end of a time interval give the change in
B
velocity, or ¢v , via vector subtraction. All of the instantaneous velocity vectors have
the same magnitude or length (constant speed), but they differ in direction. Note that
because ¢v B
is not zero, there must be an acceleration 1aB = ¢vB
>¢t2.
B
As illustrated in 䉳 Fig. 7.8, as ¢t (or ¢u) becomes smaller, ¢v points more
toward the center of the circular path. As ¢t approaches zero, the instantaneous
change in the velocity, and therefore the acceleration, points directly toward the
center of the circle. As a result, the acceleration in uniform circular motion is called
centripetal acceleration (a c), which means “center-seeking” acceleration (from the
(a) Latin centri, “center,” and petere, “to fall toward” or “to seek”).
The centripetal acceleration must be directed radially inward, that is, with no
v2 = v1 + ∆v or
component in the direction of the perpendicular (tangential) velocity, or else the
v2 – v1 = ∆v magnitude of that velocity would change (䉲 Fig. 7.9). Note that for an object in
v1
∆u ∆v 䉴 F I G U R E 7 . 9 Centripetal accel- v
eration For an object in uniform cir-
v2 cular motion, the centripetal
acceleration is directed radially ac v
inward. There is no acceleration ac
component in the tangential direc-
ac direction tion; if there were, the magnitude of
as ∆v 0 the velocity (tangential speed) would
change.
ac
(b)
uniform circular motion, the direction of the centripetal acceleration is continuously

changing. In terms of x- and y-components, ax and ay are not constant. (Can you
describe how this differs from the acceleration in projectile motion?)
The magnitude of the centripetal acceleration can be deduced from the small
shaded triangles in Fig. 7.8. (For very short time intervals, the arc length ¢s is
almost a straight line—the chord.) These two triangles are similar triangles,
because each has a pair of equal sides surrounding the same angle ¢u. (Note that
the velocity vectors have the same magnitude.) Thus, ¢v is to v as ¢s is to r,
which can be written as*
¢v ¢s
L
v r
The arc length ¢s is the distance traveled in time ¢t, thus, ¢s = v¢t so
¢v ¢s v¢t
L =
v r r
and
¢v v2
L
¢t r
Then, as ¢t approaches zero, this approximation becomes exact. The instanta-
neous centripetal acceleration, ac = ¢v>¢t, thus has a magnitude of
v2 (magnitude of centripetal acceleration

ac = (7.9)
r in terms of tangential speed)
Using Eq. 7.6 1v = rv2, the equation for centripetal acceleration can also be writ-
ten in terms of the angular speed:
v2 1rv22 (magnitude of centripetal acceleration

ac = = = rv2 (7.10)
r r in terms of angular speed)
Orbiting satellites have centripetal accelerations, and a down-to-Earth medical

application of centripetal acceleration is discussed in Insight 7.1, The Centrifuge:
Separating Blood Components.
*The subscript t will be dropped with the understanding that v is tangential speed in ufiorm circu-
lar motion.
INSIGHT 7.1 The Centrifuge: Separating Blood Components

The centrifuge is a machine with rotating parts that is used to The erythrocite sedimentation rate (ESR) has some diagnos-
separate particles of different sizes and densities suspended tic value, but clinicians usually do not want to wait a long
in a liquid (or a gas). For example, cream is separated from time to see the fractional volume of red cells (erythrocites) in
milk by centrifuging, and blood components are separated in the blood or separate them from the plasma. And so a cen-
centrifuges in medical laboratories (see Fig. 7.10). trifuge is used to speed up the process. Centrifuge tubes are
There is a much slower process to separate blood compo- pivoted and spin horizontally. The resistance of the fluid
nents. They will eventually settle in layers in a vertical test medium on the particles supplies the centripetal acceleration
tube—a process called sedimentation—under the influence of that keeps them moving in slowly widening circles as they
normal gravity alone. The viscous drag of the plasma on the travel toward the bottom of the tube. The bottom of the tube
particles is analogous to (but much greater than) the air resis- itself must exert a strong force on the contents as a whole and
tance that determines the terminal velocity of falling objects must be strong enough not to break.
(Section 4.6). Red blood cells settle in the bottom layer of the Laboratory centrifuges commonly operate at speeds suffi-
tube, because they have a greater terminal velocity than do cient to produce centripetal accelerations thousands of times
the white blood cells and platelets and reach the bottom larger than g. (See Example 7.5.) Since the principle of the
sooner. The white cells settle in the next layer and the centrifuge involves centripetal acceleration, perhaps
platelets settle on top. However, gravitational sedimentation “centripuge” would be a more descriptive name.
is generally a very slow process.
EXAMPLE 7.5 A Centrifuge: Centripetal Acceleration

A laboratory centrifuge like that shown in 䉴 Fig. 7.10 operates at a rotational
speed of 12 000 rpm. (a) What is the magnitude of the centripetal accelera-
tion of a red blood cell at a radial distance of 8.00 cm from the centrifuge’s
axis of rotation? (b) How does this acceleration compare with g?
T H I N K I N G I T T H R O U G H . Here, the angular speed and the radius are given,
so the magnitude of the centripetal acceleration can be computed directly
from Eq. 7.10. The result can be compared with g by using g = 9.80 m>s2.
SOLUTION. The data are as follows:
1p> 302rad>s
v = 11.20 * 104 rpm2 c d
Given: Find: (a) ac
rpm (b) how ac
= 1.26 * 103 rad>s compares
(using a single conversion factor) with g
r = 8.00 cm = 0.0800 m
(a) The centripetal acceleration is found from Eq. 7.10:
ac = rv2 = 10.0800 m211.26 * 103 rad>s22 = 1.27 * 105 m>s2
(b) Using the relationship 1 g = 9.80 m>s2 to express ac in terms of g, 䉱 F I G U R E 7 . 1 0 Centrifuge Centrifuges are
used to separate particles of different sizes and
ac = 11.27 * 105 m>s22 a b = 1.30 * 104 g ( = 13 000 g!2
1g
densities suspended in liquids. For example, red
9.80 m>s2 and white blood cells can be separated from
each other and from the plasma that makes up
the liquid portion of the blood in the centrifuge
tube. When spinning, the tubes are horizontal.
FOLLOW-UP EXERCISE. What angular speed in revolutions per minute would give a centripetal acceleration of 1 g at the
radial distance in this Example, and, taking gravity into account, what would be the resultant acceleration?
CENTRIPETAL FORCE
For an acceleration to exist, there must be a net force. Thus, for a centripetal (inward)
acceleration to exist, there must be a centripetal force (net inward force). Expressing
the magnitude of this force in terms of Newton’s second law 1Fnet = maB2 and insert-
B
ing the expression for centripetal acceleration from Eq. 7.9 for magnitude,
mv2
Fc = ma c = (magnitude of centripetal force) (7.11)
r
The centripetal force, like the centripetal acceleration, is directed radially toward
the center of the circular path.
CONCEPTUAL EXAMPLE 7.6 Breaking Away

A ball attached to a string is swung with uniform motion in a first law states that if no force acts on an object in motion, the
horizontal circle above a person’s head (䉴 Fig. 7.11a). If the object will continue to move in a straight line. This factor
string breaks, which of the trajectories shown in Fig. 7.11b rules out trajectories b, d, and e.
(viewed from above) would the ball follow? It should be evident from the previous discussion that at
any instant (including the instant when the string breaks),
REASONING AND ANSWER. When the string breaks, the cen- the isolated ball has a horizontal, tangential velocity. The
tripetal force goes to zero. There is no force in the outward downward force of gravity acts on it, but this force affects
direction, so the ball could not follow trajectory a. Newton’s only its vertical motion, which is not visible in Fig. 7.11b.
a b ?
F
String c
breaks ?
d
?
e
Overhead ?
view
(a) (b)
䉱 F I G U R E 7 . 1 1 Centripetal force (a) A ball is swung in a horizontal circle. (b) If the string breaks and the centripetal
force goes to zero, what happens to the ball? See Example text for description.
The ball thus flies off tangentially and is essentially a hori- Viewed from above, the ball would appear to follow the
zontal projectile (with vxo = v, vyo = 0, and ay = - g). path labeled c.
F O L L O W - U P E X E R C I S E . If you swing a ball in a horizontal circle about your head, can the string be exactly horizontal? (See
Fig. 7.11a.) Explain your answer. [Hint: Analyze the forces acting on the ball.]
Keep in mind that, in general, a net force applied at an angle to the direction of
motion of an object produces changes in the magnitude and direction of the veloc-
ity. However, when a net force of constant magnitude is continuously applied at
an angle of 90° to the direction of motion (as is centripetal force), only the direction
of the velocity changes. This is because there is no force component parallel to the
velocity. Also notice that because the centripetal force is always perpendicular to
the direction of motion, this force does no work. (Why?) Therefore, a centripetal
force does not change the kinetic energy or speed of the object.
Note that the centripetal force in the form Fc = mv 2>r is not a new individual
force, but rather the cause of the centripetal acceleration, and is supplied by either
a real force or the vector sum of several forces.
The force supplying the centripetal acceleration for satellites is gravity. In
Conceptual Example 7.6, it was the tension in the string. Another force that often
supplies centripetal acceleration is friction. Suppose that an automobile moves
into a level, circular curve. To negotiate the curve, the car must have a cen-
tripetal acceleration, which is supplied by the force of friction between the tires
and the road.
However, this static friction (why static?) has a maximum limiting value. If
the speed of the car is high enough or the curve is sharp enough, the friction will
not be sufficient to supply the necessary centripetal acceleration, and the car will
skid outward from the center of the curve. If the car moves onto a wet or icy
spot, the friction between the tires and the road may be reduced, allowing the
car to skid at an even lower speed. (Banking a curve helps vehicles negotiate the
curve without slipping.)
EXAMPLE 7.7 Where the Rubber Meets the Road: Friction and Centripetal Force
A car approaches a level, circular curve with a radius of 45.0 m. Recall from Section 4.6 that the maximum frictional force is
If the concrete pavement is dry, what is the maximum speed at given by fsmax = ms N (Eq. 4.7), where N is the magnitude of
which the car can negotiate the curve at a constant speed? the normal force on the car and is equal in magnitude to the
weight of the car, mg, on the level road (why?). Thus the mag-
T H I N K I N G I T T H R O U G H . The car is in uniform circular
nitude of the maximum static frictional force is equal to the
magnitude of the centripetal force 1Fc = mv2>r2. From this the
motion on the curve, so there must be a centripetal force. This
force is supplied by static friction, so the maximum static fric-
maximum speed can be found. To find fsmax , the coefficient of
tional force provides the centripetal force when the car is at its
friction between rubber and dry concrete is needed, and from
maximum tangential speed.
Table 4.1, ms = 1.20. Then,
SOLUTION.
fsmax = Fc
Given: r = 45.0 m Find: v (maximum speed)
ms = 1.20 (from Table 4.1) mv2
ms N = ms mg =
r
To go around the curve at a particular speed, the car must
have a centripetal acceleration, and therefore a centripetal So
v = 1ms rg = 411.202145.0 m219.80 m>s22 = 23.0 m>s
force must act on it. This inward force is supplied by static
friction between the tires and the road. (The tires are not slip-
ping or skidding relative to the road.) (about 83 km>h, or 52 mi>h).
FOLLOW-UP EXERCISE. Would the centripetal force be the same for all types of vehicles as in this Example? Explain.
The proper safe speed for driving on a highway curve is an important considera-
tion. The coefficient of friction between tires and the road may vary, depending on
weather, road conditions, the design of the tires, the amount of tread wear, and so on.
When a curved road is designed, safety may be promoted by banking, or inclining, the
roadway. This design reduces the chances of skidding because the normal force
exerted on the car by the road then has a component toward the center of the curve
that reduces the need for friction. In fact, for a circular curve with a given banking
angle and radius, there is one speed for which no friction is required at all. This condi-
tion is used in banking design. (See Conceptual Question 12 at the end of the chapter.)
Let’s look at one more example of centripetal force, this time with two objects in
uniform circular motion. Example 7.8 will help give a better understanding of the
motions of satellites in circular orbits, discussed in a later section.
EXAMPLE 7.8 Strung Out: Centripetal Force and Newton’s Second Law
Suppose that two masses, m1 = 2.5 kg and m2 = 3.5 kg, are the masses are T1 = 4.5 N and T2 = 2.9 N, which are the
connected by light strings and are in uniform circular motion respective tensions in the strings. Find the magnitude of the
on a horizontal frictionless surface as illustrated in 䉲 Fig. 7.12, centripetal acceleration and the tangential speed of (a) mass
where r1 = 1.0 m and r2 = 1.3 m. The tension forces acting on m2 and (b) mass m1.
䉴 F I G U R E 7 . 1 2 Centripetal v2
force and Newton’s second law m2
See Example text for description. v1
m1
v2
m2
r1 T2
1.0 m
T2
r2 v1
1.3 m m1
T1
T H I N K I N G I T T H R O U G H . The centripetal forces on the masses net force on a mass is equal to the mass’s centripetal force
are supplied by the tensions (T1 and T2) in the strings. By iso- 1Fc = ma c2. The tangential speeds can then be found, since the
lating the masses, ac for each mass can be found, because the radii are known 1a c = v2>r2.
SOLUTION.
Given: r1 = 1.0 m and r2 = 1.3 m Find: ac (centripetal acceleration) and v (tangential speed)
m1 = 2.5 kg and m2 = 3.5 kg (a) m2
T1 = 4.5 N (b) m1
T2 = 2.9 N
(a) By isolating m2 in the figure, it can seen that the cen- (b) The situation is a bit different for m1. In this case, two radial
tripetal force is provided by the tension in the string. (T2 is the forces are acting on m1 : the string tensions T1 (inward) and - T2
only force acting on m2 toward the center of its circular path.) (outward). By Newton’s second law, in order to have a cen-
Thus, tripetal acceleration, there must be a net force, which is given by
the difference in the two tensions, so we expect T1 7 T2, and
T2 = m2 a c2
m1 v21
and Fnet1 = + T1 + 1-T22 = m1 ac1 =
r1
T2 2.9 N where the radial direction (toward the center of the circular path)
a c2 = = = 0.83 m>s2
m2 3.5 kg is taken to be positive. Then
T1 - T2 4.5 N - 2.9 N
where the acceleration is toward the center of the circle. ac1 = = 0.64 m>s2
The tangential speed of m2 can be found from ac = v 2>r:
=
m1 2.5 kg
v2 = 2ac2 r2 = 410.83 m>s2211.3 m2 = 1.0 m>s

and
v1 = 2ac1 r1 = 410.64 m>s2211.0 m2 = 0.80 m>s
FOLLOW-UP EXERCISE. Notice in this Example that the centripetal acceleration of m2 is greater than that of m1 yet r2 7 r1 , and
a c r 1>r. Is something wrong here? Explain.
INTEGRATED EXAMPLE 7.9 Center-Seeking Force: One More Time

A 1.0-m cord is used to suspend a 0.50-kg tetherball from the are not directly toward the circle’s center located on the pole.
top of the pole. After being hit several times, the ball goes (mg and Ty are equal and opposite, because there is no accelera-
around the pole in uniform circular motion with a tangential tion in the y-direction.) The answer is obviously (2), with a com-
speed of 1.1 m>s at an angle of 20° relative to the pole. (a) The ponent of the tension force, Tx , supplying the centripetal force.
force that supplies the centripetal acceleration is (1) the (B) QUANTITATIVE REASONING AND SOLUTION. Tx supplies the
weight of the ball, (2) a component of the tension force in the centripetal force, and the given data are for the dynamical form
string, (3) the total tension in the string. (b) What is the mag- of the centripetal force, that is, Tx = Fc = mv2>r (Eq. 7.11).
nitude of the centripetal force?
(A) CONCEPTUAL REASONING. The cen- Given: L = 1.0 m Find: Fc (magnitude of the
tripetal force, being a “center-seeking” vt = 1.1 m>s centripetal force)
force, is directed perpendicularly m = 0.50 kg
toward the pole, about which the ball u = 20°
20°
is in circular motion. As suggested in Ty L
the problem-solving procedures pro- T As pointed out previously, the magnitude of the centripetal
vided in Section 1.7, it is almost always force may be found using Eq. 7.11:
helpful to sketch a diagram, such as
r mv2
that in 䉴 Fig. 7.13. Fc = Tx =
Immediately, it can be seen that (1) r
Tx
and (3) are not correct, as these forces But the radial distance r is needed. From the figure, this quan-
tity can be seen to be r = L sin 20°, so
䉴 F I G U R E 7 . 1 3 Ball on a string
mg
mv2 10.50 kg211.1 m>s22
11.0 m210.3422
Fc = = = 1.8 N
See Example text for description. L sin 20°
FOLLOW-UP EXERCISE. (a) What is the magnitude of the tension T in the string? (b) What is the period of the ball’s rotation?
DID YOU LEARN?

➥ A “center-seeking” or centripetal force is required for uniform circular motion.
➥ The tangential velocity of an object in uniform circular motion continually changes
direction. By definition there must be an acceleration—a centripetal (center-seeking)
acceleration.
7.4 Angular Acceleration

➥ How is the tangential acceleration (at) related to the angular acceleration (a)?
➥ The linear and angular kinematic equations for constant accelerations have similar
form.What are the analogous quantities?
As you might have guessed, there is another type of acceleration besides linear,
and that is angular acceleration. This quantity is the time rate of change of angular
velocity. In circular motion, if there is an angular acceleration, the motion is not
uniform, because the speed and>or direction would be changing. Analogous to
the linear case, the magnitude of the average angular acceleration (A q ) is given by
¢v
q =
a
¢t
where the bar over the alpha indicates that it is an average value, as usual. Taking
to = 0, and if the angular acceleration is constant, so that a
q = a, then
v - vo
a = (constant angular acceleration)
t
SI unit of angular acceleration: radians per second squared (rad>s2)
and rearranging,
(constant angular
v = vo + at (7.12)
acceleration only)
No boldface vector symbols with overarrows are used in Eq. 7.12, because, plus
and minus signs will be used to indicate angular directions, as described earlier. As in
the case of linear motion, if the angular acceleration increases the angular velocity,
both quantities have the same sign, meaning that their vector directions are the same
(that is, a is in the same direction as v as given by the right-hand rule). If the angular
acceleration decreases the angular velocity, then the two quantities have opposite
signs, meaning that their vectors are opposed (that is, a is in the direction opposite to
v as given by the right-hand rule, or is an angular deceleration, so to speak).
EXAMPLE 7.10 A Rotating CD: Angular Acceleration

A CD accelerates uniformly from rest to its operational speed T H I N K I N G I T T H R O U G H . (a) Once given the initial and final
of 500 rpm in 3.50 s. (a) What is the angular acceleration of the angular velocities, the constant (uniform) angular acceleration
CD during this time? (b) What is the angular acceleration of can be calculated (Eq. 7.12), since the amount of time during
the CD after this time? (c) If the CD comes uniformly to a stop which the CD accelerates is known. (b) Keep in mind that the
in 4.50 s, what is its angular acceleration during this part of operational angular speed is constant. (c) Everything is given
the motion? for Eq. 7.12, but a negative result should be expected. Why?
SOLUTION.
Given: vo = 0 Find: (a) a (during startup)
1p>302 rad>s
v = 1500 rpm2c d = 52.4 rad>s
(b) a (in operation)
rpm (c) a (in coming to a stop)
t1 = 3.50 s (starting up)
t2 = 4.50 s (coming to a stop)
(a) Using Eq. 7.12, the acceleration during startup is (b) After the CD reaches its operational speed, the angular
v - vo 52.4 rad>s - 0 velocity remains constant, so a = 0.
a = = = 15.0 rad>s2
t1 3.50 s
in the direction of the angular velocity.
7.4 ANGULAR ACCELERATION 237
(c) Again using Eq. 7.12, but this time with vo = 500 rpm and where the minus sign indicates that the angular acceleration
v = 0. is in the direction opposite that of the angular velocity (which
v - vo 0 - 52.4 rad>s is taken as + ).
a = = = - 11.6 rad>s2
t2 4.50 s
B B
FOLLOW-UP EXERCISE. (a) What are the directions of the V and A vectors in part (a) of this Example if the CD rotates clockwise
when viewed from above? (b) Do the directions of these vectors change in part (c)? Explain.
As with arc length and angle 1s = ru2 and tangential and angular speeds 1v = rv2, v
there is a relationship between the magnitudes of the tangential acceleration and v = rv
the angular acceleration. The tangential acceleration (at) is associated with
v2
changes in tangential speed and hence continuously changes direction. The mag- ac =
r
nitudes of the tangential and angular accelerations are related by a factor of r. For ac = rv 2
circular motion with a constant radius r,
¢v ¢1rv2 r¢v
at = = = = ra
¢t ¢t ¢t
so
a t = ra (magnitude of tangential acceleration) (7.13)
The tangential acceleration (at) is written with a subscript t to distinguish it from

the radial, or centripetal, acceleration (ac). Centripetal acceleration is necessary for
circular motion, but tangential acceleration is not. For uniform circular motion,
there is no angular acceleration 1a = 02 or tangential acceleration, as can be seen
(a) Uniform circular motion
(α = 0)
from Eq. 7.13. There is only centripetal acceleration (䉴 Fig. 7.14a).
However, when there is an angular acceleration a (and therefore a tangential a
at
acceleration of magnitude a t = ra), there is a change in both the angular and tan-
gential velocities. As a result, the centripetal acceleration a c = v 2>r = rv2 must
ac
increase or decrease if the object is to maintain the same circular orbit (that is, if r is a
to stay the same). When there are both tangential and centripetal accelerations, the
instantaneous acceleration is their vector sum (Fig. 7.14b).
The tangential acceleration vector and the centripetal acceleration vector are
perpendicular to each other at any instant, and the acceleration is aB = at tN + ac rN ,
where tN and rN are unit vectors directed tangentially and radially inward, respec-
tively. You should be able to find the magnitude of aB and the angle it makes rela-
tive to aB t by using trigonometry (Fig. 7.14b).
Other equations for angular kinematics can be derived, as was done for the lin-
ear equations in Section 2.4. That development will not be shown here; the set of
angular equations with their linear counterparts for constant accelerations is listed
in 䉲 Table 7.2. A quick review of Section 2.4 (with a change of symbols) will show
you how the angular equations are derived. (b) Nonuniform circular motion
( a = at + ac )
䉱 F I G U R E 7 . 1 4 Acceleration and
TABLE 7.2 Equations for Linear and Angular circular motion (a) In uniform circu-
Motion with Constant Acceleration* lar motion, there is centripetal accel-
eration, but no angular acceleration
Linear Angular (a = 0) or tangential acceleration
(at = ra = 0). (b) In nonuniform
x = vqt qt
u = v (1) circular motion, there are angular
v + vo v + vo and tangential accelerations, and
vq = q =
v (2) the total acceleration is the vector
2 2
sum of the tangential and cen-
v = vo + at v = vo + at (3) tripetal accelerations.
1 1
x = xo + vo t + at2 u = uo + vo t + at2 (4)
2 2
v2 = v2o + 2a1x - xo2 v2 = v2o + 2a1u - uo2 (5)
*The first equation in each column is general, that is, not limited to
situations where the acceleration is constant.
EXAMPLE 7.11 Even Cooking: Rotational Kinematics

A microwave oven has a 30-cm-diameter rotating plate for (a) To find the angular distance u in radians, use Eq. 4 from
even cooking. The plate accelerates from rest at a uniform rate of Table 7.2 with uo = 0:
0.87 rad>s2 for 0.50 s before reaching its constant operational u = vo t + 12 at2 = 0 + 12 10.87 rad>s 2210.50 s22 = 0.11 rad
speed. (a) How many revolutions does the plate make before
reaching its operational speed? (b) What are the operational Since 2p rad = 1 rev,
u = 10.11 rad2a b = 0.018 rev

angular speed of the plate and the operational tangential 1 rev
speed at its rim? 2p rad
T H I N K I N G I T T H R O U G H . This Example involves the use of so the plate reaches its operational speed in only a small frac-
the angular kinematic equations (Table 7.2). In (a), the angular tion of a revolution.
distance u will give the number of revolutions. For (b), first (b) From Table 7.2, it can be seen that Eq. 3 gives the angular
find v and then v = rv. speed, and
SOLUTION. Listing the given data and what is to be found: v = vo + at = 0 + 10.87 rad>s2210.50 s2 = 0.44 rad>s
Then, Eq. 7.6 gives the tangential speed at the rim radius:
Given: d = 30 cm, Find: (a) u (in revolutions)
r = 15 cm = 0.15 m (b) v and v (angular v = rv = 10.15 m210.44 rad>s2 = 0.066 m>s
(radius) and tangential F O L L O W - U P E X E R C I S E . (a) When the oven is turned off, the
vo = 0 (at rest) speeds, plate makes half a revolution before stopping. What is the
a = 0.87 rad>s2 respectively) plate’s angular acceleration during this period? (b) How long
t = 0.50 s does it take to stop?
m1
DID YOU LEARN?
➥ The tangential acceleration (at) and angular acceleration (a) are directly
proportional, at = ra (similar to speeds: vt = rv).
F12 ➥ The linear quantities x, v, and a are respectively analogous to the angular quantities
u, v and a.
r
F21
7.5 Newton’s Law of Gravitation
m2 ➥ What is meant by an inverse-square relationship?

(a) Point masses ➥ How does the acceleration due to gravity vary with altitude above the Earth’s surface?
Another of Isaac Newton’s many accomplishments was the formulation of the

universal law of gravitation. This law is very powerful and fundamental. With-
out it, for example, we would not understand the cause of tides or know how to
m1 put satellites into particular orbits around the Earth. This law allows us to analyze
the motions of planets, comets, stars, and even galaxies. The word universal in the
name indicates that it is believed to apply everywhere in the universe. (This term
highlights the importance of the law, but for brevity, it is common to refer simply
F12 to Newton’s law of gravitation or the law of gravitation.)
Newton’s law of gravitation in mathematical form gives a simple relationship for
r the gravitational interaction between two particles, or point masses, m1 and m2 sepa-
rated by a distance r (䉳 Fig. 7.15a). Basically, every particle in the universe has an
F21 attractive gravitational interaction with every other particle because of their masses.
The forces of mutual interaction are equal and opposite, forming a force pair as
B B
described by Newton’s third law (Section 4.4), that is, F12 = - F21 in Fig. 7.15a.
m2
䉳 F I G U R E 7 . 1 5 Universal law of gravitation

(a) Any two particles, or point masses, are
(b) Homogeneous spheres gravitationally attracted to each other with a
force that has a magnitude given by Newton’s
Gm1 m2 universal law of gravitation. (b) For homoge-
F12 = F21 = neous spheres, the masses may be considered
r2
to be concentrated at their centers.
7.5 NEWTON’S LAW OF GRAVITATION 239
The gravitational attraction, or force (F), decreases as the square of the distance
(r 2) between two point masses increases; that is, the magnitude of the gravita-
tional force and the distance separating the two particles are related as follows:
1 Moon
F r 2 ac << g
r
(This type of relationship is called an inverse-square, that is, F is inversely propor-

tional to r 2.)
Newton’s law also correctly postulates that the gravitational force, or attraction
of a body, depends on the body’s mass—the greater the mass, the greater the attrac-
tion. However, because gravity is a mutual interaction between masses, it should
be directly proportional to both masses, that is, to their product 1F r m1 m22.
Hence, Newton’s law of gravitation has the form F r m1 m2>r 2. Expressed as
a=g
an equation with a constant of proportionality, the magnitude of the mutually

attractive gravitational force (Fg) between two masses is given by
Gm1 m2
Fg = (Newton’s law of gravitation) (7.14)
r2
where G is a constant called the universal gravitational constant and has a value of Earth
G = 6.67 * 10-11 N # m2>kg 2

䉱 F I G U R E 7 . 1 6 Gravitational
This constant is often referred to as “big G” to distinguish it from “little g,” the insight? Newton developed his law
acceleration due to gravity. Note from Eq. 7.14 that Fg approaches zero only when of gravitation while studying the
orbital motion of the Moon. Accord-
r becomes infinitely large. That is, the gravitational force has, or acts over, an ing to legend, his thinking was
infinite range. spurred when he observed an apple
How did Newton come to his conclusions about the force of gravity? Legend falling from a tree. He supposedly
has it that his insight came after he observed an apple fall from a tree to the wondered whether the force caus-
ground. Newton had been wondering what supplied the centripetal force to keep ing the apple to accelerate toward
the ground could extend to the
the Moon in orbit and might have had this thought: “If gravity attracts an apple Moon and cause it to “fall” or accel-
toward the Earth, perhaps it also attracts the Moon, and the Moon is ‘falling’, or erate toward Earth, that is, supply
accelerating toward the Earth, under the influence of gravity” (䉴 Fig. 7.16). its orbital centripetal acceleration.
Whether or not the legendary apple did the trick, Newton assumed that the
Moon and the Earth were attracted to each other and could be treated as point
masses, with their total masses concentrated at their centers (Fig 7.15b). The
inverse-square relationship had been speculated on by some of his contempo-
raries. Newton’s achievement was demonstrating that the relationship could be
deduced from one of Johannes Kepler’s laws of planetary motion (Section 7.6).
Newton expressed Eq. 7.14 as a proportion 1Fg r m1 m2 >r 22 because he did not
know the value of G. It was not until 1798 (seventy-one years after Newton’s
death) that the value of the universal gravitational constant was experimentally
determined by an English physicist, Henry Cavendish. Cavendish used a sensi-
tive balance to measure the gravitational force between separated spherical
masses (as illustrated in Fig. 7.15b). If F, r, and the m’s are known, G can be com-
puted from Eq. 7.14.
As mentioned earlier, Newton considered the nearly spherical Earth and Moon
to be point masses located at their respective centers. It took him some years,
using mathematical methods he developed, to prove that this is the case only for
spherical, homogeneous objects.* The concept is illustrated in 䉲 Fig. 7.17.
*For a homogeneous sphere, the equivalent point mass is located at the center of mass. However,
this is a special case. The center of gravitational force and the center of mass of a configuration of parti-
cles or an object do not generally coincide.
䉴 F I G U R E 7 . 1 7 Uniform spherical masses

(a) Gravity acts between any two particles. The
resultant gravitational force exerted on an object
outside a homogeneous sphere by two particles
at symmetric locations within the sphere is
directed toward the center of the sphere.
(b) Because of the sphere’s symmetry and uni-
form distribution of mass, the net effect is as
though all the mass of the sphere were concen-
trated as a particle at its center. For this special
case, the gravitational center of force and center
of mass coincide, but this is generally not true for
other objects. (Only a few of the red force arrows
are shown because of space considerations.)
(a) (b)
EXAMPLE 7.12 Greater Gravitational Attraction?

The gravitational attractions of the Sun and the Moon give T H I N K I N G I T T H R O U G H . To see if this is true, the gravita-
rise to ocean tides. It is sometimes said that since the Moon is tional attractions of the Moon and the Sun on the Earth can be
closer to the Earth than the Sun, the Moon's gravitational easily calculated using Newton’s law of gravitation. The masses
attraction is much stronger, and therefore has a greater influ- and distances are given on the inside back cover of the book.
ence on ocean tides. Is this true? (Assume that the bodies are solid, homogenous spheres.)
SOLUTION. No data are given, so this must be available from references:

Given: From tables on the inside back cover: Find: FEM (gravitational force, Earth–Moon)
24
mE = 6.0 * 10 kg (mass of the Earth) FES (gravitational force, Earth–Sun)
mM = 7.4 * 1022 kg (mass of the Moon)
mS = 2.0 * 1030 kg (mass of the Sun)
rEM = 3.8 * 108 m (average distance between)
rES = 1.5 * 108 km (average distance between)
The average distances are taken to be the distance from the center of one to the center of the other. Using Eq. 7.14, remembering to
change kilometers to meters:
Gm1 m2 GME mM
FEM = =
r2 r2EM
16.67 * 10-11 N # m2 >kg 2216.0 * 1024 kg217.4 * 1022 kg2

13.8 * 108 m22
=
= 2.1 * 1020 N (Earth–Moon)
GmE ms 16.67 * 10-11 N # m2>kg 2216.0 * 1024 kg212.0 * 1030 kg2

FES =
11.5 * 1011 m22
=
r 2ES
= 3.6 * 1022 N (Earth–Sun)
So, the gravitational attraction of the Sun on the Earth is much what displaced, leaving the least attracted water on the oppo-
greater than that of the Moon on the Earth, on the order of 100 site side of the Moon where another tidal bulge is formed. As
times greater. But it is well known that the Moon has the the Moon revolves about the Earth, the tidal bulges tag along,
major influence on tides. How is this with less gravitational and there are two high tides (bulges) and two low tides daily.
attraction? Basically, it is because the gravitational differential (Actually, the two high tides are 12 h, 25 min apart.)
of the Moon with less gravitational attraction is greater. That Even though the Sun has greater gravitational attraction,
is, the ocean water on the side of the Earth toward the Moon the differential distances from the Sun to the water-Earth-
is closer and the gravitational attraction forms a tidal bulge. water are miniscule, and so the Sun has little effect on the
The Earth is less attracted toward the Moon, but it is some- daily tides.
F O L L O W - U P E X E R C I S E . The gravitational attraction of the Earth on the Moon provides the centripetal force that keeps the Moon
revolving in its orbit. It is sometimes said the Moon is “falling” (accelerating) toward the Earth. What is the magnitude of the
Moon’s acceleration in “falling” toward the Earth? And with this acceleration, why doesn’t the Moon get closer to the Earth?
The acceleration due to gravity at a particular distance from a planet can also be
investigated by using Newton’s second law of motion and the law of gravitation.
The magnitude of the acceleration due to gravity, which will generally be written
as ag at a distance r from the center of a spherical mass M, is found by setting the
force of gravitational attraction due to that spherical mass equal to mag. This is the
net force on an object of mass m at a distance r:
GmM
ma g =
r2
Then, the acceleration due to gravity at any distance r from the planet’s center is
GM
ag = (7.15)
r2
Notice that ag is proportional to 1/r2, so the farther away an object is from the
planet, the smaller its acceleration due to gravity and the smaller the attractive
force (mag) on the object. The force is directed toward the center of the planet.
Equation 7.15 can be applied to the Moon or any planet. For example, taking
the Earth to be a point mass ME located at its center and RE as its radius, we obtain
the acceleration due to gravity at the Earth’s surface 1a gE = g2 by setting the dis-
tance r to be equal to RE :
GME
agE = g = (7.16)
R 2E
This equation has several interesting implications. First, it reveals that taking g
to be constant everywhere on the surface of the Earth involves the assumption
that the Earth has a homogeneous distribution of mass and that the distance from
the center of the Earth to any location on its surface is the same. These two
assumptions are not exactly true. Therefore, taking g to be a constant is an approx-
imation, but one that works pretty well for most situations.
Also, you can see why the acceleration due to gravity is the same for all free-
falling objects—that is, independent of the mass of the object. The mass of the
object doesn’t appear in Eq. 7.16, so all objects in free fall accelerate at the same rate.
Finally, if you’re observant, you’ll notice that Eq. 7.16 can be used to compute
the mass of the Earth. All of the other quantities in the equation are measurable
and their values are known, so ME can readily be calculated. This is what
Cavendish did after he determined the value of G experimentally.
The acceleration due to gravity does vary with altitude. At a distance h above
the Earth’s surface, r = RE + h. The acceleration is then given by
GME
ag = (7.17)
1RE + h22
When comparing accelerations due to gravity or gravitational forces, you will often find it
convenient to work with ratios. For example, comparing ag with g (Eqs. 7.15 and 7.16) for the
Earth gives
ag GME/r 2 RE 2 ag RE 2
a b a b
RE
= = = or =
g GME/R 2E r2 r g r
Note how the constants cancel out. Taking r = RE + h you can easily compute a g /g, or
the acceleration due to gravity at some altitude h above the Earth compared with g on
the Earth’s surface (9.80 m>s 2).
Because RE is very large compared with everyday altitudes above the Earth’s surface,
the acceleration due to gravity does not decrease very rapidly with height. At an altitude
of 16 km (10 mi, about twice as high as modern jet airliners fly), a g >g = 0.99, and thus ag
is still 99% of the value of g at the Earth’s surface. At an altitude of 320 km (200 mi), ag is
91% of g. This is the approximate altitude of an orbiting space shuttle. (So “floating”
astronauts in an Earth-orbiting space station do have weight. The so-called “weightless”
condition is discussed in Section 7.6.)
EXAMPLE 7.13 Geosynchronous Satellite Orbit

Some communication and weather satellites are launched T H I N K I N G I T T H R O U G H . To remain above one location at the
into circular orbits above the Earth’s equator so they are equator, the period of the satellite’s revolution must be the
synchronous (from the Greek syn-, same, and chronos, time) same as the Earth’s period of rotation—24 h. Also, the cen-
with the Earth’s rotation. That is, they remain “fixed” or tripetal force keeping the satellite in orbit is supplied by the
“hover” over one point on the equator. At what altitude are gravitational force of the Earth, Fg = Fc . The distance between
these geosynchronous satellites? the center of the Earth and the satellite is r = RE + h . (RE is
the radius of the Earth and h is the height or altitude of the
satellite above the Earth’s surface.)
SOLUTION. Listing the known data,

Given: T 1period2 = 24 h = 8.64 * 104 s Find: h (altitude)
r = RE + h
From solar system data inside back cover:
RE = 6.4 * 103 km = 6.4 * 106 m
ME = 6.0 * 1024 kg
Setting the magnitudes of gravitational force and the motional 16.67 * 10-11 N # m2>kg 2216.0 * 1024 kg218.64 * 104 s22
centripetal force equal 1Fg = Fc2, where m is the mass of the r3 =
satellite, and putting the values in terms of angular speed, 4p2
= 76 * 1021 m3
Fg = Fc
m1rv22 And taking the cube root:
GmME mv2
= = = mrv2 3
r2 r r r = 276 * 1021 m3
and = 4.2 * 107 m
T 2 So,
= GME a b = a bT
GME GME 2
r3 = 2
v 2p 4p2 h = r - RE = 4.2 * 107m - 0.64 * 107 = 3.6 * 107m
using the relationship v = 2p>T. Then substituting values: = 3.6 * 104 km 1 = 22 000 mi2
F O L L O W - U P E X E R C I S E . Show that the period of a satellite in orbit close to the Earth’s surface 1h V RE2 may be approximated
by T2 L 4RE and compute T. (Neglect air resistance.)
Another aspect of the decrease of g with altitude concerns potential energy. In

Section 5.5, it was learned that U = mgh for an object at a height h above some
zero reference point, since g is essentially constant near the Earth’s surface. This
potential energy is equal to the work done in raising the object a distance h above
the Earth’s surface in a uniform gravitational field.
But what if the change in altitude is so large that g cannot be considered con-
stant while work is done in moving an object, such as a satellite? In this case, the
equation U = mgh doesn’t apply. In general, it can be shown (using mathematical

methods beyond the scope of this book) that the gravitational potential energy
(U) of two point masses separated by a distance r is given by
Gm1m2
U = - (7.18)
r
The minus sign in Eq. 7.18 arises from the choice of the zero reference point (the
point where U = 0), which is r = q (infinity).
In terms of the Earth and a mass m at an altitude h above the Earth’s surface,
Gm1 m2 GmME
U = - = - (7.19)
r RE + h
where r is the distance separating the Earth’s center and the mass. This means that
on the Earth we can visualize ourselves as being in a negative gravitational poten-
tial energy well (䉲 Fig. 7.18) that extends to infinity, because the force of gravity
has an infinite range. As h increases, so does U. That is, U becomes less negative, or
gets closer to zero (that is, more positive), corresponding to a higher position in
the potential energy well.
Thus, when gravity does negative work (an object moves higher in the well) or
gravity does positive work (an object falls lower in the well), there is a change in
potential energy. As with finite potential energy wells, this change in energy is one
of the most important things in analyzing situations such as these.
U0
RE r
U=0 ∞
1
U∝–
r
GmME
U=–
RE
RE
Earth
U0
䉱 F I G U R E 7 . 1 8 Gravitational potential energy well On the Earth, we can visualize

ourselves as being in a negative gravitational potential energy well. As with an actual well
or hole in the ground, work must be done against gravity to get higher in the well. The
potential energy of an object increases as the object moves higher in the well. This means
that the value of U becomes less negative. The top of the Earth’s gravitational well is at
infinity, where the gravitational potential energy is, by choice, zero.
EXAMPLE 7.14 Different Orbits: Change in Gravitational Potential Energy

Two 50-kg satellites move in circular orbits about the Earth at T H I N K I N G I T T H R O U G H . The potential energies of the satel-
altitudes of 1000 km (about 620 mi) and 37 000 km (about lites are given by Eq. 7.19. Since an increase in altitude (h)
23 000 mi), respectively. The lower one monitors particles about results in a less negative value of U, the satellite with the
to enter the atmosphere, and the higher, geosynchronous one greater h is higher in the gravitational-potential energy well
takes weather pictures from its stationary position with respect and has more gravitional potential energy.
to the Earth’s surface over the equator (see Example 7.13).
What is the difference in the gravitational potential energies of
the two satellites in their respective orbits?
SOLUTION. Listing the data so we can better see what’s given (with two significant figures),
Given: m = 50 kg Find: ¢U (difference in potential energy)
h1 = 1000 km = 1.0 * 106 m
h2 = 37 000 km = 37 * 106 m
ME = 6.0 * 1024 kg (from the inside the back
cover of the book)
RE = 6.4 * 106 m
The difference in the gravitational potential energy can be computed directly from Eq. 7.19. Keep in mind that the potential
energy is the energy of position, so we compute the potential energies for each position or altitude and subtract one from the
other. Thus,
- a- b = GmME a b
GmME GmME 1 1
¢U = U2 - U1 = - -
RE + h 2 RE + h 1 RE + h 1 RE + h 2
= 16.67 * 10-11 N # m2>kg 22150 kg216.0 * 1024 kg2
1 1
* B - R
6.4 * 106 m + 1.0 * 106 m 6.4 * 106 m + 37 * 106 m
= + 2.2 * 109 J
Because ¢U is positive, m2 is higher in the gravitational potential energy well than m1. Note that even though both U1 and U2 are
negative, U2 is “more positive,” or “less negative,” and closer to zero. Thus, it takes more energy to get a satellite farther from the
Earth.
F O L L O W - U P E X E R C I S E . Suppose that the altitude of the higher satellite in this Example were doubled, to 72 000 km. Would the
difference in the gravitational potential energies of the two satellites then be twice as great? Justify your answer.
Substituting the gravitational potential energy (Eq. 7.18) into the equation for
the total mechanical energy gives the equation a different form than it had in
Chapter 5. For example, the total mechanical energy of a mass m1 moving at a
distance r from mass m2 is
1 Gm1 m2
E = K + U = 2 m1 v 2 - (7.20)
r
This equation and the principle of the conservation of energy can be applied to the
Earth’s motion about the Sun by neglecting other gravitational forces. The Earth’s
orbit is not quite circular, but slightly elliptical. At perihelion (the point of the
Earth’s closest approach to the Sun), the mutual gravitational potential energy is
less (a larger negative value) than it is at aphelion (the point farthest from the Sun).
Therefore, as can be seen from Eq. 7.20 in the form 12 m1 v 2 = E + Gm1 m2>r, where
E is constant, the Earth’s kinetic energy and orbital speed are greatest at perihelion
(the smallest value of r) and least at aphelion (the greatest value of r). Or, in gen-
eral, the Earth’s orbital speed is greater when it is nearer the Sun than when it is
farther away.
Mutual gravitational potential energy also applies to a group, or configuration,

of more than two masses. That is, there is gravitational potential energy due to the
several masses in a configuration, because work was needed to be done in bring-
ing the masses together. Suppose that there is a single fixed mass m1, and another
mass m2 is brought close to m1 from an infinite distance (where U = 0). The work
done against the attractive force of gravity is negative (why?) and equal to the
change in the mutual potential energy of the masses, which are now separated by
a distance r12; that is, U12 = - Gm1 m2>r12.
If a third mass m3 is brought close to the other two fixed masses, there are then
two forces of gravity acting on m3, so U13 = - Gm1 m3>r13 and U23 = - Gm2 m3>r23.
The total gravitational potential energy of the configuration is therefore
U = U12 + U13 + U23
Gm1 m2 Gm1 m3 Gm2 m3
= - - - (7.21)
r12 r13 r23
A fourth mass could be brought in to further prove the point, but this develop-
ment should be sufficient to suggest that the total gravitational potential energy of
a configuration of particles is equal to the sum of the individual potential energies
for all pairs of particles.
EXAMPLE 7.15 Total Gravitational Potential Energy: Energy of Configuration

Three masses are in a configuration as shown in 䉴 Fig. 7.19. What is their total gravita- y
tional potential energy?
THINKING IT THROUGH. Equation 7.21 applies, but be sure to keep your masses and
their distances distinct. m3 = 2.0 kg (0, 4.0 m)
SOLUTION. From the figure, the data are:
Given: m1 = 1.0 kg Find: U (total gravitational

m2 = 2.0 kg potential energy)
m1 = 1.0 kg m2 = 2.0 kg
m3 = 2.0 kg
r12 = 3.0 m; r13 = 4.0 m; r23 = 5.0 m x
(0, 0) (3.0 m, 0)
(3–4–5 right triangle)
Eq. 7.21 can be used directly, since only three masses are used in this Example. (Note
that Eq. 7.21 can be extended to any number of masses.) Then,
䉱 F I G U R E 7 . 1 9 Total gravitational
U = U12 + U13 + U23 potential energy See Example text for
description.
Gm1 m2 Gm1 m3 Gm2 m3
= - - -
r12 r13 r23
= 16.67 * 10-11 N # m2 >kg 22
11.0 kg212.0 kg2 11.0 kg212.0 kg2 12.0 kg212.0 kg2
* c- - - d
3.0 m 4.0 m 5.0 m
= - 1.3 * 10-10 J
FOLLOW-UP EXERCISE. Explain what the negative potential energy in this Example means in physical terms.
Many of the effects of gravity are familiar to us. When lifting an object, it may
be thought of as being heavy, but work is being done against gravity. Gravity
causes rocks to tumble down and causes mudslides. But gravity is often put to
use. For example, fluids from bottles used for intravenous infusions flow because
of gravity. An extraterrestrial application of gravity is given in Insight 7.2, Space
Exploration: Gravity Assists.
INSIGHT 7.2 Space Exploration: Gravity Assists

After a seven-year, 3.5-billion-km (2.2-billion-mi) journey, the When the spacecraft approaches from “behind” the planet
Cassini-Huygens spacecraft arrived at Saturn in July 2004, hav- and leaves in “front” (relative to the planet’s motional direc-
ing made two Venus flybys, a Jupiter flyby, and one Earth tion), the gravitational interaction gives rise to a change in
flyby (Fig. 1).* Why was the spacecraft launched toward momentum—that is, there is a greater magnitude afterward
B
Venus, an inner planet, in order to go to Saturn, an outer and the direction is different. Then there is a ¢p in the general
B B
planet? “forward” direction of the spacecraft. Since ¢p r F, a force
Although space probes can be launched from the Earth acting on the craft gives it a “kick” of energy in that direction.
with current rocket technology, there are limitations—in par- So positive net work is done and there is an increase in kinetic
ticular, fuel versus payload: the more fuel, the smaller the energy (Wnet = ¢K 7 0 by the work–energy theorem). The
payload. Using rockets alone, planetary spacecraft are realisti- spacecraft leaves with more energy, a greater speed, and a
cally limited to visiting Venus, Mars, and Jupiter. The other new direction. (If the swing-by occurred in the opposite direc-
planets could not be reached by a spacecraft of reasonable tion, the spacecraft would slow down.)
size without taking decades to get there. Momentum and energy are conserved in this elastic colli-
So how did Cassini get to Saturn in 2004, almost seven years sion, and the planet gets an equal and opposite change in
after its 1997 launch? This was accomplished by using gravity momentum, giving a retarding effect. But because the planet’s
in a clever scheme called gravity assist. By using gravity assists, mass is so much larger than that of the spacecraft, the effect
missions to all of the planets in our solar system are possible. on the planet is negligible.
Rocket energy is needed to get a spacecraft to the first planet, To help you grasp the idea of a gravity assist, consider the
and after that, the energy is more or less “free.” Basically, dur- analogous roller derby “slingshot maneuver” illustrated in
ing a planetary flyby (or swing-by), there is an exchange of Fig. 2. The skaters interact, and skater S comes out of the
energy between the planet and the spacecraft, which enables “flyby” with increased speed. Here, the change in momentum
the spacecraft to increase its speed relative to the Sun. (This of the “slinger,” skater J, would probably be noticeable, but
phenomenon is sometimes called a slingshot effect.) that would not be so for Jupiter or any other planet.
Let’s take a brief look at the physics of this ingenious use of More recently, Jupiter was called in again for a gravity
gravity. Imagine the Cassini spacecraft making a swing-by of assist in 2007 for the New Horizons spacecraft on its way
Jupiter. Recall from Section 6.2 that a collision is an interaction toward the first close-up observation of the dwarf planet
of objects in which there is an exchange of momentum and Pluto and one of its moons. Launched in 2006, New Horizons
energy. Technically, in a swing-by, a spacecraft is having a is expected to complete its 5-billion-km (3-billion-mi) jour-
“collision” with a planet. ney in 2015.
*Giovanni Cassini (1625–1712) was a French–Italian astronomer

who studied Saturn, discovering four of its moons and that Saturn’s
rings are separated into two parts by a narrow gap, now called the
Cassini Division. The Cassini-Huygens spacecraft released a Huygens
probe to Saturn’s moon Titan, which was discovered by the Dutch sci- p2
entist Christiaan Huygens (1629–1695). pJ
S2
Venus flyby Venus flyby
1997 1999 Saturn arrival
2004
J
∆p
S
Jupiter flyby
2000
p2 p1 p1
Earth flyby S1
1999
䉱 F I G U R E 1 Cassini-Huygens spacecraft trajectory See text 䉱 F I G U R E 2 Skating swing-by Anaglous to a plane-

for description. tary swing-by is a roller derby “slingshot maneuver.”
Skater J slings skater S, who comes out of the “flyby”
with greater speed than she had before (S1 S, and S2
sequence). In this case, the change in momentum on
skater J, the slinger, would probably be noticeable, but it
is not for planets. (Why?)
7.6 KEPLER’S LAWS AND EARTH SATELLITES 247
DID YOU LEARN?

➥ When a force is proportional to 1/r2, this type of relationship is called an inverse-
square, as in the law of gravitation, Fg = Gm1 m2>r2, which means that Fg
approaches zero as r approaches infinity.
➥ The distance from the center of the Earth to some altitude or height (h) is RE + h, so
ag = GM>r2 = GME >(RE + h)2.
7.6 Kepler’s Laws and Ear th Satellites

➥ Which of Kepler’s laws tells that a planet’s speed varies in different parts of its orbit?
➥ What is meant by the Earth’s escape speed?
The force of gravity determines the motions of the planets and satellites and holds
the solar system (and galaxy) together. A general description of planetary motion
had been set forth shortly before Newton’s time by the German astronomer and
mathematician Johannes Kepler (1571–1630). Kepler formulated three empirical
laws from observational data gathered during a twenty-year period by the Danish
astronomer Tycho Brahe (1546–1601).
Kepler went to Prague to assist Brahe, who was the official mathematician at
the court of the Holy Roman Emperor. Brahe died the next year, and Kepler suc-
ceeded him, inheriting his records of the positions of the planets. Analyzing these
data, Kepler announced the first two of his three laws in 1609 (the year Galileo
built his first telescope). These laws were applied initially only to Mars. Kepler’s
third law came ten years later.
Interestingly enough, Kepler’s laws of planetary motion, which took him about
fifteen years to deduce from observed data, can now be derived theoretically with
a page or two of calculations. These three laws apply not only to planets, but also
to any system composed of a body revolving about a much more massive body to
which the inverse-square law of gravitation applies (such as the Moon, artificial
Earth satellites, and solar-bound comets).
Kepler’s first law (the law of orbits):
Planets move in elliptical orbits, with the Sun at one of the focal points.
An ellipse, shown in 䉲 Fig. 7.20a, has, in general, an oval shape, resembling a
flattened circle. In fact, a circle is a special case of an ellipse in which the focal
points, or foci (plural of focus), are at the same point (the center of the circle).
Although the orbits of the planets are elliptical, most do not deviate very much
from circles (Mercury and the dwarf planet Pluto are notable exceptions; see
“Eccentricity”, Appendix III.) For example, the difference between the perihelion
and aphelion of the Earth (its closest and farthest distances from the Sun, respec-
tively) is about 5 million km. This distance may sound like a lot, but it is only a
little more than 3% of 150 million km, which is the average distance between the
Earth and the Sun.
Kepler’s second law (the law of areas):
A line from the Sun to a planet sweeps out equal areas in equal lengths of time.
This law is illustrated in Fig. 7.20b. Since the time to travel the different orbital dis-
tances (s1 and s2) is the same such that the areas swept out (A1 and A2) are equal,
this law tells you that the orbital speed of a planet varies in different parts of its
orbit. Because a planet’s orbit is elliptical, its orbital speed is greater when it is
closer to the Sun than when it is farther away. The conservation of energy was
used in Section 7.5 (Eq. 7.20) to deduce this relationship for the Earth.
Kepler’s third law (the law of periods):
The square of the orbital period of a planet is directly proportional to the cube of the
average distance of the planet from the Sun; that is, T 2 r r 3.
䉴 F I G U R E 7 . 2 0 Kepler’s first and To sketch an ellipse,

second laws of planetary motion use two thumb tacks y
(a) In general, an ellipse has an oval (to represent foci), a
shape. The sum of the distances piece of string, and a
from the focal points F to any point pencil.
on the ellipse is constant: y=b
r1 + r2 = 2a. Here, 2a is the length
of the line joining the two points on r1 r2
the ellipse at the greatest distance
from its center, called the major axis. x = –a x
x=a
(The line joining the two points clos- F F
est to the center is b, the minor axis.)
Planets revolve about the Sun in
elliptical orbits for which the Sun is y = –b
at one of the focal points and noth-
ing is at the other. (b) A line joining 2a
the Sun and a planet sweeps out
equal areas in equal times. Since
A 1 = A 2, a planet travels faster Planet
along s1 than along s2.
Sun s1 A1 A2 s2
Sun
(a) (b)
Kepler’s third law is easily derived for the special case of a planet with a circular
orbit, using Newton’s law of gravitation. Since the centripetal force is supplied by
the force of gravity, the expressions for these forces can be set equal:
centripetal gravitational
force force
mp v 2 Gmp MS
=
r r2
and
GMS
v =
C r
In these equations, mp and MS are the masses of the planet and the Sun, respec-
tively, and v is the planet’s orbital speed. But v = 2pr>T1circumference>period =
distance>time2, so
2pr GMS
=
T C r
Squaring both sides and solving for T2 gives
4p2 3
T2 = a br
GMS
or
T2 = Kr3 (7.22)
The constant K for solar-system planetary orbits is easily evaluated from orbital data
(for T and r) for the Earth: K = 2.97 * 10-19 s 2>m3. As an exercise, you might wish to
convert K to the more useful units of y 2/km3. (Note: This value of K applies to all the
planets in our solar system, but does not apply to planet satellites as Example 7.16
will show.)
If you look inside the back cover and in Appendix III, you will find the masses of
the Sun and the planets of the solar system. How were these masses determined?
The following Example shows how Kepler’s third law can be used to do this.
EXAMPLE 7.16 By Jove!

The planet Jupiter (Roman name Jove) is the largest in the
solar system, both in volume and mass. Jupiter has 63 known
moons, the four largest having been discovered by Galileo in
1610. Two of these moons, Io and Europa, are shown in
䉴 Fig. 7.21. Given that Io is an average distance of
4.22 * 105 km from Jupiter and has an orbital period of
1.77 days, compute the mass of Jupiter.
T H I N K I N G I T T H R O U G H . Given the values for Io’s distance
from the planet (r) and period (T), this would appear to be an
application of Kepler’s third law, and it is. However, keep in
mind that the MS in Eq. 7.22 is the mass of the Sun, which the
planets orbit. The third law can be applied to any satellite, as
long as the M is that of the body being orbited by the satellite.
In this case, it will be MJ , the mass of Jupiter.
䉱 F I G U R E 7 . 2 1 Jupiter and moons Two of Jupiter’s moons

discovered by Galileo, Europa and Io, are shown here. Europa
is on the left, and Io on the right over the Great Red Spot. Io and
Europa are comparable in size to our Moon. The Great Red
Spot, roughly twice the size of the Earth, is believed to be a
huge storm, similar to a hurricane on the Earth.
SOLUTION.
Given: r = 4.22 * 105 km = 4.22 * 108 m Find: MJ (mass of Jupiter)

T = 1.77 days 18.64 * 104 s>day2 = 1.53 * 105 s
With r and T known, K can be found in Eq. 7.22 (written KI, indicating it is for Io-Jupiter)*
T2 11.53 * 105 s22
KI = = 3.11 * 10-16 s 2>m3
14.22 * 108 m23
=
r3
4p2
Then, writing KI explicitly, KI = , and
GMJ
4p2 4p2
MJ = = 1.90 * 1027 kg
16.67 * 10 N # m >kg 2213.11 * 10-16 s 2>m32
= 2
GKI -11
*Note that this is different from the K for planets orbiting about the Sun.
FOLLOW-UP EXERCISE. Compute the mass of the Sun from Earth’s orbital data.
EARTH’S SATELLITES
We are only a little more than half a century into the space age. Since the 1950s,
numerous uncrewed satellites have been put into orbit about the Earth, and now
astronauts regularly spend weeks or months in orbiting space laboratories.
Putting a spacecraft into orbit about the Earth (or any planet) is an extremely
complex task. However, a basic understanding of the method may be obtained
from fundamental principles of physics. First, suppose that a projectile could be
given the initial speed required to take it just to the top of the Earth’s potential
energy well. At the exact top of the well, which is an infinite distance away
1r = q2, the potential energy is zero. By the conservation of energy and Eq. 7.18,
initial final
Ko + Uo = K + U
or
initial final
1 2 GmME
2 mv esc - = 0 + 0
RE
where vesc is the escape speed—that is, the initial speed needed to escape from the
surface of the Earth. The final energy is zero, since the projectile stops at the top of
the well (at very large distances, and it is barely moving, K L 0), and U = 0 there.
Solving for vesc gives
2GME
vesc = (7.23)
A RE
Since g = GME >R 2E (Eq. 7.17), it is convenient to write
vesc = 12gRE (7.24)
Although derived here for the Earth, this equation may be used generally to
find the escape speeds for other planets and our Moon (using their accelera-
tions due to gravity and radii). The escape speed for Earth turns out to be
11 km>s, or about 7 mi>s.
A tangential speed less than the escape speed is required for a satellite to orbit.
Consider the centripetal force on a satellite in circular orbit about the Earth. Since
the centripetal force on the satellite is supplied by the gravitational attraction
between the satellite and the Earth, the quantities are equal and:
mv 2 GmME
Fc = =
r r2
Then
GME
v = (7.25)
A r
where r = RE + h. For example, suppose that a satellite is in a circular orbit at an
altitude of 500 km (about 300 mi); its tangential speed must be
GME GME 16.67 * 10-11 N # m2>kg 2216.0 * 1024 kg2

v =
16.4 * 106 m + 5.0 * 105 m2
= =
C r C RE + h C
= 7.6 * 103 m>s = 7.6 km>s 1about 4.7 mi>s2
This speed is about 27 000 km>h, or 17 000 mi>h. As can be seen from Eq. 7.25, the
required circular orbital speed decreases with altitude (greater r).
In practice, a satellite is given a tangential speed by a component of the thrust
from a rocket stage (䉴Fig. 7.22a). The inverse-square relationship of Newton’s
law of gravitation means that the satellite orbits that are possible about a mas-
sive planet or star are ellipses, of which a circular orbit is a special case. This con-
dition is illustrated in Fig. 7.22b for the Earth, using the previously calculated
values. If a satellite is not given a sufficient tangential speed, it will fall back to
the Earth (and possibly be burned up while falling through the atmosphere). If
the tangential speed reaches the escape speed, the satellite will leave its orbit
and go off into space.
Finally, the total energy of an orbiting satellite in circular orbit is
GmME
E = K + U = 12 mv 2 - (7.26)
r
Final Satellite 䉳 F I G U R E 7 . 2 2 Satellite orbits

Separation separation deployed (a) A satellite is put into orbit by
v giving it a tangential speed suffi-
cient for maintaining an orbit at a
Third-stage particular altitude. The higher the
Separation ignition Fc orbit, the smaller the tangential
Second-stage speed. (b) At an altitude of 500 km,
ignition a tangential speed of 7.6 km>s is
required for a circular orbit. With a
tangential speed between 7.6 km>s
and 11 km>s (the escape speed), the
satellite would move out of the cir-
cular orbit. Since it would not have
the escape speed, it would “fall”
around the Earth in an elliptical
orbit, with the Earth’s center at one
focal point. A tangential speed less
Liftoff First-stage ignition than 7.6 km>s would also give an
elliptical path about the center of
the Earth, but because the Earth is
not a point mass, a certain mini-
mum speed would be needed to
(a) keep the satellite from striking the
Earth’s surface.
Escape speed: 11 km/s
(40 000 km/h)
h = 500 km
Elliptical “orbit”:
< 7.6 km/s
Circular orbit: 7.6 km/s

(27 000 km/h)
Elliptical orbit:
>7.6 km/s
(b)
Substituting the expression for v from Eq. 7.25 into the kinetic energy term in Eq.
7.26 gives
GmME GmME
E = -
2r r
Thus,
GmME (total energy of an

E = - (7.27)
2r Earth-orbiting satellite)
Note that the total energy of the satellite is negative: more work is required to put a
satellite into a higher orbit, where it has more potential and total energy. The total
TABLE 7.3 Relationship of Radius, Speed, and Energy for Circular Orbital Motion
Increasing r (larger orbit) Decreasing r (smaller orbit)
v decreases increases
v decreases increases
K decreases increases
U increases (smaller negative value) decreases (larger negative value)
E 1 = K + U2 increases (smaller negative value) decreases (larger negative value)
energy E increases as its numerical value becomes smaller—that is, less negative—
as the satellite goes to a higher orbit toward the zero potential at the top of the
well. That is, the farther a satellite is from Earth, the greater its total energy. The
relationship of speed and energy to orbital radius is summarized in 䉱 Table 7.3.
To help understand why the total energy increases when its value becomes less
negative, think of a change in energy from, say, 5.0 J to 10 J. This change would be
an increase in energy. Similarly, a change from -10 J to -5.0 J would also be an
increase in energy, even though the absolute value has decreased:
¢U = U - Uo = - 5.0 J - 1- 10 J2 = + 5.0 J
Also note from the development of Eq. 7.27 that the kinetic energy of an orbit-
ing satellite, K = 12 mv 2 = GmME >2r, is equal to the absolute value of the satellite’s
total energy:
GmME
K = = ƒEƒ (7.28)
2r
The absolute value is taken because the kinetic energy is always positive.
Adjustments in satellite altitude (r) are made by applying forward or reverse
thrusts. For example, a reverse thrust, provided by the engines of docked cargo
ships, was used to put the Russian space station Mir into lower orbits and ulti-
mately led to its final destruction in March 2001. A final thrust sent the station into
a decaying orbit and into our atmosphere. Most of the 120-ton Mir burned up in
the atmosphere; however, some pieces did fall into the Pacific Ocean.
The advent of the space age and the use of orbiting satellites have brought us
the terms weightlessness and zero gravity, because astronauts appear to “float”
about in orbiting spacecraft (䉴 Fig. 7.23a). However, these terms are misnomers. As
mentioned earlier in the chapter, gravity is an infinite-range force, and the Earth’s
gravity acts on a spacecraft and astronauts, supplying the centripetal force neces-
sary to keep them in orbit. Gravity there is not zero, so there must be weight.*
A better term to describe the floating effect of astronauts in orbiting spacecraft
would be apparent weightlessness. The astronauts “float” because both the astro-
nauts and the spacecraft are centripetally accelerating (or “falling”) toward the
Earth at the same rate. To help you understand this effect, consider the analogous
situation of a person standing on a scale in an elevator (Fig. 7.23b). The “weight”
measurement that the scale registers is actually the normal force N of the scale on
the person. In a nonaccelerating elevator 1a = 02, N = mg = w, and N is equal to
the true weight of the individual. However, suppose the elevator is descending
with an acceleration a, where a 6 g. As the vector diagram in the figure shows,
mg - N = ma
*Another term used to describe astronaut “floating” is microgravity, implying that it is caused by an
apparent large reduction in gravity. This too is a misnomer. Using Eq. 7.18, at a typical satellite altitude
of 300 km, it can be shown that the reduction in the acceleration due to gravity is about 10%.
䉴 F I G U R E 7 . 2 3 Apparent
weightlessness (a) An astronaut
“floats” in a spacecraft, seemingly
in a weightless condition. (He is 1 2 3 4 5 6 7 8 9
not being held up.) (b) In a station-

ary elevator (top), a scale reads the
passenger’s true weight. The
兺F = 0
weight reading is the reaction
w = N = mg
force N of the scale on the person.
true weight
If the elevator is descending with
an acceleration a 6 g (middle), the
reaction force and apparent scale
weight are less than the true
weight. If the elevator were in free mg N
Not accelerating
fall (a = g; bottom), the reaction (a = 0)
force and indicated weight would
be zero, since the scale would be (a)
falling as fast as the person.
1 2 3 4 5 6 7 8 9
and the apparent weight w¿ is 兺F = ma

a mg N = ma
w¿ = N = m1g - a2 6 mg w' = N = m(g a)
less than
where the downward direction is taken as positive in this instance. With a true weight
scale
downward acceleration a, we see that N is less than mg, hence the scale
indicates that the person weighs less than his or her true weight. Note that mg N
the apparent acceleration due to gravity is g¿ = g - a. Descending with
Now suppose the elevator were in free fall, with a = g. As you can see, acceleration a < g
N (and thus the apparent weight w¿ ) would be zero. Essentially, the scale is
accelerating, or falling, at the same rate as the person. The scale may indi-
cate a “weightless” condition 1N = 02, but gravity still acts, as would be
noted by the sudden stop at the bottom of the shaft. (See Insight 7.3, 1 2 3 4 5 6 7 8 9
“Weightlessness”: Effects on the Human Body.)

Space has been called the final frontier. Someday, instead of brief stays in
Earth-orbiting spacecraft, there may be permanent space colonies with w' = N = 0
“artificial” gravity in the future. One proposal is to have a huge, rotating a=g ”weightless“
space colony in the form of a wheel—somewhat like an automobile tire,
with the inhabitants living inside the tire. As you know, centripetal force is
scale
necessary to keep an object in rotational circular motion. On the rotating
Earth, that force is supplied by gravity, and we refer to it as weight. We
mg N=0
exert a force on the ground, and the normal force (by Newton’s third law) Descending
exerted upward on our feet is what is actually sensed and gives the feeling with a = g
(b)
of “having our feet on solid ground.”
In a rotating space colony, the situation is somewhat reversed. The
rotating colony would supply the centripetal force on the inhabitants, and
the centripetal force would be perceived as a normal force acting on the
soles of the feet, providing artificial gravity. Rotation at the proper speed
would produce a simulation of “normal” gravity 1a c L g = 9.80 m>s22
within the colony wheel. Note that in the colonists’ world, “down” would
be outward, toward the periphery of the space station, and “up” would
always be inward, toward the axis of rotation (䉲 Fig. 7.24).
DID YOU LEARN

➥ Kepler’s second law (law of areas) states that a planet sweeps out equal
orbital areas equal times. Since planets have elliptical orbits, this means
the speed varies in different parts of the orbit.
➥ The initial speed of a projectile needed to raise it to the top of the Earth’s
gravitational potential energy well is termed the escape speed. (This is an
infinite distance, so such an initial speed is not practical.)
INSIGHT 7.3 “Weightlessness”: Effects on the Human Body

Astronauts spend weeks and months in orbiting spacecraft The circulatory system is affected, too. On Earth, gravity
and space stations. Although gravity acts on them, the astro- causes blood to pool in our feet. When we stand, the blood
nauts experience long durations of “zero gravity” (zero-g),* pressure in our feet (about 200 mm Hg) is much greater than
due to the centripetal motion. On Earth, gravity provides the that in our heads (60–80 mm Hg), because of the downward
force that causes our muscles and bones to develop to the force of gravity. (See Section 9.2 for a discussion of the measure-
proper strength so we may function in our environment. ment of blood pressure.) In the zero-g experienced by astro-
That is, our muscles and bones must be strong enough for us nauts, this force is not present, and the blood pressure equalizes
to be able to walk, lift objects, and so on. And we exercise throughout the body at about 100 mm Hg. This condition
and eat properly to maintain our ability to function against causes fluid to flow from the legs to the head, and gives rise to
the pull of gravity. the so-called puffy face and bird leg syndromes. The veins in
However, in a zero-g environment, muscle atrophy occurs the neck and face stand out more than usual and the eyes
quickly, because the body perceives no need for muscles. That become red and swollen. An astronaut’s legs become thinner,
is, muscles lose mass if there is no need for them to respond to because the blood flow to them is no longer gravity-assisted,
gravity. In zero-g, muscle mass may deplete as much as 5% and it is difficult for the heart to pump blood to them (Fig. 1).
per week. Bone loss also occurs at a rate of about 1% per Even more serious, the condition of above-normal blood
month. Models show that the total bone loss could reach pressure in the head is interpreted by the brain as indicating
40 -60%. The bone loss raises the calcium level in the blood, that there is too much blood in the body, and blood produc-
which may lead to kidney stones. tion is signaled to slow down. Astronauts can lose up to 22%
of their blood as a result of the equalized body pressure in
*This term will be used here for description, with the understand- zero-g. Also, with equalized blood pressure, the heart doesn’t
ing that it is apparent zero-g. work as hard, and the heart muscles may atrophy.
䉴 F I G U R E 7 . 2 4 Space colony and

artificial gravity (top) It has been
suggested that a space colony could
be housed in a huge, rotating wheel
as in this artist’s conception. The
rotation would supply the “artificial
gravity” for the colonists. (bottom)
(a) In the frame of reference of
someone in a rotating space colony,
centripetal force, coming from the
normal force N of the floor, would
be perceived as weight sensation or
artificial gravity. We are used to feel-
ing N upward on our feet to balance
gravity. Rotation at the proper
speed would simulate normal grav-
ity. To an outside observer, a
dropped ball would follow a tan-
gential straight-line path, as shown.
(b) A colonist on board the space
colony would observe the ball to fall
downward as in a normal gravita-
tional situation.
Fc
v
mg ?
(a) (b)
All of these phenomena explain why astronauts undergo rigorous physical fitness
programs before going into space and exercise in space using elastic restraints. On
returning to Earth, their bodies have to readjust to a normal “9.8 m>s2 g” environ-
ment. Each of the bodily losses requires a different recovery time. Blood volume is
typically restored in a few days with astronauts drinking lots of liquids. Most muscles
are regenerated in a month or so, depending on the length of stay in zero-g. Bone
recovery takes much longer. Astronauts spending three to six months in space may
require two or three years to regain the lost bone, if it is regained at all. Exercise and
nutrition are very important in all the recovery processes.
There is much to learn about the effects of zero-g—or even reduced-g. Uncrewed
spacecraft have visited Mars, with the aim of one day sending astronauts to the Red
Planet. This task would involve perhaps a six-month trip in zero-g and, on arrival, a
Martian surface gravity that is only 38% of the Earth’s gravity. No one yet understands
completely the effects that such a space journey might have on an astronaut’s body.
䉴 F I G U R E 1 Puffy face syndrome In zero-g, without a gravity

gradient the blood pressure equalizes throughout the body and
fluid flows from the legs to the head, giving rise to the so-called
puffy face syndrome. An astronaut’s legs become thinner (bird
leg syndrome) because the blood flow to them is no longer grav-
ity-assisted and it is difficult for the heart to pump blood to them. On Earth In space
PULLING IT TOGETHER Swinging and Releasing a Ball

A boy swings a 0.500-kg ball in a horizontal circle at a con-
stant speed as illustrated in 䉴 Fig. 7.25. The cord is 1.20 m long
T
and makes an angle of 5.00° below the horizontal of the hand.
u
Assume the boy’s hand remains still and neglect air resis-
u T cos u
tance. (a) What is the string tension? (b) What are the ball’s
tangential speed, angular speed, frequency, and period? T
w
(c) What are the ball’s tangential, angular, and centripetal accel-
erations? (d) If the boy’s hand is 2.20 m above the ground and
he releases the ball, how long will it take to hit the ground, and w
how far horizontally from the release point will it travel?
䉱 F I G U R E 7 . 2 5 Around it goes The ball is swung in a hori-
THINKING IT THROUGH. This example involves angular and zontal circle at a constant speed. The angle u is greatly exag-
tangential kinematics, Newton’s second law, centripetal force, gerated for clarity. A free-body diagram is shown on the right.
and projectile motion. (a) The tension is the force acting on the See Example text for description.
ball via the cord. Since the ball’s weight is known, Newton’s
laws and a free-body diagram should enable the tension to be acceleration would be expected. However, there will be a cen-
found. (b) The horizontal component of the tension is the centripetal acceleration due to the ball’s directional change. This
tripetal force, which is related to the ball’s speed. From this, the can be determined from the tangential speed. (d) Once it is
angular speed, frequency, and period can be determined. released, the ball becomes a two-dimensional projectile, so a
(c) Since tangential speed is constant, no tangential or angular quick look at Section 3.4 may be in order.
SOLUTION.
Given: m = 0.500 kg Find: (a) T (string tension)

L = 1.20 m (b) v (tangential speed), v (angular speed), f (frequency), T (period)
u = 5.00° (c) at (tangential acceleration), a (angular acceleration), ac (centripetal acceleration)
h = 2.20 m (d) t (time), x (distance)
(a) The free-body diagram in Fig. 7.25 shows two forces act- Since there is no vertical acceleration, summing the vertical
ing on the ball: the string tension (T) and the downward pull forces enables the tension to be found:
of gravity (weight, w). You should be able to show that the
horizontal and vertical components of the tension force are a Fy = T sin u - w = may = 0
T cos u and T sin u, respectively. (continued on next page)
1 1 cycle
and and the frequency is f = = = 1.54 Hz.
T 0.648 s
mg 10.500 kg219.80 m>s 22
T = = = 56.2 N (c) The ball’s angular and tangential accelerations are both
sin u sin 5.00° zero because the ball has no change in angular or tangential
(b) Summing the forces towards the center of the dotted circle speeds. However, its centripetal acceleration is not zero,
(and designating this as the positive direction) gives the cen- because the tangential velocity is changing due to a continual
tripetal force on the ball, which can then be used to find its directional change. Thus
tangential speed. Noting that the circle radius is r = L cos u,
we have v2 v2
ac = =
r L cos u
Fc = T cos u = mac
111.6 m>s22
v2 v2 = 113 m>s2
11.20 m2 cos 5.00°
=
= m = m
r L cos u
Solving for v, This is more than 11g!
TL (d) When the ball is released, it becomes a projectile with an
v = cos u initial horizontal velocity with components of vxo = 11.6 m>s
Am
and vyo = 0 m>s. The initial height of the ball is slightly less
156.2 N211.20 m2 than the 2.20 m since h = 2.20 m - L sin 5.00° =
= cos 5.00° = 11.6 m>s
C 0.500 kg 2.20 m - 0.105 m = 2.095 m. Choosing the coordinate origin
on the ground directly below the release point, the vertical
From this, the angular speed can be found since it is inversely
motion is described by
related to the radius:
v v 11.6 m>s y = yo + vyot - 12 gt2 = h + 0 - 12 gt2
11.20 m2 cos 5.00°
v = = = = 9.70 rad>s
r L cos u
Since the ground is at y = 0, the time in the air after release is
The period (T) is the time for one complete orbit, or for an
angular displacement of 2p radians. Recall that the angular 2h 212.095 m2
t = = = 0.654 s
speed is v = ¢u>¢t, so Ag C 9.80 m>s2
¢u 2p rad Once the ball is in projectile motion there is no horizontal
v = =
¢t T acceleration. Hence the horizontal travel distance is
and
x = vxot
= 111.6 m>s210.654 s2 = 7.59 m
2p rad
T = = 0.648 s
9.70 rad>s
■ The radian (rad) is a measure of angle; 1 rad is the angle of a Angular Kinematic Equations for uo = 0 and to = 0 (see Table
circle subtended by an arc length (s) equal to the radius (r): 7.2 for linear analogues):
u = vt (in general, not limited to constant acceleration) (7.5)
y
s = ru
v + vo
( u in radians) v = (2, Table 7.2)
2
r
u=
s=r v = vo + at u constant acceleration only (7.12)
1 rad 1 2
x u = uo + vo t + 2 at (4, Table 7.2)
2
v = vo2 + 2a1u - uo2 (5, Table 7.2)
■ Tangential speed (vt) and angular speed 1v2 for circular

motion are directly proportional, with the radius r being the
Arc Length (angle in radians): constant of proportionality:
s = ru (7.3) vt = rv (7.6)
vt = rv (v in rad/s)
m1
s m1
F12
u = vt
r F12 r
r
F21
F21
v
m2
■ The frequency ( f ) and period (T) are inversely related:
m2
1 (a) Point masses (b) Homogeneous spheres
f = (7.7)
T
■ Acceleration due to gravity at an altitude h:
■ Angular speed (with uniform circular motion) in terms of
period (T) and frequency ( f ): GME
ag = (7.17)
2p 1RE + h22
v = = 2pf (7.8)
T ■ Gravitational potential energy of two particles:
■ In uniform circular motion, a centripetal acceleration (ac) is Gm1 m2
required and is always directed toward the center of the cir- U = - (7.18)
cular path, and its magnitude is given by: r
v2 U0
ac = = rv2 (7.10)
r
RE r
v U=0 ∞
ac v
ac
1
ac U∝–
r
v
■ A centripetal force, Fc, (the net force directed toward the

center of a circle) is a requirement for circular motion, the
magnitude of which is U=–
GmME
RE
2
mv
Fc = ma c = (7.11) RE
r Earth
■ Angular acceleration (A) is the time rate of change of angu- U0
lar velocity and is related to the tangential acceleration (at)
in magnitude by ■ Kepler’s first law (law of orbits): Planets move in elliptical
orbits, with the Sun at one of the focal points.
at = ra (7.13)
Planet
a
at
Sun
ac
a
■ Kepler’s second law (law of areas): A line from the Sun to a

planet sweeps out equal areas in equal lengths of time.
s1 A1 A2 s2
Sun
Nonuniform circular motion

( a = at + ac)
■ Kepler’s third law (law of periods):
■ According to Newton’s law of gravitation, every particle
T2 = Kr3 (7.22)
attracts every other particle in the universe with a force that is
proportional to the masses of both particles and inversely pro- (K depends on the mass of the object orbited; for objects
portional to the square of the distance between them: orbiting the Sun, K = 2.97 * 10-19 s 2>m3.)
Gm1 m2 ■ Escape speed (from the Earth) is
Fg = 2
r 2GME
vesc = = 22gRE (7.23, 7.24)
1G = 6.67 * 10-11 N # m2>kg 22 (7.14) A RE
Escape speed: 11 km/s

(40 000 km/h)
■ Earth’s satellites are in a negative potential energy well; the
higher the object in the well, the greater the object’s poten-
h = 500 km tial energy and the less its kinetic energy.
■ Energy of a satellite orbiting Earth:
Elliptical “orbit”: GmME
< 7.6 km/s
E = - (7.27)
2r
K = ƒEƒ (7.28)
Circular orbit: 7.6 km/s
(27 000 km/h)
Elliptical orbit:
>7.6 km/s

7.1 ANGULAR MEASURE 7.4 ANGULAR ACCELERATION

1. The radian unit is a ratio of (a) degree>time, (b) length, (c) 11. The unit of angular acceleration is (a) s -2, (b) rpm,
length>length (d) length>time. (c) rad2>s, (d) s2.
2. For the polar coordinates of a particle traveling in a cir- 12. The angular acceleration in circular motion (a) is equal in
cle, the variables are (a) both r and u, (b) only r, (c) only u, magnitude to the tangential acceleration divided by the
(d) none of the preceding. radius, (b) increases the angular velocity if both angular
velocity and angular acceleration are in the same direc-
3. Which of the following is the greatest angle: (a) 3p>2 rad, tion, (c) has units of s -2, (d) all of the preceding.
(b) 5p>8 rad, or (c) 220°? 13. In circular motion, the tangential acceleration (a) does
not depend on the angular acceleration, (b) is constant,
(c) has units of s -2, (d) none of these.
7.2 ANGULAR SPEED AND VELOCITY 14. For uniform circular motion, (a) a = 0, (b) v = 0,
(c) r = 0, (d) none of the preceding.
4. Viewed from above, a turntable rotates counterclock-
wise. The angular velocity vector is then (a) tangential to
the turntable’s rim, (b) out of the plane of the turntable, 7.5 NEWTON’S LAW OF GRAVITATION
(c) counterclockwise, (d) none of the preceding.
15. The gravitational force is (a) a linear function of distance,
5. The frequency unit of hertz is equivalent to (a) that of the (b) an inverse function of distance, (c) an inverse func-
period, (b) that of the cycle, (c) radian>s, (d) s -1. tion of distance squared, (d) sometimes repulsive.
6. The unit of angular speed is (a) rad, (b) s -1 (c) s, 16. The acceleration due to gravity of an object on the
(d) rad>rpm. Earth’s surface (a) is a universal constant, like G, (b) does
not depend on the Earth’s mass, (c) is directly propor-
7. The particles in a uniformly rotating object all have the tional to the Earth’s radius, (d) does not depend on the
same (a) angular acceleration, (b) angular speed, (c) tan- object’s mass.
gential velocity, (d) both (a) and (b).
17. Compared with its value on the Earth’s surface, the
value of the acceleration due to gravity at an altitude of
one Earth radius is (a) the same, (b) two times as great,
7.3 UNIFORM CIRCULAR MOTION AND (c) one-half as great, (d) one-fourth as great.
CENTRIPETAL ACCELERATION
8. Uniform circular motion requires (a) centripetal acceler- 7.6 KEPLER’S LAWS AND EARTH
ation, (b) angular speed, (c) tangential velocity, (d) all of
SATELLITES
the preceding.
18. A new planet is discovered and its period determined. The
9. In uniform circular motion, there is a (a) constant veloc- new planet’s distance from the Sun could then be found
ity, (b) constant angular velocity, (c) zero acceleration, by using Kepler’s (a) first law, (b) second law, (c) third law.
(d) nonzero tangential acceleration.
19. As a planet moves in its elliptical orbit, (a) its speed is
10. If the centripetal force on a particle in uniform circular constant. (b) its distance from the Sun is constant, (c) it
motion is increased, (a) the tangential speed will remain moves faster when it is closer to the Sun, (d) it moves
constant, (b) the tangential speed will decrease, (c) the slower when it is closer to the Sun.
radius of the circular path will increase, (d) the tangen- 20. When a satellite is put into a higher circular orbit, its kinetic
tial speed will increase and>or the radius will decrease. energy (a) increases, (b) decreases, (c) remains the same.
7.1 ANGULAR MEASURE

1. Why does 1 rad equal 57.3°? Wouldn’t it be more conve-
nient to have an even number of degrees?
“Centrifugal
2. A wheel rotates about a rigid axis through its center. Do force” ??
all points on the wheel travel the same distance? How
about the same angular distance? f
7.2 ANGULAR SPEED AND VELOCITY f

3. Do all points on a wheel rotating about a fixed axis
through its center have the same angular velocity? The
same tangential speed? Explain.
䉱 F I G U R E 7 . 2 7 A center-fleeing force?
4. When “clockwise” or “counterclockwise” is used to See Conceptual Question 11.
describe rotational motion, why is a phrase such as
“viewed from above” added?
terms of Newton’s laws for a ground-based observer, this
5. Imagine yourself standing on the edge of an operating pseudo, or false, force doesn’t really exist. Analyze the sit-
merry-go-round. How would your tangential speed be uation in the figure to show that this is the case (that is,
affected if you walked toward the center? (Watch out for that the force does not exist). [Hint: Start with Newton’s
the horses going up and down.) first law.]
6. A car’s speedometer is set to read in relationship to the 12. Many curves have banked turns, which allow the cars to
angular speed of the rear wheels. If for winter the tires travel faster around the curves than if the road were flat.
are changed to larger diameter all-weather tires, would Actually, cars could also make turns on these banked
this affect the speedometer reading? Explain. How about curves if there were no friction at all. Explain this state-
the odometer? ment using the free-body diagram shown in 䉲Fig. 7.28.
7.3 UNIFORM CIRCULAR MOTION AND N

CENTRIPETAL ACCELERATION u
N cos u
7. The spin cycle of a washing machine is used to extract
water from recently washed clothes. Explain the physi-
cal principle(s) involved. N sin u
8. Can a car be moving with a constant speed of 100 km>h
u
and still be accelerating? Explain.
9. The apparatus illustrated in 䉲 Fig. 7.26 is used to demon-
strate forces in a rotating system. The floats are in jars of
mg
water. When the arm is rotated, which way will the
floats move? Does it make a difference which way the 䉱 F I G U R E 7 . 2 8 Banking safety See Conceptual
arm is rotated? Question 12.
7.4 ANGULAR ACCELERATION

13. A car increases its speed when it is on a circular track.
Does the car have centripetal acceleration? How about
angular acceleration? Explain.
14. Is it possible for a car traveling on a circular track to have
angular acceleration, but not centripetal acceleration?
Explain.
䉱 F I G U R E 7 . 2 6 When set in motion, a rotating system 15. Is it possible for a car traveling on a circular track to have
See Conceptual Question 9. a change in tangential acceleration and no change
centripetal acceleration?
10. On the rotating Earth, at what location(s) would a person
have (a) the greatest and (b) the least centripetal accelera-
tion? How does the centripetal acceleration for a person 7.5 NEWTON’S LAW OF GRAVITATION
at 40° N latitude compare to that of a person at 40° S lati- 16. Astronauts in a spacecraft orbiting the Earth or out for a
tude? (What supplies the centripetal acceleration?) “space walk” (䉲 Fig. 7.29) are seen to “float” in midair.
11. When rounding a curve in a fast-moving car, we experi- This phenomenon is sometimes referred to as
ence a feeling of being thrown outward (䉴 Fig. 7.27). It is weightlessness or zero gravity (zero-g). Are these terms cor-
sometimes said that this effect occurs because of an out- rect? Explain why an astronaut appears to float in or
ward centrifugal (center-fleeing) force. However, in near an orbiting spacecraft.
19. If the cup in 䉲 Fig. 7.30 were dropped, no water would

run out. Explain.
䉱 F I G U R E 7 . 2 9 Out for a walk Why does this astronaut

seem to “float”? See Conceptual Question 16.
䉱 F I G U R E 7 . 3 0 Let it go See Conceptual Question 19.

17. If the mass of the Moon were doubled, how would this
affect its orbit?
18. Weighing yourself at a park in Ecuador through which 20. Can you determine the mass of the Earth simply by mea-
the equator runs, you would find that you weigh slightly suring the gravitational acceleration near the Earth’s sur-
less than normal. Why is this? face? If yes, give the details.
EXERCISES
Integrated Exercises (IEs) are two-part exercises. The first part typically requires a conceptual answer choice based on
physical thinking and basic principles. The following part is quantitative calculations associated with the conceptual
choice made in the first part of the exercise.
1. The Cartesian coordinates of a point on a circle are 11. You ordered a 12-in.
●
(1.5 m, 2.0 m). What are the polar coordinates 1r, u2 of

●●
pizza for a party of five.

?
this point? For the pizza to be distrib-
2. ● The polar coordinates of a point are (5.3 m, 32°). What uted evenly, how should it
are the point’s Cartesian coordinates? be cut in triangular pieces?
(䉴 Fig. 7.31)?
3. ● Convert the following angles from degrees to radians, to
two significant figures: (a) 15°, (b) 45°, (c) 90°, and (d) 120°.
4. ● Convert the following angles from radians to degrees:
(a) p>6 rad, (b) 5p>12 rad, (c) 3p>4 rad, and (d) p rad.
5. ● Express the following angles in degrees, radians, 䉴 FIGURE 7.31
and/or revolutions (rev) as appropriate: (a) 105°, Tough pizza to cut
See Exercise 11.
(b) 1.8 rad, and (c) 5/7 rev.
6. ● You measure the length of a distant car to be sub-
tended by an angular distance of 1.5°. If the car is actu- 12. IE ● ● To attend the 2000 Summer Olympics, a fan flew
ally 5.0 m long, approximately how far away is the car? from Mosselbaai, South Africa (34°S, 22°E) to Sydney,
Australia (34°S, 151°E). (a) What is the smallest angular
7. ● How large an angle in radians and degrees does the
distance the fan has to travel: (1) 34°, (2) 12°, (3) 117°, or
diameter of the Moon subtend to a person on the Earth?
(4) 129°? Why? (b) Determine the approximate shortest
8. ●● The hour, minute, and second hands on a clock are flight distance, in kilometers.
0.25 m, 0.30 m, and 0.35 m long, respectively. What are 13. IE ● ● A bicycle wheel has a small pebble embedded in
the distances traveled by the tips of the hands in a its tread. The rider sets the bike upside down, and acci-
30-min interval? dentally bumps the wheel, causing the pebble to move
9. ●●A car with a 65-cm-diameter wheel travels 3.0 km. How through an arc length of 25.0 cm before coming to rest. In
many revolutions does the wheel make in this distance? that time, the wheel spins 35°. (a) The radius of the
10. ●● Two gear wheels with radii of 25 cm and 60 cm have wheel is therefore (1) more than 25.0 cm, (2) less than
interlocking teeth. How many radians does the smaller 25.0 cm, (3) equal to 25.0 cm. (b) Determine the radius of
wheel turn when the larger wheel turns 4.0 rev? the wheel.
EXERCISES 261
14. ●● At the end of her routine, an ice skater spins 26. IE ● ● The Earth rotates on its axis once a day and revolves
through 7.50 revolutions with her arms always fully around the Sun once a year. (a) Which is greater, the rotat-
outstretched at right angles to her body. If her arms are ing angular speed or the revolving angular speed? Why?
60.0 cm long, through what arc length distance do the (b) Calculate both angular speeds in rad>s.
tips of her fingers move during her finish? 27. ●● A little boy jumps onto a small merry-go-round
15. ●●● (a) Could a circular pie be cut such that all of the (radius of 2.00 m) in a park and rotates for 2.30 s through
wedge-shaped pieces have an arc length along the outer an arc length distance of 2.55 m before coming to rest. If
crust equal to the pie’s radius? (b) If not, how many such he landed (and stayed) at a distance of 1.75 m from the
pieces could you cut, and what would be the angular central axis of rotation of the merry-go-round, what was
dimension of the final piece? his average angular speed and average tangential speed?
16. ● ● ● Electrical wire with a diameter of 0.50 cm is wound 28. ● ● ● The driver of a car sets the cruise control and ties the
on a spool with a radius of 30 cm and a height of 24 cm. steering wheel so that the car travels at a uniform speed of
(a) Through how many radians must the spool be turned 15 m>s in a circle with a diameter of 120 m. (a) Through
to wrap one even layer of wire? (b) What is the length of what angular distance does the car move in 4.00 min?
this wound wire? (b) What arc length does it travel in this time?
17. ● ● ● A yo-yo with an axle diameter of 1.00 cm has a 29. ● ● ● In a noninjury, noncontact skid on icy pavement on
90.0-cm length of string wrapped around it many times an empty road, a car spins 1.75 revolutions while it skids
in such a way that the string completely covers the sur- to a halt. It was initially moving at 15.0 m>s, and because
face of its axle, but there are no double layers of string. of the ice it was able to decelerate at a rate of only
The outermost portion of the yo-yo is 5.00 cm from the 1.50 m>s2. Viewed from above, the car spun clockwise.
center of the axle. (a) If the yo-yo is dropped with the Determine its average angular velocity as it spun and
string fully wound, through what angle does it rotate by slid to a halt.
the time it reaches the bottom of its fall? (b) How much
arc length has a piece of the yo-yo on its outer edge trav-
eled by the time it bottoms out? 7.3 UNIFORM CIRCULAR MOTION AND
CENTRIPETAL ACCELERATION
30. ● An Indy car with a speed of 120 km>h goes around a
7.2 ANGULAR SPEED AND VELOCITY level, circular track with a radius of 1.00 km. What is the
18. ● A computer DVD-ROM has a variable angular speed centripetal acceleration of the car?
from 200 rpm to 450 rpm. Express this range of angular 31. ● A wheel of radius 1.5 m rotates at a uniform speed. If a
speed in radians per second. point on the rim of the wheel has a centripetal accelera-
19. ● A race car makes two and a half laps around a circular tion of 1.2 m>s2, what is the point’s tangential speed?
track in 3.0 min. What is the car’s average angular 32. ● A rotating cylinder about 16 km long and 7.0 km in
speed? diameter is designed to be used as a space colony. With

20. ● What are the angular speeds of the (a) second hand, what angular speed must it rotate so that the residents
(b) minute hand, and (c) hour hand of a clock? Are the on it will experience the same acceleration due to gravity
speeds constant? as on Earth?
33. ● ● An airplane pilot is going to demonstrate flying in a
21. ● What is the period of revolution for (a) a 9500-rpm
centrifuge and (b) a 9500-rpm computer hard disk drive? tight vertical circle. To ensure that she doesn’t black out
at the bottom of the circle, the acceleration must not
22. ●● Determine which has the greater angular speed: par- exceed 4.0g. If the speed of the plane is 50 m>s at the bot-
ticle A, which travels 160° in 2.00 s, or particle B, which tom of the circle, what is the minimum radius of the cir-
travels 4p rad in 8.00 s. cle so that the 4.0g limit is not exceeded?
23. ●● The tangential speed of a particle on a rotating 34. ● ● Imagine that you swing about your head a ball
wheel is 3.0 m>s. If the particle is 0.20 m from the axis of attached to the end of a string. The ball moves at a con-
rotation, how long will the particle take to make one stant speed in a horizontal circle. (a) Can the string be
revolution? exactly horizontal? Why or why not? (b) If the mass of
24. ●● A merry-go-round makes 24 revolutions in a 3.0-min the ball is 0.250 kg, the radius of the circle is 1.50 m, and
ride. (a) What is its average angular speed in rad>s? it takes 1.20 s for the ball to make one revolution, what is
(b) What are the tangential speeds of two people 4.0 m the ball’s tangential speed? (c) What centripetal force are
and 5.0 m from the center, or axis of rotation? you imparting to the ball via the string?
35. ● ● In Exercise 34, if you supplied a tension force of
25. ●● In Exercise 13, suppose the wheel took 1.20 s to stop
after it was bumped. Assume as you face the plane of the 12.5 N to the string, what angle would the string make
wheel, it was rotating counterclockwise. During this relative to the horizontal?
time, determine (a) the average angular speed and tan- 36. ● ● A car with a constant speed of 83.0 km>h enters a cir-
gential speed of the pebble, (b) the average angular cular flat curve with a radius of curvature of 0.400 km. If
speed and tangential speed of a piece of grease on the the friction between the road and the car’s tires can sup-
wheel’s axle (radius 1.50 cm), and (c) the direction of ply a centripetal acceleration of 1.25 m>s2, does the car
their respective angular velocities. negotiate the curve safely? Justify your answer.
37. IE ● ● A student is to swing a bucket of water in a vertical at the highest point of the loop in order to stay in the
circle without spilling any (䉲 Fig. 7.32). (a) Explain how loop? [Hint: What force must act on the block at the top of
this task is possible. (b) If the distance from his shoulder the loop to keep the block on a circular path?] (b) At what
to the center of mass of the bucket of water is 1.0 m, what vertical height on the inclined plane (in terms of the
is the minimum speed required to keep the water from radius of the loop) must the block be released if it is to
coming out of the bucket at the top of the swing? have the required minimum speed at the top of the loop?
42. ● ● ● For a scene in a movie, a stunt driver drives a
1.50 * 103 kg SUV with a length of 4.25 m around a circu-
lar curve with a radius of curvature of 0.333 km
(䉲 Fig. 7.34). The vehicle is to be driven off the edge of a
gully 10.0 m wide, and land on the other side 2.96 m
below the initial side. What is the minimum centripetal
acceleration the SUV must have in going around the cir-
cular curve to clear the gully and land on the other side?
䉱 F I G U R E 7 . 3 2 Weightless water? See Exercise 37.

0.333 km
38. ●● In performing a “figure 8” maneuver, a figure skater
wants to make the top part of the 8 approximately a cir-
cle of radius 2.20 m. He needs to glide through this part
of the figure at approximately a constant speed, taking v
4.50 s. His skates digging into the ice are capable of pro-
2.96 m
viding a maximum centripetal acceleration of 3.25 m>s2.
Will he be able to do this as planned? If not, what 10.0 m
adjustment can he make if he wants this part of the fig-
ure to remain the same size (assume the ice conditions 䉱 F I G U R E 7 . 3 4 Over the gully See Exercise 42.
and skates don’t change)?
39. ● ● A light string of length of 56.0 cm connects two small
43. ●●● Consider a simple pendulum of length L that has a
square blocks, each with a mass of 1.50 kg. The system is
small mass (the bob) of mass m attached to the end of its
placed on a slippery (frictionless) sheet of horizontal ice
string. If the pendulum starts out horizontally and is
and spun so that the two blocks rotate uniformly about
released from rest, show that (a) the speed at the bottom
their common center of mass, which itself does not
of the swing is vmax = 12gL and (b) the tension in the
move. They are supposed to rotate with a period of
string at that point is three times the weight of the bob,
0.750 s. If the string can exert a force of only 100 N before
or Tmax = 3mg. [Hint: Use conservation of energy to
it breaks, determine whether this string will work.
determine the speed at the bottom and centripetal force
40. IE ● ● A jet pilot puts an aircraft with a constant speed ideas and a free-body diagram to determine the tension
into a vertical circular loop. (a) Which is greater, the nor- at the bottom.]
mal force exerted on the seat by the pilot at the bottom of
the loop or that at the top of the loop? Why? (b) If the
speed of the aircraft is 700 km>h and the radius of the 7.4 ANGULAR ACCELERATION
circle is 2.0 km, calculate the normal forces exerted on
44. ●A CD originally at rest reaches an angular speed of
the seat by the pilot at the bottom and top of the loop.
40 rad>s in 5.0 s. (a) What is the magnitude of its angular
Express your answer in terms of the pilot’s weight.
acceleration? (b) How many revolutions does the CD
41. ● ● ● A block of mass m slides down an inclined plane into make in the 5.0 s?
a loop-the-loop of radius r (䉲 Fig. 7.33). (a) Neglecting
45. ● A merry-go-round accelerating uniformly from rest
friction, what is the minimum speed the block must have
achieves its operating speed of 2.5 rpm in 5 revolutions.
What is the magnitude of its angular acceleration?
m
46. ● ● A flywheel rotates with an angular speed of 25 rev>s.
As it is brought to rest with a constant acceleration, it
r turns 50 rev. (a) What is the magnitude of the angular
h acceleration? (b) How much time does it take to stop?
47. IE ● ● A car on a circular track accelerates from rest.
(a) The car experiences (1) only angular acceleration,
(2) only centripetal acceleration, (3) both angular and
centripetal accelerations. Why? (b) If the radius of the
䉱 F I G U R E 7 . 3 3 Loop-the-loop See Exercise 41. track is 0.30 km and the magnitude of the constant
EXERCISES 263
angular acceleration is 4.5 * 10-3 rad>s2, how long does 55. ●● Four identical masses of 2.5 kg each are located at the
the car take to make one lap around the track? (c) What corners of a square with 1.0-m sides. What is the net
is the total (vector) acceleration of the car when it has force on any one of the masses?
completed half of a lap? 56. ● ● The average density of the Earth is 5.52 g>cm .
3
48. ● ● Show that for a constant acceleration, Assuming this is a uniform density, compute the value
1v2 - v2o2 of G.
u = uo + 57. ● ● A 100-kg object is taken to a height of 300 km above
2a
the Earth’s surface. (a) What is the object’s mass at this
49. ● ● The blades of a fan running at low speed turn at
height? (b) What is the object’s weight at this height?
250 rpm. When the fan is switched to high speed, the
58. ● ● A man has a mass of 75 kg on the Earth’s surface.
rotation rate increases uniformly to 350 rpm in 5.75 s.
(a) What is the magnitude of the angular acceleration of How far above the surface of the Earth would he have to
the blades? (b) How many revolutions do the blades go go to “lose” 10% of his body weight?
through while the fan is accelerating? 59. ● ● It takes 27 days for the Moon to orbit the Earth in a
50. ● ● In the spin-dry cycle of a modern washing machine,

nearly circular orbit of radius 3.80 * 105 km. (a) Show in
a wet towel with a mass of 1.50 kg is “stuck to” the symbol notation that the mass of the Earth can be found
inside surface of the perforated (to allow the water out) using these data. (b) Compute the Earth’s mass and com-
washing cylinder. To have decent removal of water, pare with the value given inside the back cover of the book.
damp>wet clothes need to experience a centripetal accel- 60. IE ● ● Two objects are attracting each other with a certain
eration of at least 10g. Assuming this value, and that the gravitational force. (a) If the distance between the objects
cylinder has a radius of 35.0 cm, determine the constant is halved, the new gravitational force will (1) increase by
angular acceleration of the towel required if the washing a factor of 2, (2) increase by a factor of 4, (3) decrease by a
machine takes 2.50 s to achieve its final angular speed. factor of 2, (4) decrease by a factor of 4. Why? (b) If the
51. ● ● ● A pendulum swinging in a circular arc under the
original force between the two objects is 0.90 N, and the
influence of gravity, as shown in 䉲 Fig. 7.35, has both cen- distance is tripled, what is the new gravitational force
tripetal and tangential components of acceleration. (a) If between the objects?
the pendulum bob has a speed of 2.7 m>s when the cord 61. ● ● During the Apollo lunar explorations of the late
makes an angle of u = 15° with the vertical, what are the 1960s and early 1970s, the main section of the spaceship
magnitudes of the components at this time? (b) Where is remained in orbit about the Moon with one astronaut in
the centripetal acceleration a maximum? What is the it while the other two astronauts descended to the sur-
value of the tangential acceleration at that location? face in the landing module. If the main section orbited
about 50 mi above the lunar surface, determine that sec-
tion’s centripetal acceleration.
62. ● ● Referring to Exercise 61, determine the (a) the gravi-
tational potential energy, (b) the total energy, and (c) the
energy needed to “escape” the Moon for the main sec-
u L = 0.75 m tion of the lunar exploration mission in orbit. Assume
the mass of this section is 5000 kg.
63. ● ● ● The diameter of the Moon’s (nearly circular) orbit
about the Earth is 3.6 * 105 km and it takes 27 days for

ac one orbit. What is (a) the Moon’s tangential speed, (b) its
a
kinetic energy, (c) the system potential energy and
system total energy?
at Pendulum
64. ● ● ● (a) What is the mutual gravitational potential
bob
energy of the configuration shown in 䉲 Fig. 7.36 if all the
䉱 F I G U R E 7 . 3 5 A swinging pendulum See Exercise 51. masses are 1.0 kg? (b) What is the gravitational force per
unit mass at the center of the configuration?
52. ● ● ● A simple pendulum of length 2.00 m is released
from a horizontal position. When it makes an angle of y

30° from the vertical, determine (a) its angular accelera-
tion, (b) its centripetal acceleration, and (c) the tension m2
in the string. Assume the bob’s mass is 1.50 kg.
7.5 NEWTON’S LAW OF GRAVITATION

0.80 m 0.80 m
53. ●From the known mass and radius of the Moon (see the
tables inside the back cover of the book), compute the
value of the acceleration due to gravity, gM, at the sur- m1 m3
face of the Moon. x
54. ● The gravitational forces of the Earth and the Moon are ( –0.40 m, 0 ) ( 0.40 m, 0)
attractive, so there must be a point on a line joining their
centers where the gravitational forces on an object can- 䉱 F I G U R E 7 . 3 6 Gravitational potential, gravitational
cel. How far is this distance from the Earth’s center? force, and center of mass See Exercise 64.
65. IE ● ● ● A deep space probe mission is planned to explore 68. ●● In the year 2056, Martian Colony I wants to put a
the composition of interstellar space. Assuming the three Mars-synchronous communication satellite in orbit
most important objects in the solar system for this pro- about Mars to facilitate communications with the new
ject are the Sun, the Earth, and Jupiter, (a) what would be bases being planned on the Red Planet. At what distance
the distance of the Earth relative to Jupiter that would above the Martian equator would this satellite be
result in the lowest escape speed needed if the probe is placed? (To a good approximation, the Martian day is
to be launched from the Earth: (1) the Earth should be as the same length as that of the Earth’s.)
close as possible to Jupiter, (2) the Earth should be as far 69. ● ● The asteroid belt that lies between Mars and Jupiter
as possible from Jupiter, or (3) the distance of the Earth may be the debris of a planet that broke apart or that was
relative to Jupiter doesn’t matter? (b) Estimate the least not able to form as a result of Jupiter’s strong gravita-
escape speed for this probe, assuming planetary circular tion. An average asteroid has a period of about 5.0 y.
orbits, and only the Earth, Sun, and Jupiter are impor- Approximately how far from the Sun would this “fifth”
tant. (See data in Appendix III.) Comment on which of planet have been?
the three objects, if any, determines most of the escape 70. ● ● Using a development similar to Kepler’s law of peri-
speed. ods for planets orbiting the Sun, find the required alti-
tude of geosynchronous satellites above the Earth. [Hint:
7.6 KEPLER’S LAWS AND EARTH The period of such satellites is the same as that of the
SATELLITES Earth.]
71. ● ● Venus has a rotational period of 243 days. What
66. ●An instrument package is projected vertically upward
to collect data near the top of the Earth’s atmosphere (at would be the altitude of a synchronous satellite for this
an altitude of about 900 km). (a) What initial speed is planet (similar to geosynchronous satellite on the Earth)?
required at the Earth’s surface for the package to reach 72. ● ● ● A small space probe is put into circular orbit about a
this height? (b) What percentage of the escape speed is newly discovered moon of Saturn. The moon’s radius is
this initial speed? known to be 550 km. If the probe orbits at a height of
67. ● What is the orbital speed of a geosynchronous satel- 1500 km above the moon’s surface and takes 2.00 Earth
lite? (See Example 7.13.) days to make one orbit, determine the moon’s mass.
The Multiconcept Exercises require the use of more than one fundamental concept for their understanding and solu-
tion. The concepts may be from this chapter, but may include those from previous chapters.
73. Just an instant before reaching the very bottom of a semi- many systems have two or more stars. If there are two, it
circular section of a roller coaster ride, the automatic is a binary star system. The simplest possible case is that of
emergency brake inadvertently goes on. Assume the car two identical stars in a circular orbit about their common
has a total mass of 750 kg, the radius of that section of center of mass midway between them (small black dot in
the track is 55.0 m, and the car entered the bottom after 䉲 Fig. 7.37). Using telescopic measurements, it is some-
descending vertically (from rest) 25.0 m on a frictionless times possible to measure the distance, D, between the
straight incline. If the braking force is a steady 1700 N, star’s centers and the time (orbital period), T, for one orbit.
determine (a) the car’s centripetal acceleration (includ- Assume uniform circular motion and the following data.
ing direction), (b) the normal force of the track on the car, The stars have the same mass, the distance between
(c) the tangential acceleration of the car (including direc- them is one billion km A 1.0 * 109 km B , and the time each
tion), and (d) the total acceleration of the car. takes for one orbit is 10.0 Earth years. Determine the
74. A car accelerates uniformly from rest, and is initially mass of each star.
pointed north. It then travels in a quarter circle taking
D
10.0 s and reaching a final speed of 30.0 m>s traveling
due east. (a) What is the radius of its path? At 5.0 s from
the start, determine (b) the car’s tangential acceleration
(including direction), (c) the car’s centripetal accelera-
tion (including direction), and (d) the car’s velocity.
75. A simple pendulum consists of a light string 1.50 m long
B A
with a small 0.500-kg mass attached. The pendulum
starts out at 45° below the horizontal and is given an
initial downward speed of 1.50 m>s. At the bottom of the
arc, determine (a) the centripetal acceleration of the bob
and (b) the tension in the string.
76. To see in principle how astronomers determine stellar
masses, consider the following. Unlike our solar system, 䉱 F I G U R E 7 . 3 7 Binary stars See Exercise 76.
77. As an example of the effect of a meteor impact on a space these launch conditions, at what distance will its speed
probe, consider the following idealized situation. be equal to the escape speed?
Assume the probe is a uniform sphere of iron with a 79. A newly discovered asteroid is being tracked by
radius of 0.550 m. It is initially at rest when it is struck astronomers for possible crossing points with Earth’s
head on by a 100-g meteoroid traveling at 1.2 km>s. (a) If orbit. They have determined that its elliptical orbit
the meteor embeds itself into the probe, calculate the brings it as close as 16.0 million km to the Sun and its
center of mass speed of the combined system afterward. farthest distance from the Sun is 317 million km. Ignore
(b) Determine the percentage of kinetic energy left after any gravitational affects from objects other than the Sun.
the collision. (c) How would the energy analysis differ if Its speed at closest approach to the Sun is 126 km/s
the meteor struck the probe off center? Hint: The density (a) How many times larger is the system’s (asteroid plus
of iron is given in Table 9.2. Sun) gravitational potential energy (magnitude) when
78. The acceleration due to gravity near a planet’s surface is the asteroid is at its maximum distance, compared to its
known to be 3.00 m>s2. If the escape speed from the minimum distance? (b) What is the system’s total energy
planet is 8.42 km>s, (a) determine its radius. (b) Find the if the asteroid’s mass is 3.35 * 1013 kg? (c) What is the
mass of the planet. (c) If a probe is launched from its sur- asteroid’s minimum speed? (d) How long does it take to
face with a speed twice the escape speed and then coasts orbit the Sun? Hint: You will need the Sun’s mass. [Note:
outward, neglecting other nearby astronomical bodies, the R in Kepler’s third law refers to the average distance
what will be its speed when it is very far from the from the Sun if the orbit is not circular, that is,
planet? (Neglect any atmospheric effects also.) (d) Under (Rmax + Rmin)>2.]
Rotational Motion
8 and Equilibrium
8.1 Rigid bodies, translations,
and rotations (267)
8.2 Torque, equilibrium,

and stability (270)
■ center of gravity
8.3 Rotational dynamics (280)

■ moment of inertia
■ torque
8.4 Rotational work

and kinetic energy (288)
■ translational and
rotational motions PHYSICS FACTS
8.5
■
Angular momentum (291)
conservation of angular
momentum—no net torque
✦ If it were not for torques supplied
by our muscles, we would be with-
out body mobility.
✦ Antilock brakes are used on cars
because the rolling stopping dis-
tance is less than that of a locked-
I t’s always a good idea to keep
your equilibrium—but it’s more
important in some situations than
brake stopping distance.
in others. When looking at the
✦ The Earth’s rotational axis, which is chapter-opening photo, your first
tilted 23 12 °, precesses (rotates
about the vertical) with a period of reaction is probably to wonder
26 000 years. As a result, Polaris,
how does this tightrope walker tra-
toward which the axis currently
points, has not always been, nor versing the Niagara River at Horse-
will it always be, the North Star.
✦ Only one side of the Moon is seen
shoe Falls keep from falling.
from the Earth because the Moon’s Presumably, the pole must help—
period of rotation is the same as its
period of revolution. but in what way? You’ll find out in
✦ The planet Uranus’ spin axis is this chapter.
almost in the plane of its orbit. As a
result, Uranus rotates on its side It might be said that the
while revolving around the Sun.
tightrope walker is in equilibrium.
✦ Some figure skaters in jumps reach
Translational equilibrium ( gFi = 0)
B
rotational speeds on the order of
7 rev/s, or 420 rpm (revolutions per
minute). Some automobiles engines
was discussed in Section 4.5, but
have idle speeds of 600–800 rpm. here there is another consideration,
8.1 RIGID BODIES, TRANSLATIONS, AND ROTATIONS 267
namely, rotation. Should the walker start to fall (and it is hoped he doesn’t), there
would be a sideways rotation about the wire (ropes are rarely used anymore). To
avoid this calamity, another condition must be met: rotational equilibrium, which
will be considered in this chapter.
The tightrope walker is striving to avoid rotational motion. But rotational
motion is very important in physics, because rotating objects are all around us:
wheels on vehicles, gears and pulleys in machinery, planets in our solar system,
and even many bones in the human body. (Can you think of bones that rotate in
sockets?)
Fortunately, the equations describing rotational motion can be written as
almost direct analogues of those for translational (linear) motion. In Section 7.4,
this similarity was pointed out with respect to the linear and angular kinematic
equations. With the addition of equations describing rotational dynamics, you will
be able to analyze the general motions of real objects that can rotate, as well as
translate.
8.1 Rigid Bodies, Translations, and Rotations

➥ How are rigid body translational motion and rotational motion characterized in
terms of object particles?
➥ What is the instantaneous axis of a rolling object?
➥ What are the conditions for rolling without slipping?
In previous chapters, it was convenient to consider motion with the understand-

ing that an object can be represented by a particle located at the center of mass of
the object. Rotation, or spinning, was not a consideration, because a particle, or
point mass, has no physical dimensions. Rotational motion becomes relevant
when analyzing the motion of a solid, extended object or rigid body, which is the
focus of this chapter.
A rigid body is an object or a system of particles in which the distances between
particles are fixed (remain constant).
A quantity of liquid water is not a rigid body, but the ice that would form if the
water were frozen would be. The discussion of rigid body rotation is therefore
restricted to solids. Actually, the concept of a rigid body is an idealization. In reality,
the particles (atoms and molecules) of a solid vibrate constantly. Also, solids can
undergo elastic (and inelastic) deformations in collisions (Section 6.4). Even so, most
solids can be considered rigid bodies for purposes of analyzing rotational motion.
A rigid body may be subject to either or both of two types of motions:
translational and rotational. Translational motion is basically the linear motion
studied in previous chapters. If an object has only (pure) translational motion,
every particle in it has the same instantaneous velocity, which means that the object is
not rotating (䉲 Fig. 8.1a).
An object may have only (pure) rotational motion (motion about a fixed axis),
and all of the particles of the object have the same instantaneous angular velocity and
travel in circles about the axis of rotation (Fig. 8.1b).*
*The words rotation and revolution are commonly used synonymously. In general, this book uses
rotation when the axis of rotation goes through the body (for example, the Earth’s rotation on its axis, in
a period of 24 h) and revolution when the axis is outside the body (for example, the revolution of the
Earth about the Sun, in a period of 365 days).
268 8 ROTATIONAL MOTION AND EQUILIBRIUM
䉴 F I G U R E 8 . 1 Rolling—a combi- Translational + Rotational = Rolling

nation of translational and rotational v v = rv 2v
motions (a) In pure translational
motion, all the particles of an object
have the same instantaneous veloc-
ity. (b) In pure rotational motion, all
the particles of an object have the
v v
same instantaneous angular veloc- + =
ity. (c) Rolling is a combination of r
translational and rotational
motions. Summing the velocity vec-
tors for these two motions shows
that the point of contact (for a
sphere) or the line of contact (for a v v = rv v =0
Point of
cylinder) is instantaneously at rest.
(a) (b) contact (c)
(d) The line of contact for a cylinder
(or, for a sphere, a line through the
point of contact) is called the
instantaneous axis of rotation. Note Instantaneous
v axis of rotation
that the center of mass of a rolling
object on a level surface moves lin-
early and remains over the point or
line of contact. (Is the v vector in the
right direction?)
(d)
General rigid body motion is a combination of both translational and rotational

motions. When you throw a ball, the translational motion is described by the
motion of its center of mass (as in projectile motion). But the ball may also spin, or
rotate, and it usually does. A common example of rigid body motion involving
both translation and rotation is rolling, as illustrated in Fig. 8.1c. The combined
motion of any point or particle is given by the vector sum of the particle’s instanta-
neous velocity vectors. (Three points or particles are shown in the figure—one at
the top, one in the middle, and one at the bottom of the object.)
At each instant, a rolling object rotates about an instantaneous axis of rotation
through the point of contact of the object with the surface it is rolling on (for a
sphere) or along the line of contact of the object with the surface (for a cylinder;
Fig. 8.1d). The location of this axis changes with time. However, note in Fig. 8.1c
vCM = rω that the point or line of contact of the body with the surface is instantaneously at
rest (and thus has zero velocity), as can be seen from the vector addition of the
ω combined motions at that point. Also, the point on the top has twice the tangential
P´ P
speed (2v) of the middle (center-of-mass) point (v), because the top point is twice
r v CM
as far away from the instantaneous axis of rotation as the middle point. (With a
u
radius r, for the middle point, rv = v, and for the top point, 2rv = 2v).
s r
When an object rolls without slipping, for example, when a ball (or cylinder)
rolls in a straight line on a flat surface, it turns through an angle u, and a point (or
P P´
line) on the object that was initially in contact with the surface moves through an
s = ru
arc distance s (䉳 Fig. 8.2). And from Section 7.1, s = ru (Eq. 7.3). The center of mass
of the ball is directly over the point of contact and moves a linear distance s. Then
䉱 F I G U R E 8 . 2 Rolling without
slipping As an object rolls without s ru
slipping, the length of the arc vCM = = = rv
t t
between two points of contact on
the circumference is equal to the lin- where v = u>t. In terms of the speed of the center of mass and the angular speed v
ear distance traveled. (Think of
the condition for rolling without slipping is
paint coming off a roller.) This dis-
tance is s = ru. The speed of the
center of mass is vCM = rv. vCM = rv (rolling, no slipping) (8.1)
8.1 RIGID BODIES, TRANSLATIONS, AND ROTATIONS 269
The condition for rolling without slipping is also expressed by
s = ru (rolling, no slipping) (8.1a)
where s is the distance the object rolls (the distance the center of mass moves).
By carrying Eq. 8.1 one step further, an expression for the time rate of change of
the velocity can be obtained. Assuming the object started from rest 1vo = 02, then
¢vCM>¢t = vCM>t = 1rv2>t, which yields an equation for accelerated rolling
without slipping:
vCM rv
a CM = = = ra
t t
or
aCM = ra (accelerated rolling without slipping) (8.1b)
where a = v>t (for a constant a with vo assumed to be zero).

Essentially, an object will roll without slipping if the coefficient of static friction
between the object and surface is great enough to prevent slippage. There may be a
combination of rolling and slipping motions—for example, the slipping of a car’s
wheels when traveling through mud or ice. If there is rolling and slipping, then
there is no clear relationship between the translational and rotational motions, and
vCM = rv does not hold.
INTEGRATED EXAMPLE 8.1 Rolling without Slipping

A cylinder rolls on a horizontal surface without slipping. (a) At any point in time, the
tangential speed of the top of the cylinder is (1) v, (2) rv. (3) v+ rv, or (4) zero. (b) The
cylinder has a radius of 12 cm and a center-of-mass speed of 0.10 m>s as it rolls without
slipping. If it continues to travel at this speed for 2.0 s, through what angle does the
cylinder rotate during this time?
(A) CONCEPTUAL REASONING. Since the cylinder rolls without slipping, the relationship
vCM = rv applies. As was shown in Fig. 8.1, the speed at the point of contact is zero,
v for the center (of mass), and 2v at the top. With v = 0 at point of contact, and v = rv,
the answer is (3), v + rv = v + v = 2v.
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . Since the radius and the translational
speed are known, the angular speed can be calculated from the nonslipping condition,
vCM = rv. With this relationship and the time, the angle of rotation may be calculated.
Listing the data:
Given: r = 12 cm = 0.12 m Find: u (angle of rotation)
vCM = 0.10 m>s
t = 2.0 s
Using vCM = rv to find the angular speed,
vCM 0.10 m>s
v = = = 0.83 rad>s
r 0.12 m
Then,
u = vt = 10.83 m>s212.0 s2 = 1.7 rad
The cylinder makes a little over one quarter of a rotation. (Right? Check it yourself.)
F O L L O W - U P E X E R C I S E . How far does the CM of the cylinder travel linearly in part (b) of
this Example? Find the distance by using two different methods: translational and rota-
tional. (Answers to all Follow-Up Exercises are given in Appendix VI at the back of the book.)
DID YOU LEARN

➥ In pure translational motion, every particle of an object has the same instantaneous
velocity; in pure rotational motion, every particle has the same instantaneous
angular velocity.
➥ The instantaneous axis of a rolling object is the point or line of contact about which
the object rotates at each instant.
➥ The conditions for rolling without slipping are vCM = rv or s = r u. For accelerated
rotational motion without slipping, aCM = ra.
u 8.2 Torque, Equilibrium, and Stability

F
Force
u line of
➥ What is necessary for rational motion?
r ➥ What is necessary for mechanical equilibrium?
r⊥ action
➥ What is necessary for stable equilibrium?
Axis
TORQUE
t = r⊥F
As with translational motion, a force is necessary to produce a change in rotational
(a) Counterclockwise torque
motion. However, the rate of change of rotational motion depends not only on the
magnitude of the force, but also on the perpendicular distance of its line of action
from the axis of rotation, r⬜ (䉳 Fig. 8.3a, b). The line of action of a force is an imagi-
Force nary line extending through the force vector arrow—that is, an extended line
line of along which the force acts. (Note that if force is applied at the axis of rotation, r⬜ is
action zero and there is no rotation about that axis.)
r⊥
u Figure 8.3 shows that r⬜ = r sin u, where r is the straight-line distance between
Axis
r u F⊥ the axis of rotation and the force line of action and u is the angle between the line
B
F of r (or radial vector Br ) and the force vector F. The perpendicular distance r⬜ is
called the moment arm or lever arm.
t = r⊥F The product of the force and the lever arm is called torque (T B
), from the Latin
torquere, meaning “to twist.” The magnitude of the torque provided by the force is
(b) Smaller clockwise torque
t = r⬜ F = rF sin u (8.2)
SI unit of torque: meter * newton 1m # N2

F
r⊥ = 0 (The symbolism r⬜ F is commonly used to denote torque, but also note from
Fig. 8.3b that r⬜ F = rF⬜ ). The SI units of torque are meter * newton 1m # N2 the
Axis same as the units of work, W = Fd (N # m or J). However, the units of torque are
of
rotation usually written in reverse order as m # N to avoid confusion. But keep in mind that
torque is not work, and its unit is not the joule.
Rotational acceleration is not always produced when a force acts on a stationary
(c) Zero torque rigid body. From Eq. 8.2, it can be seen that when the force acts through the axis of
rotation such that u = 0, then t = 0 (Fig. 8.3c). Also, when u = 90°, the torque is at a
䉱 F I G U R E 8 . 3 Torque and maximum and the force acts perpendicularly to r. The angular acceleration depends
moment arm (a) The perpendicular on where a perpendicular force is applied (and therefore on the length of the lever
distance r⬜ from the axis of rotation arm). As a practical example, think of applying a force to a heavy glass door that
to the line of action of a force is
called the moment arm (or lever arm) swings in and out. Where you apply the force makes a great difference in how easily
and is equal to r sin u (where u is the the door opens or rotates (through the hinge axis). Have you ever tried to open such
angle between the line of r, orBradial a door and inadvertently pushed on the side near the hinges? This force produces a
vector Br , and the force vector F). The small torque and thus little or no rotational acceleration.
magnitude of the torque (t), or Torque in rotational motion can be thought of as the analogue of force in transla-
twisting force, that produces rota-
tional motion is t = r⬜ F. (b) The tional motion. An unbalanced or net force changes translational motion, whereas an
same force in the opposite direction unbalanced or net torque changes rotational motion. Torque is a vector. Its direction
with a smaller moment arm pro- is always perpendicular to the plane of the force and moment arm and is given by a
duces a smaller torque in the oppo- right-hand rule similar to that for angular velocity given in Section 7.2. If the fingers of
site direction. Note that r⬜ F = rF⬜
or 1r sin u2F = r1F sin u2. (c) When
the right hand are curled around the axis of rotation in the direction that the torque
a force acts through the axis of rota- would produce a rotational (angular) acceleration, the extended thumb points in the
tion, r⬜ = 0 and t = 0. direction of the torque. A sign convention, as in the case of linear motion, can be used
to represent torque directions, as will be discussed shortly.
EXAMPLE 8.2 Lifting and Holding: Muscle Torque at Work

In the human body torques produced by the contraction of With the axis of rotation through the elbow joint and the mus-
muscles cause some bones to rotate at joints. For example, cle attached 4.0 cm from the joint, what are the magnitudes of
when you lift something with your forearm, a torque is the muscle torques for cases (a) and (b) in Fig. 8.4 if the mus-
applied on the lower arm by the biceps muscle (䉴 Fig. 8.4). cle exerts a force of 600 N?
8.2 TORQUE, EQUILIBRIUM, AND STABILITY 271
䉴 F I G U R E 8 . 4 Human torque
In our bodies, torques produced by
the contraction of muscles cause
bones to rotate at joints. Here the
bicep muscles supply the force. See
Example text for description. F
F F
r⊥
30° 30°
r 120°
r⊥
Line of force
r⊥
(a) Starting to lift (b) Holding
T H I N K I N G I T T H R O U G H . As in many rotational situations, it would be greater than 90°, that is, 30° + 90°= 120°. In Fig.
B B
is important to know the orientations of the r and F vectors so 8.4b, the angle is 90°. This Example demonstrates an impor-
that the angle between them can be found to determine the tant point, namely that u is the angle between the radial vector
B B B
lever arm. Note in the inset in Fig. 8.4a that if the tails of the r r and the force vector F.
B
and F vectors were put together, the angle between them
SOLUTION. First listing the data given here and in the figure:
Given: r = 4.0 cm = 0.040 m Find: (a) ta (muscle torque magnitude) for Fig. 8.4a
F = 600 N (b) tb (muscle torque magnitude) for Fig. 8.4b
ua = 30° + 90° = 120° (note the 90° right angle
in the boxed figure.)
ub = 90°
B
(a) In this case, r is directed along the forearm, so the angle (b) Here, the distance r and the line of action of the force are
perpendicular 1ub = 90°2, and r⬜ = r sin 90° = r. Then,
B
between the r and F vectors is ua = 120°. Using Eq. 8.2,
B
ta = rF sin1120°2 = 10.040 m21600 N210.8662 = 21 m # N tb = r⬜ F = rF = 10.040 m21600 N2 = 24 m # N
at the instant in question. The torque is greater in (b). This is to be expected because the
maximum value of the torque 1tmax2 occurs when u = 90°.
F O L L O W - U P E X E R C I S E . In part (a) of this Example, there must have been a net torque, since the ball was accelerated upward by
a rotation of the forearm. In part (b), the ball is just being held and there is no rotational acceleration, so there is no net torque on
the system. Identify the other torque(s) in each case.
CONCEPTUAL EXAMPLE 8.3 My Aching Back

A person bends over as shown in 䉲 Fig. 8.5a. For most of us, the center of gravity of the
human body is in or near the chest region. When bending over, the force of gravity on
the person’s upper torso, acting through its center of gravity, gives rise to a torque that
tends to produce rotation about an axis at the base of the spine that could cause us to
fall over—but this doesn’t usually happen. So why don’t we fall when bending over
like this? (Consider only the upper torso.)
REASONING AND ANSWER. If this were the only torque acting, we would indeed fall
when bending forward. But since there is no rotation and fall, another force must be pro-
ducing a torque such that the net torque is zero. Where does this other torque come from?
Obviously from inside the body through a complicated combination of back muscles.
Representing the vector sum of all the back muscle forces as the net force Fb (as
shown in Fig. 8.5b), it can be seen that the back muscles exert a force that counterbal-
ances the torque on the torso’s center of gravity.
F O L L O W - U P E X E R C I S E . Suppose the person was bent over holding a heavy object he
had just picked up. How would this affect the back muscle force?
CG
w
Fb
Pivot
(a) (b)
䉱 F I G U R E 8 . 5 Torque but no rotation (a) When bending over, a person’s weight, act-
ing through the upper torso’s center of gravity, gives rise to a counterclockwise torque
that tends to produce rotation about an axis at the base of the spine. (b) However, the
back muscles attached between the shoulders combine to produce a force, Fb , and the
resulting clockwise torque counterbalances that of gravity, such that the net torque is zero.
F F
Before considering rotational dynamics with net torques and rotational

motions, let’s look at a situation in which the forces and torques acting on an
(a) object are balanced, and the object is in equilibrium.
EQUILIBRIUM
F F
In general, equilibrium means that forces and torques are in balance. Unbalanced
forces produce translational accelerations, but balanced forces produce the condi-
tion called translational equilibrium. Similarly, unbalanced torques produce rota-
tional accelerations, but balanced torques produce rotational equilibrium.
According to Newton’s first law of motion, when the sum of the forces acting
F
on a body is zero, the body remains either at rest (static) or in motion with a con-
stant velocity. In either case, the body is said to be in translational equilibrium
(b) (Section 4.5). Stated another way, the condition for translational equilibrium is that
the net force on a body is zero; that is, Fnet = gFi = 0. It should be apparent that
B B
F
this condition is satisfied for the situations illustrated in 䉳 Fig. 8.6a and b. Forces
with lines of action through the same point are called concurrent forces. When
these forces vectorially add to zero, as in Fig. 8.6a and b, the body is in transla-
tional equilibrium.
But what about the situation pictured in Fig. 8.6c? Here, gFi = 0, but the oppos-
B
ing forces will cause the object to rotate, and it will clearly not be in a state of static
equilibrium. (Such a pair of equal and opposite forces that do not have the same line
of action is called a couple.) Thus, the condition gFi = 0 is a necessary, but not
B
F
sufficient, condition for static equilibrium.
Since Fnet = gFi = 0 is the condition for translational equilibrium, you might
B B
(c)
predict (and correctly so) that T net = gTi = 0 is the condition for rotational equilib-
B B
䉱 F I G U R E 8 . 6 Equilibrium and
forces Forces with lines of action rium. That is, if the sum of the torques acting on an object is zero, then the object is
through the same point are said to in rotational equilibrium—it remains rotationally at rest or rotates with a con-
be concurrent. The resultants of the stant angular velocity.
concurrent forces acting on the Thus, there are actually two equilibrium conditions. Taken together, they define
objects in B(a) and (b) are zero mechanical equilibrium. A body is said to be in mechanical equilibrium when the con-
(Fnet = gFi = 0), and the objects are
B
in static equilibrium, because the ditions for both translational and rotational equilibrium are satisfied:
net torque and net force are zero. In
Fnet = gFi = 0
B B
(c), the object is in translational equi- (for translational equilibrium) (8.3)
librium, but it will undergo angular
net = gTi = 0
acceleration; thus, the object is not in T
B B
(for rotational equilibrium)
rotational equilibrium.
A rigid body in mechanical equilibrium may be either at rest or moving with a

constant linear and>or angular velocity. An example of the latter is an object
rolling (rotating) without slipping on a level surface, with the center of mass of the
object having a constant velocity. Of greater practical interest is static
equilibrium, the condition that exists when a rigid body remains at rest—that is, a
body for which v = 0 and v = 0. There are many instances in which we do not
want things to move, and this absence of motion can occur only if the equilibrium
conditions are satisfied. It is particularly comforting to know, for example, that a
bridge over which cars are crossing is in static equilibrium and not subject to
translational or rotational motions.
Let’s consider examples of static translational equilibrium and static rotational
equilibrium separately and then an example in which both apply.
EXAMPLE 8.4 Translational Static Equilibrium: No Translational Acceleration or Motion

A picture hangs motionless on a wall as shown in 䉲 Fig. 8.7a. If the picture has a mass of 3.0 kg, what are the magnitudes of the
tension forces in the wires?
䉴 F I G U R E 8 . 7 Translational sta-
tic equilibrium (a) Since the pic-
T2 sin u2
ture hangs motionless on the wall, T2 T1 sin u1
the sum of the forces acting on it T1
must be zero. The forces are con- T1 T2
current, with their lines of action
passing through a common point u1 ⫽45° u2 ⫽50°
at the nail. (b) In the free-body 50° 45°
diagram, all the forces are repre-
sented as acting at the common T2 cos u2 T1 cos u1
point. T1 and T2 have been moved
to this point for convenience.
Note, however, that the forces mg
shown are acting on the picture.
Free-body diagram
of picture
(a) (b)
T H I N K I N G I T T H R O U G H . Since the picture remains motion- With the system in static equilibrium, the net force on the
picture is zero; that is, gFi = 0. Thus, the sums of the rectan-
B
less, it must be in static equilibrium, so applying the condi-
gular components are also zero: gFx = 0 and gFy = 0. Then
B B
tions for mechanical equilibrium should give equations that
yield the tensions. Note that all the forces (tension and weight (using ; for directions),
gFx: +T1 cos u1 - T2 cos u2 = 0
forces) are concurrent; that is, their lines of action pass
(1)
through a common point, the nail. Because of this, the condi-
tion for rotational equilibrium 1g Ti = 02 is automatically sat- gFy: +T1 sin u1 + T2 sin u2 - mg = 0
B
(2)
isfied. With respect to the axis of rotation, the moment arms
1r⬜2 of the forces are zero, and therefore the torques are zero.
Then, solving for T2 in Eq. 1 (or T1 if you like),
Thus, only translational equilibrium needs to be considered. cos u1
T2 = T1 a b (3)
SOLUTION. cos u2
Then substituting Eq. 3 into Eq. 2 so as to eliminate T2 and

Given: u1 = 45°, u2 = 50° Find: T1 and T2 solving for T1 with a little algebra,
m = 3.00 kg
T1 csin 45° + a b sin 50° d - mg =
cos 45°
It is helpful to isolate the forces acting on the picture in a free- cos 50°
body diagram, as was done in Section 4.5 for force problems
T1 c0.707 + a b10.7662 d - 13.00 kg219.80 m>s22 = 0
(Fig. 8.7b). The diagram shows the concurrent forces acting 0.707
through their common point. Note that all the force vectors 0.643
have been moved to that point, which is taken as the origin of
the coordinate axes. The weight force mg acts downward. (continued on next page)
and Then, using Eq. 2 to find T2 ,

29.4 N cos u1
T2 = T1 a b = 19.0 N a b = 20.9 N
T1 = = 19.0 N 0.707
1.55 cos u2 0.643
F O L L O W - U P E X E R C I S E . Analyze the situation in Fig. 8.7 that would result if the wires were at equal angles and were shortened
such that the angles were decreased, but kept equal. Carry your analysis to the limit where the angles approach zero. Is the
answer realistic?
As pointed out earlier, torque is a vector and therefore has direction. Similar to
linear motion (Section 2.2), in which plus and minus signs were used to express
opposite directions (for example, +x and - x), torque directions can be designated
as being plus or minus, depending on the rotational acceleration they tend to pro-
duce. The rotational “directions” are taken as clockwise or counterclockwise
around the axis of rotation. A torque that tends to produce a counterclockwise
rotation will be taken as positive 1+2, and a torque that tends to produce a clock-
wise rotation will be taken as negative 1- 2. (See the right-hand rule in Section 7.2.)
To illustrate, let’s apply this convention to the situation in Example 8.5.
EXAMPLE 8.5 Rotational Static Equilibrium: No Rotational Motion

Three masses are suspended from a meterstick as shown in of m3. (Note that the lever arms are measured from the pivot
䉲 Fig. 8.8a. How much mass must be suspended on the right point, the center of the meterstick.)
side for the system to be in static equilibrium? (Neglect the
SOLUTION. From the figure the data are (using cgs units for
mass of the meterstick.)
convenience):
T H I N K I N G I T T H R O U G H . As the free-body diagram (Fig. 8.8b) Given: m1 = 25 g Find: m3 (unknown mass)
shows, the translational equilibrium condition will be satis-
B r1 = 50 cm
fied with the upward normal force N balancing the down-
ward weight forces, so long as the stick remains horizontal. m2 = 75 g
B
But N is not known if m3 is unknown, so applying the condi- r2 = 30 cm
tion for rotational equilibrium should give the required value r3 = 35 cm
r1 䉳 F I G U R E 8 . 8 Rotational
r2 r3 static equilibrium For the meter-
stick to be in rotational equilib-
0 20 cm 50 cm 85 cm 100 cm rium, the sum of the torques
acting about any selected axis
must be zero. (b) Here the axis is
taken to be through point A. (The
mass of the meterstick is consid-
25 g 75 g ? ered negligible.)
m1 m2 m3
(a)
N
(b)
negative
positive
x
A
m1g
m2g Sign convention
m3g
Free-body diagram of meterstick
Because the condition for translational equilibrium 1gFi = 02

B
on the right side produces a torque that would tend to rotate
B
is satisfied (there is no Fnet in the y-direction), N - Mg = 0, or the stick clockwise. The condition for rotational equilibrium
N = Mg where M, is the total mass. This is true no matter by is applied by summing the torques about an axis. Here,
what the total mass may be—that is, regardless of how much this axis is conveniently taken to be through the center of the
mass is added for m3. However, unless the proper mass for m3 stick at the 50-cm position, or point A in Fig. 8.8b. Then, not-
is placed on the right side, the stick will experience a net ing that N passes through the axis of rotation 1r⬜ = 02 and
torque and begin to rotate. produces no torque,
Notice that the masses on the left side produce torques that
would tend to rotate the stick counterclockwise, and a mass
gti : t1 + t2 + t3 = + r1 F1 + r2 F2 - r3 F3 (using sign convention for torque vectors)

= r11m1 g2 + r21m2 g2 - r31m3 g2 = 0
Noting that the g’s cancel and solving for m3 ,
m1 r1 + m2 r2 125 g2150 cm2 + 175 g2130 cm2
m3 = = = 100 g
r3 35 cm
(The mass of the stick was neglected. If the stick is uniform, however, its mass will not affect the equilibrium, as long as the pivot
point is at the 50-cm mark. Why?)
F O L L O W - U P E X E R C I S E . The axis of rotation could have been taken through any point along the stick. That is, if a system is in
static rotational equilibrium, the condition g Ti = 0 holds for any axis of rotation. Show that the preceding statement is true for
B
the system in this Example by taking the axis of rotation through the left end of the stick 1x = 02.
In general, the conditions for both translational and rotational equilibrium need to
be written explicitly to solve a statics problem. Example 8.6 is one such case.
EXAMPLE 8.6 Static Equilibrium: No Translation, No Rotation

A ladder with a mass of 15 kg rests against a smooth
wall (䉴 Fig. 8.9a). A painter who has a mass of 78 kg
stands on the ladder as shown in the figure. What is
the magnitude of frictional force that must act on
the bottom of the ladder to keep it from slipping?
T H I N K I N G I T T H R O U G H . Here there are a variety of Nw
forces and torques. However, the ladder will not Nw
slip as long as the conditions for static equilibrium
are satisfied. Summing both the forces and torques
to zero should enable us to solve for the necessary
frictional force. Also, as will be seen, choosing a
convenient axis of rotation, such that one or more t’s
are zero in the summation of the torques, can sim-
wm
plify the torque equation. wm
y = 5.6 m
SOLUTION.
w w
Given: m/ = 15 kg Find: fs (force of static
mm = 78 kg friction)
Distances given
Ng
in figure Ng
Because the wall is smooth, there is negligible fric-
tion between it and the ladder, and only the normal fs
reaction force of the wall (Nw) acts on the ladder at fs
x1 = 1.0 m Free-body diagram
this point (Fig. 8.9b). x2 = 1.6 m of ladder
In applying the conditions for static equilibrium, (a) (b)
any axis of rotation may be chosen. (The conditions
must hold for all parts of a system that is in static 䉱 F I G U R E 8 . 9 Static equilibrium For the painter’s sake, the ladder has
equilibrium; that is, there can’t be motion in any to be in static equilibrium; that is, both the sum of the forces and the sum
part of the system.) Note that choosing an axis at of the torques must be zero. See Example text for description.
the end of the ladder where it touches the ground (continued on next page)
eliminates the torques due to fs and Ng , since the moment substituting the given values for the masses and distances
arms are zero. Then writing the equations for force compo- yields,
nents and torque (using mg for w): 1m/ g2x1 + 1mm g2x2
Nw =
g Fx: Nw - fs = 0 y
g Fy: Ng - mm g - m/ g = 0 115 kg219.8 m>s2211.0 m2 + 178 kg219.8 m>s 2211.6 m2
=
and 5.6 m
gti: 1m/ g2x1 + 1mm g2x2 + 1 -Nw y2 = 0

2
= 2.4 * 10 N
The weight of the ladder is considered to be concentrated at Then from the g Fx equation,
its center of gravity. Solving the third equation for Nw and fs = Nw = 2.4 * 102 N
F O L L O W - U P E X E R C I S E . In this Example, would the frictional force between the ladder and the ground (call it fs ) remain the
1
same if there were friction between the wall and the ladder (call it fs2)? Justify your answer.
As the preceding Examples have shown, a good procedure to follow in working prob-
lems involving static equilibrium is as follows:
1. Sketch a space diagram of the problem.
2. Draw a free-body diagram, showing and labeling all external forces and, if neces-
sary, resolving the forces into x- and y-components.
3. Apply the equilibrium conditions. Sum the forces: gFi = 0, usually in component
B
form; g Fx = 0 and gFy = 0. Sum the torques: gTi = 0. Remember to select an

B B B
appropriate axis of rotation to reduce the number of terms as much as possible.

B
Use ⫾ sign conventions for both F and T.
B
4. Solve for the unknown quantities.
CONCEPTUAL EXAMPLE 8.7 No Net Torque: The Iron Cross

The static “iron cross” gymnastic position is one of the most strenuous and difficult to
perform (䉳 Fig. 8.10a). What makes it so difficult?
REASONING AND ANSWER. The gymnast must be extremely strong in order to achieve
and maintain such a static position because it requires a huge muscle force to suspend
his body from the rings. This can be shown by considering an analogous situation of a
weight suspended on a rope tied at each end (Fig. 8.10b). The closer the rope (or gym-
nast’s arms) gets to the horizontal position, the more force is needed to keep the weight
suspended.
From the figure, it can be seen that the vertical components of the tension force (T) in
(a) the rope must balance the downward weight force. (T is analogous to the arm muscle
forces.) That is,
2T sin u = mg
T T Note that for the rope (or gymnast’s arms) to become horizontal, the angle must
u T sin u approach zero 1u : 02. Then, what happens to the tension force T? As u : 0, then
T : q, making it very difficult to achieve a small u, and impossible for the rope (and
m gymnast’s arms to be exactly horizontal).
FOLLOW-UP EXERCISE. What is the least stressful position for a gymnast on the rings?
mg
(b) STABILITY AND CENTER OF GRAVITY
䉱 F I G U R E 8 . 1 0 The iron cross The equilibrium of a particle or a rigid body can be either stable or unstable in a
(a) The static iron cross gymnastic gravitational field. For rigid bodies, these categories of equilibria are conveniently
position is one of the most strenu- analyzed in terms of the center of gravity of the body. Recall from Section 6.5 that
ous and difficult to perform. (b) An
analogous situation of a weight sus- the center of gravity is the point at which all the weight of an object may be con-
pended on a rope tied at both ends. sidered to be acting as if the object were a particle. When the acceleration due to
See Conceptual Example 8.7. gravity is constant, the center of gravity and the center of mass coincide.
䉳 F I G U R E 8 . 1 1 Stable and unsta-

ble equilibria (a) When an object is
in stable equilibrium, any small dis-
placement from an equilibrium
position results in a force or torque
Ball that tends to return the object to that
displaced position. A ball in a bowl (left)
returns to the bottom after being
displaced. Analogously, the center
CG
Restoring of gravity (CG) of an extended
force object (right) can be thought of as
component being on an inverted potential
energy “bowl”: A small displace-
Pivot point ment raises the CG, increasing the
(a) object’s potential energy. When
released, the object will rotate about
the pivot point back to its stable
equilibrium position. (b) For an
object in unstable equilibrium, any
Ball small displacement from its equilib-
displaced rium position results in a force or
torque that tends to take the object
CG farther away from that position. The
ball on top of an overturned bowl
(left) is in unstable equilibrium. For
an extended object (right), the CG
Displacing can be thought of as being on an
force component inverted potential energy bowl. A
small displacement lowers the CG,
decreasing the object’s potential
Pivot point
energy.
(b)
If an object is in stable equilibrium, any small displacement results in a restoring

force or torque, which tends to return the object to its original equilibrium position. As
illustrated in 䉱 Fig. 8.11a, a ball in a bowl is in stable equilibrium. Analogously,
the center of gravity of an extended body on the right is in stable equilibrium.
Any slight displacement raises its center of gravity (CG), and a restoring gravi-
tational force tends to return it to the position of minimum potential energy. This
force actually produces a restoring torque that is due to a component of the
weight force and that tends to rotate the object about a pivot point back to its
original position.
For an object in unstable equilibrium, any small displacement from equilibrium
results in a torque that tends to rotate the object farther away from its equilibrium posi-
tion. This situation is illustrated in Fig. 8.11b. Note that the center of gravity of the
object is at the top of an overturned, or inverted, potential energy bowl; that is, the
potential energy is at a maximum in this case. Small displacements or slight dis-
turbances have profound effects on objects that are in unstable equilibrium. It
doesn’t take much to cause such an object to change its position.
Yet even if the angular displacement of an object in stable equilibrium is quite
substantial, the object will still be restored to its equilibrium position. As you
might have surmised, the condition for stable equilibrium is:
An object is in stable equilibrium as long as its center of gravity after a small displacement
still lies above and inside the object’s original base of support.That is,the line of action of
the weight force through the center of gravity intersects the original base of support.
When this is the case, there will always be a restoring gravitational torque
(䉲 Fig. 8.12a). However, when the center of gravity or center of mass falls outside
the base of support, over goes the object—because of a gravitational torque that
rotates it away from its equilibrium position (Fig. 8.12b).
CG
CG
Balanced on a broad Disturbance produces Balanced carefully on Disturbance produces

base of support restoring torque narrow base of support (point) displacing torque
(a) Stable Equilibrium (b) Unstable Equilibrium
䉱 F I G U R E 8 . 1 2 Examples of sta-
ble and unstable equilibria (a)
When the center of gravity is above Rigid bodies with wide bases and low centers of gravity are therefore most sta-
and inside an object’s base of sup- ble and least likely to tip over. This relationship is evident in the design of high-
port, the object is in stable equilib-
rium. There is a restoring torque
speed race cars, which have wide wheel bases and centers of gravity close to the
when the object is displaced. Note ground (䉲 Fig. 8.13a). SUVs, on the other hand, can roll over more easily. Why?
how the line of action of the weight And how about the acrobat in Fig. 8.13b?
intersects the original base of sup- The location of the center of gravity of the human body has an effect on certain
port after the displacement. (b) physical abilities. For example, women can generally bend over and touch their
When the center of gravity lies out-
side the base of support, the object
toes or touch their palms to the floor more easily than can men, who often fall over
is unstable. (There is a displacing trying. On the average, men have higher centers of gravity (larger shoulders) than
torque.) do women (larger pelvises), so it is more likely that a man’s center of gravity will
be outside his base of support when he bends over. Conceptual Example 8.8 gives
another real-life example of equilibrium and stability.
䉴 F I G U R E 8 . 1 3 Stable and unsta-

ble (a) Race cars are very stable
because of their wide wheel bases
and low center of gravity. (b) The
acrobat’s base of support is very nar-
row: the small area of head-to-head
contact. As long as his center of grav-
ity remains above the head area, he is
in equilibrium, but a displacement of
only a few centimeters would proba-
bly be enough to topple him. (Why
he is in a spread-eagle position will
become clearer in Section 8.3.)
(a) (b)
CONCEPTUAL EXAMPLE 8.8 The Center-of-Gravity Challenge

A female student issues a challenge to a male student. She place the top of his head against the wall, reach over to bring
states that she can perform a simple physical feat that he can’t. the chair directly in front of him, and place one hand on each
She places a straight-back chair (like most kitchen chairs) with side of the chair (䉴 Fig. 8.14a). Finally, without moving his feet,
its back against a wall. He is to face the wall and stand next to he is to stand up while lifting the chair. The female student
the chair with his toes touching the wall, then step two foot- demonstrates this and easily stands up.
lengths backward. (That is, he is to bring the toe of one foot Most males can’t perform this feat, but most females can.
behind the heel of the other foot twice and end up with his feet Why?
together, away from the wall.) Next, he is to lean forward and
REASONING AND ANSWER. When the male

student bends over and tries to lift the chair, he
is in unstable equilibrium (but fortunately,
using his head he doesn’t fall over). That is, the
center of gravity of the male student>chair sys-
tem falls outside (in front of) the system’s base
of support—his feet. Males tend to have a
higher center of gravity (larger shoulders and
narrower pelvis) than do females (narrow
shoulders and larger pelvis).
When the female student bends over and
lifts the chair, the center of gravity of the
female student> chair system does not fall out-
side the system’s base of support (her feet). She
is in stable equilibrium and so is able to stand
up from the bent position while she lifts the
chair.
But wait! The male student applies physics
and swings the chair back (Fig. 8.14b). The
combined center of gravity is now over his (a) (b)
base of support, and he can stand while hold-
ing the chair.
䉱 F I G U R E 8 . 1 4 The challenge (a) The male student leans forward with his
FOLLOW-UP EXERCISE. Why might some head on the wall. He is to lift the chair and stand up—but he can’t. Yet the
males be able to stand while lifting the chair, female student can easily perform this simple feat. (b) But wait. He applies
and some females not be able to do this? principles of physics and swings the chair back, and he can stand. Why?
Another classic example of equilibrium is the Leaning Tower of Pisa

(䉴 Fig. 8.15a), from which Galileo allegedly performed his “free-fall” experiments.
(See Chapter 2, Insight 2.1, Galileo Galilei and the Leaning Tower of Pisa.) The
tower started leaning before its completion in 1350 CE because of the soft subsoil
beneath it. In 1990, the lean was about 5.5° from the vertical (about 5 m, or 17 ft, at
the top) with an average increase in the lean of about 1.2 mm a year.
Attempts have been made to stop the lean increase. In 1930, cement was
injected under the base, but the lean continued to increase. In the 1990s, major
actions were taken. The tower was cabled back and counterweights were added to
the high side (Fig. 8.15b). Drilling was done diagonally below the foundation on
the high side so as to create cavities from which soil could be removed. The tower
settled back to about a 5° lean, or a shift of about 40 cm at the top. Moral of the
story: Keep that center of gravity above the base of support.
(a)
EXAMPLE 8.9 Stack Them Up: Center of Gravity

Uniform, identical bricks 20 cm long are stacked so that 4.0 cm of each brick extends
beyond the brick beneath, as shown in 䉲 Fig. 8.16a. How many bricks can be stacked in
this way before the stack falls over?
T H I N K I N G I T T H R O U G H . As each brick is added, the center of mass (or center of gravity)
of the stack moves to the right. The stack will be stable as long as the combined center of
mass (CM) is over the base of support—the bottom brick. All of the bricks have the same
mass, and the center of mass of each is located at its midpoint. So the horizontal location (b)
of the stack’s CM must be computed as bricks are added, until the CM extends beyond
䉱 F I G U R E 8 . 1 5 Hold it stable!
the base. The location of the CM was discussed in Section 6.5 (see Eq. 6.19).
(a) The Leaning Tower of Pisa.
SOLUTION. Although leaning, it is still in stable
equilibrium. Why? (b) Tons of lead
Given: brick length = 20 cm Find: maximum number of bricks that counterweight were used in an
yields stability displacement of each effort to help correct the tower’s lean
brick = 4.0 cm and keep it in stable equilibrium.
20 cm
Taking the origin to be at the center of the bottom brick, the horizontal coordinate of the
4.0 cm 4 center of mass (or center of gravity) for the first two bricks in the stack is given by Eq.
6.19, where m1 = m2 = m and x2 is the displacement of the second brick:
3 mx1 + mx2 m1x1 + x22 x1 + x2 0 + 4.0 cm

XCM2 = = = = = 2.0 cm
m + m 2m 2 2
2 The masses of the bricks cancel out (since they are all the same). For three bricks,
m1x1 + x2 + x32 0 + 4.0 cm + 8.0 cm
XCM3 = = = 4.0 cm
1
3m 3
For four bricks,

m1x1 + x2 + x3 + x42 0 + 4.0 cm + 8.0 cm + 12.0 cm
x1= 0 x2 x3 x4 XCM4 = = = 6.0 cm
4m 4
(a)
and so on.
This series of results shows that the center of mass of the stack moves horizontally 2.0
cm for each brick added to the bottom one. For a stack of six bricks, the center of mass is 10
cm from the origin and directly over the edge of the bottom brick
(2.0 cm * 5 added bricks = 10 cm, which is half the length of the bottom brick), so the
stack is just at unstable equilibrium. The stack may not topple if the sixth brick is positioned
very carefully, but it is doubtful that this could be done in practice. A seventh brick would
definitely cause the stack to fall off the bottom brick. Why? (As shown in Fig. 8.16b, you can
try it yourself and stack them up using books. Don’t let the librarian catch you.)
F O L L O W - U P E X E R C I S E . If the bricks in this Example were stacked so that, alternately,
4.0 cm and 6.0 cm extended beyond the brick beneath, how many bricks could be
stacked before the stack toppled?
For another case of stability, see Insight 8.1, Stability in Action.
DID YOU LEARN?

➥ A net torque is necessary for rotational motion, and torque is the product of the
applied force and the lever arm.
(b) B B
➥ The conditions for mechanical equilibrium are Fnet = ©Fi = 0 and Tnet = ©Ti = 0.
B B
䉱 F I G U R E 8 . 1 6 Stack them up! ➥ As long as the center of gravity of an object lies above and inside the original base
(a) How many bricks can be of support, the object is in stable equilibrium.
stacked like this before the stack
falls? See Example 8.9. (b) Try a
similar experiment with books.
8.3 Rotational Dynamics

➥ What does the moment of inertia measure?

➥ What is the rotational form of Newton’s second law?
tnet = r⊥F = rF⊥ = mr2a
Line of MOMENT OF INERTIA
action
Torque is the rotational analogue of force in linear motion, and a net torque pro-
duces rotational motion. To analyze this relationship, consider a constant net force
r⊥ u u F⊥ acting on a particle of mass m about the given axis (䉳 Fig. 8.17). The magnitude of
r F the torque on the particle is
Axis
tnet = r⬜ F = rF⬜ = rma⬜ = mr 2a (torque on a particle) (8.4)
䉱 F I G U R E 8 . 1 7 Torque on a
particle The magnitude of the where a⬜ = at = ra is the tangential acceleration (at , Eq. 7.13). For the rotation of a
torque on a particle of mass m is rigid body about a fixed axis, this equation can be applied to each particle in the
t = mr2a. See text for desccription. object and the results summed over the entire body (n particles) to find the total
8.3 ROTATIONAL DYNAMICS 281
torque. Since all the particles of a rotating rigid body have the same angular accel- Axis
eration, we can simply add the individual torque magnitudes: m1 m2
tnet = gti = t1 + t2 + t3 + Á + tn
= m1 r 21 a + m22 r2 a + m3 r 23 a + Á + mn r2n a
= 1m1 r 21 + m2 r 22 + m3 r 23 + Á + mn r 2n2a
= 1gmi r i 2a
2 x1 x2
(8.5)
But for a rigid body, the masses 1miœs2 and the distances from the axis of rotation
(a) m1 = m2 = 30 kg
x1 = x2 = 0.50 m
1ri' s2 do not change. Therefore, the quantity in the parentheses in Eq. 8.5 is con-
(b) m1 = 40 kg, m2 = 10 kg
stant, and it is called the moment of inertia, I (for a given axis): x1 = x2 = 0.50 m
(c) m1 = m2 = 30 kg
I = gmi r i
2
(moment of inertia) (8.6)
x1 = x2 = 1.5 m
SI unit of moment of inertia: kilogram-meters squared 1kg # m22 Axis

The magnitude of the net torque can be conveniently written as m1 m2
tnet = Ia (net torque on a rigid body) (8.7)
This is the rotational form of Newton’s second law (TB B

net = IA, in vector form). Keep in
mind that, as a net force is necessary to produce a translational acceleration, a net x2
torque 1tnet2 is necessary to produce an angular acceleration. x1 = 0
By comparing the rotational form of Newton’s second law (T B B
net = IA) with the
translational form 1Fnet = ma2, where m is a measure of translational inertia, it can (d) m1 = m2 = 30 kg
B B
x1 = 0, x2 = 3.0 m
be seen that the moment of inertia I is a measure of rotational inertia, or a body’s
tendency to resist change in its rotational motion. Although I is constant for a rigid (e) m1 = 40 kg, m2 = 10 kg
body and is the rotational analogue of mass, unlike the mass of a particle, the x1 = 0, x2 = 3.0 m
moment of inertia of a body is referenced to a particular axis and can have differ-
䉱 F I G U R E 8 . 1 8 Moment of inertia
ent values for different axes. The moment of inertia depends on
The moment of inertia also depends on the mass distribution of the body the distribution of mass relative to a
relative to its axis of rotation. It is easier (that is, it takes less torque) to give an particular axis of rotation and, in
object an angular acceleration about some axes than about others. The following general, has a different value for
Example illustrates this point. each axis. This difference reflects the
fact that objects are easier or more
difficult to rotate about certain axes.
See Example 8.10.
EXAMPLE 8.10 Rotational Inertia: Mass Distribution and Axis of Rotation

Find the moment of inertia about the axis indicated for each (b) I = 140 kg210.50 m22 + 110 kg210.50 m22 = 12.5 kg # m2
(c) I = 130 kg211.5 m22 + 130 kg211.5 m22 = 135 kg # m2
of the one-dimensional dumbbell configurations in 䉴 Fig. 8.18.
(Neglect the mass of the connecting bar, and give your
answers to three significant figures for comparison.) (d) I = 130 kg210 m22 + 130 kg213.0 m22 = 270 kg # m2
(e) I = 140 kg210 m22 + 110 kg213.0 m22 = 90.0 kg # m2
THINKING IT THROUGH. This is a direct application of Eq. 8.6
for cases with different masses and distances. It will show This Example clearly shows how the moment of inertia
that the moment of inertia of an object depends on the axis of depends on mass and its distribution relative to a particular axis
rotation and on the mass distribution relative to the axis of of rotation. In general, the moment of inertia is larger the farther
rotation. The sum for I will include only two terms (two the mass is from the axis of rotation. This principle is important
masses). in the design of flywheels, which are used in automobiles to
keep the engine running smoothly between cylinder firings.
SOLUTION.
The mass of a flywheel is concentrated near the rim, giving a
Given: Values of m and r in the Find: I (moment large moment of inertia, which resists changes in motion.
figure. of inertia)
F O L L O W - U P E X E R C I S E . In parts (d) and (e) of this Example,
With I = m1r21 + m2r22: would the moments of inertia be different if the axis of rota-
(a) I = 130 kg210.50 m22 + 130 kg210.50 m22 = 15.0 kg # m2 tion went through m2? Explain.
INSIGHT 8.1 Stability in Action

When riding a bicycle and going around a curve or making a torque would tend to rotate the bicycle in such a way that the
turn on a level surface, the rider instinctively leans into the wheels would slide inward underneath the rider. However, if
curve (Fig. 1). Why? You might think that leaning over, rather the rider leans inward at the proper angle (Fig. 2b), both the
B
than remaining upright, is more likely to cause a spill. However, line of action of R and the weight force act through the center
leaning really does increase stability—it’s all a matter of torques. of gravity, and there is no rotational instability (as the gentle-
When a vehicle goes around a level circular curve, a cen- man on the bicycle well knew).
tripetal force is needed to keep the vehicle on the road, as was There is still a torque on the leaning rider, however. Indeed,
learned in Section 7.3. This force is generally supplied by the when the rider leans into the curve, the weight force gives rise
force of static friction between the tires and the road. As illus- to a torque about an axis through the point of contact with the
B
trated in Fig. 2a, the force R of the ground on the bicycle pro- ground. This torque, along with the turning of the handle-
vides the required centripetal force 1Rx = Fc = fs2 to round the
B B B
bars, causes the bicycle to turn. If the bicycle were not mov-
curve, and the normal force 1Ry = N2.
B B
ing, there would be a rotation about this axis, and the bicycle
Suppose the rider tried to remain upright while going and rider would fall over.
around the curve with these forces operative, as shown in The need to lean into a curve is readily apparent in bicycle
B
Fig. 2a. Note that the line of action of R does not go through and motorcycle races on level tracks. Things can be made eas-
the system’s center of gravity (indicated by a dot). With an axis ier for the riders if tracks or roadways are banked to provide a
of rotation through the center of gravity, a counterclockwise natural lean (Section 7.3).
w
u w
N N R
R
Fc = f s Fc = fs
䉱 F I G U R E 1 Leaning into a curve When rounding (a) (b)
a curve or making a turn, a bicycle rider must lean
into the curve. (This rider could have told you why.) 䉱 F I G U R E 2 Make that turn See text for description.
INTEGRATED EXAMPLE 8.11 Balancing Act: Locating the Center of Gravity

(a) A rod with a movable ball, like that shown in 䉴 Fig. 8.19, is CG
more easily balanced if the ball is in a higher position. Is this
because, when the ball in a higher position, (1) the system has
a higher center of gravity and more stability; (2) the center of
gravity is off the vertical, and there is less torque and a
smaller angular acceleration; (3) the center of gravity is closer
to the axis of rotation; or (4) the moment of inertia about the
axis of rotation is larger? (b) Suppose the distance of the ball
r2
from the finger for the farthest position in Fig. 8.19 is
r2 = 60 cm and the distance to the closest position is
r1 = 20 cm. When the rod rotates, how many times greater is
䉳 FIGURE 8.19
the angular acceleration of the rod with the ball at the closest Greater stability with a
r1
position than that with the ball at the farthest position? higher center of gravity?
(Neglect the mass of the rod.) See Example text for
description.
(A) CONCEPTUAL REASONING. With the ball at any position some quantity (or quantities) that is not given cancels.
and the rod vertical, the system is in unstable equilibrium. As Note that the mass of the ball is not given, which would be
learned in Section 8.2, rigid bodies with wide bases and low needed to compute the gravitational torque 1t2. Also, the
centers of gravity are more stable, so answer (1) isn’t correct. angle u is not given. So it is best to start with basic equations
Any slight movement would cause the rod to rotate about an and see what happens.
axis through its point of contact. With the center of gravity
(CG) at a higher position and off the vertical, there would be Given: r1 = 20 cm Find: How many times greater the
a greater lever arm (and thus a greater torque), so (2) is also r2 = 60 cm rod’s angular acceleration is
incorrect. With the ball in a higher position, the center of with the ball at r1 compared
gravity is farther from the axis of rotation, which makes (3) to when it is at r2
incorrect. This leaves (4) by a process of elimination, but let’s
The magnitude of the angular acceleration is given by Eq. 8.7,
a = tnet >I. So attention turns to the torque tnet and the
justify it as the correct answer.
Moving the CG farther from the axis of rotation has an
moment of inertia I. From the basic equations of the chapter,
interesting consequence: a greater moment of inertia, or
tnet = r⬜ F = rF sin u (Eq. 8.2), or tnet = rmg sin u where
resistance to change in rotational motion, and hence a
F = mg in this case, with m being the mass of the ball. Simi-
smaller angular acceleration. However, with the ball in a
larly, I = mr2 (Eq. 8.6). Thus,
higher position, as the rod starts to rotate there is a greater
torque. The net result is that the increased moment of inertia tnet rmg sin u g sin u
produces an even greater resistance to rotational motion and a = = 2
=
I mr r
hence a smaller angular acceleration. [Note that the torque
1t = rF sin u2 varies as r, whereas the moment of inertia (Note that the angular acceleration a is inversely proportional
1I = mr22 varies as r2, and so has a larger increase with to the lever arm r; that is, the longer the lever arm, the smaller
increasing r. What effect does sin u have?] the angular acceleration.) The sin u is still there, but note what
Then, the smaller the angular acceleration, the more time happens when the ratio of the angular accelerations is
there is to adjust your hand under the rod to balance it by formed:
bringing the finger and the axis of rotation under the center a1 g sin u>r1 r2 60 cm a1
of gravity. The torque is then zero and the rod is again in = = = = 3 or a2 =
a2 g sin u>r2 r1 20 cm 3
equilibrium, albeit unstable. And so, the answer is (4).
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . Being asked Hence, the angular acceleration of the rod with the ball at the
how many times greater or less something is compared to upper position is one-third that with the ball at the lower
something else usually implies the use of a ratio in which position.
F O L L O W - U P E X E R C I S E . When walking on a thin bar or rail, such as a railroad rail, you have probably found that it helps to hold
your arms outstretched. Similarly, tightrope walkers often carry long poles, as in the chapter-opening photo. How does this pos-
ture and the pole help maintain balance?
As Integrated Example 8.11 shows, the moment of inertia is an important con-

sideration in rotational motion. By changing the axis of rotation and the relative
mass distribution, the value of I can be changed and the motion affected. You were
probably told to do this when playing softball or baseball as a child. When at bat,
children are often instructed to “choke up” on the bat—to move their hands far-
ther up on the handle.
Now you know why. In doing so, the child moves the axis of rotation of the bat
closer to the more massive end of the bat (or its center of mass). Hence, the
moment of inertia of the bat is decreased (smaller r in the mr2 term). Then, when a
swing is taken, the angular acceleration is greater. The bat gets around quicker,
and the chance of hitting the ball before it goes past is greater. A batter has only a
fraction of a second to swing, and with u = 12 at2, a larger a allows the bat to rotate
more quickly (swing faster).
PARALLEL AXIS THEOREM

Calculations of the moments of inertia of most extended rigid bodies require math
that is beyond the scope of this book. The results for some common shapes are given
in 䉲 Fig. 8.20. The rotational axes are generally taken along axes of symmetry—that
is, axes running through the center of mass that give a symmetrical mass
distribution. One exception is the rod with an axis of rotation through one end
(Fig. 8.20c). This axis is parallel to an axis of rotation through the center of mass of
Axis Axis Axis Axis
R
M
R
L L
1 1
I = MR2 I= 12
ML2 I= 3
ML2 I = MR2
(a) Particle (b) Thin rod (c) Thin rod (d) Thin cylindrical
shell, hoop, or ring
Axis Axis Axis
R R1
R
R2 R
I = 12 MR2 Axis 1
I= M(R12 + R22)
2 I = 25 MR2 I= 2
3
MR2
(e) Solid cylinder or disk (f) Annular cylinder (g) Solid sphere about (h) Thin spherical shell
any diameter
Axis Axis Axis
a
b
L 1
L 1
1
I = 12 M(a2 + b2) I= 12
ML2 I= ML2
3
(i) Rectangular plate (j) Thin rectangular (k) Thin rectangular

sheet sheet
䉱 F I G U R E 8 . 2 0 Moments of inertia of some uniform density objects with common shapes
M CM
the rod (Fig. 8.20b). The moment of inertia about such a parallel axis is given by a
useful theorem called the parallel axis theorem, namely,
I = ICM + Md 2 (8.8)
d
I = ICM + Md2 where I is the moment of inertia about an axis that is parallel to one through the
center of mass and at a distance d from it, ICM is the moment of inertia about an
䉱 F I G U R E 8 . 2 1 Parallel axis axis through the center of mass, and M is the total mass of the body (䉳 Fig. 8.21).
theorem The moment of inertia For the axis through the end of the rod (Fig. 8.20c), the moment of inertia is
about an axis parallel to another
obtained by applying the parallel axis theorem to the thin rod in Fig. 8.20b:
through the center of mass of a body
is I = ICM + Md2, where M is the L 2
total mass of the body and d is the I = ICM + Md 2 = 1
12 ML2 + M a b = 1
12 ML2 + 1
4 ML2 = 1
3 ML2
distance between the two axes. 2
APPLICATIONS OF ROTATIONAL DYNAMICS

The rotational form of Newton’s second law allows us to analyze dynamic rota-
tional situations. Examples 8.12 and 8.13 illustrate how this is done. In such situa-
tions, it is very important to make certain that all the data are properly listed to
help with the increasing number of variables.
EXAMPLE 8.12 Opening the Door: Torque in Action

A student opens a uniform 12-kg door by applying a constant force of 40 N at a perpen-
dicular distance of 0.90 m from the hinges (䉴 Fig. 8.22). If the door is 2.0 m in height and
1.0 m wide, what is the magnitude of its angular acceleration? (Assume that the door
rotates freely on its hinges.)
T H I N K I N G I T T H R O U G H . From the given information, the applied net torque can be cal-
culated. To find the angular acceleration of the door, the moment of inertia is needed.
This can be calculated, since the door’s mass and dimensions are known.
SOLUTION. From the information given in the problem, r

F
Given: M = 12 kg Find: a (magnitude of angular
F = 40 N acceleration)
r⬜ = r = 0.90 m
h = 2.0 m (door height)
w = 1.0m (door width)
The rotational form of Newton’s second law can be applied, tnet = Ia, where I is about
the hinge axis. tnet can be found from the given data, so the problem boils down to
determining the moment of inertia of the door.
Looking at Fig. 8.20, it can be seen that case (k) applies to a door (treated as a uniform
rectangle) rotating on hinges, so I = 13 ML2, where L = w, the width of the door. Then,
tnet = Ia
or 䉱 F I G U R E 8 . 2 2 Torque in action
tnet r⬜ F 3rF 310.90 m2140 N2 See Example text for description.
a = = 9.0 rad>s2
112
= 1 = 2
= 2
I 2 Mw kg211.0 m2
3 ML
F O L L O W - U P E X E R C I S E . In this Example, if the constant torque were applied through an angular distance of 45° and then
removed, how long would the door take to swing completely open (90°)? Neglect friction.
In problems involving pulleys in Section 4.5, the mass (and hence the inertia) of
the pulley was neglected in order to simplify things. Now you know how to
include those quantities and can treat pulleys more realistically, as seen in the next
Example.
EXAMPLE 8.13 Pulleys Have Mass, Too—Taking Pulley Inertia into

Account
A block of mass m hangs from a string wrapped around a frictionless, disk-shaped pul-
ley of mass M and radius R, as shown in 䉲 Fig. 8.23. If the block descends from rest
under the influence of gravity, what is the magnitude of its linear acceleration? (Neglect
the mass of the string.)
T H I N K I N G I T T H R O U G H . Real pulleys have mass and rotational inertia, which affect
their motion. The suspended mass (via the string) applies a torque to the pulley. Here
the rotational form of Newton’s second law can be used to find the angular acceleration
of the pulley and then its tangential acceleration, which is the same in magnitude as the
linear acceleration of the block. (Why?) No numerical values are given so the answer
will be in symbol form.
SOLUTION. The linear acceleration of the block depends on the angular acceleration of
the pulley, so we look at the pulley system first. The pulley is treated as a disk and thus has
a moment of inertia I = 12 MR 2 (Fig. 8.20e). A torque due to the tension force in the string
(T) acts on the pulley. With t = Ia (considering only the upper dashed box in Fig. 8.23),
tnet = r⬜ F = RT = Ia = A 12 MR2 B a
R such that
M 2T
T a =
MR
The linear acceleration of the block and the angular acceleration of the pulley are
related by a = Ra, where a is the tangential acceleration, and
2T
a = Ra = (1)
M
T But T is unknown. Looking at the descending mass (the lower dashed box) and sum-
ming the forces in the vertical direction (choosing down as positive) gives
mg - T = ma
a m or
T = mg - ma (2)
mg Using Eq. 2 to eliminate T from Eq. 1 yields
2T 21mg - ma2
a = =
M M
And solving for a,
y
2mg
12m + M2
a = (3)
negative positive
Note that if M : 0 (as in the case of ideal, massless pulleys in previous chapters), then
x I : 0 and a = g (from Eq. 3). Here, however, M Z 0, so a 6 g. (Why?)
F O L L O W - U P E X E R C I S E . Pulleys can be analyzed even more realistically. In this Example,
friction was neglected, but practically, a frictional torque 1tf2 exists and should be
䉱 F I G U R E 8 . 2 3 Pulley with included. What would be the form, as in Eq. 3, of the angular acceleration in this case?
inertia Taking the mass, or rota- Show that your result is dimensionally correct.
tional inertia, of a pulley into
account allows a more realistic
description of the motion. The
directional sign convention for
torque is shown. See Example 8.13. In pulley problems, as before, the mass of the string will be neglected—an
approach that still gives a good approximation if the string is relatively light. Tak-
ing the mass of the string into account would give a continuously varying mass
hanging on the pulley, thus producing a variable torque. Such problems are
beyond the scope of this book.
Suppose you had masses suspended from each side of a pulley. Here, you
would have to compute the net torque. If the values of the masses are unknown,
so which way the pulley would rotate cannot be determined, then simply assume
a direction. As in the linear case, if the result came out with the opposite sign, it
would indicate that you had assumed the wrong direction.
For problems such as those of Examples 8.13 and 8.14, dealing with coupled rotational
and translational motions, keep in mind that with no string slippage, the magnitudes of
the accelerations are usually related by a = ra, while v = rv relates the magnitudes of
the velocities at any instant of time. Applying Newton’s second law (in rotational or lin-
ear form) to different parts of the system gives equations that can be combined by using
such relationships. Also, for rolling without slipping, a = ra and v = rv, relate the
angular quantities to the linear motion of the center of mass.
Another application of rotational dynamics is the analysis of motion of objects

that can roll.
CONCEPTUAL EXAMPLE 8.14 Applying a Torque One More Time: Which Way Does the Yo-Yo Roll?
The string of a yo-yo sitting on a level surface is pulled as of the pull, so the answer is (a). (Get a yo-yo and try it if
shown in 䉲 Fig. 8.24. Will the yo-yo roll (a) toward the person you’re a nonbeliever.)
or (b) away from the person? There is more interesting physics in our yo-yo situation.
The pull force is not the only force acting on the yo-yo; there
䉴 FIGURE 8.24 are three others. Do they contribute torques? Let’s identify
Pulling the yo-yo’s these forces. There’s the weight of the yo-yo and the normal
string See Concep- F force from the surface. Also, there is a horizontal force of sta-
tual Example text for tic friction between the yo-yo and the surface. (Otherwise the
description. r yo-yo would slide rather than roll.) But these three forces act
through the line of contact or through the instantaneous axis
of rotation, so they produce no torques here. (Why?)
Instantaneous axis What would happen if the angle of the string or pull force
of rotation were increased (relative to the horizontal), as illustrated in
䉲 Fig. 8.25a? The yo-yo would still roll to the right. As can be
REASONING AND ANSWER. Apply the physics just studied to seen in Fig. 8.25b, at some critical angle uc the line of force
the situation. Note that the instantaneous axis of rotation is goes through the axis of rotation, and the net torque on the
along the line of contact of the yo-yo with the surface. If you yo-yo becomes zero, so the yo-yo does not roll.
B
had a stick standing vertically in place of the r vector and If this critical angle is exceeded (Fig. 8.25c), the yo-yo will
pulled on a string attached to the top of the stick in the direc- begin to roll counterclockwise, or to the left. Note that the line
B
tion of F, which way would the stick rotate? Of course, it of action of the force is on the other side of the axis of rotation
would rotate clockwise (about its instantaneous axis of rota- from that in Fig. 8.26a and that the lever arm 1r⬜2 has changed
tion). The yo-yo reacts similarly; that is, it rolls in the direction directions, resulting in a reversed net torque direction.
F F
uc
u u
䉴 F I G U R E 8 . 2 5 The angle
makes a difference (a) With the r⊥
line of action to the left of the
instantaneous axis, the yo-yo rolls
to the right. (b) At a critical angle Instantaneous Line
Force line
uc the line of action passes through axis of rotation of action
of action Axis Axis
the axis, and the yo-yo is in equi-
librium. (c) When the line of (a) Rolls to right (b) U = Uc, (c) U > Uc,
action is to the right of the axis,
in rotational equilibrium rolls to left
the yo-yo rolls to the left. does not roll (r⊥ to the right)
F O L L O W - U P E X E R C I S E . Suppose you set the yo-yo string at the critical angle, with the string over a round, horizontal bar at the
appropriate height, and you suspend a weight on the end of the string to supply the force for the equilibrium condition. What
will happen if you then pull the yo-yo toward you, away from its equilibrium position, and release it?
DID YOU LEARN?

➥ The moment of inertia is a measure of rotational inertia—a body’s tendency to
resist rotational motion.
B
➥ The rotational form of Newton’s second law is Tnet = IA
B B B
(analogous to Fnet = ma).
8.4 Rotational Work and Kinetic Energy

➥ How do the equations for rotational work and power compare to their translational
analogues?
➥ What is the total kinetic energy of a rolling object (without slipping)?
This section gives the rotational analogues of various equations of linear motion
associated with work and kinetic energy for constant torques. Because their devel-
opment is similar to that given for their linear counterparts, detailed discussion is
not needed. As in Section 5.1, it is understood that W is the net work if more than
one force or torque acts on an object.
Rotational Work We can go directly from work done by a force to work done by
a torque, since the two are related 1t = r⬜ F2. For rotational motion, the rotational
work, W = Fs, done by a single force F acting tangentially along an arc length s is
W = Fs = F1r⬜ u2 = tu
where u is in radians. Thus, for a single torque acting through an angle of rotation u,
W = tu (rotational work for a single force) (8.9)
In this book, both the torque 1t2 and angular displacement 1u2 vectors are almost
always along the fixed axis of rotation, so you will not need to be concerned about
parallel components, as you were for translational work. The torque and angular
displacement may be in opposite directions, in which case the torque does nega-
tive work and slows the rotation of the body. Negative rotational work is analo-
gous to F and d being in opposite directions for translational motion.
Rotational Power An expression for the instantaneous rotational power (P),

the rotational analogue of power (the time rate of doing work), is easily obtained
from Eq. 8.9:
u
= ta b = tv
W
P = (rotational power) (8.10)
t t
THE WORK—ENERGY THEOREM AND KINETIC ENERGY

The relationship between the net rotational work done on a rigid body (more than
one force acting) and the change in rotational kinetic energy of the body can be
derived as follows, starting with the equation for rotational work:
Wnet = tu = Iau
Since we assume the torques are due only to constant forces, a is constant. But
from rotational kinematics in Chapter 7, it is known that for a constant angular
acceleration, v2 = v2o + 2au, and
v2 - v2o
Wnet = I a b = 1
2 Iv2 - 1
2 Iv2o
2
From Eq. 5.6 (work–energy), Wnet = ¢K. Therefore,

1 1
Wnet = 2 Iv2 - 2 Iv2o = K - Ko = ¢K (8.11)
Then the expression for rotational kinetic energy is

1
K = 2 Iv2 (rotational kinetic energy) (8.12)
Thus, the net rotational work done on an object is equal to the change in the rotational kinetic
energy of the object (with zero linear kinetic energy). Consequently, to change the rota-
tional kinetic energy of an object, a net torque must be applied.
8.4 ROTATIONAL WORK AND KINETIC ENERGY 289
TABLE 8.1 Translational and Rotational Quantities and Equations

Translational Rotational
t = rF sin u
B
Force: F Torque (magnitude):
Mass (inertia): m Moment of inertia: I = gmi r2i
B
Tnet = IA
B B B
Newton’s second law: Fnet = ma Newton’s second law:
Work: W = Fd Work: W = tu
Power: P = Fv Power: P = tv
Kinetic energy: K = 12 mv2 Kinetic energy: K = 12 Iv2
Work–energy theorem: Wnet = 12 mv2 - 1 2
2 mv o = ¢K Work–energy theorem: Wnet = 12 Iv2 - 12 Iv2o = ¢K
B B B B
Linear momentum: p = mv Angular momentum: L = IV
It is possible to derive the expression for the kinetic energy of a rotating rigid
body (about a fixed axis) directly. Summing the instantaneous kinetic energies of
the body’s individual particles relative to the fixed axis gives
gmi v i = 12 1 gmi r i 2v2 = 12 Iv2
1 2 2
K = 2
where, for each particle of the body, vi = ri v. So, Eq. 8.12 doesn’t represent a new
form of energy; rather, it is simply another expression for kinetic energy, in a form
that is more convenient for rigid body rotation.
A summary of translational and rotational analogues is given in 䉱 Table 8.1.
(The table also contains angular momentum, which will be discussed in Section 8.5.)
When an object has both translational and rotational motion, its total kinetic
energy may be divided into parts to reflect the two kinds of motion. For example,
for a cylinder rolling without slipping on a level surface, the motion is purely rota-
tional relative to the instantaneous axis of rotation (the point or line of contact),
which is instantaneously at rest. The total kinetic energy of the rolling cylinder is
1 2
K = 2 Ii v
where Ii is the moment of inertia about the instantaneous axis. This moment of
inertia about the point of contact (the axis) is given by the parallel axis theorem
(Eq. 8.8), Ii = ICM + MR2, where R is the radius of the cylinder. Then
K = 1
2 Ii v
2
= 1
2 1ICM + MR22v2 = 1
2 ICM v
2
+ 1
2 MR 2v2
But since there is no slipping, vCM = Rv, and the total K is
1 2 1 2
K = 2 ICM v + 2 Mv CM (rolling, no slipping) (8.13)
total rotational translational
= +
K Kr Kt
Note that although a cylinder was used as an example here, this is a general result
and applies to any object that is rolling without slipping.
Thus, the total kinetic energy of such an object is the sum of two contributions: the
translational kinetic energy of the object’s center of mass and the rotational kinetic energy
of the object relative to a horizontal axis through its center of mass.
EXAMPLE 8.15 Division of Energy: Rotational and Translational

A uniform, solid 1.0-kg cylinder rolls without slipping at a speed of 1.8 m>s on a flat
surface. (a) What is the total kinetic energy of the cylinder? (b) What percentage of this
total is rotational kinetic energy?
T H I N K I N G I T T H R O U G H . The cylinder has both rotational and translational kinetic
energies, so Eq. 8.13 applies, and its terms are related by the condition of rolling with-
out slipping.
SOLUTION.
Given: M = 1.0 kg Find: (a) K (total kinetic energy)

vCM = 1.8 m>s
1 * 100%2 (percentage of
Kr
ICM = 12 MR 2 (from Fig. 8.20e) (b)
K
rotational energy)
(a) The cylinder rolls without slipping, so the condition vCM = Rv applies. Then the
total kinetic energy (K) is the sum of the rotational kinetic energy Kr and the transla-
tional kinetic energy Kt of the center of mass, KCM (Eq. 8.13):
A Ba
vCM 2 1
b + 2 Mv CM = 14 MvCM + 12 Mv CM
2 2 2 2
K = 12 ICM v2 + 12 MvCM = 1 1
2 2 MR
2
R
= 34 Mv CM = 34 11.0 kg211.8 m>s22 = 2.4 J
2
(b) The rotational kinetic energy Kr of the cylinder is the first term of the preceding
equation, so, forming a ratio in symbol form,
1 2
4 Mv CM
= 13 1* 100%2 = 33%
Kr
= 3 2
K Mv CM
4
Thus, the total kinetic energy of the cylinder is made up of rotational and translational
parts, with one-third being rotational.
Note that in part (b) the radius of the cylinder was not needed, nor was the mass.
Because a ratio was used, these quantities canceled. However, don’t think that this
exact division of energy is a general result. It is easy to show that the percentage is dif-
ferent for objects with different moments of inertia. For example, you should expect a
rolling sphere to have a smaller percentage of rotational kinetic energy than a cylinder
has, because the sphere has a smaller moment of inertia 1I = 25 MR22.
F O L L O W - U P E X E R C I S E . Potential energy can be brought into the act by applying the
conservation of energy to an object rolling up or down an inclined plane. In this Exam-
ple, suppose that the cylinder rolled up a 20° inclined plane without slipping. (a) At
what vertical height (measured by the vertical distance of its CM) on the plane does the
cylinder stop? (b) To find the height in part (a), you probably equated the initial total
kinetic energy to the final gravitational potential energy. That is, the total kinetic energy
was reduced by the work done by gravity. However, a frictional force also acts (to pre-
vent slipping). Is there not work done here, too?
EXAMPLE 8.16 Rolling Down

A uniform cylindrical hoop is released from rest at a height of SOLUTION.
0.25 m near the top of an inclined plane (䉲 Fig. 8.26). If the Given: h = 0.25 m Find: vCM (speed of CM)
hoop rolls down the plane without slipping and no energy is
ICM = MR2 (from
lost due to friction, what is the linear speed of the cylinder’s
Fig. 8.20d)
center of mass at the bottom of the incline?
T H I N K I N G I T T H R O U G H . Here, gravitational potential energy
The total mechanical energy of the cylinder is conserved:
is converted into kinetic energy—both rotational and transla- Eo = E
tional. The conservation of (mechanical) energy applies, since
Since vo = 0 at the top of the incline and assuming that U = 0
Wf (frictional work) is zero.
at the bottom, this becomes
R Uo = K
1 2 1 2
Mgh = 2 ICMv + 2 Mv CM
v initially at bottom of incline
at rest
vCM
h = 0.25 m Using the rolling condition vCM = Rv gives
vCM 2 1
Mgh = 12 1MR 22a b + 2 Mv CM = MvCM
2 2
䉱 F I G U R E 8 . 2 6 Rolling motion and energy When an object R

rolls down an inclined plane, potential energy is converted to Solving for vCM,
translational and rotational kinetic energy. This makes the
rolling slower than frictionless sliding. vCM = 1gh = 419.8m>s2210.25m2 = 1.6 m>s
8.5 ANGULAR MOMENTUM 291
F O L L O W - U P E X E R C I S E . Suppose the inclined plane in this Example were frictionless and the hoop slid down the plane instead
of rolling. How would the speed at the bottom compare in this case? Why are the speeds different?
A FIXED RACE
As Example 8.16 shows for an object rolling down an incline without slipping,
vCM is independent of M and R. The masses and radii cancel, so all objects of a par-
ticular shape (with the same equation for the moment of inertia) roll with the same
speed, regardless of their size or density. But the rolling speed does vary with the
moment of inertia, which varies with an object’s shape. Therefore, rigid bodies
with different shapes roll with different speeds. For example, if you released a
cylindrical hoop, a solid cylinder, and a uniform sphere at the same time from the
top of an inclined plane, the sphere would win the race to the bottom, followed by
the cylinder, with the hoop coming in last—every time!
You can try this as an experiment with a couple of cans of food or other cylindrical
containers—one full of some solid material (in effect, a rigid body) and one empty
and with the ends cut out—and a smooth, solid ball. Remember that the masses and
the radii make no difference. You might think that an annular cylinder (a hollow
cylinder with inner and outer radii that are appreciably different—Fig. 8.20f) would
be a possible front-runner, or “front-roller,” in such a race, but it wouldn’t be. The
rolling race down an incline is fixed even when you vary the masses and the radii.
Another aspect of rolling is discussed in Insight 8.2, Slide or Roll to a Stop?
Antilock Brakes.
DID YOU LEARN?

➥ Rotational work is W = tu (compared to translational W = Fd ), and rotational
power is P = tv (compared to translational W = Fv).
➥ The total kinetic energy of a rolling object is the sum of two contributions: the
translational kinetic energy of the object’s center of mass and the rotational kinetic
2
energy relative to a horizontal axis through the CM that is, K = 12 ICM v2 + 12 MvCM.
8.5 Angular Momentum

➥ How is angular momentum related to torque, and what is the linear analogy?
➥ When is angular momentum conserved?
Another important quantity in rotational motion is angular momentum. Recall

from Section 6.1 how the linear momentum of an object is changed by a force.
Analogously, changes in angular momentum are associated with torques. As has
been learned, torque is the product of a moment arm and a force. In a similar man-
ner, angular momentum (L) is the product of a moment arm (r) and a linear
momentum (p). For a particle of mass m, the magnitude of the linear momentum is
p = mv, where v = rv. The magnitude of the angular momentum is
2 (single-particle
L = r⬜ p = mr⬜ v = mr⬜ v (8.14)
angular momentum)
SI unit of angular momentum: kilogram-meters squared per second 1kg # m2>s2,

where v is the speed of the particle r⬜ is the moment arm, and v is the angular
speed.
INSIGHT 8.2 Slide or Roll to a Stop? Antilock Brakes

While driving, in an emergency you may instinctively jam on In Example 2.8, a vehicle’s stopping distance was given by
the brakes, trying to come to a quick stop—that is, to stop in v2o
the shortest distance. But with the wheels locked, the car x =
skids or slides, often out of control. In this case, the force of 2a
sliding friction is acting on the wheels. By Newton’s second law, the net force in the horizontal direc-
To prevent skidding, you may have learned to pump the tion is F = f = mN = mmg = ma, and the stopping accelera-
brakes in order to roll rather than slide to a stop, particularly tion is then a = mg. Thus,
on wet or icy roads. Most newer automobiles have a comput- v2o
erized antilock braking system (ABS) that pumps the brakes x = (1)
2mg
automatically. When the brakes are applied firmly and the car
begins to slide, sensors in the wheels note the sliding motion, But, as was noted in Section 4.6, the coefficient of sliding
and a computer takes control of the braking system. It (kinetic) friction is generally less than that of static friction;
momentarily releases the brakes and then varies the brake that is, mk 6 ms. The general difference between rolling stops
fluid pressure with a pumping action (up to thirteen times per and sliding stops can be seen by using the same initial veloc-
second) so that the wheels will continue to roll without ity vo for both cases. Then, using Eq. 1 to form a ratio,
slipping. mk mk
or xroll = a bxslide
xroll
In the absence of sliding, both rolling friction and static fric- =
tion act. In many cases, however, the force of rolling friction is xslide ms ms
small, and only static friction need be taken into account. The From Table 4.1, the value of mk for rubber on wet concrete is
ABS works to keep static friction near the maximum, 0.60, and the value of ms for these surfaces is 0.80. Using these
fs L fsmax which you can’t do easily by foot. values for a comparison of the stopping distances gives
Does sliding instead of rolling make a big difference in an
xroll = a bx = 10.752xslide
0.60
automobile’s stopping distance? We can calculate the differ-
ence by assuming that rolling friction is negligible. Although 0.80 slide
the external force of static friction does no work to dissipate Thus, the car comes to a rolling stop in 75% of the distance
energy in slowing a car (this is done internally by friction on required for a sliding stop—for example, 15 m instead of 20
the brake pads), it does determine whether the wheels roll or m. Although this distance may vary for different conditions, it
slide. could be an important, perhaps lifesaving, difference.
B
For circular motion, r⬜ = r, since v is perpendicular to Br . For a system of parti-
cles making up a rigid body, all the particles travel in circles, and the magnitude of
the total angular momentum is
L = 1gmi r i 2v = Iv (rigid body angular momentum)

2
(8.15)
which, for rotation about a fixed axis, is (in vector notation)
B B
L = IV (8.16)
Thus, L is in the direction of the angular velocity vector 1V 2. This direction is

B B
given by the right-hand rule (Section 8.2).

For linear motion, the change in the total linear momentum of a system is
related to the net external force by Fnet = ¢P>¢t. Angular momentum is analo-
B B
gously related to net torque (in magnitude form):
I¢v ¢1Iv2 ¢L
tnet = Ia = = =
¢t ¢t ¢t
That is,
¢L
tnet = (8.17)
¢t
Thus, the net torque is equal to the time rate of change of angular momentum. In other
words, a net torque results in a change in angular momentum.
CONSERVATION OF ANGULAR MOMENTUM

Equation 8.17 was derived using tnet = Ia, which applies to a rigid system of par-
ticles or a rigid body having a constant moment of inertia. However, Eq. 8.17 is a
general equation that also applies to even nonrigid systems of particles. In such a
system, there may be a change in the internal mass distribution and a change in
the moment of inertia.
net = ¢L>¢t = 0, and
B
If the net torque on a system is zero, then, by Eq. 8.17, T
B
B B BB
¢L = L - L o = IV - Io V
B
o = 0
or in magnitude,
Iv = Iovo (8.18)
Thus, the condition for the conservation of angular momentum is as follows:

In the absence of an external, unbalanced torque, the total (vector) angular momen-
tum of a system is conserved (remains constant).
Just as the internal forces cannot change a systems’s linear momentum, neither
can internal torques change a system’s angular momentum.
For a rigid body with a constant moment of inertia (that is, I = Io), the angular
speed remains constant 1v = vo2 in the absence of a net torque. But it is possible
for the moment of inertia to change in some systems, giving rise to a change in the
angular speed, as the following Example illustrates.
EXAMPLE 8.17 Pull It Down: Conservation of Angular Momentum

A small ball at the end of a string that passes
through a tube is swung in a circle, as illus-
trated in 䉴 Fig. 8.27. When the string is
pulled downward through the tube, the r1 r2
angular speed of the ball increases. (a) Is the
increase in angular speed caused by a v1
v2
torque due to the pulling force? (b) If the
ball is initially moving at 2.8 m>s in a circle
with a radius of 0.30 m, what will be its tan-
gential speed if the string is pulled down to
reduce the radius of the circle to 0.15 m?
(Neglect the mass of the string.)
T H I N K I N G I T T H R O U G H . (a) A force is
applied to the ball via the string, but con-
sider the axis of rotation. (b) In the absence
of a net torque, the angular momentum is
conserved (Eq. 8.18), and the tangential
speed is related to the angular speed by
v = rv.
䉱 F I G U R E 8 . 2 7 Conservation of angular momentum When the string is pulled down-
ward through the tube, the revolving ball speeds up. See Example text for description.
SOLUTION.
Given: r1 = 0.30 m Find: (a) Cause of the increase in angular speed

r2 = 0.15 m (b) v2 (final tangential speed)
v1 = 2.8 m>s
(a) The change in the angular velocity, or an angular accelera- moment of inertia of the ball (I = mr2, from Fig. 8.20a),
tion, is not caused by a torque due to the pulling force. The decreases. Because of the absence of an external torque, the
force on the ball, as transmitted by the string (tension), acts angular momentum 1Iv2 of the ball is conserved, and if I is
through the axis of rotation, and therefore the torque is zero. reduced, v must increase.
Because the rotating portion of the string is shortened, the
(b) Because the angular momentum is conserved, we can and

equate the magnitudes of the angular momenta:
v2 = a bv1 = a b 2.8 m>s = 5.6 m>s
r1 0.30 m
Io vo = Iv r2 0.15 m
Then, using I = mr2 and v = v>r gives
When the radial distance is shortened, the ball speeds up
mr1 v1 = mr2 v2 (accelerates).
F O L L O W - U P E X E R C I S E . Let’s look at the situation in this Example in terms of work and energy. If the initial speed is the same
and the vertical pulling force is 7.8 N, what is the final speed of the 0.10-kg ball?
Example 8.17 should help you understand Kepler’s law of equal areas
(Section 7.6) from another viewpoint. A planet’s angular momentum is con-
served to a good approximation by neglecting the weak gravitational torques
from other planets. (The Sun’s gravitational force on a planet produces little or
no torque. Why?) When a planet is closer to the Sun in its elliptical orbit and so
has a shorter moment arm, its speed is greater, by the conservation of angular
momentum. [This is the basis of Kepler’s second law (the law of areas).] Simi-
larly, when an orbiting satellite’s altitude varies during the course of an elliptical
orbit about a planet, the satellite speeds up or slows down in accordance with
the same principle.
REAL-LIFE ANGULAR MOMENTUM

A popular demonstration of the conservation of angular momentum is shown in
䉲 Fig. 8.28a. A person sitting on a stool that rotates holds weights with his arms
outstretched and is started slowly rotating. An external torque to start this rotation
䉳 F I G U R E 8 . 2 8 Change in moment of inertia

(a) When the student spins slowly with masses in
outstretched arms, his moment of inertia is rela-
tively large. (The masses are farther from the axis of
rotation.) Note that he is isolated, with no external
torques (neglecting friction) acting on him, so his
angular momentum, L = Iv is conserved. Pulling
his arms inward decreases his moment of inertia.
(Why?) Consequently, v must increase, and he goes
into a dizzying spin. (b) Ice skaters similarly change
their moment of inertia to increase v in doing spins.
(c) The same principle helps explain the violence of
the winds that spiral around the center of a hurri-
cane. As air rushes in toward the low pressure cen-
ter of the storm, the air’s rotational speed must
increase for angular momentum to be conserved.
(a)
(b) (c)
must be supplied by someone else, because the person on the stool cannot initiate
the motion by himself. (Why not?) Once rotating, if the person brings his arms
inward, the angular speed increases and he spins much faster. Extending his arms
again slows him down. Can you explain this phenomenon?
If L is constant, what happens to v when I is made smaller by reducing r? The
angular speed must increase to compensate and keep L constant. Ice skaters and
ballerinas perform dizzying spins by pulling in and raising their arms to reduce
their moment of inertia (Fig. 8.28b). Similarly, a diver spins during a high dive by
tucking in the body and limbs, greatly decreasing his or her moment of inertia.
The enormous wind speeds of tornadoes and hurricanes represent another exam-
ple of the same effect (Fig. 8.28c).
Angular momentum also plays a role in ice-skating jumps in which the skater
spins in the air, such as a triple axel or triple lutz. A torque applied on the jump
gives the skater angular momentum, and the arms and legs are drawn into the
body, which, as in spinning on one’s toes, decreases the moment of inertia and
increases the angular speed so that multiple spins can be made during the jump.
To land with a smaller rate of spin, the skater opens the arms and projects the non-
landing leg. You may have noticed that most jump landings proceed in a curved
arc, which allows the skater to gain control.
EXAMPLE 8.18 A Skater Model

Real-life situations are generally complicated, but some can be approximately analyzed
by using simple models. Such a model for a skater’s spin is shown in 䉴 Fig. 8.29, with a
cylinder and rods representing the skater. In (a), the skater goes into the spin with the
“arms” out, and in (b) the “arms” are over the head to achieve a faster spin by the con-
servation of angular momentum. If the initial spin rate is 1 revolution per 1.5 s, what is 20 cm
the angular speed when the arms are tucked in?
R
THINKING IT THROUGH. The body and arms of a skater are approximated by the cylin-
der and rods, for which the moments of inertia are known (Fig. 8.20). Special attention
5 kg 5 kg
must be given to finding the moment of inertia of the arms around the axis of rotation
(through the cylinder). This can be done by applying the parallel axis theorem (Eq. 8.8).
With the angular momentum conserved, L = Lo or Iv = Io vo. Knowing the initial 80 cm
75 kg
angular speed, and given quantities to evaluate the moments of inertia (Fig. 8.29), the
final angular speed can be found.
SOLUTION. Listing the given data (see Fig. 8.29):
Given: vo = 11 rev>1.5 s212p rad>rev2 = 4.2 rad>s Find: v (final angular speed)
Mc = 75 kg 1cylinder or body2
Mr = 5.0 kg 1one rod or arm2
R = 20 cm = 0.20 m (a) Arms extended (not to scale)
L = 80 cm = 0.80 m
Momenta of inertia (from Fig. 8.20). R
cylinder: Ic = 12 Mc R2 rod: Ir = 12
1
Mr L2
Let’s first compute the moments of inertia of the system using the parallel axis theorem,
I = ICM + Md 2 (Eq. 8.8).
Before: The Ic of the cylinder is straightforward (Fig. 8.20e):
Ic = 12 McR2 = 12 175 kg210.20 m22 = 1.5 kg # m2
Referencing the moment of inertia of a horizontal rod (Fig. 8.29a) to the cylinder’s axis
of rotation using the parallel axis theorem:
Ir = ICM1rod2 + Md 2
+ Mr1R + L>222
1 2 (where the parallel axis through the CM of the rod is
= 12 MrL a distance of R + L>2 from the axis of rotation)
12 15.0 kg210.80 m22 + 15.0 kg210.20 m + 0.40 m22 = 2.1 kg # m2
= 1 (b) Arms overhead
And, Io = Ic + 2Ir = 1.5 kg # m2 + 212.1 kg # m22 = 5.7 kg # m2 䉱 F I G U R E 8 . 2 9 Skater model.

(continued on next page) Change in moment of inertia and
spin. See Example 8.18.
After: In Fig. 8.29b, treating an arm mass as if its center of mass is now only about
20 cm from the axis of rotation, the moment of inertia of each arm is I = Mr R 2
(Fig. 8.20b), and,
I = Ic + 21Mr R22 = 1.5 kg # m2 + 215.0 kg # m2210.20 m22 = 1.9 kg # m2
Then with the conservation of angular momentum, L = Lo or Iv = Iovo and
5.7 kg # m2
v = a bvo = a b14.2 rad>s2 = 13 rad>s
Ia
Ib 1.9 kg # m2
So the angular speed increases by a factor of 3.
F O L L O W - U P E X E R C I S E . Suppose a skater with 75% of the mass of the skater in the
Exercise did a spin. What would be the spin rate v in this case? (Consider all masses to
be reduced by 75%.)
B
Angular momentum, L, is a vector, and when it is conserved or constant, its
magnitude and direction must remain unchanged. Thus, when no external torques
B
act, the direction of L is fixed in space. This is the principle behind passing a foot-
ball accurately, as well as that behind the movement of a gyrocompass (䉲 Fig. 8.30).
A football is normally passed with a spiraling rotation. This spin, or gyroscopic
action, stabilizes the ball’s spin axis in the direction of motion. Similarly, rifle bul-
lets are set spinning by the rifling in the barrel for directional stability.
B
The L vector of a spinning gyroscope in the compass is set in a particular direc-
tion (usually north). In the absence of external torques, the compass direction
remains fixed, even though its carrier (an airplane or ship, for example) changes
䉴 F I G U R E 8 . 3 0 Constant direc- Vertical

tion of angular momentum When v axis
angular momentum is conserved,
Horizontal
its direction is constant in space. v
axis
(a) This principle can be demon-
strated by a passed football. Spin
(b) Gyroscopic action also occurs in axis
a gyroscope, a rotating wheel that is L L
universally mounted on gimbals
(rings) so that it is free to turn about
any axis. When the frame moves,
the wheel maintains its direction.
This is the principle of the
gyrocompass.
(a) v
(b)
directions. You may have played with a toy gyroscope that Polaris
is set spinning and placed on a pedestal. In a “sleeping”
L vp vp
condition, the gyro stands straight up with its angular
momentum vector fixed in space for some time. The gyro’s
center of gravity is on the axis of rotation, so there is no net
23 2 °
1
torque due to its weight. Lspin
v
However, the gyroscope eventually slows down because
B
of friction, causing L to tilt. In watching this motion, you N
may have noticed that the spin axis revolves, or precesses, mg
about the vertical axis. It revolves tilted over, so to speak Lorbital
r⊥
(Fig. 8.30b). Since the gyroscope precesses, the angular
B
momentum vector L is no longer constant in direction, indi-
B
cating that a torque must be acting to produce a change ( ¢L) v S
with time.
As can be seen from the figure, the torque arises from (a) (b)
the vertical component of the weight force, since the cen-
ter of gravity no longer lies directly above the point of support or on the vertical 䉱 F I G U R E 8 . 3 1 Precession An
external torque causes a change in
axis of rotation. The instantaneous torque is such that the gyroscope’s axis moves angular momentum. (a) For a spin-
or precesses about the vertical axis. ning gyroscope, this change is direc-
In a similar manner, the Earth’s rotational axis precesses. The Earth’s spin axis tional, and the axis of rotation
is tilted 23 12° with respect to a line perpendicular to the plane of its revolution precesses at angular acceleration vp
about the Sun; the axis precesses about this line (䉴 Fig. 8.31). The precession is due about a vertical line. (The torque
due to the weight force would
to slight gravitational torques exerted on the Earth by the Sun and the Moon. point out of the page as drawn
The period of the precession of the Earth’s axis is about 26 000 years, so the pre- B
here, as would ¢L) Note that
cession has little day-to-day effect. However, it does have an interesting long-term although there is a torque that
effect. Polaris will not always be (nor has it always been) the North Star—that is, would topple a nonspinning gyro-
the star toward which the Earth’s axis of rotation points. About 5000 years ago, scope, a spinning gyroscope doesn’t
fall. (b) Similarly, the Earth’s axis
Alpha Draconis was the North Star, and 5000 years from now it will be Alpha precesses because of gravitational
Cephei, which is at an angular distance of about 68° away from Polaris on the cir- torques caused by the Sun and the
cle described by the precession of the Earth’s axis. Moon. We don’t notice this motion
There are some other long-term torque effects on the Earth and the Moon. Did because the period of precession is
you know that the Earth’s daily spin rate is slowing down and hence the days are about 26 000 years.
getting longer? Also, that the Moon is receding, or getting farther away, from the
Earth? This is due primarily to ocean tidal friction, which gives rise to a slowing
torque. As a result, the Earth’s spin angular momentum, and therefore its rate of
rotation, is changing. The slowing rate of rotation causes the average day to be
longer. This century will be about 25 s longer than the previous century.
But this is an average rate. At times, the Earth’s rotation speeds up for relatively
short periods. This increase is thought to be associated with the rotational inertia
of the liquid layer of the Earth’s core. (See the Chapter 13 Insight 13.1, Earth-
quakes, Seismic Waves, and Seismology.)
The tidal torque on the Earth results chiefly from the Moon’s gravitational
attraction, which is the main cause of ocean tides. This torque is internal to the
Earth–Moon system, so the total angular momentum of that system is conserved.
Since the Earth is losing angular momentum, the Moon must be gaining angular
momentum to keep the total angular momentum of the system constant. The
Earth loses rotational (spin) angular momentum, and the Moon gains orbital
angular momentum. As a result, the Moon drifts slightly farther from Earth and
its orbital speed decreases. The Moon moves away from the Earth at about 4 cm
per year. Thus, the Moon moves in a slowly widening spiral.
Finally, a common example in which angular momentum is an important consid-
eration is the helicopter. What would happen if a helicopter had a single rotor? Since
the motor supplying the torque is internal, the angular momentum would be con-
B
served. Initially, L = 0; hence, to conserve the total angular momentum of the system
(rotor plus body), the separate angular momenta of the rotor and body would have
to be in opposite directions to cancel. Thus, on takeoff, the rotor would rotate one
way and the helicopter body the other, which is not a desirable situation.
Front Rear
rotor rotor
L –L
(top view)
(a) ∑L =0
Direction of main rotor
Thrust of Reaction force of main

tail rotor rotor on body of copter
on copter
(b)
䉱 F I G U R E 8 . 3 2 Different rotors
See text for description. To prevent this situation, helicopters have two rotors. Large helicopters have
two overlapping rotors (䉱 Fig. 8.32a). The oppositely rotating rotors cancel each
other’s angular momenta, so the helicopter body does not have to rotate to pro-
vide canceling angular momentum. The rotors are offset at different heights so
that they do not collide.
Small helicopters with a single overhead rotor have small “antitorque” tail
rotors (Fig. 8.32b). The tail rotor produces a thrust like a propeller and supplies the
torque to counterbalance the torque produced by the overhead rotor. The tail rotor
also helps in steering the craft. By increasing or decreasing the tail rotor’s thrust,
the helicopter turns (rotates) one way or the other.
DID YOU LEARN?

➥ The net torque is equal to the time rate of change of angular momentum,
Tnet = ¢L>¢t. The net force is equal to the time rate of change of linear
B B
momentum, Fnet = ¢P>¢t.

B B
➥ The angular momentum is conserved in the absence of a net torque,

Tnet = ¢L>¢t = 0, and ¢L = 0. There is no change in the angular momentum, so
B B B
it is conserved.
PULLING IT TOGETHER Making a “Lazy Susan” Lazier?

“Lazy Susan” is the name for a small rotatable disk placed in Before the mass lands, (a) what are the frequency and angular
the center of a table for the easy delivery of appetizers or speed of the lazy Susan? (b) What is the moment of inertia of
condiments to those sitting around the table. Assume such a the disk about its central axis? (c) What is its kinetic energy
lazy Susan has a frictionless axis though its center on which to and angular momentum? (d) After the small mass lands,
rotate, and consists of a circular piece of wood (density of where is the new center of mass located? (e) What is the sys-
700 kg>m3) that is 60.0 cm in diameter and 1.00 cm thick. It is tem’s final kinetic energy and angular momentum? Is either
set spinning initially so that it makes one complete revolution conserved? Explain.
in 5.00 s. A small 100-g mass (piece of food) is dropped from T H I N K I N G I T T H R O U G H . This example demonstrates the con-
just above the perimeter of the rotating disk and it sticks to cepts of angular momentum, moment of inertia, rotational
the surface. (Neglect the speed of the mass as it hits the disk.) kinetic energy, and center of mass. (a) The frequency and
angular speed are inversely related to the period, which is toward the perimeter. The exact location can be determined by
given, and thus can be directly determined. (b) The moment of recalling the definition of center of mass from Section 8.4.
inertia depends on the mass and radius of the disk. The mass (e) Angular momentum is conserved because the net torque on
is determined by its volume and density. (c) Once the moment the system is zero (why?). This enables the determination of the
of inertia and angular speed are known, the rotational kinetic final (slower) angular speed and the final kinetic energy. It is
energy and angular momentum follow directly. (d) The sticky expected that the final kinetic energy will be less than the initial
mass will move the center of mass from the center of the disk kinetic energy due to the inelastic collision that takes place.
SOLUTION.
Given: m = 100 g = 0.100 kg Find: (a) f (frequency) and v (angular speed)

d = 60.0 cm = 0.600 m (diameter) (b) I (moment of inertia)
h = 1.00 cm = 0.0100 m (thickness) (c) Ko (initial kinetic energy) and Lo (angular momentum)
r = 700 kg>m3 (d) location of center of mass
T = 5.00 s (period) (e) Kf (final kinetic energy) and Lf (final angular momentum)
Are they conserved?
(a) Both the initial frequency and angular speed can be deter- (d) The center of mass is initially at the geometric center of
mined from the period (T), which is the time for one complete the disk. With the small mass on the perimeter, the center of
rotation of the disk. Thus the initial frequency is mass moves radially outward from the center toward the
1 perimeter a distance xcm . This is determined by treating the
f = uniform disk as if all of its mass (M) were at its center of mass
1x = 02 in combination with a point mass (m) located at
T
=
1 rev
= 0.200 Hz 1x = R = 0.300 m2:
10.100 kg210.300 m2
5.00 s
mR + M102
and the initial angular speed is xCM = = = 1.44 * 10-2 m
m + M 0.100 kg + 1.98 kg
vo = 2pf
= 2p10.200 Hz2 = 1.26 rad>s (e) Because the system of the disk and small mass has a net
torque of zero on it (each has an equal but opposite torque on
(b) The moment of inertia (I) of a circular disk is 12 MR2 (see it at the collision), the net torque on the system is zero. Thus
Fig. 8.20 ). To find the mass from the density, the disk volume the angular momentum of the system stays constant, and
is needed: Lo = Lf or Io vo = If vf . This can be used to find the final angu-
lar speed if the final moment of inertia is known. This is just
V = pR2 h
the original disk moment of inertia plus a term for a point
= p10.300 m2210.0100 m2 = 2.83 * 10-3 m3 mass located on the perimeter. Therefore,
Then the mass is, If = Io + mR2
= 8.91 * 10-2 kg # m2 + 10.100 kg210.300 m22
M = rV
= 1700 kg>m3212.83 * 10-3 m32
= 9.81 * 10-2 kg # m2
= 1.98 kg Now the final angular speed can be found using angular
momentum conservation:
Finally, the initial moment of inertia is
Io
vf = v
Io = 12 MR 2 If o
= 12 11.98 kg210.300 m22 18.91 * 10-2 kg # m22
11.26 rad>s2
19.81 * 10-2 kg # m22
=
= 8.91 * 10-2 kg # m2
= 1.14 rad>s
(c) The initial rotational kinetic energy and angular momentum and the final rotational kinetic energy is
are
Kf = 12 If vf2
= 12 19.81 * 10-2 kg # m2211.14 rad>s22
Ko = 12 Io vo2
= 12 18.91 * 10-2 kg # m2211.26 rad>s22 = 6.37 * 10-2 J
= 7.07 * 10-2 J
This amounts to a system loss of about 10%, so the kinetic
and energy is not conserved. A loss of kinetic energy is expected,
since this was an inelastic collision. Some of the energy was
Lo = Io vo
converted into the heating of the small ball and also perhaps
= 18.91 * 10-2 kg # m2211.26 rad>s2 sound. However, since there are no external torques, the
= 0.112 kg # m2>s angular momentum is conserved (Section 8.5 ).
■ In pure translational motion, all of the particles that make up ■ Mechanical equilibrium requires that the net force, or sum-
a rigid object have the same instantaneous velocity. mation of the forces, be zero (translational equilibrium) and
that the net torque, or summation of the torques, be zero
Translational
v
(rotational equilibrium).
Conditions for translational and rotational mechanical

equilibrium, respectively:
Fnet = gFi = 0 and Tnet = gTi = 0
v B B B B
(8.3)
■ An object is in stable equilibrium as long as its center of
gravity, upon small displacement, lies above and inside the
v
object’s original base of support.
■ In pure rotational motion (about a fixed axis), all of the

particles that make up a rigid object have the same instanta-
CG
neous angular velocity.
Rotational
v = rv
Balanced on a broad Disturbance produces
base of support restoring torque
Stable Equilibrium
r
■ Moment of inertia (I) is the rotational analogue of mass and
is given by
I = gmi r i
2
v = rv
(8.6)
Rotational form of Newton’s second law:

Condition for rolling without slipping: B B
Tnet = IA (8.7)
vCM = rv (8.1)
Parallel axis theorem:
(or s = ru or aCM = ra)
I = ICM + Md 2 (8.8)
Rolling
2v
M CM
v
d
I = ICM + Md2
Point of
v =0 Rotational work:
contact
W = tu (8.9)
B
■ Torque (T), the rotational analogue of force, is the product Rotational power:
of a force and a moment arm, or lever arm.
P = tv (8.10)
Torque (magnitude):
t = r⬜ F = rF sin u (8.2) Work–energy theorem (rotational):
(Direction given by right-hand rule.) Wnet = 12 Iv2 - 12 Iv2o = ¢K (8.11)
Rotational kinetic energy:

Force
line of
action
r⊥
K = 12 Iv2 (8.12)
u
r u F⊥ Kinetic energy of a rolling object (no slipping):
Axis
F
2
t = r⊥F K = 12 ICMv2 + 12 Mv CM (8.13)
■ Angular momentum: The product of a moment arm and Torque as change in angular momentum (vector form):
linear momentum or the product of a moment of inertia and B
angular velocity. ¢L
Tnet =
B
(8.17)
¢t
Angular momentum of a particle in circular motion
Conservation of angular momentum (with Tnet = 0):
B
(magnitude):
2
L = r⬜ p = mr⬜ v = mr⬜ v (8.14) L = Lo or Iv = Io vo (8.18)
Angular momentum of a rigid body:

B B
L = IV (8.16)
8.1 RIGID BODIES, TRANSLATIONS, (b) energy due to rotation, (c) rate of change of linear
AND ROTATIONS momentum, or (d) force that is tangent to a circle?
11. In general, the moment of inertia is greater when
1. In pure rotational motion of a rigid body, (a) all the parti-
(a) more mass is farther from the axis of rotation,
cles of the body have the same angular velocity, (b) all
(b) more mass is closer to the axis of rotation, (c) it makes
the particles of the body have the same tangential veloc-
no difference.
ity, (c) acceleration is always zero, (d) there are always
two simultaneous axes of rotation. 12. A solid sphere (radius R) and an annular cylinder
(radius 2R) with equal masses are released simultane-
2. For an object with only rotational motion, all particles of
ously from the top of a frictionless inclined plane. Then,
the object have the same (a) instantaneous velocity,
(a) the sphere reaches the bottom first, (b) the cylinder
(b) average velocity, (c) distance from the axis of rotation,
reaches the bottom first, (c) they reach the bottom
(d) instantaneous angular velocity.
together.
3. The condition for rolling without slipping is
(a) ac = rv2, (b) vCM = rv, (c) F = ma, (d) ac = v2>r.
13. The moment of inertia about an axis parallel to the axis
through the center of mass depends on (a) the mass of
4. A rolling object (a) has an axis of rotation through the axis the rigid body, (b) the distance between the axes, (c) the
of symmetry, (b) has a zero velocity at the point or line of moment of inertia about the axis through the center of
contact, (c) will slip if s = ru, (d) all of the preceding. mass, (d) all of the preceding.
5. For the tires on your rolling, but skidding car,
(a) vCM = rv, (b) vCM 7 rv, (c) vCM 6 rv, (d) none of
the preceding. 8.4 ROTATIONAL WORK AND KINETIC
ENERGY
14. From W = tu, the unit of rotational work is the (a) watt,
8.2 TORQUE, EQUILIBRIUM, AND
(b) N # m, (c) kg # rad>s2, (d) N # rad.
STABILITY
15. A bowling ball rolls without slipping on a flat surface.
6. It is possible to have a net torque when (a) all forces act The ball has (a) rotational kinetic energy, (b) translational
through the axis of rotation, (b) g Fi = 0, (c) an object is
B
kinetic energy, (c) both translational and rotational
in rotational equilibrium, (d) an object remains in unsta- kinetic energy, (d) neither translational nor rotational
ble equilibrium. kinetic energy.
7. If an object in unstable equilibrium is displaced slightly, 16. A rolling cylinder on a level surface has (a) rotational
(a) its potential energy will decrease, (b) the center of kinetic energy, (b) translational kinetic energy, (c) both
gravity is directly above the axis of rotation, (c) no gravi- translational and rotational kinetic energies.
tational work is done, (d) stable equilibrium follows.
8. Torque has the same units as (a) work, (b) force,
8.5 ANGULAR MOMENTUM
(c) angular velocity, (d) angular acceleration.
17. The units of angular momentum are (a) N # m,
(b) kg # m>s2, (c) kg # m2>s, (d) J # m.
8.2 ROTATIONAL DYNAMICS
18. The Earth’s orbital speed is greatest about (a) March 21,
9. The moment of inertia of a rigid body (a) depends on the (b) June 21, (c) Sept. 21, (d) Dec. 21.
axis of rotation, (b) cannot be zero, (c) depends on mass 19. The angular momentum may be increased by
distribution, (d) all of the preceding. (a) decreasing the moment of inertia, (b) decreasing the
10. Which of the following best describes the physical quan- angular velocity, (c) increasing the product of the angu-
tity called torque: (a) rotational analogue of force, lar momentum and moment of inertia, (d) none of these.
8.1 RIGID BODIES, TRANSLATIONS, 8.3 ROTATIONAL DYNAMICS

AND ROTATIONS 10. (a) Does the moment of inertia of a rigid body depend in
1. Suppose someone in your physics class says that it is any way on the center of mass of the body? Explain.
possible for a rigid body to have translational motion (b) Can a moment of inertia have a negative value? If so,
and rotational motion at the same time. Would you explain what this would mean.
agree? If so, give an example. 11. Why does the moment of inertia of a rigid body have dif-
2. For a rolling cylinder, what would happen if the tangen- ferent values for different axes of rotation? What does
tial speed v were less than rv? Is it possible for v to be this mean physically?
greater than rv? Explain. 12. Two cylinders of equal mass sitting on a horizontal sur-
3. If the top of your automobile tire is moving with a speed face are made from materials with different densities.
of v, what is the reading of your speedometer? (a) Which cylinder will have the greater moment of iner-
tia about an axis passing horizontally through the cen-
ter? (b) Which cylinder will have the greater moment of
8.2 TORQUE, EQUILIBRIUM, AND inertia about an axis along the surface of contact?
STABILITY 13. Here is an interesting experiment you can try for your-
4. A small force and a large force produce torques. Can you self at home. Prepare a hard-boiled egg and have a raw
tell which one will have the larger torque? Explain. egg available. Set them both spinning on the kitchen
table. Stop both eggs quickly, and the release both. You
5. In cutting large trees, loggers first notch or make a V-cut
will notice the hard-boiled one remains at rest, whereas
on the side of the tree in the desired direction of fall and
the raw one starts spinning again. Explain.
then cut from the other side. Why is this? Is there any
14. Why does jerking a paper towel from a roll cause the
danger for the logger to be on the opposite side of the
paper to tear more easily than pulling it smoothly? Will
tree of the direction of fall?
the amount of paper on the roll affect the results?
6. Explain the balancing acts in 䉲 Fig. 8.33. Where are the
15. Tightrope walkers are continually in danger of falling
centers of gravity?
(unstable equilibrium). Commonly, a performer carries a
long pole while walking the tight rope, as shown in the
chapter-opening photo. What is the purpose of the pole?
(In walking along a narrow board or rail, you probably
extend your arms for the same reason.)
16. A solid cylinder and an annular cylinder of equal mass
are rolling on the floor with the same speed. (a) If the
solid cylinder’s radius is equal to the annular cylinder’s
inner radius, which cylinder would be harder to stop?
Explain. (b) Would it make any stopping difference if the
solid cylinder’s radius were equal to the annular cylin-
der’s outer radius? Justify your answer explicitly.
8.4 ROTATIONAL WORK AND KINETIC

ENERGY
17. Can you increase the rotational kinetic energy of a wheel
䉱 F I G U R E 8 . 3 3 Balancing acts Left: A toothpick on the rim without changing its translational kinetic energy? Explain.
of the glass supports a fork and spoon. Right: Toy birds balance 18. In order to produce fuel-efficient vehicles, automobile
on their beaks. See Conceptual Question 6. manufacturers want to minimize rotational kinetic
energy and maximize translational kinetic energy when
7. “Popping a wheelie” is a motorcycle stunt in which the a car is traveling. If you were the designer of wheels of a
front end of the cycle rises up from the ground on a fast certain diameter, how would you design them?
start, and can remain there for some distance. Explain 19. What is required to produce a change in rotational
the physics involved in this stunt. kinetic energy?
8. A yo-yo is thrown downward with a rotational spin. 8.5 ANGULAR MOMENTUM

Reaching the bottom of the string, it climbs back 20. A child stands on the edge of a rotating playground
upward. Is the rotational direction reversed at the bot- merry-go-round (the hand-driven type). He then starts
tom? Explain. to walk toward the center of the merry-go-round. This
9. In the cases of both stable and unstable equilibrium, a small can result in a dangerous situation. Why?
displacement of the center of gravity causes gravitational 21. The release of vast amounts of carbon dioxide may result
work to be done. (See the balls and bowls in Fig. 8.11.) in an increase in the Earth’s average temperature through
However, there is another type of equilibrium in which the the so-called greenhouse effect and cause melting of the
displacement of the center of mass involves no gravita- polar ice caps. If this occurred and the ocean level rose sub-
tional work and the displaced center of gravity essentially stantially, what effect would it have on the Earth’s rotation?
moves in a straight line. This is called neutral equilibrium. 22. In the classroom demonstration illustrated in 䉴 Fig. 8.34,
Give an example of an object in neutral equilibrium. a person on a rotating stool holds a rotating bicycle
EXERCISES 303
wheel by handles attached to the wheel. When the wheel

is held horizontally, she rotates one way (clockwise as
viewed from above). When the wheel is turned over, she
rotates in the opposite direction. Explain why this
occurs. [Hint: Consider angular momentum vectors.]
v v
L
䉱 F I G U R E 8 . 3 4 Faster rotation See Conceptual Question 22.
23. Cats usually land on their feet when they fall, even if
held upside down when dropped (䉴 Fig. 8.35). While a
cat is falling, there is no external torque and its center of
mass falls as a particle. How can cats turn themselves
over while falling?
24. Two ice skaters that weigh the same skate toward each
other with the same mass and same speed on parallel 䉱 F I G U R E 8 . 3 5 A double rotation
paths. As they pass each other, they link arms. (a) What is See Conceptual Question 23.
the velocity of their center of mass after they link arms?
(b) What happens to their initial, translational kinetic
energies?
EXERCISES
8.1 RIGID BODIES, TRANSLATIONS, 3.60 m>s. The bowler estimates that it makes about 7.50
AND ROTATIONS complete revolutions in 2.00 seconds. Is it rolling with-
out slipping? Prove your answer, assuming that the
1. ●A wheel rolls uniformly on level ground without slip-
bowler’s quick observation limits answers to two
ping. A piece of mud on the wheel flies off when it is at
significant figures.
the 9 o’clock position (rear of wheel). Describe the subse-
quent motion of the mud. 5. ●● A ball with a radius of 15 cm rolls on a level surface,
2. ● A rope goes over a circular pulley with a radius of and the translational speed of the center of mass is
6.5 cm. If the pulley makes 4 revolutions without the rope 0.25 m>s. What is the angular speed about the center of
slipping, what length of rope passes over the pulley? mass if the ball rolls without slipping?
3. ● A wheel rolls 5 revolutions on a horizontal surface 6. IE ● ● (a) When a disk rolls without slipping, should the
without slipping. If the center of the wheel moves 3.2 m, product rv be (1) greater than, (2) equal to, or (3) less
what is the radius of the wheel? than vCM ? (b) A disk with a radius of 0.15 m rotates
4. ● ● A bowling ball with a radius of 15.0 cm travels through 270° as it travels 0.71 m. Does the disk roll with-
down the lane so that its center of mass is moving at out slipping? Prove your answer.
7. ● ● ● A bocce ball with a diameter of 6.00 cm rolls with- has a mass of 100 kg and is 1.6 m tall and 0.80 m in depth
out slipping on a level lawn. It has an initial angular and width, what is the minimum force needed to make
speed of 2.35 rad>s and comes to rest after 2.50 m. the crate start tipping? (Assume the center of mass of the
Assuming constant deceleration, determine (a) the mag- crate is at its center and static friction great enough to
nitude of its angular deceleration and (b) the magnitude prevent slipping.)
of the maximum tangential acceleration of the ball’s sur- 16. ● ● Show that the balanced meterstick in Example 8.5 is
face (tell where that part is located). in static rotational equilibrium about a horizontal axis
8. ● ● ● A cylinder with a diameter of 20 cm rolls with an through the 100-cm end of the stick.
angular speed of 0.050 rad>s on a level surface. If the 17. IE ● ● Telephone and electrical lines are allowed to sag
cylinder experiences a uniform tangential acceleration of between poles so that the tension will not be too great
0.018 m>s2 without slipping until its angular speed is when something hits or sits on the line. (a) Is it possible
1.2 rad>s, through how many complete revolutions does to have the lines perfectly horizontal? Why or why not?
the cylinder rotate during the time it accelerates? (b) Suppose that a line were stretched almost perfectly
horizontally between two poles that are 30 m apart. If a
8.2 TORQUE, EQUILIBRIUM, AND 0.25-kg bird perches on the wire midway between the
STABILITY poles and the wire sags 1.0 cm, what would be the ten-
sion in the wire? (Neglect the mass of the wire.)
9. ● In Fig. 8.4a, if the arm makes a 37° angle with the hori-
zontal and a torque of 18 m # N is to be produced, what 18. ● ● In 䉲 Fig. 8.37, what is the force Fm supplied by the
force must the biceps muscle supply? deltoid muscle so as to hold up the outstretched arm if
the mass of the arm is 3.0 kg? (Fj is the joint force on the
10. ● The drain plug on a car’s engine has been tightened to a
bone of the upper arm—the humerus.)
torque of 25 m # N. If a 0.15-m-long wrench is used to
change the oil, what is the minimum force needed to
loosen the plug?
11. ● In Exercise 10, due to limited work space, you must
crawl under the car. The force thus cannot be applied per-
pendicularly to the length of the wrench. If the applied
force makes a 30° angle with the length of the wrench,
what is the force required to loosen the drain plug?
12. ● How many different positions of stable equilibrium Fm
and unstable equilibrium are there for a cube? Consider
each surface, edge, and corner to be a different position. 9.4° 15°
Fj
13. IE ● ● Two children are sitting on opposite ends of a uni-
form seesaw of negligible mass. (a) Can the seesaw be 18 cm mg
balanced if the masses of the children are different? 26 cm
How? (b) If a 35-kg child is 2.0 m from the pivot point
(or fulcrum), how far from the pivot point will her 30-kg 䉱 F I G U R E 8 . 3 7 Arm in static equilibrium See
playmate have to sit on the other side for the seesaw to Exercise 18.
be in equilibrium?
14. ● A uniform meterstick pivoted at its center, as in
19. ●●In Figure 8.4b, determine the force exerted by the
Example 8.5, has a 100-g mass suspended at the 25.0-cm bicep muscle, assuming that the hand is holding a ball
position. (a) At what position should a 75.0-g mass be with a mass of 5.00 kg. Assume that the mass of the
suspended to put the system in equilibrium? (b) What forearm is 8.50 kg with its center of mass located
mass would have to be suspended at the 90.0-cm posi- 20.0 cm away from the elbow joint (the black dot in the
tion for the system to be in equilibrium? figure). Assume also that the center of mass of the ball
15. ● ● A worker applies a horizontal force to the top edge of in the hand is 30.0 cm away from the elbow joint.
a crate to get it to tip forward (䉲 Fig. 8.36). If the create (The muscle contact is 4.00 cm from the elbow joint;
Example 8.2.)
䉳 FIGURE 8.36 20. ● ● A bowling ball (mass 7.00 kg and radius 17.0 cm) is
Tip it over See released so fast that it skids without rotating down the
F Exercise 15.
lane (at least for a while). Assume the ball skids to the
right and the coefficient of sliding friction between the
ball and the lane surface is 0.400. (a) What is the direc-
tion of the torque exerted by the friction on the ball
about the center of mass of the ball? (b) Determine the
magnitude of this torque (again about the ball’s center
1.6 m of mass).
21. ● ● A variation of Russell traction (䉴 Fig. 8.38) supports
the lower leg in a cast. Suppose that the patient’s leg
and cast have a combined mass of 15.0 kg and m1 is
4.50 kg. (a) What is the reaction force of the leg muscles
to the traction? (b) What must m2 be to keep the leg
0.8 m horizontal?
EXERCISES 305
1.6 m
xCG
m2
CG
30 kg 25 kg
m1
䉱 F I G U R E 8 . 4 1 Locating the center of gravity See

䉱 F I G U R E 8 . 3 8 Static traction See Exercise 21. Exercise 24.
22. ●● In doing physical therapy for an injured knee joint, a Why? (b) Locate the center of gravity of the person rela-
person raises a 5.0-kg weighted boot as shown in tive to the horizontal dimension.
䉲 Fig. 8.39. Compute the torque due to the boot for each 25. ● ● (a) How many uniform, identical textbooks of width
position shown. 25.0 cm can be stacked on top of each other on a level sur-
face without the stack falling over if each successive book
is displaced 3.00 cm in width relative to the book below it?
(b) If the books are 5.00 cm thick, what will be the height
of the center of mass of the stack above the level surface?.
90⬚ 26. ● ● If four metersticks were stacked on a table with
10 cm, 15 cm, 30 cm, and 50 cm, respectively, hanging
40 cm over the edge, as shown in 䉲 Fig. 8.42, would the top
meterstick remain on the table?
60⬚
0 50 cm 100 cm
0 70 cm
30⬚ 0 85 cm
m ⫽ 5.0 kg 0⬚ 0 90 cm
䉱 F I G U R E 8 . 3 9 Torque in physical
therapy See Exercise 22.
23. ●● An artist wishes to construct a birds and bees mobile,

as shown in 䉲 Fig. 8.40. If the mass of the bee on the
lower left is 0.10 kg and each vertical support string has
a length of 30 cm, what are the masses of the other birds 䉱 F I G U R E 8 . 4 2 Will they fall off? See Exercise 26.
and bees? (Neglect the masses of the bars and strings.)
27. ●●A 10.0-kg solid uniform cube with 0.500-m sides rests
on a level surface. What is the minimum amount of work
necessary to put the cube into an unstable equilibrium
position?
30 cm 15 cm 28. ● ● While standing on a long board resting on a scaffold, a
70-kg painter paints the side of a house, as shown in
䉲 Fig. 8.43. If the mass of the board is 15 kg, how close to the
m4
end can the painter stand without tipping the board over?
25 cm 15 cm
m3
40 cm 20 cm
m1 = 0.10 kg m2
1.5 m 2.5 m 1.5 m
䉱 F I G U R E 8 . 4 0 Birds and bees See Exercise 23.
24. IE ● ● The location of a person’s center of gravity relative

to his or her height can be found by using the arrange-
ment shown in 䉴 Fig. 8.41. The scales are initially
adjusted to zero with the board alone. (a) Would you
expect the location of the center of gravity to be (1) mid-
way between the scales, (2) toward the scale at the per-
son’s head, or (3) toward the scale at the person’s feet? 䉱 F I G U R E 8 . 4 3 Not too far! See Exercise 28.
29. ●● A mass is suspended by two cords as shown in 36. ● For the system of masses shown in 䉲 Fig. 8.46, find the
䉲 Fig. 8.44. What are the tensions in the cords? moment of inertia about (a) the x-axis, (b) the y-axis,
and (c) an axis through the origin and perpendicular to
䉳 FIGURE 8.44 the page (z-axis). Neglect the masses of the connecting
A lot of tension See rods.
cord 1 Exercises 29 and 30.
2.00 kg y 3.00 kg
45°
30° cord 2
3.00 m x
O
m
1.5 kg
30. If the cord attached to the vertical wall in Fig. 8.44

●● 1.00 kg
were horizontal (instead of at a 30° angle), what would 4.00 kg
the tensions in the cords be?
31. ● ● A force is applied to a cord wrapped around a solid 5.00 m
2.0-kg cylinder as shown in 䉲 Fig. 8.45. Assuming the
cylinder rolls without slipping, what is the force of fric- 䉱 F I G U R E 8 . 4 6 Moments of inertia about
tion acting on the cylinder? different axes See Exercise 36.
50 N 䉳 F I G U R E 8 . 4 5 No 37. ●●A 2000-kg Ferris wheel accelerates from rest to an

slipping See Exercise 31. angular speed of 20 rad>s in 12 s. Approximate the Ferris
wheel as a circular disk with a radius of 30 m. What is
15 cm the net torque on the wheel?
38. IE ● ● Two objects of different masses are joined by a
light rod. (a) Is the moment of inertia about the center of
32. IE ● ● ● In a circus act, a uniform board (length 3.00 m, mass the minimum or the maximum? Why? (b) If the
mass 35.0 kg) is suspended from a bungie-type rope at two masses are 3.0 kg and 5.0 kg and the length of the
one end, and the other end rests on a concrete pillar. rod is 2.0 m, find the moments of inertia of the system
When a clown (mass 75.0 kg) steps out halfway onto the about an axis perpendicular to the rod, through the cen-
board, the board tilts so the rope end is 30° from the hor- ter of the rod and the center of mass.
izontal and the rope stays vertical. (a) In which situation 39. ● ● Two masses are suspended from a pulley as shown
will the rope tension be larger: (1) the board without the in 䉲 Fig. 8.47 (the Atwood machine revisited; see
clown on it, (2) the board with the clown on it, or (3) you Chapter 4, Exercise 55). The pulley itself has a mass of
can’t tell from the data given? (b) Calculate the force
exerted by the rope in both situations.
䉳 F I G U R E 8 . 4 7 The
33. IE ● ● ● The forces acting on Einstein and the bicycle
Atwood machine revisited
(Fig. 2 of the Insight 8.1, Stability in Action) are the total See Exercise 39.
weight of Einstein and the bicycle (mg) at the center of
gravity of the system, the normal force (N) exerted by the
road, and the force of static friction ( fs) acting on the tires
due to the road. (a) If Einstein is to maintain balance, R
should the tangent of the lean angle u(tan u) be (1) greater tf
than, (2) equal to, or (3) less than fs>N? (b) The angle u in a
T1
the picture is about 11°. What is the minimum coefficient T2
of static friction ms between the road and the tires? (c) If the
radius of the circle is 6.5 m, what is the maximum speed of
Einstein’s bicycle? [Hint: The net torque about the center of T1
gravity must be zero for rotational equilibrium.] T2
8.3 ROTATIONAL DYNAMICS m1 a

34. A fixed 0.15-kg solid-disk pulley with a radius of
● a m2 m1g
0.075 m is acted on by a net torque of 6.4 m # N. What is
the angular acceleration of the pulley?
35. ● What net torque is required to give a uniform 20-kg
solid ball with a radius of 0.20 m an angular acceleration m2g
of 20 rad>s2?
EXERCISES 307
0.20 kg, a radius of 0.15 m, and a constant torque of 䉳 FIGURE 8.50

0.35 m # N due to the friction between the rotating pulley Unwinding with gravity
and its axle. What is the magnitude of the acceleration of See Exercise 44.
the suspended masses if m1 = 0.40 kg and m2 = 0.80 kg?
(Neglect the mass of the string.)
40. ●● To start her lawn mower, Julie pulls on a cord that is T T
wrapped around a pulley. The pulley has a moment of
inertia about its central axis of I = 0.550 kg # m2 and a
radius of 5.00 cm. There is an equivalent frictional torque R
impeding her pull of tf = 0.430 m # N. To accelerate the M
pulley at a = 4.55 rad>s2, (a) how much torque does
Julie need to apply to the pulley? (b) How much tension
must the rope exert? 45. ● ● ● A planetary space probe is in the shape of a cylin-
41. ●● For the system shown in 䉲 Fig. 8.48, m1 = 8.0 kg, der. To protect it from heat on one side (from the Sun’s
m2 = 3.0 kg, u = 30°, and the radius and mass of the rays), operators on the Earth put it into a “barbecue
pulley are 0.10 m and 0.10 kg, respectively. (a) What is mode,” that is, they set it rotating about its long axis. To
the acceleration of the masses? (Neglect friction and the do this, they fire four small rockets mounted tangentially
string’s mass.) (b) If the pulley has a constant frictional as shown in 䉲 Fig. 8.51 (the probe is shown coming
torque of 0.050 m # N when the system is in motion, what toward you). The object is to get the probe to rotate com-
is the acceleration of the masses? [Hint: Isolate the forces. pletely once every 30 s, starting from no rotation at all.
The tensions in the strings are different. Why?] They wish to do this by firing all four rockets for a cer-
tain length of time. Each rocket can exert a thrust of
50.0 N. Assume the probe is a uniform solid cylinder
with a radius of 2.50 m and a mass of 1000 kg and
a neglect the mass of each rocket engine. Determine the
T1 amount of time the rockets need to be fired.
m1
T2 䉳 FIGURE 8.51
Space probe in the
m2 “barbecue mode”
See Exercise 45.
u
䉱 F I G U R E 8 . 4 8 Inclined plane and pulley See Exercise 41.
42. ●● A meterstick pivoted about a horizontal axis through

the 0-cm end is held in a horizontal position and let go.
(a) What is the initial tangential acceleration of the 100-
cm position? Are you surprised by this result? (b) Which 46. IE ● ● ● A ball of radius R and mass M rolls down an
position has a tangential acceleration equal to the accel- incline of angle u. (a) For the ball to roll without slipping,
eration due to gravity? should the tangent of the maximum angle of incline
1tan u2 be equal to (1) 3 ms >2, (2) 5 ms >2, (3) 7 ms >2, or
(4) 9 ms >2? Here, ms is the coefficient of static friction.
43. ●● Pennies are placed every 10 cm on a meterstick. One
end of the stick is put on a table and the other end is held
(b) If the ball is made of wood and the surface is also
horizontally with a finger, as shown in 䉲 Fig. 8.49. If the
wood, what is the maximum angle of incline? [Hint: See
finger is pulled away, what happens to the pennies?
Table 4.1.]
8.4 ROTATIONAL WORK AND KINETIC

ENERGY
47. ●A constant retarding torque of 12 m # N stops a rolling
wheel of diameter 0.80 m in a distance of 15 m. How
much work is done by the torque?
48. ● A person opens a door by applying a 15-N force per-
䉱 F I G U R E 8 . 4 9 Money left behind? See Exercise 43.
pendicular to it at a distance 0.90 m from the hinges. The
door is pushed wide open (to 120°) in 2.0 s. (a) How
44. ● ● ● A uniform 2.0-kg cylinder of radius 0.15 m is sus- much work was done? (b) What was the average power
pended by two strings wrapped around it (䉴 Fig. 8.50). delivered?
As the cylinder descends, the strings unwind from it. 49. IE ● In Fig. 8.23, a mass m descends a vertical distance
What is the acceleration of the center of mass of the from rest. (Neglect friction and the mass of the string.)
cylinder? (Neglect the mass of the string.) (a) From the conservation of mechanical energy, will the
linear speed of the descending mass be (1) greater than, incline before coming to rest? (b) Do the radii of the balls
(2) equal to, or (3) less than 12gh ? Why? (b) If m = 1.0 kg, make a difference? (c) After stopping, the balls roll back
M = 0.30 kg, and R = 0.15 m, what is the linear speed of down the incline. By the conservation of energy, both
the mass after it has descended a vertical distance of 2.0 m balls should have the same speed when reaching the bot-
from rest? tom of the incline. Show this explicitly.
50. ● A constant torque of 10 m # N is applied to the rim of a 62. ● ● ● In a tumbling clothes dryer, the cylindrical drum
10-kg uniform disk of radius 0.20 m. What is the angular (radius 50.0 cm and mass 35.0 kg) rotates once every sec-
speed of the disk about an axis through its center after it ond. (a) Determine the rotational kinetic energy about its
rotates 2.0 revolutions from rest? central axis. (b) If it started from rest and reached that
51. ● A 2.5-kg pulley of radius 0.15 m is pivoted about an speed in 2.50 s, determine the average net torque on the
axis through its center. What constant torque is required dryer drum.
for the pulley to reach an angular speed of 25 rad>s after 63. ● ● ● A steel ball rolls down an incline into a loop-the-
rotating 3.0 revolutions, starting from rest? loop of radius R (䉲 Fig. 8.52a). (a) What minimum speed
52. ● ● A solid ball of mass m rolls along a horizontal surface must the ball have at the top of the loop in order to stay
with a translational speed of v. What percent of its total on the track? (b) At what vertical height (h) on the
kinetic energy is translational? incline, in terms of the radius of the loop, must the ball
53. ● ● Estimate the ratio of the translational kinetic energy
be released in order for it to have the required minimum
of the Earth as it orbits the Sun to the rotational kinetic speed at the top of the loop? (Neglect frictional losses.)
energy it has about its N–S axis. (c) Figure 8.52b shows the loop-the-loop of a roller
coaster. What are the sensations of the riders if the roller
54. ● ● You wish to accelerate a small merry-go-round from
coaster has the minimum speed or a greater speed at the
rest to a rotational speed of one-third of a revolution per
top of the loop? [Hint: In case the speed is below the
second by pushing tangentially on it. Assume the merry-
minimum, seat and shoulder straps hold the riders in.]
go-round is a disk with a mass of 250 kg and a radius of
1.50 m. Ignoring friction, how hard do you have to push r 䉳 FIGURE 8.52
tangentially to accomplish this in 5.00 s? (Use energy Loop-the-loop and
methods and assume a constant push on your part.) R rotational speed See
h
55. ● ● A pencil 18 cm long stands vertically on its point end Exercise 63.
on a horizontal table. If it falls over without slipping, with
what tangential speed does the eraser end strike the table?
(a)
56. ● ● A uniform sphere and a uniform cylinder with the
same mass and radius roll at the same velocity side by

side on a level surface without slipping. If the sphere
and the cylinder approach an inclined plane and roll up
it without slipping, will they be at the same height on the
plane when they come to a stop? If not, what will be the
percentage difference of the heights?
57. ● ● A hoop starts from rest at a height 1.2 m above the
base of an inclined plane and rolls down under the influ-

ence of gravity. What is the linear speed of the hoop’s
center of mass just as the hoop leaves the incline and (b)
rolls onto a horizontal surface? (Neglect friction.)
58. ● ● A cylindrical hoop, a cylinder, and a sphere of equal
8.5 ANGULAR MOMENTUM
radius and mass are released at the same time from the
top of an inclined plane. Using the conservation of 64. ● What is the angular momentum of a 2.0-g particle
mechanical energy, show that the sphere always gets to moving counterclockwise (as viewed from above) with
the bottom of the incline first with the fastest speed and an angular speed of 5p rad>s in a horizontal circle of
that the hoop always arrives last with the slowest speed. radius 15 cm? (Give the magnitude and direction.)
59. ● ● For the following objects, which all roll without slip- 65. ● A 10-kg rotating disk of radius 0.25 m has an angular
ping, determine the rotational kinetic energy about the momentum of 0.45 kg # m2>s What is the angular speed
center of mass as a percentage of the total kinetic energy: of the disk?
(a) a solid sphere, (b) a thin spherical shell, and (c) a thin 66. ● ● Compute the ratio of the magnitudes of the Earth’s
cylindrical shell. orbital angular momentum and its rotational angular
60. ● ● An industrial flywheel with a moment of inertia of momentum. Are these momenta in the same direction?
4.25 * 102 kg # m2 rotates with a speed of 7500 rpm. 67. ● ● The Earth revolves about the Sun and spins on its axis,
(a) How much work is required to bring the flywheel to which is tilted 23 1冫2° to its orbital plane. (a) Assuming a cir-
rest? (b) If this work is done uniformly in 1.5 min, how cular orbit, what is the magnitude of the angular momen-
much power is required? tum associated with the Earth’s orbital motion about the
61. ● ● ● A hollow, thin-shelled ball and a solid ball of equal Sun? (b) What is the magnitude of the angular momentum
mass are rolled up an inclined plane (without slipping) associated with the Earth’s rotation on its axis?
with both balls having the same initial velocity at the 68. ● ● The period of the Moon’s rotation is the same as the
bottom of the plane. (a) Which ball rolls higher on the period of its revolution: 27.3 days (sidereal). What is the
angular momentum for each rotation and revolution?

(Because the periods are equal, we see only one side of Comet
the Moon from Earth.)
69. IE ● ● Circular disks are used in automobile clutches and vo
transmissions. When a rotating disk couples to a station- b d
ary one through frictional force, the energy from the v
rotating disk can transfer to the stationary one. (a) Is the
angular speed of the coupled disks (1) greater than, Sun
(2) less than, or (3) the same as the angular speed of the
original rotating disk? Why? (b) If a disk rotating at
800 rpm couples to a stationary disk with three times the
moment of inertia, what is the angular speed of the 䉱 F I G U R E 8 . 5 4 A comet “collision” See Exercise 73.
combination?
70. ● ● An ice skater has a moment of inertia of 100 kg # m
2 approach (d) in terms of the impact parameter and the
when his arms are outstretched and a moment of inertia velocities (vo at large distances and v at closest
of 75 kg # m2 when his arms are tucked in close to his approach). Assume that the radius of the Sun is negligi-
chest. If he starts to spin at an angular speed of 2.0 rps ble compared to d. (As the figure shows, the tail of a
(revolutions per second) with his arms outstretched, comet always “points” away from the Sun.)
what will his angular speed be when they are tucked in? 74. ● ● ● While repairing his bicycle, a student turns it upside
71. ● ● An ice skater spinning with outstretched arms has an down and sets the front wheel spinning at 2.00 rev>s.
angular speed of 4.0 rad>s. She tucks in her arms, decreas- Assume the wheel has a mass of 3.25 kg and all of the
ing her moment of inertia by 7.5%. (a) What is the result- mass is located on the rim, which has a radius of 41.0 cm.
ing angular speed? (b) By what factor does the skater’s To slow the wheel, he places his hand on the tire, thereby
kinetic energy change? (Neglect any frictional effects.) exerting a tangential force of friction on the wheel. It takes
(c) Where does the extra kinetic energy come from? 3.50 s to come to rest. Use the change in angular momen-
72. ● ● A billiard ball at rest is struck (bold arrow in
tum to determine the force he exerts on the wheel.
䉲 Fig. 8.53) by a cue with an average force of 5.50 N last-
Assume the frictional force of the axle is negligible.
ing for 0.050 s. The cue contacts the ball’s surface so that 75. IE ● ● ● A kitten stands on the edge of a lazy Susan (a
the lever arm is half the radius of the ball, as shown. If turntable). Assume that the lazy Susan has frictionless
the cue ball has a mass of 200 g and a radius of 2.50 cm, bearings and is initially at rest. (a) If the kitten starts to
determine the angular speed of the ball immediately walk around the edge of the lazy Susan, the lazy Susan
after the blow. (Neglect friction.) will (1) remain lazy and stationary, (2) rotate in the direc-
tion opposite that in which the kitten is walking, or
1.25 cm (3) rotate in the direction the kitten is walking. Explain.
(b) The mass of the kitten is 0.50 kg, and the lazy Susan
has a mass of 1.5 kg and a radius of 0.30 m. If the kitten
walks at a speed of 0.25 m>s, relative to the ground, what
will be the angular speed of the lazy Susan? (c) When the
kitten has walked completely around the edge and is
䉱 F I G U R E 8 . 5 3 Cueing low See Exercise 72. back at its starting point, will that point be above the
same point on the ground as it was at the start? If not,
73. ●●● A comet approaches the Sun as illustrated in where is the kitten relative to the starting point? (Specu-
䉴 Fig.8.54 and is deflected by the Sun’s gravitational late on what might happen if everyone on the Earth sud-
attraction. This event is considered a collision, and b is denly started to run eastward. What effect might this
called the impact parameter. Find the distance of closest have on the length of a day?)
76. IE A small heavy object of mass m is attached to a thin Use the same procedure to show that F = mg tan u.
string to make a simple pendulum whose length is L. (c) Prove the same result for F as in part (b) using the
When the object is pulled aside by a horizontal force F it is torque condition, summing the torques about the
in static equilibrium and the string makes a constant string’s tied end. Explain why you cannot use this
angle u from the vertical. (a) The tension in the string method to determine the string tension.
should be (1) the same as, (2) greater than, or (3) less than
77. A bowling ball with a diameter of 21.6 cm is rolling down
the object’s weight, mg. (b) Use the force condition for sta-
a level alley surface at 12.7 m>s without slipping. Assume
tic equilibrium (along with a free-body diagram of the
mg the ball is uniform and made of plastic with a density of
object) to prove that the string tension is T = 7 mg. 800 kg>m3. (a) What is the angular speed of the ball?
cos u
(b) Calculate the speed (relative to the alley surface) of a 81. In a “modern art” exhibit, a multicolored empty indus-
point on top of the ball directly above the contact point trial wire spool is suspended from two light wires as
on the floor. (c) What is the ball’s linear kinetic energy? shown in 䉲 Fig. 8.56. The spool has a mass of 50.0 kg,
(d) If it now starts to roll up a 30° incline, how far up the with an outer diameter of 75.0 cm and an inner axle
incline will it travel before it stops? diameter of 18.0 cm. One wire (#1) is attached tangen-
78. A solid cylindrical 10-kg roll of roofing paper with a tially to the axle and makes a 10° angle with the vertical.
radius of 15 cm, starting from rest rolls down a roof with The other wire (#2) is attached tangentially to the outer
a 20° incline (䉲 Fig. 8.55). (a) If the cylinder rolls 4.0 m edge and makes an unknown angle u with the vertical.
without slipping, what is the angular speed about its Determine the tension in each wire and the angle u.
center when leaving the roof? (b) If the roof edge of the
house is 6.0 m above level ground, how far from the #1 䉳 FIGURE 8.56
edge of the roof does the cylindrical roll land? (Figure #2
10⬚ Modern art See
not to scale.) Exercise 81.
u
15 cm 䉳 FIGURE 8.55
Watch out below See
Exercise 78.
4.0 m
20⬚
82. A flat, solid cylindrical grinding wheel with a diameter

of 20.2 cm is spinning at 3000 rpm when its power is
suddenly turned off. A workman continues to press his
6.0 m tool bit toward the wheel’s center at the wheel’s circum-
ference so as to continue to grind as the wheel coasts to a
stop. If the wheel has a moment of inertia of 4.73 kg # m2,
(a) determine the necessary torque that must be exerted
by the workman to bring it to rest in 10.5 s. Ignore any
friction at the axle. (b) If the coefficient of kinetic friction
between the tool bit and the wheel surface is 0.85, how
79. A flat cylindrical grinding wheel is spinning at 2000 rpm hard must the workman push on the bit?
(clockwise when viewed head-on) when its power is sud- 83. A uniform sphere of mass 2.50 kg and radius 15.0 cm is
denly turned off. Normally, if left alone, it takes 45.0 s to released from rest at the top of an incline that is 5.25 m
coast to rest. Assume the grinder has a moment of inertia long and makes an angle of 35° with the horizontal.
of 2.43 kg # m2. (a) Determine its angular acceleration dur- Assuming it rolls without slipping, (a) determine its total
ing this process. (b) Determine the tangential acceleration kinetic energy at the bottom of the incline. (b) Determine
of a point on the grinding wheel if the wheel is 7.5 cm in its rotational kinetic energy at the bottom of the incline.
diameter. (c) The slowing down is caused by a frictional (c) What type of friction, static or kinetic, is acting on the
torque on the axle of the wheel. The axle is 1.00 cm in surface of the sphere? Explain. (d) Determine the force of
diameter. Determine the frictional force on the axle. friction in part (d).
(d) How much work was done by friction on the system? 84. A stationary ice skater with a mass of 80.0 kg and a
80. Modern bowling alleys have automatic ball returns. The moment of inertia (about her central vertical axis) of
ball is lifted to a height of 2.00 m at the end of the alley and, 3.00 kg # m2 catches a baseball with her outstretched arm.
starting from rest, rolls down a ramp. It continues to roll The catch is made at a distance of 1.00 m from the central
horizontally and eventually rolls up a ramp at the other end axis. The ball has a mass of 145 g and is traveling at
that is 0.500 m off the ground. Assuming the mass of the 20.0 m>s before the catch. (a) What linear speed does the
bowling ball is 7.00 kg and its radius is 16.0 cm, determine system (skater + ball) have after the catch? (b) What is the
(a) the rotation rate of the ball during the middle horizontal angular speed of the system (skater + ball) after the catch?
travel, (b) its linear speed during the middle horizontal (c) What percentage of the ball’s initial kinetic energy is
travel, and (c) the final rotation rate and linear speed. lost during the catch? Neglect friction with the ice.
9 Solids and Fluids
9.1 Solids and elastic
moduli (312)
■ stress
■ strain
9.2 Fluids: pressure and

Pascal’s principle (317)
■ force per unit area
■ pressure transmitted
undiminished
9.3 Buoyancy and Archimedes’

principle (328)
■ upward buoyant force
■ weight of fluid displaced
PHYSICS FACTS
9.4 Fluid dynamics and

Bernoulli’s equation (333)
■ equation of continuity
✦ The Mariana Trench in the Pacific
Ocean is the deepest known point
on the Earth. It is about 11 km
(6.8 mi) below sea level. At this
S hown in the chapter-opening
photo are solid cliffs, water, and
unseen air that makes gliding possi-
depth, the ocean water exerts a
■ work-energy
pressure of about 108 MPa
ble. We walk on the solid surface of
(15 900 lb/in2), or more than the Earth and in our daily lives use
1000 atmospheres of pressure.
*9.5Surface tension, viscosity, ✦ Legend has it that Archimedes,
solid objects of all sorts, from scissors
and Poiseuille’s law (338) who is credited with the principle to computers. But we are sur-
of buoyancy, was given the prob-
lem of determining whether the rounded by fluids—liquids and
king’s gold crown was pure gold or
gases—some of which are indis-
contained some silver. According
to a Roman account, the solution pensable. Without water, survival
came to him when he got into a
full bath. On immersing, he
would be for only a few days at most.
noticed that water overflowed the By far the most abundant substance
tub. Quantities of pure gold and
silver equal in weight to the king’s in our bodies is water, and it is in the
crown were each put into bowls
filled with water, and the silver
watery environment of our cells that
caused more water to overflow. all chemical processes on which life
When the crown was tested, more
water overflowed than for the depends take place. Also, the
pure gold, which implied some sil-
gaseous oxygen in the air is essential
ver content.
for life processes.
312 9 SOLIDS AND FLUIDS
On the basis of general physical distinctions, matter is commonly divided into

three phases: solid, liquid, and gas. A solid has a definite shape and volume. A liquid
has a fairly definite volume, but assumes the shape of its container. A gas takes on
the shape and volume of its container. Solids and liquids are sometimes called
condensed matter. In this chapter, a different classification scheme will be used and
matter will be considered in terms of solids and fluids. Liquids and gases are
referred to collectively as fluids. A fluid is a substance that can flow; liquids and
gases qualify, but solids do not.
A simplistic description of solids is that they are made up of particles called atoms
that are held rigidly together by interatomic forces. In Section 8.1, the concept of an
ideal rigid body was used to describe rotational motion. Real solid bodies are not
absolutely rigid and can be elastically deformed by external forces. Elasticity usually
brings to mind a rubber band or spring that will resume its original dimensions even
after being greatly deformed. In fact, all materials—even very hard steel—are elastic
to some degree. But, as will be learned, such deformation has an elastic limit.
Fluids, however, have little or no elastic response to a force. Instead, the force
merely causes an unconfined fluid to flow. This chapter pays particular attention to
the behavior of fluids, shedding light on such questions as how hydraulic lifts work,
why icebergs and ocean liners float, and what “10W-30” on a can of motor oil
means. You’ll also discover why the person in the chapter-opening photo can soar,
with the aid of a suitably shaped piece of plastic.
Because of their fluidity, liquids and gases have many properties in common, and
it is convenient to study them together. But there are important differences as well.
For example, liquids are not very compressible, whereas gases are easily compressed.
9.1 Solids and Elastic Moduli

➥ What do stress and strain measure?

➥ What is an elastic modulus?
➥ What are the types of elastic moduli associated with materials?
As stated previously, all solid materials are elastic to some degree. That is, a body
that is slightly deformed by an applied force will return to its original dimensions
or shape when the force is removed. The deformation may not be noticeable for
many materials, but it’s there.
You may be able to visualize why materials are elastic if you think in terms of
the simplistic model of a solid in 䉳 Fig. 9.1. The atoms of the solid substance are
imagined to be held together by springs. The elasticity of the springs represents
the resilient nature of the interatomic forces. The springs resist permanent defor-
mation, as do the forces between atoms. The elastic properties of solids are com-
monly discussed in terms of stress and strain. Stress is a measure of the force
causing a deformation. Strain is a relative measure of the deformation a stress
causes. Quantitatively, stress is the applied force per unit cross-sectional area:
䉱 F I G U R E 9 . 1 A springy solid
The elastic nature of interatomic
forces is indicated by simplistically F
stress = (9.1)
representing them as springs, A
which, like the forces, resist
deformation. SI unit of stress : newton per square meter 1N>m22
9.1 SOLIDS AND ELASTIC MODULI 313
Lo A 䉳 F I G U R E 9 . 2 Tensile and com-

pressional stresses Tensile and com-
ΔL pressional stresses are due to forces
applied normally to the surface area
F F of the ends of bodies. (a) A tension,
or tensile stress, tends to increase
(a) Tensile stress the length of an object. (b) A com-
pressional stress tends to shorten
the length. 1¢L = L - Lo2 can be
Lo positive, as in (a), or negative, as in
A (b). The sign is not needed in Eq. 9.2,
so the absolute value, ƒ ¢L ƒ , is used.
ΔL
F F
(b) Compressional stress
Here, F is the magnitude of the applied force normal (perpendicular) to the cross-
sectional area. Equation 9.1 shows that the SI units for stress are newtons per
square meter 1N>m22.
As illustrated in 䉱 Fig. 9.2, a force applied to the ends of a rod gives rise to
either a tensile stress (an elongating tension, ¢L 7 0) or a compressional stress (a
shortening tension, ¢L 6 0), depending on the direction of the force. In both
these cases, the tensile strain is the ratio of the change in length 1¢L = L - Lo2 to
the original length (Lo), without regard to the sign, so the absolute value, |¢L|,
is used. Then,
ƒ change in length ƒ ƒ ¢L ƒ ƒ L - Lo ƒ
strain = = = (9.2)
original length Lo Lo
Strain is a positive unitless quantity
Thus the strain is the fractional change in length. For example, if the strain is 0.05,
the length of the material has changed by 5% of the original length.
As might be expected, the resulting strain depends on the applied stress. For
relatively small stresses, this is a direct proportion, that is, stress r strain. For rela-
tively small stresses, this is a direct (or linear) proportion. The constant of propor-
tionality, which depends on the nature of the material, is called the elastic
modulus, that is,
stress = elastic modulus * strain
or
stress
elastic modulus = (9.3)
strain
SI unit of elastic modulus: newton per square meter 1N>m22
The elastic modulus is the stress divided by the strain, and the elastic modulus has
the same units as stress. (Why?)
Three general types of elastic moduli (plural of modulus) are associated with
stresses that produce changes in length, shape, and volume. These are called
Young’s modulus, the shear modulus, and the bulk modulus, respectively.
CHANGE IN LENGTH: YOUNG’S MODULUS

䉲Figure 9.3 is a typical graph of the tensile stress versus the strain for a metal rod.
The curve is a straight line up to a point called the proportional limit. Beyond this
point, the strain begins to increase more rapidly to another critical point called the
elastic limit. If the tension is removed at this point, the material will return to its
original length. If the tension is applied beyond the elastic limit and then
removed, the material will recover somewhat, but will retain some permanent
deformation.
䉴 F I G U R E 9 . 3 Stress versus strain Stress Fracture

A plot of stress versus strain for a Elastic
typical metal rod is a straight line limit
up to the proportional limit. Then
elastic deformation continues until Elastic behavior
the elastic limit is reached. Beyond (stress proportional to strain)
that, the rod will be permanently
deformed and will eventually frac- Compression Tension
ture or break.
Strain
The straight-line part of the graph shows a direct proportionality between

stress and strain. This relationship, first formalized by the English physicist
Robert Hooke in 1678, is known as Hooke’s law. (It is the same general relationship
as that given for a spring in Section 5.2—see Fig. 5.5.) The elastic modulus for a
tension or a compression is called Young’s modulus (Y):*
F>A
= Ya b
F ¢L
or Y = (9.4)
A Lo ¢L>Lo
stress strain
SI unit of Young’s modulus: newton per square meter 1N>m22

The units of Young’s modulus are the same as those of stress, newtons per square
meter 1N>m22, since the strain is unitless. Some typical values of Young’s modulus
are given in 䉲 Table 9.1.
TABLE 9.1 Elastic Moduli for Various Materials (in N>m2)

Substance Young’s modulus (Y) Shear modulus (S) Bulk modulus (B)
Solids
Aluminum 7.0 * 1010 2.5 * 1010 7.0 * 1010
10 10
Bone (limb) Tension: 1.5 * 10 1.2 * 10
9
Compression: 9.3 * 10
Brass 9.0 * 1010 3.5 * 1010 7.5 * 1010
Copper 11 * 1010 3.8 * 1010 12 * 1010
10 10
Glass 5.7 * 10 2.4 * 10 4.0 * 1010
Iron 15 * 1010 6.0 * 1010 12 * 1010
Nylon 5.0 * 109 8.0 * 108
Steel 20 * 1010 8.2 * 1010 15 * 1010
Liquids
Alcohol, ethyl 1.0 * 109
Glycerin 4.5 * 109
Mercury 26 * 109
Water 2.2 * 109
*Thomas Young (1773–1829) was an English physician and physicist who also demonstrated the
wave nature of light. See Young’s double-slit experiment, Section 24.1.
9.1 SOLIDS AND ELASTIC MODULI 315
To obtain a conceptual or physical understanding of Young’s modulus, let’s

solve Eq. 9.4 for ¢L:
¢L = a b
FLo 1 1
or ¢L r
A Y Y
Hence, the larger the Young’s modulus of a material, the smaller its change in
length (with other parameters being equal).
EXAMPLE 9.1 Pulling My Leg: Under a Lot of Stress

The femur (upper leg bone) is the longest and strongest bone in the body. Taking a typi-
cal femur to be approximately circular in cross-section with a radius of 2.0 cm, how
much force would be required to extend a patient’s femur by 0.010% while in horizon-
tal traction?
T H I N K I N G I T T H R O U G H . Equation 9.4 should apply, but where does the percentage
increase fit in? This question can be answered as soon as it is recognized that the ¢L>Lo
term is the fractional increase in length. For example, if you had a spring with a length
of 10 cm (Lo) and you stretched it 1.0 cm 1¢L2, then ¢L>Lo = 1.0 cm>10 cm = 0.10. This
ratio can readily be changed to a percentage, and the spring’s length was increased by
10%. So the percentage increase is really just the value of the ¢L>Lo term (multiplied by
100%).
Given: r = 2.0 cm = 0.020 m Find: F (tensile force)
¢L>Lo = 0.010% = 1.0 * 10-4
Y = 1.5 * 1010 N>m2 (for bone, from Table 9.1)
Using Eq. 9.4,
F = Y1¢L>Lo2A = Y1¢L>Lo2pr2
= 11.5 * 1010 N>m2211.0 * 10-42p10.020 m22 = 1.9 * 103 N
Before
How much force is this? Quite a bit—in fact, more than 400 lb. The femur is a pretty
strong bone.
F O L L O W - U P E X E R C I S E . A total mass of 16 kg is suspended from a 0.10-cm-diameter
steel wire. (a) By what percentage does the length of the wire increase? (b) The tensile φ
F
or ultimate strength of a material is the maximum stress the material can support before
breaking or fracturing. If the tensile strength of the steel wire in (a) is 4.9 * 108 N>m2, fs
how much mass could be suspended before the wire would break? (Answers to all After
Follow-Up Exercises are given in Appendix VI at the back of the book.) (a)
A
Most types of bone are composed of protein collagen fibers that are tightly bound
together and overlapping. Collagen has great tensile strength, and the calcium
salts within the collagen give bone great compressional strength. Collagen also
makes up cartilage, tendons, and skin, which have good tensile strength. Before
x
φ A F
CHANGE IN SHAPE: SHEAR MODULUS
h
Another way an elastic body can be deformed is by a shear stress. In this case, the
deformation is due to an applied force that is tangential to the surface area F
After
(䉴 Fig. 9.4a). A change in shape results without a change in volume. The shear strain
is given by x>h, where x is the relative displacement of the faces and h is the dis- (b)
tance between them.
The shear strain may be defined in terms of the shear angle f. As Fig. 9.4b 䉱 F I G U R E 9 . 4 Shear stress and
strain (a) A shear stress is produced
shows, tan f = x>h. But the shear angle is usually quite small, so a good when a force is applied tangentially
approximation is tan f L f L x>h , where f is in radians.* (If f = 10°, to a surface area. (b) The strain is
measured in terms of the relative
displacement of the object’s faces, or
*See the Chapter 7 Learn by Drawing 7.1, The Small-Angle Approximation. the shear angle f.
for example, there is only 1.0% difference between f and tan f). The shear
modulus (S), sometimes called the modulus of rigidity, is then
F>A F>A
S = L (9.5)
x>h f
SI unit of shear modulus: newton per square meter 1N>m22

Note in Table 9.1 that the shear modulus is generally less than Young’s modulus.
In fact, S is approximately Y>3 for many materials, which indicates a greater
response to a shear stress than to a tensile stress. Note also that the inverse rela-
tionship f L 1>S is similar to that pointed out previously for Young’s modulus.
A shear stress may be of the torsional type, resulting from the twisting action of
a torque. For example, a torsional shear stress may shear off the head of a bolt that
is being tightened.
Liquids do not have shear moduli (or Young’s moduli)—hence the gaps in
Table 9.1. A shear stress cannot be effectively applied to a liquid or a gas because
fluids deform continuously in response. That is, fluids cannot support a shear.
CHANGE IN VOLUME: BULK MODULUS

Suppose that a force directed inward acts over the entire surface of a body
(䉲 Fig. 9.5). Such a volume stress is often applied by pressure transmitted by a fluid.
An elastic material will be compressed by a volume stress; that is, the material will
show a change in volume, but not in general shape, in response to a pressure
change ¢p. [Pressure (p) is force per unit area, Section 9.2.] The change in pressure
is equal to the volume stress, or ¢p = F>A. The volume strain is the ratio of the vol-
ume change 1¢V2 to the original volume (Vo). The bulk modulus (B) is then
F>A ¢p
B = = - (9.6)
- ¢V>Vo ¢V>Vo
SI unit of bulk modulus: newton per square meter 1N>m22
The minus sign is introduced to make B a positive quantity, since ¢V = V - Vo is
negative for an increase in external pressure (when ¢p is positive). Similarly to the
previous moduli relationships, ¢V r 1>B.
Bulk moduli of selected solids and liquids are listed in Table 9.1. Gases also
have bulk moduli, since they can be compressed. For a gas, it is common to talk
about the reciprocal of the bulk modulus, which is called the compressibility (k):
1
k = (compressibility for gases) (9.7)
B
The change in volume ¢V is thus directly proportional to the compressibility k.
Solids and liquids are relatively incompressible and thus have small values of
compressibility. Conversely, gases are easily compressed and have large com-
pressibilities, which vary with pressure and temperature.
䉴 F I G U R E 9 . 5 Volume stress and F F

strain (a) A volume stress is applied
when a normal force acts over an
entire surface area, as shown here F A F
for a cube. This type of stress most
commonly occurs in gases. (b) The
resulting strain is a change in F A F F F
A
volume.
F F
F F
(a) (b)
9.2 FLUIDS: PRESSURE AND PASCAL’S PRINCIPLE 317
EXAMPLE 9.2 Compressing a Liquid: Volume Stress and Bulk Modulus

By how much should the pressure on a liter of water be Note that - ¢V>Vo is the fractional change in the volume. With
changed to compress it by 0.10%? Vo = 1000 cm3, the change (reduction) in volume is
- ¢V = 0.0010 Vo = 0.001011000 cm32 = 1.0 cm3

T H I N K I N G I T T H R O U G H . Similarly to the fractional change in
length, ¢L>Lo , the fractional change in volume is given by
- ¢V>Vo , which may be expressed as a percentage. The pres- However, the change in volume is not needed. The fractional
sure change can then be found from Eq. 9.6. Compression change, as listed in the given data, can be used directly in
implies a negative ¢V. Eq. 9.6 to find the increase in pressure:
SOLUTION.
¢p = B a b = 12.2 * 109 N>m22(0.0010) = 2.2 * 106 N>m2
- ¢V
Given: - ¢V>Vo = 0.0010 (or 0.10%) Find: ¢p
Vo
Vo = 1.0 L = 1000 cm3
BH2O = 2.2 * 109 N>m2 (This increase is about twenty-two times normal atmospheric
(from Table 9.1) pressure. Not too compressible.)
FOLLOW-UP EXERCISE. If an extra 1.0 * 106 N>m2 of pressure above normal atmospheric pressure is applied to a half liter of
water, what is the change in the water’s volume?
DID YOU LEARN?

➥ Stress is a measure of the force causing a deformation. Strain is a relative measure
of the deformation a stress causes.
➥ An elastic modulus is stress divided by strain, with units of N>m2.
➥ The three major moduli are Young’s modulus (change in length), shear modulus
(change in area), and bulk modulus (change in volume).
9.2 Fluids: Pressure and Pascal’s Principle

➥ How is applied pressure transmitted in an enclosed fluid?

➥ What is the difference between absolute pressure and gauge pressure?
A force can be applied to a solid at a point of contact, but this won’t work with a
fluid, since a fluid cannot support a shear. With fluids, a force must be applied
over an area. Such an application of force is expressed in terms of pressure, or the
force per unit area:
F
p = (9.8a)
A
SI unit of pressure: newton per square meter 1N>m22, or pascal (Pa)
Pressure has SI units of newton per square meter 1N>m22, or pascal (Pa), in F
honor of the French scientist and philosopher Blaise Pascal (1623–1662), who stud- u
ied fluids and pressure. By definition,* F cos u
A
1 Pa = 1 N>m2 F⊥
The force in Eq. 9.8a is understood to be acting normally (perpendicularly) to the

surface area. F may be the perpendicular component of a force that acts at an angle
to a surface (䉴 Fig. 9.6). F⊥ F cos u
As Figure 9.6 shows, in the more general case, p= =
A A
F⬜ F cos u
p = = (9.8b)
A A 䉱 F I G U R E 9 . 6 Pressure Pressure
is usually written p = F>A, where it
Pressure is a scalar quantity (with magnitude only), even though the force produc- is understood that F is the force or
ing it is a vector. component of force normal to the
surface. In general, then,
*Notice that the unit of pressure is equivalent to energy per volume, N>m2 = N # m>m3 = J>m3, an p = 1F cos u2>A.
energy density.
In the British system, a common unit of pressure is pound per square inch (lb>in2,
or psi). Other units, some of which will be introduced later, are used in special
applications. Before going on, here’s a “solid” example of the relationship
between force and pressure.
CONCEPTUAL EXAMPLE 9.3 Force and Pressure: Taking a Nap on a Bed of Nails
Suppose you are getting ready to take a nap, and you have a If there were only one nail, the person’s weight would not be
choice of lying stretched out on your back on (a) a bed of supported by the nail, and with such a small area, the pressure
nails, (b) a hardwood floor, or (c) a couch. Which one would would be very great—a situation in which the lone nail would
you choose for the most comfort, and why? pierce the skin. However, when a bed of nails is used, the same
force (weight) is distributed over hundreds of nails, which gives
REASONING AND ANSWER. The comfortable choice is quite
a relatively large effective area of contact. The pressure is then
apparent—the couch. But here, the conceptual question is
reduced to a level at which the nails do not pierce the skin.
why.
When you are lying on a hardwood floor, the area in contact
First let’s look at the prospect of lying on a bed of nails, an
with your body is appreciable and the pressure is reduced, but
old trick that originated in India and used to be demonstrated
it still may be a bit uncomfortable. Parts of your body, such as
in carnival sideshows (See Fig. 9.28). There is really no trick
your neck and the small of your back, are not in contact with a
here, just physics—namely, force and pressure. It is the force
per unit area, or pressure 1p = F>A2, that determines whether
surface, but they would be on a couch. On a soft couch, the
body sinks into it and the contact surface is greater, therefore
a nail will pierce the skin. The force is determined by the
reduced pressure and more comfort. So (c) is the answer.
weight of the person lying on the nails. The area is deter-
mined by the effective area of the nails in contact with the skin F O L L O W - U P E X E R C I S E . What are a couple of important con-
(neglecting one’s clothes). siderations in constructing a bed of nails to lie on?
Now, let’s take a quick review of density, which is an important consideration

in the study of fluids. Recall from Section 1.4 that the density 1r2 of a substance is
defined as mass per unit volume (Eq. 1.1):
mass
density =
volume
m
r =
V
SI unit of density: kilogram per cubic meter 1kg>m32
1common cgs unit: gram per cubic centimeter, or g>cm32
The densities of some common substances are given in 䉲 Table 9.2.
Water has a density of 1.00 * 103 kg>m3 (or 1.00 g>cm3) from the original defin-
ition of the kilogram (Section 1.2). Mercury has a density of 13.6 * 103 kg>m3
(or 13.6 g>cm3). Hence, mercury is 13.6 times as dense as water. Gasoline, however,
is less dense than water. See Table 9.2. (Note: Be careful not to confuse the symbol
for density, r (Greek rho), with that for pressure, p.)
TABLE 9.2 Densities of Some Common Substances (in kg>m3)

Solids Density (R) Liquids Density (R) Gases* Density (R)
Aluminum 2.7 * 103 Alcohol, ethyl 0.79 * 103 Air 1.29

3 3
Brass 8.7 * 10 Alcohol, methyl 0.82 * 10 Helium 0.18
3 3
Copper 8.9 * 10 Blood, whole 1.05 * 10 Hydrogen 0.090
3 3
Glass 2.6 * 10 Blood plasma 1.03 * 10 Oxygen 1.43
3 3
Gold 19.3 * 10 Gasoline 0.68 * 10 Water vapor (100 °C) 0.63
3 3
Ice 0.92 * 10 Kerosene 0.82 * 10
3
Iron (and steel) 7.8 * 10 (general value) Mercury 13.6 * 103
Lead 11.4 * 103 Seawater 14° C2 1.03 * 103
Silver 10.5 * 103 Water, fresh 14° C2 1.00 * 103
3
Wood, oak 0.81 * 10
*At 0 °C and 1 atm, unless otherwise specified.
INSIGHT 9.1 Osteoporosis and Bone Mineral Density (BMD)

Bone is a living, growing tissue. Your body is continuously tak- puted, commonly in grams per cubic centimeter, after the bone
ing up old bone (resorption) and making new bone tissue. In is weighed to determine mass. If you burn the bone, weigh the
the early years of life, bone growth is greater than bone loss. remaining ash, and divide by the volume of the overall bone
This continues until a peak bone mass is reached as a young (bone tissue), you get the bone tissue mineral density, which is
adult. After this, bone growth is slowly outpaced by bone loss. commonly called the bone mineral density (BMD).
Bones naturally become less dense and weaker with age. To measure the BMD of bones in vivo, types of radiation
Osteoporosis (“porous bone”) occurs when bones deteriorate transmission through the bone are measured, which is related
to the point where they are easily fractured (Fig. 1). to the amount of bone mineral present. Also, a “projected”
area of the bone is measured. Using these measurements, a
projected BMD is computed in units of mg>cm2. Figure 2
illustrates the magnitude of the effect of bone density loss
with aging.
The diagnosis of osteoporosis relies primarily on the mea-
surement of BMD. The mass of a bone, measured by a BMD
test (also called a bone densitometry test), generally correlates to
the bone strength. It is possible to predict fracture risk, much
as blood pressure measurements can help predict stroke risk.
Bone density testing is recommended for all women age 65
and older, and for younger women at an increased risk of
䉱 F I G U R E 1 Bone mass loss An X-ray micrograph of the osteoporosis. This testing also applies to men. Osteoporosis is
bone structure of the vertebrae of a 50-year-old (left) and a often thought to be a woman’s disease, but 20% of osteoporo-
70-year-old (right). Osteoporosis, a condition characterized by sis cases occur in men. A BMD test cannot predict the certainty
bone weakening caused by loss of bone mass, is evident for of developing a fracture, but only predicts the degree of risk.
the vertebrae on the right. So how is BMD measured? This is where the physics comes
in. Various instruments, divided into central devices and
Osteoporosis and low bone mass affect an estimated 24 mil- peripheral devices, are used. Central devices are used primarily
lion Americans, most of whom are women. Osteoporosis to measure the bone density of the hip and spine. Peripheral
results in an increased risk of bone fractures, particularly of devices are smaller, portable machines that are used to mea-
the hip and the spine. Many women take calcium supple- sure the bone density in such places as the heel or finger.
ments to help prevent this. The most widely used central device relies on dual energy
To understand how bone density is measured, let’s first dis- X-ray absorptiometry (DXA), which uses X-ray imaging to
tinguish between bone and bone tissue. Bone is the solid mater- measure bone density. (See Section 20.4 for a discussion of
ial composed of a protein matrix, most of which has calcified. X-rays.) The DXA scanner produces two X-ray beams of dif-
Bone tissue includes the marrow spaces within the matrix. ferent energy levels. The amount of X-rays that pass through
(Marrow is the soft, fatty, vascular tissue in the interior cavi- a bone is measured for each beam; the amounts vary with the
ties of bones and is a major site of blood cell production.) The density of bone. The calculated bone density is based on the
marrow volume varies with the bone type. difference between the two beams. The procedure is nonin-
If the volume of an intact bone is measured (for example, by trusive and takes 10–20 min, and the X-ray exposure is usu-
water displacement), then the bone tissue density can be com- ally about one-tenth of that of a chest X-ray (Fig. 3).
䉴 F I G U R E 2 Bone density loss with aging

An illustration of how normal bone density
loss for a female hip bone increases with age
(scale on right). Osteopenia refers to
decreased calcification or bone density. A per-
son with osteopenia is at risk for developing
osteoporosis, a condition that causes bones to
become brittle and prone to fracture.

A common peripheral device uses quantative ultrasound

(QUS). Instead of using X-rays, a bone density screening is
made using high-frequency sound waves (ultrasound). (See
Section 14.1 for a discussion of ultrasound.) QUS measure-
ments are usually done on the heel. The test takes only a
minute or two, and the devices are now found in some phar-
macies or drugstores. Its purpose is to tell you if you are “at
risk” and may need a more thorough DXA test.
䉴 F I G U R E 3 Osteoporosis bone scanning

A technician runs an X-ray bone scan to check
for osteoporosis. X-ray images are displayed on
the monitor; these images can confirm the pres-
ence of osteoporosis. Such bone densitometry
tests can also be used to diagnoses rickets,
a children’s disease characterized by softening
of the bones.
Density is a measure of the compactness of the matter of a substance—the

greater the density, the more matter or mass in a given volume. Notice that den-
sity quantifies the amount or mass per unit volume. For an important density con-
sideration, see Insight 9.1, Osteoporosis and Bone Mineral Density (BMD).
PRESSURE AND DEPTH

If you have gone scuba diving, you well know that pressure increases with
depth, having felt the increased pressure on your eardrums. An opposite effect
is commonly felt when you fly in a plane or ride in a car going up a mountain.
With increasing altitude, your ears may “pop” because of reduced external air
pressure.
How the pressure in a fluid varies with depth can be demonstrated by consid-
ering a container of liquid at rest. Imagine that you can isolate a rectangular col-
umn of water, as shown in 䉲 Fig. 9.7. Then the force on the bottom of the container
below the column (or the hand) is equal to the weight of the liquid making up the
column: F = w = mg. Since density is r = m>V, the mass in the column is equal
to the density times the volume; that is, m = rV. (The liquid is assumed incom-
pressible, so r is constant.)
䉴 F I G U R E 9 . 7 Pressure and
depth The extra pressure at a depth w = r(Ah)g
h in a liquid is due to the weight of
the liquid above: p = rgh , where r p = w = r gh
A
is the density of the liquid
(assumed to be constant). This is A A
shown for an imaginary rectangu-
lar column of liquid.
h h
mg mg
The volume of the isolated liquid column is equal to the height of the column
times the area of its base, or V = hA. Thus,
F = w = mg = rVg = rghA
With p = F>A, the pressure at a depth h due to the weight of the column is
p = rgh (9.9)
This is a general result for incompressible liquids. The pressure is the same every-
where on a horizontal plane at a depth h (with r and g constant). Note that Eq. 9.9 is
independent of the base area of the rectangular column. The whole cylindrical column
of the liquid in the container in Fig. 9.7 could have been taken with the same result.
The derivation of Eq. 9.9 did not take into account pressure being applied to the
open surface of the liquid. This factor adds to the pressure at a depth h to give a
total pressure of
(incompressible liquid
p = po + rgh (9.10)
at constant density)
where po is the pressure applied to the liquid surface (that is, the pressure applied
at h = 0). For an open container, po = pa , atmospheric pressure, or the weight
(force) per unit area due to the gases in the atmosphere above the liquid’s surface.
The average atmospheric pressure at sea level is sometimes used as a unit, called
an atmosphere (atm):
1 atm = 101.325 kPa = 1.01325 * 105 N>m2 = 14.7 lb>in2
The measurement of atmospheric pressure will be described shortly.
EXAMPLE 9.4 A Scuba Diver: Pressure and Force

(a) What is the total pressure on the back of a scuba diver in a T H I N K I N G I T T H R O U G H . (a) This is a direct application of
lake at a depth of 8.00 m? (b) What is the force on the diver’s Eq. 9.10 in which po is taken as the atmospheric pressure pa.
back due to the water alone? (Take the surface of the back to (b) Knowing the area and the pressure due to the water, the
be a rectangle 60.0 cm by 50.0 cm.) force can be found from the definition of pressure, p = F>A.
SOLUTION.
Given: h = 8.00 m Find: (a) p (total pressure)
A = 60.0 cm * 50.0 cm (b) F (force due to water)
= 0.600 m * 0.500 m = 0.300 m2
rH O = 1.00 * 103 kg>m3 (from Table 9.2)
2
pa = 1.01 * 105 N>m2

(a) The total pressure is the sum of the pressure due to the water and the atmospheric pressure (pa). By Eq. 9.10, this is
p = pa + rgh
= 11.01 * 105 N>m22 + 11.00 * 103 kg>m3219.80 m>s2218.00 m2
= 11.01 * 105 N>m22 + 10.784 * 105 N>m22 = 1.79 * 105 N>m2 1or Pa2
1expressed in atmospheres2 L 1.8 atm
This is also the inward pressure on the diver’s eardrums.
(b) The pressure pH O due to the water alone is the rgh portion of the preceding equation, so pH = 0.784 * 105 N>m2.
2 2O
Then, pH = F>A, and
2O
F = pH2O A = 10.784 * 105 N>m2210.300 m22
= 2.35 * 104 N 1or 5.29 * 103 lb—about 2.6 tons!2
F O L L O W - U P E X E R C I S E . You might question the answer to part (b) of this Example—how could the diver support such a force?
To get a better idea of the forces our bodies can support, what would be the force on the diver’s back at the water surface from
atmospheric pressure alone? How do you suppose our bodies can support such forces or pressures?
PASCAL’S PRINCIPLE
F
When the pressure (for example, air pressure) is increased on the entire open sur-
A B face of an incompressible liquid at rest, the pressure at any point in the liquid or
h on the boundary surfaces increases by the same amount. The effect is the same if
pA pA 2 pressure is applied to any surface of an enclosed fluid by means of a piston
rgh (䉳 Fig. 9.8). The transmission of pressure in fluids was studied by Pascal, and the
pA + h observed effect is called Pascal’s principle:
2
C pA + r gh Pressure applied to an enclosed fluid is transmitted undiminished to every point in
the fluid and to the walls of the container.
D
For an incompressible liquid, the change in pressure is transmitted essentially
instantaneously. For a gas, a change in pressure will generally be accompanied by a
䉱 F I G U R E 9 . 8 Pascal’s principle change in volume or temperature (or both), but after equilibrium has been reestab-
The pressure applied at point A is lished, Pascal’s principle remains valid.
fully transmitted to all parts of the Common practical applications of Pascal’s principle include the hydraulic
fluid and to the walls of the con- braking systems used on automobiles. Through tubes filled with brake fluid, a
tainer. There is also pressure due to
the weight of the fluid above at dif- force on the brake pedal transmits a force to the wheel brake cylinder. Similarly,
ferent depths (for instance, rgh>2 at hydraulic lifts and jacks are used to raise automobiles and other heavy objects
C and rgh at D). (䉲 Fig. 9.9).
Using Pascal’s principle, it can be shown how such systems allow us not only to
transmit force from one place to another, but also to multiply that force. The input
pressure pi supplied by compressed air for a garage lift, for example, gives an
input force Fi on a small piston area Ai (Fig. 9.9). The full magnitude of the pres-
sure is transmitted to the output piston, which has an area Ao . Since pi = po , it fol-
lows that
Fi Fo
=
Ai Ao
and
Fo = a bF (hydraulic force multiplication)

Ao
(9.11)
Ai i
With Ao larger than Ai , then Fo will be larger than Fi. The input force is greatly
multiplied if the input piston has a relatively small area.
Fi Oil
Ai
Ao
p Piston
Ao
To reservoir
p p
Fo =
( )F
Ai i
Fluid Fo
Valve Valve
(a) (b)
䉱 F I G U R E 9 . 9 The hydraulic lift and shock absorbers (a) Because the input and output
pressures are equal (Pascal’s principle), a small input force gives a large output force pro-
portional to the ratio of the piston areas. (b) A simplified exposed view of one type of shock
absorber. (See Example 9.5 for description.)
EXAMPLE 9.5 The Hydraulic Lift: Pascal’s Principle

A garage lift has input and lift (output) pistons with diame- or
ters of 10 cm and 30 cm, respectively. The lift is used to hold 0.10 m 2 1.4 * 104 N
Fi = a b Fo =
Fo
up a car with a weight of 1.4 * 104 N. (a) What is the magni- = = 1.6 * 103 N
0.30 m 9 9
tude of the force on the input piston? (b) What pressure is
applied to the input piston? The input force is one-ninth of the output force; in other words,
the force was multiplied by 9 (that is, Fo = 9Fi).
T H I N K I N G I T T H R O U G H . (a) Pascal’s principle, as expressed
(Note that we didn’t really need to write the complete expres-
in the hydraulic Eq. 9.11, has four variables, and three are
sions for the areas. The area of a circle is proportional to the
given (areas via diameters). (b) The pressure is simply
square of the diameter of the circle. If the ratio of the piston
p = F>A.
diameters is 3 to 1, the ratio of their areas must therefore be 9
SOLUTION. to 1, and this ratio is used directly in Eq. 9.11.)
Given: di = 10 cm = 0.10 m Find: (a) Fi (input force) (b) Then applying Eq. 9.8a:
do = 30 cm = 0.30 m (b) pi (input pressure) Fi Fi Fi 1.6 * 103 N
Fo = 1.4 * 104 N pi =
p 1di>22 p10.10 m22>4
= 2
= 2
=
Ai pri
(a) Rearranging Eq. 9.11 and using A = pr2 = pd2>4 for the = 2.0 * 105 N>m2 1= 200 kPa2
circular piston 1r = d>22 gives
This pressure is about 30 lb>in2, a common pressure used in
pd2i >4 di 2
Fi = a b Fo = a 2 b Fo = a b Fo
Ai automobile tires and about twice atmospheric pressure
Ao pdo>4 do (which is approximately 100 kPa, or 15 lb>in2.)
F O L L O W - U P E X E R C I S E . Pascal’s principle is used in shock absorbers on automobiles and on the landing gear of airplanes. (The
polished steel piston rods can be seen above the wheels on aircraft.) In these devices, a large force (the shock produced on hitting
a bump in the road or on an airport runway at high speed) must be reduced to a safe level by removing energy. Basically, fluid is
forced by the motion of a large-diameter piston through small channels in the piston on each stroke cycle (Fig. 9.9b).
Note that the valves allow for fluid through the channel, which creates resistance to the motion of the piston (effectively the
reverse of the situation in Fig. 9.9a). The piston goes up and down, dissipating the energy of the shock. This is called damping
(Section 13.2). Suppose that the input piston of a shock absorber on a jet plane has a diameter of 8.0 cm. What would be the diam-
eter of an output channel that would reduce the force by a factor of 10?
As Example 9.5 shows, forces produced by pistons relate directly to their diame-
ters: Fi = 1di>do22 Fo or Fo = 1do >di22 Fi . By making do W di , huge factors of force
multiplication can be obtained, as is typical for hydraulic presses, jacks, and earth-
moving equipment. (The shiny input piston rods are often visible on front loaders
and backhoes.) Inversely, force reductions may be obtained by making di 7 do , as
in Follow-Up Exercise 9.5.
However, don’t think that you are getting something for nothing with large
force multiplications. Energy is still a factor, and it can never be multiplied by a
machine. (Why not?) Looking at the work involved and assuming that the work
output is equal to the work input, Wo = Wi (an ideal condition—why?). Then,
with W = Fx. (Eq. 5.1),
Fo xo = Fi xi
or
Fo = a bF
xi
xo i
where xo and xi are the output and input distances moved by the respective pistons.
Thus, the output force can be much greater than the input force only if the input
distance is much greater than the output distance. For example, if Fo = 10Fi , then
xi = 10xo , and the input piston must travel 10 times the distance of the output pis-
ton. Force is multiplied at the expense of distance.
PRESSURE MEASUREMENT
Pressure can be measured by mechanical devices that are often spring loaded
(such as a tire gauge). Another type of instrument, called a manometer, uses a liq-
uid—usually mercury—to measure pressure. An open-tube manometer is illustrated
in 䉲 Fig. 9.10a. One end of the U-shaped tube is open to the atmosphere, and the
other is connected to the container of gas whose pressure is to be measured. The
liquid in the U-tube acts as a reservoir through which pressure is transmitted
according to Pascal’s principle.
The pressure of the gas (p) is balanced by the weight of the column of liquid (of
height h, the difference in the heights of the columns) and the atmospheric pres-
sure (pa) on the open liquid surface:
p = pa + rgh (9.12)
The pressure p is called the absolute pressure.

You may have measured pressure using pressure gauges; a tire gauge used to
measure air pressure in automobile tires is a common example (Fig. 9.10b). Such
gauges, quite appropriately, measure gauge pressure. A pressure gauge registers
only the pressure above (or below) atmospheric pressure. Hence, to get the absolute
pressure (p), you have to add the atmospheric pressure (pa) to the gauge pressure (pg):
p = pa + pg
For example, suppose your tire gauge reads a pressure of 200 kPa 1L 30 lb>in22.
The absolute pressure within the tire is then p = pa + pg = 101 kPa + 200 kPa =
301 kPa, where normal atmospheric pressure is about 101 kPa 114.7 lb>in22, as will
be shown shortly.
The gauge pressure of a tire keeps the tire rigid or operational. In terms of the
more familiar pounds per square inch (psi, or lb>in2), a tire with a gauge pressure
of 30 psi has an absolute pressure of about 45 psi (30 + 15, with atmospheric
pressure = 15 psi). Hence, the pressure on the inside of the tire is 45 psi, and that
Scale
Pressure of Vacuum
air in tire
pa
Gas
under
pressure h Spring h
p Reference
point
Atmospheric pa
pressure
Mercury
Scale
p = pa + rgh pa = rgh
pg = p − pa
(absolute pressure) (barometric pressure)
(gauge pressure)
(a) Open-tube manometer (b) Tire gauge (c) Barometer
䉱 F I G U R E 9 . 1 0 Pressure measurement (a) For an open-tube manometer, the pressure of the gas
in the container is balanced by the pressure of the liquid column and atmospheric pressure acting
on the open surface of the liquid. The absolute pressure of the gas equals the sum of the atmos-
pheric pressure (pa) and rgh the gauge pressure. (b) A tire gauge measures gauge pressure, the dif-
ference between the pressure in the tire and atmospheric pressure: pgauge = p - pa . Thus, if a tire
gauge reads 200 kPa 130 lb>in22, the actual pressure within the tire is 1 atm higher, or 300 kPa.
(c) A barometer is a closed-tube manometer that is exposed to the atmosphere and thus reads only
atmospheric pressure.
on the outside is 15 psi. The ¢p of 30 psi keeps the tire inflated. If you open the
valve or get a puncture, the internal and external pressures equalize and you have
a flat!
Atmospheric pressure can be measured with a barometer. The principle of a
mercury barometer is illustrated in Fig. 9.10c. The device was invented by Evange-
lista Torricelli (1608–1647), Galileo’s successor as professor of mathematics at an
academy in Florence. A simple barometer consists of a tube filled with mercury
that is inverted into a reservoir. Some mercury runs from the tube into the reser-
voir, but a column supported by the air pressure on the surface of the reservoir
remains in the tube. This device can be considered a closed-tube manometer, and the
pressure it measures is just the atmospheric pressure, since the gauge pressure
(the pressure above atmospheric pressure) is zero.
The atmospheric pressure is then equal to the pressure due to the weight of the
column of mercury, or
p = rgh (9.13)
A standard atmosphere is defined as the pressure supporting a column of mer-
cury exactly 76 cm in height at sea level and at 0 °C. (For a common biological
atmospheric effect because of pressure changes, see Insight 9.2, An Atmospheric
Effect: Possible Earaches.)
Changes in atmospheric pressure can be observed as changes in the height of a
column of mercury. These changes are due primarily to high- and low-pressure air
masses that travel across the country. Atmospheric pressure is commonly reported
in terms of the height of the barometer column, and weather forecasters say that
the barometer is rising or falling. That is,
1 atm 1about 101 kPa2 = 76 cm Hg = 760 mm Hg
= 29.92 in. Hg 1about 30 in. Hg2
In honor of Torricelli, a pressure supporting 1 mm of mercury is given the name
torr:
1 mm Hg K 1 torr
and
1 atm = 760 torr*
*In the SI, one atmosphere has a pressure of 1.013 * 105 N>m2, or about 105 N>m2. Meteorologists
use yet another nonstandard unit of pressure called the millibar (mb). A bar is defined to be 105 N>m2,
and because 1 bar = 1000 mb, 1 atm = 1 bar = 1000 mb. Small changes in atmospheric pressure are
more easily reported using the millibar.
INSIGHT 9.2 An Atmospheric Effect: Possible Earaches

Variations in atmospheric pressure can have a common physi- sure is not relieved, you may soon have an earache. The pres-
ological effect: changes in pressure in the ears with changes in sure is relieved by a “pushing” of air through the Eustachian
altitude. This “plugging up” and “popping” of the ears is fre- tube into the throat, which produces a popping sound. We
quently experienced in ascents and descents on mountain often swallow or yawn to assist this process. Similarly, when
roads or on airplanes. The eardrum, so important to your descending, the higher outside pressure at lower altitudes
hearing, is a membrane that separates the middle ear from the needs to be equalized with the lower pressure in the middle
outer ear. [See Fig. 1 in the Chapter 14 (Sound) Insight 14.2, ear. Swallowing allows air to flow into the middle ear in this
The Physiology and Physics of the Ear and Hearing, to view case.
the anatomy of the ear.] The middle ear is connected to the Nature takes care of us, but it is important to understand
throat by the Eustachian tube, the end of which is normally what is going on. If you have a throat infection, the opening
closed. The tube opens during swallowing or yawning to per- of the Eustachian tube to the throat might be swollen, par-
mit air to escape, so the internal and external pressures are tially blocking the tube. You may be tempted to hold your
equalized. nose and blow with your mouth closed in order to clear your
However, when climbing relatively quickly in an airplane ears. Don’t do it! You could blow mucus into the inner ear and
or in a car in a mountainous region, the air pressure outside cause a painful inner-ear infection. Instead, swallow hard sev-
the ear may be less than that in the middle ear. This difference eral times and give some big yawns to help open the
in pressure forces the eardrum outward. If the outward pres- Eustachian tube and equalize the pressure.
䉴 F I G U R E 9 . 1 1 Aneroid Because mercury is highly toxic, it is

barometer Changes in atmospheric sealed inside a barometer. A safer and less
pressure on a sensitive metal expensive device that is widely used to
diaphragm are reflected on the dial measure atmospheric pressure is the
face of the barometer. A fair weather aneroid (“without fluid”) barometer.
prediction is generally associated
with high barometric pressures, and In an aneroid barometer, a sensitive
rainy weather with low barometric metal diaphragm on an evacuated
pressures. container (something like a drum-
head) responds to pressure
changes, which are indicated on a
dial. This is the kind of barometer
you frequently find in homes in dec-
orative wall mountings (䉳 Fig. 9.11).
Since air is compressible, the atmos-
pheric density and pressure are greatest
at the Earth’s surface and decrease with alti-
tude. We live at the bottom of the atmosphere,
but don’t notice its pressure very much in our daily
activities. Remember that our bodies are composed largely of fluids, which exert a
matching outward pressure. Indeed, the external pressure of the atmosphere is so
important to our normal functioning that it is taken with us wherever we can. For
example, the pressurized suits worn by astronauts in space or on the Moon are
needed to provide an external pressure similar to that on the Earth’s surface.
A very important gauge pressure reading is discussed in Insight 9.3, Blood
Pressure, Intraocular Pressure, and Measurement. Read this before going on to
Example 9.6.
INSIGHT 9.3 Blood Pressure and Intraocular Pressure

BLOOD PRESSURE Pulmonary
Aorta
Basically, a pump is a machine that transfers mechanical energy trunk
Left
to a fluid, thereby increasing the pressure and causing the fluid atrium
Right
to flow. One pump that is of interest to everyone is the heart, a
atrium
muscular pump that drives blood throughout the body’s circu-
latory network of arteries, capillaries, and veins. With each
pumping cycle, the human heart’s interior chambers enlarge
and fill with freshly oxygenated blood from the lungs (Fig. 1).
The human heart contains two pairs of chambers: two ven-
tricles and two atria. When the ventricles contract, blood is
Right Left
forced out through the arteries. Smaller and smaller arteries ventricle ventricle
branch off from the main ones, until the very small capillaries
are reached. There, oxygen and nutrients being carried by the (a) Intake (b) Output
blood are exchanged with the surrounding tissues, and car-
bon dioxide (a waste gas) is picked up. The blood then flows
into the veins to the lungs to expel carbon dioxide, and then
back to the heart to complete the circuit.
When the ventricles contract, forcing blood into the arterial
system, the pressure in the arteries increases sharply. The
maximum pressure achieved during ventricular contraction is
called the systolic pressure. When the ventricles relax, the arter-
ial pressure drops, and the lowest pressure before the next
contraction, called the diastolic pressure, is reached. (These
pressures are named after two parts of the pumping cycle, Systolic
systole and diastole.)
Pressure
䉴 F I G U R E 1 The heart as a pump The human heart Diastolic

is analogous to a mechanical force pump. Its pump-
ing action, consisting of (a) intake and (b) output,
gives rise to variations in blood pressure. Time
The walls of the arteries have considerable elasticity and the blood pumped from the heart. Their elasticity may dimin-
expand and contract with each pumping cycle. This alternat- ish with age, however. Cholesterol deposits can narrow and
ing expansion and contraction can be felt as a pulse in an roughen the arterial passageways, impeding the blood flow
artery near the surface of the body. For example, the radial and giving rise to a form of arteriosclerosis, or hardening of
artery near the surface of the wrist is commonly used to mea- the arteries. Because of these defects, the driving pressure
sure a person’s pulse. The pulse rate is equal to the ventricu- must increase to maintain a normal blood flow. The heart
lar contraction rate, and hence the pulse rate indicates the must work harder, which places a greater demand on the
heart rate. heart muscles. A relatively slight decrease in the effective
Taking a person’s blood pressure involves measuring the cross-sectional area of a blood vessel has a rather large effect
pressure of the blood on the arterial walls, usually in the (an increase) on the flow rate, as will be shown in Section 9.4.
arm. This is done with a sphygmomanometer. (The Greek
word sphygmo means “pulse.”) An inflatable cuff is INTRAOCULAR PRESSURE
wrapped around the arm and inflated to shut off the blood Another commonly measured pressure is intraocular
flow temporarily. The cuff pressure is slowly released, and pressure (IOP), or pressure of the eye. Elevated intraocular
the artery is monitored with a stethoscope (Fig. 2). Soon pressure can cause the damage of the optic nerve. Glaucoma
blood is just forced through the constricted artery. This flow is associated with elevated eye pressure and can cause the
is turbulent and gives rise to a specific sound with each loss of vision.
heartbeat. When the sound is first heard, the systolic pres- The process of measuring intraocular pressure is referred to
sure is noted on the gauge. When the turbulent beats disap- as tonometry. There are various types of devices, called
pear because blood begins to flow smoothly, the diastolic tonometers, used to make this measurement. One of the most
pressure is taken. common instruments is a hand-held device called the Tono-
Blood pressure is commonly reported by giving the systolic Pen AVIA®.
and diastolic pressures, separated by a slash—for example, After the eye has been numbed with anesthetic drops, the
120>80 (mm Hg, read as “120 over 80”). (The gauge in Fig. 2 is tonometer's tip is gently placed against the front surface
an aneroid type; older types of sphygmomanometers used a (cornea) of the eye (Fig. 3). The cornea bends under the force
mercury column to measure blood pressure.) Normal blood applied by the tip of the Tono-Pen. Once the cornea passes a
pressure ranges are 120–139 for systolic and 80–89 for dias- flattened stage, it becomes slightly indented and a pressure
tolic. (Blood pressure is a gauge pressure. Why?) transducer in the Pen measures the force required to reach
High blood pressure is a common health problem. The elas- this state. The result is displayed on a digital readout in mm
tic walls of the arteries expand under the hydraulic force of Hg. The procedure is painless and takes only a few seconds.
Normal eye pressures range from 10 to 20 mm Hg.
䉳 FIGURE 2
Measuring blood
pressure The pres-
sure is indicated on
the gauge in mil-
limeters Hg.
䉱 F I G U R E 3 Intraocular pressure. Intraocular (eye) pressure

being measured with a Tono-Pen AVIA®. See text for
description.
EXAMPLE 9.6 An IV: A Gravity Assist

An IV (intravenous injection) is a type of gravity assist quite T H I N K I N G I T T H R O U G H . The fluid gauge pressure at the bot-
different from that discussed for space probes in Section 7.5. tom of the IV tube must be greater than the pressure in the
Consider a hospital patient who receives an IV under gravity vein and can be computed from Eq. 9.9. (The liquid is
flow, as shown in 䉲 Fig. 9.12. If the blood gauge pressure in assumed to be incompressible.)
the vein is 20.0 mm Hg, above what height should the bottle
be placed for the IV blood transfusion to function properly? (continued on next page)
SOLUTION.
Given: pv = 20.0 mm Hg (vein gauge pressure) Find: h (height for pv 7 20 mm Hg)
r = 1.05 * 103 kg>m3
(whole blood density from Table 9.2)
First, the common medical unit of mm Hg (or torr) needs to be changed to the SI unit of
pascal (Pa, or N>m2):
pv = 120.0 mm Hg23133 Pa>1mm Hg24 = 2.66 * 103 Pa
Then, for p 7 pv ,
p = rgh 7 pv
or
2.66 * 103 Pa
= 0.259 m 1L 26 cm2
pv
h 7
11.05 * 103 kg>m3219.80 m>s22
=
rg
The IV bottle needs to be at least 26 cm above the injection site. 䉱 F I G U R E 9 . 1 2 What height is
needed? See Example text for
FOLLOW-UP EXERCISE. The normal (gauge) blood pressure range is commonly reported as description.
120>80 (in millimeters Hg). Why is the blood pressure of 20 mm Hg in this Example so low?
DID YOU LEARN

➥ By Pascal’s principle, pressure applied to an enclosed fluid is transmitted to every
point in the fluid and the walls of the container.
➥ Absolute pressure is the measured pressure plus atmospheric pressure. Gauge
pressure is that measured above (or below) atmospheric pressure, so absolute
pressure is gauge pressure plus atmospheric pressure.
9.3 Buoyancy and Archimedes’ Principle

➥ What is meant by buoyant force?

➥ What does Archimedes’ principle tell us?
➥ Under what conditions will an object float in a fluid?
When placed in a fluid, an object will either sink or float. This is most commonly
observed with liquids; for example, objects float or sink in water. But the same
effect occurs in gases: A falling object sinks in the atmosphere, while other objects
float (䉳 Fig. 9.13).
Things float because they are buoyant, or are buoyed up. For example, if you
immerse a cork in water and release it, the cork will be buoyed up to the surface
and float there. From your knowledge of forces, you know that such motion
requires an upward net force on an object. That is, there must be an upward force
acting on the object that is greater than the downward force of its weight. The
forces are equal when the object floats in equilibrium. The upward force resulting
from an object being wholly or partially immersed in a fluid is called the buoyant
force.
䉱 F I G U R E 9 . 1 3 Fluid buoyancy How the buoyant force comes about can be seen by considering a buoyant
The air is a fluid in which objects object being held under the surface of a fluid (䉴 Fig. 9.14a). The pressures on the
such as this dirigible float. The upper and lower surfaces of the block are p1 = rf gh1 and p2 = rf gh2 , respectively,
helium inside the blimp is less where rf is the density of the fluid. Thus, there is a pressure difference
¢p = p2 - p1 = rf g1h2 - h12 between the top and bottom of the block, which
dense than the surrounding air, and
displacing its volume of air, the
blimp is supported by the resulting gives an upward force (the buoyant force) Fb. This force is balanced by the applied
buoyant force. force and the weight of the block.
9.3 BUOYANCY AND ARCHIMEDES’ PRINCIPLE 329
It is not difficult to derive an expression for the magnitude of the buoyant force.
Pressure is force per unit area. Thus, if both the top and bottom areas of the block
are A, the magnitude of the net buoyant force in terms of the pressure difference is
Fb = p2 A - p1 A = 1¢p2A = rf g1h2 - h12A F
Since 1h2 - h12A is the volume of the block and hence the volume of fluid dis-
h1
p1 = rf gh1
placed by the block, Vf , the expression for Fb may be written as h2
p2 = rf gh2
Fb = rf gVf mg
Fb
But rf Vf is simply the mass of the fluid displaced by the block, mf. Thus, the
expression for the buoyant force becomes Fb = mf g: The magnitude of the buoy-
ant force is equal to the weight of the fluid displaced by the block (Fig. 9.14b). This
general result is known as Archimedes’ principle: Δp = rf g(h2 – h1)
(a)
A body immersed wholly or partially in a fluid experiences a buoyant force equal in
magnitude to the weight of the volume of fluid that is displaced:
0
Fb = mf g = rf gVf (9.14)
5 8.0 N
Archimedes (287–212 BCE), a Greek scientist, was given the task of determining
whether a gold crown made for the king was pure gold or contained a quantity of
silver. Legend has it that the solution came to him upon immersing himself in a 10
Newtons
full bath. (See the Physics Facts at the beginning of the chapter.) It is said that he
was so excited he jumped out and ran home through the streets of the city
(unclothed) shouting “Eureka! Eureka!” (Greek for “I have found it”).
Archimedes’ solution to the problem involved density and volume and it may
have gotten him thinking about buoyancy. 10 N
INTEGRATED EXAMPLE 9.7 Lighter Than Air: Buoyant Force

A spherical helium-filled weather balloon has a radius of 1.10 m. (a) Does the buoyant
Newtons
force on the balloon depend on the density of (1) helium, (2) density of air, or (3) the
weight of the rubber “skin”? 3rair = 1.29 kg>m3 and rHe = 0.180 kg>m3.4 (b) Compute
the magnitude of the buoyant force on the balloon. (c) The balloon’s rubber skin has a (b)
mass of 1.20 kg. When released, what is the magnitude of the balloon’s initial accelera-
tion if it carries a payload with a mass of 3.52 kg? 䉱 F I G U R E 9 . 1 4 Buoyancy and
Archimedes’ principle (a) A buoyant
(A) CONCEPTUAL REASONING. The upward buoyant force has nothing to do with the force arises from the difference in
helium or rubber skin and is equal to the weight of the displaced air, which can be pressure at different depths. The
found from the balloon’s volume and the density of air. So the answer is (2). pressure on the bottom of the sub-
(B, C) QUANTITATIVE REASONING AND SOLUTION. merged block (p2) is greater than
that on the top (p1), so there is a
Given: rair = 1.29 kg>m3 Find: (b) Fb (buoyant force)
(buoyant) force directed upward.
3 (c) a (initial acceleration) (Shifted for clarity.)
rHe = 0.180 kg>m
(b) Archimedes’ principle: The
ms = 1.20 kg buoyant force on the object is equal
mp = 3.52 kg to the weight of the volume of fluid
displaced. (The scale is set to read
r = 1.10 m
zero when the container is empty.)
(b) The volume of the balloon is
V = 14>32pr3 = 14>32p11.10 m23 = 5.58 m3
Then the buoyant force is equal to the weight of the air displaced:
Fb = mair g = 1rair V2g = 11.29 kg>m3215.58 m3219.80 m>s22 = 70.5 N
(c) Draw a free-body diagram. There are three weight forces downward—those of the
helium, the rubber skin, and the payload—and the upward buoyant force. Sum these
forces to find the net force, and then use Newton’s second law to find the acceleration.

The weights of the helium, rubber skin, and payload are as follows:
wHe = mHe g = 1rHe V2g = 10.180 kg>m3215.58 m3219.80 m>s22 = 9.84 N
ws = ms g = 11.20 kg219.80 m>s 22 = 11.8 N
wp = mp g = 13.52 kg219.80 m>s 22 = 35.5 N
Summing the forces (taking upward as positive),
Fnet = Fb - wHe - ws - wp = 70.5 N - 9.84 N - 11.8 N - 35.5 N = 13.4 N
and with the masses found from the weights:
Fnet Fnet 13.4 N
a = = = = 2.34 m>s2
mtotal mHe + ms + mp 1.00 kg + 1.20 kg + 3.52 kg
F O L L O W - U P E X E R C I S E . As the balloon rises, it eventually stops accelerating and rises
at a constant velocity for a short time, then starts sinking toward the ground. Explain
this behavior in terms of atmospheric density and temperature. [Hint: Temperature and
air density decrease with altitude. The pressure of a quantity of gas is directly propor-
tional to temperature.]
EXAMPLE 9.8 Your Buoyancy in Air

Air is a fluid and our bodies displace air. And so, a buoyant placed, which is the same as the volume of the person. The
force is acting on each of us. Estimate the magnitude of the question is, how do we find the volume of the person?
buoyant force on a 75-kg person due to the air displaced. The mass is given, and if the density of the person were
known, the volume could be found 1r = m>V2 or V = m>r.
T H I N K I N G I T T H R O U G H . The key word here is estimate, Here is where the estimate comes in. Most people can barely
because not much data are given. We know that the buoyant float in water, so the density of the human body is about that
force is Fb = ra gV, where ra is the density of air (which can of water, r = 1000 kg>m3. Using this estimate, the buoyant
be found in Table 9.2), and V is the volume of the air dis- force can also be estimated.
SOLUTION.
Given: m = 75 kg Find: Fb (buoyant force)

ra = 1.29 kg>m3 (Table 9.2)
rp = 1000 kg>m3 (estimated density of person)
First, let’s find the volume of the person:
m 75 kg
Vp = = = 0.075 m3
rp 1000 kg>m3
Then,
Fb = ra gVp = 11.29 kg>m3219.8 m>s2210.075 m32
= 0.95 N 1L 1.0 N or 0.225 lb2
This amount is not much when you weigh yourself. But it does mean that your weight is L 0.2 lb more than the scale reading.
F O L L O W - U P E X E R C I S E . Estimate the buoyant force on a helium-filled weather balloon that has a diameter on the order of a
meteorologist’s arm span (arms held horizontally), and compare with the result in the Example.
INTEGRATED EXAMPLE 9.9 Weight and Buoyant Force: Archimedes’ Principle

A container of water with an overflow tube, similar to that reading then be (1) exactly 48 N, (2) between 40 N and 48 N,
shown in Fig. 9.14b, sits on a scale that reads 40 N. The water (3) exactly 40 N, or (4) less than 40 N? (b) Suppose you
level is just below the exit tube in the side of the container. pushed down on the wooden cube with your finger such that
(a) An 8.0-N cube of wood is placed in the container. The the top surface of the cube was even with the water level.
water displaced by the floating cube runs out the exit tube How much force would have to be applied if the wooden
into another container that is not on the scale. Will the scale cube measured 10 cm on a side?
9.3 BUOYANCY AND ARCHIMEDES’ PRINCIPLE 331
(A) CONCEPTUAL REASONING. By Archimedes’ principle, the block on the water is transmitted to the bottom of the container
block is buoyed upward with a force equal in magnitude to the (Pascal’s principle) and is registered on the scale. (Make a
weight of the water displaced. Since the block floats, the sketch showing the forces on the cube.)
upward buoyant force must balance the weight of the cube and
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . Here three
so has a magnitude of 8.0 N. Thus, a volume of water weighing
forces are acting on the stationary cube: the buoyant force upward
8.0 N is displaced from the container as 8.0 N of weight is added
and the weight and the force applied by the finger downward.
to the container. The scale still reads 40 N, so the answer is (3).
The weight of the cube is known, so to find the applied finger
Note that the upward buoyant force and the block’s
force, we need to determine the buoyant force on the cube.
weight act on the block. The reaction force (pressure) of the
Given: / = 10 cm = 0.10 m (side length of cube) Find: Ff (downward applied force necessary
w = 8.0 N (weight of cube) to put cube even with water level)
The summation of the forces acting on the cube is the fluid is that of water, which is known (1.0 * 103 kg>m3,
©Fy = + Fb - w - Ff = 0, where Fb is the upward buoyant Table 9.2), so
Fb = rf gVf = 11.0 * 103 kg>m3219.8 m>s2210.10 m23 = 9.8 N
force and Ff is the downward force applied by the finger.
Hence, Ff = Fb - w. As we know, the magnitude of the buoy-
ant force is equal to the weight of the water the cube dis- Thus,
places, which is given by Fb = rf gVf (Eq. 9.14). The density of Ff = Fb - w = 9.8 N - 8.0 N = 1.8 N
F O L L O W - U P E X E R C I S E . In part (a), would the scale still read 40 N if the object had a density greater than that of water? In
part (b), what would the scale read?
BUOYANCY AND DENSITY

It is commonly said that helium and hot-air balloons float because they are lighter
than air. To be technically correct, it should be said that the balloons are less dense
than air. An object’s density will tell you whether it will sink or float in a fluid, as
long as you also know the density of the fluid. Consider a solid uniform object
that is totally immersed in a fluid. The weight of the object is
wo = mo g = ro Vo g
The weight of the volume of fluid displaced, or the magnitude of the buoyant force, is
Fb = wf = mf g = rf Vf g
If the object is completely submerged, Vf = Vo . Dividing the second equation by the
first gives
or Fb = a b wo (object completely submerged)

Fb rf rf
= (9.15)
wo ro ro
Thus, if ro is less than rf , then Fb will be greater than wo , and the object will be
buoyed to the surface and float. If ro is greater than rf , then Fb will be less than wo ,
and the object will sink. If ro equals rf , then Fb will be equal to wo , and the object
will remain in equilibrium at any submerged depth (as long as the density of the
fluid is constant). If the object is not uniform, so that its density varies over its vol-
ume, then the density of the object in Eq. 9.15 is the average density.
Expressed in words, these three conditions are as follows:
An object will float in a fluid if the average density of the object is less than the den-
sity of the fluid 1ro 6 rf2
An object will sink in a fluid if the average density of the object is greater than the
density of the fluid 1ro 7 rf2
An object will be in equilibrium at any submerged depth in a fluid if the average den-
sity of the object and the density of the fluid are equal 1ro = rf2
See 䉴 Fig. 9.15 for an example of the last condition. 䉱 F I G U R E 9 . 1 5 Equal densities
A quick look at Table 9.2 will tell you whether an object will float in a fluid, and buoyancy This soft drink con-
regardless of the shape or volume of the object. The three conditions just stated tains colored gelatin beads that
remain suspended for months with
also apply to a fluid in a fluid, provided that the two are immiscible (do not mix). virtually no change. What is the
For example, you might think that cream is “heavier” than skim milk, but that’s density of the beads compared to
not so: Since cream floats on milk, it is less dense than milk. the density of the drink?
DEMONSTRATION 2 Buoyancy and Density

This demonstration of buoyancy shows that the overall density of a
can of Diet Coke is less than that of water while the density of a can
of a Classic Coke is greater. Consider the following questions: Does
one can have a greater volume of metal? higher gas pressure inside?
more fluid volume? Do calories make a difference? Investigate the
possibilities to determine the reason(s) for the different densities.
The can of Classic Coke sinks and the can of Diet Coke
floats.
In general, the densities of objects or fluids will be assumed to be uniform and

constant in this book. (The density of the atmosphere does vary with altitude, but
is relatively constant near the surface of the Earth.) In any event, in practical appli-
cations it is the average density of an object that often matters with regard to float-
ing and sinking. For example, an ocean liner is, on average, less dense than water,
even though it is made of steel. Most of its volume is occupied by air, so the liner’s
average density is less than that of water. Similarly, the human body has air-filled
spaces, so most of us float in water. The surface depth at which a person floats
depends on his or her density. (Why?)
In some instances, the overall density of an object is purposefully varied. For
example, a submarine submerges by flooding its tanks with seawater (called
“taking on ballast”), which increases its average density. When the sub is ready to
surface, the water is pumped out of the tanks, so the average density of the sub
becomes less than that of the surrounding seawater.
Similarly, many fish control their depths by using their swim bladders or gas
bladders. A fish changes or maintains buoyancy by regulating the volume of gas in
the gas bladder. Maintaining neutral buoyancy (neither rising nor sinking) is
important because it allows the fish to stay at a particular depth for feeding. Some
fish may move up and down in the water in search of food. Instead of using up
energy to swim up and down, the fish alters its buoyancy to rise and sink.
This is accomplished by adjusting the quantities of gas in the gas bladder. Gas
is transferred from the gas bladder to the adjoining blood vessels and back again.
Deflating the bladder decreases the volume and increases the average density, and
the fish sinks. Gas is forced into the surrounding blood vessels and carried away.
Conversely, to inflate the bladder, gases are forced into the bladder from the
blood vessels, thereby increasing the volume and decreasing the average density,
and the fish rises. These processes are complex, but Archimedes’ principle is being
applied in a biological setting.
EXAMPLE 9.10 Float or Sink? Comparison of Densities

A uniform solid cube of material 10.0 cm on each side has a Both of these forces are related to the cube’s volume, so we
mass of 700 g. (a) Will the cube float in water? (b) If so, how can write them in terms of that volume and equate them.
much of its volume would be submerged?
SOLUTION. It is sometimes convenient to work in cgs units
THINKING IT THROUGH. (a) The question is whether the den- in comparing small quantities. For densities in grams per
sity of the material the cube is made of is greater or less than cubic centimeter, divide the values in Table 9.2 by 103, or drop
that of water, so we compute the cube’s density. (b) If the cube the “ * 103” from the values given for solids and liquids, and
floats, then the buoyant force and the cube’s weight are equal. replace with “ * 10-3” for gases.
9.4 FLUID DYNAMICS AND BERNOULLI’S EQUATION 333
Given: m = 700 g Find: (a) Whether the cube will float in water
L = 10.0 cm (b) The percentage of the volume submerged
rH O = 1.00 * 103 kg>m3 if the cube does float
2
= 1.00 g>cm3 (Table 9.2)
(a) The density of the cube is cube displaces. Equating the expressions for weight and
700 g buoyant force gives
m m
= 0.700 g>cm3 6 rH O = 1.00 g>cm3
110.0 cm23
rc = = 3 =
Vc L 2 rH2O gVH2O = rc gVc
Since rc is less than rH2O the cube will float. or
(b) The weight of the cube is wc = rc gVc . When the cube is VH2O rc 0.700 g>cm3
= = = 0.700
floating, it is in equilibrium, which means that its weight is Vc rH 1.00 g>cm2
2O
balanced by the buoyant force. That is, Fb = rH O gVH2O,
2
where VH2O is the volume of water the submerged part of the Thus, VH2O = 0.70Vc , and 70% of the cube is submerged.
FOLLOW-UP EXERCISE. Most of an iceberg floating in the ocean is submerged (䉲 Fig. 9.16). The visible portion is the proverbial
“tip of the iceberg.” What percentage of an iceberg’s volume is seen above the surface? (Note: Icebergs are frozen fresh water
floating in salty sea water.)
A quantity called specific gravity is related to density. It is commonly used for

liquids, but also applies to solids. The specific gravity (sp. gr.) of a substance is
equal to the ratio of the density of the substance 1rs2 to the density of water 1rH O2
2
at 4 °C, the temperature for maximum density:
rs
sp. gr. =
rH O
2
Because it is a ratio of densities, specific gravity has no units. In cgs units,

rH O = 1.00 g>cm3, so
2
1rs in g>cm3 only2

rs
sp. gr. = = rs
1.00
That is, the specific gravity of a substance is equal to the numerical value of its
density in cgs units. For example, if a liquid has a density of 1.5 g>cm3, its specific
gravity is 1.5, which tells you that it is 1.5 times as dense as water. (As pointed out
earlier, to get density values for solids and liquids in grams per cubic centimeter,
divide the value in Table 9.2 by 103.)
DID YOU LEARN?

➥ Buoyant force is the upward force resulting from an object being wholly or partially 䉱 F I G U R E 9 . 1 6 The tip of the
submerged in a fluid. iceberg The vast majority of an ice-
➥ Archimedes’ principle allows the buoyant force to be measured: The magnitude of berg’s bulk is underneath the
water, as illustrated here in a false
the buoyant force on an object is equal to the weight of the volume of fluid
photo. See Example 9.10 Follow-Up
displaced. Exercise.
➥ An object will float if its average density is less than that of the fluid.
9.4 Fluid Dynamics and Bernoulli’s Equation

➥ What are the characteristics of ideal fluid flow?

➥ What does the equation of continuity tell about incompressible fluid flow?
➥ On what is Bernoulli’s equation based?
In general, fluid motion is difficult to analyze. For example, think of trying to

describe the motion of a particle (a molecule, as an approximation) of water in a rush-
ing stream. The overall motion of the stream may be apparent, but a mathematical
description of the motion of any one particle of it may be virtually impossible
Streamlines because of eddy currents (small whirlpool motions), the gushing of water over rocks,
v1 frictional drag on the stream bottom, and so on. A basic description of fluid flow is
conveniently obtained by ignoring such complications and considering an ideal
fluid. Actual fluid flow can then be approximated with reference to this simpler theo-
retical model.
In this simplified approach to fluid dynamics, it is customary to consider four
characteristics of an ideal fluid. In such a fluid, flow is (1) steady, (2) irrotational,
(3) nonviscous, and (4) incompressible.
v2
Condition 1: Steady flow means that all the particles of a fluid have the same velocity
as they pass a given point.
Steady flow might be called smooth or regular flow. The path of steady flow can
be depicted in the form of streamlines (䉳 Fig. 9.17a). Every particle that passes a
Paddle wheel
particular point moves along a streamline. That is, every particle moves along the
same path (streamline) as particles that passed by earlier. Streamlines never cross;
(a)
if they did, a particle would have alternative paths and abrupt changes in its
velocity, in which case the flow would not be steady.
Steady flow requires low velocities. For example, steady flow is approximated
by the flow relative to a canoe that is gliding slowly through still water. When the
flow velocity is high, eddies tend to appear, especially near boundaries, and the
flow becomes turbulent, as in Fig. 9.17b.
Streamlines also indicate the relative magnitude of the velocity of a fluid. The
velocity is greater where the streamlines are closer together. Notice this effect in
Fig. 9.17a. The reason for it will be explained shortly.
Condition 2: Irrotational flow means that a fluid element (a small volume of the fluid)
has no net angular velocity. This condition eliminates the possibility of whirlpools and
eddy currents. (Nonturbulent flow.)
Consider the small paddle wheel in Fig. 9.17a. With a zero net torque, the wheel
does not rotate. Thus, the flow is irrotational.
Condition 3: Nonviscous flow means that viscosity is negligible.
Viscosity refers to a fluid’s internal friction, or resistance to flow. (For example,
honey has a much greater viscosity than water.) A truly nonviscous fluid would flow
freely with no internal energy loss. Also, there would be no frictional drag between
(b) the fluid and the walls containing it. In reality, when a liquid flows through a pipe,
the speed is lower near the walls because of frictional drag and is higher toward the
䉱 F I G U R E 9 . 1 7 Streamline flow center of the pipe. (Viscosity is discussed in more detail in Section 9.5.)
(a) Streamlines never cross and are Condition 4: Incompressible flow means that the fluid’s density is constant.
closer together in regions of greater
fluid velocity. The stationary paddle Liquids can usually be considered incompressible. Gases, by contrast, are quite
wheel indicates that the flow is irro- compressible. Sometimes, however, gases approximate incompressible flow—for
tational, or without whirlpools and example, air flowing relative to the wings of an airplane traveling at low speeds.
eddy currents. (b) The smoke from
an extinguished candle begins to Theoretical or ideal fluid flow is not characteristic of most real situations, but the
rise in nearly streamline flow, but analysis of ideal flow provides results that approximate, or generally describe, a vari-
quickly becomes rotational and ety of applications. Usually, this analysis is derived, not from Newton’s laws, but
turbulent. instead from two basic principles: conservation of mass and conservation of energy.
EQUATION OF CONTINUITY
If there are no losses of fluid within a uniform tube, the mass of fluid flowing into
the tube in a given time must be equal to the mass flowing out of the tube in the
same time (by the conservation of mass). For example, in 䉴 Fig. 9.18a, the mass
1¢m12 entering the tube during a short time 1¢t2 is
¢m1 = r1 ¢V1 = r11A 1 ¢x12 = r11A 1v1 ¢t2
where A1 is the cross-sectional area of the tube at the entrance and, in a time ¢t, a
fluid particle moves a distance equal to v1 ¢t. Similarly, the mass leaving the tube
in the same interval is (Fig. 9.18b)
¢m2 = r2 ¢V2 = r21A 2 ¢x22 = r21A 2v2 ¢t2
䉳 F I G U R E 9 . 1 8 Flow continuity
Ideal fluid flow can be described in
terms of the conservation of mass
A2
by the equation of continuity. See
text for description.
v1
A1 y2
F1 = p1A1 Δ m1
Density, 1
Δ x1 = v1 Δ t y1
(a) Mass enters tube
Δ m2 v2
Density, 2
F2 = p2A2
Δ x2 = v2 Δ t
y2
y1
(b) Mass exits tube
Since the mass is conserved, ¢m1 = ¢m2 , and it follows that
r1A 1v1 = r2A 2v2 or rAv = constant (9.16)
This general result is called the equation of continuity.

For an incompressible fluid, the density r is constant, so
A 1v1 = A 2v2 or Av = constant (for an incompressible fluid) (9.17)
This is sometimes called the flow rate equation. Av is called the volume rate of flow,
and is the volume of fluid that passes by a point in the tube per unit time. (The
units of Av are m2 # m>s = m3>s, volume per time.)
Note that the flow rate equation shows that the fluid speed is greater where the
cross-sectional area of the tube is smaller. That is,
v2 = a bv
A1
A2 1
and v2 is greater than v1 if A2 is less than A1. This effect is evident in the common
experience that the speed of water is greater from a hose fitted with a nozzle than
䉱 F I G U R E 9 . 1 9 Flow rate By the
that from the same hose without a nozzle (䉴 Fig. 9.19). flow rate equation, the speed of a
The flow rate equation can be applied to the flow of blood in your body. Blood fluid is greater when the cross-
flows from the heart into the aorta. It then makes a circuit through the circulatory sectional area of the tube through
system, passing through arteries, arterioles (small arteries), capillaries, and which the fluid is flowing is smaller.
Think of a hose that is equipped
venules (small veins) and back to the heart through veins. The speed is lowest in
with a nozzle such that the cross-
the capillaries. Is this a contradiction? No: The total area of the capillaries is much sectional area of the hose is made
larger than that of the arteries or veins, so the flow rate equation is still valid. smaller.
EXAMPLE 9.11 Blood Flow: Cholesterol and Plaque

High cholesterol in the blood can cause fatty deposits called T H I N K I N G I T T H R O U G H . The flow rate equation (Eq. 9.17)
plaques to form on the walls of blood vessels. Suppose a plaque applies, but note that no values of area or speed are given.
reduces the effective radius of an artery by 25%. How does this This indicates that we should use ratios.
partial blockage affect the speed of blood through the artery? (continued on next page)
SOLUTION. Taking the unclogged artery to have a radius r1, Rearranging and canceling,
that the plaque then reduces the effective radius to r2. r1 2
Given: r2 = 0.75r1 (for a 25% reduction) Find: v2 v2 = a b v1
r2
Writing the flow rate equation in terms of the radii, From the given information, r1>r2 = 1>0.75, so
A 1v1 = A 2v2 v2 = 11>0.7522 v1 = 1.8v1
1pr212v1 = 1pr222v2 Hence, the speed through the clogged artery increases by 80%.
F O L L O W - U P E X E R C I S E . By how much would the effective radius of an artery have to be reduced to have a 50% increase in the
speed of the blood flowing through it?
EXAMPLE 9.12 Speed of Blood in the Aorta

Blood flows at a rate of 5.00 L>min through an aorta with a Let’s first find the cross-sectional area of the circular aorta.
A = pr2 = 13.14211.00 * 10-2 m22 = 3.14 * 10-4 m2
radius of 1.00 cm. What is the speed of blood flow in the
aorta?
Then the (volume) flow rate needs to be put into standard units.
THINKING IT THROUGH. It is noted that the flow rate is a vol-
ume flow rate, which implies the use of the flow rate equation 5.00 L>min = 15.00 L>min2110-3 m3>L211 min>60 s2
= 8.33 * 10-5 m3>s
(Eq. 9.17), Av = constant. Since the constant is in terms of
volume>time, the given flow rate is the constant.
Using the flow rate equation,
Given: Flow rate = 5.00 L>min Find: v (blood constant 8.33 * 10-5 m3>s
v = = = 0.265 m>s
r = 1.00 cm = 1.00 * 10-2 m speed) A 3.14 * 10-4 m2
F O L L O W - U P E X E R C I S E . Constrictions of the arteries occur with hardening of the arteries. If the radius of the aorta in this Exam-
ple were constricted to 0.900 cm, what would be the percentage change in blood flow?
BERNOULLI’S EQUATION
The conservation of energy or the general work–energy theorem leads to another
relationship that has great generality for fluid flow. This relationship was first
derived in 1738 by the Swiss mathematician Daniel Bernoulli (1700–1782) and is
named for him. Bernoulli’s result was
Wnet = ¢K + ¢U
1p1 - p22 = 12 ¢m1v 22 - v 212 + ¢mg1y2 - y12

¢m
r
where ¢m is a mass increment as in the derivation of the continuity equation.
Note that in working with a fluid, the terms in Bernoulli’s equation are work or
energy per unit volume 1J>m32. That is, W = F¢x = p1A¢x2 = p¢V and therefore
p = W>¢V (work>volume). Similarly, with r = m>V, we have 12 rv2 = 12 mv 2>V
(energy>volume) and rgy = mgy>V (energy>volume).
Canceling each ¢m and rearranging gives the common form of Bernoulli’s
equation:
p1 + 12 rv 21 + rgy1 = p2 + 12 rv22 + rgy2 (9.18)
or
p + 12 rv 2 + rgy = constant
Bernoulli’s equation, or principle, can be applied to many situations. For exam-

ple, for a fluid at rest 1v2 = v1 = 02. Bernoulli’s equation becomes
p2 - p1 = rg1y1 - y22
This is the pressure–depth relationship derived earlier (Eq. 9.10). Also, if there
is horizontal flow 1y1 = y22, then p + 12 rv 2 = constant, which indicates that
䉳 F I G U R E 9 . 2 0 Flow rate and

High High
pressure pressure pressure Taking the horizontal dif-
Low Low ference in flow heights to be negligi-
speed pressure ble in a constricted pipe, we obtain,
High for Bernoulli’s equation,
speed
v1 p + 12 rv2 = constant. In a region of
A1 v2 smaller cross-sectional area, the
flow speed is greater (see flow rate
Smaller equation); from Bernoulli’s equa-
A2 cross-sectional tion, the pressure in that region is
Larger cross-sectional area area lower than in other regions.
the pressure decreases if the speed of the fluid increases (and vice versa). This
effect is illustrated in 䉱 Fig. 9.20, where the difference in flow heights through the
pipe is considered negligible (so the rgy term drops out).
Chimneys and smokestacks are tall in order to take advantage of the more consis-
tent and higher wind speeds at greater heights. The faster the wind blows over the
top of a chimney, the lower the pressure, and the greater the pressure difference
between the bottom and top of the chimney. Thus, the chimney “draws” exhaust out
more efficiently. Bernoulli’s equation and the continuity equation 1Av = constant2
also tell you that if the cross-sectional area of a pipe is reduced so that the speed of
the fluid passing through it is increased, then the pressure is reduced.
The Bernoulli effect (as it is sometimes called) gives a simplistic explanation for
the lift of an airplane. Ideal airflow over an airfoil or wing is shown in 䉴 Fig. 9.21.
(Turbulence is neglected.) The wing is curved on the top side and is
High speed, low pressure
angled relative to the incident streamlines. As a result, the streamlines
above the wing are closer together than those below, which causes a
higher air speed and lower pressure above the wing. With a higher
pressure on the bottom of the wing, there is a net upward force, or lift.
This rather common explanation of lift is termed simplistic because Low speed, high pressure
Bernoulli’s effect does not apply to the situation. Bernoulli’s principle
requires the conditions of both ideal fluid flow and energy conservation
䉱 F I G U R E 9 . 2 1 Airplane lift—Bernoulli’s
within the system, neither of which is satisfied in aircraft flying condi- principle in action Because of the shape and
tions. It is perhaps better to rely on Newton’s laws, which always must be orientation of an airfoil or airplane wing, the air
satisfied. Basically, the wing deflects the airflow downward, giving rise to streamlines are closer together, and the air
a downward change in the airflow momentum and a downward force speed is greater above the wing than below it.
(Newton’s second law). This results in an upward reaction force on the By Bernoulli’s principle, the resulting pressure
difference supplies part of the upward force
wing (Newton’s third law). When this upward force exceeds the weight called the lift. (But, Bernoulli’s principle is not
of the plane, there is enough lift for takeoff and flight. applicable, see text.)
EXAMPLE 9.13 Flow Rate from a Tank: Bernoulli’s Equation

A cylindrical tank containing water has a small hole punched
in its side below the water level, and water runs out
(䉴 Fig. 9.22). What is the approximate initial flow rate of water
out of the tank in terms of the heights shown?
T H I N K I N G I T T H R O U G H . Equation 9.17 1A 1v1 = A 2v22 is the y2 y1
flow rate equation, where Av has units of m3>s, or
volume>time. The v terms can be related by Bernoulli’s equa- y2
tion, which also contains y, and can be used to find differ- y1
ences in height. The areas are not given, so relating the v
terms might require some sort of approximation, as will be
seen. (Note that the approximate initial flow rate is wanted.)
SOLUTION. 䉱 F I G U R E 9 . 2 2 Fluid flow from a tank The flow
Given: No specific values are Find: An expression for the rate is given by Bernoulli’s equation. See Example
given, so symbols approximate initial text for description.
will be used. water flow rate from
the hole (continued on next page)
Bernoulli’s equation, tank and A1 is that of the hole. Since A2 is much greater than
A1 , then v1 is much greater than v2 (initially, v2 L 0). So, to a
p1 + 12 rv21 + rgy1 = p2 + 12 rv22 + rgy2
good approximation,
v21 = 2g1y2 -y12 v1 = 22g1y2 - y12
can be used. Note that y2 - y1 is just the height of the surface
or
of the liquid above the hole. The atmospheric pressures acting
on the open surface and at the hole, p1 and p2 , respectively, The flow rate (volume>time) is then
are essentially equal and cancel from the equation, as does the
flow rate = A 1v1 = A 1 22g(y2 - y1)
density, so
v 21 - v22 = 2g1y2 - y12
Given the area of the hole and the height of the liquid above
it, the initial speed of the water coming from the hole and the
By the equation of continuity (the flow rate equation, Eq. flow rate can be found. (What happens as the water level
9.17), A 1v1 = A 2v2 , where A2 is the cross-sectional area of the falls?)
F O L L O W - U P E X E R C I S E . What would be the percentage change in the initial flow rate from the tank in this Example if the diame-
ter of the small circular hole were increased by 30.0%?
CONCEPTUAL EXAMPLE 9.14 A Stream of Water: Smaller and Smaller

You have probably observed that a steady stream of water flowing increases. Then, by Bernoulli’s principle, the liquid pressure
out of a kitchen faucet gets smaller the farther the water falls from inside the stream decreases. (See Fig. 9.20.) A pressure difference
the faucet. Why does that happen? between that inside stream and the atmospheric pressure on the
outside is thus created. As a result, there is an increasing inward
REASONING AND ANSWER. This effect can be explained by force as the stream falls, so it becomes smaller. Eventually, the
Bernoulli’s principle. As the water falls, it accelerates and its speed stream may get so thin that it breaks up into individual droplets.
FOLLOW-UP EXERCISE. The equation of continuity can also be used to explain this stream effect. Give this explanation.
DID YOU LEARN

➥ Ideal fluid flow is steady, irrrotational, nonviscous, and incompressible.
➥ For an incompressible fluid, the volume flow rate (Av) is constant.
➥ Bernoulli’s equation is based on the conservation of energy or the general
work–energy theorem.
*9.5 Sur face Tension, Viscosity, and Poiseuille’s Law

➥ What is surface tension?

➥ What is viscosity and how does it arise?
➥ In Poiseuille’s law, what is the effect of the radius of the flow tube?
SURFACE TENSION
The molecules of a liquid exert small attractive forces on each other. Even though
molecules are electrically neutral overall, there is often some slight asymmetry of
charge that gives rise to attractive forces between them (called van der Waals forces).*
Within a liquid, any molecule is completely surrounded by other molecules, and the
net force is zero (䉴Fig. 9.23a). However, for molecules at the surface of the liquid,
there is no attractive force acting from above the surface. (The effect of air molecules
is small and considered negligible.) As a result, net forces act upon the molecules of
the surface layer, due to the attraction of neighboring molecules just below the sur-
face. This inward pull on the surface molecules causes the surface of the liquid to
contract and to resist being stretched or broken, a property called surface tension.
If a sewing needle is carefully placed on the surface of a bowl of water, the sur-
face acts like an elastic membrane under tension. There is a slight depression in
*After Johannes van der Waals (1837–1923), a Dutch scientist who first postulated an intermolecular
force.
*9.5 SURFACE TENSION, VISCOSITY, AND POISEUILLE’S LAW 339
Drop
of F F
liquid Fy Fy
Fx Fx
mg
(a) (b) (c)
䉱 F I G U R E 9 . 2 3 Surface tension (a) The net force on a molecule in the interior of a liquid is zero, because the mol-
ecule is surrounded by other molecules. However, a nonzero fluid force acts on a molecule at the surface, due to the
attractive forces of the neighboring molecules just below the surface. (b) For an object such as a needle to form a
depression on the surface, work must be done, since more interior molecules must be brought to the surface to
increase its area. As a result, the surface area acts like a stretched elastic membrane, and the weight of the object is
supported by the upward components of the surface tension. (c) Insects such as this water strider can walk on water
because of the upward components of the surface tension, much as you might walk on a large trampoline. Note the
depressions in the surface of the liquid where the legs touch it.
the surface, and molecular forces along the depression act at an angle to the sur-
face (Fig. 9.23b). The vertical components of these forces balance the weight (mg)
of the needle, and the needle “floats” on the surface. Similarly, surface tension
supports the weight of a water strider (Fig. 9.23c).
The net effect of surface tension is to make the surface area of a liquid as small
as possible. That is, a given volume of liquid tends to assume the shape that has
the least surface area. As a result, drops of water and soap bubbles have spherical
shapes, because a sphere has the smallest surface area for a given volume
(䉲 Fig. 9.24). In forming a drop or bubble, surface tension pulls the molecules
together to minimize the surface area. (See Insight 9.4, The Lungs and Baby’s First
Breath for an example of surface tension in respiration.)
VISCOSITY
All real fluids have an internal resistance to flow, or viscosity, which can be con-
sidered to be friction between the molecules of a fluid. In liquids, viscosity is
caused by short-range cohesive forces, and in gases, it is caused by collisions
between molecules. (See the discussion of air resistance in Section 4.6.) The vis-
cous drag for both liquids and gases depends on their speeds and may be directly
proportional to it in some cases. However, the relationship varies with the condi-
tions; for example, the drag is approximately proportional to either v2 or v3 in tur-
bulent flow.
䉳 F I G U R E 9 . 2 4 Surface tension
at work Because of surface tension,
(a) water droplets and (b) soap bub-
bles tend to assume the shape that
minimizes their surface area—that
of a sphere.
(a) (b)
INSIGHT 9.4 The Lungs and Baby’s First Breath

Respiration, or breathing, is vital to life. It is a fascinating pro- sure of an internal fluid. That is, the wall tension is directly
cedure that supplies oxygen to the blood and carries away proportional to the spherical radius. So when the alveoli
carbon dioxide—and a lot of physics is involved. inflate, there is greater tension. Once they are inflated, exhala-
The process of respiration involves the lowering of the tion is accomplished when the diaphragm relaxes and the
diaphragm to increase the volume of the thoracic cavity. wall tension of the alveoli acts to force the air out. Also, there
Figure 1 shows a bell jar model of respiration. By the ideal gas is a fluid coating on the alveoli, which is a surfactant—a sub-
law (Section 10.3), the lowering of the diaphragm and the stance that lowers surface tension. A reduction in surface ten-
increasing of the volume of the thoracic cavity lower the pression makes it easier to inflate the alveoli on inhalation.
sure (p r 1>V), and air is inhaled. The inhalation process The pulmonary disease emphysema, most common in long-
inflates the alveoli—small balloonlike structures in the lungs, term smokers, results from an enlargement of the alveoli as
as illustrated in Fig. 2a. (Figure 2b shows an illustration of a some are destroyed and others either enlarge or combine
damaged lung, the cause and effects of which will be dis- (Fig. 2b). Normally it would take twice the pressure to inflate
cussed shortly.) a membrane with twice the radius. The enlarged alveoli pro-
The oxygen exchange with the blood takes place across the vide less recoil on exhalation, and a person with emphysema
membrane surfaces of the alveoli. The total membrane surface has difficulty breathing as well as reduced oxygen exchange.
in the lungs is on the order of 100 m2, with a thickness of less Now, about a baby’s first breath. Most everyone knows that
than a millionth of a meter ( 61 mm, micrometer), making the it is much more difficult to blow up a balloon for the first time
gas exchange very efficient. The behavior of the alveoli may than to blow it up again. This is because the applied pressure
be described by Laplace’s law and surface tension.* does not create much tension in the balloon to start the
Laplace’s law states that the larger a spherical membrane, stretching process. It takes a greater tension increase to
the greater the wall tension required to withstand the pres- expand a small balloon than to expand a large balloon. Con-
sider the tension ratios for a 3-cm expansion in radius—say,
from 1 cm to 4 cm A 41 = 4 B and 10 cm to 13 cm A 13
10 = 1.3 B .
*Pierre-Simon de Laplace (1749–1827) was a French astronomer
and mathematician.
In a newborn baby, the alveoli are small and collapsed and
must be inflated with an initial inhalation. The traditional
practice to accomplish this are slaps on the baby’s bottom to
make the newborn cry and inhale.
(a) (b)
(a) Inhalation (b) Exhalation
䉱 F I G U R E 2 Alveoli (a) Inhalation inflates the alveoli,
䉱 F I G U R E 1 Bell jar model of respiration (a) Lowering the balloonlike structures of the lungs. There are between 300
diaphragm (rubber sheet) and increasing the volume of the million and 400 million alveoli in each lung. (b) Pulmonary
thoracic cavity lowers the pressure and air is inhaled into the disease can cause the enlargement of the alveoli as some
lungs (balloons). (b) When the diaphragm moves upward, are destroyed and others enlarge or combine. As a result,
the process is reversed and air is exhaled. there is less oxygen exchange and a shortness of breath.
Internal friction causes the layers of a fluid to move relative to each other in
response to a shear stress. This layered motion, called laminar flow, is characteristic
of steady flow for viscous liquids at low velocities (䉴 Fig. 9.25a). At higher veloci-
ties, the flow becomes rotational, or turbulent, and difficult to analyze.
Since there are shear stresses and shear strains (deformation) in laminar flow,
the viscous property of a fluid can be described by a coefficient, like the elastic
moduli discussed in Section 9.1. Viscosity is characterized by a coefficient of
viscosity, h (the Greek letter eta), commonly referred to as simply the viscosity.
A F A v 䉳 F I G U R E 9 . 2 5 Laminar flow
(a) A shear stress causes layers of a
Fluid fluid to move over each other in
h laminar flow. The shear force and
the flow rate depend on the viscos-
v=0 ity of the fluid. (b) For laminar flow
through a pipe, the speed of the
Parallel
fluid is less near the walls of the
planes
(a) pipe than near the center because of
frictional drag between the walls
and the fluid.
Velocity of fluid
p1 r p2
v
v
L
(b)
The coefficient of viscosity is, in effect, the ratio of the shear stress to the rate of
change of the shear strain (since motion is involved). Unit analysis shows that the
SI unit of viscosity is the pascal-second 1Pa # s2 This combined unit is called the
poiseuille (Pl), in honor of the French scientist Jean Poiseuille (1797–1869), who
studied the flow of liquids, particularly blood. (Poiseuille’s law on flow rate will
be presented shortly.) The cgs unit of viscosity is the poise (P). A smaller multiple,
the centipoise (cP), is widely used because of its convenient size; 1 P = 102 cP.
The viscosities of some fluids are listed in 䉴 Table 9.3. The greater
the viscosity of a liquid, which is easier to visualize than that of a TABLE 9.3 Viscosities of Various Fluids*
gas, the greater the shear stress required to get the layers of the liq-
uid to slide along each other. Note, for example, the large viscosity Viscosity (H)
of glycerin compared to that of water.* Fluid [Poiseuille (Pl)]
As you might expect, viscosity, and thus fluid flow, varies with
temperature, which is evident from the old saying, “slow as Liquids
molasses in January.” A familiar application is the viscosity grad- Alcohol, ethyl 1.2 * 10-3
ing of motor oil used in automobiles. In winter, a low-viscosity, or Blood, whole (37 °C) 1.7 * 10-3
relatively thin, oil should be used (such as SAE grade 10W
Blood plasma (37 °C) 2.5 * 10-3
or 20W), because it will flow more readily, particularly when the
engine is cold at startup. In summer, a higher viscosity, or thicker, Glycerin 1.5 * 10-3
oil is used (SAE 30, 40, or even 50).† Mercury 1.55 * 10-3
Seasonal changes in the grade of motor oil are not necessary if Oil, light machine 1.1
you use the multigrade, year-round oils. These oils contain addi-
Water 1.00 * 10-3
tives called viscosity improvers, which are polymers whose mole-
cules are long, coiled chains. An increase in temperature causes the Gases
molecules to uncoil and intertwine. Thus, the normal decrease in Air 1.9 * 10-5
viscosity is counteracted. The action is reversed on cooling, and the
Oxygen 2.2 * 10-5
oil maintains a relatively small viscosity range over a large temper-
ature range. Such motor oils are graded, for example, as SAE *At 20 °C unless otherwise indicated.
10W-30 (“ten-W-thirty”).
*If you want to think about a substance with a very large viscosity, consider glass. It has been said
that the glass in the stained glass windows of medieval churches has “flowed” over time, such that the
panes are now thicker at the bottom than at the top. However a more recent analysis indicates that
window glass may flow over incredibly long periods that exceed the limits of human history. On
human time scales, such a flow would not be evident. [See E. D. Zanotto, American Journal of Physics, 66
(May 1998), 392–395.]
†
SAE stands for Society of Automotive Engineers, an organization that designates the grades of motor
oils based on their viscosity.
POISEUILLE’S LAW
Viscosity makes analyzing fluid flow difficult. For example, when a fluid flows
through a pipe, there is frictional drag between the liquid and the walls, and the
fluid speed is greater toward the center of the pipe (Fig. 9.25b). In practice, this
effect makes a difference in a fluid’s average flow rate Q = Av = ¢V>¢t (see
Eq. 9.17), which describes the volume 1¢V2 of fluid flowing past a given point
during a time ¢t. The SI unit of flow rate is cubic meters per second 1m3>s2. The
flow rate depends on the properties of the fluid and the dimensions of the pipe, as
well as on the pressure difference 1¢p2 between the ends of the pipe.
Jean Poiseuille studied flow in pipes and tubes, assuming a constant viscosity
and steady or laminar flow. He derived the following relationship, known as
Poiseuille’s law, for the flow rate:
¢V pr 4 ¢p
Q = = (9.19)
¢t 8hL
Here, r is the radius of the pipe and L is its length.

As expected, the flow rate is inversely proportional to the viscosity 1h2 and the
length of the pipe. Also as expected, the flow rate is directly proportional to the
pressure difference ¢p between the ends of the pipe. Somewhat surprisingly, how-
ever, the flow rate is proportional to r4, which makes it more highly dependent on
the radius of the tube than might have been thought.
An application of fluid flow in a medical IV was examined in Example 9.6.
However, Poiseuille’s law, which incorporates the flow rate, affords more reality
to this application, as the next Example shows.
EXAMPLE 9.15 Poiseuille’s Law: A Blood Transfusion

A hospital patient needs a blood IV (intravenous) transfusion, T H I N K I N G I T T H R O U G H . This is an application of Poiseuille’s
which will be administered through a vein in the arm via a law (Eq. 9.19) to find the pressure needed at the inlet of the
gravity IV. The physician wishes to have 500 cc of whole needle that will provide the required flow rate (Q). Note that
blood delivered over a period of 10 min by an 18-gauge nee- ¢p = pin - pout (inlet pressure minus outlet pressure). Know-
dle with a length of 50 mm and an inner diameter of 1.0 mm. ing the inlet pressure, the required height of the bag can be
At what height above the arm should the bag of blood be found, as in Example 9.6. (Caution: There are a lot of nonstan-
hung? (Assume a venous blood pressure of 15 mm Hg.) dard units here, and some quantities are assumed to be
known from tables.)
SOLUTION. First writing the given (and known) quantities and converting to standard SI units:
Given: ¢V = 500 cc = 500 cm3 11 m3>106cm32 Find: h (height of bag)
= 5.00 * 10-4 m3
¢t = 10 min = 600 s = 6.00 * 102 s
L = 50 mm = 5.0 * 10-2 m
d = 1.0 mm, or r = 0.50 mm = 5.0 * 10-4 m
pout = 15 mm Hg = 15 torr (133 Pa>torr) = 2.0 * 103 Pa
h = 1.7 * 10-3 Pl (whole blood, from Table 9.3)
The flow rate is
5.00 * 10-4 m3
= 8.33 * 10-7 m3>s
¢V
Q = =
¢t 6.00 * 102 s
Inserting this number into Eq. 9.19 and solving for ¢p:
8hLQ 811.7 * 10-3 Pl215.0 * 10-2 m218.33 * 10-7 m3>s2
¢p = = = 2.9 * 103 Pa
pr4 p15.0 * 10-4 m24
With ¢p = pin - pout ,

pin = ¢p + pout = 12.9 * 103 Pa2 + 12.0 * 103 Pa2 = 4.9 * 103 Pa
Then, to find the height of the bag that will deliver this amount of pressure, we use pin = rgh (where rwhole blood = 1.05 * 103 kg>m3
from Table 9.2). Thus,
pin 4.9 * 103 Pa
h = = 0.48 m
11.05 * 103 kg>m3219.80 m>s22
=
rg
Hence, for the prescribed flow rate, the bag of blood should be hung about 48 cm above the needle in the arm.
F O L L O W - U P E X E R C I S E . Suppose the physician wants to follow up the blood transfusion with 500 cc of saline solution at the
same rate of flow. At what height should the saline bag be placed? (The isotonic saline solution administered by IV is a 0.85%
aqueous salt solution, which has the same salt concentration as do body cells. To a good approximation, saline has the same den-
sity as water.)
Gravity flow IVs are still used, but with modern technology, the flow rates of
IVs are now often controlled and monitored by machines (䉴 Fig. 9.26).
DID YOU LEARN?

➥ Surface tension arises from the inward pull on the surface molecules of a liquid,
which causes the surface to contract and resist being stretched or broken.
➥ Viscosity, the internal resistance to fluid flow, is caused by short-range cohesive
forces between molecules in a liquid and by molecular collisions in a gas.
➥ Poiseuille’s law indicates the flow rate in a pipe or tube depends highly on the
radius, r4.
䉴 F I G U R E 9 . 2 6 IV technology
The mechanism of intravenous injec-
tion is still a gravity assist, but IV
flow rates are now commonly con-
trolled and monitored by machines.
PULLING IT TOGETHER Sunken Treasure

A Spanish galleon is about to be boarded by bloodthirsty pirates T H I N K I N G I T T H R O U G H . This example uses the concepts of
in the shallows of a Caribbean island. To save a box of treasure Newton’s first law, buoyant force, and density, along with
on board, the captain orders his crew to secretly toss the box work and energy. (a) Constant velocity means that the box has
overboard, planning to come back for it later. The rectangular no net force on it. The free-body diagram should show this.
box is waterproof and measures 40.0 cm by 25.0 cm by 30.0 cm. There is an upward buoyant force and a downward pull of
It is made of wood and has mostly gold pieces inside, resulting gravity (weight). Since the box sinks with a constant velocity,
in an average box density three times that of seawater. there must be a third upward force to make the net force zero.
Sinking below the surface, the box moves at a constant ver- (b) The box’s weight can be found from the volume and den-
tical velocity of 1.15 m>s for 12.0 m (that’s 2 fathoms for sity, the buoyant force from Archimedes’ principle, and the
pirates) before hitting the bottom. (a) Draw the free-body dia- water drag force is the difference. (c) All the forces are con-
gram for the box, (b) determine the magnitudes of the forces stant; thus, work can be determined by using the definition in
on the box, and (c) calculate the work done by each force and Section 5.1. (d) Change in gravitational potential energy is
the net work done on the box. (d) Calculate the change in the discussed in Section 5.4. (e) The box’s kinetic energy is con-
box’s gravitational potential energy. (e) What is the change in stant; thus, by the work–energy theorem, the net work will be
the box’s total energy and what happens to it? zero.
SOLUTION.
Given: l * w * h = 40.0 cm * 25.0 cm * 30.0 cm Find: (a) free-body diagram
= 0.400 m * 0.250 m * 0.300 m (box dimensions) (b) wbox , Fb , Fdrag (forces on box)
r = 3rsw = 3.09 * 103 kg>m3 (density; rsw from Table 9.2) (c) Wgrav , Wb , Wdrag (work done by each force)
v = 1.15 m>s downward (box’s velocity) (d) ¢Ug (change in potential energy)
¢y = - 12.0 m, d = 12.0 (box’s vertical movement) (e) ¢E (change in total energy and what
happened to it)
(a) Since the box’s acceleration is zero, the net force on it must be zero. The upward Fdrag
buoyant force 1Fb2 is less than the downward pull of gravity (weight). Thus there
must be a water (fluid) drag force 1Fdrag2 upward to help cancel the weight force.
Fb
See 䉴 Fig. 9.27.
(b) The weight of the box depends on its volume and density. Its volume is ΣFi = 0
Vbox = l * w * h = 10.400 m210.250 m210.300 m2 = 3.00 * 10-2 m3
a=0
Thus its mass is

mbox = rq boxVbox = 13.09 * 103 kg>m3213.00 * 10-2 m32 = 92.7 kg w
and its weight is
䉱 F I G U R E 9 . 2 7 Free-body diagram for
wbox = mbox g = 192.7 kg219.80 m>s22 = 908 N the sinking chest The chest sinks with a
constant velocity so the sum of the forces
The buoyant force, by Archimedes’s principle, is equal in magnitude to the weight on it is zero. See Example text for
of the seawater displaced. Since the box is fully submerged, the volume of displaced description.
seawater is the same as the volume of the box. Therefore,
Fb = wsw = rswVsw g = 11.03 * 103 kg>m3213.00 * 10-2 m3219.80 m>s22 = 303 N
From the free-body diagram, the upward fluid drag force is the difference between
these two forces, or
Fdrag = wbox - wsw = 908 N - 303 N = 605 N
(c) The work done by a constant force is given by W = Fd cos u, where u is the angle between the displacement and the force
direction (Section 5.1 ). The weight is in the same direction as that of the box’s displacement, so u = 0° and
Wgrav = Fd cos u = 1908 N2112.0 m2 cos 0° = + 1.09 * 104 J
Both the buoyant force and the fluid drag force will do negative work because they are exactly opposite the direction of the dis-
placement. Hence the work done by the buoyancy force is
Wbuoy = Fd cos u = 1303 N2112.0 m2 cos 180° = - 3.64 * 103 J
and the work done by the fluid drag force is
Wdrag = Fd cos u = 1605 N2112.0 m2 cos 180° = - 7.26 * 103 J
The net work done on the box is zero, consistent with the work–energy theorem, since its kinetic energy does not change.
Wnet = a Wi = Wgrav + Wbuoy + Wdrag = 1.09 * 104 J - 3.64 * 103 J - 7.27 * 103 J = 0
i
(For the operation, the 1.09 * 104 J is converted to 10.9 * 103 J. Why?)
(d) The change in the box’s gravitational potential energy is
¢Ug = mbox g¢y = 192.7 kg219.80 m>s221 -12.0 m2 = - 1.09 * 104 J
(e) The box’s kinetic energy does not change, but its potential energy decreases, thus its total energy decreases, since
¢E = ¢K + ¢U = 0 + 1- 1.09 * 104 J2 = - 1.09 * 104 J
This energy is gained by the seawater in the form of increased thermal energy (it is slightly warmed) and turbulence (kinetic
energy of the water).
■ In the deformation of elastic solids, stress is a measure of Strain is a relative measure of the deformation a stress
the force causing the deformation: causes:
F change in length ¢L |L - Lo|
stress = (9.1) strain = = = (9.2)
A original length Lo Lo
Lo A
ΔL
F
F F
h1
(a) Tensile stress
p1 = rf gh1
h2
Lo p2 = rf gh2
mg
A Fb
ΔL
F F
(b) Compressional stress
■ An elastic modulus is the ratio of stress to strain. ■ An object will float in a fluid if the average density of the
object is less than the density of the fluid. If the average den-
Young’s modulus: sity of the object is greater than the density of the fluid, the
F>A object will sink.
Y = (9.4)
¢L>Lo ■ For an ideal fluid, the flow is (1) steady, (2) irrotational,
(3) nonviscous, and (4) incompressible. The following equa-
Shear modulus: tions describe such a flow:
F>A F>A Equation of continuity:
S = L (9.5)
x>h f
r1A 1v1 = r2A 2v2 or rAv = constant (9.16)
A Flow rate equation (for an incompressible fluid):
A 1v1 = A 2v2 or Av = constant (9.17)
Before
Before Bernoulli’s equation (for an incompressible fluid):
p1 + 12 rv21 + rgy1 = p2 + 12 rv22 + rgy2
x
φ A F
φ
F
h or
fs F After
p + 12 rv 2 + rgy = constant (9.18)
After
Bulk modulus:
High High
pressure pressure
F>A ¢p Low Low
B = = - (9.6) speed
High pressure
- ¢V>Vo ¢V>Vo speed
v1 v2
A1
■ Pressure is the force per unit area: Smaller
A2 cross-sectional
F Larger cross-sectional area area
p = (9.8a)
A
■ Bernoulli’s equation is a statement of the conservation of
Pressure–depth relationship (for an incompressible fluid at
energy for a fluid.
constant density):
■ Surface tension: The inward pull on the surface molecules
p = po + rgh (9.10)
of a liquid that causes the surface to contract and resist
■ Pascal’s principle. Pressure applied to an enclosed fluid is being stretched or broken.
transmitted undiminished to every point in the fluid and to
the walls of the container. Drop
of F F
Fi liquid Fy Fy
Fx Fx
Ai
mg
Ao
p
To reservoir (a) (b)

p p
Fluid Fo
Valve Valve
■ Viscosity: A fluid’s internal resistance to flow. All real fluids
■ Archimedes’ principle. A body immersed wholly or par- have a nonzero viscosity.
tially in a fluid is buoyed up by a force equal in magnitude ■ Poiseuille’s law (flow rate in pipes and tubes for fluids with
to the weight of the volume of fluid displaced.
constant viscosity and steady or laminar flow):
Buoyant force: pr 4 ¢p
¢V
Q = = (9.19)
Fb = mf g = rf gVf (9.14) ¢t 8hL

9.1 SOLIDS AND ELASTIC MODULI 11. A glass containing an ice cube is filled to the brim and
the cube floats on the surface. When the ice cube melts,
1. The pressure on an elastic body is described by (a) a
(a) water will spill over the sides of the glass, (b) the
modulus, (b) work, (c) stress, (d) strain.
water level decreases, (c) the water level is at the top of
2. Shear moduli are not zero for (a) solids, (b) liquids, the glass without any spill.
12. Comparing an object’s average density 1ro2 to that of a
(c) gases, (d) all of these.
3. A relative measure of deformation is (a) a modulus, fluid 1rf2. what is the condition for the object to float:
(b) work, (c) stress, (d) strain. (a) ro 6 rf , or (b) rf 6 ro?
4. The volume stress for the bulk modulus is (a) ¢p, (b) ¢V, 13. A block of material of known density (rb) floats two-
(c) Vo , (d) ¢V>Vo . thirds submerged in a liquid of unknown density (ro).
Using Archimedes’ principle, the unknown liquid den-
sity is (a) ru = 23 rb , (b) ru = 32 rb , (c) ru = 13 rb ,
9.2 FLUIDS: PRESSURE AND PASCAL’S
(d) ru = 3 rb .
PRINCIPLE
5. For a liquid in an open container, the total pressure at 9.4 FLUID DYNAMICS AND
any depth depends on (a) atmospheric pressure, (b) liq-
uid density, (c) acceleration due to gravity, (d) all of the
preceding. 14. If the speed at some point in a fluid changes with time,
6. For the pressure–depth relationship for a fluid 1p = rgh2.
the fluid flow is not (a) steady, (b) irrotational, (c) incom-
pressible, (d) nonviscous.
it is assumed that (a) the pressure decreases with depth,
(b) a pressure difference depends on the reference point, 15. An ideal fluid is not (a) steady, (b) compressible, (c) irro-
(c) the fluid density is constant, (d) the relationship tational, (d) nonviscous.
applies only to liquids. 16. Bernoulli’s equation is based primarily on (a) Newton’s
laws, (b) conservation of momentum, (c) a nonideal
7. When measuring automobile tire pressure, what type of
fluid, (d) conservation of energy.
pressure is this: (a) gauge, (b) absolute, (c) relative, or
(d) all of the preceding? 17. According to Bernoulli’s equation, if the pressure on the
liquid in Fig. 9.20 is increased, (a) the flow speed always
increases, (b) the height of the liquid always increases,
9.3 BUOYANCY AND ARCHIMEDES’ (c) both the flow speed and the height of the liquid may
PRINCIPLE increase, (d) none of the preceding.
8. A wood block floats in a swimming pool. The buoyant

force exerted on the block by water depends on (a) the *9.5 SURFACE TENSION, VISCOSITY,
volume of water in the pool, (b) the volume of the wood AND POISEUILLE’S LAW
block, (c) the volume of the wood block under water, 18. Water droplets and soap bubbles tend to assume the
(d) all of the preceding shape of a sphere. This effect is due to (a) viscosity, (b) sur-
9. If a submerged object displaces an amount of liquid of face tension, (c) laminar flow, (d) none of the preceding.
greater weight than its own and is then released, the 19. Some insects can walk on water because (a) the density of
object will (a) rise to the surface and float, (b) sink, water is greater than that of the insect, (b) water is viscous,
(c) remain in equilibrium at its submerged position. (c) water has surface tension, (d) none of the preceding.
10. A rock is thrown into a lake. While sinking, the buoyant 20. The viscosity of a fluid is due to (a) forces causing fric-
force (a) is zero, (b) decreases, (c) increases, (d) remains tion between the molecules, (b) surface tension, (c) den-
constant. sity, (d) none of the preceding.
9.1 SOLIDS AND ELASTIC MODULI 3. Ancient stonemasons sometimes split huge blocks of
rock by inserting wooden pegs into holes drilled in the
1. Which has a greater Young’s modulus, a steel wire or a
rock and then pouring water on the pegs. Can you
rubber band? Explain.
explain the physics that underlies this technique? [Hint:
2. Why are scissors sometimes called shears? Is this a Think about sponges and paper towels.]
descriptive name in the physical sense?
9.2 FLUIDS: PRESSURE AND PASCAL’S 12. A water dispenser for pets contains an inverted plastic
PRINCIPLE bottle, as shown in 䉲 Fig. 9.30. (The water is dyed blue
for contrast.) When a certain amount of water is drunk
4. 䉲 Figure 9.28 shows a famous “bed of nails” trick. The from the bowl, more water flows automatically from the
woman lies on a bed of nails with a cinder block on her bottle into the bowl. The bowl never overflows. Explain
chest. A person hits the anvil with a sledgehammer. The the operation of the dispenser. Does the height of the
nails do not pierce the woman’s skin. Explain why. water in the bottle depend on the surface area of the
water in the bowl?
䉳 F I G U R E 9 . 3 0 Pet
barometer See Conceptual
Question 12.
䉱 F I G U R E 9 . 2 8 A bed of nails See Conceptual Question 4.
5. Automobile tires are inflated to about 30 lb>in2, whereas 9.3 BUOYANCY AND ARCHIMEDES’
thin bicycle tires are inflated to 90 to 115 lb>in2—at least PRINCIPLE
three times as much pressure! Why? 13. (a) What is the most important factor in constructing a
6. (a) Why is blood pressure usually measured at the arm? life jacket that will keep a person afloat? (b) Why is it so
(b) Suppose the pressure reading were taken on the calf easy to float in Utah’s Great Salt Lake?
of the leg of a standing person. Would there be a differ- 14. An ice cube floats in a glass of water. As the ice melts,
ence, in principle? Explain. how does the level of the water in the glass change?
7. What kind of pressure does a sphygmomanometer Would it make any difference if the ice cube were hol-
measure? low? Explain.
8. What is the principle of drinking through a straw? (Liq- 15. Ocean-going ships in port are loaded to the so-called
uids aren’t “sucked” up.) Plimsoll mark, which is a line indicating the maximum
9. What is the absolute pressure inside a flat tire? safe loading depth. However, in New Orleans, located at
10. (a) Two dams form artificial lakes of equal depth. How- the mouth of the Mississippi River, where the water is
ever, one lake backs up 15 km behind the dam, and the brackish (partly salty and partly fresh), ships are loaded
other backs up 50 km behind. What effect does the dif- until the Plimsoll mark is somewhat below the water
ference in length have on the pressures on the dams? (b) line. Why?
Dams are usually thicker at the bottom. Why? 16. A heavy object is dropped into a lake. As it descends
11. Water towers (storage tanks) are generally bulb shaped, below the surface, does the pressure on it increase? Does
as shown in 䉲 Fig. 9.29. Wouldn’t it be better to have a the buoyant force on the object increase?
cylindrical storage tank of the same height? Explain. 17. Ocean liners weigh thousands of tons. How are they
made to float?
䉳 F I G U R E 9 . 2 9 Why 18. Two blocks of equal volume, one iron and one alu-
a bulb-shaped water minum, are dropped into a body of water. Which block
tower? See Conceptual will experience the greater buoyant force? Why?
Question 11.
19. An inventor comes up with an idea for a perpetual
motion machine, as illustrated in 䉲 Fig. 9.31. It contains a
䉳 FIGURE 9.31
Hg H2O Perpetual motion? See
?
sealed chamber with mercury (Hg) in one half and water equation, explain how this concavity supplies extra
(H2O) in the other. A cylinder is mounted in the center and downward force to the car in addition to that supplied
is free to rotate. The inventor reasons that since mercury is by the front and rear wings. (b) What is the purpose of
much denser than water (13.6 g>cm3 to 1.00 g>cm3), the the “spoiler” on the back of the racer?
weight of the mercury displaced by half the cylinder is 25. Here are two common demonstrations of Bernoulli
much greater than the water displaced by the other half. effects: (a) If you hold a narrow strip of paper in front of
Therefore, the buoyant force on the mercury side is your mouth and blow over the top surface, the strip will
greater than that on the water side—more than thirteen rise (䉲 Fig. 9.33a). (Try it.) Why? (b) A plastic egg is sup-
times greater. The difference in forces and torques should ported vertically by a stream of air from a tube
cause the cylinder to rotate—perpetually. Would you (Fig. 9.33b). The egg will not move away from the mid-
invest any money in this invention? Why or why not? stream position. Why not?
9.4 FLUID DYNAMICS AND

20. The speed of blood flow is greater in arteries than in cap-
illaries. However, the flow rate equation
1Av = constant2 seems to predict that the speed should
be greater in the smaller capillaries. Can you resolve this
apparent inconsistency?
21. When driving your car on an interstate at the posted
speed limit (of course) and an 18-wheeler quickly passes
you going in the opposite direction, you feel an force
toward the truck. Why is this?
22. A pump spray bottle or “atomizer” operates by the (a) (b)
Bernoulli principle. Explain how this works.
䉱 F I G U R E 9 . 3 3 Bernoulli effects See Conceptual
23. Whea a large on-coming truck passes you on a highway, Question 25.
you may feel your car sway toward the truck. Why
is this?
*9.5 SURFACE TENSION, VISCOSITY,
24. (a) If an Indy racer had a flat bottom, it would be highly AND POISEUILLE’S LAW
unstable (like an airplane wing) due to the lift it gets
when it moves at a high speed. To increase friction and 26. A motor oil is labeled 10W-40. What do the numbers 10
stability, the bottom has a concave section called the and 40 measure? How about the W?
Venturi tunnel (䉲 Fig. 9.32). (a) In terms of Bernoulli’s 27. Why are clothes generally washed in hot water and with
detergent?
䉱 F I G U R E 9 . 3 2 Venturi tunnel and spoiler See Conceptual

Question 24.
EXERCISES 349
EXERCISES
Use as many significant figures as you need to show small changes.
9.1 SOLIDS AND ELASTIC MODULI

1. ● A tennis racket has nylon strings. If one of the strings
2.0 cm
with a diameter of 1.0 mm is under a tension of 15 N, Brass
how much is it lengthened from its original length of F
Copper 2.0 cm
40 cm?
2. ● Suppose you use the tip of one finger to support a
1.0-kg object. If your finger has a diameter of 2.0 cm,
what is the stress on your finger? 䉱 F I G U R E 9 . 3 4 Bimetallic rod and mechanical stress See
Exercise 11.
3. ● A 2.5-m nylon fishing line used to hold up a 8.0-kg
fish has a diameter of 1.6 mm. How much is the line 12. IE ● ● Two same-size metal posts, one aluminum and
elongated? one copper, are subjected to equal shear stresses.
4. ● A 5.0-m-long rod is stretched 0.10 m by a force. What (a) Which post will show the larger deformation angle,
is the strain in the rod? (1) the copper post or (2) the aluminum post? Or (3) Is
5. ● A 250-N force is applied at a 37° angle to the surface of the angle the same for both? Why? (b) By what factor is
the end of a square bar. The surface is 4.00 cm on a side. the deformation angle of one post greater than the other?
What are (a) the compressional stress and (b) the shear 13. ● ● A 85.0-kg person stands on one leg and 90% of the
stress on the bar? weight is supported by the upper leg connecting the
knee and hip joint—the femur. Assuming the femur is
6. ●● A 4.0-kg object is supported by an aluminum wire of
0.650 m long and has a radius of 2.00 cm, by how much
length 2.0 m and diameter 2.0 mm. How much will the
is the bone compressed?
wire stretch?
14. ● ● Two metal plates are held together by two steel riv-
7. ●● A copper wire has a length of 5.0 m and a diameter of ets, each of diameter 0.20 cm and length 1.0 cm. How
3.0 mm. Under what load will its length increase by much force must be applied parallel to the plates to
0.30 mm? shear off both rivets?
8. ●● A metal wire 1.0 mm in diameter and 2.0 m long 15. IE ● ● (a) Which of the liquids in Table 9.1 has the great-
hangs vertically with a 6.0-kg object suspended from it. est compressibility? Why? (b) For equal volumes of ethyl
If the wire stretches 1.4 mm under the tension, what is alcohol and water, which would require more pressure
the value of Young’s modulus for the metal? to be compressed by 0.10%, and how many times more?
9. IE ● ● When railroad tracks are installed, gaps are left 16. ● ● How much pressure would be required to compress
between the rails. (a) Should a greater gap be used if the a quantity of mercury by 0.010%?
rails are installed on (1) a cold day or (2) a hot day? Or 17. ● ● ● A brass cube 6.0 cm on each side is placed in a pres-
(3) does the temperature not make any difference? Why? sure chamber and subjected to a pressure of
(b) Each steel rail is 8.0 m long and has a cross-sectional 1.2 * 107 N>m2 on all of its surfaces. By how much will
area of 0.0025 m2. On a hot day, each rail thermally each side be compressed under this pressure?
expands as much as 3.0 * 10-3 m If there were no gaps
18. ● ● ● A cylindrical eraser of negligible mass is dragged
between the rails, what would be the force on the ends of
across a paper at a constant velocity to the right by its
each rail?
pencil. The coefficient of kinetic friction between eraser
10. ●● A rectangular steel column 120.0 cm * 15.0 cm2 sup- and paper is 0.650. The pencil pushes down with 4.20 N.
ports a load of 12.0 metric tons. If the column is 2.00 m in The height of the eraser is 1.10 cm and its diameter is
length before being stressed, what is the decrease in 0.760 cm. Its top surface is displaced horizontally
length? 0.910 mm relative to the bottom. Determine the shear
11. IE ● ● A bimetallic rod as illustrated in 䉴 Fig. 9.34 is com- modulus of the eraser material.
posed of brass and copper. (a) If the rod is subjected to a 19. ● ● ● A 45-kg traffic light is suspended from two steel cables
compressive force, will the rod bend toward the brass or of equal length and radii 0.50 cm. If each cable makes a 15°
the copper? Why? (b) Justify your answer mathemati- angle with the horizontal, what is the fractional increase in
cally if the compressive force is 5.00 * 104 N. their length due to the weight of the light?
9.2 FLUIDS: PRESSURE AND PASCAL’S 29. ●● What is the fractional decrease in pressure when a
PRINCIPLE barometer is raised 40.0 m to the top of a building?
(Assume that the density of air is constant over that
20. IE ● In his original barometer, Pascal used water instead distance.)
of mercury. (a) Water is less dense than mercury, so the
30. ● ● To drink a soda (assume same density as water)
water barometer would have (1) a higher height than,
through a straw requires that you lower the pressure at
(2) a lower height than, or (3) the same height as the mer-
the top of the straw. What does the pressure need to be at
cury barometer. Why? (b) How high would the water
the top of a straw that is 15.0 cm above the surface of the
column have been?
soda in order for the soda to reach your lips?
21. ● If you dive to a depth of 10 m below the surface of a
31. ● ● During a plane flight, a passenger experiences ear
lake, (a) what is the pressure due to the water alone?
pain due to a head cold that has clogged his Eustachian
(b) What is the absolute pressure at that depth?
tubes. Assuming the pressure in his tubes remained at
22. IE ● In an open U-tube, the pressure of a water column 1.00 atm (from sea level) and the cabin pressure is main-
on one side is balanced by the pressure of a column of tained at 0.900 atm, determine the air pressure force
gasoline on the other side. (a) Compared to the height of (including its direction) on one eardrum, assuming it has
the water column, the gasoline column will have (1) a a diameter of 0.800 cm.
higher height, (2) a lower height, or (3) the same height.
32. ● ● Here is a demonstration Pascal used to show the
Why? (b) If the height of the water column is 15 cm,
importance of a fluid’s pressure on the fluid’s depth
what is the height of the gasoline column?
(䉲 Fig. 9.36): An oak barrel with a lid of area 0.20 m2 is
23. ● A 75.0-kg athlete performs a single-hand handstand. If filled with water. A long, thin tube of cross-sectional area
the area of the hand in contact with the floor is 125 cm2, 5.0 * 10-5 m2 is inserted into a hole at the center of the
what pressure is exerted on the floor? lid, and water is poured into the tube. When the water
24. ● A rectangular fish tank measuring 0.75 m * 0.50 m is reaches 12 m high, the barrel bursts. (a) What was the
filled with water to a height of 65 cm. What is the gauge weight of the water in the tube? (b) What was the pres-
pressure on the bottom of the tank? sure of the water on the lid of the barrel? (c) What was
25. ● (a) What is the absolute pressure at a depth of 10 m in the net force on the lid due to the water pressure?
a lake? (b) What is the gauge pressure?
26. ● ● The gauge pressure in both tires of a bicycle is
690 kPa. If the bicycle and the rider have a combined
mass of 90.0 kg, what is the area of contact of each tire
with the ground? (Assume that each tire supports half
the total weight of the bicycle.)
27. ● ● In a sample of seawater taken from an oil spill, an oil
layer 4.0 cm thick floats on 55 cm of water. If the density
of the oil is 0.75 * 103 kg>m3, what is the absolute pres-
sure on the bottom of the container?
28. IE ● ● In a lecture demonstration, an empty can is used to
demonstrate the force exerted by air pressure (䉲Fig. 9.35).
A small quantity of water is poured into the can, and the
water is brought to a boil. Then the can is sealed with a
rubber stopper. As you watch, the can is slowly crushed
with sounds of metal bending. (Why is a rubber stopper
used as a safety precaution?) (a) This is because of (1) ther-
mal expansion and contraction, (2) a higher steam pres-
sure inside the can, (3) a lower pressure inside the can as
steam condenses. Why? (b) Assuming the dimensions of
the can are 0.24 m * 0.16 m * 0.10 m and the inside of 䉱 F I G U R E 9 . 3 6 Pascal and the
the can is in a perfect vacuum, what is the total force bursting barrel See Exercise 32.
exerted on the can by the air pressure?
33. ●●The door and the seals on an aircraft are subject to a
tremendous amount of force during flight. At an altitude
of 10 000 m (about 33 000 ft), the air pressure outside the
airplane is only 2.7 * 104 N>m2 while the inside is still at
normal atmospheric pressure, due to pressurization of
the cabin. Calculate the force due to the air pressure on a
door of area 3.0 m2.
34. ● ● The pressure exerted by a person’s lungs can be mea-
sured by having the person blow as hard as possible into
one side of a manometer. If a person blowing into one side
of an open tube manometer produces an 80-cm difference
between the heights of the columns of water in the
䉱 F I G U R E 9 . 3 5 Atmospheric pressure See Exercise 28. manometer arms, what is the gauge pressure of the lungs?
EXERCISES 351
35. ●●In a head-on auto collision, the driver, who had his air sions are length 918 m, width 43.0 m, and depth 4.25 m.
bags disconnected, hits his head on the windshield, frac- (a) When filled with water, what is the weight of the
turing his skull. Assuming the driver’s head has a mass of water? (b) What is the pressure on the bridge floor?
4.0 kg, the area of the head to hit the windshield to be 41. ● ● ● A hypodermic syringe has a plunger of area 2.5 cm2
25 cm2, and an impact time of 3.0 ms, with what speed and a 5.0 * 10-3-cm2 needle. (a) If a 1.0-N force is
does his head hit the windshield? (Take the compressive applied to the plunger, what is the gauge pressure in the
fracture strength of the cranial bone to be 1.0 * 108 Pa.) syringe’s chamber? (b) If a small obstruction is present at
36. ● ● A cylinder has a diameter of 15 cm (䉲 Fig. 9.37). The the end of the needle, what force does the fluid exert on
water level in the cylinder is maintained at a constant it? (c) If the blood pressure in a vein is 50 mm Hg, what
height of 0.45 m. If the diameter of the spout pipe is force must be applied on the plunger so that fluid can be
0.50 cm, how high is h, the vertical stream of water? injected into the vein?
(Assume the water to be an ideal fluid.) 42. ● ● ● A funnel has a cork blocking its drain tube. The cork
has a diameter of 1.50 cm and is held in place by static
䉳 FIGURE 9.37 friction with the sides of the drain tube. When water is
How high a fountain?
added to a height of 10.0 cm above the cork, it comes fly-
See Exercise 36.
ing out of the tube. Determine the maximum force of sta-
tic friction between the cork and drain tube. Neglect the
weight of the cork.
9.3 BUOYANCY AND ARCHIMEDES’

PRINCIPLE
h
43. IE ● (a) If the density of an object is exactly equal to the
density of a fluid, the object will (1) float, (2) sink, (3) stay
at any height in the fluid, as long as it is totally immersed.
(b) A cube 8.5 cm on each side has a mass of 0.65 kg. Will
37. ●●In 1960, the U.S. Navy’s bathyscaphe Trieste (a sub- the cube float or sink in water? Prove your answer.
mersible) descended to a depth of 10 912 m (about
44. ● A rectangular boat, as illustrated in 䉲 Fig. 9.39, is over-
35 000 ft) into the Mariana Trench in the Pacific Ocean.
loaded such that the water level is just 1.0 cm below the
(a) What was the pressure at that depth? (Assume that
top of the boat. What is the combined mass of the people
seawater is incompressible.) (b) What was the force on a
and the boat?
circular observation window with a diameter of 15 cm?
38. ● ● The output piston of a hydraulic press has a cross-
sectional area of 0.25 m2. (a) How much pressure on the
input piston is required for the press to generate a force
of 1.5 * 106 N? (b) What force is applied to the input pis-
ton if it has a diameter of 5.0 cm?
39. ● ● A hydraulic lift in a garage has two pistons: a small
one of cross-sectional area 4.00 cm2 and a large one of
0.30 m 2.0 m
cross-sectional area 250 cm2. (a) If this lift is designed to
raise a 3500-kg car, what minimum force must be
applied to the small piston? (b) If the force is applied 4.5 m
through compressed air, what must be the minimum air
pressure applied to the small piston?
䉱 F I G U R E 9 . 3 9 An overloaded boat See Exercise 44.
40. ● ● The Magdeburg water bridge is a channel bridge
over the River Elbe in Germany (䉲 Fig. 9.38). Its dimen-
45. ●● An object has a weight of 8.0 N in air. However, it
apparently weighs only 4.0 N when it is completely sub-
merged in water. What is the density of the object?
46. ● ● When a 0.80-kg crown is submerged in water, its
apparent weight is measured to be 7.3 N. Is the crown

pure gold?
47. ● ● A steel cube 0.30 m on each side is suspended from a
scale and immersed in water. What will the scale read?

48. ● ● A solid ball has a weight of 3.0 N. When it is sub-
merged in water, it has an apparent weight of 2.7 N.

What is the density of the ball?
49. ● ● A wood cube 0.30 m on each side has a density of
700 kg>m3 and floats levelly in water. (a) What is the dis-
tance from the top of the wood to the water surface?
(b) What mass has to be placed on top of the wood so
䉱 F I G U R E 9 . 3 8 Water bridge See Exercise 40. that its top is just at the water level?
50. ●● (a) Given a piece of metal with a light string attached,

a scale, and a container of water in which the piece of
metal can be submersed, how could you find the volume
of the piece without using the variation in the water
level? (b) An object has a weight of 0.882 N. It is sus-
pended from a scale, which reads 0.735 N when the piece
is submerged in water. What are the volume and density
of the piece of metal?
51. ● ● An aquarium is filled with a liquid. A cork cube,
10.0 cm on a side, is pushed and held at rest completely

submerged in the liquid. It takes a force of 7.84 N to hold
it under the liquid. If the density of cork is 200 kg>m3,
find the density of the liquid.
52. ● ● A block of iron quickly sinks in water, but ships con- 䉱 F I G U R E 9 . 4 1 Dunking a sphere See Exercise 56.
structed of iron float. A solid cube of iron 1.0 m on each
side is made into sheets. To make these sheets into a hol-
low cube that will not sink, what should be the minimum
length of the sides of the sheets? 9.4 FLUID DYNAMICS AND
53. ● ● Plans are being made to bring back the zeppelin, a BERNOULLI’S EQUATION
lighter-than-air airship like the Goodyear blimp that car-
ries passengers and cargo, but is filled with helium, not 57. ● An ideal fluid is moving at 3.0 m>s in a section of a
flammable hydrogen as was used in the ill-fated pipe of radius 0.20 m. If the radius in another section is
Hindenburg. One design calls for the ship to be 110 m 0.35 m, what is the flow speed there?
long and to have a total mass (without helium) of 30.0 58. IE ● (a) If the radius of a pipe narrows to half of its origi-
metric tons. Assuming the ship’s “envelope” to be cylin- nal size, will the flow speed in the narrow section
drical, what would its diameter have to be so as to lift the (1) increase by a factor of 2, (2) increase by a factor of 4,
total weight of the ship and the helium? (3) decrease by a factor of 2, or (4) decrease by a factor of
54. ● ● A girl floats in a lake with 97% of her body beneath 4? Why? (b) If the radius widens to three times its origi-
the water. What are (a) her mass density and (b) her nal size, what is the ratio of the flow speed in the wider
weight density? section to that in the narrow section?
55. ● ● ● A spherical navigation buoy is tethered to the lake
59. ●● Water flows through a horizontal tube similar to that
floor by a vertical cable (䉲 Fig. 9.40). The outside diame- in Fig. 9.20. However in this case, the constricted part of
ter of the buoy is 1.00 m. The interior of the buoy consists the tube is half the diameter of the larger part. If the
of an aluminum shell 1.0 cm thick, and the rest is solid water speed is 1.5 m>s in the larger parts of the tube, by
plastic. The density of aluminum is 2700 kg>m3 and the how much does the pressure drop in the constricted
density of the plastic is 200 kg>m3. part? Express the final answer in atmospheres.
The buoy is set to float exactly halfway out of the water.
Determine the tension in the cable. 60. ●● The speed of blood in a major artery of diameter
1.0 cm is 4.5 cm>s. (a) What is the flow rate in the artery?
1.00 m (b) If the capillary system has a total cross-sectional area
䉳 FIGURE 9.40 of 2500 cm2, the average speed of blood through the cap-
Inner It’s a buoy See
illaries is what percentage of that through the major
plastic Exercise 55.
artery? (c) Why must blood flow at low speed through
sphere
the capillaries?
61. ●● The blood flow speed through an aorta with a
radius of 1.00 cm is 0.265 m>s. If hardening of the arter-
ies causes the aorta to be constricted to a radius of
0.800 cm, by how much would the blood flow speed
increase?
1.00-cm-thick 62. ●● Using the data and result of Exercise 61, calculate the
aluminum shell
pressure difference between the two areas of the aorta.
Cable (Blood density: r = 1.05 * 103 kg>m3.)
63. ●● In a dramatic lecture demonstration, a physics pro-
fessor blows hard across the top of a copper penny that
56. ● ● ● 䉴 Figure 9.41 shows a simple laboratory experiment. is at rest on a level desk. By doing this at the right speed,
Calculate (a) the volume and (b) the density of the sus- he can get the penny to accelerate vertically, into the
pended sphere. (Assume that the density of the sphere is airstream, and then deflect it into a tray, as shown in
uniform and that the liquid in the beaker is water.) 䉴 Fig. 9.42. Assuming the diameter of a penny is 1.90 cm
(c) Would you be able to make the same determinations and its mass is 2.50 g, what is the minimum airspeed
if the liquid in the beaker were mercury? (See Table 9.2.) needed to lift the penny off the tabletop? Assume the air
Explain. under the penny remains at rest.
EXERCISES 353
capable of producing a gauge pressure of 140 kPa, at

what rate (in L>s) can water be pumped to the house
assuming the hose has a radius of 5.00 cm?
68. ● ● ● A Venturi meter can be used to measure the flow
speed of a liquid. A simple such device is shown in
䉲 Fig. 9.44. Show that the flow speed of an ideal fluid is
given by
2g¢h
v1 =
A 1A 21>A 222 - 1
䉱 F I G U R E 9 . 4 2 A big blow See Exercise 63.

Δh
64. ● ● The spout heights in the container in 䉲 Fig. 9.43 are
10 cm, 20 cm, 30 cm, and 40 cm. The water level is main-

tained at a 45-cm height by an outside supply. (a) What v1
A1 v2 A1
is the speed of the water out of each hole? (b) Which v1
water stream has the greatest range relative to the base
of the container? Justify your answer. A2
䉱 F I G U R E 9 . 4 4 A flow speed meter See Exercise 68.
*9.5 SURFACE TENSION, VISCOSITY,

AND POISEUILLE’S LAW
69. ●● The pulmonary artery, which connects the heart to
the lungs, is about 8.0 cm long and has an inside diame-
ter of 5.0 mm. If the flow rate in it is to be 25 mL>s, what
is the required pressure difference over its length?
70. ●● A hospital patient receives a quick 500-cc blood trans-
fusion through a needle with a length of 5.0 cm and an
inner diameter of 1.0 mm. If the blood bag is suspended
0.85 m above the needle, how long does the transfusion
䉱 F I G U R E 9 . 4 3 Streams as projectiles See Exercise 64. take? (Neglect the viscosity of the blood flowing in the
plastic tube between the bag and the needle.)
71. ●● A nurse needs to draw 20.0 cc of blood from a patient
65. ●● In Conceptual Example 9.14, it was explained why a and deposit it into a small plastic container whose inte-
stream of water from a faucet necks down into a smaller rior is at atmospheric pressure. He inserts the needle end
cross-sectional area as it descends. Suppose at the top of of a long tube into a vein where the average gauge pres-
the stream it has a cross-sectional area of 2.0 cm2, and a sure is 30.0 mm Hg. This allows the internal pressure in
vertical distance 5.0 cm below the cross-sectional area of the vein to push the blood into the collection container.
the stream is 0.80 cm2. What is (a) the speed of the water The needle is 0.900 mm in diameter and 2.54 cm long.
and (b) the flow rate? The long tube is wide and smooth enough that we can
66. ● ● Water flows at a rate of 25 L>min through a horizon- assume its resistance is negligible, and that all the resis-
tal 7.0-cm-diameter pipe under a pressure of 6.0 Pa. At tance to blood flow occurs in the narrow needle. How
one point, calcium deposits reduce the cross-sectional long does it take him to collect the sample?
area of the pipe to 30 cm2. What is the pressure at this 72. ●● What is the difference in volume (due only to pres-
point? (Consider the water to be an ideal fluid.) sure changes, and not temperature or other factors)
67. ● ● ● As a fire-fighting method, a homeowner in the deep between 1000 kg of water at the surface (assume 4 °C) of
woods rigs up a water pump to bring water from a lake the ocean and the same mass at the deepest known
that is 10.0 m below the level of the house. If the pump is depth, 8.00 km? (Mariana Trench, assume 4 °C also.)
73. A rock is suspended from a string in air. The tension in Determine the ratio of the input side area to that of the lift-
the string is 2.94 N. When the rock is then dunked into a ing side (output) area?
liquid and the string is allowed to go slack, it sinks and 78. A spherical object has an outside diameter of 48.0 cm. Its
comes to rest on a spring with a spring constant of outer shell is composed of aluminum and is 2.00 cm
200 N>m. The spring’s final compression is 1.00 cm. If thick. The remainder is uniform plastic with a density of
the density of the rock is 2500 kg>m3, what is the density 800 kg>m3. (a) Determine the object’s average density.
of the liquid? (b) Will this object float by itself in fresh water? Explain
74. An unevenly weighted baton (cylindrical in shape) con- your reasoning. (c) If it does float, how much of it is
sists of two sections: a denser (lower) section and a less above the water surface? If it doesn’t float, determine the
dense (upper) section. When placed in water, it is force required to keep it from sinking if it is entirely
upright and barely floats. The baton has a diameter of submerged.
2.00 cm; its lower part is made of steel with a density of 79. As a medical technologist, you are attending to a worker
7800 kg>m3, and the upper part is made of wood with a who has been wounded by an accidental industrial
density of 810 kg>m3. The steel part has a length of explosion. After measuring her arterial blood pressure to
5.00 cm. Find the length of the wooden section. be 132>86, you determine her major wound to be a small
75. (a) Referring to the metal rod in Figure 9.2a (under ten- circular puncture of an artery, with an estimated diame-
sile stress), show that Eq. 9.4 can be rewritten to resem- ter of 0.25 mm. Determine (a) the maximum speed at
ble a Hooke’s law type of spring relationship for the rod. which blood is flowing out of the puncture and (b) the
That is, show that it can be written as F = k ¢L, where k maximum rate (in cc>min) at which she is losing blood
is the “effective” spring constant for the rod. Express k through it.
symbolically in terms of the rod’s cross-sectional area A, 80. An engineer is designing a water filter that works by
its Young’s modulus Y, and its unstressed length Lo and forcing water through a circular plate that has many
show that it has the proper SI units. (b) Now consider a identical holes in it. The plate is to be welded into a pipe
thin rod of iron that is subjected to a tensile force of so the water stream and the plate have the same 2.54-cm
2.00 * 103 N. If it has a cross-section of radius 1.00 cm diameter. (See 䉲 Figure 9.45.) Before it hits the filter, the
and an unstressed length of 25.0 cm, determine its effec- water is to be traveling at 75.0 cm>s. The holes are
tive spring constant. (c) By how much does this rod planned to be circular and 0.100 mm in diameter. (a) If
stretch when this force is applied? (d) How much work the holes are to cover 65% of the total plate area, how
is done by this stretching force? [Hint: Remember the many of them will you need? (b) Assuming that initially
expression for work done on a spring.] none of the holes are plugged, what is the flow speed
76. The ocean can be as deep as 10 km. (a) Assuming the just after the water leaves the holes? (c) Later, if 25% of
density for seawater is constant at the value given in the holes are completely plugged with gunk and the
Table 9.2, what is the absolute pressure at such depths? water speed before the filter has not changed, what will
(b) What would be the percentage change in volume of a be the flow speed of the water upon leaving the filter
cube of aluminum that measured 1.00 m on a side when area? (d) Compare the flow rate in this system (in liters
at the ocean surface? (c) By how much did the aluminum per minute) before and after some of the holes plug up.
cube’s volume change?
77. In preparation for its tire rotation, a car weighing
2.25 tons is placed on a hydraulic garage lift. The
mechanic then raises the car 30.0 cm. (a) Calculate the
work done on the car when it is lifted. (b) Assuming no
frictional losses in the hydraulic fluid, how much work
was done by the lift on the input side? (c) What was the
force on the input side if its piston moved 52.5 cm? (d) 䉱 F I G U R E 9 . 4 5 Filter that water See Exercise 80.
Temperature
10 and Kinetic Theory
10.1 Temperature and
heat (356)
■ internal energy
10.2 The Celsius and Fahrenheit

temperature scales (358)
■ temperature conversions
10.3 Gas laws, absolute

temperature, and the Kelvin
temperature scale (362)
■ ideal gas law
■ absolute zero
10.4 Thermal expansion (368)
G
PHYSICS FACTS
■ linear expansion lobal warning has becoming
✦ The Celsius and Fahrenheit tem-
perature scales have equal read- a very popular topic lately
ings at -40 degrees, that is,
due to the increasing evidence
10.5 The kinetic theory -40 °C = - 40 °F.
of gases (372) ✦ The lowest possible temperature that human activities have acceler-
is absolute zero 1 -273.15 °C2.
■ temperature and kinetic energy ated the increase of the atmos-
There is no known upper limit on
■ internal energy of
monatomic gas
temperature. pheric temperature of the Earth.
✦ The Golden Gate Bridge over San
Increasing global temperatures
Francisco Bay varies in length by
almost 1 m between summer and will cause polar ice caps to melt
*10.6 Kinetic theory, diatomic winter (thermal expansion).
gases, and the equipartition
and sea levels to rise. The NASA
✦ While the normal average human
theorem (376) body temperature is 37 °C satellite photographs (chapter-
■ internal energy of diatomic gas (98.6 °F), the normal skin tempera-
ture is only 33 °C (91 °F). The skin
opener photographs) taken since
temperature depends on air tem- 1979 clearly show the shrinking of
perature and time spent in that
environment. the Arctic ice caps. The photo-
✦ Almost all substances have posi- graph on the top was taken in 1979
tive coefficients of thermal expan-
sion (expanding on heating). A few and the one on the bottom in 2005.
have negative coefficients (con-
traction on heating). Water con-
Temperature and heat are fre-
tracts on heating from 0 °C to 4 °C. quent subjects of conversation, but
if you had to explain what the words
356 10 TEMPERATURE AND KINETIC THEORY
really mean, you might find yourself at a loss. We use various types of
thermometers to measure temperatures, which provide an objective equivalent
for our sensory experience of hot and cold. A temperature change generally
results from the addition or removal of heat. Temperature, therefore, is related
to heat. But how? And what is heat? In this chapter, you’ll find that the answers
to such questions lead to an understanding of some far-reaching physical
principles.
An early theory of heat considered it to be a fluid-like substance called
caloric (from the Latin word calor, meaning “heat”) that could be made to flow
into and out of a body. Even though this theory has been abandoned, we still
speak of heat as “flowing” from one object to another. Heat is now known to be
energy in transit, and temperature and thermal properties are explained by con-
sidering the atomic and molecular behavior of substances. This and the next
two chapters examine the nature of temperature and heat in terms of micro-
scopic (molecular) theory and macroscopic observations. Here, you’ll explore
the nature of heat and the ways temperature is measured. You’ll also encounter
the gas laws, which explain not only the pressure increase of a hot automobile
tire, but also more important phenomena, such as how our lungs supply us
with the oxygen we need to live.
10.1 Temperature and Heat

➥ What is the difference between temperature and heat?

➥ What types of energy make up the internal energy of a diatomic gas?
➥ Does higher temperature mean a system having more internal energy?
A good way to begin studying thermal physics is with definitions of temperature

and heat. Temperature is a relative measure, or indication, of hotness or coldness.
A hot stove is said to have a high temperature and an ice cube to have a low tem-
perature. An object that has a higher temperature than another object is said to be
hotter, or the other object is said to be colder. Note that hot and cold are relative
terms, like tall and short. We can perceive temperature by touch. However, this
temperature sense is somewhat unreliable, and its range is too limited to be useful
for scientific purposes.
Heat is related to temperature and describes the process of energy transfer from
one object to another. That is, heat is the net energy transferred from one object to
another because of a temperature difference. Heat is energy in transit, so to speak. Once
transferred, the energy becomes part of the total energy of the molecules of the
object or system, that is, its internal energy. So heat (energy) transfers between
objects can result in internal energy changes.*
On a microscopic level, temperature is associated with molecular motion. In
kinetic theory (Section 10.5), which treats gas molecules as point particles, tem-
perature is shown to be a measure of the average translational kinetic energy of
the molecules. However, diatomic gas, besides having such translational
“temperature” kinetic energy, also may have kinetic energy due to rotations
*Note: Some of the energy may go into doing work and not into internal energy (Section 12.2).
10.1 TEMPERATURE AND HEAT 357
Total
internal energy
Kinetic energy Potential energy

of molecules of molecules (due
to intermolecular
and intramolecular
forces)
Vibrational Rotational
Random translational
energy of energy of
energy of molecules
molecules molecules
Translational motion Vibrational motion Rotational motion

(a) (b) (c)
䉱 F I G U R E 1 0 . 1 Diatomic molecular motions The total internal energy is

made up of kinetic and intermolecular and intramolecular potential energy.
The kinetic energy has the following forms: (a) translational kinetic energy.
(b) linear vibrational kinetic energy and (c) rotational kinetic energy.
and vibrations, as well as potential energy due to intermolecular and

intramolecular interactions. The total internal energy is the sum of all such
energies (䉱Fig. 10.1).
Note that a higher temperature does not necessarily mean that one system has a
greater internal energy than another. For example, in a classroom on a cold day,
the air temperature is relatively high compared to that of the outdoor air. But all
that cold air outside the classroom has far more internal energy than does the
warm air inside, simply because there is so much more of it. If this were not the
case, heat pumps would not be practical (Section 12.5). In other words, the internal
energy of a system also depends on its mass, or the number of molecules in the
system.
When heat is transferred between two objects, regardless of whether they are
touching, the objects are said to be in thermal contact. When there is no longer a net
heat transfer between objects in thermal contact, they have come to the same tem-
perature and are said to be in thermal equilibrium.
DID YOU LEARN?

➥ Heat is the net energy transferred between objects due to temperature differences,
and temperature is an indication of the average translational kinetic energy of the
molecules.
➥ The total internal energy of a diatomic gas may consist of translational kinetic
energy, vibrational kinetic energy, rotational kinetic energy, and potential energy
due to attractive forces between the atoms.
➥ The mass or number of molecules in a system is a factor in determining the total
internal energy.
Iron
Brass
Scale
(a) Initial condition (b) Heated condition
䉱 F I G U R E 1 0 . 2 Thermal expansion (a) A bimetallic strip is made of two strips of differ-

ent metals bonded together. (b) When such a strip is heated, it bends because of unequal
expansions of the two metals. Here, brass expands more than iron, so the deflection is
toward the iron. The deflection of the end of a strip could be used to measure temperature.
10.2 The Celsius and Fahrenheit Temperature Scales

➥ How is a temperature scale constructed?

➥ How do you convert a temperature on the Fahrenheit scale to a temperature on the
Celsius scale?
➥ How do you convert a temperature on the Celsius scale to a temperature on the
Fahrenheit scale?
A measure of temperature is obtained by using a thermometer, a device con-

structed to make use of some property of a substance that changes with tempera-
(a) ture. Fortunately, many physical properties of materials change sufficiently with
temperature to be used as the bases for thermometers. By far the most obvious
and commonly used property is thermal expansion (Section 10.4), a change in the
dimensions or volume of a substance that occurs when the temperature changes.
Almost all substances expand with increasing temperature, and do so to differ-
ent extents. They also contract with decreasing temperature. (Thermal expansion
refers to both expansion and contraction; contraction is considered a negative
expansion.) Because some metals expand more than others, a bimetallic strip (a
strip made of two different metals bonded together) can be used to measure tem-
perature changes. As heat is added, the composite strip will bend away from the
side made of the metal that expands more (䉱 Fig. 10.2). Coils formed from such
strips are used in dial thermometers and in common household thermostats (䉳 Fig.
10.3).
A common thermometer is the liquid-in-glass type, which is based on the ther-
mal expansion of a liquid. A liquid in a glass bulb expands into a glass stem, rising
(b)
in a capillary bore (a thin tube). Mercury and alcohol (usually dyed red to make it
䉱 F I G U R E 1 0 . 3 Bimetallic coil more visible) are the liquids used in most liquid-in-glass thermometers. These
Bimetallic coils are used in (a) dial substances are chosen because of their relatively large thermal expansion and
thermometers (the coil is in the cen-
ter) and (b) household thermostats because they remain liquids over normal temperature ranges.
(the coil is to the right). Thermostats Thermometers are calibrated so that a numerical value can be assigned to a
are used to regulate a heating or given temperature. For the definition of any standard scale or unit, two fixed
cooling system, turning off and on as reference points are needed. The ice point and the steam point of water at stan-
the temperature of the room dard atmospheric pressure are two convenient fixed points. More commonly
changes. The expansion and contrac-
tion of the coil causes the tilting of a known as the freezing and boiling points, these are the temperatures at which
glass vial containing mercury, which pure water freezes and boils, respectively, under a pressure of 1 atm (standard
makes and breaks electrical contact. pressure).
10.2 THE CELSIUS AND FAHRENHEIT TEMPERATURE SCALES 359
The two most familiar temperature scales are the Fahrenheit temperature Fahrenheit Celsius
scale* (used in the United States) and the Celsius temperature scale† (used in the
212 °F 100 °C
rest of the world). As shown in 䉴 Fig. 10.4, the ice and steam points have values of
Steam Steam
32 °F and 212 °F, respectively, on the Fahrenheit scale and 0 °C and 100 °C, respec- point point
tively, on the Celsius scale. On the Fahrenheit scale, there are 180 equal intervals,
or degrees (°F), between the two reference points; on the Celsius scale, there are
100 degrees (°C). Therefore, since 180>100 = 9>5 = 1.8, one Celsius degree is
almost twice as large as one Fahrenheit degree. 180 °F 100 °C
A relationship for converting between the two scales can be obtained from a
graph of Fahrenheit temperature (TF) versus Celsius temperature (TC), such as the
one in 䉲 Fig. 10.5. The equation of the straight line (in slope-intercept form,
y = mx + b) is TF = 1180>1002TC + 32, and Ice Ice
point point
32 °F 0 °C
TF = 95 TC + 32
(Celsius-to-Fahrenheit
or (10.1)
conversion)
TF = 1.8TC + 32
– 40 °F – 40 °C
where 95 = 1.8 is the slope of the line and 32 is the intercept on the vertical axis.
Thus, to change from a Celsius temperature (TC) to its equivalent Fahrenheit tem-
perature (TF), you simply first multiply the Celsius reading by 95 and then add 32. 䉱 F I G U R E 1 0 . 4 Celsius and
The equation can be solved for TC to convert from Fahrenheit to Celsius: Fahrenheit temperature scales
Between the ice and steam fixed
TC = 59 1TF - 322
points, there are 100 degrees on the
(Fahrenheit-to-Celsius conversion) (10.2) Celsius scale and 180 degrees on the
Fahrenheit scale. Thus, a Celsius
Therefore, to change from a Fahrenheit temperature (TF) to its equivalent Celsius degree is 1.8 times as large as a
Fahrenheit degree.
temperature (TC), you first subtract 32 from the Fahrenheit reading and then mul-
tiply by 59 .
TF
212
5
Temperature (°F)
9/
=
00
0/1
18 ∆TF = 180 °F
=
o pe
Sl
32 ∆TC = 100 °C
TC
0 100
Temperature (°C)
䉱 F I G U R E 1 0 . 5 Fahrenheit versus Celsius A plot of

Fahrenheit temperature versus Celsius temperature
gives a straight line of the general form y = mx + b, where
TF = 95 TC + 32.
*Daniel Gabriel Fahrenheit (1686–1736), a German instrument maker, constructed the first alcohol
thermometer (1709) and mercury thermometer (1714). The freezing and boiling points of water were
measured to be 32 °F and 212 °F.
†
Anders Celsius (1701–1744), a Swedish astronomer, invented the Celsius temperature scale with a
100-degree interval between the freezing and boiling point of water (0 °C and 100 °C).
EXAMPLE 10.1 Converting Temperature Scale Readings:

Fahrenheit and Celsius
What are (a) the typical room temperature of 20 °C and a cold temperature of - 15 °C
on the Fahrenheit scale, and (b) another cold temperature of - 10 °F and normal body
temperature, 98.6 °F, on the Celsius scale?
THINKING IT THROUGH. These are direct applications of Eqs. 10.1 and 10.2.
SOLUTION.
Given: (a) TC = 20 °C and Find: for each temperature,

TC = - 15 °C (a) TF
(b) TF = - 10 °F and (b) TC
TF = 98.6 °F
(a) Equation 10.1 is for changing Celsius readings to Fahrenheit:
20 °C: TF = 95 TC + 32 = C 95 1202 + 32 D °F = 68 °F
(This typical room temperature of 20 °C is a good one to remember.)
- 15 °C: TF = 95 TC + 32 = C 95 1- 152 + 32 D °F = 5.0 °F
(b) Equation 10.2 changes Fahrenheit to Celsius:
- 10 °F: TC = 59 1TF - 322 = C 59 1- 10 - 322 D °C = - 23 °C

98.6 °F: TC = 59 1TF - 322 = C 59 198.6 - 322 D °C = 37.0 °C
Note that one Celsius degree is 1.8 times (almost twice) as large as one Fahrenheit
degree. For example, a body temperature of 40.0 °C represents an elevation of 3.0 °C
over normal body temperature. However, on the Fahrenheit scale, this is an increase of
3.0 * 1.8 °F = 5.4 °F, or a temperature of 198.6 + 5.42 °F = 104.0 °F.
F O L L O W - U P E X E R C I S E . Convert the following temperatures: (a) - 40 °F to Celsius and
(b) -40 °C to Fahrenheit. (Answers to all Follow-Up Exercises are given in Appendix VI at
the back of the book.)
Because Eqs. 10.1 and 10.2 are so similar, it is easy to miswrite them. Since they are
equivalent, you need to know only one of them—say, Celsius to Fahrenheit, Eq. 10.1,
TF = 95 TC + 32. Solving this equation for TC algebraically gives Eq. 10.2. A good way to
make sure that you have written the conversion equation correctly is to test it with a
known temperature, such as the boiling point of water. For example, TF = 212 °F, so
TC = 59 1TF - 322 = C 59 1212 - 322 D °C = 59 11802 °C = 100 °C
Thus, we know the equation is correct.
Liquid-in-glass thermometers are adequate for many temperature measure-

ments, but problems arise when highly accurate determinations are needed. A
material may not expand uniformly over a wide temperature range. When cali-
brated to the ice and steam points, an alcohol thermometer and a mercury ther-
mometer have the same readings at those points, but because alcohol and mercury
have different expansion properties, the thermometers will not have exactly the
same reading at an intermediate temperature, such as room temperature. For very
sensitive temperature measurements and to define intermediate temperatures
precisely, some other type of thermometer must be used. One such thermometer, a
gas thermometer, is discussed in the next section.
10.2 THE CELSIUS AND FAHRENHEIT TEMPERATURE SCALES 361
DID YOU LEARN?

➥ Two fixed temperature reference points, such as the ice and steam points of water,
are typically used to construct a temperature scale.
➥ To change TF to TC , subtract 32 from TF and then multiple by 5>9, that is,
TC = 59 (TF - 32).
➥ To change TC to TF, first multiply TC by 9>5 and then add 32, that is, TF = 95 TC + 32.
INSIGHT 10.1 Human Body Temperature

We commonly take “normal” human body temperature to be (lower-than-normal body temperature) can be beneficial. A
98.6 °F or 37.0 °C. The source of this value is a study of decrease in body temperature slows down the body’s chemi-
human temperature readings done in 1868—more than 135 cal reactions, and cells use less oxygen than they normally do.
years ago. A more recent study, conducted in 1992, notes that This effect is applied in some surgeries (Fig. 1). A patient’s
the 1868 study used thermometers that were not as accurate body temperature may be lowered significantly to avoid dam-
as modern electronic (digital) thermometers. The new study age to the brain and to the heart, which must be stopped dur-
has some interesting results. ing some procedures.
The normal human body temperature from oral measure-
ments varies among individuals over a range of about 96 °F
to 101 °F, with an average temperature of 98.2 °F. After stren-
uous exercise, the oral temperature can rise as high as 103 °F.
When the body is exposed to cold, oral temperatures can fall
below 96 °F. A rapid drop in temperature of (2 to 3) °F pro-
duces uncontrollable shivering. The skeletal muscles contract
and so do the tiny muscles attached to the hair follicles. The
result is “goose bumps.”
Your body temperature is typically lowest in the morning,
after you have slept and your digestive processes are at a low
point. Normal body temperature generally rises during the
day to a peak and then recedes. The 1992 study also indicated
that women have a slightly higher average body temperature
than do men (98.4 °F versus 98.1 °F).
What about the extremes? A fever temperature is typically
between 102 °F and 104 °F. A body temperature above 106 °F
is extremely dangerous. At such temperatures, the enzymes
that take part in certain chemical reactions in the body begin to
be inactivated, and a total breakdown of body chemistry can
result. On the cold side, decreased body temperature results in F I G U R E 1 Lower than normal During some surgeries, the
memory lapses and slurred speech, muscular rigidity, erratic patient’s body temperature is lowered to slow down the
heartbeats, and loss of consciousness. Below 78 °F, death body’s chemical reactions and to reduce the need for blood
occurs due to heart failure. However, mild hypothermia to supply oxygen to the tissues.
INSIGHT 10.2 Warm-Blooded Versus Cold-Blooded

With few exceptions, all mammals and birds are warm- glands only in their feet. Pigs and whales have no sweat
blooded and all fish, reptiles, amphibians, and insects are glands. Pigs generally rely on wallowing in mud for cooling,
cold-blooded. The difference is that warm-blooded creatures and whales can change water depths for temperature changes
try to maintain their bodies at a relatively constant tempera- or seasonally migrate.
ture, while cold-blooded creatures take on the temperature of Also, some animals have fur coats for warmth in the winter
their surroundings (Fig. 1). and shed them to cool off in the summer. Warm-blooded ani-
Warm-blooded creatures maintain a relatively constant mals can shiver to activate certain muscles to increase metab-
body temperature by generating their own heat when in a olism and thereby generate heat. Some birds (and some
cold environment and by cooling themselves when in a hot people) migrate between colder and warmer regions.
environment. To generate heat, warm-blooded animals con- The body temperature of cold-blooded creatures changes
vert food into energy. To stay cool on hot days, they sweat, with the temperature of their environment. They are very
pant, or get wet and thereby remove heat by water evapora- active in warm environments and are sluggish when it is cold.
tion. Primates (humans, apes, monkeys, and so on) have
sweat glands all over their bodies. Dogs and cats have sweat (continued on next page)
This is because their muscle activity depends on chemical

reactions that vary with temperature. Cold-blooded creatures
often bask in the sun to warm up to increase their metabo-
lism. Fish can change water depths or seasonally migrate.
Frogs, toads, and lizards hibernate during winter. To stay
warm, honeybees crowd together and rapidly flap their
wings to generate heat.
Some animals do not fall into the strict definitions of being
warm-blooded or cold-blooded. Bats, for example, are mam-
mals that cannot maintain a constant body temperature, and
they cool off when not active. Some warm-blooded animals,
such as bears, groundhogs, and gophers, hibernate in winter.
During the hibernation period, they live off stored body fat;
their body temperatures may drop as much as 10 °C 118 °F2.
F I G U R E 1 Warm-blooded and cold-

blooded The infrared images show that
cold-blooded creatures take on the tempera-
ture of their surroundings. Both the gecko
and the scorpion are at the same temperature
(color) as the air surrounding them. Notice
the difference between these cold-blooded
creatures and the warm-blooded humans
holding them.
10.3 Gas Laws, Absolute Temperature, and the Kelvin

Temperature Scale
➥ What are the three common forms of the ideal gas law?
➥ How is absolute zero determined?
➥ How do you convert a temperature on the Celsius scale to a temperature on the
Kelvin scale?
Whereas different liquid-in-glass thermometers show slightly different readings

for temperatures other than fixed points because of the liquids’ different expan-
sion properties, a thermometer that uses a gas gives the same readings regardless
of the gas used. The reason is that at very low densities all gases exhibit the same
expansion behavior.
The variables that describe the behavior of a given quantity (mass) of gas are
pressure, volume, and temperature (p, V, and T). When temperature is held con-
stant, the pressure and volume of a quantity of gas are related as follows:
pV = constant or p1 V1 = p2 V2 (at constant temperature) (10.3)
That is, the product of pressure and volume is a constant. This relationship is
known as Boyle’s law, after Robert Boyle (1627–1691), the English chemist who dis-
covered it.
When the pressure is held constant, the volume of a quantity of gas is related to
the absolute temperature (to be defined shortly):
V V1 V2
= constant or = (at constant pressure) (10.4)
T T1 T2
10.3 GAS LAWS, ABSOLUTE TEMPERATURE, AND THE KELVIN TEMPERATURE SCALE 363
That is, the ratio of the volume to the temperature is a constant. This relationship
is known as Charles’s law, named for the French scientist Jacques Charles
(1746–1823), who took early hot-air balloon flights and was therefore quite inter-
ested in the relationship between the volume and temperature of a gas. A popular
demonstration of Charles’s law is shown in 䉴 Fig. 10.6.
Low-density gases obey these laws, which may be combined into a single rela-
tionship. Since pV = constant and V>T = constant for a given quantity of gas,
pV>T must also equal a constant. This relationship is the ideal gas law:
pV p1 V1 p2 V2
= constant or = (ideal gas law, ratio form) (10.5)
T T1 T2
That is, the ratio pV>T at one time (t1) is the same as at another time (t2), or at any
other time, as long as the quantity (number of molecules or mass) of gas does not
change.
This relationship can be written in a more general form that applies not just to a
(a) (b)
given quantity of a single gas, but to any quantity of any low-pressure, dilute gas.
With a quantity of gas determined by the number of molecules (N) in the gas (that 䉱 F I G U R E 1 0 . 6 Charles’s law in
is, pV>T r N), it follows that action Demonstrations of the rela-
tionship between the volume and
pV the temperature of a quantity of gas.
= NkB or pV = NkB T (ideal gas law) (10.6) A weighted balloon, initially at
T room temperature, is placed in a
beaker of water. (a) When ice is
where kB is a constant of proportionality known as the Boltzmann’s constant: placed in the beaker and the tem-
kB = 1.38 * 10-23 J>K.* perature falls, the balloon’s volume
decreases. (b) When the water is
The K stands for temperature on the Kelvin scale, discussed shortly. Note that
heated and the temperature rises,
the mass of the sample does not appear explicitly in Eq. 10.6. However, the num- the balloon’s volume increases.
ber of molecules N in a sample of a gas is proportional to the total mass of the gas.
The ideal gas law, sometimes called the perfect gas law, applies to real gases with
low pressures and densities, and describes the behavior of most gases fairly accu-
rately at normal densities.
DEMONSTRATION 3 Boyle’s Shaving Cream

A demonstration that shows Boyle’s law, p r 1>V, the inverse relationship of pressure (p)
and volume (V).
In a dramatic volume
As the pressure is reduction, the shav-
reduced, the swirl of ing cream cannot
cream grows in vol- stand up to the sud-
ume from the expan- den increase in pres-
sion of the air sure when the
bubbles trapped in chamber is vented to
the cream. the atmosphere.
*Named after the Austrian physicist Ludwig Boltzmann (1844–1906), who made important contri-
butions in determining this constant.
MACROSCOPIC FORM OF THE IDEAL GAS LAW

Equation 10.6 is a microscopic (micro means extremely small) form of the ideal gas law
in that it refers specifically to the number of molecules, N. However, the law can be
rewritten in a macroscopic (macro means large) form, which involves quantities that can
be measured with everyday laboratory equipment. The ideal gas law in this form is
pV = nRT (ideal gas law) (10.7)
using nR rather than NkB for convenience since n r N. Here, n is the number of moles
(mol) of the gas, a quantity defined next, and R is called the universal gas constant:
R = 8.31 J>(mol # K)
A mole (abbreviated mol) of a substance is defined as the quantity that contains
Avogadro’s number NA of molecules:
NA = 6.02 * 1023 molecules>mol
Thus, n and N in the two forms of the ideal gas law are related by N = nNA. From
Eq. 10.7, it can be shown that 1 mol of any gas occupies 22.4 L at 0 °C and 1 atm. These
conditions, 0 °C and 1 atm, are known as standard temperature and pressure (STP).
It is important to note what these equations for the macroscopic (Eq. 10.7) and
microscopic (Eq. 10.6) forms of the ideal gas law represent. For the macroscopic
form of the ideal gas law, the constant R = pV>1nT2 has units of J>1mol # K2.
For the microscopic form of the law, kB = pV>1NT2, with units of J>1molecule # K2.
Note that the difference between the macroscopic and microscopic forms of the
ideal gas law is moles versus the number of molecules, and gas quantities are usu-
ally measured in moles in the laboratory.
To use Eq. 10.7, we need to know the number of moles of a quantity of gas. This
is done by finding the molar mass, M, of a compound or element. Molar mass is the
mass of one mole of substance, so M = mNA, where m is the molecular mass or the
mass of one molecule. Because molecular masses are so small in relation to the SI
standard kilogram, another unit, the atomic mass unit (u), is used:
1 atomic mass unit 1u2 = 1.66054 * 10-27 kg*
The molecular mass is determined from the chemical formula and the atomic
masses of the atoms. (The latter are listed in Appendix IV and are commonly
rounded to the nearest one half.) For example, water, H2O, with two hydrogen
atoms and one oxygen atom, has a molecular mass of 21mH2 + 11mO2 = 211.0 u2 +
1116.0 u2 = 18.0 u, because the atomic mass of each hydrogen atom is 1.0 u and that
of an oxygen atom is 16.0 u. Then, one mole of water has a molar mass of
118 u211.66054 * 10-27 kg>u216.02 * 1023 >mol2 = 0.0180 kg>mol = 18.0 g>mol.
Similarly, the oxygen we breathe, O2, has a molecular mass of 2 * 16.0 u = 32.0 u.
Hence, one mole of oxygen has a mass of 32.0 g.
The reverse calculation can also be made. For example, suppose you want to
know the mass of a water molecule (H2O). As was just seen, the molar mass of
water is 18.0 g, or 18.0 g>mol. The molecular mass (m) is then given by
M 1molar mass2 118.0 g>mol2
mH2O = =
NA 6.02 * 1023 molecules>mol
= 2.99 * 10-23 g>molecule = 2.99 * 10-26 kg>molecule
ABSOLUTE ZERO AND THE KELVIN TEMPERATURE SCALE

The product of the pressure and the volume of a sample of ideal gas is directly
proportional to the temperature of the gas: pV r T. This relationship allows a gas
to be used to measure temperature in a constant volume gas thermometer. Holding
the volume of the gas constant, which can be done easily in a rigid container,
means that p r T (䉴 Fig. 10.7). Then using a constant volume gas thermometer,
*The atomic mass unit is based on assigning a carbon-12 atom the value of exactly 12 u.
Pa Pa 䉳 F I G U R E 1 0 . 7 Constant volume
gas thermometer Such a thermome-
ter indicates temperature as a func-
tion of pressure, since, for a
low-density gas, p r T. (a) At some
initial temperature, the pressure
reading has a certain value.
(b) When the gas thermometer is
heated, the pressure (and tempera-
ture) reading is higher, because, on
average, the gas molecules are mov-
ing faster.
(a) Initial temperature (b) Heat added
one reads the temperature in terms of pressure. A plot of pressure versus tempera-
ture gives a straight line in this case (䉲 Fig. 10.8a).
As can be seen in Fig. 10.8b, measurements of real gases (plotted data points) devi-
ate from the values predicted by the ideal gas law at very low temperatures. This is
because the gases liquefy at such temperatures. However, the relationship is linear
over a large temperature range, and it looks as though the pressure might reach zero
with decreasing temperature if the gas were to ramain in its gaseous state.
The absolute minimum temperature for an ideal gas is therefore inferred by
extrapolating, or extending the straight line to the axis, as in Fig. 10.8b. This tem-
perature is found to be -273.15 °C and is designated as absolute zero. Absolute
zero is believed to be the lower limit of temperature, but it has never been
attained. In fact, there is a law of thermodynamics that says it never can be
achieved (Section 12.5).* There is no known upper limit to temperature. For exam-
ple, the temperatures at the centers of some stars are estimated to be greater than
100 million degrees Celsius.
Absolute zero is the foundation of the Kelvin temperature scale, named after
the British scientist Lord Kelvin who proposed it in 1848.† On this scale, - 273.15 °C
Pressure Gas A
Pressure Gas B
×
× Gas C
×
×
–273.15 °C ×
Temperature ×
–200 °C –100 °C 0 °C 100 °C –273.15 °C 0 °C Temperature
0K
(a) (b)
䉱 F I G U R E 1 0 . 8 Pressure versus temperature (a) A low-density gas kept at a constant volume gives a straight line
on a graph of p versus T, that is, p = 1NkB>V2T. When the line is extended to the zero pressure value, a temperature of
- 273.15 °C is obtained, which is taken to be absolute zero. (b) Extrapolation of lines for all low-density gases indicates
the same absolute zero temperature. The actual behavior of gases deviates from this straight-line relationship at low
temperatures because the gases start to liquefy.
*At the time of this writing, the lowest overall average thermodynamic temperature that scientists
have been able to attain is 450 * 10-12 K, that is, 450 pK (picokelvins) above absolute zero.
†
Lord Kelvin, born William Thomson (1824–1907), developed devices to improve telegraphy and the
compass and was involved in the laying of the first transatlantic cable. When he received his title, it is
said that he considered choosing Lord Cable or Lord Compass, but decided on Lord Kelvin, after a river
that runs near the University of Glasgow in Scotland, where he was a professor of physics for fifty years.
Kelvin Celsius Fahrenheit is taken as the zero point—that is, as 0 K (䉳 Fig. 10.9). The size of a single
Steam point: unit of Kelvin temperature is the same as that of the degree Celsius, so
373 K 100 °C 212 °F temperatures on these scales are related by
T = TC + 273.15 (Celsius-to-Kelvin conversion) (10.8)

Ice point:
273 K 0 °C 32 °F where T is the temperature in kelvins (for example, a temperature of
300 kelvins). The kelvin is abbreviated as K (not degrees Kelvin, °K). For
general calculations, it is common to round the 273.15 in Eq. 10.8 to 273,
that is,
T = TC + 273 (for general calculations) (10.8a)
The absolute Kelvin scale is the official SI temperature scale; however, the
Celsius scale is used in most parts of the world for everyday temperature
Absolute zero: readings. The absolute temperature in kelvins is used primarily in scien-
0K −273 °C −459 °F tific applications.
TK = TC + 273 PROBLEM-SOLVING HINT
Keep in mind that Kelvin temperatures must be used with any form of the ideal gas law.
䉱 F I G U R E 1 0 . 9 The Kelvin tem- It is a common mistake to use Celsius or Fahrenheit temperatures. Suppose you used a
perature scale The lowest tempera- Celsius temperature of T = 0 °C in the gas law. You would have pV = 0, which makes
ture on the Kelvin scale
(corresponding to - 273.15 °C) is no sense, since neither p nor V is zero at the ice point of water.
absolute zero. A unit interval on the Note that there can be no negative temperatures on the Kelvin scale if absolute zero is
Kelvin scale, called a kelvin and the lowest possible temperature. That is, the Kelvin scale doesn’t have an arbitrary zero
abbreviated K, is equivalent to a temperature somewhere within the scale as on the Fahrenheit and Celsius scales—zero
temperature change of 1 °C, thus, K is absolute zero, period.
TK = TC + 273.15. (The constant is
usually rounded to 273 for conve-
nience.) For example, a temperature
of 0 °C is equal to 273 kelvins.
EXAMPLE 10.2 Deepest Freeze: Absolute Zero on the Fahrenheit Scale

What is absolute zero on the Fahrenheit scale? Temperatures on the Kelvin scale are related directly to Cel-
sius temperatures by T = TC + 273.15 (Eq. 10.8 for accuracy),
T H I N K I N G I T T H R O U G H . This requires the conversion of 0 K
so first we convert 0 K to a Celsius value:
to the Fahrenheit scale. But first a conversion to the Celsius
scale is in order. (Why?) TC = T - 273.15 = 10 - 273.152 °C = - 273.15 °C
Then, converting to Fahrenheit (Eq. 10.1) gives
C 95 1- 273.152 + 32 D °F = - 459.67 °F
SOLUTION.
Given: T = 0K Find: TF TF = 95 TC + 32 =
Thus, absolute zero is about -460 °F.
F O L L O W - U P E X E R C I S E . There is an absolute temperature scale associated with the Fahrenheit temperature scale called the
Rankine scale. A Rankine degree is the same size as a Fahrenheit degree, and absolute zero is taken as 0 °R (zero degree Rankine).
Write the conversion equations between (a) the Rankine and the Fahrenheit scales, (b) the Rankine and the Celsius scales, and
(c) the Rankine and the Kelvin scales.
Initially, gas thermometers were calibrated by using the ice and steam points. The
Kelvin scale uses absolute zero and a second fixed point adopted in 1954 by the
International Committee on Weights and Measures. This second fixed point is the
triple point of water, at which water coexists simultaneously in equilibrium as a
solid (ice), liquid (water), and gas (water vapor). The triple point occurs at a unique
set of values for temperature and pressure—a temperature of 0.01 °C and a pressure
of 4.58 mm Hg 1611.73 Pa2—and provides a reproducible reference temperature for
the Kelvin scale. The temperature of the triple point on the Kelvin scale was
assigned a value of 273.16 K. The SI kelvin unit is then defined in terms of the tem-
perature at the triple point of water.*
The Kelvin temperature scale has special significance. As will be seen in Section
10.5, the absolute temperature is directly proportional to the internal energy of an
ideal gas and so can be used as an indication of that energy. There are no negative
values on the absolute scale. Negative absolute temperatures would imply negative
internal energy for the gas, a meaningless concept.
Now let’s use various forms of the ideal gas law, which requires absolute
temperatures.
EXAMPLE 10.3 The Ideal Gas Law: Using Absolute Temperatures

A quantity of ideal gas in a rigid container is initially at room T H I N K I N G I T T H R O U G H . The law requires absolute tempera-
temperature 120 °C2 and a particular pressure (p1). If the gas tures, so we need to change the Celsius temperatures to
is heated to a temperature of 60 °C, by what factor does the kelvins. A “factor” of change implies a ratio 1p2>p12, so the
pressure change? ideal gas law in ratio form (Eq. 10.5) should apply. Note that
the container is rigid, which means that V1 = V2.
SOLUTION.
Given: T1 = 20 °C = 120 + 2732 K = 293 K Find: p2>p1 (pressure ratio or factor)
T2 = 60 °C = 160 + 2732 K = 333 K
V1 = V2
Since the factor by which the pressure changes is wanted, So, p2 is 1.14 times p1; that is, the pressure increases by a factor
p2>p1 is written as a ratio. For example, if p2>p1 = 2, then of 1.14, or 14%. (What would the factor be if the Celsius tem-
p2 = 2p1, or the pressure would change (increase) by a factor of peratures were incorrectly used? It would be much larger:
2. Using the ideal gas law in ratio form (Eq. 10.5) 160 °C2>120 °C2 = 3, or p2 = 3p1. Wrong.)
p2V2>T2 = p1V1>T1, we have, with V1 = V2 ,
≤p = a bp = 1.14p1
T2 333 K
p2 = ¢
T1 1 293 K 1
F O L L O W - U P E X E R C I S E . If the gas in this Example is heated from an initial temperature of 20 °C (room temperature) so that the
pressure increases by a factor of 1.26, what is the final Celsius temperature?
EXAMPLE 10.4 The Ideal Gas Law: How Much Oxygen?

A patient receiving breathing therapy purchased a filled M9 needs to be calculated first. That implies that the macroscopic
(medical size 9) oxygen (O2) tank. The tank has a volume of form of the ideal gas law (Eq. 10.7) should be used. Then the
2.5 L and is filled with pure oxygen to an absolute pressure of molar mass of oxygen (O2) can be used to calculate the mass.
100 atm at 20 °C. What is the mass of the oxygen in the tank? In addition, the units of pressure and volume should be in
standard SI units, and temperature should be in kelvin.
THINKING IT THROUGH. Since the mass of oxygen (O2) is to
be determined, the number of moles of the gas in the tank
SOLUTION.
Given: T = 20 °C = 120 + 2732 K = 293 K Find: m (mass of oxygen)
p = 100 atm = 1100211.01 * 105 Pa2 = 1.01 * 107 Pa
V = 2.5 L = 0.0025 m3 1because 1 m3 = 1000 L2
Using the macroscopic form of the ideal gas law (Eq. 10.7) and Oxygen (O2) has a molecular mass of 2 * 16.0 u = 32.0 u, so
solving for the number of moles, n, its molar mass is 32.0 g>mol. Therefore, the mass of oxygen in
pV 11.01 * 107 Pa210.0025 m32 the tank is
38.31 J>1mol # K241293 K2 m = 110.4 mol2132.0 g>mol2 = 333 g = 0.333 kg

n = = = 10.4 mol
RT
FOLLOW-UP EXERCISE. When the patient breathes the oxygen, it is depressurized to 1 atm. What is the volume of the oxygen at
this pressure?
*The 273.16 value given here for the triple point temperature and the -273.15 value, as determined
in Fig. 10.8, indicate different things. The - 273.15 °C is taken as 0 K. The 273.16 K (or 0.01 °C) is a dif-
ferent reading on a different temperature scale.
DID YOU LEARN?

pV
➥ The common forms of the ideal gas law are the ration form = constant, the
T
microscopic form, pV = NkB T, and the macroscopic form, pV = nRT.
➥ Absolute zero is determined by extrapolating or extending the pressure versus
temperature plot of an ideal gas until the pressure is zero.
➥ To convert a Celsius temperature to a Kelvin temperature, simply add 273 to TC , that
is, T = TC + 273.
10.4 Thermal Expansion

➥ What is the fundamental cause of thermal expansion?

➥ If water is heated from 0 °C to 4 °C, will it expand or contract?
➥ Which temperature scale(s) can be used in calculating the change in temperature ( ¢T)
in thermal expansion when the unit of the coefficient of thermalexpansion is 1>°C?
Changes in the dimensions and volumes of materials are common thermal effects.
As you learned earlier, thermal expansion provides a means of measuring temper-
ature. The thermal expansion of gases is generally described by the ideal gas law
and is very obvious. Less dramatic, but by no means less important, is the thermal
expansion of liquids and solids (discussed in Section 9.1).
Thermal expansion results from a change in the average distance separating the
atoms of a substance as it is heated. The atoms are held together by bonding
forces, which can be simplistically represented as springs in a simple model of a
solid. (See Fig. 9.1.) With increased temperature, the atoms vibrate back and forth
over greater distances. With wider vibrations in all dimensions, the solid expands
as a whole.
The change in one dimension of a solid (length, width, or thickness) is called
linear expansion. For small temperature changes, linear expansion (or contraction)
is approximately proportional to ¢T, or T - To (䉲 Fig. 10.10a). The fractional
change in length is 1L - Lo2>Lo or ¢L>Lo , where Lo is the original length of the solid
䉲 F I G U R E 1 0 . 1 0 Thermal expan-
sion (a) Linear expansion is propor-
tional to the temperature change; at the initial temperature.* This ratio is related to the change in temperature by
that is, the change in length ¢L is
proportional to ¢T, and ¢L
¢L>Lo = a¢T, where a is the ther-
= a¢T or ¢L = aLo ¢T (linear expansion) (10.9)
Lo
mal coefficient of linear expansion.
(b) For isotropic expansion, the where a is the thermal coefficient of linear expansion. Note that the unit of a is
thermal coefficient of area expan- inverse temperature: inverse degree Celsius (1>°C, or °C-1). Values of a for some
sion is approximately 2a. (c) The
thermal coefficient of volume materials are given in 䉴 Table 10.1.
expansion for solids is about 3a.
To
Lo
∆L
Ao Vo
T = To + ∆T
L
∆A ∆V
∆ L = ␣ ∆T ∆A = 2␣ ∆T ∆V = 3␣ ∆T
Lo Ao Vo
(a) Linear expansion (b) Area expansion (c) Volume expansion
*A fractional change may also be expressed as a percent change. For example, by analogy, if you
invested $100 ($o) and made $10 1¢$2, then the fractional change would be ¢$>$o = 10>100 = 0.10, or
an increase (percent change) of 10%
10.4 THERMAL EXPANSION 369
TABLE 10.1 Values of Thermal Expansion Coefficients (in °C-1) for Some Materials at 20 °C
Coefficient of Linear Coefficient of Volume
Material Expansion (A) Material Expansion ( B )
Aluminum 24 * 10-6 Alcohol, ethyl 1.1 * 10-4

Brass 19 * 10-6 Gasoline 9.5 * 10-4
Brick or concrete 12 * 10-6 Glycerin 4.9 * 10-4
Copper 17 * 10-6 Mercury 1.8 * 10-4
Glass, window 9.0 * 10 -6
Water 2.1 * 10-4
Glass, Pyrex 3.3 * 10-6
Gold 14 * 10-6 Air (and most other 3.5 * 10-3
gases at 1 atm)
Ice 52 * 10-6
Iron and steel 12 * 10-6
A solid may have different coefficients of linear expansion for different direc-
tions, but for simplicity it will be assumed that the same coefficient applies to all
directions (in other words, that solids show isotropic expansion). Also, the coeffi-
cient of expansion may vary slightly for different temperature ranges. Since this
variation is negligible for most common applications, a will be considered to be
constant and independent of temperature.
Equation 10.9 can be rewritten to give the final length (L) after a change in LEARN BY DRAWING 10.1
temperature:
L - Lo = aLo ¢T so L = Lo + aLo ¢T thermal area
or
expansion
L = Lo11 + a¢T2 (10.10)
Equation 10.10 can be used to compute the thermal expansion of areas of flat Lo Ao = Lo2
objects. Since area (A) is length squared (L2) for a square,
A = L2 = L2o11 + a¢T22 = A o11 + 2a¢T + a2 ¢T22
where Ao is the original area. Because the values of a for solids are much less than
1 1~10-52, as shown in Table 10.1), the second-order term (containing Lo
a2 L 110-522 = 10-10 V 10-5) can be dropped with negligible error. As a first- ∆ A3
order approximation, then, and with the understanding that the change in area ∆A1 }∆L
¢A = A - A o , we have
A = A o11 + 2a¢T2 or
¢A
= 2a¢T (area expansion) (10.11)
Ao Lo Ao ∆A2
Thus, the thermal coefficient of area expansion (Fig. 10.10b) is twice as large as
the coefficient of linear expansion. (That is, it is equal to 2a). This relationship is
valid for all flat shapes. (See Learn by Drawing 10.1, Thermal Area Expansion.)
}
Lo ∆L
Similarly, a first-order expression for thermal volume expansion is
∆A = ∆A1 + ∆A2 + ∆A3
∆A1 = ∆A2 = Lo ∆L
V = Vo11 + 3a¢T2 or
¢V
= 3a¢T (volume expansion) (10.12) = Lo (␣Lo ∆T ) = ␣Ao ∆T
Vo
Since ∆A3 is very small
compared to ∆A1 and ∆A2,
The thermal coefficient of volume expansion (Fig. 10.10c) is equal to 3a (for
∆A 2␣Ao ∆T
isotropic solids).
(a) (b)
䉱 F I G U R E 1 0 . 1 1 Expansion gaps (a) Expansion gaps are built into bridge roadways to
prevent contact stresses produced by thermal expansion. (b) These loops in oil pipelines
serve a similar purpose. As hot oil passes through them, the pipes expand, and the loops
take up the extra length. The loops also accommodate expansions resulting from day–night
and seasonal temperature variations.
Keep in mind that the equations for thermal expansions are approximations
(why?), so they may apply only in certain situations.
The thermal expansion of materials is an important consideration in construc-
tion. Seams are put in concrete highways and sidewalks to allow room for expan-
sion and to prevent cracking. Expansion gaps in large bridges and between
railroad rails are necessary to prevent damage (䉱 Fig. 10.11a). The Golden Gate
Bridge across San Francisco Bay varies in length by about 1 m between summer
and winter. Similarly, expansion loops are found in oil pipelines (Fig. 10.11b). The
height of the Eiffel Tower in Paris varies 0.36 cm for each degree Celsius change.
The thermal expansion of steel beams and girders can cause tremendous pres-
sures, as the following Example shows.
EXAMPLE 10.5 Temperature Rising: Thermal Expansion and Stress

A steel beam is 5.0 m long at a temperature of 20 °C (68 °F). T H I N K I N G I T T H R O U G H . (a) This is a direct application of
On a hot day, the temperature rises to 40 °C (104 °F). (a) What Eq. 10.9. (b) As the constricted beam expands, it applies a
is the change in the beam’s length due to thermal expansion? stress, and hence a force, to the supports. For linear expan-
(b) Suppose that the ends of the beam are initially in contact sion, Young’s modulus (Section 9.1) should come into play.
with rigid vertical supports. How much force will the
expanded beam exert on the supports if the beam has a cross-
sectional area of 60 cm2?
SOLUTION.
Given: Lo = 5.0 m Find: (a) ¢L (change in length)

To = 20 °C (b) F (force)
T = 40 °C
a = 12 * 10-6 °C-1 (from Table 10.1)
1m 2
A = 160 cm22a b = 6.0 * 10-3 m2
100 cm
(a) Using Eq. 10.9 to find the change in length with ¢T = T - To = 40 °C - 20 °C = 20 °C, we have
¢L = aLo ¢T = 112 * 10-6 °C-1215.0 m2120 °C2 = 1.2 * 10-3 m = 1.2 mm
This may not seem like much of an expansion, but it can give rise to a great deal of force if the beam is constrained and kept from
expanding, as part (b) will show.
10.4 THERMAL EXPANSION 371
(b) By Newton’s third law, if the beam is kept from expanding, the force the beam exerts on its constraint supports is equal to the
force exerted by the supports to prevent the beam from expanding by a length ¢L. This is the same as the force that would be
required to compress the beam by that length. Using Young’s modulus and Eq. 9.4 with Y = 20 * 1010 N>m2 (Table 9.1), the
stress on the beam is
F Y¢L 120 * 1010 N>m2211.2 * 10-3 m2
= = = 4.8 * 107 N>m2
A Lo 5.0 m
The force is then
F = 14.8 * 107 N>m22A = 14.8 * 107 N>m2216.0 * 10-3 m22
= 2.9 * 105 N 1about 65 000 lb, or 32.5 tons!2
F O L L O W - U P E X E R C I S E . Expansion gaps between identical steel beams laid end to end are specified to be 0.060% of the length of
a beam at the installation temperature. With this specification, what is the temperature range for noncontact expansion?
CONCEPTUAL EXAMPLE 10.6 Larger or Smaller? Area Expansion

A circular piece is cut from a flat metal sheet (䉴 Fig. 10.12a). If
the sheet is then heated in an oven, the size of the hole will
(a) become larger, (b) become smaller, (c) remain unchanged.
REASONING AND ANSWER. It is a common misconception to
think that the area of the hole will shrink because the metal
expands inwardly around it. To counter this misconception,
think of the piece of metal removed from the hole rather than Circular piece
of the hole itself. This piece would expand with increasing replaced
temperature. The metal in the heated sheet reacts as if the
piece that was removed were still part of it. (Think of putting (a) Metal plate with (b) Metal plate without
hole hole
the piece of metal back into the hole after heating, as in
Fig. 10.12b, or consider drawing a circle on an uncut metal 䉱 F I G U R E 1 0 . 1 2 A larger or smaller hole? See
sheet and heating it.) So the answer is (a). Example text for description.
F O L L O W - U P E X E R C I S E . A student is trying to fit a bearing onto a shaft. The inside diameter of the bearing is just slightly smaller
than the outside diameter of the shaft. Should the student heat the bearing or the shaft in order to fit the shaft inside the bearing?
Fluids (liquids and gases), like solids, normally expand with increasing temper-
ature. Because fluids have no definite shape, only volume expansion (and not lin-
ear or area expansion) is meaningful. The expression is
¢V
= b¢T (fluid volume expansion) (10.13)
Vo
where b is the coefficient of volume expansion for fluids. Note in Table 10.1 that
the values of b for fluids are typically larger than the values of 3a for solids.
Unlike most liquids, water exhibits an anomalous expansion in volume near its
ice point. The volume of a given amount of water decreases as it is cooled from
room temperature, until its temperature reaches 4 °C (䉲 Fig. 10.13a). Below 4 °C,
the volume increases, and therefore the density decreases (Fig. 10.13b). This
means that water has its maximum density 1r = m>V2 at 4 °C (actually, 3.98 °C).
When water freezes, its molecules form a hexagonal (six-sided) lattice pattern.
(This is why snowflakes have hexagonal shapes.) It is the open structure of this
lattice that gives water its unusual property of expanding on freezing and being
less dense as a solid than as a liquid. (This is why ice floats in water and frozen
water pipes burst—water expands by about 9% on freezing.)
This property has an important environmental effect: Bodies of water such as
lakes and ponds freeze at the top first, and the ice that forms floats. As a lake cools
toward 4 °C, water near the surface contracts and becomes denser, and sinks. The
warmer, less dense water near the bottom rises. However, once the colder water
䉴 F I G U R E 1 0 . 1 3 Thermal expan- V ␳
sion of water Water exhibits nonlin-
Volume of 1 kg of water (× 10 –3 m3)

1.043 43
ear expansion behavior near its ice 1.000 000
point. (a) Above 4 °C (actually,
3.98 °C), water expands with
Density (kg/m3 × 10 3)
0.999 95
increasing temperature, but from
4 °C down to 0 °C, it expands with
0.999 90
decreasing temperature. (b) As a
result, water has its maximum den-
sity near 4 °C. 0.999 85
0.999 80
1.000 13
0.999 75
1.000 00
T 0.999 70 T
0 5 10 100 0 5 10
Temperature (°C) Temperature (°C)
(a) (b)
on top reaches temperatures below 4 °C, it becomes less dense and remains at the
surface, where it freezes. If water did not have this property, lakes and ponds
would freeze from the bottom up, which would destroy much of their animal and
plant life (and would make ice skating a lot less popular). There would also be no
oceanic ice caps at the polar regions. Instead, there would be a thick layer of ice at
the bottom of the ocean, covered by a layer of water.
DID YOU LEARN?

➥ Thermal expansion is caused by the change in the average distance between the
atoms in a substance when the temperature is changed.
➥ Water exhibits an anomalous expansion in volume between 0 °C and 4 °C; that is,
water will actually contract if heated from 0 °C to 4 °C.
➥ Either the Kelvin or Celsius temperature scales can be used for the ¢T in thermal
expansion.The Fahrenheit temperature scale cannot be used.
10.5 The Kinetic Theory of Gases

➥ What fundamental physical quantity does the absolute temperature of a gas

determine?
➥ If a sample of gas with more massive molecules and another with less massive mole-
cules are at the same temperature, which gas molecules will have a higher rms speed?
➥ For a given monatomic gas, what is the relationship between its absolute tempera-
ture and its total internal energy?
If the molecules of a sample of gas are viewed as colliding particles, the laws of
mechanics can be applied to each molecule of the gas. Then the gas’s microscopic
characteristics, such as velocity and kinetic energy, can be described in terms of
molecular motion. Because of the large number of particles involved, however, a
statistical approach is employed for such a microscopic description.
One of the major accomplishments of theoretical physics was to do exactly
that—derive the ideal gas law from mechanical principles. This derivation led to a
new interpretation of temperature in terms of the translational kinetic energy of
the gas molecules. As a theoretical starting point, the molecules of an ideal gas are
viewed as point masses in random motion with relatively large distances separat-
ing them so molecular collisions can be neglected.
In this section, we consider primarily the kinetic theory of monatomic (single-
atom) gases, such as He, and learn about the internal energy of such a gas. In the
10.5 THE KINETIC THEORY OF GASES 373
next section, the internal energy of diatomic (two-atom molecules) gases, such as
O2, will be considered.
According to the kinetic theory of gases, the molecules of an ideal gas undergo
perfectly elastic collisions (discussed in Section 6.4) with the walls of its container.
From Newton’s laws of motion, the force on the walls of the container can be cal-
culated from the change in momentum of the gas molecules when they collide
with the walls (䉴 Fig. 10.14). If this force is expressed in terms of pressure
1force>area2, the following equation is obtained (see Appendix II for derivation):
pV = 13 Nmv 2rms (10.14) y
Here, V is the volume of the container or gas, N is the number of gas molecules in vy
v
the closed container, m is the mass of a gas molecule, and vrms is the average speed
of the molecules, but a special kind of average. It is obtained by averaging the
–vx
squares of the speeds and then taking the square root of the average—that
is, 2v -2 = vrms . As a result, vrms is called the root-mean-square (rms) speed. F
Solving Eq. 10.6 for pV and equating the resulting expression with Eq. 10.14 shows
how temperature came to be interpreted as a measure of translational kinetic energy: vy
pV = NkB T = 13 Nmv 2rms or v
1 2
2 mv rms = 32 kB T (for ideal gases) (10.15)
vx
Thus, the absolute temperature of a gas is directly proportional to its average ran- x
dom kinetic energy (per molecule), since K = 12 mv2rms = 32 kB T.
Wall
∆p m∆v
F= =
∆t ∆t
INTEGRATED EXAMPLE 10.7 Molecular Speed: Relation to (Force = time rate of change
Absolute Temperature of momentum)
A helium molecule (He) in a helium balloon is at 20 °C. (a) If it is heated to 40 °C, its rms
䉱 F I G U R E 1 0 . 1 4 Kinetic theory
speed will (1) double, (2) increase by less than a factor of 2, (3) be half as much,
of gases The pressure a gas exerts
(4) decrease by less than a factor of 2. Explain. (b) Calculate the rms speeds at these two on the walls of a container is due to
temperatures. (Take the mass of the helium molecule to be 6.65 * 10-27 kg). the force resulting from the change
( A ) C O N C E P T U A L R E A S O N I N G According to Eq. 10.15, the absolute temperature is pro- in momentum of the gas molecules
portional to the square of the rms speed, T r v 2rms , or the rms speed is proportional to that collide with the wall. The wall
the square root of the absolute temperature, vrms r 1T. Therefore, a higher tempera- exerts a force (action) on the mole-
ture will increase the rms speed, thus (3) and (4) are not possible. cule to change its momentum. The
When the temperature increases from 20 °C to 40 °C, the absolute temperature molecule exerts a reaction force on
increases only from 1273 + 202 K = 293 K to 1273 + 402 K = 313 K, not even close to
the wall. The force exerted by an
individual molecule is equal to the
doubling. Furthermore, even if the absolute temperature were to double, the square root time rate of change of momentum;
of it would not double either (but it would still increase). Thus, the answer is (2) increase > ¢t = m¢v > ¢t,
B B B
that is, F = ¢p
by less than a factor of 2. B B
where p = mv. The sum of the
( B ) Q U A N T A T I V E R E A S O N I N G A N D S O L U T I O N All the data needed to solve for the average instantaneous normal components
speed in Eq. 10.15 are given. The Celsius temperatures must be changed to kelvins. of the collision forces gives rise to
the average pressure on the wall.
Given: m = 6.65 * 10-27 kg Find: vrms (rms speed)
T1 = 20 °C = 1273 + 202 K = 293 K
T2 = 40 °C = 1273 + 402 K = 313 K
Rearranging Eq. 10.15,
3kB T 311.38 * 10-22 J>K21293 K2
For 20 °C: vrms = =
C m C 6.65 * 10-27 kg
= 1.35 * 103 m>s = 1.35 km>s 13020 mph2
311.38 * 10-23 J>K21313 K2

For 40 °C: vrms = = 1.40 * 103 m>s
C 6.65 * 10-27 kg
= 1.40 km>s 13130 mph2
F O L L O W - U P E X E R C I S E . In this Example, if the rms speed is to double its value at 20 °C,
what would be the new Celsius temperature?
Interestingly, Eq. 10.15 predicts that at absolute zero 1T = 0 K2, all translational
molecular motion of a gas would cease. According to classical theory, this would
correspond to absolute zero energy. However, modern quantum theory says that
there would still be some zero-point motion and a corresponding minimum zero-
point energy. Basically, absolute zero is the temperature at which all the energy that
can be removed from an object has been removed.
INTERNAL ENERGY OF MONATOMIC GASES

Because the “particles” in an ideal monatomic gas do not vibrate or rotate, as
explained previously, the total translational kinetic energy of all the molecules is
equal to the total internal energy of the gas. That is, the gas’s internal energy is all
“temperature” energy (Section 10.1). With N molecules in a system, Eq. 10.15 can
be used, expressing the energy per molecule, to write an equation for the total
internal energy U:
U = N A 12 mv2rms B = 32 NkB T = 32 nRT (for monatomic gases) (10.16)
Thus, it can be seen that the internal energy of an ideal monatomic gas is directly
proportional to its absolute temperature. (In Section 10.6, it will be learned that
this is true regardless of the molecular structure of the gas. However, the expres-
sion for U will be a bit different for gases that are not monatomic.) This means that
if the absolute temperature of a gas is doubled, for example, from 200 K to 400 K,
then the internal energy of the gas is also doubled.
DIFFUSION
We depend on our sense of smell to detect odors, such as the smell of smoke from
something burning. That you can smell something from a distance implies that
molecules get from one place to another in the air—from the source to your nose.
This process of random molecular mixing in which particular molecules move
from a region where they are present in higher concentration to one where they
are in lower concentration is called diffusion. Diffusion also occurs readily in liq-
uids; think about what happens to a drop of ink in a glass of water (䉲 Fig. 10.15). It
even occurs to some degree in solids.
The rate of diffusion for a particular gas depends on the rms speed of its mole-
cules. Even though gas molecules have large average speeds (Example 10.8), their
average positions change slowly, and the molecules do not fly from one side of a
䉱 F I G U R E 1 0 . 1 5 Diffusion in liquids Random molecular motion would eventually dis-

tribute the dye throughout the water. Here there is some distribution due to mixing, and
the ink colors the water after a few minutes. The distribution would take more time by dif-
fusion only.
10.5 THE KINETIC THEORY OF GASES 375
䉳 F I G U R E 1 0 . 1 6 Separation by
gaseous diffusion The molecules of
O2 both gases diffuse through the
porous barrier, but because oxygen
CO2 molecules have the greater average
speed, more of them pass through.
Thus, over time, there is a greater
Porous barrier
concentration of oxygen molecules
on the other side of the barrier.
Equal volumes Diffusion
of O2 and CO2 through barrier
room to the other. Instead, there are frequent collisions, and as a result, the mole-
cules “drift” rather slowly. For example, suppose someone opened a bottle of
ammonia on the other side of a closed room. It would take some time for the
ammonia to diffuse across the room until you could smell it.
The kinetic theory of gases says that the average translational kinetic energy
(per molecule) of a gas is proportional to the absolute temperature of the gas:
1 2 3
2 mv rms = 2 kB T. So on the average, the molecules of different gases (having differ-
ent masses) move at different speeds at a given temperature. As you might expect,
because they move faster, less massive gas molecules diffuse faster than do more
massive gas molecules.
For instance, at a particular temperature, molecules of oxygen (O2) move faster
on the average than do the more massive molecules of carbon dioxide (CO2), so
oxygen can diffuse through a barrier faster than carbon dioxide can. Suppose that
a mixture of equal volumes of oxygen and carbon dioxide is contained on one side
of a porous barrier (䉱 Fig. 10.16). After a while, some O2 molecules and some CO2
molecules will have diffused through the barrier, but more oxygen than carbon
dioxide. Purer oxygen can be obtained by repeating the separation process many
times. Separation by gaseous diffusion is a key process in obtaining enriched ura-
nium, which was used in the first atomic bomb and in early nuclear reactors that
generate electricity (Section 30.2).
Fluid diffusion is very important to organisms. In plant photosynthesis, carbon
dioxide from the air diffuses into leaves, and oxygen and water vapor diffuse out.
The diffusion of a liquid across a permeable membrane with a concentration gra-
dient (a concentration difference) is called osmosis, a process that is vital in living
cells. Osmotic diffusion is also important to kidney functioning: Tubules in the
kidneys concentrate waste matter from the blood in much the same way that oxy-
gen is removed from mixtures. (See the accompanying Insight 10.3, Physiological
Diffusion in Life Processes, for other examples of diffusion.)
Osmosis is the tendency for the solvent of a solution, such as water, to diffuse
across a semipermeable membrane from the side where the solvent is at higher
concentration to the side where it is at lower concentration. When pressure is
applied to the side with the lower concentration, the diffusion is reversed—a
process called reverse osmosis. Reverse osmosis is used in desalination plants to
provide freshwater from seawater in dry coastal regions and in drinking water
purification.
DID YOU LEARN?

➥ The absolute temperature of a gas is directly proportional to the average
translational kinetic energy of the gas molecules.That is, if the absolute
temperature is doubled, the average kinetic energy of the molecules will also
double.
➥ The two samples of gas molecules have the same average translational kinetic
energy because they are at the same temperature.Therefore, the less massive
molecules will have a higher rms speed because K = 12 mv2rms.
➥ The total internal energy of a monatomic gas is directly proportional to the
absolute temperature, U = 32 nRT. When the absolute temperature doubles, the gas
will have twice as much total internal energy.
INSIGHT 10.3 Physiological Diffusion in Life Processes

Diffusion plays a central role in many life processes. For sugar), the blood continuously brings in fresh supplies of the
example, consider a cell membrane in the lung. Such a mem- substances to maintain the concentration gradient needed for
brane is permeable to a number of substances, any of which diffusion to the cells. The continuous production of carbon
will diffuse through the membrane from a region where its dioxide (CO2) and metabolic wastes in the cells produces con-
concentration is high to a region where its concentration is centration gradients in the opposite direction for these sub-
low. Most important, the lung membrane is permeable to oxy- stances. They therefore diffuse out of the cells into the blood,
gen (O2), and the transfer of O2 across the membrane occurs to be carried away from the tissues by the circulatory system.
because of a concentration gradient. During periods of physical exertion, cellular activity
The blood carried to the lungs is low in O2, having given up increases. More O2 is used up and more CO2 is produced,
the oxygen during its circulation through the body to tissues thereby increasing the concentration gradients and the diffu-
requiring O2 for metabolism. Conversely, the air in the lungs sion rates. How do the lungs respond to an increased demand
is high in O2 , because there is a continuous exchange of fresh for O2 to the blood? As you might expect, the rate of diffusion
air in the breathing process. As a result of this concentration depends on the surface area and thickness of the lung mem-
difference, or gradient, O2 diffuses from the space within the brane. Deeper breathing during exercise causes the alveoli
lungs into the blood that flows through the lung tissue, and (small air sacs in the lungs) to increase in volume. The alveo-
the blood leaving the lungs is high in O2 . lar surface area increases accordingly, and the thickness of the
Exchanges between the blood and the tissues occur across membrane wall decreases, allowing more rapid diffusion.
capillary walls, and diffusion again is a major factor. The Also, the heart works harder during exercise, and the blood
chemical composition of arterial blood is regulated to main- pressure is raised. The increased pressure forces open capil-
tain the proper concentrations of particular solutes (substances laries that are normally closed during rest or mild activity. As
dissolved in the blood solution), so diffusion takes place in the a result, the total exchange area between the blood and cells is
appropriate directions across capillary walls. For example, as increased. Each of these changes helps expedite the exchange
cells take up O2 and nutrients, including glucose (blood of gases during exercise.
*10.6 Kinetic Theory, Diatomic Gases,

and the Equipar tition Theorem
➥ What is a diatomic molecule, and what are some examples?

➥ What is the essence of the equipartition theorem?
➥ How does the equipartition theorem apply to monatomic and diatomic gases?
In the real world, most gases are not monatomic gases. Monatomic gases are ele-
ments known as noble or inert gases, because they do not readily combine with
other atoms. These elements are found on the far right side of the periodic table:
helium, neon, argon, krypton, xenon, and radon.
However, the mixture of gases we breathe (collectively known as “air”) consists
mainly of diatomic molecules of nitrogen (N2, 78% by volume) and oxygen (O2,
21% by volume). Each of these gases has two identical atoms chemically bonded
together to form a single molecule. How do we deal with these more complicated
molecules in terms of the kinetic theory of gases? [There are even more compli-
cated gas molecules consisting of more than two atoms, such as carbon dioxide
(CO2). However, because of the complexity of such gas molecules, our discussion
will be limited to diatomic molecules.]
THE EQUIPARTITION THEOREM

As was learned in Section 10.5, the translational kinetic energy of a gas is deter-
mined by the gas’s temperature. Thus, for any type of gas, regardless of how
many atoms make up its molecules, it is always true that the average translational
kinetic energy per molecule is still proportional to the temperature of the
gas (Eq. 10.15): 12 mv 2rms = 32 kB T (for all gases).
Recall that for monatomic gases, the total internal energy U consists solely of
translational kinetic energy. For diatomic molecules, this is not true, because a
*10.6 KINETIC THEORY, DIATOMIC GASES, AND THE EQUIPARTITION THEOREM 377
diatomic molecule is free to rotate and vibrate in addition to moving linearly.

Therefore, these extra forms of energy must be taken into account. The expression
given in Eq. 10.16 (U = 32 NkB T) for monatomic gases, which assumes that the total
energy is due only to translational kinetic energy, therefore does not hold for
diatomic gases.
Scientists wondered exactly how the expression for the internal energy of a
diatomic gas might differ from that for a monatomic gas. In looking at the
derivation of Eq. 10.16 from the kinetic theory, they realized that the factor of 3
in that equation was due to the fact that the gas molecules had three indepen-
dent linear ways (dimensions) of moving. Thus, for each molecule, there were
three independent ways of possessing kinetic energy: with x, y, and z linear
motion. Each independent way a molecule has for possessing energy is called a
degree of freedom.
According to this scheme, a monatomic gas has only three degrees of freedom,
since its molecules can move only linearly and can possess kinetic energy in three
dimensions.
On the basis of the understanding of monatomic gases and their three degrees
of freedom, the equipartition theorem was proposed. (As the name implies, the
total energy of a gas or molecule is “partitioned,” or divided, equally for each
degree of freedom.) That is,
On average, the total internal energy U of an ideal gas is divided equally among each
degree of freedom its molecules possess. Furthermore, each degree of freedom con-
tributes 12 NkB T (or 12 nRT) to the total internal energy of the gas.
THE INTERNAL ENERGY OF A DIATOMIC GAS

To use the equipartition theorem to calculate the internal energy of a diatomic gas
such as oxygen, it must be realized that U now includes all the available degrees of
freedom. A diatomic molecule could rotate (see Fig. 10.1), thus having rotational
kinetic energies about three independent axes of rotations (three more degrees of
freedom). A diatomic gas might also vibrate, thus having vibrational kinetic and
potential energies (two additional degrees of freedom). Altogether, a diatomic
molecule should have seven degrees of freedom.
Consider a symmetric diatomic molecule—for example, O2. A classical model
describes such a diatomic molecule as though the molecules were particles con- z
nected by a rigid rod (䉴 Fig. 10.17). The rotational moment of inertia, I, has the vz
same value about each of the axes (x and y) that pass perpendicularly through the
center of the rod. The moment of inertia about the z-axis is essentially zero.
(Why?) Thus, only two degrees of freedom are associated with the rotational
kinetic energies of diatomic molecules.
Furthermore, quantum theory predicts (and experiment verifies) that for
normal (room) temperatures, the vibrational kinetic energy and potential
energy are much smaller than the translational and rotational kinetic energies vy y
and therefore can be ignored. Thus, the total internal energy of a diatomic gas is x vx
composed of the internal energies due to the three linear degrees of freedom
and the two rotational degrees of freedom, for a total of five degrees of free-
dom. Hence,
䉱 F I G U R E 1 0 . 1 7 Model of a
U = Ktrans + Krot = 3 A 1
2 nRT B + 2A 1
2 nRT B diatomic gas molecule A dumbbell-
(for diatomic gases) (10.17) like molecule can rotate about three
5 5
= 2 nRT = 2 NkB T
axes. The moment of inertia, I, about
the x- and y-axes is the same. The
masses (molecules) on the ends of
Thus, a given monatomic sample of gas at normal room temperature has 40% less the rod are point-like particles, so
internal energy than a similar diatomic sample at the same temperature. Or, the moment of inertia about the
equivalently, the monatomic sample possesses only 60% of the internal energy of z-axis Iz is negligible compared to Ix
the diatomic sample. and Iy.
EXAMPLE 10.8 Monatomic versus Diatomic: Are Two Atoms Better Than One?
More than 99% of the air we breathe consists of diatomic T H I N K I N G I T T H R O U G H . (a) We have to consider the number
gases, mainly nitrogen (N2, 78%) and oxygen (O2, 21%). There of degrees of freedom in a monatomic gas and a diatomic gas
are traces of other gases, one of which is radon (Rn), a in computing the internal energy U. (b) Only three linear
monatomic gas arising from radioactive decay of uranium in degrees of freedom contribute to the translational kinetic
the ground. (a) Calculate the total internal energy of 1.00-mol energy portion (Utrans) of the internal energy.
samples each of oxygen and radon at room temperature
(20 °C). (b) For each sample, calculate the amount of internal
energy associated with molecular translational kinetic energy.
SOLUTION. Listing the data and converting to kelvins because internal energy is expressed in terms of absolute temperature:
Given: n = 1.00 mol Find: (a) U (for O2 and Rn samples)
T = 120 + 2732 K = 293 K (b) Utrans (for O2 and Rn at 20 °C)
(a) Let’s compute the total internal energy of the (monatomic) radon sample first, using Eq. 10.16:
URn = 32 nRT = 32 11.00 mol238.31 J>1mol # K241293 K2 = 3.65 * 103 J

The (diatomic) oxygen will also include internal energy stored as two extra degrees of freedom, due to rotation. Thus, we have
UO2 = 52 nRT = 52 11.00 mol238.31 J>1mol # K241293 K2 = 6.09 * 103 J
As we have seen, even though each sample has the same number of molecules and the same temperature, the oxygen sample has
about 67% more total internal energy.
(b) For (monatomic) radon, all the internal energy is in the For (diatomic) oxygen, only 32 nRT of the total internal energy
form of translational kinetic energy; hence, the answer is the A 52 nRT B is in the form of translational kinetic energy, so the
same as in part (a): answer is the same as for radon; that is, Utrans = 3.65 * 103 J
Utrans = URn = 3.65 * 103 J for both gas samples.
F O L L O W - U P E X E R C I S E . (a) In this Example, how much energy is associated with the rotational motion of the oxygen molecules?
(b) Which sample has the higher rms speed? (Note: The mass of one radon atom is about seven times the mass of an oxygen mole-
cule.) Explain your reasoning.
DID YOU LEARN?

➥ A diatomic molecule consists of two atoms in a molecule. For example, oxygen gas
(O2) and nitrogen gas (N2) both have two atoms (O and N, respectively) in the gas
molecule.
➥ The equipartition theorem states that each degree of freedom of gases contributes
equally to the total internal energy by an amount equal to 12 nRT.
➥ A monatomic gas has three (3) degrees of translational freedom, so its total internal
energy is U = 3 * 12 nRT = 32 nRT. A diatomic gas has three (3) degrees of
translational freedom plus two (2) degrees of rotational freedom, so its total internal
energy is U = 5 * 21 nRT = 25 nRT.
PULLING IT TOGETHER Temperatures, Ideal Gases, and Internal Energies

A quantity of 2.0 moles of an ideal monatomic gas at a pres- monatomic gas relationship (Eq. 10.16), and temperature con-
sure of 1.5 atm is confined in a volume of 0.040 m3. It is then version between the Kelvin and Fahrenheit scales. In order to
heated so its internal energy doubles. What is its final Fahren- find the final temperature, the initial absolute temperature is
heit temperature? first found using the ideal gas law. Then the internal energy of
monatomic gas formula can be used to find the final absolute
T H I N K I N G I T T H R O U G H . This example involves the macro- temperature. Finally, the absolute temperature is converted to
scopic ideal gas law (Eq. 10.7), the internal energy of the Fahrenheit scale.
SOLUTION. Eq. 10.7 and Eq. 10.16 can be used. Also, pressure needs to be converted from atm to Pa.
Given: n = 2.0 mol Find: T2(F) (final Fahrenheit temperature)
V1 = V2 = 0.040 m3 1confined2
p1 = 1.5 atm = 11.5211.01 * 105 Pa2 = 1.515 * 105 Pa
U2 = 2U1
Rearranging Eq. 10.7 to find T1: Converting T2 to Celsius scale: T2(C) = 1729.2 - 2732 °C =
pV 11.515 * 10 Pa210.040 m 2
5 3 456.2 °C .
12.0 mol238.31 J>1mol # K24

T1 = = = 364.6 K Finally, converting T2(C) to the Fahrenheit scale:
nR
T2(F) = 95 T2(C) + 32 = C 95 1456.22 + 32 D °F = 853 °F
According to Eq. 10.16, U = 32 nRT, doubling U will require T
to double as well.
U2 = 2U1 , so T2 = 2T1 = 21364.6 K2 = 729.2 K
■ Celsius–Fahrenheit conversions: ■ Absolute zero (0 K) corresponds to - 273.15 °C.

9 Pressure
TF = 5 TC + 32 or TF = 1.8TC + 32 (10.1)
TC = 59 1TF - 322 (10.2)
–273.15 °C
Fahrenheit Celsius
Temperature
212 °F 100 °C
–200 °C –100 °C 0 °C 100 °C
Steam Steam
point point (a)
Celsius–Kelvin conversion:
180 °F 100 °C T = TC + 273.15 (10.8)
T = TC + 273 1for general calculations2 (10.8a)
Ice Ice
point point ■ Thermal coefficients of expansion relate the fractional
32 °F 0 °C
change in dimension(s) to a change in temperature.
Thermal expansion of solids:
–40 °F –40 °C
= a¢T or L = Lo11 + a¢T2
¢L
linear: (10.9, 10.10)
Lo
To
■ Heat is the net energy transferred from one object to another Lo
∆L
because of temperature differences. Once transferred, the
energy becomes part of the internal energy of the object (or T = To + ∆T
system). L
■ The ideal (or perfect) gas law relates the pressure, volume,
= 2a¢T or A = A o11 + 2a¢T2
¢A
and absolute temperature of an ideal gas. area: (10.11)
Ao
Ideal (or perfect) gas law (always use absolute temperatures):
p1 V1 p2 V2 Ao
= or pV = NkB T (10.5, 10.6)
T1 T2
∆A
= 3a¢T or V = Vo11 + 3a¢T2

¢V
or volume: (10.12)
Vo
pV = nRT (10.7)
where kB = 1.38 * 10-23 J>K and R = 8.31 J>1mol # K2

Vo
∆V
Thermal volume expansion of fluids: Results of kinetic theory of gases:

¢V pV = 13 Nmv2rms (10.14)
= b ¢T (10.13) 1 2 3
Vo 2 mv rms = 2 kB T (for ideal gases) (10.15)
■ According to the kinetic theory of gases, the absolute tem- U = 32 NkB T = 32 nRT (for monatomic gases) (10.16)
perature of a gas is directly proportional to the average U = 52 NkB T = 52 nRT (for diatomic gases) (10.17)
translational kinetic energy per molecule.

10.1 TEMPERATURE AND HEAT 8. Are the units of the thermal coefficient of linear expan-
AND sion (a) m>°C, (b) m2>°C, (c) m # °C, or (d) 1>°C?
10.2 THE CELSIUS AND FAHRENHEIT 9. Which of the following describes the behavior of water
TEMPERATURE SCALES density in the temperature range of 0 °C to 4 °C:
(a) increases with increasing temperature, (b) remains
1. Temperature is associated with molecular (a) kinetic
constant, (c) decreases with increasing temperature, or
energy, (b) potential energy, (c) momentum, (d) all of the
(d) none of the preceding?
preceding.
2. What types of energies can make up the internal energy
of a diatomic gas: (a) rotational kinetic energy, (b) trans-
10.5 THE KINETIC THEORY OF GASES
lational kinetic energy, (c) vibrational kinetic energy, or
(d) all of the preceding? 10. If the average kinetic energy of the molecules in an ideal
3. An object at a higher temperature (a) must, (b) may, or gas initially at 20 °C doubles, what is the final temperature
(c) must not have more internal energy than another of the gas: (a) 10 °C, (b) 40 °C, (c) 313 °C, or (d) 586 °C?
object at a lower temperature. 11. If the temperature of a quantity of ideal gas is raised
from 100 K to 200 K, is the internal energy of the gas
(a) doubled, (b) halved, (c) unchanged, or (d) none of the
10.3 GAS LAWS, ABSOLUTE
preceding?
TEMPERATURE, AND
12. Two different gas samples are at the same temperature.
THE KELVIN TEMPERATURE SCALE
The more massive gas molecules will have (a) a higher,
4. The temperature used in the ideal gas law must be (b) a lower, or (c) the same rms speed as that of the less
expressed on which scale: (a) Celsius, (b) Fahrenheit, massive gas molecules.
(c) Kelvin, or (d) any of the preceding?
5. If a low-pressure gas at constant volume were to reach
absolute zero, (a) its pressure would reach zero, (b) its *10.6 KINETIC THEORY, DIATOMIC
pressure would reach infinity, (c) its mass would disap- GASES, AND THE EQUIPARTITION
pear, or (d) its mass would be infinite. THEOREM
6. When the temperature of a quantity of gas is increased,
13. Which of the following is a diatomic molecule: (a) He,
(a) the pressure must increase, (b) the volume must
(b) N2, (c) CO2, or (d) Ne?
increase, (c) both the pressure and volume must
increase, (d) none of the preceding. 14. A diatomic gas such as O2 near room temperature has
an internal energy of (a) 32 nRT, (b) 52 nRT, (c) 72 nRT, or
(d) none of the preceding.
10.4 THERMAL EXPANSION
15. On average, is the total internal energy of a gas divided
7. What is the predominant cause of thermal expansion: equally among (a) each molecule, (b) each degree of free-
(a) atom sizes change, (b) atom shapes change, or (c) the dom, (c) translational motion, rotational motion, and
distances between atoms change? vibrational motion, or (d) none of the preceding?
10.1 TEMPERATURE AND HEAT contact with it. Does heat always flow from a body with
AND more internal energy to one with less internal energy?
10.2 THE CELSIUS AND FAHRENHEIT Explain.
TEMPERATURE SCALES
2. What is the hottest (highest temperature) item in a
1. Heat flows spontaneously from a body at a higher tem- home? [Hint: Think about this one, and maybe a light
perature to one at a lower temperature that is in thermal will come on.]
3. The tires of commercial jumbo jets are inflated with pure 10.4 THERMAL EXPANSION
nitrogen, not air. Why? [Hint: air contains moisture.]
11. A cube of ice sits on a bimetallic strip at room tempera-
4. When temperature changes during the day, which scale, ture (䉲 Fig. 10.19). What will happen if (a) the upper strip
Celsius or Fahrenheit, will read a smaller change? Explain. is aluminum and the lower strip is brass, and (b) the
5. What types of energy make up the internal energy of upper strip is iron and the lower strip is copper? (c) If the
monatomic gases? How about diatomic gases? cube is made of a hot metal rather than ice and the two
strips are brass and copper, which metal should be on
10.3 GAS LAWS, ABSOLUTE top to keep the cube from falling off?
TEMPERATURE, AND
THE KELVIN TEMPERATURE SCALE
6. A type of constant volume gas thermometer is shown in
䉲 Fig. 10.18. Describe how it operates.
Bimetallic strip
Reference
mark
h
䉱 F I G U R E 1 0 . 1 9 Which way will the cube go?
See Conceptual Question 11.
12. A solid metal disk rotates freely, so the conservation of

Temperature angular momentum applies (Chapter 8). If the disk is
bath heated while it is rotating, will there be any effect on the
rate of rotation (the angular speed)?
Gas
13. A demonstration of thermal expansion is shown in
䉲 Fig. 10.20. (a) Initially, the ball fits through the ring
made of the same metal. When the ball is heated (b), it
Flexible does not fit through the ring (c). If both the ball and the
tube ring are heated, the ball again fits through the ring.
Explain what is being demonstrated.
䉱 F I G U R E 1 0 . 1 8 A type of constant volume gas 14. A circular ring of iron has a tight-fitting iron bar across its
thermometer See Conceptual Question 6. diameter, as illustrated in 䉲 Fig. 10.21. If the arrangement is
heated in an oven to a high temperature, will the circular
ring be distorted? What if the bar is made of aluminum?
7. Describe how a constant pressure gas thermometer
might be constructed. 䉳 FIGURE 10.21
8. In terms of the ideal gas law, what would a temperature Stress out of shape?
of absolute zero imply? How about a negative absolute See Conceptual
temperature? Question 14.
9. Excited about a New Year’s Eve party in Times Square,
you pump up ten balloons in your warm apartment and
take them to the cold square. However, you are very dis-
appointed with your decorations. Why?
10. Which has more molecules, 1 mole of oxygen or 1 mole
of nitrogen? Explain.
䉴 F I G U R E 1 0 . 2 0 Ball-
and-ring expansion See
Conceptual Question 13
and Exercise 45.
(a) (b) (c)

15. We often use hot water to loosen tightly sealed metal lids 18. Natural gas is odorless; to alert people to gas leaks, the
on glass jars. Explain why this works. gas company inserts an additive that has a distinctive
scent. When there is a gas leak, the additive reaches your
nose before the gas does. What can you conclude about
10.5 THE KINETIC THEORY OF GASES the masses of the additive molecules and gas molecules?
16. Gas sample has twice as much average translational
kinetic energy as gas sample B. What can be said about
*10.6 KINETIC THEORY, DIATOMIC
the absolute temperatures of the gas samples?
GASES, AND THE EQUIPARTITION
17. Equal volumes of helium gas (He) and neon gas (Ne) at
THEOREM
the same temperature (and pressure) are on opposite
sides of a porous membrane (䉲 Fig. 10.22). Describe what 19. If a monatomic gas and a diatomic gas have the same
happens after a period of time, and why. average kinetic energy per molecule will they have the
same temperature? Explain.
20. Why does a diatomic gas have more internal energy than
a monatomic gas with the same number of molecules at
the same temperature?
21. A monatomic gas and a diatomic gas both have n moles
and are at temperature T. What is the difference in their
internal energies? Express your answer in n, R, and T.
He gas Ne gas
䉱 F I G U R E 1 0 . 2 2 What happens as time passes? See Con-

ceptual Question 17.
EXERCISES*
10.1 TEMPERATURE AND HEAT 7. The highest and lowest recorded air temperatures in
●
AND the United States are, respectively, 134 °F (Death Valley,

10.2 THE CELSIUS AND FAHRENHEIT California, 1913) and - 80 °F (Prospect Creek, Alaska,
TEMPERATURE SCALES 1971). What are these temperatures on the Celsius scale?
8. ● ● During open-heart surgery it is common to cool the
1. ● A person running a fever has a body temperature of
patient’s body down to slow body processes and gain an
40 °C. What is this temperature on the Fahrenheit scale?
extra margin of safety. A drop of 8.5 °C is typical in these
2. ● Convert the following to Celsius readings: (a) 80 °F,
types of operations. If a patient’s normal body tempera-
(b) 0 °F, and (c) -10 °F. ture is 98.2 °F, what is her final temperature in both Cel-
3. ● Convert the following to Fahrenheit readings: sius and Fahrenheit?
(a) 120 °C (b) 12 °C and (c) - 5 °C 9. ● ● In the troposphere (the lowest part of the atmos-
4. ● Which is the lower temperature: (a) 245 °C or 245 °F? phere), the temperature decreases rather uniformly with
(b) 200 °C or 375 °F? altitude at a so-called “lapse” rate of about 6.5 °C>km.
5. ● The coldest inhabited village in the world is What are the temperatures (a) near the top of the tropo-
Oymyakon, a town located in eastern Siberia, where it sphere (which has an average thickness of 11 km) and
gets as cold as - 94 °F. What is this temperature on the (b) outside a commercial aircraft flying at a cruising alti-
Celsius scale? tude of 34 000 ft? (Assume that the ground temperature
6. ● The highest and lowest recorded air temperatures in is normal room temperature.)
the world are, respectively, 58 °C (Libya, 1922) and 10. IE ● ● The temperature drops from 60 °F during the day
- 89 °C (Antarctica, 1983). What are these temperatures to 35 °F during the night. (a) The corresponding temper-
on the Fahrenheit scale? ature drop on the Celsius scale is (1) greater than, (2) the
*Assume all temperatures to be exact, and neglect significant figures for small changes in dimension.
EXERCISES 383
same as, or (3) less than . Explain. (b) Compute the tem- 22. ●● Show that 1.00 mol of ideal gas under STP occupies a
perature drop on the Celsius scale. volume of 0.0224 m3 = 22.4 L.
11. IE ● ● There is one temperature at which the Celsius and 23. ●● What volume is occupied by 160 g of oxygen under a
Fahrenheit scales have the same reading. (a) To find that pressure of 2.00 atm and a temperature of 300 K?
temperature, would you set (1) 5TF = 9TC (2) 9TF = 5TC 24. ●● An athlete has a large lung capacity, 7.0 L. Assuming
or (3) TF = TC? Why? (b) Find the temperature. air to be an ideal gas, how many molecules of air are in
12. ● ● (a) The largest temperature drop recorded in the the athlete’s lungs when the air temperature in the lungs
United States in one day occurred in Browning, Montana, is 37 °C under normal atmospheric pressure?
in 1916, when the temperature went from 7 °C to - 49 °C.
25. ●● Is there a temperature that has the same numerical
What is the corresponding change on the Fahrenheit scale?
value on the Kelvin and the Fahrenheit scales? Justify
(b) On the Moon, the average surface temperature is
your answer.
127 °C during the day and -183 °C during the night.
What is the corresponding change on the Fahrenheit scale? 26. ●● A husband buys a helium-filled anniversary balloon
for his wife. The balloon has a volume of 3.5 L in the
13. ● ● ● Astronomers know that the temperatures of stellar
warm store at 74 °F. When he takes it outside, where the
interiors are “extremely high.” By this they mean they
temperature is 48 °F, he finds it has shrunk. By how
can convert from Fahrenheit to Celsius temperature
much has the volume decreased?
using a rough rule of thumb:
27. ●● An automobile tire is filled to an absolute pressure of
T1in °C2 L 12 T1in °F2 3.0 atm at a temperature of 30 °C. Later it is driven to a
(a) Determine the exact fraction (it isn’t 12 ) and (b) the place where the temperature is only - 20 °C. What is the
percentage error astronomers make by using 12 at high absolute pressure of the tire at the cold place? (Assume
temperatures. that the air in the tire behaves as an ideal gas and the
14. IE ● ● ● Fig. 10.5 is a plot of Fahrenheit temperature volume is constant.)
versus Celsius temperature. (a) Is the value of the 28. ●● On a warm day (92 °F), an air-filled balloon occupies
y-intercept found by setting (1) TF = TC , (2) TC = 0, a volume of 0.200 m3 and has a pressure of 20.0 lb>in2. If
or (3) TF = 0? Why? (b) Compute the value of the the balloon is cooled to 32 °F in a refrigerator while its
y-intercept. (c) What would be the slope and y-intercept pressure is reduced to 14.7 lb>in2, what is the volume of
if the graph were plotted the opposite way (Celsius ver- the air in the container? (Assume that the air behaves as
sus Fahrenheit)? an ideal gas.)
29. ●● A steel-belted radial automobile tire is inflated to a
gauge pressure of 30.0 lb>in2 when the temperature is
10.3 GAS LAWS, ABSOLUTE 61 °F. Later in the day, the temperature rises to 100 °F.
TEMPERATURE, AND Assuming the volume of the tire remains constant, what is
THE KELVIN TEMPERATURE SCALE the tire’s pressure at the elevated temperature? [Hint:
15. ● Convert the following temperatures to absolute tem- Remember that the ideal gas law uses absolute pressure.]
peratures in kelvins: (a) 0 °C, (b) 100 °C, (c) 20 °C, and 30. ●● A scuba diver takes a tank of air on a deep dive. The
(d) - 35 °C. tank’s volume is 10 L and it is completely filled with air at
16. ● Convert the following temperatures to Celsius: (a) 0 K, an absolute pressure of 232 atm at the start of the dive.
(b) 250 K, (c) 273 K, and (d) 325 K. The air temperature at the surface is 94 °F and the diver
17. ● (a) Derive an equation for converting Fahrenheit tem-
ends up in deep water at 60 °F. Assuming thermal equi-
peratures directly to absolute temperatures in kelvins. librium and neglecting air loss, determine the absolute
(b) Which is the lower temperature, 300 °F or 300 K? internal pressure of the air when it is cold.
18. ● When lightning strikes, it can heat the air around it to
31. IE ● ● (a) If the temperature of an ideal gas increases and
more than 30 000 K, five times the surface temperature its volume decreases, will the pressure of the gas
of the Sun. (a) What is this temperature on the Fahren- (1) increase, (2) remain the same, or (3) decrease? Why?
heit and Celsius scales? (b) The temperature is some- (b) The Kelvin temperature of an ideal gas is doubled
times reported to be 30 000 °C. Assuming that 30 000 K and its volume is halved. How is the pressure affected?
is correct, what is the percentage error of this Celsius 32. ●● If 2.4 m3 of a gas initially at STP is compressed to
value? 1.6 m3 and its temperature is raised to 30 °C, what is its
19. ● How many moles are in (a) 40 g of water, (b) 245 g of final pressure?
CO2 (carbon dioxide), (c) 138 g of N2 (nitrogen), and 33. IE ● ● The pressure on a low-density gas in a cylinder is
(d) 56 g of O2 (oxygen) at STP? kept constant as its temperature is increased. (a) Does
20. IE ● (a) In a constant volume gas thermometer, if the the volume of the gas (1) increase, (2) decrease, or
pressure of the gas decreases, has the temperature of the (3) remain the same? Why? (b) If the temperature is
gas (1) increased, (2) decreased, or (3) remained the same? increased from 10 °C to 40 °C, what is the percentage
Why? (b) The initial absolute pressure of a gas is 1000 Pa change in the volume of the gas?
at room temperature (20 °C). If the pressure increases to 34. ●●● A diver releases an air bubble of volume 2.0 cm3
1500 Pa, what is the new Celsius temperature? from a depth of 15 m below the surface of a lake, where
21. ● If the pressure of an ideal gas is doubled while its the temperature is 7.0 °C. What is the volume of the bub-
absolute temperature is halved, what is the ratio of the ble when it reaches just below the surface of the lake,
final volume to the initial volume? where the temperature is 20 °C?
35. ●●● (a) Show that for the Kelvin temperature range 46. ●● When exposed to sunlight, a hole in a sheet of copper
5 expands in diameter by 0.153% compared to its diameter
T W 273 K, T L TC L 9 TF at 68 °F. What is the Celsius temperature of the copper
(b) For room temperature, what percentage error would sheet in the sun?
result from using this estimation to determine the Kelvin 47. ● ● One morning, an employee at a rental car company
temperature? (c) For a typical stellar interior tempera- fills a car’s steel gas tank to the top and then parks the
ture of 10 million °F, what is the percentage error in the car a short distance away. (a) That afternoon, when the
Kelvin temperature? (Carry as many significant figures temperature increases, will any gas overflow? Why?
as needed.) (b) If the temperatures in the morning and afternoon are,
respectively, 10 °C and 30 °C and the gas tank can hold
25 gal in the morning, how much gas will be lost?
10.4 THERMAL EXPANSION (Neglect the expansion of the tank.)
48. ● ● A copper block has an internal spherical cavity with a
36. ● A steel beam 10 m long is installed in a structure at
10-cm diameter (䉲 Fig. 10.23). The block is heated in an
20 °C. What is the beam’s change in length when the
oven from 20 °C to 500 K. (a) Does the cavity get larger or
temperature reaches (a) - 25 °C and (b) 45 °C?
smaller? (b) What is the change in the cavity’s volume?
37. IE ● An aluminum tape measure is accurate at 20 °C.
(a) If the tape measure is placed in a freezer, would it 䉳 FIGURE 10.23
read (1) high, (2) low, or (3) the same? Why? (b) If the A hole in a block See
temperature of the freezer is -5.0 °C, what would be the Exercise 48.
stick’s percentage error because of thermal contraction?
10 cm
38. ● Concrete highway slabs are poured in lengths of
5.00 m. How wide should the expansion gaps between
the slabs be at a temperature of 20 °C to ensure that there
will be no contact between adjacent slabs over a temper-
ature range of - 25 °C to 45 °C?
39. ● A man’s gold wedding ring has an inner diameter of 49. ● ● ● A brass rod has a circular cross-section of radius 5.00
2.4 cm at 20 °C. If the ring is dropped into boiling water, cm. The rod fits into a circular hole in a copper sheet with a
what will be the change in the inner diameter of the clearance of 0.010 mm completely around it when both the
ring? rod and the sheet are at 20 °C. (a) At what temperature will
40. ●● A circular steel plate of radius 15 cm is cooled from the clearance be zero? (b) Would such a tight fit be possible
350 °C to 20 °C. By what percentage does the plate’s area if the sheet were brass and the rod were copper?
decrease? 50. ● ● ● An aluminum rod is measured with a steel tape at
20 °C, and the length of the rod is found to be 75 cm.
41. ●● What temperature change would cause a 0.20%
What length will the tape indicate when both the rod
increase in the volume of a quantity of water that was
and the tape are at (a) -10 °C? (b) 50 °C? [Hint: Both the
initially at 20 °C?
rod and tape will either expand or shrink as temperature
42. ●● A piece of copper tubing used in plumbing has a changes. Keep as many significant figures as needed to
length of 60.0 cm and an inner diameter of 1.50 cm at express the answer.]
20 °C. When hot water at 85 °C flows through the tube,
51. ● ● ● Table 10.1 states that the (experimental) coefficient
what are (a) the tube’s new length and (b) the change in
of volume expansion b for air (and most other ideal
gases at 1 atm and 20 °C) is 3.5 * 10-3>°C. Use the defin-
its cross-sectional area? Does the latter affect the flow
speed?
ition of the volume expansion coefficient to show that
43. ●● A pie plate is filled up to the brim with pumpkin pie this value can, to a very good approximation, be pre-
filling. The pie plate is made of Pyrex and its expansion dicted from the ideal gas law and that the result holds
can be neglected. It is a cylinder with an inside depth of for all ideal gases, not just air.
2.10 cm and an inside diameter of 30.0 cm. It is prepared 52. ● ● ● A Pyrex beaker that has a capacity of 1000 cm3 at
at a room temperature of 68 °F and placed in an oven at 20 °C contains 990 cm3 of mercury at that temperature. Is
400 °F. When it taken out, 151 cc of the pie filling has there some temperature at which the mercury will com-
flowed out and over the rim. Determine the coefficient of pletely fill the beaker? Justify your answer. (Assume that
volume expansion of the pie filling, assuming it is a fluid. no mass is lost by vaporization and include the expan-
44. IE ● ● A circular piece is cut from an aluminum sheet at sion of the beaker.)
room temperature. (a) When the sheet is then placed in
an oven, will the hole (1) get larger, (2) get smaller, or
10.5 THE KINETIC THEORY OF GASES
(3) remain the same? Why? (b) If the diameter of the hole
is 8.00 cm at 20 °C and the temperature of the oven is 53. If the average kinetic energy per molecule of a
●
150 °C, what will be the new area of the hole? monatomic gas is 7.0 * 10-21 J, what is the Celsius tem-
45. IE ● ● In Fig. 10.20, the steel ring of diameter 2.5 cm is perature of the gas?
0.10 mm smaller in diameter than the steel ball at 20 °C. 54. ● What is the average kinetic energy per molecule in a
(a) For the ball to go through the ring, should you heat monatomic gas at (a) 10 °C and (b) 90 °C?
(1) the ring, (2) the ball, or (3) both? Why? (b) What is the 55. IE ● If the Celsius temperature of a monatomic gas is
minimum required temperature? doubled, (a) will the internal energy of the gas (1) double,
(2) increase by less than a factor of 2, (3) be half as much, or 65. IE ● ● ● During the race to develop the atomic bomb in
(4) decrease by less than a factor of 2? Why? (b) If the tem- World War II, it was necessary to separate a lighter iso-
perature is raised from 20 °C to 40 °C, what is the ratio of tope of uranium (U-235 was the fissionable one needed
the final internal energy to initial internal energy? for bomb material) from a heavier variety (U-238). The
56. ● What is the rms speed of the molecules in low-density uranium was converted into a gas, uranium hexafluo-
oxygen gas at 0 °C? (The mass of an oxygen molecule, ride (UF6), and the two uranium isotopes were sepa-
O2, is 5.31 * 10-26 kg). rated by gaseous diffusion using the difference in their
57. ● (a) What is the average kinetic energy per molecule of
rms speeds. As a two-component molecular mixture at
a monatomic gas at a temperature of 25 °C? (b) What is room temperature, which of the two types of molecules
the rms speed of the molecules if the gas is helium? would be moving faster, on average: (1) 235UF6 or
(A helium molecule consists of a single atom of mass (2) 238UF6. Or (3) would they move equally fast?
6.65 * 10-27 kg). Explain. (b) Determine the ratio of their rms speeds,
58. ● ● (a) Estimate the total amount of translational kinetic
light molecule to heavy molecule. Treat the molecules
energy in a small classroom at normal room temperature. as ideal gases and neglect rotations and/or vibrations
Assume the room measures 4.00 m by 10.0 m by 3.00 m. of the molecules. The masses of the three atoms in
(b) If this energy were all harnessed, how high would it be atomic mass units are 238 and 235 for the two uranium
able to lift an elephant with a mass of 1200 kg? isotopes and 19 for fluorine.
59. ● ● A quantity of an ideal gas is at 0 °C. An equal quan-
tity of another ideal gas is at twice the absolute tempera-

*10.6 KINETIC THEORY, DIATOMIC
ture. What is its Celsius temperature?
GASES, AND THE EQUIPARTITION
60. IE ● ● A sample of oxygen (O2) and another sample of
THEOREM
nitrogen (N2) are at the same temperature. (a) The rms
speed of the nitrogen sample is (1) greater than, (2) the 66. ●What is the total internal energy of 1.00 mol of 30 °C
same as, or (3) less than the rms speed of the oxygen He gas and O2 gas, respectively?
sample. Explain. (b) Calculate the ratio of the rms speed 67. ● If 1.0 mol of a monatomic gas has a total internal
in the nitrogen sample to in the oxygen sample. energy of 5.0 * 103 J at a certain temperature, what is the
61. ● ● If 2.0 mol of oxygen gas is confined in a 10-L bottle total internal energy of 1.0 mol of a diatomic gas at the
under a pressure of 6.0 atm, what is the average kinetic same temperature?
energy of an oxygen molecule? 68. ● ● For an average molecule of N2 gas at 10 °C, what are
62. ● ● If the temperature of an ideal gas increases from 300 K to its (a) translational kinetic energy, (b) rotational kinetic
600 K, what happens to the rms speed of the gas molecules? energy, and (c) total energy? Repeat for He gas at the
63. ● ● If the temperature of an ideal gas is raised from 25 °C same temperature.
to 100 °C, how much faster is the new rms speed of the 69. ● ● A diatomic gas has a certain total kinetic energy at
gas molecules? 25 °C. If a monatomic gas of the same number of mole-
64. ● ● If the rms speed of the molecules in an ideal gas at cules as the diatomic gas has the same total kinetic
20 °C increases by a factor of 2, what is the new Celsius energy, what is the Celsius temperature of the
temperature? monatomic gas?
70. IE (a) When cooled, the densities of most objects 73. 2.00 mol of a monatomic gas at atmospheric pressure has
(1) increase, (2) decrease, (3) stay the same. (b) By what a total internal energy of 7.48 * 103 J. What is the vol-
percentage does the density of a bowling ball change ume occupied a rigid cylinder by the gas?
(assuming it is a uniform sphere) when it is taken from 74. An ideal gas in a cylinder is at 20 °C and 2.0 atm. If it is
room temperature (68 °F) into the cold night air in heated so its rms speed increases by 20%, what is its new
Nome, Alaska 1- 40 °F2. Assume the ball is made out of pressure?
a material that has a linear coefficient of expansion a of
75.2 * 10-6>°C.
75. IE The escape speed from the Earth is about 11 000 m>s
(Section 7.5). Assume that for a given type of gas to even-
71. When a full copper kettle is tipped vertically at room tually escape the Earth’s atmosphere, its average molec-
temperature (68 °F), water initially pours out of its spout ular speed must be about 10% of the escape speed.
at 100 cm3>s (cubic centimeters per second). By what (a) Which gas would be more likely to escape the Earth:
percentage will this change if the kettle instead contains (1) oxygen, (2) nitrogen, or (3) helium? (b) Assuming a
boiling water at 212 °F? Assume that the only significant temperature of -40 °F in the upper atmosphere, deter-
change is due to the change in size of the spout. mine the rms speed of a molecule of oxygen. Is it enough
72. An ideal gas sample occupies a container of volume to escape the Earth? (Data: The mass of an oxygen mole-
0.75 L at STP. Find (a) the number of moles and (b) the cule is 5.34 * 10-26 kg, that of a nitrogen molecule is
number of moleculesin in the sample. (c) If the gas is car- 4.68 * 10-26 kg, and that of a helium molecule is
bon monoxide (CO), what is the sample’s mass? 6.68 * 10-27 kg.
11 Heat
11.1 Definition and units
of heat (387)
■ units of heat
■ mechanical equivalent of heat
11.2 Specific heat and

calorimetry (389)
■ specific heat
■ conservation of energy
11.3 Phase changes and latent

heat (393)
■ latent heat of fusion
■ latent heat of vaporization
PHYSICS FACTS
11.4
■
Heat transfer (400)
conduction, convection,
radiation
✦ With a skin temperature of 34 °C
(93.2 °F), a person sitting in a room
at 23 °C (73.4 °F) will lose about 100 J
of heat per second, which is the
H eat is crucial to our existence.
Our bodies must balance
heat loss and gain to stay within
power output approximately equal
to that of a 100-W lightbulb. This is
the narrow temperature range nec-
why a closed room full of people essary for life. This thermal balance
tends to get very warm.
✦ A couple of inches of fiberglass in
is delicate and any disturbance can
the attic can cut heat loss by as have serious consequences. Sick-
much as 90% (see Example 11.7).
✦ If the Earth did not have an atmos-
ness can disrupt the balance, and
phere (hence no greenhouse as a result, our bodies produce a
effect), its average surface tempera-
ture would be 30 °C (86 °F) lower chill or fever.
than it is now. That would freeze liq-
uid water and basically eliminate life
To maintain our health, we exer-
as we know it. cise by doing mechanical work
✦ Most metals are excellent thermal con-
ductors. However, stainless steel is a
such as lifting weights and riding
relatively poor conductor; it conducts bicycles. Our bodies convert food
only about 5% as much as copper.
✦ During a race on a hot day, a profes-
energy (chemical potential) to
sional cyclist can lose as much as 7 L mechanical work; however, this
of water in 3 h to evaporation in
getting rid of the heat generated by process is not perfect. That is, the
this vigorous activity.
body cannot convert all the food
11.1 DEFINITION AND UNITS OF HEAT 387
energy into mechanical work—in fact, it converts less than 20% depending on
which muscle groups are doing the work. The rest becomes heat transferred to the
environment. The leg muscles are the largest and most efficient in performing
mechanical work; for example, cycling and running are relatively efficient
processes. The arm and shoulder muscles are less efficient; hence, snow shoveling
is a low-efficiency exercise. The body must have special cooling mechanisms to
get rid of excess thermal energy generated during intense exercise. The most effi-
cient mechanism is through perspiring, or the evaporation of water. The Olympic
marathon champion Gezahgne Abera tries to promote cooling and evaporation
by pouring water over his head, as shown in the chapter-opening photograph.
On a larger scale, heat exchanges are important to our planet’s ecosystem. The
average temperature of the Earth, so critical to our environment and to the survival
of the organisms that inhabit it, is maintained through a heat exchange balance.
Each day, a vast quantity of solar energy reaches our planet’s atmosphere and sur-
face. Scientists are concerned that a buildup of atmospheric “greenhouse” gases, a
product of our industrial society, could significantly raise the Earth’s average temper-
ature. This change would undoubtedly have a negative effect on life on the Earth.
On a more practical level, most of us know to be very careful while handling
anything that has recently been in contact with a flame or other source of heat. Yet
while the copper bottom of a steel pot on the stove can be very hot, the steel pot
handle is only warm to the touch. Sometimes direct contact isn’t necessary for
heat to be transmitted, but how was heat transferred? And why was the steel han-
dle not nearly as hot as the pot? The answer has to do with thermal conduction, as
you will learn.
In this chapter, what heat is and how it is measured will be discussed. Also
studied are the various mechanisms by which heat is transferred from one object
to another. This knowledge will allow you to explain many everyday phenomena,
as well as provide a basis for understanding the conversion of thermal energy into
useful mechanical work.
11.1 Definition and Units of Heat

➥ What is heat?
➥ What are four common units of heat?
➥ What is the mechanical equivalent of heat?
Like work, heat is related to a transfer of energy. In the 1800s, it was thought that
heat described the amount of energy an object possessed, but this is not true.
Rather, heat is the name used to describe a type of energy transfer. “Heat,” or “heat
energy,” is the energy added to, or removed from, the total internal energy of an
object due to temperature differences.
Heat then is energy in transit, and is measured in the standard SI unit, the joule
(J). However, other nonstandard, commonly used units of heat are also defined.
An important one is the kilocalorie (kcal) (䉲 Fig. 11.1a):
One kilocalorie (kcal) is defined as the amount of heat needed to raise the tempera-
ture of 1 kg of water by 1 °C (from 14.5 °C to 15.5 °C).
388 11 HEAT
䉴 F I G U R E 1 1 . 1 Units of heat ΔT = 1 °C
(a) A kilocalorie raises the tempera- ΔT = 1 °C ΔT = 1 °F
ture of 1 kg of water by 1 °C.
(b) A calorie raises the temperature
of 1 g of water by 1 °C. (c) A Btu
raises the temperature of 1 lb of 1 kg
water by 1 °F. (Not drawn to scale.) water 1 lb
1g water
water
(a) 1 kilocalorie (kcal) (b) 1 calorie (cal) (c) 1 British thermal unit (Btu)
or Calorie (Cal)
(This kilocalorie is technically known as the “15° kilocalorie.”) The temperature

range is specified because the energy needed varies slightly with temperature—a
variation so small that it can be ignored for our purposes.
For smaller quantities, the calorie (cal) is sometimes used 11 kcal = 1000 cal2.
One calorie is the amount of heat needed to raise the temperature of 1 g of water
by 1 °C (from 14.5 °C to 15.5 °C) (Fig. 11.1b).
A familiar use of the larger unit, the kilocalorie, is for specifying the energy val-
ues of foods. In this context, the word is usually shortened to Calorie (Cal). That is,
people on diets really count kilocalories. This quantity refers to the food energy
that is available for conversion to heat to be used for mechanical movement, to
maintain body temperature, or to increase body mass. The capital C distinguishes
the larger kilocalorie, from the smaller calorie. They are sometimes referred to as
䉱 F I G U R E 1 1 . 2 It’s a joule! In “big Calorie” and “little calorie.” (In some countries, the joule is used for food
Australia, diet drinks are labeled as
being “low joule.” In Germany, the values—see 䉳 Fig. 11.2.)
labeling is a bit more specific: “Less A unit of heat sometimes used in industry is the British thermal unit (Btu). One
than 4 kilojoules (1 kcal) in Btu is the amount of heat needed to raise the temperature of 1 lb of water by 1 °F
0.3 Liter.” How does this labeling (from 63 °F to 64 °F; Fig. 11.1c), and 1 Btu = 252 cal = 0.252 kcal. If you buy an air
compare to that for diet drinks in conditioner or an electric heater, you will find that it is rated in Btu, which is really
the United States?
Btu per hour—in other words, a power rating. For example, window air condi-
tioners range from 4000 to 25 000 Btu>h. This specifies the rate at which the appli-
ance can transfer heat.
Thermometer
THE MECHANICAL EQUIVALENT OF HEAT
The idea that heat is actually a transfer of energy is the result of work by many sci-
entists. Some early observations were made by the American Benjamin Thompson
(Count Rumford), 1753–1814, while he was supervising the boring of cannon bar-
Weight Weight rels in Germany. Rumford noticed that water put into the bore of the cannon (to
Insulation
prevent overheating during drilling) boiled away and had to be replenished. The
theory of heat at that time pictured it as a “caloric fluid,” which flowed from hot
Paddle objects to colder ones. Rumford did several experiments to detect “caloric fluid”
wheel
by measuring changes in the weights of heated substances. Since no weight
Water change was detected, he concluded that the mechanical work done by friction was
actually responsible for the heating of the water.
This conclusion was later proven quantitatively by the English scientist James
䉱 F I G U R E 1 1 . 3 Joule’s apparatus Joule (after whom the unit of energy is named; see Section 5.6). Using the appara-
for determining the mechanical
tus illustrated in 䉳 Fig. 11.3, Joule demonstrated that when a given amount of
equivalent of heat As the weights
descend, the paddle wheels churn mechanical work was done, the water was heated, as indicated by an increase in
the water, and the mechanical its temperature. He found that for every 4186 J of work done, the temperature of
energy, or work, is converted into the water rose 1 °C per kg, or that 4186 J was equivalent to 1 kcal:
heat energy, raising the temperature
of the water. For every 4186 J of 1 kcal = 4186 J = 4.186 kJ or 1 cal = 4.186 J
work done, the temperature of the
water rises 1 °C per kilogram. Thus, This relationship is called the mechanical equivalent of heat. Example 11.1 illus-
4186 J is equivalent to 1 kcal. trates an everyday use of these conversion factors.
11.2 SPECIFIC HEAT AND CALORIMETRY 389
EXAMPLE 11.1 Working Off That Birthday Cake: Mechanical Equivalent of Heat to the Rescue
At a birthday party, a student eats a piece of cake (food T H I N K I N G I T T H R O U G H . Power is the rate at which the stu-
energy value of 200 Cal). To prevent this energy from being dent does work, and the watt (W) is its SI unit (1 W = 1 J>s;
stored as fat, she takes a stationary bicycle workout class right Section 5.6). To find the time it will take to do this work, the
after the party. This exercise requires the body to do work at food energy content is expressed in joules and the definition
an average rate of 200 watts. How long must the student bicy- of average power, P = W>t 1work>time2 is used.
cle to achieve her goal of “working off” the cake’s energy?
SOLUTION. The work required to “burn up” the energy content of the cake is at least 200 Cal. Listing the data given and con-
verting to SI units (remember that Cal means kcal):
Given: W = 1200 kcal2a b = 8.37 * 105 J

4186 J
Find: t (time to “burn up” 200 Cal)
kcal
P = 200 W = 200 J>s
Rearranging the equation for average power,
W 8.37 * 105 J
t = = = 4.19 * 103 s = 69.8 min = 1.16 h
P 200 J>s
F O L L O W - U P E X E R C I S E . If the 200 Cal in this Example were used to increase the student’s gravitational potential energy, how
high would she rise? (Assume her mass is 60 kg.) (Answers to all Follow-Up Exercises are given in Appendix VI at the back of the book.)
DID YOU LEARN?

➥ Heat is a form of energy. It is the energy added to or removed from an object due to
temperature differences.
➥ Four common units of heat are the joule, kilocalorie (Cal), calorie, and Btu (British
thermal unit).
➥ The mechanical equivalent of heat relates mechanical energy (work) and heat
energy.The relationship is 1 cal = 4.186 J or 1 kcal = 4186 J.
11.2 Specific Heat and Calorimetry

➥ How is specific heat of a substance defined?

➥ If two different substances (with different specific heats) have equal mass and
receive equal amounts of heat, which substance will experience a larger tempera-
ture change?
➥ What is the fundamental physical principle that supports the calorimetry technique?
SPECIFIC HEATS OF SOLIDS AND LIQUIDS

Recall from Chapter 10 that when heat is added to a solid or liquid, the energy may
go toward increasing the average molecular kinetic energy (temperature change) and
also toward increasing the potential energy associated with the molecular bonds
(phase change). Different substances have different molecular configurations and
bonding patterns. Thus, if equal amounts of heat are added to equal masses of differ-
ent substances, the resulting temperature changes will not generally be the same.
Specific heat capacity, or simply specific heat (c) is defined as the heat (transfer)
required to raise (or lower) the temperature of 1 kg of a substance by 1 °C. The SI
units of specific heat are J>1kg # K2 or J>1kg # °C2, because 1 K = 1 °C. Specific heat
is a characteristic of the substance type. The specific heats of some common sub-
stances are given in 䉲 Table 11.1. Specific heats vary slightly with temperature, but
they can be considered constant for our purpose.
The amount of heat (Q) required to change the temperature of a substance of
mass m by a temperature difference of ¢T1Tf - Ti2 is then
Q
Q = cm¢T or c = (specific heat) (11.1)
m¢T
390 11 HEAT
TABLE 11.1 Specific Heats of Various Substances (Solids and Liquids)

at 20 °C and 1 atm
Specific Heat (c)
Substance J>(kg # °C) kcal>(kg # °C) or cal>(g # °C)
Solids
Aluminum 920 0.220
Copper 390 0.0932
Glass 840 0.201
Ice 1-10 °C2 2100 0.500
Iron or steel 460 0.110
Lead 130 0.0311
Soil (average) 1050 0.251
Wood (average) 1680 0.401
Human body (average) 3500 0.84
Liquids
Ethyl alcohol 2450 0.585
Glycerin 2410 0.576
Mercury 139 0.0332
Water (15 °C) 4186 1.000
Gas
Steam (Water vapor, 100 °C) 2000 0.48
The larger the specific heat of a substance, the more heat must be transferred to
or taken from it (per kilogram of mass) to change its temperature by a given
amount. That is, a substance with a higher specific heat requires more heat for a
given temperature change and mass than one with a lower specific heat. Table 11.1
shows that metals have specific heats considerably lower than that of water. Thus
it takes only a small amount of heat to produce a relatively large temperature
increase in a metal object, compared to the same mass of water.
Compared to most common materials, water has a very large specific heat of
4186 J>1kg # °C2, or 1.00 kcal>1kg # °C2. You have been the victim of the high spe-
cific heat of water if you have ever burned your mouth on a baked potato or the
hot cheese on a pizza. These foods have high water content, and due to water’s
high specific heat, they don’t cool off as quickly as some other drier foods do. The
large specific heat of water is also responsible for the mild climate of places near
large bodies of water. (See Section 11.4 for more details.)
Note from Eq. 11.1 that when there is a temperature increase, ¢T is positive
1Tf 7 Ti2, then Q is positive. This condition corresponds to energy being added to a
system or object. Conversely, ¢T and Q are negative when energy is removed from a
system or object. This sign convention will be used throughout this book.
EXAMPLE 11.2 Birthday Cake Revisited: Specific Heat for a Warm Bath?
At the birthday party in Example 11.1, a student ate a piece of T H I N K I N G I T T H R O U G H . The heat energy from the cake is
cake (200 Cal). To get an idea of the magnitude of this amount used to heat water from 20 °C to 45 °C. Using the mechanical
of energy there is in that piece of cake, the student would like equivalent of heat and Eq. 11.1, the mass of water can be
to know how much water at 20 °C can be brought to 45 °C found.
(enough for a bath?). Can you help her out?
11.2 SPECIFIC HEAT AND CALORIMETRY 391
SOLUTION. Listing the data given and converting to SI units (remember that Cal means kcal):
Q = 1200 kcal2a b = 8.37 * 105 J

4186 J
Given: Find: m (mass of water)
kcal
Ti = 20 °C
Tf = 45 °C
c = 4186 J>1kg # °C2 (from Table 11.1)
From Eq. 11.1, Q = cm¢T = cm1Tf - Ti2. Solving for m gives
Q 8.37 * 105 J
34186 J>1kg # °C24145 °C - 20 °C2
m = = = 8.00 kg
c¢T
This mass of water occupies more than 2 gallons—quite a bit, but it may not be enough for a bath. It does show that “a piece of
cake” contains a fair amount of energy.
FOLLOW-UP EXERCISE. In this Example, how would the answer change if the water was initially at a temperature of 5 °C rather
than 20 °C?
INTEGRATED EXAMPLE 11.3 Cooking Class 101: Studying Specific Heats While Learning
How to Boil Water
To prepare pasta, you bring a pot of water from room temper- (A) CONCEPTUAL REASONING. The temperature increase is the
ature (20 °C) to its boiling point 1100 °C2. The pot itself has a same for the water and the pot. Thus, the required heat is
mass of 0.900 kg, is made of steel, and holds 3.00 kg of water. affected by the product of mass and specific heat. There is
(a) Which of the following is true: (1) the pot requires more 3.00 kg of water to heat. This is more than three times the
heat than the water, (2) the water requires more heat than the mass of the pot. From Table 11.1, the specific heat of water is
pot, or (3) they require the same amount of heat? (b) Deter- about nine times larger than that of steel. Both factors
mine the required heat for both the water and the pot, and the together indicate that the water will require significantly
ratio Qw >Qpot. more heat than the pot, so the answer is (2).
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . The heat needed can be found using Eq. 11.1, after looking up the specific heats in
Table 11.1. The temperature change is easily determined from the initial and final values.
Listing the data given:
Given: mpot = 0.900 kg Find: Qw , Qpot and Qw>Qpot (the required heat for
mw = 3.00 kg the water and the pot, and the heat ratio)
cpot = 460 J>kg # °C (from Table 11.1)
cw = 4186 J>kg # °C (from Table 11.1)
¢T = Tf - Ti = 100 °C - 20 °C = 80 °C
In general, the amount of heat is given by Q = cm¢T. The temperature increase 1¢T2 for both objects is 80 °C. Thus, the heat for
the water is
Qw = cw mw ¢Tw
= 34186 J>1kg # °C2413.00 kg2180 °C2 = 1.00 * 106 J
and the heat required for the pot is
Qpot = cpot mpot ¢Tpot
= 3460 J>1kg # °C2410.900 kg2180 °C2 = 3.31 * 104 J
Therefore,
Qw 1.00 * 106 J
= = 30.2
Qpot 3.31 * 104 J
Hence the water requires more than thirty times the heat required for the pot, because it has more mass and a greater specific heat.
F O L L O W - U P E X E R C I S E . (a) In this Example, if the pot were the same mass but instead made out of aluminum, would the heat
ratio (water to pot) be smaller or larger than the answer for the steel pot? Explain. (b) Verify your answer.
392 11 HEAT
CALORIMETRY
Calorimetry is a technique that quantitatively measures heat exchanges. Such
measurements are made by using an instrument called a calorimeter (cal-oh-RIM-i-
ter), usually an insulated container that allows little heat exchange with the envi-
ronment (ideally none). A simple laboratory calorimeter is shown in 䉳 Fig. 11.4.
The specific heat of a substance can be determined by measuring the masses
and temperature changes of the objects involved and using Eq. 11.1.* Usually the
unknown is the unknown specific heat, c. Typically, a substance of known mass
and temperature is put into a quantity of water in a calorimeter. The water is at a
different temperature from that of the substance, usually a lower one. The princi-
ple of the conservation of energy is then applied to determine the substance’s spe-
䉱 F I G U R E 1 1 . 4 Calorimetry
cific heat, c. This procedure is called the method of mixtures. Example 11.4 illustrates
apparatus The calorimetry cup
the use of this procedure. Such heat exchanges are simply the applications of the
conservation of energy. The total of all the heat losses 1Q 6 02 must have the same
(center, with black insulating ring)
goes into the larger container. The
cover with the thermometer and absolute value as all the heat gains 1Q 7 02. This means the algebraic sum of all
stirrer is seen at the right. Metal shot the heat transfers must equal zero, or g Qi = 0, assuming negligible heat exchange
or pieces of metal are heated in the
with the environment.
small cup (with the handle) in the
steam generator on the tripod.
EXAMPLE 11.4 Calorimetry Using the Method of Mixtures

Students in a physics lab are to determine the specific heat of and negative signs. In calorimetry problems, it is important to
copper experimentally. They place 0.150 kg of copper shot identify and label all of the quantities with proper signs. Iden-
into boiling water and let it stay for a while, so as to reach a tification of the heat gains and losses is crucial. You will prob-
temperature of 100 °C. Then they carefully pour the hot shot ably use this method in the laboratory.
into a calorimeter cup (Fig. 11.4) containing 0.200 kg of water
SOLUTION. The subscripts Cu, w, and Al will be used to
at 20.0 °C. The final temperature of the mixture in the cup is
refer to the copper, water, and aluminum calorimeter cup,
measured to be 25.0 °C. If the aluminum cup has a mass of
respectively. The subscripts h, i, and f will refer to the temper-
0.0450 kg, what is the specific heat of copper? (Assume that
ature of the hot metal shot, the water (and cup) initially at
there is no heat exchange with the surroundings.)
room temperature, and the final temperature of the system,
T H I N K I N G I T T H R O U G H . The conservation of heat energy is respectively. With this notation,
involved: g Qi = 0, taking into account the correct positive
Given: mCu = 0.150 kg Find: cCu (specific heat)

mw = 0.200 kg
cw = 4186 J>1kg # °C2 (from Table 11.1)
mAl = 0.0450 kg
cAl = 920 J>1kg # °C2 (from Table 11.1)
Th = 100 °C (initial temperature of Cu shot),
Ti = 20.0 °C, and
Tf = 25.0 °C
If there is no heat exchange with the surroundings, the system’s total energy is conserved, g Qi = 0, and
g Qi = Qw + QAl + QCu = 0
Substituting the relationship in Eq. 11.1 for these heats,
cw mw ¢Tw + cAl mAl ¢TAl + cCu mCu ¢TCu = 0
or
cw mw1Tf - Ti2 + cAl mAl1Tf - Ti2 + cCu mCu1Tf - Th2 = 0
Here, the water and aluminum cup, initially at Ti , are heated to Tf , so ¢Tw = ¢TAl = 1Tf - Ti2. The copper initially at Th is
cooled to Tf , so ¢TCu = 1Tf - Th2 and this is a negative quantity, indicating a temperature drop for the copper. Solving for cCu ,
1cw mw + cAl mAl21Tf - Ti2
mCu1Tf - Th2
cCu = -
534186 J>1kg # °C2410.200 kg2 + 3920 J>1kg # °C2410.0450 kg26125.0 °C - 20.0 °C2
10.150 kg2125.0 °C - 100 °C2
= -
#
= 390 J>1kg °C2
*In this section, calorimetry will not involve phase changes, such as ice melting or water boiling.
These effects are discussed in Section 11.3.
11.3 PHASE CHANGES AND LATENT HEAT 393
Notice that the proper use of signs resulted in a positive answer for cCu , as required. If, for example, the QCu term had not had
the correct sign, the answer would have been negative—a big clue that you had an initial sign error.
F O L L O W - U P E X E R C I S E . In this example, what would the final equilibrium temperature be if the calorimeter (water and cup)
initially had been at a warmer 30 °C?
SPECIFIC HEAT OF GASES

When heat is added to or removed from most materials, they expand or contract.
During expansion, for example, the materials would then do work on the environ-
ment. For most solids and liquids, this work is negligible, because the volume
changes are very small (Section 10.4). This is why this effect wasn’t included in our
discussion of specific heat of solids and liquids.
However, for gases, expansion and contraction can be significant. It is therefore
important to specify the conditions under which heat is transferred when referring
to a gas. If heat is added to a gas at constant volume (a rigid container), the gas
does no work. (Why?) In this case, all of the heat goes into increasing the gas’s
internal energy and, therefore, to increasing its temperature. However, if the same
amount of heat is added at constant pressure (a nonrigid container allowing a vol-
ume change), a portion of the heat is converted to work as the gas expands. Thus,
not all of the heat will go into the gas’s internal energy. This process results in a
smaller temperature change than occurred during the constant volume process.
To designate the physical quantities that are held constant while heat is added
to or removed from a gas, a subscript notation will be used: cp means specific heat
under conditions of constant pressure (p), and cv means specific heat under condi-
tions of constant volume (v). The specific heat for water vapor (H2O) given in
Table 11.1 is the specific heat under constant pressure (cp).
An important result is that for a particular gas, cp is always greater than cv. This is
true because for a specific mass of gas, c r Q>¢T. Since for a given Q, ¢Tv is as large
as it can be, cv will be less than cp. In other words, ¢Tv 7 ¢Tp. Specific heats of gases
play an important role in adiabatic thermodynamic processes. (See Section 12.3.)
(a) Solid
DID YOU LEARN?

➥ The specific heat of a substance is defined as the amount of heat required to
change the temperature of 1 kg substance by 1 °C or 1 K.
➥ For equal amounts of heat and mass, a substance with a smaller specific heat will
experience a larger temperature change.
➥ The fundamental physical principle behind calorimetry is the conservation of
energy.The total heat lost by components in an isolated system must equal the (b) Liquid
heat gained by other commponents within the system so ©Qi = 0 or there is no
net heat exchange within the system.
11.3 Phase Changes and Latent Heat

(c) Gas
➥ What are the three common phases of matter? 䉱 F I G U R E 1 1 . 5 Three phases of

➥ When a substance undergoes a phase change, what happens to its matter (a) The molecules of a solid
temperature? are held together by bonds; conse-
quently, a solid has a definite shape
➥ How is the latent heat of fusion (vaporization) of a substance defined? and volume. (b) The molecules of a
liquid can move more freely, so a
Matter normally exists in one of three traditional phases: solid, liquid, or gas liquid has a definite volume and
(䉴 Fig. 11.5). However, this division into three common phases is only approximate assumes the shape of its container.
since there are other phases, such as a plasma phase and a superconducting phase. (c) The molecules of a gas interact
The phase that a substance is in depends on the substance’s internal energy (as weakly and are separated by rela-
tively large distances; thus, a gas
indicated by its temperature) and the pressure on it. However, it is likely that you has no definite shape or volume,
think of adding or removing heat as the way to change the phase of a substance. unless it is confined in a container.
394 11 HEAT
In the solid phase, molecules are held together by attractive forces, or bonds
(Fig. 11.5a). Adding heat causes increased motion about the molecular equilib-
rium positions. If enough heat is added to provide sufficient energy to break the
intermolecular bonds, most solids undergo a phase change and become liquids.
The temperature at which this phase change occurs is called the melting point.
The temperature at which a liquid becomes a solid is called the freezing point.
In general, these temperatures are the same for a given substance, but they can
differ slightly.
In the liquid phase, molecules of a substance are relatively free to move and a
liquid assumes the shape of its container (Fig. 11.5b). In certain liquids, there
may be some locally ordered structure, giving rise to liquid crystals, such as
those used in LCDs (liquid crystal displays) of calculators and computer dis-
plays (Section 24.4).
Adding even more heat increases the motion of the molecules of a liquid. When
they have enough energy to become separated, the liquid changes to the gaseous
(vapor) phase. This change may occur slowly, by evaporation, or rapidly, at a par-
ticular temperature called the boiling point. The temperature at which a gas con-
denses into a liquid is the condensation point.
Some solids, such as dry ice (solid carbon dioxide), mothballs, and certain air
fresheners, change directly from the solid to the gaseous phase at standard pres-
sure. This process is called sublimation. Like the rate of evaporation, the rate of
sublimation increases with the temperature of the surrounding medium. A phase
change from a gas to a solid is called deposition. Frost, for example, is solidified
water vapor (gas) deposited on grass, car windows, and other objects. Frost is not
frozen dew (liquid water), as is sometimes mistakenly assumed.
LATENT HEAT
In general, when heat is transferred to a substance, its temperature increases as
the average kinetic energy per molecule increases. However, when heat is added
(or removed) during a phase change, the temperature of the substance does not
change. For example, if heat is added to a quantity of ice at -10 °C, the tempera-
ture of the ice increases until it reaches its melting point of 0 °C. At this point, the
addition of more heat does not increase the ice’s temperature, but causes it to
melt, or change phase. (The heat must be added slowly so that the ice and
melted water remain in thermal equilibrium, otherwise, the ice water can warm
above 0 °C even though the ice remains at 0 °C.) Only after the ice is completely
melted does adding more heat cause the temperature of the water to rise.
A similar situation occurs during the liquid–gas phase change at the boiling
point. Adding more heat to boiling water only causes more vaporization. A tem-
perature increase occurs only after the water is completely boiled, resulting in
superheated steam. Keep in mind that ice can be colder than 0 °C and steam can be
hotter than 100 °C.
During a phase change, the heat goes into breaking the attractive bonds and
separating molecules rather than into increasing the temperature (increasing the
potential, rather than kinetic, energies). The heat required for a phase change is
called the latent heat (L), which is defined as the magnitude of the heat needed
per unit mass to induce a phase change:
ƒQƒ
L = (latent heat) (11.2)
m
where m is the mass of the substance. Latent heat has the SI unit of joule per kilo-
gram 1J>kg2, or kilocalorie per kilogram kcal>kg.
The latent heat for a solid–liquid phase change is called the latent heat of
fusion (Lf), and that for a liquid–gas phase change is called the latent heat of
vaporization (Lv.) These quantities are often referred to as simply the heat of fusion
TABLE 11.2 Temperatures of Phase Changes and Latent Heats for Various Substances (at 1 atm)
Lf Lv
Substance Melting Point J>kg kcal>kg Boiling Point J>kg kcal>kg
Alcohol, ethyl - 114 °C 1.0 * 105 25 78 °C 8.5 * 105 204

5 5
Gold 1063 °C 0.645 * 10 15.4 2660 °C 15.8 * 10 377
5
Helium* — — — -269 °C 0.21 * 10 5
Lead 328 °C 0.25 * 105 5.9 1744 °C 8.67 * 105 207
Mercury - 39 °C 0.12 * 105 2.8 357 °C 2.7 * 105 65
5 5
Nitrogen - 210 °C 0.26 * 10 6.1 -196 °C 2.0 * 10 48
5 5
Oxygen - 219 °C 0.14 * 10 3.3 -183 °C 2.1 * 10 51
Tungsten 3410 °C 1.8 * 105 44 5900 °C 48.2 * 105 1150
Water 0 °C 3.33 * 105 80 100 °C 22.6 * 105 540
*Not a solid at a pressure of 1 atm; melting point is -272 °C at 26 atm.
and the heat of vaporization. The latent heats of some substances, along with their
melting and boiling points, are given in 䉱 Table 11.2. (The latent heat for the less
common solid–gas phase change is called the latent heat of sublimation and is sym-
bolized by Ls.) As you might expect, the latent heat (in joules per kilogram) is the
amount of energy per kilogram given up when the phase change is in the opposite
direction, that is, from liquid to solid or gas to liquid.
A more useful form of Eq. 11.3 is given by solving for Q and including a
positive>negative sign for the two possible directions of heat flow:
Q = mL (signs with latent heat) (11.3)
This equation is more practical for problem solving because in calorimetry prob-
lems, you are typically interested in applying conservation of energy in the form
of g Qi = 0. The positive>negative sign 12 must be explicitly expressed because
heat can flow either into 1+2 or out of 1- 2 the object or system of interest.
When solving calorimetry problems involving phase changes, you must be
careful to use the correct sign for those terms, in agreement with our sign conven-
tions (䉲 Fig. 11.6). For example, if water is condensing from steam into liquid
droplets, removal of heat is involved, necessitating the choice of the negative sign.
Latent heat of fusion 䉳 F I G U R E 1 1 . 6 Phase changes

Qf > 0 and latent heats (a) At 0 °C,
3.33 * 105 J must be added to 1 kg
3.33 105 J/kg of ice or removed from 1 kg of liq-
(80 kcal/kg) uid water to change its phase. (b) At
100 °C, 22.6 * 105 J must be added
Qf < 0
to 1 kg of liquid water or removed
Ice, 0 °C Water, 0 °C
from 1 kg of steam to change its
(a)
phase.
Latent heat of vaporization
Qv > 0
22.6 105 J/kg
(540 kcal/kg)
Qv < 0
Water, 100 °C Steam, 100 °C
(b)
396 11 HEAT
Recall in Section 11.2 that there were no phase changes, the expression for heat
Q = cm¢T automatically gave the correct sign for Q from the sign of ¢T. But there is no
¢T during a phase change. Choosing the correct sign is up to you.
For water, the latent heats of fusion and vaporization are
Lf = 3.33 * 105 J>kg

Lv = 22.6 * 105 J>kg
The accompanying Learn by Drawing 11.1, From Cold Ice to Hot Steam is
numerically expressed in Example 11.5 and shows explicitly the two types of
heat terms (specific heat and latent heat) that must be employed in the general
situation when any of the objects undergo a temperature change and a phase
change.
LEARN BY DRAWING 11.1 at a temperature of about -10 °C.) At the phase change
(0 °C and 100 °C) heat is added without a temperature
change. Once each phase change is complete, adding more
from cold ice to hot steam heat causes the temperature to increase. The slopes of the
It can be helpful to focus on the fusion and vaporization of lines in the drawings are not all the same, which indicates
water graphically. To heat a piece of cold ice at -10 °C all that the specific heats of the various phases are not the
the way to hot steam at 110 °C, five separate specific heat same. (Why do different slopes mean different specific
and latent heat calculations are necessary. (Most freezers are heats?) The numbers come from Example 11.5.
(1) (2) (3)
100 100 100

Warming ice Melting ice Warming water
Temperature (ºC)
Temperature (ºC)
Temperature (ºC)
50 50 50
water
Ice + water at 0 ºC Ice + water at 0 ºC
0 0 0
Ice Ice Ice

0 500 0 500 0 500 1000
Added heat (kJ) Added heat (kJ) Added heat (kJ)
(4) (5)
100 100
Heating steam
Temperature (ºC)
Temperature (ºC)
Steam
Water + Steam
50 50
Vaporizing water
Qtotal = Q1 + Q2 + Q3 + Q4 + Q5
Ice + water at 0 ºC Ice + water at 0 ºC
0 0
Ice Ice
0 500 1000 3000 0 500 1000 3000
Added heat (kJ) Added heat (kJ)
EXAMPLE 11.5 From Cold Ice to Hot Steam

Heat is added to 1.00 kg of cold ice at - 10 °C. How much heat to its boiling point (specific heat of water), (4) vaporizing
is required to change the cold ice to hot steam at 110 °C? water to steam (water vapor) at 100 °C (latent heat, a phase
change), and (5) heating steam (specific heat of steam). The
T H I N K I N G I T T H R O U G H . Five steps are involved: (1) heating key idea here is that the temperature does not change during
ice to its melting point (specific heat of ice), (2) melting ice to a phase change. [Refer to Learn by Drawing 11.1, From Cold
water at 0 °C (latent heat, a phase change), (3) heating water Ice to Hot Steam.]
SOLUTION.
Given: m = 1.00 kg Find: Qtotal (total heat required)
Ti = - 10 °C
Tf = 110 °C
Lf = 3.33 * 105 J>kg (from Table 11.2)
Lv = 22.6 * 105 J>kg (from Table 11.2)
cice = 2100 J>1kg # °C2 (from Table 11.1)
cwater = 4186 J>1kg # °C2 (from Table 11.1)
csteam = 2000 J>1kg # °C2 (from Table 11.1)
1. Q1 = cice m¢T1 = 32100 J>1kg # °C2411.00 kg230 °C - 1 -10 °C24 1heating ice2
= + 2.10 * 104 J
2. Q2 = + mLv = 11.00 kg213.33 * 105 J>kg2 = + 3.33 * 105 J 1melting ice2
3. Q3 = cwater m¢T2 = 34186 J>1kg # °C2411.00 kg21100 °C - 0 °C2 1heating water2
= + 4.19 * 105 J
4. Q4 = + mLv = 11.00 kg2122.6 * 105 J>kg2 = + 2.26 * 106 J 1vaporizing water2
5. Q5 = csteam m¢T3 = 32000 J>1kg # °C2411.00 kg21110 °C - 100 °C2 1heating steam2
= + 2.00 * 104 J
The total heat required is
Qtotal = g Qi = 2.10 * 104 J + 3.33 * 105 J + 4.19 * 105 J + 2.26 * 106 J + 2.00 * 104 J
= 3.05 * 106 J
The latent heat of vaporization is, by far, the largest. It is actually greater than the sum of the other four terms.
FOLLOW-UP EXERCISE. How much heat must a freezer remove from liquid water (initially at 20 °C) to create 0.250 kg of ice at
- 10 °C?
Note that the latent heat must be computed at each phase change. It is a common error
to use the specific heat equation with a temperature interval that includes a phase change.
Also, a complete phase change cannot be assumed until you have checked for it numeri-
cally. (See Example 11.6.)
Technically, the freezing and boiling points of water (0 °C and 100 °C, respec-
tively) apply only at 1 atm of pressure. Phase change temperatures generally vary
with pressure. For example, the boiling point of water decreases with decreasing
pressure. At high altitudes, where there is lower atmospheric pressure, the boiling
point of water is lowered. For example, at Pikes Peak, Colorado, at an elevation of
about 4300 m, the atmospheric pressure is about 0.79 atm and water boils at about
94 °C rather than at 100 °C. The lower temperature lengthens the cooking time of
food. Conversely, some cooks use a pressure cooker to reduce cooking time—by
increasing the pressure, a pressure cooker raises the boiling point.
The freezing point of water actually decreases with increasing pressure. This
inverse relationship is characteristic of only a very few substances, including
water (Section 10.4), that expand when they freeze.
398 11 HEAT
EXAMPLE 11.6 Practical Calorimetry: Using Phase Changes to Save a Life

Organ transplants are becoming commonplace. Many times, ice will reach. If it gets to the freezing point, it will begin to
the procedure involves removing a healthy organ from a melt, and a phase change must be considered. If all of it melts,
deceased person and flying it to the recipient. During that time, then additional heat required to warm that water to a temper-
to prevent its deterioration, the organ is packed in ice in an ature above 0 °C must be considered. Thus, care must be
insulated container. Assume that a human liver has a mass of taken, since it cannot be assumed that all the ice melts, or even
0.500 kg and is initially at 29 °C. The specific heat of the liver is that the ice reaches its melting point. Hence, the calorimetry
3500 J>1kg # °C2. The liver is surrounded by 2.00 kg of ice ini- equation (conservation of energy) cannot be written down
tially at - 10 °C. Calculate the final equilibrium temperature. until the terms in it are determined. First a review of the
possible heat transfers is needed. Only then can the final tem-
T H I N K I N G I T T H R O U G H . Clearly, the liver will cool, and the
perature be determined.
ice will warm. However, it is not clear what temperature the
SOLUTION. Listing the data given and the information obtained from tables,
Given: ml = 0.500 kg Find: Tf (the final temperature of the system)
mice = 2.00 kg
cl = 3500 J>1kg # °C2
cice = 2100 J>1kg # °C2 (from Table 11.1)
Lf = 3.33 * 105 J>1kg # °C2 (from Table 11.2)
The amount of heat required to bring the ice from -10 °C to 0 °C is
Qice = cice mice ¢Tice = 32100 J>1kg # °C2412.00 kg230 °C - 1- 10 °C24 = + 4.20 * 104 J
Since this heat must come from the liver, the maximum heat available from the liver needs to be calculated; that is, if its tempera-
ture drops all the way from 29 °C to 0 °C:
Ql,max = c1 m1 ¢Tl,max = 33500 J>1kg # °C2410.500 kg210 °C - 29 °C2 = - 5.08 * 104 J
This is enough heat to bring the ice to 0 °C. If 4.20 * 104 J of heat flows into the ice (bringing it to 0 °C), the liver is still not at 0 °C
Then how much ice melts? This depends on how much more heat can be transferred from the liver.
How much more heat Q¿ would be transferred from the liver if its temperature were to drop to 0 °C? This value is just the
maximum amount minus the heat that went into warming the ice, or
Q¿ = ƒ Ql,max ƒ - 4.20 * 104 J
= 5.08 * 104 J - 4.20 * 104 J = 8.8 * 103 J
Compare this with the magnitude of the heat needed to melt the ice completely 1 ƒ Qmelt ƒ 2 to decide whether this can, in fact, hap-
pen. The heat required to melt all the ice is
ƒ Qmelt ƒ = + miceLice = + 12.00 kg213.33 * 105 J>kg2 = + 6.66 * 105 J
Since this amount of heat is much larger than the amount available from the liver, only part of the ice melts. In the process, the
temperature of the liver has dropped to 0 °C, and the remainder of the ice is at 0 °C. Since everything in the “calorimeter” is at
the same temperature, heat flow stops, and the final system temperature, Tf , is 0 °C. Thus the final result is that the liver is in a
container with ice and some liquid water, all at 0 °C. Since the container is a very good insulator, it will prevent any inward heat
flow that might raise the liver’s temperature. It is therefore expected that the liver will arrive at its destination in good shape.
FOLLOW-UP EXERCISE. (a) In this Example, how much of the ice melts? (b) If the ice originally had been at its melting point
10 °C2, what would the equilibrium temperature have been?
Notice that in Example 11.6 numbers were not plugged directly into the g Qi = 0 equa-
tion, which is equivalent to assuming that all the ice melts. In fact, if this step had been
done, we would have been on the wrong track. For calorimetry problems involving phase
changes, a careful step-by-step numerical “accounting” procedure should be followed
until all of the pieces of the system are at the same temperature. At that point, the prob-
lem is over, because no more heat exchanges can happen.
EVAPORATION
The evaporation of water from an open container becomes evident only after a rel-
atively long period of time. This phenomenon can be explained in terms of the
kinetic theory (Section 10.5). The molecules in a liquid are in motion at different
speeds. A faster-moving molecule near the surface may momentarily leave the liq-
uid. If its speed is not too large, the molecule will return to the liquid, because of
the attractive forces exerted by the other molecules. Occasionally, however, a mole-
cule has a large enough speed to leave the liquid entirely. The higher the tempera-
ture of the liquid, the more likely this phenomenon is to occur.
The escaping molecules take their energy with them. Since those molecules with
greater-than-average energy are the ones most likely to escape, the average molecular
energy, and thus the temperature of the remaining liquid, will be reduced. That is,
evaporation is a cooling process for the object from which the molecules escape. You have
probably noticed this phenomenon when drying off after a bath or shower. You can
read more about this in Insight 11.1, Physiological Regulation of Body Temperature.
DID YOU LEARN?

➥ The three common phases of matter are the solid phase, the liquid phase, and the
gaseous phase.
➥ The temperature of matter during a phase change remains constant. If the phase
change is between solid and liquid or vice versa, the temperature will remain at the
melting or freezing point; if the phase change is between liquid and gas or vice
versa, the temperature will remain at the boiling or condensation point.
➥ The latent heat of fusion (vaporization) of a substance is the heat required to
change 1 kg of the substance from solid to liquid (liquid to gas) or vice versa while
the temperature remains constant at the freezing (boiling) point.
INSIGHT 11.1 Physiological Regulation of Body Temperature

Being warm-blooded, humans must maintain a narrow range loss at a rate that keeps our body temperature in the safe
of body temperature. (See Chapter 10 Insight 10.1, Human range. However, when the ambient temperature becomes too
Body Temperature.) The generally accepted value for the aver- high, these mechanisms cannot do the job completely. To
age normal body temperature is 37.0 °C 198.6 °F2. However, it avoid heat stroke, as a last resort, the body produces heavy
can be as low as 35.5 °C 195.9 °F2 in the early morning hours perspiration (the most efficient mechanism). The evaporation
on a cold day and as high as 39.5 °C 1103 °F2 during intense of water from the skin removes a lot of heat, thanks to the
exercise on a hot day. For females, the body temperature at rest large value of the latent heat of vaporization of water.
rises very slightly after ovulation as a result of a rise in the hor- Evaporation draws heat from our skin and thus cools our
mone progesterone. This information can be used to predict on bodies. The removal of a minimum of 2.26 * 106 J of heat
which day ovulation will occur in the next cycle. from the body is required to evaporate each kilogram (liter) of
When the ambient temperature is lower than the body tem- water.* For the body of a 75-kg person (which is mainly com-
perature, the body loses heat. If the body loses too much heat, posed of water), the heat loss due to evaporation of 1 kg of
a circulatory mechanism causes a reduction of blood flow to water could lower the body temperature as much as
the skin in order to reduce heat loss. A physiological response
Q 2.26 * 106 J
to this mechanism is to increase heat generation (and thus
34186 J>1kg # °C24175 kg2
¢T = = = 7.2 °C
warm the body) through shivering by “burning” the body’s cm
reserves of carbohydrate or fat. If the body temperature drops During a race on a hot day, a professional cyclist can lose as
below 33 °C 191.4 °F2, hypothermia may result, which can much as 7.0 kg of water in 3.5 h via evaporation. This heat
cause severe thermal injuries to organs and even death. transfer through perspiration is the mechanism that enables
At the other extreme, if the body is subjected to ambient the body to keep its temperature in the safe range.
temperatures higher than the body temperature, along with On a summer day, a person may stand in front of a fan and
intense exercise, the body can overheat. Heat stroke is a pro- remark how “cool” the blowing air feels. But the fan is merely
longed elevation of body temperature above 40 °C 1104 °F2. blowing hot air from one place to another. The air feels cool
If the body becomes overheated, the blood vessels to the skin because it is relatively drier (has low humidity compared to
dilate, carrying more warm blood to the skin, enabling the the body’s perspiration), and therefore its flow promotes
interior of the body and organs to remain cooler. (The per- evaporation, which removes heat.
son’s face may turn red.)
Usually, radiation, conduction, natural convection (dis- *The actual latent heat of vaporization of perspiration is about
cussed later in Section 11.4), and possibly slight evaporation 2.42 * 106 J>1kg # °C2—greater than the 2.26 * 106 J>1kg # °C2 value
of perspiration on the skin are sufficient to maintain a heat used here for temperature at 100 °C. Why?
400 11 HEAT
T1 T2 11.4 Heat Transfer

ΔT LEARNING PATH QUESTIONS
➥ What are the three mechanisms of heat transfer?

Surface
➥ Should house insulation materials have high or low thermal conductivity?
area A
➥ Which mechanism of heat transfer does not require a transport medium?
d CONDUCTION
You can keep a pot of coffee hot on an electric stove because heat is conducted
ΔQ through the bottom of the coffeepot from the hot metal burner. The process of
Δt conduction results from molecular interactions. Molecules at a higher-temperature
region on an object move relatively rapidly. They collide with, and transfer some of
their energy to, the less energetic molecules in a nearby cooler part of the object. In
this way, energy is conductively transferred from a higher-temperature region to a
lower-temperature region—transfer as a result of a temperature difference.
Heat flow Solids can be divided into two general categories: metals and nonmetals. Met-
als are generally good conductors of heat, or thermal conductors. Metals have a
ΔQ kA ΔT
= large number of electrons that are free to move around (not permanently bound to
Δt d
a particular molecule or atom). These free electrons (rather than the interaction
between adjacent atoms) are primarily responsible for the good heat conduction
䉱 F I G U R E 1 1 . 7 Thermal conduc-
tion Heat conduction is character- in metals. Nonmetals, such as wood and cloth, have relatively few free electrons.
ized by the time rate of heat flow The absence of this transfer mechanism makes them poor heat conductors relative
( ¢Q>¢t) in a material with a tem- to metals. A poor heat conductor is called a thermal insulator.
perature difference across it of ¢T. In general, the ability of a substance to conduct heat depends on the substance’s
For a slab of material, ¢Q> ¢t is phase. Gases are poor thermal conductors; their molecules are relatively far apart, and
directly proportional to the cross-
sectional area (A) and the thermal collisions are therefore infrequent. Liquids and solids are better thermal conductors
conductivity (k) of the material; it is than gases, because their molecules are closer together and can interact more readily.
inversely proportional to the thick- Heat conduction is usually described using the time rate of heat flow 1¢Q>¢t2
ness of the slab (d). in a material for a given temperature difference 1¢T2, as illustrated in 䉳 Fig. 11.7.
Experiment has established that the rate of heat flow through a substance depends
on the temperature difference between its boundaries. Heat conduction also
depends on the size and shape of the object as well as its composition.
Experimentally, it was found that the heat flow rate ( ¢Q>¢t in J>s or W)
through a slab of material is directly proportional to the material’s surface area (A)
and the temperature difference across its ends 1¢T2, and is inversely proportional
to its thickness (d). That is,
¢Q A¢T
r
¢t d
Using a constant of proportionality k allows us to write the relation as an equation:
¢Q kA¢T
= (conduction only) (11.4)
¢t d
The constant k, called the thermal conductivity, characterizes the heat-conducting

ability of a material and depends only on the type of material. The greater the
value of k for a material, the better it will conduct heat, all other factors being
equal. The units of k are J>1m # s # °C2 = W>1m # °C2. The thermal conductivities of
䉱 F I G U R E 1 1 . 8 Copper- various substances are listed in 䉴 Table 11.3. These values vary slightly with tem-
bottomed pots Copper is used on perature, but can be considered constant over normal temperature ranges.
the bottoms of some stainless steel
Compare the relatively large thermal conductivities of the good thermal conduc-
pots and saucepans. The high ther-
mal conductivity of copper ensures tors, the metals, with the relatively small thermal conductivities of some good thermal
the rapid and even spread of heat insulators, such as Styrofoam and wood. Some stainless steel cooking pots have cop-
from the burner; the low thermal per bottoms (䉳 Fig. 11.8). Being a good conductor of heat, the copper conducts heat
conductivity of stainless steel faster to the food being cooked and also promotes the distribution of heat over the bot-
retains the heat in the pot and keeps
tom of a pot for even cooking. Conversely, Styrofoam is a good insulator, mainly
the handle not too hot to touch. (The
thermal conductivity of stainless because it contains small, trapped pockets of air, thus reducing conduction and con-
steel is only 12% of that of copper.) vection losses (discussed in the next section). When you step on a tile floor with one
11.4 HEAT TRANSFER 401
TABLE 11.3 Thermal Conductivities of Some Substances

Thermal Conductivity, k
Substance J>(m # s # °C)orW >(m # °C) kcal>(m # s # °C)
Metals
Aluminum 240 5.73 * 10-2
Copper 390 9.32 * 10-2
Iron 80 1.9 * 10-2
Stainless steel 16 3.8 * 10-3
Silver 420 10 * 10-2
Liquids
Transformer oil 0.18 4.3 * 10-5
Water 0.57 14 * 10-5
Gases
Air 0.024 0.57 * 10-5
Hydrogen 0.17 4.1 * 10-5
Oxygen 0.024 0.57 * 10-5
Other Materials
Brick 0.71 17 * 10-5
Concrete 1.3 31 * 10-5
Cotton 0.075 1.8 * 10-5
Fiberboard 0.059 1.4 * 10-5
Floor tile 0.67 16 * 10-5
Glass (typical) 0.84 20 * 10-5
Glass wool 0.042 1.0 * 10-5
Goose down 0.025 0.59 * 10-5
Human tissue (average) 0.20 4.8 * 10-5
Ice 2.2 53 * 10-5
Styrofoam 0.042 1.0 * 10-5
Wood, oak 0.15 3.6 * 10-5
Wood, pine 0.12 2.9 * 10-5
Vacuum 0 0
bare foot and on an adjacent rug with the other bare foot, you feel that the tile is
“colder” than the rug. However, both the tile and rug are actually at the same tempera-
ture. But the tile is a much better thermal conductor, so it transfers heat from your foot
more efficiently than the rug, making your foot on tile feel colder.
EXAMPLE 11.7 Thermal Insulation: Helping Prevent Heat Loss

A room with a pine ceiling that measures 3.0 m by 5.0 m and T H I N K I N G I T T H R O U G H . Here there are two materials, so Eq.
is 2.0 cm thick has a layer of glass wool insulation above it 11.4 is applied for two different thermal conductivities (k). We
that is 6.0 cm thick (䉲 Fig. 11.9a). On a cold day, the tempera- want to find ¢Q>¢t for the combination so that ¢Q can be
ture inside the room at ceiling height is 20 °C, and the temper- found for ¢t = 1.0 h. The situation is a bit complicated, because
ature in the attic above the insulation layer is 8.0 °C. the heat flows through two materials. But at a steady rate, the
Assuming that the temperatures remain constant and heat heat flows must be the same through both. (Why?) To find the
loss is due to conduction only, how much energy does the energy saved in 1.0 h, the heat conducted in this time both with-
layer of insulation save in 1.0 h? out and with the layer of insulation needs to be calculated.
(continued on next page )
402 11 HEAT
T2 = 8 °C Heat flow
䉴 F I G U R E 1 1 . 9 Insulation and thermal con-
ductivity (a), (b) Attics should be insulated to
prevent loss of heat by the mechanism of con- d2 6.0 cm T2
duction. See Example 11.7 and Insight 11.2,
2.0 cm
Physics, the Construction Industry, and
d1 T1 = 20 °C
Energy Conservation. (c) This thermogram of d2 k2
a house allows us to visualize the house’s
heat loss. Blue represents the areas that have
the lowest rate of heat leaking; white, pink, T
d1 k1
and red indicate areas with increasingly
larger heat losses. (Red areas have the most T1
loss.) What recommendations would you
make to the owner of this house to save both
money and energy? (Compare this figure
with Fig. 11.15.) (a) (b)
(c)
SOLUTION. Listing the data, computing some of the quantities in Eq. 11.4, and making conversions:
Given: A = 3.0 m * 5.0 m = 15 m2 Find: Energy saved in 1.0 h
d1 = 2.0 cm = 0.020 m
d2 = 6.0 cm = 0.060 m
¢T = T1 - T2 = 20 °C - 8.0 °C = 12 °C
¢t = 1.0 h = 3.6 * 103 s
k1 = 0.12 J>1m # s # °C21wood, pine2
f (from Table 11.3)
k2 = 0.042 J>1m # s # °C21glass wool2
(In working such problems with several given quantities, it is especially important to label all the data correctly.)
First, let’s consider how much heat would be conducted in 1.0 h through the wooden ceiling without insulation. Since ¢t is
known, Eq. 11.4* can be rearranged to find ¢Qc (heat conducted through the wooden ceiling alone, assuming the same ¢T):
30.12 J>1m # s # °C24115 m22112 °C2

r 13.6 * 103 s2 = 3.9 * 106 J
k1 A¢T
¢Qc = ¢ ≤ ¢t = b
d1 0.020 m
Now we need to find the heat conducted through the ceiling and the insulation layer together. Let T be the temperature at the
interface of the materials and T1 and T2 be the warmer and cooler temperatures, respectively (Fig. 11.9b). Then
¢Q1 k1 A1T1 - T2 ¢Q2 k2 A1T - T22

= and =
¢t d1 ¢t d2
T is not known, but when the conduction is steady, the flow rates are the same for both materials; that is, ¢Q1> ¢t = ¢Q2> ¢t, or
k1 A1T1 - T2 k2 A1T - T22

=
d1 d2
The A’s cancel, and solving for T gives
k1 d2 T1 + k2 d1 T2 30.12 J>1m # s # °C2410.060 m2120 °C2 + 30.042 J>1m # s # °C2410.020 m218.0 °C2
30.12 J>1m # s # °C2410.060 m2 + 30.042 J>1m # s # °C2410.020 m2
T = = = 18.7 °C
k1 d2 + k2 d1
*Equation 11.4 can be extended to any number of layers or slabs of materials: ¢Q>¢t = A1T2 - T12> g 1di>ki2. (See Insight 11.2, Physics, the
Construction Industry, and Energy Conservation, involving insulation in building construction.)
Since the flow rate through the wood and the insulation is the same, we can use the expression for either material to calculate
it. Let’s use the expression for the wood ceiling. Here, care must be taken to use the correct ¢T. The temperature at the
wood–insulation interface is 18.7 °C; thus,
¢Twood = ƒ T1 - T ƒ = ƒ 20 °C - 18.7 °C ƒ = 1.3 °C
Therefore, the heat flow rate is
k1 A ƒ ¢Twood ƒ 30.12 J>1m # s # °C24115 m2211.3 °C2
= 1.2 * 102 J>s 1or W2
¢Q1
= =
¢t d1 0.020 m
In 1.0 h, the heat loss with insulation in place is
* ¢t = 11.2 * 102 J>s213600 s2 = 4.3 * 105 J

¢Q1
¢Q1 =
¢t
This value represents a decreased heat loss of
¢Qc - ¢Q1 = 3.9 * 106 J - 4.3 * 105 J = 3.5 * 106 J
3.5 * 106 J
This amount represents an energy savings of * 1100%2 = 90%.
3.9 * 106 J
FOLLOW-UP EXERCISE. Verify that the heat flow rate through the insulation is the same as that through the wood 11.2 * 102 J>s2
in this Example.
INSIGHT 11.2 Physics, the Construction Industry, and Energy Conservation

Many homeowners have found it cost-effective to provide
their homes with better insulation, especially in the last few
years due to the higher energy costs. To quantify the insulat-
ing properties of various materials, the insulation and con-
struction industries do not use thermal conductivity, k.
Rather, they use a quantity called thermal resistance, which is
related to the inverse of k.
To see how these two quantities are related, consider
Eq. 11.4 rewritten as
= a b A¢T = ¢ ≤ A¢T
¢Q k 1
¢t d Rt
where the thermal resistance is Rt = d>k. Note that Rt depends
not only on the material’s properties (expressed in the ther-
mal conductivity, k), but also on its thickness, d. Rt is a mea-
sure of how “resistant” to heat flow a certain thickness of F I G U R E 1 Differences in R-values For insulation blankets
made of identical materials, the R-values are proportional to
material is. The heat flow rate, ¢Q> ¢t, is inversely related to
the materials’ thickness.
the thermal resistance: More thermal resistance results in less
heat flow. More resistance is attained using thicker material
with a low conductivity.
For homeowners, the lesson is clear. To reduce heat flow mostat setting on their heating system (also lowering ¢T but
(and thus minimize heat loss in the winter and heat gain in the by decreasing Tinterior).
summer), they should reduce areas of low thermal resistance, Insulation and building materials are classified according
such as windows, or at least increase the windows’ resistance to their R-values, that is, their thermal resistance values. In the
by switching to double or triple panes. Similarly, increasing the United States, the units of Rt are ft 2 # h # °F>Btu. While these
thermal resistance by adding or upgrading insulation to walls units may seem awkward, the important point is that they are
is the way to go. Lastly, changing interior temperature require- proportional to the thermal resistance of the material. Thus,
ments (changing ¢T = ƒ Texterior - Tinterior ƒ ) can make a big dif- wall insulation with a value of R-30 (meaning
ference. In the summer, homeowners should raise the Rt = 30 ft 2 # h # °F>Btu) is about 2.3 times (or 30>13) as resis-
thermostat setting on their air conditioning (lowering ¢T by tive as insulation with a value of R-13. A photo showing vari-
increasing Tinterior), and in winter, they should lower the ther- ous types of insulation is shown in Fig. 1.
404 11 HEAT
DAY
NIGHT
Air current Air current
Sea breeze Land breeze
Land warmer than water Water warmer than land
䉱 F I G U R E 1 1 . 1 0 Convection cycles During the day, natural convections give rise to sea
breezes near large bodies of water. At night, the pattern of circulation is reversed, and the land
breezes blow. The temperature differences between land and water are the result of their spe-
cific heat differences. Water has a much larger specific heat, so the land warms up more
quickly during the day. At night, the land cools more quickly, while the water remains
warmer, because of its larger specific heat.
CONVECTION
In general, compared with solids, liquids and gases are not good thermal conduc-
tors. However, the mobility of molecules in fluids permits heat transfer by another
process—convection. (A fluid is a substance that can flow, and hence includes
both liquids and gases.) Convection is heat transfer as a result of mass transfer,
which can be natural or forced.
Natural convection occurs in liquids and gases. For example, when cold water is in
contact with a hot object, such as the bottom of a pot on a stove, heat is transferred to
the water adjacent to the pot by conduction. Since the water at the bottom is
Hot air
register warmer, its density is lower, causing it to rise to the top. The top water, being cooler,
has a higher density, so it sinks to the bottom. This sets up a natural convection.
Cold air
Such convections are also important in atmospheric processes, as illustrated in
register
䉱 Fig. 11.10. During the day, the ground heats up more quickly than do large bodies
of water, as you may have noticed if you have been to the beach. This phenomenon
occurs mainly because the water has a higher specific heat than land. The air in con-
tact with the warm ground is heated and expands, becoming less dense. As a result,
the warm air rises (air currents) and, to fill the space, other air moves horizontally
(winds)—creating a sea breeze near a large body of water. Cooler air descends, and
a thermal convection cycle is set up, which transfers heat away from the land. At
night, the ground loses its heat more quickly than the water, and the surface of the
water is warmer than the land. As a result, the current is reversed. Since the prevail-
Furnace Blower
ing jet streams over the Northern Hemisphere flow mostly from west to east, west
coasts usually have a milder climate than east coasts. The winds move the Pacific
䉱 F I G U R E 1 1 . 1 1 Forced convec- ocean air with more constant temperature toward the west coasts.
tion Houses are commonly heated In forced convection, the fluid is moved mechanically. Common examples of
by forced convection. Registers or forced convection systems are forced-air heating systems in homes (䉳 Fig. 11.11),
gratings in the floors or walls allow the human circulatory system, and the cooling system of an automobile engine.
heated air to enter and cooler air to
return to the heat source. (Can you The human body loses a great deal of heat when the surroundings are colder than
explain why the registers are the body. The internally generated heat is transferred close to the surface of the
located near the floor?) skin by blood circulation. From the skin, the heat is conducted to the air or lost by
radiation (the other heat transfer mechanism, to be discussed shortly). This circu-
latory system is highly adjustable; blood flow can be increased or decreased to
specific areas depending on needs.
Coolant is circulated (pumped) through most automobile cooling systems. (Some
smaller engines are air-cooled.) The coolant carries engine heat to the radiator (a
form of heat exchanger), where forced-air flow produced by the fan and car move-
ment carries it away. The radiator of an automobile is actually misnamed—most of
the heat is transferred from it by forced convection rather than by radiation.
CONCEPTUAL EXAMPLE 11.8 Foam Insulation: Better Than Air?

Foam insulation is sometimes blown into the space between convection within the wall space. In the winter, the air near the
the inner and outer walls of a house. Since air is a better ther- warm inner wall is heated and rises, thus setting up a convec-
mal insulator than foam (Table 11.3), why is the foam insulation cycle in the space and transferring heat to the cold outer
tion needed: (a) to prevent loss of heat by conduction, (b) to wall. In the summer, with air conditioning, the heat loss cycle
prevent loss of heat by convection, or (c) for fireproofing? is reversed. Foam blocks the movement of air and thus stops
such convection cycles. Hence, the answer is (b).
REASONING AND ANSWER. Foams will generally burn, so
(c) isn’t likely to be the answer. Air is a poor thermal conduc- F O L L O W - U P E X E R C I S E . Thermal underwear and thermal
tor, even poorer than foam (Styrofoam—see Table 11.3), so the blankets are loosely knit with lots of small holes. Wouldn’t
answer can’t be (a). However, as a gas, the air is subject to they be more effective if the material were closely knit?
RADIATION
Conduction and convection require some material as a trans-
port medium. The third mechanism of heat transfer needs no Conduction
medium; it is called radiation, which refers to energy transfer
by electromagnetic waves (Section 20.4). Heat is transferred to
the Earth from the Sun through empty space by radiation. Visi- Convection
ble light and other forms of electromagnetic radiation are com-
monly referred to as radiant energy.
You have experienced heat transfer by radiation if you’ve
ever stood near an open fire (䉴 Fig. 11.12). You can feel the heat Radiation
on your exposed hands and face. This heat transfer is not due
to convection or conduction, since heated air rises and air is a
poor conductor. Visible radiation is emitted from the burning
䉱 F I G U R E 1 1 . 1 2 Heating by conduction, convection,
material, but most of the heating effect comes from the invisi- and radiation The hands on top of the flame are warmed
ble infrared radiation. You feel this radiation because it is by the convection of rising hot air (and some radiation).
absorbed by water molecules in your skin. The water mole- The gloved hand is warmed by conduction. The hands to
cule has an internal vibration whose frequency coincides with the right of the flame are warmed by radiation.
that of infrared radiation, which is therefore readily absorbed.
(This effect is called resonance absorption. The electromagnetic wave drives the mol-
ecular vibration, and energy is transferred to the molecule, somewhat like push-
ing a swing. See Chapter 13 on oscillations for more details.) Heat transfer by
radiation can play a practical role in daily living (䉴 Fig. 11.13).
Infrared radiation is sometimes referred to as “heat radiation” or thermal radia-
tion. You may have noticed the reddish infrared lamps used to keep food warm in
cafeterias. Heat transfer by infrared radiation is also important in maintaining our
planet’s warmth by a mechanism known as the greenhouse effect. This important
environmental topic is discussed in Insight 11.3, The Greenhouse Effect.
Although infrared radiation is invisible to the human eye, it can be detected by
other means. Infrared detectors can measure temperature remotely (䉲 Fig. 11.14).
䉱 F I G U R E 1 1 . 1 3 A practical
application of heat transfer by radia-
tion A Tibetan teakettle is heated by
focusing sunlight, using a metal
reflector.
䉳 F I G U R E 1 1 . 1 4 Detecting SARS
Infrared thermometers were used to
measure body temperature during
the severe acute respiratory syn-
drome (SARS) outbreak in 2003.
406 11 HEAT
INSIGHT 11.3 The Greenhouse Effect

The greenhouse effect helps regulate the Earth’s long-term aver- trapping the reradiated infrared radiation inside. The glass
age temperature, which has been fairly constant for some cen- enclosure also keeps warm air from escaping upward, result-
turies. When a portion of the solar radiation (mostly visible ing in the elimination of heat loss by convection. It is quite
light) reaches and warms the Earth’s surface, the Earth, in warm in a greenhouse on a sunny day, even in winter. We
turn, reradiates energy in the form of infrared radiation (IR). have all observed this warming effect—for example, in a
As the reradiated IR radiation passes back through the atmos- closed car on a sunny but cold day.
phere, some of the radiation is absorbed by the greenhouse The problem on Earth is that human activities since the
gases there—primarily water vapor, carbon dioxide (CO2), beginning of the industrial age have been accelerating
and methane. These gases are selective absorbers: They greenhouse warming. With the combustion of hydrocarbon
absorb radiation at certain IR wavelengths but not at others fuels (gas, oil, coal, and so on), vast amounts of CO2 and
(Fig. 1a). Without this absorption, the IR radiation would go other greenhouse gases are vented into the atmosphere,
back into space and life on the Earth would probably not where they trap increasingly more IR radiation. There is
exist, because the average surface temperature would be a grave concern that the result of this trend is increasing the
cold - 18 °C, rather than the present 15 °C. Earth’s average surface temperature—global warming. Such
Why is this phenomenon called the greenhouse effect? The an increase could dramatically affect agricultural production
reason is that the atmosphere functions somewhat like the and world food supplies. It can also cause partial melting of
gases in a greenhouse. In general, visible radiation is trans- the polar ice caps. Sea levels would then rise, flooding low-
mitted but infrared radiation is selectively absorbed by the lying regions and endangering coastal ports and population
gas in a greenhouse (Fig. 1b) so the greenhouse heats up by centers.
Infrared
Sunlight Sunlight
radiation
(visible) (visible)
Infrared
radiation
Selectively
absorbed
Atmospheric Selectively
F I G U R E 1 The greenhouse effect absorbed
(a) The greenhouse gases of the gases
atmosphere, particularly water vapor,
methane, and carbon dioxide, are
selective absorbers with absorption
properties similar to those of the glass
used in greenhouses. Visible light is
transmitted and heats the Earth’s sur-
face, while some of the infrared radia-
tion that is re-emitted is absorbed and
trapped in the Earth’s atmosphere.
(b) A greenhouse operates in a similar
way. (a) (b)
Also, cameras using special infrared films take pictures consisting of contrasting
bright and dark areas that correspond to regions of higher and lower temperatures,
respectively. Special instruments that apply such thermography are used in medi-
cine and industry; the images they produce are called thermograms (䉴 Fig. 11.15).
A new application of thermograms is for security. An infrared camera takes a
picture of an individual using the unique heat pattern emitted by the facial blood
vessels. A computer then compares the picture with an earlier stored image.
The rate at which an object radiates energy has been found to be proportional to
the fourth power of the object’s absolute temperature (T4). This relationship is
expressed in an equation known as Stefan’s law,*
¢Q
P = = sAeT4 (radiation only) (11.5)
¢t
where P1¢Q>¢t2 is the power radiated in watts (W), or joules per second (J>s).
A is the object’s surface area and T is its temperature in Kelvin. The symbol s (the
*Developed by the Austrian physicist Joseph Stefan (1835–1893).

Greek letter sigma) is the Stefan–Boltzmann constant: s = 5.67 * 10-8 W>1m2 # K42.
The emissivity (e) is a unitless number between 0 and 1 that is a characteristic of
the material. Dark surfaces have emissivities close to 1, and shiny surfaces have
emissivities close to 0. The emissivity of human skin is about 0.70.
Dark surfaces not only are better emitters of radiation, but also are good
absorbers. This must be the case because to maintain a constant temperature, the
incident energy absorbed must equal the emitted energy. Thus, a good absorber is
also a good emitter. An ideal, or perfect, absorber (and emitter) is referred to as a
black body 1e = 1.02. Shiny surfaces are poor absorbers, since most of the inci-
dent radiation is reflected. This fact can be demonstrated easily, as shown in
䉴 Fig. 11.16. (Can you see why it is better to wear light-colored clothes in the sum- 䉱 F I G U R E 1 1 . 1 5 Applied ther-
mography Thermograms can be
mer and dark-colored clothes in the winter?)
used to detect breast cancer by
When an object is in thermal equilibrium with its surroundings, its temperature showing tumor regions that are
is constant; thus, it must be emitting and absorbing radiation at the same rate. higher in temperature than normal.
However, if the temperatures of the object and its surroundings are different, there
will be a net flow of radiant energy. If an object is at a temperature T and its sur-
roundings are at a temperature Ts , the net rate of energy loss or gain per unit time
(power) is given by
Pnet = sAe1T4s - T42 (11.6)
Note that if Ts is less than T, then P will be negative, indicating a net heat energy
loss, in keeping with our heat flow sign convention. Keep in mind that the tempera-
tures used in calculating radiated power are the absolute temperatures in kelvins.
You may have noticed in Section 10.1 that heat was defined as the net energy
transfer due to temperature differences. The word net here is important. It is possi-
ble to have energy transfer between an object and its surroundings, or between
objects, at the same temperature. Note that if Ts = T (that is, there is no tempera-
ture difference), there is a continuous exchange of radiant energy, but there is no
net change of the internal energy of the object.
EXAMPLE 11.9 Body Heat: Radiant Heat Transfer

Suppose that your skin has an emissivity of 0.70, a temperature of 34 °C, and a total
area of 1.5 m2. How much net energy per second will be radiated from your skin if the
ambient room temperature is 20 °C?
T H I N K I N G I T T H R O U G H . Everything is given for us to find Pnet from Eq. 11.6. The net
radiant energy transfer is between the skin and the surroundings. We must remember
to work with temperatures in kelvins.
SOLUTION.
Given: Ts = 120 + 2732 K = 293 K Find: Pnet (net power)

T = 134 + 2732 K = 307 K
e = 0.70
A = 1.5 m2
s = 5.67 * 10-8 W>1m2 # K42 1known2
䉱 F I G U R E 1 1 . 1 6 Good absorber
Using Eq. 11.6 directly, Black objects are generally good
absorbers of radiation. The bulb of
Pnet = sAe1Ts4 - T42 = 35.67 * 10-8 W>1m2 # K42411.5 m2210.70231293 K24 - 1307 K244 the thermometer on the right has
= - 90 W 1or - 90 J>s2 been painted black. Note the differ-
ence in temperature readings.
Thus, 90 J of energy is radiated, or lost (as indicated by the minus sign), each second.
That is, the human body loses heat at a rate that is close to that of a 100-W lightbulb. No
wonder a room full of people can get warm.
F O L L O W - U P E X E R C I S E . (a) In this Example, suppose the skin had been exposed to an
ambient room temperature of only 10 °C. What would the rate of heat loss be? (b) Ele-
phants have huge body masses and large daily caloric food intakes. Can you explain how
their huge ear flaps (large surface area) might help stabilize their body temperature?
408 11 HEAT
Outer glass Silver film

wall
Inner glass Hot or cold
wall liquid
Partial
vacuum
䉱 F I G U R E 1 1 . 1 7 Thermal insulation
The Thermos bottle minimizes all three 䉱 F I G U R E 1 1 . 1 8 A dark robe in the desert? Dark objects absorb more
mechanisms of heat transfer. See text for radiation than do lighter ones, and they become hotter. What’s going on
description. here? See the book for an explanation.
Note that in Example 11.9, the fourth powers of the temperatures were found first, and
then their difference was found. It is not correct to find the temperature difference and
then raise it to the fourth power: T4s - T4 Z 1Ts - T24.
Let’s look at a few more real-life examples of heat transfer. In the spring, a late
frost could kill the buds on fruit trees. To save the buds, some growers spray water
on the trees to form ice before a hard frost occurs. Using ice to save buds? Ice is a
relatively poor (and inexpensive) thermal conductor, so it has an insulating effect.
It will maintain the buds’ temperature at 0 °C, not going below that value, and
therefore protects the buds.
Another method to protect orchards from freezing is the use of smudge pots,
containers in which material is burned to create a dense cloud of smoke. At night,
when the Sun-warmed ground cools off by radiation, the cloud absorbs this heat
and reradiates it back to the ground. Thus, the ground takes longer to cool, hope-
fully without reaching freezing temperatures before the Sun comes up.
A Thermos bottle (䉱 Fig. 11.17) keeps cold beverages cold and hot ones hot. It
consists of a double-walled, partially evacuated container with silvered walls
(mirrored interior). The bottle is constructed to minimize all three mechanisms of
heat transfer. The double-walled and partially evacuated container counteracts
conduction and convection because both processes depend on a medium to trans-
fer the heat (the double walls are more for holding the partially evacuated region
than for reducing conduction and convection). The mirrored interior minimizes
loss by radiation. The stopper on top of the thermos stops convection off the top of
the liquid as well.
Look at 䉱 Fig. 11.18. Why would anyone wear a dark robe in the desert? It was
previously learned that dark objects absorb radiation (Fig. 11.16). Wouldn’t a
white robe be better? A dark robe definitely absorbs more radiant energy and
warms the air inside near the body. But note that the robe is open at the bottom.
The warm air rises (since it is less dense) and exits at the neck area, and outside
cooler air enters the robe at the bottom—natural convection air circulation.
Finally, consider some of the thermal factors involved in “passive” solar house
design used as far back as in ancient China (䉴 Fig. 11.19). The term passive means
that the design elements require no active use of energy. In Beijing, China, for
Summer solstice
Equinoxes
Winter solstice
27° 50° 76°
(a) (b)
䉱 F I G U R E 1 1 . 1 9 Aspects of passive solar design in ancient China (a) In summer, with the sun angle high, the overhangs
provide shade to the building. The brick and mud walls are thick to reduce conductive heat flow to the interior. In winter,
the sun angle is low, so the sunlight streams into the building, especially with the help of the upward curved overhangs.
The leaves of nearby deciduous trees provide additional shade in the summer but allow sunlight in when they have
dropped their leaves in the winter. (b) A photo of such a building in Beijing, China, in December.
example, the angles of the sunlight are 76°, 50°, and 27° above the horizon at the
summer solstice, the spring and fall equinoxes, and the winter solstice, respec-
tively. With a proper combination of column height and roof overhang length, a
maximum amount of sunlight is allowed into the building in the winter, but most
of the sunlight will not reach the inside of the building in the summer. The over-
hangs of the roofs are also curved upward, not just for good looks, but also for let-
ting the maximum amount of light into the building in the winter. Trees planted
on the south side of the building can also play important roles in both summer
and winter. In the summer, the leaves block and filter the sunlight; in the winter,
the dropped leaves will let sunlight through.
DID YOU LEARN?

➥ The three mechanisms of heat transfer are conduction, convection, and radiation.
➥ Insulation materials should have low thermal conductivities so as to conduct less
heat.
➥ Radiation does not require a transport medium, that is, radiation heat transfer can
occur in a vacuum.
PULLING IT TOGETHER Ice Skating and Latent Heat

The world record in 500-m speed skating is 34.03 s, set by S O L U T I O N . Eq. 11.3 and the kinetic energy formula,
Canadian skater Jeremy Wotherspoon, who has a mass of K = 12 mv2, are used. The latent heat of fusion of ice can be
82.0 kg. Assume his final speed as he crossed the finish line is looked up in Table 11.3.
his average speed of the race. In coasting to a smooth stop,
40% of the frictional heat generated by the skate blades goes Given: m = 82.0 kg Find: mice
into melting the ice (assumed to be at 0 °C). How much ice is 500 m (mass of ice
vo = = 14.69 m>s metted)
melted after he crossed the finish line? Where does the other 34.03 s
60% of the energy go? v = 0
Lf = 3.33 * 105 J>kg
T H I N K I N G I T T H R O U G H . This example involves kinetic (from Table 11.2)
energy, energy conversion, and latent heat. In order to find
the mass of ice melted, the energy lost when Jeremy coasted
to a stop needs to be determined. Then 40% of this energy loss
became heat to melt the ice using latent heat. (continued on next page)
410 11 HEAT
The change in kinetic energy (energy loss) is equal to Using Eq. 11.3,
¢K = K - Ko = 12 mv2 - 12 mo v 2 Q 3.54 * 103 J
2 182.0 kg2102 - 2 182.0 kg2114.69
1 2 1 mice = = = 0.0106 kg = 10.6 g
= m>s22 Lf 3.33 * 105 J>kg
3
= - 8.85 * 10 J
The rest of the energy loss 160%2 went into heating the skates,
The negative sign means energy is being lost. Since 40% of generating noise, etc.
this energy loss became heat used to melt ice, so
Q = 10.402 ƒ ¢K ƒ = 10.40218.85 * 103 J2 = 3.54 * 103 J
■ Heat (Q) is the energy exchanged between objects, com- ■ Heat transfer due to direct contact of objects that have dif-
monly because they are at different temperatures. ferent temperatures is called conduction. The rate of heat
flow by conduction through a slab of material is given by
ΔT = 1 °C
¢Q kA¢T
= (11.4)
¢t d
1 kg T1 T2
water
ΔT
Surface
area A
1 kilocalorie (kcal)
or Calorie (Cal)
■ The specific heat (c) tells how much heat is needed to raise ΔQ
the temperature of 1 kg of a particular material by 1 °C It is Δt
a characteristic of the type of material and is defined by
Q
c = (11.1)
m¢T Heat flow
■ Calorimetry is a technique that uses heat transfer between ■ Convection refers to heat transfer due to mass movement of
objects, most commonly to measure specific heats of materi- gas or liquid molecules. Natural convection is driven by density
als. It is based on conservation of energy, written as ©Qi = 0, differences caused by temperature differences. In forced con-
assuming no heat losses or gains to the environment. vection, the movement is driven by mechanical means.
DAY
Air current
Sea breeze
Land warmer than water
■ Latent heat (L) is the heat required to change the phase of ■ Radiation refers to heat transferred by electromagnetic radia-
an object per kilogram of mass. During the phase change, the tion between objects that have different temperatures, usually
temperature of the system does not change. Its general defi- an object and its surroundings. The rate of transfer is given by
nition is Pnet = sAe1T4s - T42 (11.6)
ƒQƒ where s is the Stefan–Boltzmann constant,
L = or Q = mL (11.2, 11.3)
ƒmƒ 5.67 * 10-8 W>1m2 # K42.
Learning Path Questions and Exercises* For instructor-assigned homework, go to www.masteringphysics.com

MULTIPLE CHOICE QUESTION
11.1 DEFINITION AND UNITS OF HEAT 11.3 PHASE CHANGES AND

1. The SI unit of heat energy is the (a) calorie, (b) kilocalo- LATENT HEAT
rie, (c) Btu, (d) joule. 8. The units of latent heat are (a) 1>°C, (b) J>1kg # °C2,
2. Which of the following is the largest unit of heat energy: (c) J>°C, (d) J>kg.
(a) calorie, (b) Btu, (c) joule, or (d) kilojoule? 9. Latent heat is always (a) part of the specific heat,
3. The mechanical equivalent of heat is (a) 1 kcal = 4.186 J, (b) related to the specific heat, (c) the same as the
(b) 1 J = 4.186 cal, (c) 1 cal = 4.186 J, (d) 1 Cal = 4.186 J. mechanical equivalent of heat, (d) none of the preceding.
10. When a substance undergoes a phase change, the added
11.2 SPECIFIC HEAT AND CALORIMETRY heat changes (a) the temperature, (b) the kinetic energy,
(c) the potential energy, (d) the mass of the substance.
4. The amount of heat necessary to change the temperature
of 1 kg of a substance by 1 °C is called the substance’s
(a) specific heat, (b) latent heat, (c) heat of combustion,
11.4 HEAT TRANSFER
(d) mechanical equivalent of heat.
5. The same amount of heat Q is added to two objects of 11. House insulation materials should have (a) high thermal
the same mass. If object 1 experienced a greater tempera- conductivity, (b) low thermal conductivity, (c) high emis-
ture change than object 2, that is, ¢T1 7 ¢T2 , then sivity, (d) low emissivity.
(a) c1 7 c2 , (b) c1 6 c2 , (c) c1 = c2 . 12. Which of the following is the dominant heat transfer
6. The fundamental physical principle for calorimetry is mechanism by which the Earth receives energy from the
(a) Newton’s second law, (b) conservation of momen- Sun: (a) conduction, (b) convection, (c) radiation, or
tum, (c) conservation of energy, (d) equilibrium. (d) all of the preceding?
7. For gases, which of the following is true about the spe- 13. Water is a poor heat conductor, but a pot of water can be
cific heat under constant pressure, cp , and specific heat heated more quickly than you might think. This fast
under constant volume, cv : (a) cp 7 cv , (b) cp = cv , or heating time is mainly due to heat (a) conduction,
(c) cp 6 cv? (b) convection, (c) radiation, (d) all of the preceding.
11.1 DEFINITION AND UNITS OF HEAT 7. A hot steel ball is dropped into a cold aluminum cup
containing some water. (Assume the system is an iso-
1. Discuss the difference between a calorie and a Calorie.
lated.) If the ball loses 400 J of heat, what can be said
2. What is the main difference between internal energy and according to calorimetry?
heat?
3. If someone says that a hot object contains more heat than
a cold one, would you agree? Why?
11.3 PHASE CHANGES AND
LATENT HEAT
11.2 SPECIFIC HEAT AND CALORIMETRY 8. You are monitoring the temperature of some cold ice
cubes 1- 5.0 °C2 in a cup as the ice and cup are heated.
4. At a lake, does the lake water or the lake beach get hotter Initially, the temperature rises, but it stops at 0 °C. After
during a summer day? Which gets colder during a win- a while, it begins rising again. Is anything wrong with
ter night? Explain. the thermometer? Explain.
5. Equal amounts of heat are added to two different objects 9. Discuss the energy conversion in the process of adding
at the same initial temperature. What factors can cause heat to an object that is undergoing a phase change.
the final temperature of the two objects to be different?
10. In general, you would get a more severe burn from
6. Many people have performed firewalking, in which a
steam at 100 °C than from the same mass of hot water at
bed of red-hot coals (temperature over 2000 °F) is
100 °C. Why?
walked on with bare feet. (You should not try this at
home!) How is this possible? [Hint: Human tissues 11. When you breathe out in the winter, you can see your
largely consist of water.] breath, like fog. Explain.
*Neglect heat losses to the external environment in the questions and exercises unless instructed
otherwise, and consider all temperatures to be exact.
412 11 HEAT
11.4 HEAT TRANSFER

12. A plastic ice cube tray and a metal ice cube tray are
removed from the same freezer, at the same initial tem-
perature. However, the metal one feels cooler to the
touch. Why?
13. Why is the warning shown on the highway road sign in
䉴 Fig. 11.20 necessary?
14. Polar bears have an excellent heat insulation system.

(Sometimes even infrared cameras cannot detect them.)
Polar bear hairs are actually hollow inside. Explain how
this helps the bears maintain their body temperature in
the cold winter. 䉱 F I G U R E 1 1 . 2 0 A cold warning See Concep-
15. Explain how the Thermos bottle shown in Fig. 11.17 can tual Question 13.
minimize all mechanisms of heat transfer.
EXERCISES*
11.1 DEFINITION AND UNITS OF HEAT 8. ● A 5.00-g pellet of aluminum reaches a final tempera-
ture of 63 °C when gaining 200 J of heat. What is its ini-
1. ● A window air conditioner has a rating of 20 000 Btu>h.
tial temperature?
What is this rating in watts?
2. ● A person goes on a 1500-Cal-per-day diet to lose
9. ● Blood can carry excess heat from the interior to the
weight. What is his daily energy allowance expressed in surface of the body, where the heat is transferred to the
joules? outside environment. If 0.250 kg of blood at a tempera-
ture of 37.0 °C flows to the surface and loses 1500 J of
3. ● A typical NBA basketball player will do about
heat, what is the temperature of the blood when it flows
3.00 * 106 J of work per hour. Express this work in
back into the interior? Assume blood has the same spe-
Calories.
cific heat as water.
4. ● ● A typical person’s normal metabolic rate (the rate at
10. IE ● ● Equal amounts of heat are added to an aluminum
which food>stored energy is consumed) is about
block and a copper block of different masses to achieve
4 * 105 J>h, and the average food energy in a Big Mac is
the same temperature increase. (a) The mass of the alu-
600 Calories. If a person lived on nothing but Big Macs,
minum block is (1) more, (2) the same, (3) less than the
how many per day would he or she have to eat to main-
mass of the copper block. Why? (b) If the mass of the
tain a constant body weight?
copper block is 3.00 kg, what is the mass of the alu-
5. ● ● A student ate a Thanksgiving dinner that totaled
minum block?
2800 Cal. He wants to use up all that energy by lifting a
20-kg mass a distance of 1.0 m. Assume that he lifts the 11. ●● A modern engine of alloy construction consists of
mass with constant velocity and no work is required in 25 kg of aluminum and 80 kg of iron. How much heat
lowering the mass. (a) How many times must he lift the does the engine absorb as its temperature increases from
mass? (b) If he can lift and lower the mass once every 20 °C to 100 °C as it warms up to operating temperature?
5.0 s, how long does this exercise take? 12. IE ● ● Equal amounts of heat are added to different
quantities of copper and lead. The temperature of the
copper increases by 5.0 °C and the temperature of the
11.2 SPECIFIC HEAT AND CALORIMETRY lead by 10 °C. (a) The lead has (1) a greater mass than the
6. ●It takes 2.0 * 106 J of heat to bring a quantity of water copper, (2) the same amount of mass as the copper,
from 20 °C to a boil. What is the mass of water? (3) less mass than the copper. (b) Calculate the mass ratio
7. IE ● The temperature of a lead block and a copper block, of the lead to the copper to prove your answer to part (a).
both 1.0 kg and at 20 °C, is to be raised to 100 °C. (a) The 13. IE ● ● Initially at 20 °C, 0.50 kg of aluminum and 0.50 kg of
copper will require (1) more heat, (2) the same heat, iron are heated to 100 °C. (a) The aluminum gains (1) more
(3) less heat than the lead. Why? (b) Calculate the differ- heat than the iron, (2) the same amount of heat as the iron,
ence between the heat required for the two blocks to (3) less heat than the iron. Why? (b) Calculate the difference
prove your answer to part (a). in heat required to prove your answer to part (a).
*Assume all temperatures to be exact.
EXERCISES 413
14. ●● A 0.20-kg glass cup at 20 °C is filled with 0.40 kg of hot 11.3 PHASE CHANGES AND
water at 90 °C. Neglecting any heat losses to the environ- LATENT HEAT
ment, what is the equilibrium temperature of the water?
25. ● How much heat is required to melt a 2.5-kg block of ice
15. ●● A 0.250-kg coffee cup at 20 °C is filled with 0.250 kg at 0 °C?
of brewed coffee at 100 °C. The cup and the coffee come
to thermal equilibrium at 80 °C. If no heat is lost to the 26. ●How much heat is required to boil away 1.50 kg of
environment, what is the specific heat of the cup mater- water that is initially at 100 °C?
ial? [Hint: Consider the coffee essentially to be water.] 27. IE ● (a) Converting 1.0 kg of water at 100 °C to steam at
16. ●● An aluminum spoon at 100 °C is placed in a Styro- 100 °C requires (1) more heat, (2) the same amount of
foam cup containing 0.200 kg of water at 20 °C. If the heat, (3) less heat than converting 1.0 kg of ice at 0 °C to
final equilibrium temperature is 30 °C and no heat is lost water at 0 °C. Explain. (b) Calculate the difference in
to the cup itself or the environment, what is the mass of heat required to prove your answer to part (a).
the aluminum spoon? 28. ● Water is boiled to add moisture to the air in the winter to
17. ●● A student doing an experiment pours 0.150 kg of help a congested person breathe better. Calculate the heat
heated copper shot into a 0.375-kg aluminum calorime- required to boil away 1.0 L of water that is initially at 50 °C.
ter cup containing 0.200 kg of water. The cup and water 29. ● An artist wants to melt some lead to make a statue.
are both initially at 25 °C. The mixture (and the cup) How much heat must be added to 0.75 kg of lead at
comes to thermal equilibrium at 28 °C. What was the ini- 20 °C to cause it to melt completely?
tial temperature of the shot?
30. ● First calculate the heat that needs to be removed to
18. ●● At what average rate would heat have to be removed
convert 1.0 kg of steam at 100 °C to water at 40 °C and
from 1.5 L of (a) water and (b) mercury to reduce the
then compute the heat that needs to be removed to lower
liquid’s temperature from 20 °C to its freezing point in
the temperature of water at 100 °C to water at 40 °C.
3.0 min?
Compare the two results. Are you surprised?
19. ●● When resting, a person gives off heat at a rate of
about 100 W. If the person is submerged in a tub contain- 31. ● How much heat is required to completely boil away
ing 150 kg of water at 27 °C and the heat from the person 0.50 L of liquid nitrogen at -196 °C? (Take the density of
goes only into the water, how many hours will it take for liquid nitrogen to be 0.80 * 103 kg>m3.)
the water temperature to rise to 28 °C? 32. IE ● ● An alcohol rub can rapidly decrease body (skin)
20. ●● To determine the specific heat of a new metal alloy, temperature. (a) This is because of (1) the cooler temper-
0.150 kg of the substance is heated to 400 °C and then ature of the alcohol, (2) the evaporation of alcohol,
placed in a 0.200-kg aluminum calorimeter cup contain- (3) the high specific heat of the human body. (b) To
ing 0.400 kg of water at 10.0 °C. If the final temperature decrease the body temperature of a 65-kg person by
of the mixture is 30.5 °C, what is the specific heat of the 1.0 °C, what mass of alcohol must be evaporated from
alloy? (Ignore the calorimeter stirrer and thermometer.) the person’s skin? Ignore the heat involved in raising the
temperature of alcohol to its boiling point (why?) and
21. IE ● ● In a calorimetry experiment, 0.50 kg of a metal at
approximate the human body as water.
100 °C is added to 0.50 kg of water at 20 °C in an alu-
minum calorimeter cup. The cup has a mass of 0.250 kg. 33. IE ● ● Heat has to be removed to condense mercury
(a) If some water splashed out of the cup when the metal vapor at a temperature of 630 K into liquid mercury.
was added, the measured specific heat will appear to be (a) This heat involves (1) only specific heat, (2) only
(1) higher, (2) the same, (3) lower than the value calcu- latent heat, or (3) both specific and latent heats. Explain.
lated for the case in which the water does not splash out. (b) If the mass of the mercury vapor is 15 g, how much
Why? (b) If the final temperature of the mixture is 25 °C, heat would have to be removed?
and no water splashed out, what is the specific heat of 34. ●● If 0.050 kg of ice at 0 °C is added to 0.300 kg of water
the metal? at 25 °C in a 0.100-kg aluminum calorimeter cup, what is
22. ●●● Lead pellets of total mass 0.60 kg are heated to the final temperature of the water?
100 °C and then placed in a well-insulated aluminum 35. ●● How much ice (at 0 °C) must be added to 0.500 kg of
cup of mass 0.20 kg that contains 0.50 kg of water ini- water at 100 °C in a 0.200-kg aluminum calorimeter cup
tially at 17.3 °C. What is the equilibrium temperature of to end up with all liquid at 20 °C?
the mixture?
36. ●● Ice (initially at 0 °C) is added to 0.75 L of tea at 20 °C
23. ●●● A student mixes 1.0 L of water at 40 °C with 1.0 L of
to make the coldest possible iced tea. If enough ice is
ethyl alcohol at 20 °C. Assuming that no heat is lost to
added so the final mixture is all liquid, how much liquid
the container or the surroundings, what is the final tem-
is in the pitcher when this condition occurs?
perature of the mixture? [Hint: See Table 11.1.]
24. ●●● We all have had the experience that a room full of 37. ●● To cool a very hot piece of 4.00-kg steel at 900 °C, the
people always feels warmer than when the room is steel is put into a 5.00-kg water bath at 20 °C . What is
empty. Ten people are in a 4.0 m * 6.0 m * 3.0 m room the final temperature of the steel-water mixture?
at 20 °C. If each person gives off heat at a rate of about 38. ●● Steam at 100 °C is bubbled into 0.250 kg of water at
100 W and there is no heat loss to the outside of the 20 °C in a calorimeter cup, where it condenses into liq-
room, what is the temperature of the room after 10 min? uid form. How much steam will have been added when
At 20 °C, the density of air is 1.2 kg>m3 and its specific the water in the cup reaches 60 °C? (Ignore the effect of
heat at constant pressure is 1005 J>1kg # °C2. the cup.)
414 11 HEAT
39. IE ● ● Evaporation of water from our skin is a very 45. IE ● A house can have a brick wall or a concrete wall
important mechanism for controlling body temperature. with the same thickness. (a) Compared with the concrete
(a) This is because (1) water has a high specific heat, wall, the brick wall will conduct heat away from the
(2) water has a high latent heat of vaporization, (3) water house (1) faster, (2) at the same rate, (3) slower. Why?
contains more heat when hot, (4) water is a good heat (b) Calculate the ratio of the rate of heat flow of the brick
conductor. (b) In a 3.5-h intense cycling race, a cyclist can wall to that of the concrete wall.
loses 7.0 kg of water through perspiration. Estimate how 46. ● Assume a goose has a 2.0-cm-thick layer of feather
much heat the cyclist loses in the process. down (on average) and a body surface area of 0.15 m2.
40. IE ● ● ● A 0.400-kg piece of ice at - 10 °C is placed in an What is the rate of heat loss (conduction only) if the
equal mass of water at 30 °C. (a) When thermal equilib- goose, with a body temperature of 41 °C, is outside on a
rium is reached between the two, (1) all the ice will melt, winter day when the air temperature is 11 °C?
(2) some of the ice will melt, (3) none of the ice will melt. 47. ● Assume that your skin has an emissivity of 0.70, a nor-
(b) How much ice melts? mal temperature of 34 °C, and a total exposed area of
41. ● ● ● One kilogram of a substance experimentally shows 0.25 m2. How much heat energy per second do you lose
the T-versus-Q graph in 䉲 Fig. 11.21. (a) What are its due to radiation if the outside temperature is 22 °C?
melting and boiling points? In SI units, what are (b) the 48. ● The U.S. five-cent coin, the nickel, has a mass of 5.1 g,
specific heats of the substance during its various phases a volume of 0.719 cm3, and a total surface area of
and (c) the latent heats of the substance at the various 8.54 cm2. Assuming that a nickel is an ideal radiator, how
phase changes? much radiant energy per second comes from the nickel,
if it is at 20 °C?
160 49. IE ● ● An aluminum bar and a copper bar of identical
cross-sectional area have the same temperature differ-
150 ence between their ends and conduct heat at the same
Temperature (°C)
rate. (a) The copper bar is (1) longer, (2) of the same
140 length, (3) shorter than the aluminum bar. Why? (b) Cal-
culate the ratio of the length of the copper bar to that of
130 the aluminum bar.
50. ●● A copper teakettle has a circular bottom 30.0 cm in
120 diameter that has a uniform thickness of 2.50 mm. It sits
on a burner whose temperature is 150 °C. (a) If the
110 teakettle is full of boiling water, what is the rate of heat
conduction through its bottom? (b) Assuming that the
100 heat from the burner is the only heat input, how much
0.20 0.40 0.60 0.80 1.0 1.2 1.4 1.6 1.8 2.0
water is boiled away in 5.0 min? Is your answer unrea-
Q (× 104 J)
sonably large? If yes, explain why.
䉱 F I G U R E 1 1 . 2 1 Temperature versus heat input See Exer- 51. ●● Assuming that the human body has a 1.0-cm-thick
cise 41. layer of skin tissue and a surface area of 1.5 m2, estimate
the rate at which heat is conducted from inside the body
to the surface if the skin temperature is 34 °C. (Assume a
42. ● ● ● In an experiment, a 0.150-kg piece of a ceramic normal body temperature of 37 °C for the temperature of
material at 20 °C is placed in liquid nitrogen at its boiling the interior.)
point to cool in a perfectly insulated flask, which allows 52. IE ● ● The emissivity of an object is 0.50. (a) Compared
the gaseous N2 to immediately escape. How many liters with a perfect blackbody at the same temperature, this
of liquid nitrogen will be boiled away during this opera- object would radiate (1) more power, (2) the same
tion? (Take the specific heat of the ceramic material to be amount of power, (3) less power. Why? (b) Calculate the
that of glass and the density of liquid nitrogen to be ratio of the power radiated by the blackbody to that radi-
0.80 * 103 kg>m3.) ated by the object.
53. ●● A lamp filament radiates energy at a rate of 100 W
when the temperature of the surroundings is 20 °C, and
11.4 HEAT TRANSFER only 99.5 W when the surroundings are at 30 °C. If the
43. The single glass pane in a window has dimensions of
●
temperature of the filament is the same in each case,
2.00 m by 1.50 m and is 4.00 mm thick. How much heat what is its temperature in Celsius?
will flow through the glass in 1.00 h if there is a tempera- 54. IE ● ● (a) If the Kelvin temperature of an object is dou-
ture difference of 2 °C between the inner and outer sur- bled, its radiated power increases by (1) 2, (2) 4, (3) 8,
faces? (Consider conduction only.) (4) 16 times. Explain. (b) If its temperature is increased
44. IE ● Assume that a tile floor and an oak floor each have from 20 °C to 40 °C, by how much does the radiated
the same temperature and thickness. (a) Compared with power change?
the oak floor, the tile floor will conduct heat away from 55. ●● A certain object with a surface temperature of 100 °C
your bare feet (1) faster, (2) at the same rate, (3) slower. is radiating heat at a rate of 200 J>s. To double the
Why? (b) Calculate the ratio of the rate of heat flow of object’s rate of radiation energy, what should be its sur-
the tile floor to that of the oak floor. face temperature in Celsius?
56. IE ● ● The thermal insulation used in building is com- 60. ● ● ● The lowest natural temperature ever recorded on
monly rated in terms of its R-value, defined as d>k, where d the Earth was at Vostok, a Russian Antarctic station,
is the thickness of the insulation in inches and k is its ther- when a temperature of - 89.4 °C 1 - 129 °F2 was recorded
mal conductivity. (See Insight 11.2 on p. 403.) In the on July 21, 1983. A typical person has a body tempera-
United States, R-values are expressed in British units. For ture of 37.0 °C, skin tissue 0.0250 m thick, and a total
example, 3.0 in. of foam plastic would have an R-value of skin surface area of 1.50 m2. (a) What would be the rate
3.0>0.30 = 10, where k = 0.30 Btu # in.>1ft 2 # h # °F2. This of heat loss of a naked human? (b) What would be the
value is expressed as R-10. (a) Better insulation has a rate of heat loss of a human wearing a 0.100-m-thick
(1) high, (2) low, or (3) zero R-value. Explain. (b) What goose down jacket and pants capable of covering the
thicknesses of (1) styrofoam and (2) brick would give an whole body?
R-value of R-10?
61. ● ● ● The wall of a house is composed of a solid concrete
57. IE ● ● A piece of pine 14 in. thick has an R-value of 19. block with an outside brick veneer and is faced on the
(a) For glass wool to have the same R-value, its thickness inside with fiberboard, as illustrated in 䉲 Fig. 11.23. If the
should be (a) thicker than, (2) the same as, (3) thinner outside temperature on a cold day is - 10 °C and the
than 14 in. Why? (b) Calculate the required thickness of inside temperature is 20 °C, how much energy is con-
such a piece of glass wool. (See Exercise 56 and ducted through the wall in 1.0 h if it measures 3.5 m by
Insight 11.2.) 5.0 m?
58. ● ● Solar heating takes advantage of solar collectors such
as the type shown in 䉲 Fig. 11.22. During daylight hours,
the average intensity of solar radiation at the top of the
Fiberboard
atmosphere is about 1400 W>m2. About 50% of this radi-
ation reaches the Earth during daylight hours. (The rest
is reflected, scattered, absorbed, and so on.) How much
heat energy would be received, on average, by the cylin-
drical collector shown in the figure during 10 h of day-
light?
Concrete
4.0 m Brick
r = 0.50 m
15.0 cm 7.0 cm
2.0
cm
䉱 F I G U R E 1 1 . 2 3 Thermal conductivity and heat loss
See Exercise 61.
䉱 F I G U R E 1 1 . 2 2 Solar collector and solar heating
62. ● ● ● Suppose you wished to cut the heat loss through
See Exercise 58.
the wall in Exercise 61 in half by installing insulation.
What thickness of Styrofoam should be placed between
For Exercises 59–64, read Example 11.7
the fiberboard and concrete block to accomplish this
and the footnote on p. 402.
goal?
59. ● ● ● A large window measures 2.0 m by 3.0 m. At what
63. ● ● ● A steel cylinder of radius 5.0 cm and length 4.0 cm
rate will heat be conducted through the window when
is placed in end-to-end thermal contact with a copper
the room temperature is 20 °C and the outside tempera-
cylinder of the same dimensions. If the free ends of the
ture is 0 °C if (a) the window consists of a single pane of
two cylinders are maintained at constant temperatures
glass 4.0 mm thick and (b) the window instead has a
of 95 °C (steel) and 15 °C (copper), how much heat will
double pane of glass (a “thermopane”), in which each
flow through the cylinders in 20 min?
pane is 2.0 mm thick, with an intervening air space of
1.0 mm? (Assume that there is a constant temperature 64. ● ● ● In Exercise 63, what is the temperature at the inter-
difference and consider conduction only.) face of the cylinders?
65. A 0.60 kg piece of ice at 14 °F is placed in 0.30 kg of water 66. A large Styrofoam cooler has a surface area of 1.0 m2 and a
at 323 K. How much liquid is left when the system thickness of 2.5 cm. If 5.0 kg of ice at 0 °C is stored inside and
reaches thermal equilibrium? the outside temperature is a constant 35 °C, how long does
it take for all the ice to melt? (Consider conduction only.)
416 11 HEAT
67. A 1600-kg automobile traveling at 55 mph brakes 70. A cyclist with a total skin area of 1.5 m2 is riding a bicy-
smoothly to a stop. Assume 40% of the heat generated in cle on a day when the air temperature is 20 °C and her
stopping the car is dissipated in the front steel brake skin temperature is 34 °C. The cyclist does work at about
disks. Each front disk has a mass of 3.0 kg. What is the 200 W (moving the pedals) but her efficiency is only
temperature rise of the front brake disks during the about 20% in terms of converting energy into mechanical
stop? work. Estimate the amount of water this cyclist must
68. A waterfall is 75 m high. If 20% of the gravitational poten- evaporate per hour (through perspiration) to get rid of
tial energy of the water went into heating the water, by the excess body heat she produces. Assume a skin emis-
how much would the temperature of the water, increase sivity of 0.70.
in going from the top of the falls to the bottom? [Hint: 71. A 200-kg cast iron machine part at 500 °C is left to cool at
Consider a kilogram of water going over the falls.] room temperature. Assume the machine part is a cube
69. A 0.030-kg lead bullet hits a steel plate, both initially at and has an emissivity of 0.780. At what rate is the
20 °C. The bullet melts and splatters on impact. (This machine part initially losing heat due to radiation? [Hint:
action has been photographed.) Assuming that 80% of The density of iron can be found in Table 9.2.]
the bullet’s kinetic energy goes into increasing its tem-
perature and then melting it, what is the minimum speed
it must have to melt on impact?
12 Thermodynamics †
12.1 Thermodynamic systems,

states, and processes (418)
12.2 The first law of

thermodynamics (420)
■ conservation of energy
12.3 Thermodynamic processes

for an ideal gas (424)
■ isothermal; isobaric;
isometric; adiabatic
12.4 The second law of

thermodynamics and
entropy (431)
PHYSICS FACTS
A
■ entropy change
✦ An automobile with a typical ther-
s the word implies,
■entropy increase
(natural process) modynamic efficiency of one-fifth thermodynamics deals with
will lose about one-third of its
energy through the exhaust, the transfer (dynamics) of heat (the
another one-third to the coolant,
and about one-tenth to the sur-
Greek word for “heat” is therme).
12.5 Heat engines and
thermal pumps (436) roundings. The development of thermo-
✦ In Europe, more than 52% of cars
■ efficiency
sold in the first half of 2007 were
dynamics started about 200 years
■ COP
diesel-powered. In the United ago out of efforts to develop heat
States, fewer than 3% of cars sold
are diesel-powered, but diesel engines. The steam engine was
sales are projected to triple to 9%
12.6 The carnot cycle and by 2013 due to diesel engine’s
one of the first such devices,
ideal heat engines (443) higher efficiency. designed to convert heat to
■ Carnot efficiency
✦ The efficiency of the human body
can be as high as 20% when large
mechanical work. Steam engines in
muscle groups, such as leg muscles factories and locomotives powered
are used, but as low as 3% to 5%
when only the small muscle groups, the Industrial Revolution, which
such as arm muscles, are used.
changed the world.
✦ The brain makes up 2% of a per-
son’s weight, but consumes 20% Automobiles are very useful
of the body’s energy. The average
†
The mathematics needed in this chapter power consumption of a typical
tools for civilized society. However,
involves natural logarithms (ln) and common
logarithms (log). You may want to review
adult is 100 W, with the brain con- with decreasing resources, soaring
suming 20 W.
these in Appendix I.
oil prices, and concern about
418 12 THERMODYNAMICS
increasing greenhouse gas emissions, automobile manufacturers have been

striving to produce cars using the highest possible thermodynamic efficiencies.
The 2010 Honda Accord Diesel is powered by a clean 2.2-liter diesel engine,
shown in the chapter-opening photograph. Its real-world fuel economy exceeds
50 mpg (miles per gallon) highway. Honda has said that the car has achieved
62.8 mpg in tests.
In this chapter, you’ll learn under what conditions, and with what efficiency,
heat can be exploited to perform work in the human body and in machines as dif-
ferent as automobile engines and home freezers. The laws governing such energy
conversions include some of the most general and far-reaching laws in all of
physics. Although our study is primarily concerned with heat and work, thermody-
namics is a broad and comprehensive science that includes a great deal more than
heat engine theory. In this chapter, the laws on which thermodynamics is based, as
well as the concept of entropy, will be presented.
12.1 Thermodynamic Systems, States, and Processes

➥ What is a thermodynamic system?

➥ What is meant by the state of a thermodynamic system?
➥ What is a thermodynamic process?
Thermodynamics is a field that describes systems with so many particles—think

of the number of molecules in a gas sample—that using ordinary dynamics
(Newton’s laws) to keep track of them is impossible. Therefore, even though the
underlying physics is the same as for other systems, we generally use alternative
(macroscopic) variables, such as pressure and temperature, to describe thermody-
namic systems as a whole. Because of this difference in language, it is important to
become familiar with the terms and definitions at the outset.
The term system, as used in thermodynamics, refers to a definite quantity of
matter enclosed by boundaries or surfaces, either real or imaginary. For example,
a quantity of gas in the piston cylinder of an engine has real boundaries, and
imaginary boundaries enclose a cubic meter of air in a room.
The interchange of energy between a system and its surroundings is very
important. This exchange may occur through a transfer of heat and>or the perfor-
mance of mechanical work. For example, if a gasoline and air mixture is ignited
inside a piston of an engine, it can expand and do work by exerting a force
through a distance on the piston.
If no heat is transferred into or out of a system, it is said to be a thermally
isolated system. However, work may be done on a thermally isolated system,
thus transferring energy to it. For example, a thermally isolated syringe (perhaps
surrounded by heavy insulation) filled with gas can be compressed by an external
force exerted on a plunger. Work is thus done on the system, and as we know,
work is a way of transferring energy.
When heat does enter or leave a system, it is usually taken in from or given up
to the surroundings, or to what is called a heat reservoir. A heat reservoir is a sys-
tem assumed to have unlimited heat capacity. Any amount of heat can be with-
drawn from or added to a heat reservoir without appreciably changing its
temperature. For example, pouring a bottle of warm water into a cold lake does
not noticeably raise the lake’s temperature. This cold lake is an example of a low-
temperature heat reservoir.
12.1 THERMODYNAMIC SYSTEMS, STATES, AND PROCESSES 419
STATE OF A SYSTEM y
Just as there are kinematic equations to describe the motion of an object, there are
equations of state to describe the conditions of thermodynamic systems. Such an
equation expresses a mathematical relationship between the thermodynamic vari-
ables that describe the system. The ideal gas law, pV = nRT (Section 10.3), is an (x, y)
example of an equation of state. This expression establishes a relationship among
the pressure (p), volume (V), absolute temperature (T), and number of moles (n, or
equivalently, N, the number of molecules, since from Section 10.3, N = nNA) of a
gas. These ideal gas quantities are examples of state variables. Clearly, different x
states have different sets of values for these variables. (a)
For a quantity of ideal gas, a set of these three variables (p, V, and T) that satis- p
fies the ideal gas law specifies its state completely as long as the system is in ther-
mal equilibrium. Such a system is said to be in a definite state. It is convenient to
plot the states according to the thermodynamic coordinates (p, V, T), much as
graphs using Cartesian coordinates (x, y, z) are plotted. A general two-dimen- (V, p)
sional illustration of such a plot is shown in 䉴 Fig. 12.1.
Just as the coordinates (x, y) specify individual points on a Cartesian graph, the
coordinates (V, p) specify individual states on the p–V graph or diagram. This is
because the ideal gas law, pV = nRT, can be solved for the unique temperature of a
gas if the gas’s pressure, volume, and number of molecules or moles in the sample V
are known. In other words, on a p–V diagram, each “coordinate” or point gives the (b)
pressure and volume of a gas directly, and the temperature of the gas indirectly.
䉱 F I G U R E 1 2 . 1 Graphing states
Thus, to describe a gas completely, only a p–V plot is necessary. In some cases, (a) On a Cartesian graph, the coor-
however, it can be instructive to refer to other plots, such as p–T or T–V plots. dinates (x, y) represent an individ-
(Notice that Fig. 12.1b could illustrate a phenomenon that you might be familiar ual point. (b) Similarly, on a p–V
with—reduction of the pressure of a gas, resulting in its expansion.) graph or diagram, the coordinates
(V, p) represent a particular state of
a system. (It is common to say p–V,
rather than V–p, because the plot is
a p vs. V graph.)
PROCESSES
A process is any change in the state, or the thermodynamic coordinates, of a sys-
tem. For instance, when an ideal gas undergoes a process, its state variables p,
V, and T will, in general, all change. Suppose a gas initially in state 1, described
by state variables (p1 , V1 , T1 ), changes to a second state, state 2. Then state 2
will, in general, be described by a different set of state variables (p2 , V2 , T2 ). A
system that has undergone a change of state has been subjected to a
thermodynamic process. p
? 2
Processes are classified as either reversible or irreversible. Suppose that a sys-
tem of gas in equilibrium (with known p, V, and T values) is allowed to expand 1 ?
quickly when the pressure on it is reduced. The state of the system will change
rapidly and unpredictably, but eventually the system will reach a different state of ?
3
equilibrium, with another set of thermodynamic coordinates. On a p–V diagram
(䉴 Fig. 12.2), the initial and final states (labeled 1 and 2, respectively) are known,
but what happened in between them is not. This type of process is called an
4
irreversible process—a process for which the intermediate steps are nonequilib-
rium states. “Irreversible” does not mean that the system can’t be taken back to V
the initial state; it means only that the process path can’t be retraced, because of
the nonequilibrium conditions that existed. An explosion is an example of an irre- 䉱 F I G U R E 1 2 . 2 Paths of
reversible and irreversible processes
versible process. If a gas quickly goes from state 1 to
If, however, the gas changes state very, very slowly, passing from one equilib- state 2, the process is irreversible,
rium state to a neighboring one and eventually arriving at the final state (see since we do not know the “path.” If,
Fig. 12.2, initial and final states 3 and 4, respectively), then the process path is however, the gas is taken through
known. In such a situation, the system could be brought back to its initial condi- many closely spaced equilibrium
states (as in going from state 3 to
tions by “traveling” the path in the opposite direction, re-creating every interme- state 4), the process is reversible in
diate state (again, in many small steps) along the way. Such a process is called a principle. Reversible means “exactly
reversible process. In practice, a perfectly reversible process cannot be achieved. retraceable.”
All real thermodynamic processes are irreversible to some degree, because they
follow complicated paths with many intermediate nonequilibrium states. How-
ever, the concept of an ideal reversible process is useful and will be the primary
tool in discussing the thermodynamics of an ideal gas.
DID YOU LEARN?

➥ A thermodynamic system is simply a quantity of matter enclosed in real or
imaginary boundaries or surfaces.
➥ The state of a thermodynamic system describes the conditions of the system. Some
common variables used to specify the state of an ideal gas are pressure (p), volume
(V), and temperature (T).
➥ A thermodynamic process changes the system from one state to another. For an
ideal gas, a process changes one set of values for pressure, volume, and
temperature (p,V,T) to another set.
12.2 The First Law of Thermodynamics

➥ What is the first law of thermodynamics?

➥ How can the internal energy of a system be changed?
➥ How can the work done by an ideal gas based on a process curve on a p–V diagram
be calculated?
Recall from Section 5.1 that work describes the transfer of energy from one object
to another by application of a force. For example, when you push on a chair ini-
tially at rest and set it into motion, some of the work done on the chair (exerting a
force through a distance) goes into increasing its kinetic energy. At the same time,
you lose stored (chemical) energy in your body in doing so. For example, when a
gas (enclosed in a cylinder and fitted with a piston) is allowed to expand, the gas
does work on the piston at the expense of some of its internal energy. From
Chapters 10 and 11, we know there is a second way to change the energy of a
system—by adding or removing heat energy. Thus, internal energy is lost by a hot
object when the heat is transferred to a cold object, which then gains internal
energy. This process changes both objects’ internal energies, but in opposite ways.
Although the actual process cannot be seen, heat transfer is really the same con-
cept as mechanical work, but on a microscopic (atomic) level. During a conduction
process, for example, energy is transferred from a hot object to a cold object,
because the faster-vibrating atoms of the hot object do work on the slower atoms
of the cold object (䉲 Fig. 12.3). This energy is then transferred farther into the vol-
ume of the cold object as more work is done on the neighboring (slower-vibrating)
atoms. This ongoing process is the “flow” or “transfer” of energy observed macro-
scopically as heat transfer.
Cold
Hot
object
object
Q
(a) (b)
䉱 F I G U R E 1 2 . 3 Heat flow (via conduction) on the atomic scale (a) Macroscopically, heat
is transferred by conduction from the hot object to the cold one. (b) On the atomic scale,
heat conduction is explained as the energy transfer from the more energetic atoms (in the
hot object) to the less energetic atoms (in the cold object). This transfer of energy from an
atom to its neighbor results in the heat transfer observed in part (a).
12.2 THE FIRST LAW OF THERMODYNAMICS 421
∆T > 0; ∆U > 0
Added ∆T < 0; ∆U < 0

Q>0
(b)
Removed
Q<0
(a)
Compression, (c) Expansion,

W<0 W>0
䉱 F I G U R E 1 2 . 4 Sign conventions for Q, W, and ¢U (a) If heat flows into a system, Q is

positive. For heat flowing out, Q is designated as negative. (b) The experimental way to tell
if a gas’s internal energy changes is to take its temperature, assuming there is no phase
change. Since internal energy is determined by temperature, a rise or fall in one of these
quantities implies a similar rise or fall in the other. (c) If a gas expands, the work W it does
is positive. If the gas is compressed, the work done by the gas is negative.
The first law of thermodynamics describes how work and heat are related to a
system’s internal energy. This law is a statement of energy conservation in terms of
thermodynamic variables. It relates the change in internal energy 1¢U2 of a system
to the work (W) done by or on that system and the heat energy transferred (Q) to or
from that system. Depending on the conditions, heat transfer Q can result in a
change in that system’s internal energy, ¢U. However, because of the heat trans-
fer, the system might do work on the environment. Thus, heat transferred to a sys-
tem can end up in change in the internal energy of the system and>or work done
by the system. This is just a statement of energy conservation. Therefore, the first
law of thermodynamics can be written as
Q = ¢U + W (the first law of thermodynamics) (12.1)
As always, it is important to remember what the symbols mean and what their
sign conventions denote (shown in 䉱 Fig. 12.4). Q is the net heat added to or removed
from the system, ¢U is the change in internal energy of the system and W is the work
done by the system (on the environment).* For example, a sample of gas may
absorb 1000 J of heat and do 400 J of work on the environment, thus leaving 600 J
as the increase in the gas’s internal energy. If the gas were to do more than 400 J of
work, less energy would go to the internal energy of the gas. The first law does not
tell you the values of ¢U or W in processes. These amounts depend, as will be
seen, on the system’s conditions or the specific process involved (constant pres-
sure, constant volume, and so on) as the heat energy is transferred (Section 12.3).
It is important to note that heat flow is not necessary for temperature to change.
When a soda bottle is opened, as shown in 䉴 Fig. 12.5, the gas inside the bottle
expands because it is at a higher pressure than the atmosphere. In doing so it does
(positive) work on the surroundings (the atmospheric gases) and its internal
energy decreases. This is because the net heat flow is zero in this process. Since
¢U = Q - W, then ¢U is negative (U decreases) if Q = 0 and W is positive.
This reduction in internal energy will result in a temperature drop which in turn
will cause the water vapor in the bottled gas to condense into a cloud of tiny liquid
*In some chemistry and engineering books, the first law of thermodynamics is written as
Q = ¢U - W¿ . The two equations are the same, but each has a different emphasis. In this expression, 䉱 F I G U R E 1 2 . 5 Temperature
W¿ means the work done by the environment on the system and is thus the negative of our work W decrease without removing heat The
(why?), or W = - W¿ . The first law was discovered by researchers interested in building heat engines gas does positive work on the outside
(Sections 12.5 and 12.6). Their emphasis was on finding the work done by the system, W, not W¿ . Since air upon the opening of the bottle.
our main concern is to understand heat engines, the historical definition is adopted: W means the work This results in a decrease of both its
done by the system. internal energy and temperature.
water droplets. You can demonstrate this high-pressure cooling by putting your
palm near your mouth and blowing air with your mouth opened wide. You feel a
gush of warm air (roughly at body temperature). However, if you repeat this with
your lips puckered so as to increase the pressure, the air will feel cooler.
Although the above discussion of the first law of thermodynamics was primar-
ily about gas systems, the law holds for any systems. Consider the application of
the first law of thermodynamics to exercise and weight loss in Example 12.1.
EXAMPLE 12.1 Energy Balancing: Exercising Using Physics

A 65-kg worker shovels coal for 3.0 h. During the shoveling, T H I N K I N G I T T H R O U G H . Since the time duration of the shov-
the worker did work at an average rate of 20 W and lost heat eling, the rate of work being done (power), and the rate of
to the environment at an average rate of 480 W. Ignoring the heat loss are known, the total work done and the heat can be
loss of water by the evaporation of perspiration from his skin, calculated. Then the change in internal energy can be found
how much fat will the worker lose? The energy value of fat using the first law of thermodynamics. This change in inter-
(Ef) is 9.3 kcal>g. nal energy (a decrease) results in a loss of fat.
SOLUTION. Listing the given values, and converting power to work and heat units:
Given: W = Pt = 120 J>s213.0 h213600 s>h2 = 2.16 * 105 J Find: m (mass of fat burned)
(W is positive because work is done by the worker)
Q = - 1480 J>s213.0 h213600 s>h2 = - 5.18 * 106 J
(Q is negative because heat is lost)
Ef = 9.3 kcal>g = 9.3 * 103 kcal>kg = 19.3 * 103 kcal>kg214186 J>kcal2
= 3.89 * 107 J>kg
From the first law of thermodynamics, Q = ¢U + W.

¢U = Q - W = - 5.18 * 106 J - 2.16 * 105 J = - 5.40 * 106 J
Thus the mass of fat loss is
ƒ ¢U ƒ 5.40 * 106 J
m = = = 0.14 kg
Ef 3.89 * 107 J>kg
That is about a third of a pound, or about 5 ounces (140 g).
F O L L O W - U P E X E R C I S E . How much fat would be lost if the worker were playing basketball for 3.0 h, doing work at a rate of
120 W and generating heat at a rate of 600 W? (Answers to all Follow-Up Exercises are given in Appendix VI at the back of the book.)
When applying the first law of thermodynamics, the proper use of signs
(shown in Fig. 12.4) cannot be overemphasized. The signs for work are easy to
remember if you keep in mind that positive work is done by a force that acts gen-
erally in the direction of the displacement, such as when a gas expands. Similarly,
Final negative work means that the force acts generally opposite to the direction of the
piston
position displacement, as when a gas contracts.
But how do you compute the work done by the gas? To answer this question,
∆x ∆V = A∆x
A consider a cylindrical piston with end area A, containing a known sample of gas
(䉳 Fig. 12.6). Let us imagine that the gas is allowed to expand over a very small dis-
Initial
F = pA piston tance ¢x. If the volume of the gas does not change appreciably, then the pressure
position remains constant. In moving the piston slowly and steadily outward, the gas does
positive work on the piston. Thus from the definition of work,
W = F¢x cos u = F¢x cos 0° = F¢x
W = F∆x = pA∆x = p∆V
In terms of pressure, P = F>A or F = pA. Substituting for F, we have
䉱 F I G U R E 1 2 . 6 Work in thermo-
dynamic terms If a gas expands by a W = pA¢x
very small amount and does so
slowly, its pressure remains con- But A¢x is the volume of a cylinder with end area A and height ¢x. Here, that vol-
stant. The small amount of work ume represents the change in volume of the gas, or ¢V = A¢x, and
done by the gas is p¢V.
W = p¢V
12.2 THE FIRST LAW OF THERMODYNAMICS 423
Note that the work done in Fig. 12.6 is positive because ¢V is positive. If the gas p
contracts, the work is negative because the volume change is negative 1¢V 6 02. area = W = p∆V
Of course, gases don’t always change their volumes by small amounts and
aren’t usually subject to constant pressure. In fact, changes in volume and pres-
sure can be significant. How is the calculation of work handled under these cir-
cumstances? The answer is seen in 䉴 Fig. 12.7. Here, we have a reversible path on a
p–V diagram. Notice that during each small step, the pressure remains approxi- p
mately constant. Therefore, for each step, we approximate the work done by p¢V.
Graphically, this quantity is just the area of a small narrow rectangle, extending
from the process curve to the V-axis. To approximate the total work, we add up
these small amounts of work or W L ©1p¢V2. To get an exact value, think of the V
area as made up of a very large number of very thin rectangles. As the number of V1 ∆V V2
rectangles becomes infinitely large, each rectangle’s thickness approaches zero. (a)
This process involves calculus and is beyond the scope of this book. However, it
should be clear that the following is true: p
The work done by a system is equal to the area under the process curve on a p–V diagram. total W = area under curve
Before discussing specific types of processes, note that there is a fundamental

difference between U and both Q and W. Any system “contains” a certain amount
of internal energy U. However, it is wrong to say that a system “possesses” certain
“amounts” of heat or work, as these quantities represent energy transfers, not total
energies. A further distinction is that both Q and W depend on the path the gas
takes from its initial to its final state, whereas ¢U does not. The heat added to, or
removed from, a system depends on the conditions under which this transfer is done
(Section 11.2). Similarly, from the area-under-the-curve representation, the work V
depends on the path (䉲 Fig. 12.8). For example, more work is done if the process takes V1 V2
place at higher pressures. This situation is represented as a larger area, with more (b)
work done by a larger force with the same volume change.
Contrast these properties with those of ¢U for an ideal gas, for example. To find 䉱 F I G U R E 1 2 . 7 Thermodynamic
¢U, we need know only the internal energies at the ends of the path. This is because work as the area under the process
curve (a) If a gas expands by a sig-
for an ideal gas (with a fixed number of moles), the internal energy depends only on nificant amount, the work done can
the absolute temperature of the gas. For that case (see Section 10.5), U r nRT; thus, be computed by treating the expan-
¢U = U2 - U1 depends only on ¢T. To summarize, the change in the internal sion in little steps, each one yielding
energy, ¢U, is independent of the process path for an ideal gas, whereas Q and W a small amount of work. The total
both depend on the path. work is determined (approximately)
by adding up the many rectangular
strips. (b) If the number of rectangu-
DID YOU LEARN? lar strips becomes large, and each
➥ The first law of thermodynamics is the law of conservation of energy applied to one becomes very thin, the calcula-
thermodynamic systems. It relates the change in internal energy ¢U, work W, and tion of the area becomes exact. The
heat Q. work done is equal to the area
➥ The internal energy of a system can be changed by either work done or between the process curve and the
heat exchange. V-axis.
➥ The work done by an ideal gas is equal to the area under the process curve on a
p–V diagram.
p 䉳 F I G U R E 1 2 . 8 Thermodynamic work
depends on the process path This graph
Ι 2 shows the work done by a gas as it expands
p2 the same amount, but by three different
processes. The work done during process I
1 ΙΙ is larger than the work during process II,
Pressure
p1
which in turn is larger than the work during
ΙΙΙ process III. Fundamentally, applying a
larger force (pressure) through the same dis-
tance (volume change) requires more work.
Process I includes the blue, green, and pink
areas; process II includes just the green and
V pink areas; and process III includes just the
V1 V2
pink area.
Volume
12.3 Thermodynamic Processes for an Ideal Gas

➥ What are the four important thermodynamic processes?

➥ What is the change in internal energy of an ideal gas after an isothermal process?
➥ What is the work done by a gas in an isometric process?
The first law of thermodynamics can be applied to several processes for a system
consisting of an ideal gas. Note that in three of the processes, one thermodynamic
variable is kept constant. Such processes have names that begin with iso- (from the
Greek isos, meaning “equal”).
ISOTHERMAL PROCESS
An isothermal process is a constant-temperature process (iso for equal, thermal for
temperature). In this case, the process path is called an isotherm, or a curve of constant
temperature. (See 䉲 Fig. 12.9.) The ideal gas law may be rewritten as p = nRT>V.
Since the gas remains at constant temperature, nRT is a constant. Therefore, p is
inversely proportional to V—that is, p r 1>V, which is a hyperbola. (Recall that a
hyperbola is written as y = a>x or y r 1>x, and it plots as a downward curve.)
In the expansion from state 1 (initial) to state 2 (final) in Fig. 12.9, heat is added
to the system, while both the pressure and volume vary in such a way as to keep
the temperature constant. Positive work is done by the expanding gas. On an
isotherm, ¢T = 0; therefore, ¢U = 0. The heat added to the gas is exactly equal to
the amount of work done by the gas, and none of the heat goes into increasing the
gas’s internal energy. See the Learn by Drawing 12.1, Leaning on Isotherms.
In terms of the first law of thermodynamics,
Q = ¢U + W = 0 + W
or
Q = W (ideal gas isothermal process) (12.2)
p
Isothermal
T 2 = T1
1 Isotherm
Pressure
T2 = T1
V
V1 V2 Volume
Q V1
V2 W=Q
䉱 F I G U R E 1 2 . 9 Isothermal (constant temperature) process All of the heat added to the

gas goes into doing work (the expanding gas moves the piston): Because ¢T = 0, then
¢U = 0, and from the first law of thermodynamics, Q = W. As always, the work is equal
to the area (shaded) under the isotherm on the p–V diagram.
12.3 THERMODYNAMIC PROCESSES FOR AN IDEAL GAS 425
The magnitude of the work done by the gas is equal to the area under the curve
(requiring calculus to compute), which can be written as follows.
V2
Wisothermal = nRT ln ¢ ≤ (ideal gas isothermal process) (12.3)
V1
Since the product nRT is a constant along a given isotherm, the work done
depends on the ratio of the endpoint volumes.
In Eq. 12.3, the function “ln” stands for natural logarithm. Recall that common logarithms
(“log”) are referenced to the base 10 (see Appendix I). For this type, the exponent of the
base 10 is the logarithm of the number in question. For example, 100 = 102, so the loga-
rithm of 100 is 2, or, in equation terms, log 100 = 2. In general, if y = 10x, then x is the
logarithm of y, or x = log y. The natural logarithm is similar, except it uses a different
base, e, which is an irrational number 1e L 2.71832. As a check, find the natural loga-
rithm of 100 on your calculator. (The answer is ln 100 = 4.605).
ISOBARIC PROCESS
A constant-pressure process is called an isobaric process (iso for equal, and bar for
pressure).* An isobaric process for an ideal gas is illustrated in 䉲 Fig. 12.10. On a
p–V diagram, an isobaric process is represented by a horizontal line called an
isobar. When heat is added to or removed from an ideal gas at constant pressure,
the ratio V>T remains constant 1V>T = nR>p = constant2. As the heated gas
expands, its temperature must increase, and the gas crosses to higher temperature
isotherms. This temperature increase means that the internal energy of the gas
increases, since ¢U r ¢T.
As can be seen from the isobar in Fig. 12.10, the area representing the work is
rectangular. Thus, the work is relatively easy to compute (length times width):
Wisobaric = p1V2 - V12 = p¢V (ideal gas isobaric process) (12.4)
p 䉳 F I G U R E 1 2 . 1 0 Isobaric (con-
Isotherms stant pressure) process The heat
T1 T2 Isobaric
added to the gas in the frictionless
p 2 = p1 piston goes into work done by the
gas and into changing the internal
T2 >T1 Isobar energy of the gas: Q = ¢U + W.
Pressure
The work is equal to the area under

the isobar (from state 1 to state 2
here, shown as shaded) on the p–V
2 diagram. Note the two isotherms.
1 They are not part of the isobaric
p process, but they show us that the
temperature rises during the iso-
V baric expansion.
V1 V2 Volume
V2 – V1
Q V1
V2 W = p(V2 – V1)
*Pressure can be measured in bars 11 bar = 1 atm2.

For example, when heat is added to or removed from a gas under isobaric con-
ditions, the gas’s internal energy changes and the gas expands or contracts, doing
positive or negative work, respectively. (See Integrated Example 12.2 for the
signs.) This relationship can be written, using the first law of thermodynamics,
with the work expression appropriate for isobaric conditions (Eq. 12.4):
Q = ¢U + W = ¢U + p¢V (ideal gas isobaric process) (12.5)
To see a detailed comparison of an isobaric process and an isothermal process,

consider the following Integrated Example.
INTEGRATED EXAMPLE 12.2 Isotherms versus Isobars: Which Area?

Isotherm at T = 273 K Two moles of a monatomic ideal gas, initially at 0 °C and 1.00 atm, are expanded to
Isotherm at T > 273 K twice their original volume, using two different processes. They are expanded isother-
mally or isobarically, both starting in the same initial state. (a) Does the gas (1) do more
work during the isothermal process, (2) do more work during the isobaric process, or
p Isobar (3) do the same work during both processes? Explain. (b) To prove your answer, deter-
mine the work done by the gas in each process.
1 3
p1 (A) CONCEPTUAL REASONING. As shown in 䉳 Fig. 12.11, both processes involve expan-
sion. The isobar is horizontal, and the isotherm is a decreasing hyperbola. Thus, the gas
does more work during the isobaric expansion (more area under the curve). Fundamen-
tally, this is because the isobaric process is done at higher (constant) pressure than the
Pressure
2 isothermal process (where the pressure drops as the gas expands according to gas law).
p2 In both cases, the work is positive. (How do we know this?) Thus, the correct answer to
part (a) is (2), that the isobaric process does more work.
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . If the volumes are known, Eqs. 12.3 and
12.4 can be used. These quantities can be calculated from the ideal gas law.
V Listing the data,
V1 V2
Volume Given: p1 = 1.00 atm = 1.01 * 105 N>m2 Find: Wisothermal and Wisobar
T1 = 0 °C = 273 K (the work done during the
䉱 F I G U R E 1 2 . 1 1 Comparing n = 2.00 mol (see Section 10.3) isothermal and isobaric
work In Integrated Example 12.2, processes)
V2 = 2V1
the gas does positive work while
expanding. It does more work
For the isotherm, use Eq. 12.3 (the natural logarithm of the volume ratio is
under isobaric conditions (from
state 1 to state 3) than under isother- ln 2 = 0.693):
mal conditions (from state 1 to
≤ = 12.00 mol238.31 J>1mol # K241273 K21ln 22
V2
state 2) because the pressure Wisothermal = nRT ln ¢
remains constant on the isobar but V1
decreases along the isotherm. (Com-
= + 3.14 * 103 J
pare areas under the curves.)
For the isobar, we need to know the two volumes. Using the ideal gas law,
nRT1 12.00 mol238.31 J>1mol # K241273 K2

V1 = = = 4.49 * 10-2 m3
p1 1.01 * 105 N>m2
and therefore
V2 = 2V1 = 8.98 * 10-2 m3
The work is given by Eq. 12.4 as
Wisobar = p1V2 - V12 = 11.01 * 105 N>m2218.98 * 10-2 m3 - 4.49 * 10-2 m32
= + 4.53 * 103 J
This amount is larger than the isothermal work, as expected from part (a).
FOLLOW-UP EXERCISE. In this Example, what is the heat flow in each process?
p 䉳 F I G U R E 1 2 . 1 2 Isometric (con-
Isometric
stant volume) process All of the
V2 = V1 heat added to the gas goes into
increasing the gas’s internal energy,
because there is no work done
2 1W = 02; thus, Q = ¢U. (Notice the
Pressure
Isomet locking nut on the piston, which
prevents any movement.) Again, the
Isotherms isotherms, although not part of the
isometric process, tell us visually
T2 that the temperature of the gas rises.
1 T2 > T1
T1
V
Volume
Q W=0 Q = ∆U
V1 = V2
ISOMETRIC PROCESS
An isometric process (short for isovolumetric, or constant-volume, process), some-
times called an isochoric process, is a constant-volume process. As illustrated in
䉱 Fig. 12.12, the process path on a p–V diagram is a vertical line, called an isomet.
No work is done, since the area under such a curve is zero. (There is no displace-
ment, as there is no change in volume.) Because the gas does not do any work, if
heat is added it must go completely into increasing the gas’s internal energy and,
therefore, its temperature. In terms of the first law of thermodynamics,
Q = ¢U + W = ¢U + 0 = ¢U
and thus
Q = ¢U (ideal gas isometric process) (12.6)
Consider the following example of an isometric process in action.
EXAMPLE 12.3 A Practical Isometric Exercise: How Not to Recycle a Spray Can
Many “empty” aerosol cans contain remnant propellant gases T H I N K I N G I T T H R O U G H . This is an isovolumetric process;
under approximately 1 atm of pressure (assume 1.00 atm) at hence, all the heat goes into increasing the gas’s internal
20 °C. They display the warning “Do not dispose of this can energy. A pressure rise is expected, which is where the danger
in an incinerator or open fire.” (a) Explain why it is dangerous lies. The change in internal energy can be calculated with
to throw such a can into a fire. (b) If there are 0.0100 moles of Eq. 10.16. The final pressure can be obtained using the ideal
monatomic gas in the can and its temperature rises to 2000 °F, gas law.
how much heat was added to the gas? (c) What is the final
pressure of the gas?
SOLUTION. Listing the data and converting given temperatures into kelvins (again, for qualitative reasoning, refer to the Learn
by Drawing 12.1, Learning on Isotherms);
Given: p1 = 1.00 atm = 1.01 * 105 N>m2 Find: (a) Explain the danger in heating the can.
V1 = V2 (b) Q (heat added to gas)
T1 = 20 °C = 293 K (c) p2 (final pressure of gas)
T2 = 2000 °F = 1.09 * 103 °C
= 1.37 * 103 K
n = 0.0100 mol
(a) When heat is added, it all goes into increasing the gas’s (why?). ¢U = Q - W = Q - 0 = Q or Q = ¢U. From
internal energy. Because at constant volume, pressure is pro- Eq. 10.16, U = 32 nRT, then
portional to temperature, the final pressure will be greater
than 1.00 atm. The danger is that the container could explode ¢U = 32 nR¢T
into metallic fragments like a grenade if the maximum design = 32 10.0100 mol238.31 J>1mol # K2411.37 * 103 K - 293 K2
pressure of the container is exceeded.
= 134 J.
(b) To calculate the heat, we use the first law of thermodynam-
ics. Recall that the work done in an isometric process is zero
(c) The final pressure of the gas is determined directly from the ideal gas law:
V1 1.37 * 103 K
≤ ¢ ≤ = 11.00 atm2 ¢ ≤ ¢
p2V2 p1V1 V1 T2
= or p2 = p1 ¢ ≤ = 4.68 atm
T2 T1 V2 T1 V1 293 K
F O L L O W - U P E X E R C I S E . Suppose the can were designed to withstand pressures up to 3.50 atm. What would be the highest Celsius
temperature it could reach without exploding?
ADIABATIC PROCESS
In an adiabatic process, no heat is transferred into or out of the system. That is,
Q = 0 (䉲 Fig. 12.13). (The Greek word adiabatos means “impassable.”) This condi-
tion is satisfied for a thermally isolated system, one completely surrounded by
“perfect” insulation. This is an ideal situation; for real-life conditions, adiabatic
processes can only be approximated. For example, nearly adiabatic processes can
take place if the changes occur rapidly enough so that there isn’t time for signifi-
cant heat to flow into or out of the system. In other words, quick processes can
approximate adiabatic conditions.
The curve for this process is called an adiabat. During an adiabatic process, all
three thermodynamic coordinates (p, V, T) change. For example, if the pressure on
a gas is reduced, the gas expands. However, no heat flows into the gas. Without a
compensating input of heat, work is done at the expense of the gas’s internal
energy. Therefore, ¢U must be negative. Since the internal energy, and thus the
temperature, both decrease, such an expansion is a cooling process. Similarly, an
adiabatic compression is a warming process (temperature increase).
From the first law of thermodynamics, an adiabatic process can be described by
Q = 0 = ¢U + W
䉴 F I G U R E 1 2 . 1 3 Adiabatic (no p
Adiabatic
heat transfer) process In an adia- Isotherms
batic process (shown here for a Q=0
cylinder with heavy insulation), no
heat is added to or removed from
1
the system; thus, Q = 0. During
Pressure
expansion (shown here), positive Adiabat

work is done by the gas at the
expense of its internal energy:
W = - ¢U. The pressure, volume, T1
and temperature all change in the
2 T 2 < T1
process. The work done by the gas T2
is the shaded area between the adia- V
V1 V2 Volume
bat and the V-axis.
Q=0 V1
V2 W = −∆U
or
¢U = - W (adiabatic process) (12.7)
For completeness, we state some other relationships in an adiabatic process. An

important factor is the ratio of the gas’s molar specific heats, defined by a dimen-
sionless quantity g = cp>cv, where cp and cv are the specific heats at constant pres-
sure and volume, respectively. For the two common types of gas molecules,
monatomic and diatomic, the values of g are about 1.67 and 1.40, respectively. The
volume and pressure at any two points on an adiabat are related by
g g
p1V1 = p2V2 (ideal gas adiabatic process) (12.8)
The work done by an ideal gas during an adiabatic process can be shown to be
p1V1 - p2V2
Wadiabatic = (ideal gas adiabatic process) (12.9)
g - 1
To clear up confusion that often occurs between isotherms and abiabats, see
Integrated Example 12.4.
INTEGRATED EXAMPLE 12.4 Adiabats versus Isotherms:

Two Different Processes That Are Often Confused
A sample of helium gas expands to triple its initial volume adia- curve is larger; thus, the gas does more work during its
batically in one case and isothermally in another. In both cases, isothermal expansion, and the correct answer is (2). Physi-
it starts from the same initial state. The sample contains 2.00 mol cally, the isothermal expansion involves more work because
of helium 1g = 1.672, initially at 20 °C and 1.00 atm. (a) Does the pressures are always higher during the isothermal expan-
the gas, (1) do more work during the adiabatic process, (2) do sion than during the adiabatic one.
more work during the isothermal process, or (3) do the same ( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . To determine
work during both processes? (b) Calculate the work done dur- the isothermal work, the ratio of the final to initial volumes is
ing each process to verify your reasoning in part (a). needed and is given. For the adiabatic work, the ratio of spe-
( A ) C O N C E P T U A L R E A S O N I N G . To determine graphically which cific heats, g, is important, as are the final pressure and vol-
process involves more work, note the areas under the process ume. The final pressure can be found using Eq. 12.8, and the
curves (see Fig. 12.13). The area under the isothermal process ideal gas law enables the determination of the final volume.
Listing the given values and converting the temperature into kelvins:
Given: p1 = 1.00 atm = 1.01 * 105 N>m2 Find: Wisothermal and Wadiabatic
n = 2.00 mol (work done during each process)
T1 = 120 + 2732 K = 293 K
V2 = 3V1
g = 1.67
The data necessary to calculate the isothermal work from Eq. 12.3 are given. The volume ratio is 3 and ln 3 = 1.10, so
V2
Wisothermal = nRT ln ¢ ≤
V1
= 12.00 mol238.31 J>1mol # K241293 K21ln 32 = + 5.35 * 105 J
For the adiabatic process, the work can be determined from Eq. 12.9, but first the final pressure and volume are needed. The final
pressure can be determined from a ratio form of Eq. 12.8:
V1 g V1 g 1 1.67
p2 = p1 ¢ ≤ = p1 ¢ ≤ = p1 a b = 0.160p1
V2 3V1 3
= 10.160211.01 * 105 N>m22 = 1.62 * 104 N>m2
The initial volume is determined from the ideal gas law:
nRT1 12.00 mol238.31 J>1mol # K241293 K2
V1 = =
p1 1.01 * 105 N>m2
= 4.82 * 10-2 m3
Therefore, V2 = 3V1 = 0.145 m3. Then, applying Eq. 12.9,

p1V1 - p2V2
Wadiabatic =
g - 1
11.01 * 105 N>m2214.82 * 10-2 m32 - 11.62 * 104 N>m2210.145 m32
=
1.67 - 1
3
= + 3.76 * 10 J
As expected, this result is less than the isothermal work.
F O L L O W - U P E X E R C I S E . In this Example, (a) calculate the final temperature of the gas in the adiabatic expansion. (b) During the
adiabatic expansion, determine the gas’s change in internal energy using its temperature change (helium is a monatomic gas).
Does it equal the negative of the work done (as calculated in the Example)? Explain.
LEARN BY DRAWING 12.1 (horizontal line), and adiabat = no heat flow (down-
ward sloping curve, steeper than an isotherm).
leaning on isotherms ■ Next, use the graphs to determine the signs of W and
¢U. W is represented by the area under the p–V curve
When you are analyzing thermodynamic processes, it is for the process represented, and its sign is determined
sometimes hard to keep track of the signs of heat flow (Q),
work (W), and internal energy change 1¢U2. One method
by whether the gas expanded (positive) or was com-
pressed (negative). The sign of ¢T will be clear from
that can help with this bookkeeping is to superimpose a the isotherms, since they serve as a temperature scale.
series of isotherms on the p–V diagram you are working For example, a rise in T implies an increase in U.
with (as in Figs. 12.9 through 12.13). This method is useful ■ Last, determine the sign of Q from the first law of ther-
even if the situation you are studying does not involve modynamics, Q = ¢U + W. From the sign of Q, it
isothermal processes. should be clear whether heat was transferred into or
Before starting, recall that an isothermal process is one in out of the system.
which the temperature remains constant:
The example in Fig. 2 shows the power of this visual
1. In an isothermal process for an ideal gas, ¢U is zero. (Why?) approach. Here, we are to decide whether heat flows into or
2. Since T is constant, pV must also be constant, since, from the out of the gas during an isobaric expansion. Expansion implies
ideal gas law (Eq. 10.3), pV = nRT = constant. You may positive work done by the gas. But what is the direction of the
recall from algebra that p = costant>V is the equation of a heat flow (or is it zero)? After sketching the isobar, it can be
hyperbola. Thus, on a p–V diagram, an isothermal process is seen that it crosses from lower-temperature isotherms to
described by a hyperbola. The farther from the axes the higher-temperature ones. Hence, there is a temperature
hyperbola is, the higher the temperature it represents (Fig. 1). increase, and ¢U is positive. From Q = ¢U + W, we see that
To take advantage of these properties, follow these steps: Q is the sum of two positive quantities, ¢U and W. Therefore,
■ Sketch a set of isotherms for a series of increasing tem- Q must be positive, which means that heat enters the gas.
peratures on the p–V diagram (Fig. 1). As an exercise, try analyzing an isometric process using
■ Then sketch the process you are analyzing—for this graphical approach. See also Integrated Examples 12.2
example, the isobar shown in Fig. 2. Isomet = constant and 12.4.
volume (vertical line), isobar = constant pressure
∆T > 0, so
p p ∆U > 0
∆V > 0, so
W> 0
4 5 1 234 5 Therefore Q= ∆U + W > 0
1 2 3
Isobaric
expansion
T4 > T3
T3 > T2
T2 > T1
pV ⬀ T T1
V V
F I G U R E 1 Isotherms on a p–V diagram F I G U R E 2 An isobaric expansion
12.4 THE SECOND LAW OF THERMODYNAMICS AND ENTROPY 431
TABLE 12.1 Important Thermodynamic Processes

Process Definition Characteristic Result of the First Law
Isothermal T = constant ¢U = 0 Q = W
Isobaric p = constant W = p¢V Q = ¢U + p¢V
Isometric V = constant W = 0 Q = ¢U
Adiabatic Q = 0 ¢U = - W
As a final summary, the characteristics and consequences of these thermody-

namic processes are listed in 䉱 Table 12.1.
DID YOU LEARN?

➥ The four important thermodynamic processes are isobaric, isothermal, isometric,
and adiabatic.
➥ The change in internal energy of an ideal gas after an isothermal process is equal to
zero because its internal energy depends only on its temperature, which is a
constant in an isothermal process.
➥ The work done by a gas in an isometric process is zero because there is no volume
change (no movement, no work).
12.4 The Second Law of Thermodynamics and Entropy

➥ If the entropy of a system increases, is the system becoming more ordered or more
disordered?
➥ Under what condition, if any, can heat be transferred from a cooler object to a
warmer object?
➥ Can heat energy be completely converted to useful mechanical work in a thermody-
namic cycle?
Suppose that a piece of hot metal is placed in an insulated container of cold water.
Heat will be transferred from the metal to the water, and the two will eventually
reach thermal equilibrium at some intermediate temperature. For a thermally iso-
lated system, the system’s total energy remains constant. Could heat have been
transferred from the cold water to the hot metal instead? This process would not
happen naturally. But if it did, the total energy of the system would still remain
constant, and this “impossible” inverse process would not violate energy conser-
vation or the first law of thermodynamics.
There must be another principle that specifies the direction in which a process
can take place. This principle is embodied in the second law of thermodynamics,
which states that certain processes do not take place, or have never been observed
to take place, even though they may be consistent with the first law.
There are several equivalent statements of the second law, which are worded
according to their application. One applicable to the aforementioned situation is
as follows:
Heat will not flow spontaneously from a cooler body to a warmer body.
An equivalent alternative statement of the second law involves thermal cycles.
A thermal cycle typically consists of several separate thermal processes after which
the system ends up back at its starting conditions. If the system is a gas, this means
the same p–V–T state from which it started. The second law, stated in terms of a
thermal cycle (operating as a heat engine; see Section 12.5), is as follows:
In a thermal cycle,heat energy cannot be completely transformed into mechanical work.
In general, the second law of thermodynamics applies to all forms of energy. It
is considered true because no one has ever found an exception to it. If it were not
true, a perpetual motion machine could have been built. Such a machine could
first transform heat completely into work and motion (mechanical energy), with
no energy loss. The mechanical energy could then be transformed back into heat
and be used to reheat the reservoir from where the heat came originally (again
with no loss). Since the processes could be repeated indefinitely, the machine
would run perpetually, just shifting energy back and forth. All of the energy is
accounted for, so this situation does not violate the first law. However, it is obvious
that real machines are always less than 100% efficient (even if there were no fric-
tion)—that is, the work output is always less than the energy input. Another state-
ment of the second law is therefore as follows:
It is impossible to construct an operational perpetual motion machine.
Attempts have been made to construct such perpetual machines, with no
success.*
It would be convenient to have some way of expressing the direction of a
process in terms of the thermodynamic properties of a system. One such property
is temperature. In analyzing a conductive heat transfer process, you need to know
the temperatures of the system and its surroundings. Knowing the temperature
difference between the two processes allows you to state the direction in which
the heat transfer will spontaneously take place. Another useful quantity, particu-
larly during the discussion of heat engines, is entropy.
ENTROPY
A quantity that indicates the natural direction of a process was first described by
Rudolf Clausius (1822–1888), a German physicist. This quantity is called entropy.
Entropy is a multifaceted concept, with various different physical interpretations:
■ Entropy is a measure of a system’s ability to do useful work. As a system loses
the ability to do work, its entropy increases.
■ Entropy determines the direction of time. It is “time’s arrow” that points out
the forward flow of events, distinguishing past events from future ones.
■ Entropy is a measure of disorder. A system naturally moves toward greater dis-
order, or disarray. The more order there is, the lower the system’s entropy.
■ The entropy of the universe is increasing.
All of these statements (and others) turn out to be equally valid interpretations
of entropy and are physically equivalent, as will be seen in the upcoming discus-
sions. First, however, the definition of the change in entropy is introduced. The
change in a system’s entropy 1¢S2 when an amount of heat (Q) is added or
removed by a reversible process at a constant temperature is
Q (change in entropy
¢S = (12.10)
T at constant temperature)
SI unit of entropy: joule per kelvin 1J>K2
The temperature T must be in kelvins. ¢S is positive if a system absorbs heat

1Q 7 02 and negative if a system loses heat 1Q 6 02. If the temperature changes
during the process, calculating the change in entropy requires advanced mathe-
matics. Our discussions will be limited to isothermal processes or those involving
small temperature changes. For the latter, entropy changes can be approximated
by average temperatures, as in Example 12.6. But first, let’s look at an example of a
change in entropy and how it is interpreted.
*Although perpetual motion machines cannot exist, (very nearly) perpetual motion is known to
exist—for example, the planets have been in motion around the Sun for about 5 billion years.
EXAMPLE 12.5 Change in Entropy: An Isothermal Process

While doing physical exercise at a temperature of 34 °C, an Since a phase change occurs, latent heat is absorbed by water:
Q = mLv = 10.400 kg2124.2 * 105 J>kg2 = 9.68 * 105 J
athlete loses 0.400 kg of water per hour by the evaporation of
perspiration from his skin. Estimate the change in entropy of
the water as it vaporizes. The latent heat of vaporization of Then
perspiration is about 24.2 * 105 J>kg.
Q +9.68 * 105 J
T H I N K I N G I T T H R O U G H . A phase change occurs at constant
¢S = = = + 3.15 * 103 J>K
T 307 K
temperature; hence, Eq. 12.10 1¢S = Q>T2 applies after con-
verting temperature to kelvins. From Eq. 11.2 1Q = mLv2 the Q is positive, because heat is added to water. The change in
amount of heat added can be computed. entropy, then, is also positive, and the entropy of the water
increases. This outcome is reasonable, because a gaseous state
SOLUTION. From the statement of the problem, is more random (disordered) than a liquid state.
Given: m = 0.400 kg Find: ¢S (change in
T = 134 + 2732 K = 307 K entropy)
Lv = 24.2 * 105 J>kg
FOLLOW-UP EXERCISE. What is the change in entropy of a 1.00-kg water sample when it freezes to form ice at 0 °C?
EXAMPLE 12.6 A Warm Spoon into Cool Water: System Entropy Increase or Decrease?
A metal spoon at 24 °C is immersed in 1.00 kg of water at that is, Qs + Qw = 0, where the subscripts s and w stand for
18 °C. The system (spoon and water) is thermally isolated and spoon and water, respectively. Qw can be determined from the
comes to equilibrium at a temperature of 20 °C. (a) Find the known water mass, specific heat, and temperature change.
approximate change in the entropy of the system. (b) Repeat Therefore, both Q values (equal but opposite signs) can be
the calculation, assuming, although this can’t happen, that the determined. Strictly speaking, Eq. 12.10 cannot be used
water temperature dropped to 16 °C and the spoon’s temper- because it is applicable only for constant temperature
ature increased to 28 °C. Comment on how entropy shows processes. However, here the temperature changes are small,
that the situation in part (b) cannot happen. so a good approximation for ¢S can be obtained by using
each object’s average temperature T.
T H I N K I N G I T T H R O U G H . The system is thermally isolated, so
there is heat exchange only between the spoon and the water,
SOLUTION. Using i and f to stand for initial and final, respectively,
Given: Ts, i = 24 °C Find: (a) ¢S (change in entropy of the system

Tw, i = 18 °C in a realistic situation)
mw = 1.00 kg (b) ¢S (change in entropy of the system
cw = 4186 J>1kg # °C2 (from Table 11.1) in an unrealistic situation)
(a) Tf = 20 °C
(b) Ts,f = 28 °C; Tw,f = 16 °C
(a) The amount of heat transferred (Q) needs to be deter- Then using these average temperatures and Eq. 12.10 to com-
mined in order to solve for ¢S. With ¢Tw = Tf - Tw, i = pute the approximate entropy changes for the water and the
20 °C - 18 °C = + 2.0 °C, the heat gained by the water is metal:
from Eq. 11.1.
Qw + 8.37 * 103 J
Qw = c w mw ¢T = 34186 J>1kg # C°211.00 kg212.0 °C2 ¢Sw L
Tw
=
292 K
= + 28.7 J>K
= + 8.37 * 103 J
Qs - 8.37 * 103 J
This quantity is also the magnitude of the heat lost by the ¢Ss L = = - 28.4 J>K
Ts 295 K
metal. Therefore,
Qs = - 8.37 * 103 J The change in the entropy of the system is the sum of these, or
The average temperatures are ¢S = ¢Sw + ¢Ss L + 28.7 J>K - 28.4 J>K = + 0.3 J>K
Tw, i + Tf 18 °C + 20 °C The entropy of the spoon decreased, because heat was lost.
Tw = = = 19 °C = 292 K
2 2 The entropy of the water increased by a greater amount, so
Ts, i + Tf 24 °C + 20 °C overall, the system’s entropy increased.
Ts = = = 22 °C = 295 K
2 2 (continued on next page)
(b) Although this situation conserves energy, it violates the sec- Qs + 8.37 * 103 J
ond law of thermodynamics. To see this violation in terms of ¢Ss L = = + 28.0 J>K
Ts 299 K
entropy, let’s repeat the foregoing calculation, using the second
set of numbers. With ¢Tw = Tf - Tw, i = 16 °C - 18 °C = The change in the entropy of the system is:
-2.0 °C, the heat lost by the water is
¢S = ¢Sw + ¢Ss L - 28.9 J>K + 28.0 J>K = - 0.9 J>K
Qw = cw mw ¢T = 34186 J>1kg # °C211.00 kg241-2.0 °C2
In this unrealistic scenario, the entropy of the metal increased,
= - 8.37 * 103 J but the entropy of the water decreased by a greater amount,
Again using the average temperatures, Tw = 17 °C = 290 K and the total system entropy decreased.
and Ts = 26 °C = 299 K, to compute the approximate entropy
changes for the water and the metal spoon:
Qw - 8.37 * 103 J
¢Sw L = = - 28.9 J>K
Tw 290 K
F O L L O W - U P E X E R C I S E . What should the initial temperatures in this Example be to make the overall system entropy change
zero? Explain in terms of heat transfers.
Note that the entropy change of the system in Example 12.6a is positive,
because the process is a natural one. That is, it is a process that is always observed
to occur. In general, the direction of any process is toward an increase in total sys-
tem entropy. That is, the entropy of an isolated system never decreases. Another way to
state this observation is to say that the entropy of an isolated system increases for every
natural process 1¢S 7 02. In coming to an intermediate temperature, the water and
spoon in Example 12.6a are undergoing a natural process. The process in 12.6b
would never be observed, and the decrease in system entropy indicates this. Simi-
larly, water at room temperature in an isolated ice cube tray will not naturally
(spontaneously) turn into ice.
However, if a system is not isolated, it may undergo a decrease in entropy. For
example, if the ice cube tray filled with water is instead put into a freezer compart-
ment, the water will freeze, undergoing a decrease in entropy. But there will be a
larger increase in entropy somewhere else in the universe. In this case, the freezer
warms the kitchen as it freezes the ice, and the total entropy of the system (ice plus
kitchen) actually increases.
Thus, a statement of the second law of thermodynamics in terms of entropy (for
natural processes) is:
The total entropy of the universe increases in every natural process.
Processes exist for which the entropy is constant. One obvious such process is
any adiabatic process, since Q = 0. In this case, ¢S = Q>T = 0. Similarly, any
reversible isothermal expansion that is followed immediately by an isothermal
compression along the same path has zero net entropy change. This last example
is true because the two heat flows are the same, but opposite in sign, and the tem-
peratures are also the same; thus, ¢S = Q>T + 1-Q>T2 = 0. With the realization
that, under some circumstances, it is possible to have ¢S = 0, the previous state-
ment of the second law of thermodynamics can be generalized to include all possi-
ble processes. This is as follows:
During any process, the entropy of the universe can only increase or remain constant
(¢S Ú 0).
To appreciate one of the many alternative (and equivalent) interpretations of

entropy, consider the foregoing statements rewritten in terms of order and disor-
der. Here, entropy is interpreted as a measure of the disorder of a system. Thus, a
larger value for entropy means more disorder (or, equivalently, less order):
All naturally occurring processes move toward a state of greater disorder or disarray.
A working definition of order and disorder may be extracted from everyday

observations. Suppose you are making a pasta salad and have chopped tomatoes
ready to toss into the cooked pasta. Before you mix the pasta and tomatoes, there is a
relative amount of order; that is, the ingredients are separate and unmixed. Upon
mixing, the separate ingredients become one dish, and there is less order (or more
disorder, if you prefer). The pasta salad, once mixed, will never separate into the
individual ingredients on its own (that is, by a natural process—one that happens
on its own). Of course, you could go in and pick out the individual tomato pieces,
but that would not be a natural process. Similarly, broken eyeglasses do not, on their
own, go back together. See Global Warming: Some Inconvenient Facts.
DID YOU LEARN?

➥ An increase in entropy is an indication of the increasing disorder of a system.
Therefore, a system is becoming more disordered if its entropy increases.
➥ Heat can be transferred from a cooler object to a warmer object, but this is not a
spontaneous process. In other words, work has to be done in order to move heat
from a cooler object to a warmer object.
➥ According to the second law of thermodynamics, heat energy cannot be
completely transformed into mechanical work.
INSIGHT 12.1 Global Warming: Some Inconvenient Facts

Human activities such as driving cars and running refrigera- The average temperatures in the Arctic have risen at twice
tors release heat and greenhouse gases into the Earth atmos- the global average, according to the multinational Arctic Cli-
phere. (See Insight 11.3, The Greenhouse Effect.) There is no mate Impact Assessment report compiled between 2000 and
doubt that these activities will cause the global temperature to 2004. This temperature increase is causing the Arctic ice to
rise. There are, however, hot debates about whether these rapidly disappear. (See the Chapter 10 opening satellite pho-
activities are causing significant and noticeable global warming. tographs that show the decreasing Arctic ice.)
In this Insight, some facts about global warming are given. The global average sea level has been rising at an average
Average temperatures have climbed 0.6 °C ⫾ 0.2 °C around rate of 11 to 22 mm>year over the past 100 years, which is sig-
the world since 1880, with much of this increase (about 0.2 °C nificantly faster than the rate averaged over the last several
to 0.3 °C) occurring over the past 25 years (the period with the thousand years. Coral reefs, which are highly sensitive to
most credible data), according to NASA’s Goddard Institute for small changes in water temperature, suffered the worst die-
Space Studies. The rate of warming is increasing. The past 25 off ever recorded in 1998.
years were the hottest in 400 years (Figure 1).
Global Temperatures
0.6
0.4
Annual Average
Temperature Anomaly (°C)
Five Year Average

0.2
– 0.2
F I G U R E 1 Global temperature This graph
shows global average temperatures as compiled
by the Hadley Centre for Climate Prediction – 0.4
and Research of the UK Meteorological Office.
Temperature anomaly is the difference from
long-term average temperatures defined by – 0.6
NCDC (National Climate Data Center). 1860 1880 1900 1920 1940 1960 1980 2000
12.5 Heat Engines and Thermal Pumps

➥ What is the working principle of a heat engine?

➥ How is the thermal efficiency of a heat engine calculated in terms of the heat
absorbed from the hot reservoir and the heat expelled to the cold reservoir?
➥ How does a refrigerator work?
A heat engine is any device that converts heat energy into work. Since the second
law of thermodynamics prohibits complete conversion of heat energy into work in
a heat engine, some of the heat input will unfortunately be lost and not go into
work. For our purposes, a heat engine is any device that takes heat from a high-
temperature source (a hot, or high-temperature, reservoir), converts some of it to
useful work, and expels the rest to its surroundings (a cold, or low-temperature,
reservoir). For example, most turbines that generate electricity (Section 20.2) are
heat engines, using heat from various sources such as oil, gas, coal, or energy
released in nuclear reactions (Section 30.2). They might be cooled by river water,
for example, thus losing heat to this low-temperature reservoir. A generalized heat
engine is represented in 䉲 Fig. 12.14a. (We will not be concerned with the mechani-
cal details of an engine, such as pistons and cylinders, only thermodynamic
processes.)
A few reminders about our sign convention are in order before starting to ana-
lyze heat engines. For engines, we are interested primarily in the work W done by
the gas. During expansion, the gas does positive work. Similarly, during a com-
pression, the work done by the gas is negative. Also, it will be assumed that the
“working substance” (the material that absorbs the heat and does the work)
behaves like an ideal gas. The fundamental physics on which heat engines are
based is the same regardless of the working substance. However, using ideal gases
makes the mathematics easier.
Adding heat to a gas can produce work. But since a continuous output is usually
wanted, practical heat engines operate in a thermal cycle, or a series of processes
that brings the system back to its original condition. Cyclic heat engines include
steam engines and internal combustion engines, such as automobile engines.
High-temperature reservoir
Qin = Qh Heat input

per cycle
Wnet = Qh – Qc p
Wnet
HEAT
ENGINE 1 2
Pressure
Mechanical work
output per cycle Wnet
Heat output
Qout = Qc
per cycle
4 3
Low-temperature reservoir V
Volume
(a) (b)
䉱 F I G U R E 1 2 . 1 4 Heat engine (a) Energy flow for a generalized cyclic heat engine. Note
that the width of the arrow representing Qh (heat flow out of hot reservoir) is equal to the
combined widths of the arrows representing Wnet and Qc (heat flow into cold reservoir),
reflecting the conservation of energy: Qh = Qc + Wnet . (b) This specific cyclic process con-
sists of two isobars and two isomets. The net work output per cycle is the area of the rec-
tangle formed by the process paths. (See Example 12.11 for the analysis of this particular
cycle.)
12.5 HEAT ENGINES AND THERMAL PUMPS 437
representing work in thermal cycles

p p p Subtracting the two areas
First sketch the expansion work ... ...then the compression work to complete the cycle. leaves the enclosed area,
which represents the net
work per cycle
Wnet
V V V
0 0 0
An idealized, rectangular thermodynamic cycle is shown in Fig. 12.14b. It con-

sists of two isobars and two isomets. When these processes occur in the sequence
indicated, the system goes through a cycle (1–2–3–4–1), returning to its original
condition. When the gas expands (during 1 to 2), it does (positive) work equal to
the area under the isobar. Doing positive work is exactly the desired output of an
engine. (Think of a car engine piston moving the crankshaft.) However, there must
be a compression of the gas (during 3 to 4) to bring it back to its initial conditions.
During this phase, the work done by the gas is negative, which is not the purpose
of an engine. In a sense, a portion of the positive work done by the gas is “can-
celled” by the negative work done during the compression.
From this discussion, it can be seen that the important quantity in engine
design is the net work (Wnet) per cycle. This quantity is represented graphically as
the area enclosed by the process curves that make up the cycle in the accompany-
ing Learn By Drawing 12.2, Representing Work in Thermal Cycles. (In Fig 12.14b,
the area is rectangular.) When the paths are not straight lines, numerical calcula-
tions of the areas may be difficult, but the concept is the same.
THERMAL EFFICIENCY
The thermal efficiency (E) of a heat engine is defined as
net work output Wnet (thermal efficiency

e = = (12.11)
heat input Qin of a heat engine)
Efficiency tells us how much useful work (Wnet) the engine does in comparison with
the input heat it receives (Qin). For example, modern automobile engines have an
efficiency of about 20% to 25%. This means that only about one-fourth of the heat
generated by igniting the air–gasoline mixture is actually converted into mechanical
work, which turns the car wheels and so on. Alternatively, you could say that the
engine wastes about three-fourths of the heat, eventually transferring it to the
atmosphere through the hot exhaust system, radiator system, and metal engine.
For one cycle of an ideal gas heat engine, Wnet is determined by applying the
first law of thermodynamics to the complete cycle. Recall that our heat sign con-
vention designates Qout as negative. For our discussion of heat engines and
pumps, all heat symbols (all Q’s) will represent magnitude only. Therefore, Qout is writ-
ten as -Qc (the negative of a positive quantity Qc to indicate flow out of the engine
into a cold reservoir). Qin is positive by our sign convention and is shown as + Qh
(to indicate flow to the engine from the hot gas ignition).
Applying the first law of thermodynamics to the expansion part of the cycle
and showing the work done by the gas as W = + Wexpansion , then
¢Uh = + Qh - Wexpansion . For the compression part of the cycle, the work done by
the gas is shown explicitly as being negative (W = - Wcompression and
¢Uc = - Qc + Wcompression). Adding these equations and realizing that for an
ideal gas, ¢Ucycle = ¢Uh + ¢Uc = 0 (why?),
0 = 1Qh - Qc2 + 1Wcompression - Wexpansion2
or
Wexpansion - Wcompression = Qh - Qc
However, Wnet = Wexp - Wcomp , and the final result is (remember Q represents
magnitude here)
Wnet = Qh - Qc
So the thermal efficiency of a heat engine can be rewritten in terms of the heat
flows as
Wnet Qh - Qc Qc (efficiency of
e = = = 1 - (12.12)
Qh Qh Qh an ideal gas heat engine)
Like mechanical efficiency, thermal efficiency is a dimensionless fraction and is

commonly expressed as a percentage. Equation 12.12 indicates that a heat engine
could have 100% efficiency if Qc were zero. This condition would mean that no
heat energy would be lost and all the input (hot reservoir) heat would be con-
verted to useful work. However, this situation is impossible according to the sec-
ond law of thermodynamics. In 1851, this observation led Lord Kelvin (who
developed the Kelvin temperature scale; Section 10.3) to state the second law in
yet another physically equivalent manner:
No cyclic heat engine can convert its heat input completely to work.
From Eq. 12.12 it can be seen that to maximize the work output per cycle of a heat
engine, Qc>Qh must be minimized, which increases the efficiency.
Almost all automobile gasoline engines use a four-stroke cycle. An approximation
of this important cycle involves the steps shown in 䉲 Fig. 12.15, along with a p–V
diagram of the thermodynamic processes that make up the cycle. This theoretical
Air–fuel Spark plug Spark Exhaust

mixture
Valve Valve
open closed
Piston
Intake Compression Ignition (3–4) Power Exhaust

stroke (1–2) stroke (2–3) stroke (4–5) (5–2) stroke (2–1)
Adiabatic Isometric Adiabatic Isometric
p p compression p heating p expansion p cooling p
Pressure
4 4 4 4 4 4
3 5 3 5 3 5 3 5 3 5 3 5
1 atm
1 2 1 2 1 2 1 2 1 2 1 2
V V V V V V
Volume
䉱 F I G U R E 1 2 . 1 5 The four-stroke cycle of a heat engine The steps of the four-stroke Otto
cycle. The piston moves up and down twice each cycle, for a total of four strokes per cycle.
See text for description.
cycle is called the Otto cycle, named for the German engineer Nikolaus Otto
(1832–1891), who built one of the first successful gasoline engines.
During the intake stroke (1–2), an isobaric expansion, the air–fuel mixture is
admitted at atmospheric pressure through the open intake valve as the piston
drops. This mixture is compressed adiabatically (quickly) on the compression
stroke (2–3). This step is followed by fuel ignition (3–4, when the spark plug fires,
giving an isometric pressure rise). Next, an adiabatic expansion occurs during the
power stroke (4–5). Following this step is an isometric cooling of the system when
the piston is at its lowest position (5–2). The final, exhaust stroke is along the iso-
baric leg of the Otto cycle (2–1). Notice that it takes two up and down motions of
the piston to produce one power stroke.
EXAMPLE 12.7 Thermal Efficiency: What You Get Out of What You Put In
The small, gasoline-powered engine of a leaf blower SOLUTION.
absorbs 800 J of heat energy from a high-temperature reser- Given: Qh = 800 J Find: e (thermal efficiency)
voir (the ignited gas–air mixture) and exhausts 700 J to a Qc = 700 J
low-temperature reservoir (the outside air, through its cool- The net work done by the engine per cycle is
ing fins). What is the engine’s thermal efficiency?
Wnet = Qh - Qc = 800 J - 700 J = 100 J
THINKING IT THROUGH. The definition of thermal efficiency
Therefore, the thermal efficiency is
of a heat engine (Eq. 12.12) can be used if Wnet is determined.
= 0.125 1or 12.5%2
(Keep in mind that the Q’s mean heat magnitudes.) Wnet 100 J
e = =
Qh 800 J
FOLLOW-UP EXERCISE. (a) What would be the net work per cycle of the engine in this Example if the efficiency were raised to
15% and the input heat per cycle were raised to 1000 J? (b) How much heat would be exhausted in this case?
Here is a more practical application of a small heat engine.
EXAMPLE 12.8 Thermal Efficiency: Pumping Water

A gasoline-powered water pump can pump 7.6 * 103 kg (about T H I N K I N G I T T H R O U G H . The definition of thermal efficiency
2000 gal) of water from a basement floor to the ground outside of a heat engine applies (Eq. 12.12). However, we need to cal-
the house in each hour while consuming 1.0 gal of gasoline. culate the heat input from the energy content of gasoline as
Assume the energy content of gasoline is 1.3 * 106 J>gal and the well as net work output from raising the water to increase its
basement floor is 3.0 m below the ground. (a) What is the ther- potential energy (Eq. 5.8).
mal efficiency of the water pump? (b) How much heat is wasted
to the environment in 1.0 h? Ignore frictional losses and assume
no change in the kinetic energy of the water.
SOLUTION. The heat input is from the energy in 1.0 gal of gasoline.
Given: Qh = 11.0 gal211.3 * 106 J>gal2 = 1.3 * 106 J Find: (a) e (thermal efficiency)
m = 7.6 * 103 kg (b) Qc (heat to environment)
¢y = 3.0 m
(a) The work output is equal to the increase in potential energy of the water:
Wnet = mg¢y = 17.6 * 103 kg219.80 m>s2213.0 m2 = 2.2 * 105 J
The thermal efficiency is then
2.2 * 105 J
= 0.17 1or 17%2
Wnet
e = =
Qh 1.3 * 106 J
(b) The heat exhausted to the environment in 1 h is
Qc = Qh - Wnet = 1.3 * 106 J - 2.2 * 105 J = 1.1 * 106 J
F O L L O W - U P E X E R C I S E . If the heat exhausted to the environment were completely absorbed by the pumped water, what would
be the temperature change of the water?
INSIGHT 12.2 Thermodynamics and the Human Body

Like the bodies of all other organisms, the human body is not
a closed system. We must consume food and oxygen to sur-
vive. Both the first and second laws of thermodynamics have
interesting implications for our bodies.
The human body metabolizes the chemical energy stored in
food and>or the fatty tissues of the body. This is quite an effi-
cient process; typically 95% of the energy content in food is
eventually metabolized. Some of this metabolized energy is
converted into work, W, to circulate blood, perform daily
tasks, and so on. The rest is lost to the environment in the
form of heat, Q. For a typical 65-kg person, about 80 J of work
per second is needed just to keep the body parts, such as the
liver, brain, and skeletal muscles, performing their functions.
The first law of thermodynamics, or the law of conserva-
tion of energy, can be written as ¢U = Q - W.
Here ¢U is the change in internal energy of the body, which
could come from two sources: the already consumed food or the
body’s stored fat. Thus we can write ¢U = ¢Ufood + ¢Ufat .
Hence ¢U is a negative quantity, because as the energy stored F I G U R E 1 Measuring energy consumed and work per-
in food and fat is converted to heat and work, our bodies have formed A cyclist is tested with a breathing device and a
less energy stored (until more food is consumed again). Since Q dynamometer so that both her power and metabolic rate
can be measured.
is heat lost to the environment, it is also a negative quantity.
The human body is an example of a biological heat engine.
The energy source is the energy metabolized from food and
fatty tissues. Some of this energy is converted into work, and
proportional to the rate of oxygen consumption, so this rate
1|¢U|>¢t2 can be measured by using breathing devices, as
the rest is expelled to the environment in the form of heat.
This situation is directly analogous to a heat engine taking in
in Fig. 1. Thus, the efficiency of the body performing different
heat from the hot reservoir, doing mechanical work, and
activities can be determined by measuring the rate of oxygen
exhausting waste heat into the environment. Thus the effi-
consumption associated with each separate activity.
ciency of the human body is
The efficiency of the human body, for the most part,
work output W depends on muscle activity and which muscles are used. The
e = = largest muscles in the body are leg muscles, so if an activity
|internal energy loss| |¢U|
uses these muscles, the efficiency associated with that activity
Since W, Q, and ¢U vary widely from one activity to is relatively high. For example, some professional bicycle rac-
another, efficiency is often determined by using the time rate ers can achieve efficiencies as high as 20%, generating more
of these quantities—that is, the work per unit time (power P), than 2 hp of power in short bursts. (A typical table saw deliv-
W>¢t, and the energy consumed per unit time (metabolic ers about 2 hp.) Arm muscles, conversely, are relatively small,
rate), |¢U|>¢t: so activities such as bench pressing have efficiencies of less
W W> ¢t P than 5%. Like any other heat engine, the human body can
1|¢U|>¢t2 1|¢U|>¢t2
e = = = never achieve 100% efficiency. When people exercise, a lot of
|¢U|
waste heat is generated; they must get rid of it through
The power exerted during a particular activity such as run- processes such as perspiring to avoid overheating. Read more
ning or cycling can be measured by a device called a about this in Chapter 11 Insight 11.1, Physiological Regulation
dynamometer. The metabolic rate has been found to be directly of Body Temperature.
THERMAL PUMPS: REFRIGERATORS, AIR CONDITIONERS,

AND HEAT PUMPS
The function performed by a thermal pump is basically the reverse of that of a heat
engine. The name thermal pump is a generic term for any device that transfers heat
from a low-temperature reservoir to a high-temperature reservoir (䉴 Fig. 12.16a),
including refrigerators, air conditioners, and heat pumps. For such a transfer to
occur, there must be work input. Since the second law of thermodynamics says that
heat will not spontaneously flow from a cold body to a hot body, the means for this
process to happen must be provided; that is, work must be done on the system.
A familiar example of a thermal pump is an air conditioner. By using input
work from electrical energy, heat is transferred from the inside of the house (low-
temperature reservoir) to the outside of the house (high-temperature reservoir), as
Heat output
Qh per cycle
Qh
Win = Qh – Qc Thermal
Win pump
THERMAL Qc
PUMP
Outside
Mechanical work
input per cycle Inside
Heat input Win
Qc per cycle (from
electrical
energy)
Low-temperature reservoir
(a) (b)
䉱 F I G U R E 1 2 . 1 6 Thermal pumps (a) An energy flow diagram for a generalized cyclic

thermal pump. The width of the arrow representing Qh, the heat transferred into the high-
temperature reservoir, is equal to the combined widths of the arrows representing Win and
Qc, reflecting the conservation of energy: Qh = Win + Qc. (b) An air conditioner is an
example of a thermal pump. Using the input work, it transfers heat (Qc) from a low-
temperature reservoir (inside the house) to a high-temperature reservoir (outside).
shown in Fig. 12.16b. A refrigerator (䉲 Fig. 12.17) uses the exact same principles and
processes. With the work performed by the compressor (Win), heat (Qc) is transferred
to the evaporator coils inside of the refrigerator. The combination of this heat and
work (Qh) is then expelled to the outside of the refrigerator through the condenser.
In essence, a refrigerator or air conditioner pumps heat up a temperature gradi-
ent, or “hill.” (Think of pumping water up an actual hill against the force of gravity.)
The cooling efficiency of this operation is based on the amount of heat extracted from
the low-temperature reservoir (the refrigerator, the freezer, or the inside of a house),
Qc , compared with the work Win needed to do so. Since a practical refrigerator oper-
ates in a cycle to provide continuous removal of heat, ¢U = 0 for the cycle. Then, by
the conservation of energy (the first law of thermodynamics), Qc + Win = Qh ,
where Qh is the heat ejected to the high-temperature reservoir, or the outside.
The measure of an air conditioner’s or a refrigerator’s performance is defined dif-
ferently from that of a heat engine, because of the difference in their functions. For the
cooling appliances, the efficiency is expressed as a coefficient of performance (COP).
䉳 F I G U R E 1 2 . 1 7 Refrigerator
Evaporator Heat (Qc) is carried away from the
(inside interior by the refrigerant as latent
refrigerator) heat. This heat energy and that of
Tc the work input (Win) are discharged
Qc Expansion
valve from the condenser to the surround-
ings (Qh). A refrigerator can be
Condenser thought of as a remover of heat (Qc)
(outside from an already cold region (its inte-
refrigerator)
rior) or as a heat pump that adds
heat (Qh) to an already warm area
Qh (the kitchen).
Compressor
Win
Since the purpose is to extract the most heat (Qc to make or keep things cold) per unit
of work input (Win), the coefficient of performance for a refrigerator or air conditioner
(COPref) is the ratio of these two quantities:
Qc Qc
COPref = = (refrigerator or air conditioner) (12.13)
Win Qh - Qc
Thus, the greater the COP, the better the performance—that is, more heat is
extracted for each unit of work done. For normal operation, the work input is less
than the heat removed, so the COP is greater than 1. The COPs of typical refrigera-
tors and air conditioners range from 3 to 5, depending on operating conditions
and design details. This range means that the amount of heat removed from the
cold reservoir (the refrigerator, freezer, or house interior) is three to five times the
amount of work needed to remove it.
Any machine that transfers heat in the opposite direction to that in which it
would naturally flow is called a thermal pump. The term heat pump is specifically
applied to commercial devices used to cool homes and offices in the summer and
to heat them in the winter. The summer operation is that of an air conditioner. In
this mode, it cools the interior of the house and heats the outdoors. Operating in
its winter heating mode, a heat pump heats the interior and cools the outdoors,
usually by taking heat energy from the cold air or ground.
For a heat pump in its heating mode, the heat output (to warm something up or
keep something warm) is the item of interest, so the COP for heat pump (hp) is
defined differently than that of a refrigerator or air conditioner. As you might
guess, it is the ratio of Qh to Win (the heating you get for the work put in), or
Qh Qh
COPhp = = (heat pump in heating mode) (12.14)
Win Qh - Qc
where, again, Qc + Win = Qh is used. Typical COPs for heat pumps range
between 2 and 4, again depending on the operating conditions and design.
Compared with electrical heating, heat pumps are very efficient. For each unit
of electric energy consumed, a heat pump typically pumps in from two to four
times as much heat as direct electric heating systems provide. Some heat pumps
use water from underground reservoirs, wells, or buried loops of pipe as a low-
temperature reservoir. These heat pumps are more efficient than the ones that use
the outside air, because water has a larger specific heat than air, and the average
temperature difference between the water and the inside air is usually smaller.
EXAMPLE 12.9 Air Conditioner/Heat Pump: Thermal Switch Hitting

A thermal pump operating as an air conditioner in summer (a) From Eq. 12.13, the COP for this thermal pump operating
extracts 1000 J of heat from the interior of a house for every as an air conditioner is
400 J of electric energy required to operate it. Determine (a)
Qc 1000 J
the air conditioner’s COP and (b) its COP if it runs as a heat COPref = = = 2.5
pump in the winter. Assume it is capable of moving the same Win 400 J
amount of heat for the same amount of electric energy, regard- (b) When the thermal pump operates as a heat pump, the rel-
less of the direction in which it runs. evant heat is the output heat, which can be calculated from
the conservation of energy:
T H I N K I N G I T T H R O U G H . The input work and input heat in
part (a) is known, so the definition of COP for a refrigerator (Eq. Qh = Qc + Win = 1000 J + 400 J = 1400 J
12.13) can be applied. For the reverse operation, it is the output
Thus, the COP for this engine operating as a heat pump in
heat that is important, so we use the definition of COP for a
winter is, from Eq. 12.14,
heat pump (Eq. 12.14).
Qh 1400 J
SOLUTION. COPhp = = = 3.5
Win 400 J
Given: Qc = 1000 J Find: (a) COPref (COP of air conditioner)
Win = 400 J (b) COPhp (COP of heat pump)
FOLLOW-UP EXERCISE. (a) Suppose you redesigned the thermal pump in this Example to perform the same operation, but with
25% less work input. What would be the new values of the two COPs? (b) Which COP would have the larger percentage increase?
12.6 THE CARNOT CYCLE AND IDEAL HEAT ENGINES 443
DID YOU LEARN?

➥ A heat engine takes heat from a hot reservoir, converts some of it into useful work,
and expels the rest into a cold reservoir.
➥ The efficiency of a heat engine is e = 1 - Qc>Qh , where Qh and Qc are heats
absorbed from the hot reservoir and expelled to the cold reservoir, respectively.
➥ A refrigerator moves heat from a cold reservoir into a hot reservoir by using work
from electrical energy.The total heat deposited into the hot reservoir is equal to the
sum of the heat removed from the cold reservoir and the work, that is,
Qh = Qc + Win.
12.6 The Carnot Cycle and Ideal Heat Engines

➥ Since a Carnot engine cannot be built, what then is the significance of a Carnot
engine?
➥ If two Carnot engines are operating at the same hot reservoir temperature but dif-
ferent cold reservoir temperatures, which Carnot engine is more thermodynamically
efficient?
➥ Is it possible to reach absolute zero?
Lord Kelvin’s statement of the second law of thermodynamics says that any cyclic
heat engine, regardless of its design, must always exhaust some heat energy
(Section 12.5). But how much heat must be lost in the process? In other words,
what is the maximum possible efficiency of a heat engine? In designing heat
engines, engineers strive to make them as efficient as possible, but there must be
some theoretical limit, and, according to the second law, it must be less than 100%.
Sadi Carnot (1796–1832), a French engineer, studied this limit. The first thing
he sought was the thermodynamic cycle an ideal heat engine would use, that is,
the most efficient cycle. Carnot found that the ideal heat engine absorbs heat
from a constant high-temperature reservoir (Th) and exhausts it to a constant low-
temperature reservoir (Tc). These processes are ideally reversible isothermal
processes and may be represented as two isotherms on a p–V diagram. But what
are the processes that complete the cycle? Carnot showed that these processes are
reversible adiabatic processes. As we saw in Section 12.3, the curves on a p–V dia-
gram are called adiabats and are steeper than isotherms (䉲 Fig. 12.18a). An irre-
versible heat engine operating between two heat reservoirs at constant
temperatures cannot have an efficiency greater than that of a reversible heat
engine operating between the same two temperatures.
p 1 T
Qh Q = (Th – Tc )∆S
Adiabat Isotherm Th
1 2
Temperature
(compression) Th
Pressure
2
Q
4 Adiabat
(expansion) Tc
4 3
Isotherm Tc
Qc 3
V S
Volume S1 S2
Entropy
(a) (b)
䉱 F I G U R E 1 2 . 1 8 The Carnot cycle (a) The Carnot cycle consists of two isotherms and
two adiabats. Heat is absorbed during the isothermal expansion and exhausted during the
isothermal compression. (b) On a T–S diagram, the Carnot cycle forms a rectangle, the area
of which is equal to Q.
Thus, the ideal Carnot cycle consists of two isotherms and two adiabats and is
conveniently represented on a T–S diagram, where it forms a rectangle (Fig.
12.18b). The area under the upper isotherm (1–2) is the heat added to the system
from the high-temperature reservoir: Qh = Th ¢S. Similarly, the area under the
lower isotherm (3–4) is the heat exhausted: Qc = Tc ¢S. Here, Qh and Qc are the
heat transfers at constant temperatures (Th and Tc , respectively). There is no heat
transfer 1Q = 02 during the adiabatic legs of the cycle. (Why?)
The difference between these heat transfers is the work output, which is equal
to the area enclosed by the process paths (the shaded areas on the diagrams):
Wnet = Qh - Qc = 1Th - Tc2¢S
Since ¢S is the same for both isotherms (see Fig. 12.18b, processes 1–2 and 3–4),
the ¢S expressions can be used to relate the temperatures and heats. That is,
since
Qh Qc
¢S = and ¢S =
Th Tc
then
Qh Qc Qc Tc
= or =
Th Tc Qh Th
This equation can be used to express the efficiency of an ideal heat engine in
terms of temperature. From Eq. 12.12, this ideal Carnot efficiency (Ec) is
Qc Tc
eC = 1 - = 1 -
Qh Th
or
Tc
eC = 1 - (Carnot efficiency, ideal heat engine) (12.15)
Th
where, as usual, the fractional efficiency is often expressed as a percentage. Note

that Tc and Th must be expressed in kelvins.
The Carnot efficiency expresses the theoretical upper limit on the thermody-
namic efficiency of a cyclic heat engine operating between two known tempera-
tures. In practice, this limit can never be achieved because no real engine processes
are reversible. A true Carnot engine cannot be built, because the necessary
reversible processes can only be approximated.
However, the Carnot efficiency does illustrate a general idea: The greater the
difference in the temperatures of the heat reservoirs, the greater the Carnot effi-
ciency. For example, if Th is twice Tc , or Tc>Th = 0.50, the Carnot efficiency is
Tc
eC = 1 - = 1 - 0.50 = 0.501* 100%2 = 50%
Th
However, if Th is four times Tc , or Tc>Th = 0.25, then

Tc
eC = 1 - = 1 - 0.25 = 0.751* 100%2 = 75%
Th
Since a heat engine can never attain 100% thermal efficiency, it is useful to com-
pare its actual efficiency e with its theoretical maximum efficiency, that of a Carnot
cycle, eC. To see the importance of this concept in more detail, study the next
Example carefully.
There are also Carnot COPs for refrigerators and heat pumps. (See Exercise 66.)
12.6 THE CARNOT CYCLE AND IDEAL HEAT ENGINES 445
EXAMPLE 12.10 Carnot Efficiency: The Dream Measure of Efficiency for Any Real Engine
An engineer is designing a cyclic heat engine to operate T H I N K I N G I T T H R O U G H . The maximum efficiency for specific
between the temperatures of 150 °C and 27 °C, (a) What is the high and low temperatures is given by Eq. 12.15. Remember,
maximum theoretical efficiency that can be achieved? (b) Sup- we must convert to absolute temperatures. In part (b), we cal-
pose the engine, when built, does 1000 J of work per cycle for culate the actual efficiency and compare it with our answer in
every 5000 J of input heat per cycle. What is its efficiency, and part (a).
how close is it to the Carnot efficiency?
SOLUTION.
Given: Th = 1150 + 2732 K = 423 K Find: (a) eC (Carnot efficiency)
Tc = 127 + 2732 K = 300 K (b) e (actual efficiency) and compare it with ec
Wnet = 1000 J
Qh = 5000 J
(a) Using Eq. 12.15 to find the maximum theoretical efficiency,
Tc 300 K
eC = 1 - = 1 - = 0.2911* 100%2 = 29.1%
Th 423 K
(b) The actual efficiency is, from Eq. 12.12,
= 0.200 1or 20.0%2

Wnet 1000 J
e = =
Qh 5000 J
Thus,
= 0.687 1or 68.7%2

e 0.200
=
eC 0.291
In other words, the heat engine is operating at 68.7% of its theoretical maximum. That’s pretty good.
F O L L O W - U P E X E R C I S E . If the operating high temperature of the engine in this Example were increased to 200 °C, what would
be the change in the theoretical efficiency?
THE THIRD LAW OF THERMODYNAMICS

Another inference might be drawn from the expression for the Carnot efficiency
(Eq. 12.15). It would seem possible to have eC equal to 100%, if only Tc could be
absolute zero. (See Section 10.3.) However, absolute zero has never been achieved,
although ultralow-temperature (cryogenic) experiments have come within 450 pK
14.5 * 10-10 K2 of it. Apparently, reducing the temperature of a system already
close to absolute zero in a finite number of steps is impossible. This is embodied in
the third law of thermodynamics which, simply stated, reads:
It is impossible to reach absolute zero in a finite number of thermal processes.
DID YOU LEARN?

➥ A Carnot engine sets the theoretical upper limit of the thermodynamic efficiency of
a heat engine operating between two temperatures. In other words, one should
never expect a heat engine to have an efficiency equal to or higher than the Carnot
efficiency.
➥ Since the Carnot efficiency is equal to eC = 1 - Tc>Th , the engine with a lower cold
reservoir temperature will have a smaller Tc>Th or greater 1 - Tc>Th , and therefore
a higher efficiency.
➥ The third law of thermodynamics states that it is impossible to reach absolute zero.
If it were possible, then the Carnot efficiency would be 100%, or heat could be
completely converted to work, which is a violation of the second law of
thermodynamics.
PULLING IT TOGETHER Ideal Gas Law, Thermodynamics, and Thermal Efficiency

Assume you have 0.100 mol of an ideal monatomic gas that energy, heat, and the ideal gas law. Care, however, needs to be
follows the cycle given in Fig. 12.14b and that the pressure taken because heat exchanges can occur during more than
and temperature at the lower left-hand corner of that figure one of the processes in the cycle. To determine heat input dur-
are 1.00 atm and 20 °C, respectively. Further assume that the ing the isobaric expansion, the change in internal energy and
pressure doubles during the isometric process and the vol- thus the change in temperature are needed. So it seems likely
ume also doubles during the isobaric expansion. What would that the temperatures at all four corners of the cycle will be
be the thermal efficiency of this cycle? needed. These can be calculated using the ideal gas law. The
four thermodynamic processes involved are two isobaric and
T H I N K I N G I T T H R O U G H . This example combines thermal effi- two isometric processes.
ciency (Eq. 12.12), thermal dynamic processes, work, internal
SOLUTION. The four corners are labeled with numbers as shown in Figure 12.14b. Listing the data given and converting to SI units,
Given: p4 = p3 = 1.00 atm = 1.01 * 105 N>m2 Find: e (thermal efficiency)
n = 0.100 mol
T4 = 20 °C = 293 K
p1 = p2 = 2.00 atm = 2.02 * 105 N>m2
V2 = V3 = 2V4 = 2V1
First, the volumes and temperatures at the corners are computed, using the ideal gas law:
nRT1 10.100 mol238.31 J>1mol # K241293 K2

V4 = V1 = = = 2.41 * 10-3 m3
p1 1.01 * 105 N>m2
Therefore,
V2 = V3 = 2V1 = 4.82 * 10-3 m3
During isometric processes, temperature (absolute in kelvins) is directly proportional to pressure (p>T = constant), and during
isobaric processes, temperature is directly proportional to volume (V>T = constant). Therefore,
T1 = 2T4 = 586 K
T2 = 2T1 = 1172 K
T3 = 12 T2 = 586 K
Now the heat transfers can be calculated. W = 0 during the 4–1 process, and for a monatomic gas, ¢U = 32 nR¢T. Therefore,
Q41 = ¢U41 = 32 nR¢T41 = 32 10.100 mol238.31 J>1mol # K241586 K - 293 K2 = + 365 J
During the 1–2 process, the gas expands and its internal energy increases. The work done by the gas is
W12 = p1 ¢V12 = 12.02 * 105 N>m2214.82 * 10-3 m3 - 2.41 * 10-3 m32 = + 487 J
Since work was done and the internal energy increased,
Q12 = ¢U12 + W12 = 32 nR¢T12 + 487 J

= 32 10.100 mol238.31 J>1mol # K2411172 K - 586 K2 + 487 J
= + 730 J + 487 J = + 1.22 * 103 J
Thus the total heat input per cycle, Qh , is
Qh = Q41 + Q12 = 1.59 * 103 J
To find the net work, we need the area enclosed by the cycle. Therefore,
Wnet = 1¢p2321¢V122 = 11.01 * 105 N>m2212.41 * 10-3 m32 = + 243 J
and the efficiency is
Wnet 243 J
e = = = 0.153 or 15.3%
Qh 1.59 * 103 J
■ The first law of thermodynamics is a statement of the conser- ■ A heat engine is a device that converts heat into work. Its
vation of energy for a thermodynamic system. Expressed in thermal efficiency e is the ratio of work output to heat
equation form, it relates the change in a system’s internal ener- input, or
gy to the heat flow and the work done by it and is written as Wnet Qh - Qc Qc
e = = = 1 - (12.12)
Q = ¢U + W (12.1) Qh Qh Qh
■ Some thermodynamic processes (for gases) are
isothermal: a process at constant temperature (T = constant)
isobaric: a process at constant pressure (p = constant)
Qin = Qh Heat input
per cycle
isometric: a process at constant volume (V = constant)
Wnet = Qh – Qc
adiabatic: a process involving no heat flow (Q = 0) Wnet
HEAT
ENGINE
p
Isothermal
Mechanical work
T2 = T1 output per cycle
Heat output
Qout = Qc
per cycle
1 Isotherm
Pressure
2 Low-temperature reservoir
T 2 = T1
■ A thermal pump is a device that transfers heat energy from

V
V1 V2 Volume a low-temperature reservoir to a high-temperature reser-
voir. The coefficient of performance (COP) is the ratio of
■ The expressions for thermodynamic work done by an ideal heat transferred to the input work. The COP differs depend-
gas during various processes are ing on whether the thermal pump is used as a heat pump or
as an air conditioner>refrigerator.
V2
Wisothermal = nRT ln ¢ ≤ (ideal gas isothermal process) (12.3)
V1 High-temperature reservoir
Wisobaric = p1V2 - V12 = p¢V (ideal gas isobaric process) (12.4)

Heat output
p1 V1 - p2 V2 Qh per cycle
Wadiabatic = (ideal gas adiabatic process) (12.9)
g - 1 Win = Qh – Qc
Win
(In the adiabatic process, g = cp>cv is the ratio of specific THERMAL
PUMP
heats at constant pressure and volume, respectively.)
Mechanical work
input per cycle
Heat input
Qc per cycle
Final
piston Low-temperature reservoir
position
∆x ∆V = A∆x ■ A Carnot cycle is a theoretical heat engine cycle consisting

A
Initial of two isotherms and two adiabats. Its efficiency is the high-
piston
F = pA
position est possible efficiency that any heat engine could have,
operating between two temperature extremes. The effi-
ciency of a Carnot cycle is
W = F∆x = pA∆x = p∆V
Tc
■ The second law of thermodynamics determines whether a eC = 1 - (12.15)
Th
process can take place naturally or, alternatively, specifies
p
the direction a process can take. 1
■ Entropy (S) is a measure of the disorder of a system. The Qh

Adiabat Isotherm Th
change in entropy of an object at constant temperature is (compression)
given by
Pressure
Q 4 Adiabat
¢S = (12.10) (expansion)
T
Isotherm Tc
The total entropy of the universe increases in every nat- Qc 3
V
ural process. Volume

12.1 THERMODYNAMIC SYSTEMS, 10. Which one of the following statements is a violation of
STATES, AND PROCESSES the second law of thermodynamics: (a) heat flows natu-
rally from hot to cold, (b) heat can be completely con-
1. On a p–V diagram, a reversible process is a process (a) verted to mechanical work, (c) the entropy of the
whose path is known, (b) whose path is unknown, (c) for universe can never decrease, or (d) it is not possible to
which the intermediate steps are nonequilibrium states, construct a perpetual motion engine?
11. An ideal gas is compressed isothermally. The change in
2. There may be an exchange of heat with the surroundings
entropy of the gas for this process is (a) positive,
for (a) a thermally isolated system, (b) a completely isolated
(b) negative, (c) zero, (d) none of the preceding.
system, (c) a heat reservoir, (d) none of the preceding.
3. Only initial and final states are known for irreversible
processes on (a) p–V diagrams, (b) p–T diagrams, (c) V–T 12.5 HEAT ENGINES AND THERMAL
diagrams, (d) all of the preceding. PUMPS
12. If the first law of thermodynamics is applied to a heat
12.2 THE FIRST LAW OF engine, the result is (a) Wnet = Qh + Qc ,
THERMODYNAMICS (b) Wnet = Qh - Qc , (c) Wnet = Qc - Qh , (d) Qc = 0.
AND 13. For a cyclic heat engine, (a) e = 1, (b) Qh = Wnet,
12.3 THERMODYNAMIC PROCESSES FOR (c) ¢U = Wnet , (d) Qh 7 Qc .
AN IDEAL GAS
14. A thermal pump (a) is rated by thermal efficiency,
4. There is no heat flow into or out of the system in an (b) requires work input, (c) has Qh = Qc , (d) has COP = 1.
(a) isothermal process, (b) adiabatic process, (c) isobaric
process, (d) isometric process. 15. Which of the following determines the thermal effi-
ciency of a heat engine: (a) Qc * Qh , (b) Qc>Qh ,
5. If the work done by a system is equal to zero, the process (c) Qh - Qc , or (d) Qh + Qc ?
is (a) isothermal, (b) adiabatic, (c) isobaric, (d) isometric.
6. According to the first law of thermodynamics, if work is
done on a system, then (a) the internal energy of the sys- 12.6 THE CARNOT CYCLE AND IDEAL
tem must change, (b) heat must be transferred from the HEAT ENGINES
system, (c) the internal energy of the system may change
16. The Carnot cycle consists of (a) two isobaric and two
and>or heat may be transferred from the system, (d) heat
isothermal processes, (b) two isometric and two adia-
must be transferred to the system.
batic processes, (c) two adiabatic and two isothermal
7. When heat is added to a system of an ideal gas during processes, (d) four arbitrary processes that return the
the process of an isothermal expansion, (a) work is done system to its initial state.
by the system, (b) the internal energy increases, (c) work
is done on the system, (d) the internal energy decreases. 17. Which of the following temperature reservoir relation-
ships would yield the lowest efficiency for a Carnot
engine: (a) Tc = 0.15Th , (b) Tc = 0.25Th , (c) Tc = 0.50Th ,
12.4 THE SECOND LAW OF or (d) Tc = 0.90Th ?
THERMODYNAMICS AND ENTROPY 18. For a heat engine that operates between two reservoirs
8. In any natural process, the overall change in the entropy of of temperatures Tc and Th , the Carnot efficiency is the
the universe could not be (a) negative, (b) zero, (c) positive. (a) highest possible value, (b) lowest possible value,
(c) average value, (d) none of the preceding.
9. For which type of thermodynamic process is the change
in entropy equal to zero: (a) isothermal, (b) isobaric, 19. If absolute zero were reached, then the Carnot efficiency
(c) isometric, or (d) none of the preceding? could be (a) 0%, (b) 50%, (c) 75%, (d) 100%.
12.1 THERMODYNAMIC SYSTEMS, 12.2 THE FIRST LAW OF

STATES, AND PROCESSES THERMODYNAMICS
AND
1. Explain why the process shown in Fig. 12.1b is not that
12.3 THERMODYNAMIC PROCESSES FOR
for an ideal gas at constant temperature.
AN IDEAL GAS
2. Does an irreversible process mean the system cannot
return to its original state? Explain. 4. On a p–V diagram, sketch a cyclic process that consists of
3. What are the four state variables, used in this chapter, an isothermal expansion followed by an isobaric com-
for ideal gases? pression, and lastly followed by an isometric process.
5. In 䉲 Fig. 12.19, the plunger of a syringe is pushed in 12.4 THE SECOND LAW OF
quickly, and the small pieces of paper in the syringe THERMODYNAMICS AND ENTROPY
catch fire. Explain this phenomenon using the first law
11. Heat is converted to mechanical energy in many applica-
of thermodynamics. (Similarly, in a diesel engine, there
tions, such as cars. Is this a violation of the second law of
are no spark plugs. How can the air–fuel mixture
thermodynamics? Explain.
ignite?)
12. Does the entropy of each of the following objects
increase or decrease? (a) Ice as it melts; (b) water vapor as
it condenses; (c) water as it is heated on a stove; (d) food
as it is cooled in a refrigerator.
13. When a quantity of hot water is mixed with a quantity of
cold water, the combined system comes to thermal equi-
librium at some intermediate temperature. How does
the entropy of the system (both liquids) change?
14. A student challenges the second law of thermodynamics
by saying that entropy does not have to increase in all
situations, such as when water freezes to ice. Is this chal-
lenge valid? Why or why not?
15. Is a living organism an open system or an isolated sys-
tem? Explain.
䉱 F I G U R E 1 2 . 1 9 Syringe fire See Conceptual Question 5.
16. A student tries to cool his dormitory room by opening
the refrigerator door. Will that work? Explain.
6. Discuss heat, work, and the change in internal energy of
your body when you shovel snow.
12.5 HEAT ENGINES AND THERMAL
7. In an adiabatic process, there is no heat exchange
PUMPS
between the system and the environment, but the tem-
perature of the ideal gas changes. How can this be? 17. What happens to the pressure and internal energy of a
Explain. cyclic heat engine after a complete cycle?
8. In an isobaric process, an ideal gas sample can do work 18. Lord Kelvin’s statement of the second law of thermody-
on the environment but its temperature also increases. namics as applied to heat engines (“No heat engine
How can this be? operating in a cycle can convert its heat input completely
to work”) refers to their operation in a cycle. Why is the
9. An ideal gas initially at temperature To, pressure po, and phrase “in a cycle” included?
volume Vo is compressed to one-half its initial volume.
As shown in 䉲 Fig. 12.20, process 1 is adiabatic, 2 is 19. If heat engine A absorbs more heat than heat engine B
isothermal, and 3 is isobaric. Rank the work done on the from a hot reservoir, will engine A necessarily do more
gas and the final temperatures of the gas, from highest to net work than engine B? Explain your reasoning.
lowest, for all three processes, and explain how you 20. The heat output of a thermal pump is greater than the
decided upon your rankings. energy used to operate the pump. Does this device vio-
late the first law of thermodynamics?
p 21. The maximum efficiency of a heat engine is 1 (or 100%).
Can the COP of a thermal pump be greater than 1?
Explain.
1
12.6 THE CARNOT CYCLE AND IDEAL
2
HEAT ENGINES
po To
3 22. Diesel engines are more efficient than gasoline engines.
Which type of engine wold you expect to run hotter?
O V Why?
Vo Vo
2 23. If you have the choice of running your heat engine
between either of the following two sets of temperatures
䉱 F I G U R E 1 2 . 2 0 Thermodynamic processes See Concep- for the cold and hot reservoirs, which would you choose,
tual Question 9. and why: between 100 °C and 300 °C, or between 50 °C
and 250 °C?
10. If ideal gas sample A receives more heat than ideal gas 24. Carnot engine A operates at a higher hot reservoir tem-
sample B, will sample A experience a higher increase in perature than Carnot engine B. Will engine A necessarily
internal energy? Explain. have a higher Carnot efficiency? Explain.
EXERCISES*
12.2 THE FIRST LAW OF

THERMODYNAMICS p
AND 1
12.3 THERMODYNAMIC PROCESSES FOR (2p1, V )
2 1
AN IDEAL GAS (2p1, V1)
Pressure
1. ● While playing in a tennis match, you lost 6.5 * 105 J of
heat, and your internal energy also decreased by
1.2 * 106 J. How much work did you do in the match? (p1, V1)
2. IE ● A rigid container contains 1.0 mol of an ideal gas that

slowly receives 2.0 * 104 J of heat. (a) The work done by
the gas is (1) positive, (2) zero, (3) negative. Why? V
Volume
(b) What is the change in the internal energy of the gas?
3. IE ● A quantity of ideal gas goes through an isothermal
process and does 400 J of net work. (a) The internal 䉱 F I G U R E 1 2 . 2 1 A p–V diagram for an ideal gas
energy of the gas is (1) higher than, (2) the same as, See Exercise 9.
(3) less than when it started. Why? (b) Is a net amount of
heat added to or removed from the system, and how 10. ●● A fixed quantity of gas undergoes the reversible
much is involved? changes illustrated in the p–V diagram in 䉲 Fig. 12.22.
4. ● An ideal gas goes through a thermodynamic process in How much work is done in each process?
which 500 J of work is done on the gas and the gas loses 300 J
of heat. What is the change in internal energy of the gas?
p
5. IE ● While doing 500 J of work, an ideal gas expands
4 5
adiabatically to 1.5 times its initial volume. (a) The tem- 1.00 × 105
perature of the gas (1) increases, (2) remains the same,
(3) decreases. Why? (b) What is the change in the inter-
Pressure (Pa)
nal energy of the gas?

2
6. IE ● An ideal gas expands from 1.0 m3 to 3.0 m3 at 0.50 × 105 3
atmospheric pressure while absorbing 5.0 * 105 J of heat
in the process. (a) The temperature of the system
(1) increases, (2) stays the same, (3) decreases. Explain. 1
(b) What is the change in internal energy of the system?
7. ●● An ideal gas is under an initial pressure of V
0 0.25 0.50 0.75 1.00
2.45 * 104 Pa and occupies a volume of 0.20 m3. The slow
Volume (m3)
addition of 8.4 * 103 J of heat to this gas causes it to
expand isobarically to a volume of 0.40 m3. (a) How much
work is done by the gas in the process? (b) Does the inter- 䉱 F I G U R E 1 2 . 2 2 A p–V diagram and work See Exercises
nal energy of the gas change? If so, by how much? 10 and 11.
8. ●● An Olympic weight lifter lifts 145 kg a vertical dis-
tance of 2.1 m. When he does so, 6.0 * 104 J of heat is 11. ●● Suppose that after the final process in Fig. 12.22 (see
transferred to air through perspiration. Does he gain or Exercise 10), the pressure of the gas is decreased isomet-
lose internal energy and how much? rically from 1.0 * 105 Pa to 0.70 * 105 Pa, and then the
gas is compressed isobarically from 1.0 m3 to 0.80 m3.
9. IE ● ● An ideal gas is taken through the reversible
What is the total work done in all of these processes,
processes shown in 䉴 Fig. 12.21. (a) Is the overall change
including 1 through 5?
in the internal energy of the gas (1) positive, (2) zero, or
(3) negative? Explain. (b) In terms of state variables p 12. IE ● ● A gas is enclosed in a cylindrical piston with a
and V, how much work is done by or on the gas, and 12.0-cm radius. Heat is slowly added to the gas while the
(c) what is the net heat transfer in the overall process? pressure is maintained at 1.00 atm. During the process,
*Take temperatures and efficiencies to be exact.

EXERCISES 451
the piston moves 6.00 cm. (a) This is an (1) isothermal,

(2) isobaric, (3) adiabatic process. Explain. (b) If the heat p
transferred to the gas during the expansion is 420 J, what
is the change in the internal energy of the gas?
Pressure (atm)
13. IE ● ● 2.0 mol of an ideal gas expands isothermally from D
a volume of 20 L to 40 L at 20 °C. (a) The work done by 2.00
C
the gas is (1) positive, (2) negative, (3) zero. Explain. T3
(b) What is the magnitude of the work?
1.00 T2
14. ●● A monatomic ideal gas 1g = 1.672 is compressed B A 400 K
adiabatically from a pressure of 1.00 * 105 Pa and vol- T1 200 K
ume of 240 L to a volume of 40.0 L. (a) What is the final V
0
pressure of the gas? (b) How much work is done on V1 V2
the gas? Volume
15. ●● An ideal gas sample expands isothermally by
tripling its volume and doing 5.0 * 104 J of work at 䉱 F I G U R E 1 2 . 2 4 A cyclic process See Exercise 18.
40 °C. (a) How many moles of gas are there in the sam-
ple? (b) Was heat added to or removed from the sample,
and how much? 12.4 THE SECOND LAW OF
16. IE ● ● ● The temperature of 2.0 mol of ideal gas is THERMODYNAMICS AND ENTROPY
increased from 150 °C to 250 °C by two different 19. ● What is the change in entropy of mercury vapor
processes. In process A, 2500 J of heat is added to the gas; 1Lv = 2.7 * 105 J>kg2 when 0.50 kg of it condenses to a
in process B, 3000 J of heat is added. (a) In which case is liquid at its boiling point of 357 °C?
more work done: (1) process A, (2) process B, or (3) the 20. IE ● 2.0 kg of ice melts completely into liquid water at
same amount of work is done? Explain. [Hint: See 0 °C. (a) The change in entropy of the ice (water) in this
Eq. 10.16.] (b) Calculate the change in internal energy process is (1) positive, (2) zero, (3) negative. Explain.
and work done for each process. (b) What is the change in entropy of the ice (water)?
17. IE ● ● ● One handred moles of a monatomic gas is com- 21. IE ● A process involves 1.0 kg of steam condensing to
pressed as shown on the p–V diagram in 䉲 Fig. 12.23. (a) water at 100 °C. (a) The change in entropy of the steam
Is the work done by the gas (1) positive, (2) zero, or (3) (water) is (1) positive, (2) zero, (3) negative. Why?
negative? Why? (b) What is the work done by the gas? (b) What is the change in entropy of the steam (water)?
(c) What is the change in temperature of the gas? (d) 22. ● During a liquid-to-solid phase change of a substance,
What is the change in internal energy of the gas? (e) its change in entropy is -4.19 * 103 J>K. If 1.67 * 106 J
How much heat is involved in the process? of heat is removed in the process, what is the freezing
point of the substance in degrees Celsius?
23. ●● In an isothermal expansion at 27 °C, an ideal gas does
p 1 60 J of work. What is the change in entropy of the gas?
5.0 × 105
24. IE ● ● One mole of an ideal gas undergoes an isothermal
compression at 0 °C, and 7.5 * 103 J of work is done in
4.0 × 105
compressing the gas. (a) Will the entropy of the gas
(1) increase, (2) remain the same, or (3) decrease? Why?
Pressure (Pa)
3.0 × 105 (b) What is the change in entropy of the gas?

2 25. IE ● ● A quantity of an ideal gas undergoes an isother-
2.0 × 105 mal expansion at 20 °C and does 3.0 * 103 J of work on
its surroundings in the process. (a) Will the entropy of
1.0 × 105 the gas (1) increase, (2) remain the same, or (3) decrease?
Explain. (b) What is the change in the entropy of the gas?
V 26. ●● In the winter, heat from a house with an inside temper-
0 0.50 1.0
ature of 18 °C leaks out at a rate of 2.0 * 104 J>s. The out-
Volume (m3) side temperature is 0 °C. (a) What is the change in entropy
per second of the house? (b) What is the total change in
entropy per second of the house–outside system?
䉱 F I G U R E 1 2 . 2 3 A variable p–V process and work
See Exercise 17. 27. IE ● ● An isolated system consists of two very large ther-
mal reservoirs at constant temperatures of 100 °C and
0 °C. Assume the reservoirs made contact and 1000 J of
heat flew from the cold reservoir to the hot reservoir
18. ● ● ● One mole of an ideal gas is taken through the cyclic spontaneously. (a) The total change in entropy of the iso-
process shown in 䉴 Fig. 12.24. (a) Compute the work lated system (both reservoirs) would be (1) positive,
involved for each of the four processes. (b) Find ¢U, W, (2) zero, (3) negative. Explain. (b) Calculate the total
and Q for the complete cycle. (c) What is T3? change in entropy of this isolated system.
28. IE ● ● Two large heat reservoirs at temperatures 200 °C 36. IE ● The heat output of a particular engine is 7.5 * 103 J
and 60 °C, respectively, are brought into thermal contact, per cycle, and the net work out is 4.0 * 103 J per cycle.
and 1.50 * 103 J of heat spontaneously flows from one (a) The heat input is (1) less than 4.0 * 103 J, (2) between
to the other with no significant temperature change. 4.0 * 103 J and 7.5 * 103 J, (3) greater than 7.5 * 103 J.
(a) The change in the entropy of the two-reservoir system Explain. (b) What is the heat input and thermal effi-
is (1) positive, (2) zero, (3) negative. Explain. (b) Calculate ciency of the engine?
the change in the entropy of the two-reservoir system. 37. ●● A gasoline engine burns fuel that releases 3.3 * 108 J
29. IE ● ● ● A system goes from state 1 to state 3 as shown on of heat per hour. (a) What is the energy input during a
the T–S diagram in 䉲 Fig. 12.25. (a) The heat transfer for 2.0-h period? (b) If the engine delivers 25 kW of power
the process going from state 2 to state 3 is (1) positive, during this time, what is its thermal efficiency?
(2) zero, (3) negative. Explain. (b) Calculate the total heat 38. IE ● ● A steam engine is to have its thermal efficiency
transferred in going from state 1 to state 3. improved from 8.00% to 10.0% while continuing to pro-
duce 4500 J of useful work each cycle. (a) Does the ratio
T of the heat output to heat input (1) increase, (2) remain
the same, or (3) decrease? Why? (b) What is the change in
Qc>Qh in this example?
3
373 39. IE ● ● An engineer redesigns a heat engine and improves
Temperature (K)
its thermal efficiency from 20% to 25%. (a) Does the ratio
273 of the heat input to heat output (1) increase, (2) remain
1 2 the same, or (3) decrease? Explain. (b) What is the
engine’s change in Qh>Qc?
40. ●● When running, a refrigerator exhausts heat to the
kitchen at a rate of 10 kW when the required input work
is done at a rate of 3.0 kW. (a) At what rate is heat
S removed from its cold interior? (b) What is the COP of
0 100 200
the refrigerator?
Entropy (J/K)
41. ●● A refrigerator with a COP of 2.2 removes 4.2 * 105 J
of heat from its interior each cycle. (a) How much heat is
䉱 F I G U R E 1 2 . 2 5 Entropy and heat See exhausted each cycle? (b) What is the total work input in
Exercises 29 and 30. joules for 10 cycles?
42. ●● An air conditioner has a COP of 2.75. What is the
30. IE ● ● ● Suppose that the system described by the T–S dia- power rating of the unit if it is to remove 1.00 * 107 J of
gram in Fig. 12.25 is returned to its original state, state 1, heat from a house interior in 20 min?
by a reversible process depicted by a straight line from 43. ●● A heat pump removes 2.2 * 103 J of heat from the
state 3 to state 1. (a) The change in entropy of the system outdoors and delivers 4.3 * 103 J of heat to the inside of
for this overall cyclic process is (1) positive, (2) zero, a house each cycle. (a) How much work is required per
(3) negative. Explain. (b) How much heat is transferred in cycle? (b) What is the COP of this pump?
the cyclic process? [Hint: See Example 12.6.]
44. ●● A steam engine has a thermal efficiency of 15.0%. If
31. ●●● A 50.0-g ice cube at 0 °C is placed in 500 mL of its heat input for each cycle is supplied by the condensa-
water at 20 °C. Estimate the change in entropy (after all tion of 8.00 kg of steam at 100 °C. (a) what is the net
the ice has melted) (a) for the ice, (b) for the water, and work output per cycle, and (b) how much heat is lost to
(c) for the ice–water system. the surroundings in each cycle?
45. ● ● ● A coal-fired power plant produces 900 MW of elec-
12.5 HEAT ENGINES AND THERMAL tric power and operates at a thermal efficiency of 25%
PUMPS (a) What is the input heat rate from the burning coal?
(b) What is the rate of heat discharge from the plant?
32. ● If an engine does 200 J of net work and exhausts 800 J (c) Water at 15 °C from a nearby river is used to cool the
of heat per cycle, what is its thermal efficiency? discharged heat. If the cooling water is not to exceed a
33. ● A gasoline engine has a thermal efficiency of 28%. If temperature of 40 °C, how many gallons per minute of
the engine absorbs 2000 J of heat per cycle, (a) what is the the cooling water is required?
net work output per cycle? (b) How much heat is 46. ● ● ● A gasoline engine has a thermal efficiency of 25.0%.
exhausted per cycle? If heat is expelled from the engine at a rate of

1.50 * 106 J>h, how long does the engine take to perform
34. ● A heat engine with a thermal efficiency of 20% does a task that requires an amount of work of 1.5 * 106 J?
500 J of net work each cycle. How much heat per cycle is
47. ● ● ● A four-stroke engine runs on the Otto cycle. It deliv-
lost to the low-temperature reservoir?
ers 150 hp at 3600 rpm. (a) How many cycles are in 1
35. ● An internal combustion engine with a thermal efficiency min? (b) If the thermal efficiency of the engine is 20%,
of 15.0% absorbs 1.75 * 105 J of heat from the hot reser- what is the heat input per minute? (c) How much heat is
voir. How much heat is lost by the engine in each cycle? wasted (per minute) to the environment?
EXERCISES 453
12.6 THE CARNOT CYCLE AND IDEAL 60. ●● In each cycle, a Carnot engine takes 800 J of heat from
HEAT ENGINES a high-temperature reservoir and discharges 600 J to a
low-temperature reservoir. What is the ratio of the tem-
48. ● A Carnot engine has an efficiency of 35% and takes in perature of the high-temperature reservoir to that of the
heat from a high-temperature reservoir at 178 °C. What is low-temperature reservoir?
the Celsius temperature of the engine’s low-temperature
reservoir? 61. IE ● ● A Carnot engine operating between reservoirs at
49. ● A steam engine operates between 100 °C and 20 °C. 27 °C and 227 °C does 1500 J of work in each cycle.
What is the Carnot efficiency of the ideal engine that (a) The change in entropy for the engine for each cycle is
operates between these temperatures? (1) negative, (2) zero, (3) positive. Why? (b) What is the
heat input of the engine?
50. ● It has been proposed that temperature differences in
the ocean could be used to run a heat engine to generate 62. ● ● The autoignition temperature of a fuel is defined as the
electricity. In tropical regions, the water temperature is temperature at which a fuel–air mixture would self-
about 25 °C at the surface and about 5 °C at very deep explode and ignite. Thus, it sets an upper limit on the
depths. (a) What would be the maximum theoretical effi- temperature of the hot reservoir in an automobile
ciency of such an engine? (b) Would a heat engine with engine. The autoignition temperatures for commonly
such a low efficiency be practical? Explain. available gasoline and diesel fuel are about 495 °F and
600 °F, respectively. What are the maximum Carnot effi-
51. ● What is the Celsius temperature of the hot reservoir of
ciencies of a gasoline engine and a diesel engine if the
a Carnot engine that is 32% efficient and has a 20 °C cold
cold reservoir temperature is 40 °C?
reservoir?
52. ● An engineer wants to run a heat engine with an effi- 63. ●● Because of limitations on materials, the maximum
ciency of 40% between a high-temperature reservoir at temperature of the superheated steam used in a turbine
300 °C and a low-temperature reservoir. What is the for the generation of electricity is about 540 °C. (a) If the
maximum Celsius temperature of the low-temperature steam condenser operates at 20 °C, what is the maxi-
reservoir? mum Carnot efficiency of a steam turbine generator?
(b) The actual efficiency of such generators is about 35%
53. ●● A Carnot engine with an efficiency of 40% operates to 40%. What does this range tell you?
with a low-temperature reservoir at 40 °C and exhausts
1200 J of heat each cycle. What are (a) the heat input per 64. ●● The working substance of a cyclic heat engine is
cycle and (b) the Celsius temperature of the high- 0.75 kg of an ideal gas. The cycle consists of two isobaric
temperature reservoir? processes and two isometric processes, as shown in
䉲 Fig. 12.26. What would be the efficiency of a Carnot
54. ●● A Carnot engine takes 2.7 * 104 J of heat per cycle
from a high-temperature reservoir at 320 °C and exhausts engine operating with the same high-temperature and
some of it to a low-temperature reservoir at 120 °C How low-temperature reservoirs?
much net work is done by the engine per cycle?
55. IE ● ● A Carnot engine takes in heat from a reservoir at p
1 2
350 °C and has an efficiency of 35%. The exhaust tempera- 250
ture is not changed and the efficiency is increased to 40%,
Pressure (kPa)
(a) The temperature of the hot reservoir is (1) lower than

(2) equal to, (3) higher than 350 °C. Explain. (b) What is
the new Celsius temperature of the hot reservoir?
150
56. ●● An inventor claims to have created a heat engine that 4 3
produces 10.0 kW of power for a 15.0-kW heat input
while operating between reservoirs at 27 °C and 427 °C. V
0 1.75 2.25
(a) Is this claim valid? (b) To produce 10.0 kW of power,
what is the minimum heat input required? Volume (× 10–2 m3)
57. ●● An inventor claims to have developed a heat engine

that, each cycle, takes in 5.0 * 105 J of heat from a high- 䉱 F I G U R E 1 2 . 2 6 Thermal efficiency See Exercise 64.
temperature reservoir at 400 °C and exhausts 2.0 * 105 J
to the surroundings at 125 °C. Would you invest your
money in the production of this engine? Explain. 65. IE ● ● Equation 12.15 shows that the greater the tempera-
ture difference between the reservoirs of a heat engine, the
58. ●● A heat engine operates at a thermal efficiency that is greater the engine’s Carnot efficiency. Suppose you had
45% of the Carnot efficiency. If the temperatures of the the choice of raising the temperature of the high-tempera-
high-temperature and low-temperature reservoirs are ture reservoir by a certain number of kelvins or lowering
400 °C and 50 °C, respectively, what are the Carnot effi- the temperature of the low-temperature reservoir by the
ciency and the thermal efficiency of the engine? same number of kelvins. (a) To produce the largest
59. ●● A heat engine’s thermal efficiency is 70.0% of the increase in efficiency, you should choose (1) to raise the
Carnot efficiency of an engine operating between tem- high-temperature reservoir, (2) to lower the low-tempera-
peratures of 80 °C and 375 °C.(a) What is the Carnot effi- ture reservoir, (3) both 1 and 2 produce the same change in
ciency of the heat engine? (b) If the heat engine absorbs efficiency, so it does not matter which you choose. Explain.
heat at a rate of 50 kW, at what rate is heat exhausted? (b) Prove your answer to part (a) mathematically.
66. ● ● ● There is a Carnot coefficient of performance (COPC) 68. ● ● ● An ideal heat pump is equivalent to a Carnot
for an ideal, or Carnot, refrigerator. (a) Show that this engine running in reverse. (a) Show that the Carnot COP
quantity is given by of the heat pump is
Tc 1
COPC = COPC =
Th - Tc eC
(b) What does this tell you about adjusting the tempera- where eC is the Carnot efficiency of the heat engine. (b) If
tures for the maximum COP of a refrigerator? (Can you a Carnot engine has an efficiency of 40%, what would be
guess the equation for the COPC for a heat pump?) the COPC when it runs in reverse as a heat pump? (See
Exercise 66.)
67. ● ● ● A salesperson tells you that a new refrigerator with
a high COP removes 2.6 * 103 J (each cycle) from the

inside of the refrigerator at 5.0 °C and expels 2.8 * 103 J
into the 30 °C kitchen. (a) What is the refrigerator’s
COP? (b) Is this scenario possible? Justify your answer.
(See Exercise 66.)
69. A heat engine with a thermal efficiency of 25% is used to 73. A Carnot engine is to produce 100 J of work per cycle. If
hoist 2.5-kg bricks to an elevation of 3.0 m. If the engine 300 J of heat is exhausted to a 27 °C cold reservoir per
expels heat to the environment at a rate of 1.2 * 106 J>h, cycle, what is the change in entropy of the hot reservoir
how many bricks can the engine hoist in 2.0 h? per cycle?
70. When cruising at 75 mi>h on a highway, a car’s engine 74. A quantity of an ideal gas at an initial pressure of 2.00
develops 45 hp. If this engine has a thermodynamic effi- atm undergoes an adiabatic expansion to atmospheric
ciency of 25% and 1 gal of gasoline has an energy content pressure. What is the ratio of the final temperature to the
of 1.3 * 108 J, what is the fuel efficiency (in miles per initial temperature of the gas?
gallon) of this car?
71. A gram of water (volume of 1.00 cm3) at 100 °C is con- 75. A 100-MW power generating plant has an efficiency of
verted to 1.67 * 103cm3 of steam at atmospheric pres- 40%. If water is used to carry off the wasted heat and the
sure. What is the change in the internal energy of the temperature of the water is not to increase by more than
water (steam)? 10 °C, what mass of water must flow through the plant
each second?
72. In a highly competitive game, a basketball player can
produce 300 W of power. Assuming the efficiency of the 76. An ice machine is to convert 10 °C water to 0 °C ice. If
player’s “engine” is 15% and heat dissipates primarily the machine has a COP of 2.0 and consumes electrical
through the evaporation of perspiration, what mass of power at a rate of 1.0 kW, how much ice can it make
perspiration is evaporated per hour? in 1.0 h?
13 Vibrations and Waves
13.1 Simple harmonic
motion (456)
■ Hooke’s law
■ frequency and period
13.2 Equations of motion (459)

■ mass–spring system
■ simple pendulum
13.3 Wave motion (468)

■ transverse and longitudinal
waves
■ wave equation
Wave properties (473) PHYSICS FACTS
T
13.4
■ interference and diffraction ✦ Disturbances set up waves. Soldiers
he chapter-opening photo-
■ reflection and refraction marching across older wooden graph depicts what a lot of
bridges are told to break step and
not march in a periodic cadence. people probably first think of when
Such a cadence might correspond
to a natural frequency of the bridge,
hearing the word wave. We’re all
13.5 Standing waves and
resonance (477) resulting in resonance and large familiar with ocean waves and their
oscillations that could damage the
■ nodes and antinodes
bridge and even cause it to collapse. smaller relatives, the ripples that
■ harmonic series This happened in 1850 in France.
About 500 soldiers marching across
form on the surface of a lake or
a suspension bridge over a river pond when something disturbs the
caused a resonant vibration that rose
to such a level that the bridge col- surface. However, the waves that
lapsed. Over 200 soldiers drowned.
are most important to us, as well as
✦ Tidal waves are not related to tides. A
more appropriate name for them is most interesting to physicists,
the Japanese name tsunami, which
either are invisible or don’t look like
means “harbor wave.” The waves are
generated by subterranean earth- water waves. Sound, for example, is
quakes and can race across the
ocean at speeds up to 960 km>h,
a wave. Perhaps most surprisingly,
with little surface evidence. When a light is a wave. In fact, all electro-
tsunami reaches the shallow coast,
friction slows the wave down, at the magnetic radiations are waves—
same time causing it to roll up into a
5- to 30-m-high wall of water that
radio waves, microwaves, X-rays,
crashes down on the shoreline. and so on. Whenever you peer
456 13 VIBRATIONS AND WAVES
through a microscope, put on a pair of glasses, or look at a rainbow, you are experi-
encing wave energy in the form of light. In Section 28.1, you’ll learn how even
k moving particles have wavelike properties. But first we need to look at the basic
m
description of waves.
x=0 In general, waves are related to vibrations or oscillations—back-and-forth
(a) Equilibrium motion—such as that of a mass on a spring or a swinging pendulum, and funda-
mental to such motions are restoring forces or torques. In a material medium, the
Fs restoring force is provided by intermolecular forces. If a molecule is disturbed,
Fa
restoring forces from interactions with its neighbors tend to return the molecule to
x = 0 x = ⫹A
its original position, and it begins to oscillate. In so doing, it affects adjacent mole-
cules, which are in turn set into oscillation, and so on. This is referred to as
(b) t = 0 Just before release
propagation. One might ask, “What is propagated by the molecules in a material?”
Fs = 0 The answer is energy. A single disturbance, which happens when you give the end
of a stretched rope a quick shake, gives rise to a wave pulse. A continuous, repeti-
tive disturbance gives rise to a continuous propagation of energy called a wave
x=0
motion. But before looking at waves in media, it is helpful to analyze the oscilla-
1
(c) t = T
4 tions of a single mass.
Fs
13.1 Simple Harmonic Motion
x = –A x = 0 LEARNING PATH QUESTIONS
1 ➥ What is the type of force necessary for an object to be in simple harmonic motion?
(d) t = 2
T
➥ At what position of mass is the total energy of a mass–spring system a maximum?
➥ At what position pf mass is the speed of a mass–spring system a maximum?
Fs = 0
The motion of an oscillating object depends on the restoring force that makes the
object go back and forth. It is convenient to begin to study such motion by consid-
x=0 ering the simplest type of force acting along the x-axis: a force that is directly pro-
3
portional to the object’s displacement from equilibrium. A common example is the
(e) t = 4
T (ideal) spring force, described by Hooke’s law (Section 5.2),
Fs = - kx (Hooke’s law) (13.1)

Fs
where k is the spring constant representing the stiffness of the spring. The negative
sign indicates that the spring force is always opposite to its displacement. That is,
x = 0 x = ⫹A the force always tends to restore the object to the spring’s equilibrium position.
(f) t = T
Suppose that an object on a horizontal frictionless surface is connected to a
spring as shown in 䉳 Fig. 13.1. When the object is displaced to one side of its equi-
䉱 F I G U R E 1 3 . 1 Simple harmonic librium position and released, it will move back and forth—that is, it will vibrate,
motion (SHM) When an object on a or oscillate. Here, an oscillation or a vibration is clearly a periodic motion—a motion
spring (a) is at its equilibrium posi-
tion 1x = 02 and then (b) is dis-
that repeats itself again and again along the same path. For linear oscillations, like
placed and released, the object those of an object attached to a spring, the path may be back and forth or up and
undergoes SHM (assuming no fric- down. For the angular oscillation of a pendulum, the path is back and forth along
tional losses). The time it takes to a circular arc.
complete one cycle is the period of The motion under the influence of the type of force described by Hooke’s law is
oscillation (T). (Here, Fs is the spring called simple harmonic motion (SHM). This is because the force is the simplest
force and Fa is the applied force.)
(c) At t = T>4, the object is back at restoring force and the motion can be described by harmonic functions (sine and
its equilibrium position; (d) at cosine functions), as will be seen later in the chapter. The change in position of an
t = T>2, it is at x = - A. (e) During object in SHM from its equilibrium position is the object’s displacement, a vector
the next half cycle, the motion is to quantity (Section 2.2). Often, the equilibrium position is chosen to be at the origin,
the right; (f) at t = T, the object is
back at its initial 1t = 02 starting
so xo = 0; then the displacement ¢x = x - xo = x. Note in Fig. 13.1 that the dis-
position as in (b). placement can be either positive or negative, which indicates direction. The maxi-
13.1 SIMPLE HARMONIC MOTION 457
TABLE 13.1 Terms Used to Describe Simple Harmonic Motion

displacement—the change in position of an object measured from its
equilibrium position 1x - xo = x with xo = 02.
amplitude (A)—the magnitude of the maximum displacement, or the maximum
distance, of an object from its equilibrium position.
period (T)—the time for one complete cycle of motion.
frequency (f)—the number of cycles per second (in hertz or inverse seconds,
where f = 1>T).
mum displacements are +A and - A (Fig. 13.1b, d). The magnitude of the maxi-
mum displacement, or the maximum distance of an object from its equilibrium
position, is called the object’s amplitude (A), a scalar quantity that expresses the
distance of both extreme displacements from the equilibrium position.
Besides the amplitude, two other important quantities used in describing an
oscillation are its period and frequency. The period (T) is the time it takes the
object to complete one cycle of motion. A cycle is a complete round trip, or motion
through a complete oscillation. For example, if an object starts at x = A
(Fig. 13.1b), then when it returns to x = A (as in Fig. 13.1f), it will have completed
one cycle in one period. If an object were initially at x = 0 when disturbed, then its
second return to this point would mark a cycle. (Why a second return?) In either
case, the object would travel a distance of 4A during one cycle. Can you show
this?
The frequency (f) is the number of cycles per second. The frequency and the
period are inversely proportional, that is,
1
f = (frequency and period) (13.2)
T
SI unit of frequency: hertz 1Hz2, or cycles per second 1cycles>s or 1>s or s -12.
The inverse relationship is reflected in the units. The period is the number of seconds
per cycle, and the frequency is the number of cycles per second. For example,
if T = 12 s>cycle, then it completes 2 cycles each second, or f = 2 cycles>s.
The standard unit of frequency is the hertz (Hz), which is 1 cycle per second.*
From Eq. 13.2, frequency has the unit inverse seconds (1>s, or s -1), since the period
is a measure of time. Although a cycle is not really a unit, it is convenient at times to
express frequency in cycles per second to help with unit analysis. This is similar to
the way the radian (rad) is used in the description of circular motion in Sections 7.1
and 7.2.
The terms used to describe SHM are summarized in 䉱 Table 13.1.
ENERGY AND SPEED OF A MASS—SPRING SYSTEM IN SHM

Recall from Section 5.4 that the potential energy stored in a spring that is stretched
or compressed a distance x from equilibrium (chosen to be xo = 0) is
U = 12 kx 2 (potential energy of a deformed spring) (13.3)
The change in potential energy of an object oscillating on a spring is related to the
work done by the spring force. An object with mass m oscillating on a spring also
has kinetic energy. The kinetic and potential energies together give the total
mechanical energy E of the system:
E = K + U = 12 mv 2 + 12 kx 2 (total energy) (13.4)
*The unit is named for Heinrich Hertz (1857–1894), a German physicist and early investigator of
electromagnetic waves.
1 1 1
E = Umax = 2
kA2 E = Kmax = 2
mv2max E = Umax = 2
kA2
v=0 vmax v=0

m
F max m F max m
x = –A x=0 x = +A
K=0 U=0 K=0
䉱 F I G U R E 1 3 . 2 Oscillations and energy For a mass oscillating in SHM on a spring (on a

frictionless surface), the total energy at the amplitude positions 1⫾A2 is all potential
energy (Umax), and E = 12 kA2, which is the total energy of the system. At the center posi-
tion 1x = 02, the total energy is all kinetic energy (E = 12 mv2max , where m is the mass of the
block). How is the total energy divided at locations somewhere between x = 0 and x = ⫾A?
When the object is at one of its maximum displacements, x = + A or -A, it is

instantaneously at rest, v = 0 (䉱 Fig. 13.2). Thus, all the energy is in the form of
potential energy (Umax) at this location; that is,
E = 12 m1022 + 12 k1⫾A22 = 12 kA2
or
(total energy of an object

E = 12 kA2 (13.5)
in SHM on a spring)
This outcome is a general result for SHM:
The total energy of an object in simple harmonic motion is directly proportional to

the square of the amplitude.
Equation 13.5 allows us to express the velocity of an object oscillating on a spring

as a function of position:
1
E = K + U or 2 kA
2
= 12 mv 2 + 12 kx 2
Solving for v2 and taking the square root:
1A2 - x 22
k
v = ⫾ (velocity of an object in SHM) (13.6)
Am
where the positive and negative signs indicate the direction of the velocity. Note
that at x = ⫾A, the velocity is zero, since the object is instantaneously at rest at its
maximum displacement from equilibrium.
Note also that when the oscillating object passes through its equilibrium posi-
tion 1x = 02, its potential energy is zero. At that instant, the energy is all kinetic,
and the object is traveling at its maximum speed vmax. The expression for the
energy in this case is
E = 12 kA2 = 12 mv 2max
and
k (maximum speed
vmax = A (13.7)
Am of mass on a spring)
In the next Example, as well as in the accompanying Learn by Drawing 13.1,

Oscillating in a Parabolic Potential Well, you can visualize the continuous trade-off
between kinetic and potential energy.
13.2 EQUATIONS OF MOTION 459
EXAMPLE 13.1 A Block and a Spring: Simple Harmonic Motion

A block with a mass of 0.50 kg sitting on a frictionless surface SOLUTION. First the given data and what is to be found are
is connected to a light spring that has a spring constant of listed, as usual. The initial position corresponds to the ampli-
180 N>m (see Fig. 13.1). If the block is displaced 15 cm from tude. (Why?) The speeds can be calculated with Eq. 13.6.
its equilibrium position and released, what are (a) the total
Given: m = 0.50 kg Find: (a) E (total energy)
energy of the system, (b) the speed of the block when it is
k = 180 N>m (b) v (speed)
10 cm from its equilibrium position, and (c) the maximum
A = 15 cm = 0.15 m (c) vmax
speed of the block?
x = ⫾10 cm = ⫾0.10 m
T H I N K I N G I T T H R O U G H . The total energy depends on the
(a) The total energy is given by Eq. 13.5:
spring constant (k) and the amplitude (A), which are given. At
x = 10 cm, the speed should be less than the maximum speed. E = 12 kA2 = 12 1180 N>m210.15 m22 = 2.0 J
(Why?)
(b) The instantaneous speed of the block at a distance 10 cm from the equilibrium position is given by Eq. 13.6. As you can see,
whether x = + 0.10 m or x = - 0.10 m makes no difference because of x2 in Eq. 13.6.
1A2 - x 22 = 310.15 m22 - 1⫾0.10 m224 = 34.5 m2>s 2 = 2.1 m>s

k 180 N>m
v =
Am A 0.50 kg
(c) The maximum speed occurs at x = 0, so Eq. 13.2 becomes
310.15 m22 - 10224 = 2.8 m>s

180 N>m
vmax =
A 0.50 kg
You could also use Eq. 13.7 directly to calculate vmax.
FOLLOW-UP EXERCISE. What is the magnitude of the acceleration when x = ⫾0.10 m? What is the maximum acceleration?
DID YOU LEARN?

➥ The type of force in SHM has to be a restoring force, such as that described by Hooke’s
law.The direction of the force has to be toward the equilibrium position at all times.
➥ The total energy of a mass–spring system is constant, so it is the same at any
position. However, the kinetic energy and potential energy that make up the total
energy do vary at different positions.
➥ The speed of a mass–spring system is at a maximum at the equilibrium position,
where the potential energy is zero.The energy is all kinetic; therefore, the speed is at
a maximum.
13.2 Equations of Motion

➥ What are the two possible trigonometric functions of the equations of motion that
could be used for simple harmonic motion?
➥ Does the oscillation period or frequency of a mass–spring system or a simple pen-
dulum depend on the amplitude of vibration?
➥ How do the motions of two objects in SHM compare if they have a phase difference
of 180°?
The equation that gives an object’s position as a function of time is referred to as

the equation of motion. For example, the equation of motion with a constant linear
acceleration is x = xo + vo t + 12 at2, where vo is the initial velocity (Section 2.4).
However, the acceleration is not constant in simple harmonic motion, so the kine-
matic equations of Section 2.4 do not apply to this case.
The equation of motion for an object in SHM can be found from a relationship
between simple harmonic and uniform circular motions. SHM can be simulated
LEARN BY DRAWING 13.1 Since E is constant and independent of x, it is plotted as a

horizontal line (shown in green). The kinetic energy is the
part of the total energy that is not potential energy; that is,
oscillating in a parabolic potential well K = E - U. It can be graphically interpreted (purple arrow)
A way to visualize the conservation of energy in simple har- as the vertical distance between the potential energy
monic motion is shown in Fig. 1. The potential energy of a parabola and the horizontal green total energy line. As the
mass–spring system can be sketched on a plot of energy (E) object oscillates on the x-axis, the energy trade-offs can be
versus position (x). Since U = 12 kx 2 r x 2, the graph is a visualized as the lengths of the two arrows change.
parabola. A general location, x1, is shown in Fig. 2. Neither the
In the absence of nonconservative forces, the total energy kinetic energy nor the potential energy is at its maximum
of the system, E, is constant. But E is the sum of the kinetic value of E there. These maximum values occur instead at
and potential energies. During the oscillations, there is a x = 0 and x = ⫾A, respectively. The motion cannot exceed
continuous trade-off between the two types of energies, but x = ⫾A, because that would imply a negative kinetic
their sum remains constant. Mathematically, this relation- energy, which is physically impossible. (Why?) The ampli-
ship is written as E = K + U. In Fig. 2, the potential energy tude positions are sometimes called the endpoints of the
U (indicated by a blue arrow) is represented by the vertical motion, because they are the locations where the speed is
distance from the x-axis. instantaneously zero and the object reverses direction.
k
m
Energy
x = −A x=0 x = +A
1 2
U = 2
kx
E
Energy
1
U = kx2 E = constant
2
K
Kmax = E Umax = E
U
−A x=0 +A −A x=0 x1 +A
F I G U R E 1 The potential energy “well” of a F I G U R E 2 Energy transfers as the spring–mass system oscil-
spring–mass system The potential energy of a lates The vertical distance from the x-axis to the parabola is the
spring that is stretched or compressed from its system’s potential energy. The remainder—the vertical distance
equilibrium position 1x = 02 is a parabola, since between the parabola and the horizontal line representing the
U r x 2. At x = ⫾A, all of the system’s energy is system’s constant total energy E—is the system’s kinetic
potential. energy (K).
by a component of uniform circular motion, as illustrated in 䉴 Fig. 13.3. As the

object moves in uniform circular motion (with constant angular speed v) in a verti-
cal plane, its shadow moves back and forth vertically, following the same path as
the object on the spring, which is in simple harmonic motion. Since the shadow
and the object have the same position at any time, it follows that the equation of
motion for the shadow of the object in circular motion is the same as the equation
of motion for the oscillating object on the spring.
From the reference circle in Fig. 13.3b, the y-coordinate (position) of the object is
given by
y = A sin u
But the object moves with a constant angular velocity of magnitude v. In terms of
the angular distance u, assuming that uo = 0 at t = 0, then u = vt, so
(SHM for yo = 0.
y = A sin vt (13.8)
initial upward motion)
Shadow
on block
y=+A
+y +y v
t at t > 0
Ligh y=0
m
fro nt A A y = A sin θ
a
dist ce u
= A sin vt
ur u
so
at t = 0
y=–A
Screen –y
–y
(a) (b)
䉱 F I G U R E 1 3 . 3 Reference circle for vertical motion (a) The shadow of an object in uni-
form circular motion has the same vertical motion as an object oscillating on a spring in
simple harmonic motion. (b) The motion can be described by y = A sin u = A sin vt
(assuming that y = 0 at t = 0).
Note that as t increases from zero, y increases in the positive direction, so the
equation describes initial upward motion.
With Eq. 13.8 as the equation of motion, the mass must be initially at yo = 0. But
what if the mass on the spring were initially at the amplitude position +A? In that
case, the sine equation would not describe the motion, because it does not describe
the initial condition—that is, yo = + A at to = 0. So another equation of motion is
needed, and y = A cos vt applies. By this equation, at to = 0, the mass is at
yo = A cos vt = A cos v102 = + A, and the cosine equation correctly describes the
initial conditions (䉲 Fig. 13.4):
(initial downward motion
y = A cos vt (13.9)
with yo = + A)
Here, the initial motion is downward, because, for times shortly after to = 0, the
value of y decreases. If the amplitude were -A, the mass would initially be at the
bottom and the initial motion would be upward.
䉳 F I G U R E 1 3 . 4 Sinusoidal equa-
tion of motion As time passes, the
oscillating object traces out a sinu-
k soidal curve on the moving paper.
In this case, y = A cos vt, because
the object’s initial displacement is
y = A cos ␻t +A +y yo = + A.
Displacement
Time m y=0
t=0
y = +A –A –y
T
1 period
Thus, the equation of motion for an oscillating object may be either a sine or a
cosine function. Both of these functions are referred to as being sinusoidal. That is,
simple harmonic motion is described by a sinusoidal function of time.
The angular speed v 1in rad>s2 of the reference circle object (Fig. 13.3) is called the
angular frequency of the oscillating object, since v = 2pf, where f is the frequency
of revolution or rotation of the object (Section 7.2). Figure 13.3 shows that the fre-
quency of the “orbiting” object is the same as the frequency of the oscillating
object on the spring. Thus, using f = 1>T, Eq. 13.8 may be written as
b
2pt (SHM for yo = 0
y = A sin12pft2 = A sina (13.10)
T initial upward motion)
Note that this equation is for initial upward motion, because after to = 0, the value
of y increases positively. For initial downward motion, y = - A sin12pft2.
INITIAL CONDITIONS AND PHASE

You may be wondering how to decide whether to use a sine or cosine function to
describe a particular case of simple harmonic motion. In general, the form of the func-
tion is determined by the initial displacement and velocity of the object: the initial con-
ditions of the system. These initial conditions are the values of the displacement and
velocity at t = 0; taken together, they tell how the system is initially set into motion.
Let’s look at four special cases. If an object in vertical SHM has an initial dis-
placement of y = 0 at t = 0 and moves initially upward, the equation of motion is
y = A sin vt (䉲 Fig. 13.5a). Note that y = A cos vt does not satisfy the initial condi-
tion, because yo = A cos vt = A cos v102 = A, since cos 0 = 1.
䉴 F I G U R E 1 3 . 5 Initial conditions y
and equations of motion The initial y = A sin ωt
+A
conditions (yo and to) determine the
form of the equation of motion—for t=T
the cases shown here, either a sine 0 t
or a cosine. For to = 0, the initial
displacements are (a) yo = 0,
(b) yo = + A, (c) yo = 0, and −A
(a)
(d) yo = - A. The equations of
motion must match the initial con-
ditions. (See book for description.) y = A cos ω t
+A
t=T
0 t
−A
(b)
y = –A sin ωt
+A
0 t
t=T
−A
(c)
y = –A cos ωt
+A
t=T
0 t
(d) −A
Suppose that the object is initially released 1t = 02 from its positive amplitude
position 1 +A2, as in the case of the object on a spring shown in Fig. 13.4. Here, the
equation of motion is y = A cos vt (Fig. 13.5b). This expression satisfies the initial
condition: yo = A cos v102 = A.
The other two cases are (1) y = 0 at t = 0, with motion initially downward (for
an object on a spring) or in the negative direction (for horizontal SHM), and
(2) y = - A at t = 0, meaning that the object is initially at its negative amplitude
position. These motions are described by y = - A sin vt and y = - A cos vt,
respectively, as illustrated in Figs. 13.5c and d.
Only these four initial conditions will be considered in our study. Should yo
have a value other than 0 or ⫾A, the equation of motion is somewhat complicated.
Note in Fig. 13.5 that if the curves are extended in the negative direction of the
horizontal axis (dashed purple lines), they all have the same shape, but have been
“shifted,” so to speak. In (a) and (b), one curve is ahead of the other by 90°, or 14
cycle. That is, the two curves are shifted by a quarter cycle with respect to one
another. The oscillations are then said to have a phase difference of 90°. In (a) and (c),
the curves are shifted 180° and are 180° out of phase. (Note in this case that the
oscillations are opposite: When one mass is going up, the other is going down.)
What about the oscillations in (a) and (d)?
DEMONSTRATION 4 Simple Harmonic Motion (SHM) and Sinusoidal Oscillation

A demostration to show that SHM can be represented by a
sinusoidal function. A “graph” of the function is generated
with an analogue of a strip chart recorder.
A salt-filled funnel oscillates, suspended from two strings.
Away we go. The salt falls on a black-painted poster board The salt trail traces out a plot of displacement versus time, or
that will be pulled in a direction perpendicular to the plane y = A sin1vt + d2. Note that in this case the phase constant
of the funnel’s oscillation. is about d = 90° and y = A cos vt. (Why?)
A figure with a 360° (or 0°) phase shift is not shown, because this would be the
same as that in (a). When two objects in SHM have the same equation of motion,
they are said to be oscillating in phase, which means that they are oscillating
together with identical motions. Objects with a 180° phase shift or difference are
said to be completely out of phase and will always be going in opposite directions
and be at opposite amplitudes at the same time.
Hence, we may write in general,
(+ for initial motion
b
2pt upward with yo = 0;
y = ⫾A sin vt = ⫾A sin12pft2 = ⫾A sina (13.8a)
T - for initial motion
downward with yo = 0)
By a similar development, Eq. 13.9 has the general form

(+ for initial motion
y = ⫾A cos vt = ⫾A cos12pft2 = ⫾A cos a b

2pt downward with yo = + A;
T - for initial motion (13.9b)
upward with yo = - A)
The next Example demonstrates the usage of the equation of motion for SHM.
EXAMPLE 13.2 An Oscillating Mass: Applying the Equation of Motion

A mass on a spring oscillates vertically with an amplitude of Since the mass is initially moving upward, the equation of
15 cm, and a frequency of 0.20 Hz. At t = 0, it is at yo = 0 and it motion should be of the form of Eq. 13.8 or Eq. 13.10. Part (b)
moves initially in the upward direction. (a) Write the equation is then the application of the equation of motion. In part (c),
of motion for this oscillation. (b) What are the position and the number of oscillations means the number of cycles, and
direction of motion of the mass at t = 3.1 s? (c) How many recall that frequency is sometimes expressed in cycles per
oscillations (cycles) does the mass make in a time of 12 s? second. Hence, multiplying the frequency by the time will
give the number of cycles or oscillations.
T H I N K I N G I T T H R O U G H . For part (a), the equation of motion
has to be a sine function because yo = 0. (Why not cosine?)
SOLUTION.
Given: A = 15 cm = 0.15 m Find: (a) equation of motion

f = 0.20 Hz (b) y (position and direction of motion)
(b) t = 3.1 s (c) n (number of oscillations or cycles)
(c) t = 12 s
(a) First, since the frequency f is given, it is convenient to use In t = 3.1 s, the mass has gone through 3.1 s>5.0 s = 0.62, or
the equation of motion in the form y = A sin 2pft (Eq. 13.10). 62%, of a period or cycle, so it is moving downward. The motion
As can be seen from the equation, at to = 0, yo = 0, so initially is up A 14 cycle B and back A 14 cycle B to yo = 0 in 12, or 50%, of the
the mass is at the zero (equilibrium) position. At t 7 0, y 7 0, cycle, and therefore downward during the next 14 cycle.]
so it is moving upward. Using the known quantities, the (c) The number of oscillations (cycles) is equal to the product
equation of motion is then of the frequency 1cycles>s2 and the elapsed time (s), both of
y = 10.15 m2 sin32p10.20 Hz2t4 = 10.15 m2 sin310.4p rad>s2t4 which are given:
(b) At t = 3.1 s, n = ft = 10.20 cycles>s2112 s2 = 2.4 cycles
y = 10.15 m2 sin310.4p rad>s213.1 s24 = 10.15 m2 sin13.9 rad2 or with f = 1>T,
= - 0.10 m
t 12 s
n = = = 2.4 cycles
So the mass is at y = - 0.10 m at t = 3.1 s. But what is its T 5.0 s
direction of motion? Let’s look at the period (T) and see what
(Note that cycle is not a unit and is used only for convenience.)
part of its cycle the mass is in. By Eq. 13.2,
Thus, the mass has gone through two complete cycles and
1 1 0.4 of another, which means that it is on its way back to yo = 0
T = = = 5.0 s
f 0.20 Hz from its amplitude position of + A. (Why?)
FOLLOW-UP EXERCISE. Find what is asked for in this Example at times (1) t = 4.5 s and (2) t = 7.5 s.
Note that in the calculation in part (b) of Example 13.2, for sin 3.9, the angle is given in
radians, not degrees. Don’t forget to set your calculator to radians (rather than degrees)
when finding the value of a trigonometric function in equations for simple harmonic or
circular motion.
On what does the period of oscillation depend? Let us compute the period of
the spring–mass system by comparing it to the reference circle in uniform circular
motion. (See Fig. 13.3.) Note that the time for the object in the reference circle to
make one complete “orbit” is exactly the time it takes for the oscillating object to
make one complete cycle. Thus, all we need is the time for one orbit around the
reference circle, and we have the period of oscillation. Because the object “orbit-
ing” the reference circle is in uniform circular motion at a constant speed equal to
the maximum speed of oscillation vmax , the object travels a distance of one circum-
ference in one period. Then t = d>v, where t = T, the circumference is d, and v is
vmax given by Eq. 13.7; that is,
d 2pA
T = =
vmax A2k>m
or
m (period of object
T = 2p (13.11)
Ak oscillating on a spring)
Because the amplitudes canceled out in Eq. 13.11, the period and frequency are
independent of the amplitude of the simple harmonic motion.
From Eq. 13.11 it can be seen that the greater the mass, the longer the period
and the greater the spring constant (or the stiffer the spring), the shorter the
period. It is the ratio of mass to stiffness that determines the period. Thus, an
increase in mass can be offset by using a stiffer spring.
Since f = 1>T,
1 k (frequency of mass
f = (13.12)
2p A m oscillating on a spring)
Thus, the greater the spring constant (the stiffer the spring), the more frequently
the system vibrates, as expected.
Also, note that since v = 2pf, we may write
k (angular frequency of mass

v = (13.13)
Am oscillating on a spring)
As another example, a simple pendulum (a small, heavy object on a string) will

undergo simple harmonic motion for small angles of oscillation. The period of a
simple pendulum oscillating through a small angle u 6 10° is given, to a good
approximation, by
L
T = 2p (period of a simple pendulum) (13.14)
Ag
where L is the length of the pendulum and g is the acceleration due to gravity. A
pendulum-driven clock that is not properly rewound and is running down
would still keep correct time, because the period would remain unchanged as
the amplitude decreased. As shown by Eq. 13.14, the period is independent of
amplitude.
An important difference between the period of the mass–spring system and

that of the pendulum is that the latter is independent of the mass of the pendulum
bob. (See Eq. 13.11 and Eq. 13.14.) Can you explain why? Think about what
supplies the restoring force for the pendulum’s oscillations. It is the gravitational
force. Hence, the acceleration (along with the velocity and period) is expected to
be independent of mass. That is, the gravitational force automatically provides the
same acceleration to different bob masses on pendulums with the same length.
Similar effects occur in free fall (Section 2.5) and with blocks sliding and cylinders
rolling down inclines (Sections 4.5 and 8.4, respectively).
Let’s take a look at two Examples related to the frequency and period of SHM.
EXAMPLE 13.3 Fun with a Pothole: Frequency and Spring Constant

A typical family automobile has a mass of 1500 kg. Assume T H I N K I N G I T T H R O U G H . (a) The frequency is given by
that the car has one spring on each wheel, that the springs are Eq. 13.12 and the spring constant can then be found. Keep in
identical, and that the mass is equally distributed over the mind that each spring will carry 1>4 of the total mass of the
four springs. (a) What is the spring constant of each spring if car. (b) Once the spring constant is found, Equation 13.1 can
the empty car bounces up and down 1.2 times each second be used again to find the new frequency. The spring constant
when hit a pothole? (b) What will be the car’s oscillation fre- is the same with or without the people in the car.
quency when four 75-kg people are in the car?
SOLUTION. The data are listed below, where m represents the mass on each individual spring.
1500 kg
Given: (a) m = = 375 kg Find: (a) k (spring constant)
4
f1 = 1.2 Hz (b) f (new frequency)
1500 kg + 4175 kg2
(b) m = = 450 kg
4
(a) Using Eq. 13.12 to solve for the spring constant, k.
k = 4p2f 2m = 4p211.2 Hz221375 kg2 = 2.13 * 104 N>m

1 k
f = so
2p A m
(b) The new frequency is found again from Eq. 13.12.
1 k 1 2.13 * 104 N>m
f = = = 1.1 Hz
2p A m 2p C 450 kg
F O L L O W - U P E X E R C I S E . Research has shown that the human body feels most comfortable if the oscillation frequency of a car is
1.0 Hz. What spring constant would you use for a half-loaded car (two 75-kg people)?
EXAMPLE 13.4 Fun with a Pendulum: Frequency and Period

A helpful older brother takes his sister to play on the swings (a) We can take the reciprocal of Eq. 13.14 to solve directly for
in the park. He pushes her from behind on each return. the frequency:
Assuming that the swing behaves as a simple pendulum with
1 1 g 1 9.80 m>s2
a length of 2.50 m, (a) what would be the frequency of the f = = = = 0.315 Hz
oscillations, and (b) what would be the interval between the T 2p A L 2p C 2.50 m
brother’s pushes? (b) The period is then found from the frequency:
T H I N K I N G I T T H R O U G H . (a) The period is given by Eq. 13.14, 1 1
and the frequency and period are inversely related: f = 1>T. T = = = 3.17 s
f 0.315 Hz
(b) Since the brother pushes from one side on each return, he
must push once every cycle that is completed, so the time The brother must push every 3.17 s to maintain a steady
between his pushes is equal to the swing’s period. swing (and to keep his sister from complaining).
SOLUTION.
Given: L = 2.50 m Find: (a) f (frequency)
(b) T (period)
F O L L O W - U P E X E R C I S E . In this Example, the older brother, a physics buff, carefully measures the period of the swing to be 3.18 s,
not 3.17 s. If the length of 2.50 m is accurate, what is the acceleration due to gravity at the location of the park? Considering this
accurate value of g, do you think the park is at sea level?
VELOCITY AND ACCELERATION IN SHM

Expressions for the velocity and acceleration of an object in SHM can also
be obtained. Using advanced mathematics, one can show that v = ¢y>¢t =
¢1A sin vt2>¢t in the limit as ¢t goes to zero gives the following expression for
the instantaneous velocity:
(vertical velocity if vo is upward at

v = vA cos vt (13.15)
to = 0, yo = 0)
The acceleration can be found by using Newton’s second law with the spring
force Fs = - ky:
Fs -ky k
a = = = - A sin vt
m m m
Since v = 1k>m,
(vertical acceleration if vo is
a = - v2A sin vt = - v2y (13.16)
upward at to = 0, yo = 0)
Note that the functions for the velocity and acceleration are out of phase with that
for the displacement. Since the velocity is 90° out of phase with the displacement,
the speed is greatest when cos vt = ⫾1 at y = 0, that is, when the oscillating object
is passing through its equilibrium position. The acceleration is 180° out of phase
with the displacement (as indicated by the negative sign on the right-hand side of
Equation 13.16). Therefore, the magnitude of the acceleration is a maximum when
sin vt = ⫾1 at y = ⫾A, that is, when the displacement is a maximum, or when the
object is at an amplitude position. At any position except the equilibrium position,
the directional sign of the acceleration is opposite that of the displacement, as it
should be for an acceleration resulting from a restoring force. At the equilibrium
position, both the displacement and acceleration are zero. (Can you see why?)
Note also that the acceleration in SHM is not constant with time. Hence, the kine-
matic equations for acceleration (Chapter 2) cannot be used, since they describe con-
stant acceleration.
DAMPED HARMONIC MOTION

Simple harmonic motion with constant amplitude implies that there are no losses
of energy, but in practical applications there are always some frictional losses.
Therefore, to maintain a constant amplitude motion, energy must be added to a
system by some external driving force, such as someone pushing a swing. Without
a driving force, the amplitude and the energy of an oscillator decrease with time,
giving rise to damped harmonic motion (䉲 Fig. 13.6a). The time required for the
oscillations to cease, or damp out, depends on the magnitude and type of the
damping force (such as air resistance).
In many applications involving continuous periodic motion, damping is
unwanted and necessitates an energy input. However, in some instances, damping
is desirable. For example, the dial in a spring-operated bathroom scale oscillates
briefly before stopping at a weight reading. If not properly damped, these oscilla-
tions would continue for some time, and you would have to wait before you could
read your weight. Shock absorbers provide damping in the suspension systems of
Constant amplitude Driving force removed

+A
0 t
–A Damped harmonic oscillation

Steady-state oscillation
with a driving force
(a) (b)
䉱 F I G U R E 1 3 . 6 Damped harmonic motion (a) When a driving force adds energy to a system in an amount
equal to the energy losses of the system, the oscillation is steady with a constant amplitude. When the driving
force is removed, the oscillations decay (that is, they are damped), and the amplitude decreases nonlinearly
with time. (b) In some applications, damping is desirable and even promoted, as with shock absorbers in auto-
mobile suspension systems. Otherwise, the passengers would be in for a bouncy ride.
automobiles (Fig. 13.6b; also see Fig. 9.9b). Without “shock absorbers” to dissipate
energy after hitting a bump, the ride would be bouncy. In California, many new
buildings incorporate damping mechanisms (giant shock absorbers) to dampen
their oscillatory motion after they are set in motion by earthquake waves.
DID YOU LEARN?

➥ The functions of the equations of motion for simple harmonic motion are sinusoidal
(harmonic), that is, they are either sine or cosine functions.
➥ The period or frequency of a mass–spring system or a simple pendulum does not
depend on the amplitude of vibration.
➥ With a phase difference of 180°, the oscillations of two objects in SHM are out of phase.
That is, when one object is at a maximum, the other is at a minimum and vice versa.
13.3 Wave Motion

➥ What are the most common four parameters that describe a wave?
➥ What is meant by a transverse wave?
➥ What is meant by a longitudinal wave?
The world is full of waves of various types; some examples are water waves, sound
waves, waves generated by earthquakes, and light waves. All waves result from a
disturbance, the source of the wave. In this chapter, the focus will be mechanical
waves—those that are propagated in some medium. (Light waves, which do not
require a propagating medium, will be considered in more detail in later chapters.)
When a medium is disturbed, energy is imparted to it. The addition of the
energy sets some of the particles in the medium vibrating. Because the particles
are linked by intermolecular forces, the oscillation of each particle affects that of its
neighbors. The added energy propagates, or spreads, by means of interactions
among the particles of the medium. An analogy to this process is shown in
䉱 F I G U R E 1 3 . 7 Energy transfer
䉳 Fig. 13.7, where the “particles” are dominoes. As each domino falls, it topples the
The propagation of a disturbance, or
a transfer of energy through space, one next to it. Thus, energy is transferred from domino to domino, and the distur-
is seen in a row of falling dominoes. bance propagates through the medium—the energy travels, not the medium.
13.3 WAVE MOTION 469
䉳 F I G U R E 1 3 . 8 Wave pulse The

a hand disturbs the stretched rope in
x1 x2 a quick up-and-down motion, and a
wave pulse propagates along the
rope. (The red arrows represent the
velocities of the hand and of pieces
b of the rope at different times and
locations.) The rope “particles”
move up and down as the pulse
passes. The energy in the pulse is
thus both kinetic (motion) and
c potential (elastic).
In this case, there is no restoring force between the dominoes, so they do not
oscillate. Therefore, the disturbance moves in space, but it does not repeat itself in
time at any one location.
Similarly, if the end of a stretched rope is given a quick shake, the disturbance
transfers energy from the hand to the rope, as illustrated in 䉱 Fig. 13.8. The forces
acting between the “particles” in the rope cause them to move in response to the
motion of the hand, and a wave pulse travels down the rope. Each “particle” goes
up and then back down as the pulse passes by. This motion of individual particles
and the propagation of the wave pulse as a whole can be observed by tying pieces
of ribbon onto the rope (at x1 and x2 in the figure). As the disturbance passes point
x1, the ribbon rises and falls, as do the rope’s “particles.” Later, the same thing
happens to the ribbon at x2, which indicates that the energy disturbance is propa-
gating or traveling along the rope.
In a continuous material medium, particles interact with their neighbors, and
restoring forces cause them to oscillate when they are disturbed. Thus, a disturbance
not only propagates through space, but also may be repeated over and over in time at
each position. Such a regular, rhythmic disturbance in both time and space is called a
wave, and the transfer of energy is said to take place by means of wave motion.
A continuous wave motion, or periodic wave, requires a disturbance from an
oscillating source (䉲 Fig. 13.9). In this case, the particles move up and down contin-
uously. If the driving source is such that a constant amplitude is maintained (the
source oscillates in simple harmonic motion), the resulting particle motion is also
simple harmonic.
λ v Crest
+A
Amplitude
–A
λ Trough
䉱 F I G U R E 1 3 . 9 Periodic wave A continuous harmonic disturbance can set up a sinu-

soidal wave in a stretched rope, and the wave travels down the rope with wave speed v.
Note that the “particles” in the rope oscillate vertically in simple harmonic motion. The
distance between two successive points that are in phase (for example, at two crests) on the
waveform is the wavelength l of the wave. Can you tell how much time has elapsed, as a
fraction of the period T, between the first (red) and last (blue) waves?
Such periodic wave motion will have sinusoidal forms (sine or cosine) in both
time and space. Being sinusoidal in space means that if you took a photograph of the
wave at any instant (“freezing” it in time), you would see a sinusoidal waveform
(such as one of the curves in Fig. 13.9). However, if you looked at a single point in
space as a wave passed by, you would see a particle of the medium oscillating up and
down sinusoidally with time, like the mass on a spring discussed in Section 13.2. (For
example, imagine looking through a thin slit at a fixed location on the moving paper
in Fig. 13.4. The wave trace would be seen rising and falling like a particle in SHM.)
WAVE CHARACTERISTICS
Specific quantities are used to describe sinusoidal waves. As with a particle in
simple harmonic motion, the amplitude (A) of a wave is the magnitude of the
maximum displacement, or the maximum distance, from the particle’s equilib-
rium position (Fig. 13.9). This quantity corresponds to the height of a wave crest or
the depth of a trough. Recall from Section 13.2 that, in SHM, the total energy of the
oscillator is proportional to the square of the amplitude. Similarly, the energy
transported by a wave is proportional to the square of its amplitude 1E r A22. Note
the difference, though: A wave is one way of transmitting energy through space,
whereas an oscillator’s energy is localized in space.
For a periodic wave, the distance between two successive crests (or troughs) is
called the wavelength (L) (see Fig. 13.9). Actually, it is the distance between any two
successive parts of the wave that are in phase (that is, that are at identical points on
the waveform). The crest and trough positions are usually used for convenience.
The frequency (f ) of a periodic wave is the number of waves per second—that
is, the number of complete waveforms, or wavelengths, that pass by a given point
during each second. The frequency of the wave is the same as the frequency of the
SHM source that created it.
A periodic wave is said to possess a period (T). The period T = 1>f is the time
for one complete waveform (a wavelength) to pass by a given point. Since a wave
moves, it also has a wave speed (v) (or velocity if the wave’s direction is speci-
fied). Any particular point on the wave (such as a crest) travels a distance of one
wavelength l in a time of one period T. Then, since v = d>t and f = 1>T,
l
v = = lf (wave speed) (13.17)
T
Note that the dimensions of v are correct 1length>time2. In general, the wave
speed depends on the nature of the medium, in addition to the source frequency f.
EXAMPLE 13.5 Dock of the Bay: Finding Wave Speed

A person on a pier observes a set of incoming water waves (a) The lapping indicates the arrival of a wave crest; hence,
that have a sinusoidal form with a distance of 5.6 m between 2.0 s is the wave period—the time it takes to travel one wave-
the crests. If a wave laps against the pier every 2.0 s, what are length (the crest-to-crest distance). Then
(a) the frequency and (b) the speed of the waves?
T H I N K I N G I T T H R O U G H . Since the period and wavelength

are known, the definition of frequency and Eq. 13.17 for wave (b) The frequency or the period can be used in Eq. 13.17 to
speed can be used. find the wave speed:
SOLUTION. The distance between crests is the wavelength, v = lf = 15.6 m210.50 s -12 = 2.8 m>s
so we have the following information:
Alternatively,
Given: l = 5.6 m Find: (a) f (frequency) l 5.6 m
v = = = 2.8 m>s
T = 2.0 s (b) v (wave speed) T 2.0 s
F O L L O W - U P E X E R C I S E . On another day, the person measures the speed of sinusoidal water waves at 2.5 m>s. (a) How far does a
wave crest travel in 4.0 s? (b) If the distance between successive crests is 6.3 m, what is the frequency of these waves?
13.3 WAVE MOTION 471
TYPES OF WAVES Direction of

wave
In general, waves may be divided into two types, based on the direction of the propagation
particles’ oscillations relative to that of the wave velocity. In a transverse wave,
the particle motion is perpendicular to the direction of the wave velocity. The
wave produced in a stretched string (Fig. 13.9) is an example of a transverse wave,
as is the wave shown in 䉴 Fig. 13.10a. A transverse wave is sometimes called a shear
wave, because the disturbance supplies a force that tends to shear the medium—to
separate layers of that medium at a right angle to the direction of the wave velo- Direction
city. Shear waves can propagate only in solids, since a liquid or a gas cannot sup- of particle
port a shear. That is, a liquid or a gas does not have sufficient restoring forces motion
between its particles to propagate a transverse wave. (a)
In a longitudinal wave, the particle oscillation is parallel to the direction of the
Direction of
wave velocity. A longitudinal wave can be produced in a stretched spring by mov- wave Compression
ing the coils back and forth along the spring axis (Fig. 13.10b). Alternating pulses of propagation
compression and relaxation travel along the spring. A longitudinal wave is some-
times called a compressional wave, because the force tends to compress the medium.
Sound waves in air are another example of longitudinal waves. A periodic distur-
bance produces compressions in the air. Between the compressions are rarefactions—
regions where the density of the air is reduced, or rarefied. A loudspeaker oscillating Direction of Relaxation
back and forth, for example, can create these compressions and rarefactions, which particle motion
travel out into the air as sound waves. (Sound is discussed in detail in Chapter 14.) (b)
Longitudinal waves can propagate in solids, liquids, and gases, because all phases
of matter can be compressed to some extent. The propagations of transverse and lon- 䉱 F I G U R E 1 3 . 1 0 Transverse and
gitudinal waves in different media give information about the Earth’s interior struc- longitudinal waves (a) In a trans-
ture, as discussed in Insight 13.1, Earthquakes, Seismic Waves, and Seismology. verse wave, the motion of the parti-
cles is perpendicular to the direction
The sinusoidal profile of water waves might make you think that they are
of the wave velocity, as shown here
transverse waves. Actually, they reflect a combination of longitudinal and trans- in a spring for a wave moving to the
verse motions (䉲 Fig. 13.11). The particle motion may be nearly circular at the sur- left. (b) In a longitudinal wave, the
face and becomes more elliptical with depth, eventually becoming longitudinal. A particle motion is parallel to (or
hundred meters or so below the surface of a large body of water, the wave distur- along) the direction of the wave
velocity. Here, a wave pulse also
bances have little effect. For example, a submarine at these depths is undisturbed
moves to the left. Can you explain
by large waves on the ocean’s surface. As a wave approaches shallower water near the motion of the wave source for
shore, the water particles have difficulty completing their elliptical paths. When both types of waves?
the water becomes too shallow, the particles can no longer move through the bot-
tom parts of their paths, and the wave breaks. Its crest falls forward to form break-
ing surf as the waves’ kinetic energy is transformed into potential energy—a
water “hill” that eventually topples over.
䉲 F I G U R E 1 3 . 1 1 Water waves
Water waves are a combination of
DID YOU LEARN?
longitudinal and transverse motions.
➥ The four parameters that describe a wave are amplitude A, period T, frequency f, (a) At the surface, the water particles
and speed v, where v = lf. move in circles, but their motions
➥ In a transverse wave, the particle motion is perpendicular to the wave velocity. become more longitudinal with
An example is a wave on a stretched string. depth. (b) When a wave approaches
➥ In a longitudinal wave, the particle oscillation is parallel to the wave velocity. the shore, the lower particles are
An example is a sound wave. forced into steeper paths until,
finally, the wave breaks or falls over
to form surf.
v Wave Surf
(a) (b)
INSIGHT 13.1 Earthquakes, Seismic Waves, and Seismology

The structure of the Earth’s interior is something of a mystery. tudinal waves can travel through solids or liquids, transverse
The deepest mine shafts and drillings extend only a few kilome- waves can travel only through solids. When an earthquake
ters into the Earth, compared with a depth of about 6400 km to occurs, P waves are detected on the side of the Earth opposite
the Earth’s center. Using waves to probe the Earth’s structure is the focus, but S waves are not. (See Fig. 3.) The absence of S
one way to investigate it further. Waves generated by earth- waves in a shadow zone leads to the conclusion that the Earth
quakes have proved to be especially useful for this purpose. must have a region near its center that is in the liquid phase.
Seismology is the study of these waves, called seismic waves. When the transmitted P waves enter and leave the liquid
Earthquakes are caused by the sudden release of built-up region, they are refracted (bent). This refraction gives rise to a
stress along cracks and faults, such as the famous San P-wave shadow zone, which indicates that only the outer part
Andreas Fault in California (Fig. 1). According to the geologi- of the core is liquid. As you will learn in Chapter 19, the com-
cal theory of plate tectonics, the outer layer of the Earth con- bination of a liquid outer core and the Earth’s rotation may be
sists of rigid plates—huge slabs of rock that move very slowly responsible for the Earth’s magnetic field.
relative to one another. Stresses continuously build up, partic-
ularly along boundaries between plates.
When slippage of plates finally occurs, the energy from this
stress-relieving event propagates outward as (seismic) waves
from a site below the surface called the focus. The point on the
Earth’s surface directly above the focus is called the epicenter,
and receives the greatest impact of a quake. Seismic waves are
of two general types: surface waves and body waves. Surface
waves, which move along the Earth’s surface, account for most
earthquake damage (Fig. 2). Body waves travel through the
Earth and are both longitudinal and transverse waves. The
compressional (longitudinal) waves are called P waves, and
the shear (transverse) waves are called S waves (Fig. 3).
P and S stand for primary and secondary and indicate the
waves’ relative speeds (actually, their arrival times at moni-
toring stations). In general, primary waves travel through
materials faster than do secondary waves, so are detected
first. An earthquake’s rating on the Richter scale is related to
the energy released in the form of seismic waves.
F I G U R E 2 Bad vibrations Earthquake damage caused by
Seismic stations around the world monitor P and S waves the surface waves of a major shock that struck Kobe, Japan,
with sensitive detecting instruments called seismographs. From in January 1995.
the data gathered, the paths of the waves through the Earth can
be mapped, and thereby learn about the interior structure of
our planet. The Earth’s interior seems to be divided into three Earthquake focus
general regions: the crust, the mantle, and the core, which itself
S P
has a solid inner region and a liquid outer region.*
The locations of these regions’ boundaries are determined in
part by shadow zones, regions where no waves of a particular S
type are detected. These zones appear because, although longi- S P
*In most places, the crust is about 24–30 km (15–20 mi) thick; the Solid
S P
mantle is 2900 km (1800 mi) thick; and the core has a radius of 3450 km inner core
one
(2150 mi). The solid inner core has a radius of about 1200 km (750 mi).
of irec adow z
Liquid
ves l
wa riva
outer core P
No
F I G U R E 1 The
S - w ct a
P t ar
sh
San Andreas Fault

dir
P
av
A small section of P
e
e
av
ha Mantle
s
the fault, which w

P-
d
va dow
rri
runs through the lo zon s N

fS e ve
San Francisco Bay wa P wa or e
area as well as ves hc
throug
across the more
rural regions of
California, is F I G U R E 3 Compressional and shear waves Earthquakes
shown here. produce waves that travel through the Earth. Because
transverse (S) waves are not detected on the opposite side
of the Earth, scientists believe that at least part of the
Earth’s core is a viscous liquid under high pressures and
temperatures. The waves bend continuously, or refract,
because their speed varies with depth.
13.4 WAVE PROPERTIES 473
13.4 Wave Proper ties

➥ What is the principle that wave action obeys that result in constructive or destruc-
tive interference?
➥ What is meant by a total destructive interference?
➥ What phenomenon explains why you can hear “around corners”—for example, why
you can hear people around the corner of a building talking but not see them?
Among the properties exhibited by all waves are superposition, interference,

reflection, refraction, dispersion, and diffraction.
SUPERPOSITION AND INTERFERENCE

When two or more waves meet or pass through the same region of a medium,
they pass through each other and each wave proceeds without being altered.
While they are in the same region, the waves are said to be interfering.
What happens during interference? That is, what does the combined waveform
look like? The relatively simple answer is given by the principle of superposition:
At any time, the combined waveform of two or more interfering waves is given by the
sum of the displacements of the individual waves at each point in the medium.
Using the principle of superposition, interference is illustrated in 䉲 Fig. 13.12. The

displacement of the combined waveform at any point is y = y1 + y2 , where y1
and y2 are the displacements of the individual pulses at that point. (Directions are
indicated by positive and negative values.) Interference, then, is the physical addi-
tion of waves. In adding waves, we must take into account the possibility that
v1 v2
y1 y2
y = y1+ y2
y = y1+ y2
(a) (b)
䉱 F I G U R E 1 3 . 1 2 Principle of superposition (a) When two waves meet, they interfere as

shown in the photographs. (b) The beige tint marks the area where the two waves, moving
in opposite directions, overlap and combine. The displacement at any point on the com-
bined wave is equal to the sum of the displacements on the individual waves: y = y1 + y2 .
1 1
2 2
3 3
4 4
(a) Total constructive interference (b) Total destructive interference
䉱 F I G U R E 1 3 . 1 3 Interference (a) When two wave pulses of the same amplitude meet
and are in phase, they interfere constructively. When the pulses are exactly superimposed
(3), total constructive interference occurs. (b) When the interfering pulses are 180° out of
phase and exactly superimposed (3), total destructive interference occurs.
they are producing disturbances in opposite directions. In other words, we must

treat the disturbances in terms of vector addition.
In the figure, the vertical displacements of the two pulses are in the same direc-
tion, and the amplitude of the combined waveform is greater than that of either
pulse. This situation is called constructive interference. Conversely, if one pulse
has a negative displacement, the two pulses tend to cancel each other when they
overlap, and the amplitude of the combined waveform is smaller than that of
either pulse. This situation is called destructive interference.
The special cases of total constructive and total destructive interference for
traveling wave pulses of the same width and amplitude are shown in 䉱 Fig. 13.13.
At the instant these interfering waves exactly overlap (crest coinciding with crest),
the amplitude of the combined waveform is twice that of either individual wave.
This case is referred to as total constructive interference. When the interfering
pulses have opposite displacements and are exactly superimposed (crest coincid-
ing with trough), the waveforms momentarily disappear; that is, the amplitude of
the combined wave is zero. This case is called total destructive interference.
The word destructive unfortunately tends to imply that the energy, as well as the
form of the waves, is destroyed. This is not the case. At the point of total destruc-
tive interference, when the net wave shape and, hence, potential energy are zero,
the wave energy is stored in the medium completely in the form of kinetic energy.
That is, the straight string has instantaneous velocity.
There are several practical applications of destructive interference. One of these
is automobile mufflers. Exhaust gases from the engine passing from a high pres-
sure in the cylinders to normal atmospheric pressure would produce loud noises.
Typically, a muffler consists of a metal jacket containing perforated pipes and
chambers. The pipes and chambers are arranged so that the pressure waves of the
exhaust gases are reflected back and forth, giving rise to destructive interference.
This greatly reduces the noise of the exhaust coming from the tail pipe.
Other applications are termed “active noise cancellation.” This involves sound
modification, particularly sound cancellation by electro-acoustical means. A par-
ticularly important application is for airline or helicopter pilots who need to hear
what’s going on around them over the engine noise (䉴 Fig. 13.14). Pilots use special
headphones mounted with a microphone that picks up the low-frequency engine
noise. A component in the headphone then creates a wave that is the inverse of the
engine noise. This is played back through the headphones, and destructive inter-
ference produces a quieter background. A pilot can then better hear the mid- and
high-frequency sounds, such as conversation and instrument warning sounds.
13.4 WAVE PROPERTIES 475
(a)
Original noise
Microphone 䉲 F I G U R E 1 3 . 1 5 Reflection
(a) When a wave (pulse) on a string
is reflected from a fixed boundary,
Combined wave
the reflected wave is inverted. (b) If
the string is free to move at the
boundary, the phase of the reflected
wave is not shifted from that of the
Inverted wave incident wave.
Incident
pulse
Speaker
(b)
䉱 F I G U R E 1 3 . 1 4 Destructive interference in action (a) Pilots use headphones mounted

with a microphone that picks up low-frequency engine noise. (b) A wave is generated
that is inverse that of the engine noise. When played back through the headphones,
destructive interference produces less engine noise. This process is called “active noise
cancellation.”
REFLECTION, REFRACTION, DISPERSION, AND Reflected

DIFFRACTION pulse
(a) Fixed boundary,
Besides meeting other waves, waves can (and do) meet objects or a boundary with pulse inverted
another medium. In such cases, several things may occur. One of these is on reflection
reflection, which occurs when a wave strikes an object or comes to a boundary
with another medium and is at least partly diverted back into the original Incident
medium. An echo is the reflection of sound waves, and mirrors reflect light waves. pulse
Two cases of reflection are illustrated in 䉴 Fig. 13.15. If the end of the string is fixed,
the reflected pulse is inverted (Fig. 13.15a). This is because the pulse causes the string
to exert an upward force on the wall, and the wall exerts an equal and opposite
downward force on the string (by Newton’s third law). The downward force creates
the downward, or inverted, reflected pulse. If the end of the string is free to move,
then the reflected pulse is not inverted. (There is no phase shift.) This is illustrated in
Fig. 13.15b, which shows the string attached to a light ring that can move freely on a
smooth pole. The ring is accelerated upward by the front portion of the incoming Reflected
pulse
pulse and then comes back down, thus creating a noninverted reflected pulse.
More generally, when a wave strikes a boundary, the wave is not completely
reflected. Instead, some of the wave’s energy is reflected and some is transmitted (b) Free (movable) boundary,
or absorbed. When a wave crosses a boundary into another medium, its speed pulse not inverted
generally changes because the new material has different characteristics. When on reflection
entering the medium obliquely (at an angle), the transmitted wave moves in a
direction different from that of the incident wave. This phenomenon is called
refraction (䉳 Fig. 13.16).
Since refraction depends on changes in the speed of the wave, you might be
wondering which physical parameters determine the wave speed. Generally,
there are two types of situations. The simplest kind of wave is one whose speed
does not depend on its wavelength (or frequency). All such waves travel at the
same speed, determined solely by the properties of the medium. These waves are
called nondispersive waves, because they do not disperse, or spread apart from one
another. An example of a nondispersive transverse wave is a wave on a string,
whose speed, as we shall see, is determined only by the tension and mass density
䉱 F I G U R E 1 3 . 1 6 Refraction The
of the string (Section 13.5). Sound is a nondispersive longitudinal wave; the speed
refraction of water waves is shown of sound (in air) is determined only by the compressibility and density of the air.
from overhead. As the crests Indeed, if the speed of sound did depend on the frequency, at the back of the sym-
approach the triangle from the phony hall you might hear the violins well before the cellos, even though the two
right, the wave speed slows because sound waves were in perfect synchronization when they left the orchestra.
it is entering shallow water. Thus,
the wave changes its direction.
When the wave speed does depend on wavelength (or frequency), the waves are
said to exhibit dispersion: waves of different frequencies spread apart from one
another. An example of dispersion is light waves. When they enter some media,
they are spread out or dispersed. This is the basis for prisms separating sunlight
into a color spectrum and for the formation of a rainbow, as will be seen in
Section 22.5. Dispersion will be most important in the study of light, but waves
other than light can also be dispersive under the right conditions.
Diffraction refers to the bending of waves around an edge of an object and is not
related to refraction. For example, if you stand along an outside wall of a building
near the corner of the street, you can hear people talking around the corner. Assum-
ing that there are no reflections or air motion (wind), this would not be possible if the
sound waves traveled in a straight line. As the sound waves pass the corner, instead
of being sharply cut off, they “wrap around” the edge, and you can hear the sound.
In general, the effects of diffraction are evident only when the size of the dif-
fracting object or opening is about the same as or smaller than the wavelength of
the waves. The dependence of diffraction on the wavelength and size of the object
or opening is illustrated in 䉲 Fig. 13.17. For many waves, diffraction is negligible
under normal circumstances. For instance, visible light has wavelengths on the
order of 10-6 m. Such wavelengths are much too small to exhibit diffraction when
they pass through common-sized openings, such as an eyeglass lens.
Reflection, refraction, dispersion, and diffraction will be considered in more
detail when we study light waves in Chapters 22 and 24.
DID YOU LEARN?

➥ The principle of superposition explains constructive and destructive wave
interference.
➥ When two waves of the same wavelength and amplitude have opposite
displacements (a 180° phase difference), the wave resulting from superposition has
zero amplitude.
➥ Wave diffraction causes sound to be “bent” around corners.
䉴 F I G U R E 1 3 . 1 7 Diffraction
Diffraction effects are greatest when
the opening (or object) is about the
same size as or smaller than the
wavelength of the waves. (a) With
an opening much larger than the
wavelength of these plane water
waves, diffraction is noticeable only
near the edges. (b) With an opening
about the same size as the wave-
length of the waves, diffraction pro-
duces nearly semicircular waves. (a) (b)
13.5 STANDING WAVES AND RESONANCE 477
Antinodes
t=0 Nodes
t= T
8
(a)
t= T
䉱 F I G U R E 1 3 . 1 8 Standing waves (a) Standing waves are formed by interfering waves 4
traveling in opposite directions. (b) Conditions of destructive and constructive interference
recur as each wave travels a distance of l>4 in a time t = T>4. The velocities of the rope’s
particles are indicated by the arrows. This motion gives rise to standing waves with sta-
tionary nodes and maximum amplitude antinodes.
t = 3T
8
t= T
2
13.5 Standing Waves and Resonance

LEARNING PATH QUESTIONS t = 5T
8
➥ How is a standing wave formed?
➥ What are the nodes and antinodes of a standing wave?
➥ Under what condition does resonance occur for a particular oscillatory system?
t = 3T
If you shake one end of a stretched rope that is fixed at the other end, waves travel 4
along the rope to the fixed end and are reflected back. The waves going to and
from the fixed end interfere with each other. In most cases, the combined wave-
forms have a changing, jumbled appearance. But if the rope is shaken at just the
right frequency, a steady waveform, or series of uniform loops, appears to stand in t = 7T
8
place along the rope. Appropriately, this phenomenon is called a standing wave
(䉱 Fig. 13.18). It arises because of interference with the reflected waves, which have
the same wavelength, amplitude, and speed as the incident waves. Since the two
identical waves travel in opposite directions, the net energy flow in the rope is
zero. In effect, the energy is standing in the loops. t=T
Some points on the rope remain stationary at all times and are called nodes. At (b)
these points, the displacements of the interfering waves are always equal and
opposite. Thus, by the principle of superposition, the interfering waves cancel
each other completely at these points, and the rope does not undergo displace-
ment there. At all other points, the rope oscillates back and forth at the same fre-
quency. The points of maximum amplitude, where constructive interference is
greatest, are called antinodes. As you can see in Fig. 13.18a, adjacent antinodes are
separated by a half-wavelength 1l>22, or one loop; adjacent nodes are also sepa-
rated by a half-wavelength.
Standing waves can be generated in a rope by more than one driving fre-
quency; the higher the frequency, the more oscillating half-wavelength loops there
are in the rope. The only requirement is that an integer number of half-
wavelengths “fit” the length of the rope. The frequencies at which standing waves
are produced are called natural frequencies, or resonant frequencies. The result-
ing standing wave patterns are called normal, or resonant, modes of vibration. In gen-
eral, all systems that oscillate have one or more natural frequencies, which depend
on such factors as mass, elasticity or restoring force, and geometry (boundary con-
ditions). The natural frequencies of a system are sometimes called its characteristic
frequencies.
␭1
L= First harmonic
2
␭2
L=2 Second harmonic
2
␭3
L =3 Third harmonic
2
䉱 F I G U R E 1 3 . 1 9 Natural frequencies A stretched string can have standing waves only

at certain frequencies. These correspond to the numbers of half-wavelength loops that will
fit along the length of string between the nodes at the fixed ends.
A stretched string or rope can be analyzed to determine its natural frequencies.

The boundary condition is that the ends are fixed; thus, there must be a node at
each end. The number of closed segments or loops of a standing wave that will fit
between the nodes at the ends (along the length of the string) is equal to an inte-
gral number of half-wavelengths (䉱 Fig. 13.19). Note that L = 1l1>22, L = 21l2>22,
L = 31l3>22, L = 41l4>22, and so on. In general,
1for n = 1, 2, 3, Á 2
ln 2L
L = n¢ ≤ or ln =
2 n
The natural frequencies of oscillation are therefore
= na b = nf1 for n = 1, 2, 3, Á
v v (natural frequencies
fn = (13.18)
ln 2L for a stretched string)
where v is the speed of waves on a string. The lowest natural frequency

(f1 = v>2L for n = 1) is called the fundamental frequency. All of the other nat-
ural frequencies are integral multiples of the fundamental frequency f1: fn = nf1
(for n = 1, 2, 3, Á ). The set of frequencies f1 , f2 = 2f1 , f3 = 3f1 , Á is called a
harmonic series: f1 (the fundamental frequency) is the first harmonic, f2 the second
harmonic, and so on.
Strings that are fixed at each end are found in stringed musical instruments
䉱 F I G U R E 1 3 . 2 0 Fundamental such as violins, pianos, and guitars. When such a string is disturbed—that is,
frequencies Performers of stringed
instruments such as the violin use plucked, struck, or bowed, the resulting vibration generally includes several
their fingers to stop or fret the higher harmonics in addition to the first harmonic. The number of harmonics
strings. By pressing a string against depends on how and where the string is disturbed. It is the combination of har-
the fingerboard, the player reduces monic frequencies that gives a particular instrument its characteristic sound qual-
the amount of its length that is free ity (more on this in Section 14.6). As Eq. 13.18 shows, all harmonic frequencies
to vibrate. This reduction increases
the harmonic frequencies of the depend on the length of the string. Different notes are obtained on a particular
string and the pitch of the tones it string of a violin by touching the string at a particular location so as to change its
produces. vibrating length (䉳 Fig. 13.20).
13.5 STANDING WAVES AND RESONANCE 479
Natural frequencies also depend on other parameters, such as mass and force,
which affect the wave speed in the string. For a stretched string, the wave speed
(v) can be shown to be
FT (wave speed
v = (13.19)
Am in a stretched string)
where FT is the tension in the string and m is the linear mass density (mass per unit
length, m = m>L). (We use FT, rather than the T in previous chapters, so as not to
confuse the tension with the period T.) Thus, Eq. 13.18 can be written as
b = = nf1 1for n = 1, 2, 3 Á 2
v n FT
fn = na (13.20)
2L 2L A m
Note that the greater the linear mass density of a string, the lower its natural fre-
quencies. As you may know, the low-note strings on a violin or guitar are thicker,
or more massive, than the high-note strings. By tightening a string, all the frequen-
cies of that string are increased. Changing the tension in the string is how violin-
ists, for example, tune their instruments before a performance.
EXAMPLE 13.6 A Piano String: Fundamental Frequency and Harmonics

A piano string with a length of 1.15 m and a mass of 20.0 g is T H I N K I N G I T T H R O U G H . The linear mass density can be
under a tension of 6.30 * 103 N. (a) What is the fundamental determined from the data. This, along with the given string
frequency of the string? (b) What are the frequencies of the tension can be used to find the fundamental frequency. With
next two harmonics? this, the harmonics can be calculated.
SOLUTION.
Given: L = 1.15 m Find: (a) f1 (fundamental frequency)

m = 20.0 g = 0.0200 kg (b) f2 and f3 (frequencies of next two harmonics)
FT = 6.30 * 103 N
(a) The linear mass density of the string is (b) Since f2 = 2f1 and f3 = 3f1 it follows that
m 0.0200 kg f2 = 2f1 = 21262 Hz2 = 524 Hz
m = = = 0.0174 kg>m
L 1.15 m
and
Then, using Eq. 13.20, we have
f3 = 3f1 = 31262 Hz2 = 786 Hz
1 FT 1 6.30 * 103 N
f1 = = = 262 Hz The second harmonic corresponds approximately to C5 on a
2L A m 211.15 m2 C 0.0174 kg>m
piano, since, by definition, the frequency doubles with each
This is approximately the frequency of middle C (C4) on a piano. octave (every eighth white key).
F O L L O W - U P E X E R C I S E . A musical note is referenced to the fundamental frequency, or first harmonic. In musical terms, the second
harmonic is called the first overtone, the third harmonic is the second overtone, and so on. If an instrument has a third overtone
with a frequency of 880 Hz, what is the frequency of the first overtone?
RESONANCE
When an oscillating system is driven at one of its natural, or resonant, frequencies,
the maximum amount of energy is transferred to the system. The natural frequen-
cies of a system are the frequencies at which the system “prefers” to vibrate, so to
speak. The condition of driving a system at a natural frequency is referred to as
resonance.
A common example of a system in mechanical resonance is someone being
pushed on a swing. Basically, a swing is a simple pendulum and has only one reso-
nant frequency for a given length 3f = 1>T = 1g>L>12p24. If you push the swing
with this frequency and in phase with its motion, its amplitude and energy
increase (䉳 Fig. 13.21). If you push at a slightly different frequency, the energy
transfer is no longer a maximum. (What do you think happens if you push with
the resonant frequency, but 180° out of phase with the swing’s motion?)
Unlike a simple pendulum, a stretched string has many natural frequencies.
Almost any driving frequency will cause a disturbance in the string. However, if
the frequency of the driving force is not equal to one of the natural frequencies, the
resulting wave will be relatively small and jumbled. By contrast, when the fre-
quency of the driving force matches one of the natural frequencies, the maximum
amount of energy is transferred to the string. A steady standing wave pattern
results, with the amplitude at the antinodes becoming relatively large.
When a large number of soldiers march over a small bridge, they are generally
ordered to break step. The reason is that the marching (stepping) frequency may
䉱 F I G U R E 1 3 . 2 1 Resonance in correspond to one of the natural frequencies of the bridge and set it into resonant
the playground The swing behaves vibration, which could cause it to collapse. This actually occurred on a suspension
like a pendulum in SHM. To trans-
fer energy efficiently, the man must
bridge in England in 1831. The bridge was weak and in need of repair, but it col-
time his pushes to the natural fre- lapsed as a direct result of the resonance vibrations induced by the marching
quency of the swing. soldiers—with some injuries.
Another incident of a bridge vibrating was the collapse of the “Galloping
Gertie,” the Tacoma Narrows Bridge (in Washington State). On November 7, 1940,
winds with speeds of 65–72 km>h 140–45 mi>h2 started the main span vibrating.
The bridge, 855 m (2800 ft) long and 12 m (39 ft) wide, had been opened to traffic
only four months earlier. The main span vibrated in two different modes: a trans-
verse mode of frequency 0.6 Hz and a torsional (twisting) mode of frequency
0.2 Hz. The main span finally collapsed (䉲 Fig. 13.22). The cause of the collapse is
quite complicated, but the energy provided by the wind was a major factor.
Mechanical resonance is not the only type of resonance. When you tune a radio,
you are changing the resonant frequency of an electrical circuit (Section 21.5) so
that it will be driven by, or will pick up, a signal at the frequency of the station you
want.
DID YOU LEARN?

➥ A standing wave is the result of the superposition or interference of two waves of
the same wavelength, amplitude, and speed, but traveling in opposite directions.
➥ Nodes are points that remain stationary at all times, and antinodes correspond to
points having maximum amplitudes.
➥ Resonance occurs when the external driving frequency is equal to the natural
frequency or resonant frequency of a system. Maximum energy is transferred to the
system, so the amplitude of the system is at a maximum.
䉴 F I G U R E 1 3 . 2 2 Galloping Ger-
tie The collapse of the Tacoma Nar-
rows Bridge on November 7, 1940,
is captured in this frame from a
movie camera.
PULLING IT TOGETHER Frequency, Density, and Diameter of a Guitar String

Suppose you want to increase the fundamental frequency of a Since the strings are of the same material (same density r),
guitar string. (a) Would you (1) loosen the string to decrease the greater the diameter of a string, the greater its volume, the
its tension, (2) tighten the string to increase its tension, (3) use greater its mass, and therefore the greater its mass per unit
another string of the same material with a smaller diameter at length (greater m). Hence, a thinner string, with a smaller m,
the same tension, or (4) use another string of the same mater- will vibrate at a higher frequency, and answer (3) is also correct.
ial with a larger diameter at the same tension? (b) In actuality, (B) QUANTITATIVE REASONING AND SOLUTION. At first, one
you want to go from the A note (220 Hz) below middle C to might think of computing the frequencies directly, using Eq.
the A note (440 Hz) above middle C. Show that a string with 13.20. But this can’t be done, because not enough data are
half the diameter of the same material will have double the given, which usually implies the use of a ratio. Since the
fundamental frequency. linear mass density, m = m>L, depends on mass, and mass in
(A) CONCEPTUAL REASONING. The fundamental frequency of turn depends on diameter, the frequency equation needs to be
a stretched string is given by Eq. 13.20: expressed in terms of the diameter of a string.
Recall that the mass of a string depends on its density r and
1for n = 1)
1 FT
f = volume V; that is, r = m>V, or m = rV. Then, the volume V of
2L A m a length (L) of wire, which may be thought of as a long cylinder
The frequency of the string is proportional to the square root of with circular cross-section A, can be determined by V = AL.
the tension force FT, so loosening the string—that is, decreas- The circular area is proportional to the square of the
ing FT—would not increase the frequency. Increasing the ten- pd2
diameter of the wire, A = ¢ ≤ . This is the key to our proof.
sion does increases the frequency, so (2) is correct. Please note 4
that doubling the tension will not double the frequency
because 12FT Z 21FT.
Find: Prove that if d2 = d1>2, then f2 = 2f1.

1 FT
Given: f = (for n = 1, Eq. 13.20)
2L A m
d2 = d1>2
The linear mass density of the wire string can be expressed in The quantities in the brackets are constant, and f r 1>d, so, in
terms of its density and volume, the latter of which is propor- ratio form,
tional to the square of the string’s diameter:
f2 d1 0.30 cm
= 2 and f2 = 2f1
pd2
rV rAL = =
m f1 d2 0.15 cm
m = = = = r¢ ≤
L L L 4
Putting this into Eq. 13.20 yields

1 FT 1 4FT 1 FT 1
f = = = ¢ ≤
2L A m 2L A rpd 2 L A rp d
■ Simple harmonic motion (SHM) requires a restoring force ■ In general, the total energy of an object in SHM is directly
directly proportional to the displacement, such as an ideal proportional to the square of the amplitude.
spring force, which is given by Hooke’s law. Total energy of a spring and mass in SHM:
Fs Fa E = 12 kA2 = 12 mv2 + 12 kx 2 (13.4–5)
Energy
x = 0 x = ⫹A
1 2
Hooke’s law: U = 2
kx
Fs = - kx (13.1)
E = constant
■ The frequency (f) and period (T) for SMH are inversely related.
K
Frequency and period for SHM: Kmax = E Umax = E
1 U
f = (13.2)
T −A x=0 x1 +A
■ The form of an equation of motion for an object in SHM ■ A wave is a disturbance in time and space; energy is trans-
depends on the object’s initial (yo) displacement. ferred or propagated by wave motion.
Equations of motion for SHM: Wave speed:
b
2pt l
y = ⫾A sin vt = ⫾A sin12pft2 = ⫾A sina (13.8a) v = = lf (13.17)
T T
+ for initial motion upward with yo = 0 λ v Crest
+A
- for initial motion downward with yo = 0 Amplitude
y
+A y = A sin ω t
t=T
0 t
–A
−A λ Trough
y = A cos ω t
■ At any time, the combined waveform of two or more inter-
+A
fering waves is given by the sum of the displacements of the
t=T
0 t individual waves at each point in the medium.
−A v1 v2
y1 y2
b
2pt
y = ⫾A cos vt = ⫾A cos12pft2 = ⫾A cosa (13.9a)
T
y = y1+ y2
+ for initial motion downward with yo = + A
- for initial motion upward with yo = - A
y = y1+ y2
Velocity of a mass oscillating on a spring:
1A2 - x 22
k
v = ⫾ (13.6)
Am ■ At natural frequencies, standing waves can form on a string
Period of a mass oscillating on a spring: as a result of the interference of two waves of identical
wavelength, amplitude, and speed traveling in opposite
m directions on a string.
T = 2p (13.11)
Ak
Angular frequency of a mass oscillating on a spring:
k l1
L= First harmonic
v = 2pf = (13.13) 2
Am
Period of a simple pendulum (small-angle approximation):
L l2
T = 2p (13.14) L=2 Second harmonic
Ag 2
Velocity of a mass in SHM: Natural frequencies for a stretched string:
b = = nf1 1for n = 1, 2, 3, Á 2
(vertical velocity if vo v n FT
v = vA cos vt (13.15) fn = na (13.20)
is upward at to = 0, yo = 0) 2L 2L A m
Acceleration of a mass in SHM:
(vertical acceleration if vo
a = - v2A sin vt = - v2y (13.16)
is upward at to = 0, yo = 0)

13.1 SIMPLE HARMONIC MOTION 2. The maximum kinetic energy of a mass–spring system
in SHM is equal to (a) A, (b) A2, (c) kA, (d) kA2>2.
1. For a particle in SHM, the force on it (F) and its displace-
ment from its equilibrium position (x) are (a) in the same 3. If the frequency of a system in SHM is doubled, the
direction, (b) opposite in direction, (c) perpendicular to period of the system is (a) doubled, (b) halved, (c) four
each other, (d) none of the preceding. times as large, (d) one-quarter as large.
4. When a particle in a horizontal SHM is at the equilib- 13.4 WAVE PROPERTIES

rium position, the kinetic energy of the system is
11. When two waves meet each other and interfere, the
(a) zero, (b) at a maximum, (c) half the maximum value,
resultant waveform is determined by (a) reflection,
(b) refraction, (c) diffraction, (d) superposition.
12. When two identical waves of the same wavelength (l)
13.2 EQUATIONS OF MOTION and amplitude (A) interfere in phase, the amplitude of
5. The equation of motion for a particle in SHM (a) is a sine the resulting wave is (a) A, (b) 2A, (c) 3A, (d) 4A.
or cosine function, (b) is a tangent or cotangent function, 13. You can often hear people talking from around a corner
(c) could be any mathematical function, (d) gives the of a building. This is due primarily to (a) reflection,
velocity of the particle as a function of time. (b) refraction, (c) interference, (d) diffraction.
6. For the SHM equation y = A sin31200p rad>s2t4, the
frequency of oscillation, f, is (a) 50 Hz, (b) 100 Hz,
(c) 200 Hz, (d) 200p Hz. 13.5 STANDING WAVES AND
7. For the SHM equation y = A sin12pt>T2, the y-position RESONANCE
of the object three-quarters of the period after the motion
14. For two traveling waves to form standing waves, the
starts is (a) + A, (b) -A, (c) A>2, (d) 0.
waves must have the same (a) wavelength, (b) ampli-
tude, (c) speed, (d) all of the preceding.
13.3 WAVE MOTION 15. The points of zero amplitude on a rope that is support-
8. Wave motion in a material medium involves (a) the ing a standing wave waveform are called (a) nodes,
propagation of a disturbance, (b) interparticle interac- (b) antinodes, (c) fundamentals, (d) resonance points.
tions, (c) the transfer of energy, (d) all of the preceding. 16. For a standing wave on a rope, the distance between two
9. For a longitudinal wave, the direction between the wave adjacent antinodes is (a) 1>4 wavelength, (b) 1>2 wave-
velocity and particle oscillation is (a) perpendicular, length, (c) one wavelength, (d) two wavelengths.
(b) parallel, (c) 45°, or (d) none of the preceding. 17. When a stretched violin string oscillates in its second
10. A water wave is (a) transverse, (b) longitudinal, (c) a harmonic mode, the standing wave in the string will
combination of transverse and longitudinal, (d) none of exhibit (a) 1>4 wavelength, (b) 1>2 wavelength, (c) one
the preceding. wavelength, (d) two wavelengths.
13.1 SIMPLE HARMONIC MOTION 9. One simple harmonic motion is described by a sine func-
tion, y = A sin1vt2, and another is described by a cosine
1. If the amplitude of a particle in SHM is doubled, how are
function, y = A cos1vt2. Discuss the differences in their
(a) the total energy and (b) the maximum speed affected?
initial position, velocity, and acceleration.
2. How does the speed of a mass in SHM change as the
mass leaves its equilibrium position? Explain.
13.3 WAVE MOTION
3. A mass–spring system in SHM has an amplitude A and
period T. How long does the mass take to travel a dis- 10. When a wave pulse travels along a rope, what travels with
tance A? How about 2A? the wave motion and what does not travel with the wave?
11. 䉲 Figure 13.23 shows pictures of two mechanical waves.
4. A tennis player uses a racket to bounce a ball up and
Identify each as being transverse or longitudinal.
down with a constant period. Is this a simple harmonic
12. What type(s) of wave(s), transverse or longitudinal, will
motion? Explain.
propagate through (a) solids, (b) liquids, and (c) gases?
13.2 EQUATIONS OF MOTION

5. If a mass–spring system were taken to the Moon, would
the period of the system change? How about the period
of a pendulum taken to the Moon? Explain.
6. If you want to increase the frequency of vibration of a
mass–spring system, would you increase or decrease the
mass? Explain.
7. If the length of a pendulum is doubled, what is the ratio
of the new period to the old one?
8. Would the period of a pendulum in an upward-
accelerating elevator be increased or decreased compared 䉱 F I G U R E 1 3 . 2 3 Transverse or longitudinal? See Concep-
with its period in a nonaccelerating elevator? Explain. tual Exercise 11.
13. Standing on a hill and looking at a tall wheat field, you 13.5 STANDING WAVES AND
see a beautiful wave traveling across the field whenever RESONANCE
a breeze blows. What type of wave is this? Explain.
17. Can harmonic sound of any frequency be generated and
heard from a violin string with a fixed tension? Explain.
13.4 WAVE PROPERTIES
18. If they have the same tension and length, will a thicker
14. What is cancelled out when destructive interference or a thinner guitar string sound higher in frequency?
occurs? What happens to the wave energy in such a situ- Why?
ation? Explain. 19. A child’s swing (a pendulum) has only one natural fre-
15. Dolphins and bats determine the location of their prey quency, f1, yet it can be driven or pushed smoothly at fre-
by emitting ultrasonic sound waves. Which wave phe- quencies of f1>2, f1>3, and 2f1. How is this possible?
nomenon is involved? 20. By rubbing the circular lip of a wide, thin wine glass
16. If sound waves were dispersive (that is, if the speed of with a moist finger, you can make the glass “sing.” (Try
sound depended on its frequency), what would be the it.) (a) What causes this? (b) What would happen to the
consequences of someone listening to an orchestra in a frequency of the sound if you added water to the glass?
concert hall?
EXERCISES
13.1 SIMPLE HARMONIC MOTION 10. IE ● ● (a) At what position is the magnitude of the force
on a mass in a mass–spring system minimum: (1) x = 0,
1. ● A particle oscillates in SHM with an amplitude A. (2) x = - A, or (3) x = + A? Why? (b) If m = 0.500 kg,
What is the total distance (in terms of A) the particle trav- k = 150 N>m, and A = 0.150 m, what are the magnitude
els in three periods? of the force on the mass and the acceleration of the mass
2. ● If it takes a particle in SHM 0.50 s to travel from the at x = 0, 0.050 m, and 0.150 m?
equilibrium position to the maximum displacement 11. IE ● ● (a) At what position is the speed of a mass in a
(amplitude), what is the period of oscillation? mass–spring system maximum: (1) x = 0, (2) x = - A, or
3. ● A 0.75-kg object oscillating on a spring completes a cycle (3) x = + A? Why? (b) If m = 0.250 kg, k = 100 N>m,
every 0.50 s. What is the frequency of this oscillation? and A = 0.10 m for such a system, what is the mass’s
4. ● A particle in simple harmonic motion has a frequency maximum speed?
of 40 Hz. What is the period of this oscillation? 12. ● ● A mass–spring system is in SHM in the horizontal
5. ● The frequency of a simple harmonic oscillator is dou- direction. If the mass is 0.25 kg, the spring constant is
bled from 0.25 Hz to 0.50 Hz. What is the change in its 12 N>m, and the amplitude is 15 cm, (a) what is the max-
period? imum speed of the mass, and (b) where does this occur?
6. ● An object of mass 0.50 kg is attached to a spring with (c) What is the speed at a half-amplitude position?
spring constant 10 N>m. If the object is pulled down 13. ● ● A horizontal spring on a frictionless level air track
0.050 m from the equilibrium position and released, has a 0.150-kg object attached to it and it is stretched
what is its maximum speed? 6.50 cm. Then the object is given an outward initial
7. ● An object of mass 1.0 kg is attached to a spring with velocity of 2.20 m>s. If the spring constant is 35.2 N>m,
spring constant 15 N>m. If the object has a maximum determine how much farther the spring stretches.
speed of 0.50 m>s, what is the amplitude of oscillation? 14. ● ● A 0.25-kg object is suspended on a light spring of
8. ● ● Atoms in a solid are in continuous vibrational spring constant 49 N>m. The spring is then compressed
motion due to thermal energy. At room temperature, to a position 15 cm above the stretched equilibrium posi-
the amplitude of these atomic vibrations is typically tion. How much more energy does the system have at
about 10-9cm, and their frequency is on the order of the compressed position than at the stretched equilib-
1012 Hz. (a) What is the approximate period of oscillation rium position?
of a typical atom? (b) What is the maximum speed of 15. ● ● A 0.25-kg object is suspended on a light spring of
such an atom? spring constant 49 N>m and the system is allowed to
9. ● ● A particle of mass 0.10 kg is attached to a spring of come to rest at its equilibrium position. The object is then
spring constant 10 N>m. If the maximum acceleration of pulled down 0.10 m from the equilibrium position and
the particle is 5.0 m>s2, what is the maximum speed of released. What is the speed of the object when it goes
the particle? through the equilibrium position?
EXERCISES 485
16. ●● A 0.350-kg block moving vertically upward collides 26. ● The equation of motion for an oscillator in vertical
with a light vertical spring and compresses it 4.50 cm SHM is given by y = 10.10 m2 sin31100 rad>s2t4. What
before coming to rest. If the spring constant is 50.0 N>m, are the (a) amplitude, (b) frequency, and (c) period of this
what was the initial speed of the block? (Ignore energy motion?
losses to sound and other factors during the collision.) 27. ● The displacement of an object is given by
17. ● ● ● A 75-kg circus performer jumps from a 5.0-m height y = 15.0 cm2 cos3120p rad>s2t4. What are the object’s
onto a trampoline and stretches it downward 0.30 m. (a) amplitude, (b) frequency, and (c) period of oscillation?
Assuming that the trampoline obeys Hooke’s law, 28. ● If the displacement of an oscillator in SHM is described
(a) how far will it stretch if the performer jumps from a by the equation y = 10.25 m2 cos31314 rad>s2t4, where y
height of 8.0 m? (b) How far will the trampoline stretch if is in meters and t is in seconds, what is the position of the
the performer stands still on it while taking a bow? oscillator at (a) t = 0, (b) t = 5.0 s, and (c) t = 15 s?
18. ● ● ● A vertical spring has a 0.200-kg mass attached to it. 29. ● ● The equation of motion of a SHM oscillator is
The mass is released from rest and falls 22.3 cm before x = 10.50 m2 sin12pf2t, where x is in meters and t is in
stopping. (a) Determine the spring constant. (b) Determine seconds. If the position of the oscillator is at
the speed of the mass when it has fallen only 10.0 cm. x = 0.25 m at t = 0.25 s, what is the frequency of the
19. ● ● ● A 0.250-kg ball is dropped from a height of 10.0 cm oscillator?
onto a spring, as illustrated in 䉲 Fig. 13.24. If the spring 30. IE ● ● The oscillations of two oscillating mass–spring
has a spring constant of 60.0 N>m, (a) what distance will systems are graphed in 䉲 Fig. 13.25. The mass in
the spring be compressed? (Neglect energy loss during System A is four times that in System B. (a) Compared
collision.) (b) On recoiling upward, how high will the with System B, System A has (1) more, (2) the same, or
ball go? (3) less energy. Why? (b) Calculate the ratio of energy
between System B and System A.
䉳 FIGURE 13.24
How far down? y (cm)
See Exercise 19. 5.0
0 t (s)
–5.0 4.0 s
System A
y (m)
0.10
0 t (s)
0.15 0.45 0.75 1.05
–0.10
System B
13.2 EQUATIONS OF MOTION 䉱 F I G U R E 1 3 . 2 5 Wave energy and equation of motion See

20. ● A 0.50-kg mass oscillates in simple harmonic motion Exercises 30, 42, and 43.
on a spring with a spring constant of 200 N>m. What are
(a) the period and (b) the frequency of the oscillation? 31. ●● Show that the total energy of a mass–spring system
21. ● The simple pendulum in a tall clock is 0.75 m long. in simple harmonic motion is given by 12 mv2A2.
What are (a) the period and (b) the frequency of this 32. ● ● Show that for a pendulum to oscillate at the same fre-
pendulum? quency as a mass on a spring, the pendulum’s length
22. ● How much mass should be at the end of a spring must be given by L = mg>k.
1k = 100 N>m2 in order to have a period of 2.0 s? 33. ● ● The velocity of a vertically oscillating mass–spring
23. ● If the frequency of a mass–spring system is 1.50 Hz system is given by v = 10.650 m>s2 sin314 rad>s2t4.
and the mass on the spring is 5.00 kg, what is the spring Determine (a) the amplitude and (b) the maximum accel-
constant? eration of this oscillator.
24. ● A breeze sets a suspended lamp into oscillation. If the 34. IE ● ● (a) If the mass in a mass–spring system is halved,
period is 1.0 s, what is the distance from the ceiling to the the new period is (1) 2, (2) 12, (3) 1> 12, (4) 1>2 times the
lamp at the lowest point? Assume that the lamp is a old period. Why? (b) If the initial period is 3.0 s and the
point mass and acts as a simple pendulum. mass is reduced to 1>3 of its initial value, what is the new
25. ● Write the general equation of motion for a mass that is period?
on a horizontal frictionless surface and is connected to a 35. IE ● ● (a) If the spring constant in a mass–spring system
spring at equilibrium (a) if the mass is initially pulled in is halved, the new period is (1) 2, (2) 12, (3) 1> 12,
the +x axis from the spring (stretched) and released, and (4) 1>2 times the old period. Why? (b) If the initial period
(b) if the mass is pushed in the -x axis toward the spring is 2.0 s and the spring constant is reduced to 1>3 of its
(compressed) and released. initial value, what is the new period?
36. ●● Students use a simple pendulum with a length of 45. ● ● ● The acceleration as a function of time of a
36.90 cm to measure the acceleration of gravity at the mass–spring system is given by

location of their school. If it takes 12.20 s for the pendu- a = 10.60 m>s22 sin312 rad>s2t4. If the spring constant is
lum to complete ten oscillations, what is the experimen- 10 N>m, what are (a) the amplitude, (b) the initial veloc-
tal value of g at the school? ity and (c) the mass of the object?
37. ●● The equation of motion of a particle in vertical SHM 46. ● ● ● A clock uses a pendulum that is 75 cm long. The
is given by y = 110 cm2 sin310.50 rad>s2t4. What are the clock is accidentally broken, and when it is repaired, the
particle’s (a) displacement, (b) velocity, and (c) accelera- length of the pendulum is shortened by 2.0 mm. Con-
tion at t = 1.0 s? sider the pendulum to be a simple pendulum. (a) Will
the repaired clock gain or lose time? (b) By how much
38. ●● What is the maximum elastic potential energy of a will the time indicated by the repaired clock differ from
simple horizontal mass–spring oscillator whose equa-
tion of motion is given by x = 10.350 m2 sin317 rad>s2t4?
the correct time (taken to be the time determined by the
original pendulum in 24 h)? (c) If the pendulum rod
The mass on the end of the spring is 0.900 kg. were metal, would the surrounding temperature make a
39. ●● Two masses oscillate on light springs. The second difference in the timekeeping of the clock? Explain.
mass is half of the first and its spring constant is twice 47. ● ● ● The velocity of a vertically oscillating 5.00-kg mass on
that of the first. Which system will have the greater a spring is given by v = 1 -0.600 m>s2 sin316 rad>s2t4.
frequency, and what is the ratio of the frequency of the (a) Determine the equation of motion (y). (b) Where does
second mass to that of the first mass? the motion start and in what direction does the object move
40. ●● During an earthquake, the floor of an apartment initially and with what speed? (c) Determine the period of
building is measured to oscillate in approximately sim- the motion. (d) Determine the maximum force on the mass.
ple harmonic motion with a period of 1.95 seconds and
an amplitude of 8.65 cm. Determine the maximum speed 13.3 WAVE MOTION
and acceleration of the floor during this motion. 48. ● A sound wave has a speed of 340 m>s in air. If this
41. IE ● ● (a) If a pendulum clock were taken to the Moon, wave produces a tone with a frequency of 1000 Hz, what
where the acceleration due to gravity is only one-sixth is its wavelength?
49. ● A wave on a rope that measures 10 m long takes 2.0 s
(assume the figure to be exact) that on the Earth, will the
period of vibration (1) increase, (2) remain the same, or to travel the whole rope. If the wavelength of the wave is
(3) decrease? Why? (b) If the period on the Earth is 2.0 s, 2.5 m, what is the frequency of oscillation of any piece of
what is the period on the Moon? the rope?
50. ● A student reading his physics book on a lake dock
42. ●● The motion of a particle is described by the curve for
notices that the distance between two incoming wave
System A in Fig. 13.25. (a) Write the equation of motion
crests is about 0.75 m, and he then measures the time of
in terms of a sine or cosine function. (b) If the spring con-
arrival between the crests to be 1.6 s. What is the approx-
stant is 20 N>m, what is the mass of the object?
imate speed of the waves?
43. ●● The motion of a 0.25-kg mass oscillating on a light 51. ● Dolphins and bats determine the location of their prey
spring is described by the curve for System B in using echolocation (see Conceptual Question 15). If it
Fig. 13.25. (a) Write the equation for the displacement of takes 15 ms for a bat to receive the ultrasonic sound
the mass as a function of time. (b) What is the spring wave reflected off a mosquito, how far is the mosquito
constant of the spring? from the bat? Take the speed of sound as 345 m>s.
44. ● ● ● The forces acting on a simple pendulum are shown 52. ● Light waves travel in a vacuum at a speed of
in 䉲 Fig. 13.26. (a) Show that, for the small angle approxi- 3.00 * 108 m>s. The frequency of blue light is about
mation 1sin u L u2, the force producing the motion has 6 * 1014 Hz. What is the approximate wavelength of the
the same form as Hooke’s law. (b) Show by analogy with light?
a mass on a spring that the period of a simple pendulum 53. ● ● A sonar generator on a submarine produces periodic
is given by T = 2p1L>g. [Hint: Think of the effective ultrasonic waves at a frequency of 2.50 MHz. The wave-
spring constant.] length of the waves in seawater is 4.80 * 10-4 m. When
the generator is directed downward, an echo reflected
from the ocean floor is received 10.0 s later. How deep is
the ocean at that point?
L 54. ● ● The range of sound frequencies audible to the human
θ ear extends from about 20 Hz to 20 kHz. If the speed of

sound in air is 345 m>s, what are the wavelength limits
of this audible range?
T
55. IE ● ● The AM frequencies on a radio dial range from
m 550 kHz to 1600 kHz, and the FM frequencies range from
s 88.0 MHz to 108 MHz. All of these radio waves travel at a
mg cos θ
speed of 3.00 * 108 m>s (speed of light). (a) Compared
mg sin θ θ with the FM frequencies, the AM frequencies have
mg (1) longer, (2) the same, or (3) shorter wavelengths. Why?
(b) What are the wavelength ranges of the AM band and the
䉱 F I G U R E 1 3 . 2 6 SHM of a pendulum See Exercise 44. FM band?
EXERCISES 487
56. ●● 䉲 Fig. 13.27a shows a snapshot of a wave traveling on 63. ● If the frequency of the fifth harmonic of a vibrating
a rope, and Fig. 13.27b describes the position as a func- string is 425 Hz, what is the frequency of the second
tion of time of a point on the rope. (a) What is the ampli- harmonic?
tude of the traveling wave? (b) What is the wavelength 64. ● A standing wave is formed in a stretched string that is
of the wave? (c) What is the period of the wave? 3.0 m long. What are the wavelengths of (a) the first har-
(d) What is the wave speed? monic and (b) the second harmonic?
y 65. ● If the wavelength of the third harmonic on a string is

v 5.0 m, what is the length of the string?
66. IE ● ● A piece of steel string is under tension. (a) If the
tension doubles, the transverse wave speed (1) doubles,
0 x (cm) (2) halves, (3) increases by 12, (4) decreases by 12.
3.0 9.0 15.0
Why? (b) If the linear mass density of a 10.0-m length of
string is 0.125 kg>m and it is under a tension of 9.00 N,
what is the transverse wave speed in the string? (c) What
(a)
y (cm) are its waves’ natural frequencies?
15 67. ●● On a violin, a correctly tuned A string has a fre-
quency of 440 Hz. If an A string produces sound at
0.60 450 Hz under a tension of 500 N, what should the ten-
0 t (s)
0.20 1.0 1.4 1.8 sion be to produce the correct frequency?
68. ●● Will a standing wave be formed in a 4.0-m length of
–15 stretched string that transmits waves at a speed of 12 m>s
(b) if it is driven at a frequency of (a) 15 Hz or (b) 20 Hz?
䉱 F I G U R E 1 3 . 2 7 How high and how fast? See Exercise 56. 69. ●● Two waves of equal amplitude and frequency of
250 Hz travel in opposite directions at a speed of 150 m>s
in a string. If the string is 0.90 m long, for which harmonic
57. ●● Assume that P and S (primary and secondary) waves mode is the standing wave set up in the string?
from an earthquake with a focus near the Earth’s surface 70. ●● A university physics professor buys 100 m of string
travel through the Earth at nearly constant but different and determines its total mass to be 0.150 kg. This string
average speeds. A monitoring station that is 1000 km is used to set up a standing wave laboratory demonstra-
from the epicenter detected the S wave to arrive at 42 s tion between two posts 3.0 m apart. If the desired second
after the arrival of the P wave. If the P wave has an aver- harmonic frequency is 35 Hz, what should be the
age speed of 8.0 km>s, what is the average speed of the S required string tension?
wave?
71. IE ● ● String A has twice the tension but half the linear
58. ● ● ● The speed of longitudinal waves traveling in a long,
mass density as string B, and both strings have the same
solid rod is given by v = 1Y>r, where Y is Young’s
length. (a) The frequency of the first harmonic on string
modulus and r is the density of the solid. If a distur-
A is (1) four times, (2) twice, (3) half, (4) 1>4 times that of
bance has a frequency of 40 Hz, what is the wavelength
string B. Explain. (b) If the lengths of the strings are
of the waves it produces in (a) an aluminum rod and
2.5 m and the wave speed on string A is 500 m>s, what
(b) a copper rod? [Hint: See Tables 9.1 and 9.2.]
are the frequencies of the first harmonic on both strings?
59. ● ● ● Fred strikes a steel train rail with a hammer at a fre-
quency of 2.50 Hz, and Wilma puts her ear to the rail 72. ●● You are setting up two standing string waves. You
1.0 km away. (a) How long after the first strike does have a length of uniform piano wire that is 3.0 m long
Wilma hear the sound? (b) What is the time interval and has a mass of 0.150 kg. You cut this into two lengths,
between the successive sound pulses she hears? [Hint: one of 1.0 m and the other of 2.0 m, and place each
See Tables 9.1 and 9.2 and Exercise 58.] length under tension. What should be the ratio of ten-
sions (expressed as short to long) so that their funda-
60. ● ● ● Refer to the wave shown in Fig. 13.27 (Exercise 56).
mental frequencies are the same?
(a) Locate the points on the rope that have a maximum
speed. Determine (b) the maximum speed, and (c) the 73. IE ● ● A violin string is tuned to a certain frequency (first
distance between successive high and low spots on harmonic or the fundamental frequency). (a) If a violinist
the string. wants a higher frequency, should the string be (1) length-
ened, (2) kept the same length, or (3) shortened? Why?
(b) If the string is tuned to 520 Hz and the violinist puts a
13.5 STANDING WAVES AND finger down on the string one-eighth of the string length
RESONANCE from the neck end, what is the frequency of the string
61. ●If the frequency of the third harmonic of a vibrating when the instrument is played this way?
string is 600 Hz, what is the frequency of the first 74. ● ● ● A tight uniform string with a length of 1.80 m is tied
harmonic? down at both ends and placed under a tension of 100 N.
62. ● The fundamental frequency of a stretched string is When it vibrates in its third harmonic (draw a sketch),
150 Hz. What are the frequencies of (a) the second har- the sound given off has a frequency of 75.0 Hz. What is
monic and (b) the third harmonic? the mass of the string?
75. ● ● ● In a common laboratory experiment on standing 76. ● ● ● A student uses a 2.00-m-long steel string with a
waves, the waves are produced in a stretched string by diameter of 0.90 mm for a standing wave experiment.
an electrical vibrator that oscillates at 60 Hz The tension on the string is tweaked so that the second
(䉲 Fig. 13.28). The string runs over a pulley, and a hanger harmonic of this string vibrates at 25.0 Hz. (a) Calculate
is suspended from the end. The tension in the string is the tension the string is under. (b) Calculate the first har-
varied by adding weights to the hanger. If the active monic frequency for this string. (c) If you wanted to
length of the string (the part that vibrates) is 1.5 m and increase the first harmonic frequency by 50%, what
this length of the string has a mass of 0.10 g, what would be the tension in the string? [Hint: See Table 9.2]
masses must be suspended to produce the first four har-
monics in that length?
䉱 F I G U R E 1 3 . 2 8 Standing waves on strings Twin vibrating

strings with standing waves. This demonstration model allows
you to vary the string’s tension, length, and type (linear mass
density). Also, the vibration frequency can be adjusted. See
Exercise 75.
77. To study the effects of acceleration on the period of oscil- 80. A 2.0-kg mass resting on a horizontal frictionless surface
lation, a student puts a grandfather clock with a 0.9929-m- is connected to a fixed spring. The mass is displaced
long pendulum inside an elevator. Find the period of the 16 cm from its equilibrium position and released. At
grandfather clock (a) when the elevator is stationary, (b) t = 0.50 s, the mass is 8.0 cm from its equilibrium posi-
when the elevator is accelerating upward at 1.50 m>s2, tion (and has not passed through it yet). (a) What is the
(c) when the elevator is accelerating downward at period of oscillation of the mass? (b) What are the speed
1.50 m>s2, (d) when the cable on the elevator breaks and of the mass and the force on the mass at t = 0.50 s?
the elevator simply falls, and (e) when the elevator is 81. A simple pendulum is set into small-angle motion, mak-
moving upward at a constant speed of 5.00 m>s . ing a maximum angle with the vertical of 5°. Its period is
78. A 0.500-kg mass is attached to a vertical spring and the 2.21 s. (a) Determine its length. (b) Determine its maxi-
system is allowed to come to equilibrium. The mass is mum speed. (c) What is the acceleration of the pendu-
then given an initial downward speed of 1.50 m>s. The lum bob when it is at the lowest position?
mass travels downward 25.3 cm before stopping and 82. Spring A 150.0 N>m2 is attached to the ceiling. The top of
returning. (a) Determine the spring constant. (b) What is spring B 130.0 N>m2 is hooked onto the bottom of
its speed after it falls 15.0 cm? (c) What is the accelera- spring A. Then a 0.250-kg mass is then attached to the
tion of the mass at the very bottom of the motion? bottom of Spring B. (a) How far will the object fall until it
79. During an earthquake, a house plant of mass 15.0 kg in a reaches equilibrium? (b) What is the period of the result-
tall building oscillates with a horizontal amplitude of ing oscillation?
10.0 cm at 0.50 Hz. What are the magnitudes of (a) the
maximum velocity, (b) the maximum acceleration, and
(c) the maximum force on the plant? (Assume SHM.)
14 Sound †
14.1 Sound waves (490)

■ infrasound
■ audible sound
■ ultrasound
14.2 The speed of sound (494)

■ temperature dependence
in air
14.3 Sound intensity and sound

intensity level (498)
■ loudness
■ decibel level
Sound phenomena (503) PHYSICS FACTS
T
14.4
■ interference ✦ Sound is (a) the physical propaga-
he band shown in the chapter-
■ beats tion of a disturbance (energy) in a opening photo is clearly giving
medium, and (b) the physiological
and psychological response gener- good vibrations! We owe a lot to
ally to pressure waves.
sound waves. Not only do they pro-
14.5 The Doppler effect (507) ✦ Humans cannot hear sounds with
■ frequency and motion frequencies below 20 Hz— vide us with one of our main
infrasound. Both elephants and rhi-
■ sonic booms
noceroses communicate by
sources of enjoyment in the form of
infrasound. Infrasound is produced music, but they also bring us a
by avalanches, meteors, tornadoes,
earthquakes, and ocean waves. wealth of vital information about
14.6 Musical instruments and
sound characteristics (514) ✦ The normal audible frequency range our environment, from the chime
of human hearing is between 20 Hz
■ standing waves
and 20 kHz. of a doorbell to the warning shrill of
■ loudness, pitch, quality
✦ The visible part of the outer ear is a police siren to the song of a bird.
called the pinna, or ear flap. Many ani-
mals move the ear flap in order to Indeed, sound waves are the basis
focus their hearing in a certain direc-
tion. Humans cannot do so—but
for our major form of communica-
some people can wiggle their ears. tion—speech. These waves can
✦ Ultrasound 1frequency 7 20 kHz2 is
used to make fetal images—“baby’s
also constitute highly irritating
first picture.” distractions (noise). But sound
†
The mathematics needed in this chapter ✦ Loud noise exposure—for example,
involves common logarithms (base 10). You from rock bands—is a common cause
waves become music, speech, or
may want to review these functions in
Appendix I.
of tinnitus, or ringing in the ears. noise only when our ears perceive
490 14 SOUND
them. Physically, sound is simply waves that propagate in solids, liquids, and gases.
Without a medium, there can be no sound; in a vacuum, as in outer space, there is
utter silence.
This distinction between the sensory and physical meanings of sound pro-
vides an answer to the old philosophical question: If a tree falls in the forest where
there is no one to hear it, is there sound? The answer depends on how sound is
defined—the answer is no if thinking in terms of sensory hearing, but yes if consid-
ering physical waves.
Since sound waves are all around us most of the time, we are exposed to many
interesting sound phenomena. Some of the most important of these will be con-
sidered in this chapter.
14.1 Sound Waves

➥ How are sound waves generated, and what type of wave is sound?
➥ What are the regions and divisions of the sound frequency spectrum?
For sound waves to exist there must be a disturbance or vibrations in some

medium. This disturbance may be the clapping of hands or the skidding of tires as
a car comes to a sudden stop. Under water, you can hear the click of rocks against
one another. If you put your ear to a thin wall, you can hear sounds from the other
side of the wall. Sound waves in gases and liquids (both are fluids, Chapter 9) are
primarily longitudinal waves. However, sound disturbances moving through
solids can have both longitudinal and transverse components. The intermolecular
interactions in solids are much stronger than in fluids and allow transverse com-
ponents to propagate.
The characteristics of sound waves can be visualized by considering those pro-
duced by a tuning fork, which is essentially a metal bar bent into a U shape
(䉲 Fig. 14.1). The prongs, or tines, vibrate when struck. The fork vibrates at its fun-
damental frequency, so a single tone is heard. (A tone is sound with a definite fre-
quency.) The vibrations disturb the air, producing alternating high-pressure
regions called condensations and low-pressure regions called rarefactions. Assum-
ing the fork vibrates continually, the disturbances propagate outward, and a series
of them can be described by a sinusoidal wave (Fig. 14.1b).
Condensations
Rarefactions
Pressure fluctuations in air
(a) (b)
䉱 F I G U R E 1 4 . 1 Vibrations make waves (a) A vibrating tuning fork disturbs the air, pro-
ducing alternating high-pressure regions (condensations) and low-pressure regions
(rarefactions), which form sound waves. (b) After being picked up by a microphone, the
pressure variations are converted to electrical signals. When these signals are displayed on
an oscilloscope, the sinusoidal waveform is evident.
14.1 SOUND WAVES 491
When the disturbances traveling through the air reach the ear, the eardrum (a (Upper limit) 1 GHz
thin membrane) is set into vibration by the pressure variations. On the other side
of the eardrum, tiny bones (the hammer, anvil, and stirrup) carry the vibrations to
the inner ear, where they are picked up by the auditory nerve.
Characteristics of the ear limit the perception of sound. Only sound waves with
Ultrasonic
frequencies between about 20 Hz and 20 kHz (kilohertz) initiate nerve impulses that
are interpreted by the human brain as sound. This frequency range is called the
audible region of the sound frequency spectrum (䉴 Fig. 14.2). Hearing is most acute 20 kHz
in the 1000 Hz–10 000 Hz range, with speech mainly in the frequency range between
Frequency
300 Hz–3400 Hz (that used for the telphone).
INFRASOUND
Sound wave frequencies lower than 20 Hz are in the infrasonic region (infra- Audible
sound). Waves in this region, which humans are unable to hear, are found in
nature. Longitudinal waves generated by earthquakes have infrasonic frequen-
cies, and these waves are used to study the Earth’s interior (see Chapter 13 Insight
13.1, Earthquakes, Seismic Waves, and Seismology). Infrasonic waves are also
generated by wind and weather patterns. Elephants and cattle have hearing 20 Hz
Infrasonic
responses in the infrasonic region and may give early warnings of earthquakes
and weather disturbances, such as tornadoes. (Elephants can detect sounds with
䉱 F I G U R E 1 4 . 2 Sound frequency
frequencies as low as 1 Hz, but the pigeon takes the infrasound hearing prize, spectrum The audible region of
being able to detect sound frequencies as low as 0.1 Hz.) It has been found that the sound for humans lies between
vortex of a tornado produces infrasound, and the frequency changes—low fre- about 20 Hz and 20 kHz. Below this
quencies when the vortex is small and higher frequencies when the vortex is large. is the infrasonic region, and above it
Infrasound can be detected miles away from a tornado, and so may be a method is the ultrasonic region. The upper
limit is about 1 GHz, because of the
for gaining increased warning times for tornado approaches. elastic limitations of materials.
Nuclear explosions produce infrasound, and after the Nuclear Test Ban Treaty
of 1963, infrasound listening stations were set up to detect possible violations. Now
these stations can be used to detect other sources such as earthquakes and tornadoes.
ULTRASOUND
Above 20 kHz in the sound frequency spectrum is the ultrasonic region (ultra-
sound). Ultrasonic waves can be generated by high-frequency vibrations in crystals.
Ultrasonic waves cannot be detected by humans, but can be by other animals. The
audible region for dogs extends to about 40 kHz, so ultrasonic or “silent” whistles
can be used to call dogs without disturbing people. Cats and bats have even higher
audible ranges, up to about 70 kHz and 100 kHz, respectively.
There are many practical applications of ultrasound. Because ultrasound can
travel for kilometers in water, it is used in sonar to detect underwater objects and
their ranges (distances), much like radar uses radio waves. Sound pulses gener-
ated by the sonar apparatus are reflected by underwater objects, and the resulting
echoes are picked up by a detector. The time required for a sound pulse to make
one round trip, together with the speed of sound in water, gives the distance or
range of the object. Sonar is also widely used by fishermen to detect schools of
fish, and in a similar manner, ultrasound is used in autofocus cameras. Distance
measurements allow focal adjustments to be made.
There are applications of ultrasonic sonar in nature. Sonar appeared in the ani-
mal kingdom long before it was developed by human engineers. On their noctur-
nal hunting flights, bats use a kind of natural sonar to navigate in and out of their
caves and to locate and catch flying insects (䉲 Fig. 14.3a). The bats emit pulses of
ultrasound and track their prey by means of the reflected echoes. The technique is
known as echolocation. The auditory system and data-processing capabilities of
bats are truly amazing. (Note the size of the bat’s ears in Fig. 14.3b.)
On the basis of the intensity of the echo, a bat can tell how big an insect is—the
smaller the insect, the less intense the echo. The direction of motion of an insect is
sensed by the frequency of the echo. If an insect is moving away from the bat, the
returning echo will have a lower frequency. If the insect is moving toward the bat,
492 14 SOUND
䉴 F I G U R E 1 4 . 3 Echolocation
(a) With the aid of their own natural
sonar systems, bats hunt flying
insects. The bats emit pulses of
ultrasonic waves, which lie within
their audible region, and use the
echoes reflected from their prey to
guide their attack. (b) Note the size Wall
of the bat’s ears—good for ultrasonic
hearing. Do you know why bats
roost hanging upside down? See
text for the answer.
Insect
(a) (b)
the echo will have a higher frequency. The change in frequency is known as the
Doppler effect, which is presented in more detail in Section 14.5. Dolphins also use
ultrasonic sonar to locate objects. This is very efficient since sound travels almost
five times as fast in water as in air.
INSIGHT 14.1 Ultrasound in Medicine

Probably the best known applications of ultrasound are in medi- as the transducer is scanned across the mother’s abdomen. A still
cine. For instance, ultrasound is used to obtain an image of a shot, or “echogram,” of a fetus is shown in Fig. 1b. A developing
fetus, avoiding potentially dangerous X-rays. Ultrasonic genera- fetus, which is surrounded by a sac containing the amniotic fluid,
tors (transducers) made of piezoelectric materials produce high- can be distinguished from other anatomical features, and the
frequency pulses that are used to scan the designated region of position, size, sex, and possible abnormalities may be detected.
the body.* When the pulses encounter a boundary between two Ultrasound can be used to assess stroke risk. Plaque deposits
tissues that have different densities, the pulses are reflected may accumulate on the inner walls of blood vessels and restrict
(Fig. 1a). These reflections are monitored by a receiving trans- blood flow. One of the major causes of stroke is the obstruction
ducer, and a computer constructs an image from the reflected sig- of the carotid artery in the neck, which directly affects the
nals. Images of the fetus are recorded several times each second blood supply to the brain. The presence and severity of such
obstructions may be detected by using ultrasound (Fig. 2). An
ultrasonic generator is placed on the neck, and the reflections
*When an electric field is applied to a piezoelectric material, it
undergoes mechanical distortion. Periodic applications allow the from blood cells moving through the artery are monitored to
generation of ultrasonic waves. Conversely, when the material expe- determine the rate of blood flow, thereby providing an indica-
riences wave pressure, an electric voltage develops. This allows the tion of the severity of any blockage. This procedure involves
detection of ultrasonic waves. shifting the frequency of the reflected waves, as described by
Computer
constructs
F I G U R E 1 Ultrasound in use
image
(a) Ultrasound generated by
transducers, which convert elec-
trical oscillations into mechanical
Probe vibrations and vice versa, is trans-
with crystal mitted through tissue and is
transducer reflected from internal structures.
The reflected waves are detected
by the transducers, and the sig-
nals are used to construct an
image, or echogram. (b) An
echogram of a well-developed
fetus.
(a) (b)
14.1 SOUND WAVES 493
The bat, the only mammal to have evolved true flight, is a much maligned and
feared creature. However, because they feed on tons of insects yearly, bats save the
environment from a lot of insecticides. “Blind as a bat” is a common expression,
yet bats have fairly good vision, which complements their use of echolocation.
Finally, do you know why bats roost and hang upside down (Fig. 14.3b)? That is
their takeoff position. Unlike birds, bats can’t launch themselves from the ground.
Their wings don’t produce enough lift to allow takeoff directly from the ground,
and their legs are so small and underdeveloped that they can’t run to build up
takeoff speed. So they use their claws to hang, and fall into flight when they are
ready to fly.
Ultrasound is used to clean teeth with ultrasonic toothbrushes. In industrial
and home applications, ultrasonic baths are used to clean metal machine parts,
dentures, and jewelry. The high-frequency (short-wavelength) ultrasound vibra-
tions loosen particles in otherwise inaccessible places. Perhaps the best known
medical application of ultrasound is to view a fetus without exposing it to harmful
X-rays. (See Insight 14.1, Ultrasound in Medicine.) Also, ultrasound is used to
diagnose gallstones and kidney stones, and can be used to break these up by a
technique called lithotripsy (Greek, “stone breaking”).
Ultrasonic frequencies extend into the megahertz (MHz) range, but the sound
frequency spectrum does not continue indefinitely. There is an upper limit
of about 109 Hz, or 1 GHz (gigahertz), which is determined by the upper limit of
the elasticity of the materials through which the sound propagates.
the Doppler effect. (More on this effect in Section 14.5, along with In cases of uncontrolled bleeding, such as blunt trauma
more on Doppler “flow meters.”) resulting from a car accident or severe wounds received in
Another widely used ultrasonic device is the ultrasonic combat, rapid hemostasis (termination of bleeding) is essen-
scalpel, which uses ultrasonic energy for both precise cutting tial. Solutions being investigated and developed include the
and coagulation. Vibrating at about 55 kHz, the scalpel makes use of diagnostic ultrasound to detect the site of bleeding and
small incisions, at the same time causing a protein clot to form high-intensity focused ultrasound (HIFU) to induce hemosta-
that seals blood vessels—“bloodless” surgery, so to speak. sis by ultrasonic cauterization. In China, ultrasound-guided
The ultrasonic scalpel has been used in gynecological proce- HIFU has been used successfully for several years and is
dures such as the removal of fibroid tumors, in tonsillec- becoming the treatment of choice for many forms of cancer.
tomies, and in many other types of surgical procedures.†
Note: Adapted from the plenary lecture given by Dr.
†
One of the most remarkable and complicated inventions of nature Lawrence A. Crum at the 18th International Congress on
is the blood clot. It can be life-saving, as when it magically forms and Acoustics in Kyoto, Japan, in the summer of 2004. Professor
stops a site of bleeding, or it can be life-threatening, as when it blocks Crum is at the Applied Physics Laboratory at the University
an artery in the heart or the brain. of Washington in Seattle, Washington.
Receiving
crystal Ultrasonic
transducer
Transmitting crystal
Transmitted
ultrasonic
wave
F I G U R E 2 Carotid artery blockage? Ultrasound is used to measure blood flow through the carotid artery to see if there is a
blockage. See text for description.
494 14 SOUND
DID YOU LEARN?

➥ Disturbances and vibrations generate sound waves, which are generally
longitudinal waves.
➥ The sound frequency spectrum has three regions: infrasonic (f 6 20 Hz), audible
(20 Hz 6 f 6 20 kHz), and ultrasonic (f 7 20 kHz).
14.2 The Speed of Sound

➥ In what type of media is the speed of sound greatest and why?

➥ On what does the speed of sound generally depend?
In general, the speed at which a disturbance moves through a medium depends on

physical quantities, elasticity and density of the medium. For example, as learned in
Section 13.5, the wave speed in a stretched string is given by v = 1FT>m, where FT
is the tension in the string and m is the linear mass density of the string.
Similar expressions describe wave speeds in solids and liquids, for which the
elasticity is expressed in terms of moduli (Section 9.1). In general, the speed of
sound in a solid and in a liquid is given by v = 1Y>r and v = 1B>r, respectively,
where Y is Young’s modulus, B is the bulk modulus, and r is the density. The
speed of sound in a gas is inversely proportional to the square root of the molecular
mass, but the complex equation will not be presented here.
Solids are generally more elastic than liquids, which in turn are more elastic
than gases. In a highly elastic material, the restoring forces between the atoms or
molecules cause a disturbance to propagate faster. Thus, the speed of sound is
generally about two to four times as fast in solids as in liquids and about ten to fif-
teen times as fast in solids as in gases such as air (䉲 Table 14.1).
The speed of sound generally depends on the temperature in a gaseous
medium. In dry air, for example, the speed of sound is 331 m>s (about 740 mi>h) at
TABLE 14.1 Speed of Sound in Various Media

(Typical Values)
Medium Speed (m >s)
Solids
Aluminum 5100
Copper 3500
Iron 4500
Glass 5200
Polystyrene 1850
Zinc 3200
Liquids
Alcohol, ethyl 1125
Mercury 1400
Water 1500
Gases
Air 10 °C2 331
Air 1100 °C2 387
Helium 10 °C2 965
Hydrogen 10 °C2 1284
Oxygen 10 °C2 316
14.2 THE SPEED OF SOUND 495
0 °C. As the temperature increases, so does the speed of sound. For normal environ-
mental temperatures, the speed of sound in air increases by about 0.6 m>s for each
degree Celsius above 0 °C. Thus, a good approximation of the speed of sound in
air for a particular (environmental) temperature is given by
v = 1331 + 0.6TC2 m>s (speed of sound in dry air) (14.1)
where TC is the air temperature in degrees Celsius.* Although not written explic-
itly, the units associated with the factor 0.6 are meters per second per Celsius
degree 3m>1s # °C24.
Let’s take a comparative look at the speed of sound in different media.
EXAMPLE 14.1 Solid, Liquid, Gas: Speed of Sound in Different Media

From their material properties, find the speed of sound in T H I N K I N G I T T H R O U G H . We know that the speed of sound in a
(a) a solid copper rod, (b) liquid water, and (c) air at room solid or a liquid depends on the elastic modulus and the density
temperature (20 °C). of the solid or liquid. These values are available in Tables 9.1
and 9.2. The speed of sound in air is given by Eq. 14.1.
SOLUTION.
Given: YCu = 11 * 1010 N>m2 Find: (a) vCu (speed in copper)

BH2O = 2.2 * 109 N>m2 (b) vH2O (speed in water)
rCu = 8.9 * 103 kg>m3 (c) vair (speed in air)
rH2O = 1.0 * 103 kg>m3
(values from Tables 9.1 and 9.2)
TC = 20 °C (for air)
(a) To find the speed of sound in a copper rod, the expression v = 1Y>r is used.
Y 11 * 1010 N>m2
vCu = = = 3.5 * 103 m>s
Ar C 8.9 * 103 kg>m3
(b) For water, v = 1B>r:
B 2.2 * 109 N>m2

vH2O = = = 1.5 * 103 m>s
Ar C 1.0 * 103 kg>m3
(c) For air at 20 ºC, by Eq. 14.1,
vair = 1331 + 0.6TC2 m>s = 3331 + 0.612024 m>s = 343 m>s = 3.43 * 102 m>s
F O L L O W - U P E X E R C I S E . In this Example, how many times faster is the speed of sound in copper (a) than in water and (b) than
in air (at room temperature)? Compare your results with the values given at the beginning of the section. (Answers to all Follow-
Up Exercises are given in Appendix VI at the back of the book.)
A generally useful approximate value for the speed of sound in air is 13 km>s (or
1
5 mi>s). Using this value, you can, for example, estimate how far away lightning is
by counting the number of seconds between the time you observe the flash and
the time you hear the associated thunder. Because the speed of light is so fast, you
see the lightning flash almost instantaneously. The sound waves of the thunder
travel relatively slowly, at about 13 km>s. For example, if the interval between the
two events is measured to be 6 s (often by counting “one thousand one, one thou-
sand two, ...”), the lightning stroke was about 2 km away (13 km>s * 6 s = 2 km,
or 15 mi>s * 6 s = 1.2 mi).
You may also have noticed the delay in the arrival of sound relative to that of
light at a baseball game. If sitting in the outfield stands, you see the batter hit the
ball before you hear the crack of the bat.
*A better approximation of these and higher temperatures is given by the expression

TC
v = ¢ 331 1 + ≤ m>s
A 273
In Table 14.1, see v for air at 100 °C, which is outside the normal environmental temperature range.
496 14 SOUND
EXAMPLE 14.2 Good Approximations?

(a) Show how good the approximations of 13 km>s and 15 mi>s T H I N K I N G I T T H R O U G H . Taking the actual speed of sound to
are for the speed of sound. Use room temperature and dry air be given by Eq. 14.1, and converting 13 km>s and 15 mi>s to m>s,
conditions. (b) Find the percent error of each compared to the comparisons can be made.
exact value.
SOLUTION. Listing what is given, along with the calculation of the speed of sound:
Given: TC = 20 °C (room temperature) Find: (a) How approximations compare to actual value
v = 1331 + 0.6TC2 m>s (b) Percent errors
= 3331 + 0.612024 m>s = 343 m>s
1
vkm = 3 km>s
1
vmi = 5 mi>s
(a) Doing the conversions to standard units:

vkm = 1
3 km>s 1103 m>km2 = 333 m>s
vmi = 1
5 mi>s 11609 m>mi2 = 322 m>s
The approximations are somewhat reasonable, with vkm being the better of the two.
(b) The percent error is given by the absolute difference of the values, divided by the accepted value times 100%. So (where units
cancel),
ƒ 343 - 333 ƒ 10
vkm = 13 km>s: % error = * 100% = * 100% = 2.9%
343 343
1 ƒ 343 - 322 ƒ 21
vmi = 5 mi>s: % error = * 100% = * 100% = 6.1%
343 343
The kilometers per second approximation is considerably better.
F O L L O W - U P E X E R C I S E . Suppose the thunderstorm and lightning occurs on a very hot day with a dry air temperature of 38 °C.
Would the percent errors in the Example increase or decrease? Justify your answer.
The speed of sound in air depends on various factors. Temperature is the most
important, but there are other considerations, such as the homogeneity and compo-
sition of the air. For example, the air composition may not be “normal” in a polluted
area. These effects will not be considered here. However, the dependence of the
speed of sound on humidity is considered conceptually in the following Example.
CONCEPTUAL EXAMPLE 14.3 Speed of Sound: Sound Traveling Far and Wide
Note that the speed of sound in dry air for a given tempera- In a volume of moist air, a large number of water (H2O)
ture is given to a good approximation by Eq. 14.1. However, molecules occupy the space normally occupied by either
the moisture content of the air (humidity) varies, and this nitrogen (N2) or oxygen (O2) molecules, which make up 98%
variation affects the speed of sound. At the same temperature, of the air. Water molecules are less massive than both nitro-
would sound travel faster in (a) dry air or (b) moist air? gen and oxygen molecules. [From Section 10.3, the molecular
(formula) masses are H2O, 18 g>mol; N2, 28 g>mol; and O2, 32
REASONING AND ANSWER. According to an old folklore say- g>mol.] Thus, the average molecular mass of a volume of
ing, “Sound traveling far and wide, a stormy day will betide.” moist air is less than that of dry air, and the speed of sound is
This saying implies that sound travels faster on a highly humid greater in moist air.
day, when a storm or precipitation is likely. But is the saying true? This situation can be looked at in another way. Since water
Near the beginning of this section, it was pointed out that molecules are less massive, they have less inertia and respond
the speed of sound in a gas is inversely proportional to the to the sound wave faster than nitrogen or oxygen molecules
square root of the molecular mass of the gas. So at constant do. The water molecules therefore propagate the disturbance
pressure, is moist air more or less dense than dry air? faster.*
F O L L O W - U P E X E R C I S E . Considering only molecular masses, would you expect the speed of sound to be greatest in nitrogen,
oxygen, or helium (at the same temperature and pressure)? Explain.
*Humidity was included here as an interesting consideration for the speed of sound in air. However, henceforth, in computing the speed of sound
in air at a certain temperature, only dry air will be considered (Eq. 14.1) unless otherwise stated.
14.2 THE SPEED OF SOUND 497
Always keep in mind that our discussion generally assumes ideal conditions
for the propagation of sound. Actually, the speed of sound depends on many
things, one of which is humidity, as the preceding Conceptual Example shows. A
variety of other properties affect the propagation of sound. As an example, let’s
ask the question, “Why do ships’ foghorns have such a low pitch or frequency?”
The answer is that low-frequency sound waves travel farther than high-frequency
ones under identical conditions.
This effect is explained by a couple of characteristics of sound waves. First, sound
waves are attenuated (that is, lose energy) because of the viscosity of the air
(Section 9.5). Second, sound waves tend to interact with oxygen and water mole-
cules in the air. The combined result of these two properties is that the total attenua-
tion of sound in air depends on the frequency of the sound: the higher the frequency,
the more the attenuation and the shorter the distance traveled. It turns out that the
attenuation increases as the square of the frequency. For example, a 200-Hz sound
will travel 16 times as far as an 800-Hz sound to obtain the same attenuation. So,
low-frequency foghorns are used. Because of this wave dependence on frequency,
you might notice that when a storm’s lightning is farther away, the thunder you gen-
erally hear is a low-frequency rumble. (See Insight 14.2, The Physiology and Physics
of the Ear and Hearing.)
INSIGHT 14.2 The Physiology and Physics of the Ear and Hearing
The ear consists of three basic parts: the outer ear, the middle
ear, and the inner ear (Fig. 1). The visible part of the ear is the Inner ear
Outer ear
pinna (or ear flap), and it collects and focuses sound waves. Pinna Hammer Semicircular
Many animals can move the ear flap in order to focus their hear- (ear flap) canals
ing in a particular direction; humans have generally lost this Anvil Oval window
ability and must turn the head. The sound enters the ear and Auditory
travels through the ear canal to the eardrum of the middle ear. nerve
The eardrum is a membrane that vibrates in response to the (to brain)
pressure variations of impinging sound waves. The vibrations
Cochlea
are transmitted through the middle ear by an intricate set of
three bones called the malleus, or hammer; the incus, or anvil; Ear Middle
and the stapes, or stirrup. These bones form a linkage to the oval canal ear Stirrup
window, the opening to the inner ear. The eardrum transmits Eardrum
(tympanum)
sound vibrations to the bones of the middle ear, which in turn Eustachian tube
transmits the vibrations through the oval window to the fluid of
the inner ear. F I G U R E 1 Anatomy of the human ear The ear converts
The inner ear consists of the semicircular canals, the cochlea, pressure waves in the air into electrical nerve impulses that
and the auditory nerve. The semicircular canals and the cochlea are interpreted as sounds by the brain. See text for description.
are filled with a water-like liquid. The liquid and the nerve
cells in the semicircular canal play no role in the process of aged by loud noises temporarily or permanently. Over time,
hearing but serve to detect rapid movements and assist in loud sounds can cause permanent injury because hair cells are
maintaining balance. lost. Because the hair cells are (resonance) frequency specific, a
The inner surface of the cochlea, a snail-shaped organ, is person may be unable to hear sounds at particular frequencies.
lined with more than 25 000 hairlike nerve cells. These nerve In a quiet room, put both thumbs in your ears firmly and
cells differ from each other slightly in length and have different listen. Do you hear a low pulsating sound? You are hearing
degrees of resiliency to the fluid waves passing through the the sound, at about 25 Hz, made by the contracting and relaxing
cochlea. Different hair cells are sensitive to particular frequen- of the muscle fibers in your hands and arms. Although in the
cies of waves. When the frequency of a compressional wave audible range, these sounds are not normally heard, because the
matches the natural frequency of hair cells, the cells resonate human ear is relatively insensitive to low-frequency sounds.
(Section 13.5) with a larger amplitude of vibration. This causes The middle ear is connected to the throat by the Eustachian
the release of electrical impulses from the nerve cells, which are tube, the end of which is normally closed. It opens during
transmitted to the auditory nerve. The auditory nerve carries swallowing and yawning to permit air to enter and leave, so
the signals to the brain, where they are interpreted as sound. that internal and external pressures are equalized. You have
The hair cells of the cochlea are very critical to hearing. Dam- probably experienced a “stopping up” of your ears with a
age to those cells can give rise to tinnitus, or “ringing in the sudden change in atmospheric pressure (for example, during
ears.” Exposure to loud noises is a common cause of tinnitus rapid ascents or descents in elevators or airplanes). Swallow-
and often leads to hearing loss as well. After a loud rock concert ing opens the Eustachian tubes and relieves the excess pres-
in an enclosed room, people often experience a temporary ring- sure difference on the middle ear. (See the Chapter 9
ing in the ears and slight loss of hearing. Hair cells can be dam- Insight 9.2, An Atmospheric Effect: Possible Earaches.)
498 14 SOUND
DID YOU LEARN?

➥ The speed of sound is greatest in solids because solids are more elastic than liquids
and gases.
➥ In general, the speed of sound depends on the elasticity and density of the
material, and temperature.
14.3 Sound Intensity and Sound Intensity Level

➥ What does sound intensity mean physically?

➥ Does doubling the sound intensity double the intensity level?
➥ How does the difference in intensity levels affect the sound intensity?
Wave motion involves the propagation of energy. The rate of energy transfer is
expressed in terms of intensity, which is the energy transported per unit time
across a unit area. Since energy divided by time is power, intensity is power
divided by area:
energy>time power E>t
cI = d
P
intensity = = =
area area A A
The standard units of intensity 1power>area2 are watts per square meter 1W>m22.
Consider a point source that sends out spherical sound waves, as shown in
䉲 Fig. 14.4. If there are no losses, the sound intensity at a distance R from the source is
P P
I = = (point source only) (14.2)
A 4pR 2
where P is the power of the source and 4pR 2 is the area of a sphere of radius R,
through which the sound energy passes perpendicularly.
The intensity of a point source of sound is therefore inversely proportional to the
square of the distance from the source (an inverse-square relationship). Two intensi-
ties at different distances from a point source of constant power can be compared
as a ratio:
I2 P>14pR 222 R21
P>14pR 212
= =
I1 R22
or
I2 R1 2
= ¢ ≤ (point source only) (14.3)
I1 R2
R
2R
A
3R
I 4A
9A
I/4 I ⬀ 12
Point source R
I/9
䉱 F I G U R E 1 4 . 4 Intensity of a point source The energy emitted from a point source

spreads out equally in all directions. Since intensity is power divided by area,
I = P>A = P>14pR22, where the area is that of a spherical surface. The intensity then
decreases with the distance from the source as 1> R2 (figure not to scale).
14.3 SOUND INTENSITY AND SOUND INTENSITY LEVEL 499
Suppose that the distance from a point source is doubled; that is, R2 = 2R1 or
R1>R2 = 12 . Then
R1 2 1 2
= ¢ ≤ = a b =
I2 1
I1 R2 2 4
and
I1
I2 =
4
Since the intensity decreases by a factor of 1>R2, doubling the distance decreases
the intensity to a quarter of its original value.
A good way to understand this inverse-square relationship intuitively is to look
at the geometry of the situation. As Fig. 14.4 shows, the greater the distance from
the source, the larger the area over which a given amount of sound energy is
spread, and thus the lower its intensity. (Imagine having to paint two walls of dif-
ferent areas. If you had the same amount of paint to use on each, you’d have to
spread it more thinly over the larger wall.) Since this area increases as the square
of the radius R, the intensity decreases accordingly—that is, as 1>R 2.
Sound intensity is perceived by the ear as loudness. On the average, the human
ear can detect sound waves (at 1 kHz) with an intensity as low as 10-12 W>m2.
This intensity (Io) is referred to as the threshold of hearing. Thus, for us to hear a
sound, it must not only have a frequency in the audible range, but also be of suffi-
cient intensity. As the intensity is increased, the perceived sound becomes louder.
At an intensity of 1.0 W>m2, the sound is uncomfortably loud and may be painful
to the ear. This intensity (Ip) is called the threshold of pain.
Note that the thresholds of pain and hearing differ by a factor of 1012 :
Ip 1.0 W>m2
= = 1012
Io 10-12 W>m2
That is, the intensity at the threshold of pain is a trillion times that at the threshold
of hearing. Within this enormous range, the perceived loudness is not directly
proportional to the intensity. That is, if the intensity is doubled, the perceived
loudness does not double. In fact, a doubling of perceived loudness corresponds
approximately to a tenfold increase in intensity. For example, a sound with
an intensity of 10-5 W>m2 would be perceived to be twice as loud as one with
an intensity of 10-6 W>m2. (The smaller the negative exponent, the larger the
intensity.)
SOUND INTENSITY LEVEL: THE BEL AND THE DECIBEL

It is convenient to compress the large range of sound intensities by using a logarith-
mic scale (base 10) to express intensity levels (not to be confused with sound intensity
in W>m2). The intensity level of a sound must be referenced to a standard intensity,
which is taken to be that of the threshold of hearing, Io = 10-12 W>m2. Then, for any
intensity I, the intensity level is the logarithm (or log) of the ratio of I to Io , that is,
log I>Io. For example, if a sound has an intensity of I = 10-6 W>m2,
10-6 W>m2
= log a -12 b = log 106 = 6 B
I
log
Io 10 W>m2
(Recall that log 10 10x = x.) The exponent of the power of 10 in the final log term is
taken to have a unit called the bel (B).* Thus, a sound with an intensity of
10-6 W>m2 has an intensity level of 6 B on this scale. That way, the intensity range
from 10-12 W>m2 to 1.0 W>m2 is compressed into a scale of intensity levels rang-
ing from 0 B to 12 B.
*The bel was named in honor of Alexander Graham Bell, who got the first patent on the telephone.
500 14 SOUND
䉴 F I G U R E 1 4 . 5 Sound intensity Sound intensity level

levels and the decibel scale The (decibels)
intensity levels of some common 180 Rocket launch
180 dB
sounds on the decibel (dB) scale.
140 Jet plane takeoff
120 Pneumatic drill

Threshold 110 Rock band with amplifiers
of pain 120 dB
100 Machine shop
90 Subway train
Exposure to sound over 80 Average factory
90 dB for long periods
may affect hearing 70 City traffic
60 Normal conversation
50 Average home
60 dB
40 Quiet library
20 Soft whisper
20 dB
0 Threshold of hearing
A finer intensity scale is obtained by using a smaller unit, the decibel (dB), which
is a tenth of a bel. The range from 0 to 12 B corresponds to 0 to 120 dB. In this case,
the equation for the relative sound intensity level, or decibel level (B), is
I
b = 10 log (14.4)
Io
where Io = 10-12 W>m2. Note that the sound intensity level (in decibels, which are
dimensionless) is not the same as the sound intensity (in watts per square meter).
The decibel intensity scale and familiar sounds at some intensity levels are
shown in 䉱 Fig. 14.5. Taking the decibel prize is the blue whale, which can produce
sounds up to 188 dB in a frequency range of 10 Hz to 40 Hz. The sounds are trans-
mitted hundreds of miles underwater. The blue whale also takes the size prize,
being the largest creature ever known to have existed on the Earth—reaching up
to a length of 33 m (108 ft) and a weight of 145 tons. By comparison, the largest
dinosaur had a length of about 22 m (72 ft) and a weight of about 36 tons.
EXAMPLE 14.4 Sound Intensity Levels: Using Logarithms

What are the intensity levels of sounds with intensities of THINKING IT THROUGH. The sound intensity levels can be
(a) 10-12 W>m2 and (b) 5.0 * 10-6 W>m2? found by using Eq. 14.4.
SOLUTION.
Given: (a) I = 10-12 W>m2 Find: (a) b (sound intensity level)

(b) I = 5.0 * 10-6 W>m2 (b) b
(a) Using Eq. 14.4,
I 10-12 W>m2
b = 10 log = 10 log ¢ -12 ≤ = 10 log 1 = 0 dB
Io 10 W>m2
The intensity 10-12 W>m2 is the same as that at the threshold of hearing. (Recall that log 1 = 0, since 1 = 100 and log 100 = 0.)
Note that an intensity level of 0 dB does not mean that there is no sound.
14.3 SOUND INTENSITY AND SOUND INTENSITY LEVEL 501
I 5.0 * 10-6 W>m2

(b) b = 10 log = 10 log ¢ ≤
Io 10-12 W>m2
= 10 log15.0 * 1062 = 101log 5.0 + log 1062 = 1010.70 + 6.02 = 67 dB
FOLLOW-UP EXERCISE. Note in this Example that the intensity of 5.0 * 10-6 W>m2 is halfway between 10-6 and 10-5 (or 60 and
70 dB), yet this intensity does not correspond to a midway value of 65 dB. (a) Why? (b) What intensity does correspond to 65 dB?
(Compute it to three significant figures.)
EXAMPLE 14.5 Intensity Level Differences: Using Ratios

(a) What is the difference in the intensity levels if the inten- THINKING IT THROUGH. (a) If the sound intensity is doubled,
sity of a sound is doubled? (b) By what factors does the I2 = 2I1, or I2>I1 = 2, then Eq. 14.4 can be used to find the
intensity increase for intensity level differences 1¢ b2 of intensity level difference, b 2 - b 1. Recall that log a - log b =
10 dB and 20 dB? log a>b. (b) Here it is important to note that these values are
intensity level differences, ¢b = b 2 - b 1, not intensity levels.
The equation developed in part (a) will work. (Why?)

Given: (a) I2 = 2I1 Find: (a) ¢b (intensity level difference)
(b) ¢ b = 10 dB (b) I2>I1 (factors of intensity increase)
¢b = 20 dB
(a) Using Eq. 14.4 and the relationship log a - log b = log a>b, for the intensity level difference
¢ b = b 2 - b 1 = 103log1I2>Io2 - log1I1>Io24 = 10 log31I2>Io2>1I1>Io24 = 10 log I2>I1
Then,
I2
¢b = 10 log = 10 log 2 = 3 dB
I1
Thus, doubling the intensity increases the intensity level by 3 dB (such as an increase from 55 dB to 58 dB).
(b) For a 10-dB difference, Since log 102 = 2,

I2 I2 I2
¢ b = 10 dB = 10 log and log = 1.0 = 102 and I2 = 100 I1
I1 I1 I1
Thus, an intensity level difference of 10 dB corresponds to
Since log 101 = 1, the intensity ratio is 10:1 because
changing (increasing or decreasing) the intensity by a factor
I2 of 10. An intensity level difference of 20 dB corresponds to
= 101 and I2 = 10 I1 changing the intensity by a factor of 100.
I1
You should be able to guess the factor that corresponds to an
Similarly, for a 20-dB difference, intensity level difference of 30 dB. In general, the factor of the
intensity change is 10¢b, where ¢ b is the difference in levels of
I2 I2 bels. Since 30 dB = 3 B and 103 = 1000, the intensity changes
¢ b = 20 dB = 10 log and log = 2.0
I1 I1 by a factor of 1000 for an intensity level difference of 30 dB.
F O L L O W - U P E X E R C I S E . A ¢ b of 20 dB and a ¢ b of 30 dB correspond to factors of 100 and 1000, respectively, in intensity

changes. Does a ¢ b of 25 dB correspond to an intensity change factor of 500? Justify your answer.
EXAMPLE 14.6 Combined Sound Levels: Adding Intensities

Sitting at a sidewalk restaurant table, a friend talks to you in be added arithmetically, since energy and power are scalar
normal conversation (60 dB). At the same time, the intensity quantities. Then the combined intensity level can be found
level of the street traffic reaching you is also 60 dB. What is from the sum of the intensities.
the total intensity level of the combined sounds?
SOLUTION. Listing the data and what is to be found:
THINKING IT THROUGH. It is tempting simply to add the two
Given: b 1 = 60 dB Find: Total b
sound intensity levels together and say that the total is 120
b 2 = 60 dB
dB. But intensity levels in decibels are logarithmic, so you
can’t add them in the normal way. However, intensities (I) can (continued on next page)
502 14 SOUND
First let’s find the intensities associated with the intensity levels:
I1 I1
b 1 = 60 dB = 10 log = 10 log ¢ -12 ≤
Io 10 W>m2
By inspection,
I1 = 10-6 W>m2
That is, I1 = 10-6 W>m2 for the 10 log term to be equal to 60 dB.
Similarly, I2 = 10-6 W>m2, since both intensity levels are 60 dB. So the total intensity is
Itotal = I1 + I2 = 1.0 * 10-6 W>m2 + 1.0 * 10-6 W>m2 = 2.0 * 10-6 W>m2
Then, converting back to intensity level,
2.0 * 10-6 W>m2
≤ = 10 log12.0 * 1062
Itotal
b = 10 log = 10 log ¢
Io 10-12 W>m2
= 101log 2.0 + log 1062 = 1010.30 + 6.02 = 63 dB
This value is a long way from 120 dB. Notice that the combined intensities doubled the intensity value, and the intensity level
increased by 3 dB, in agreement with our finding in part (a) of Example 14.5.
F O L L O W - U P E X E R C I S E . In this Example, suppose the added noise gave a total that tripled the sound intensity level of the conver-
sation. What would be the total combined intensity level in this case?
PROTECT YOUR HEARING

Hearing may be damaged by excessive noise, so our ears sometimes need protection
from continuous loud sounds (䉲 Fig. 14.6). Hearing damage depends on the sound
intensity level (decibel level) and the exposure time. The exact combinations vary
for different people, but a general guide to noise levels is given in 䉴 Table 14.2.
Studies have shown that sound levels of 90 dB and above will damage receptor
nerves in the ear, resulting in a loss of hearing. At 90 dB, it takes 8 h or less for
damage to occur. In general, if the sound level is increased by 5 dB, the safe expo-
sure time is cut in half. For example, if a sound level of 95 dB (that of a very loud
lawn mower or motorcycle) takes 4 h to damage your hearing, then a sound level
of 105 dB takes only 1 h to do damage. Because of the detrimental effects on hear-
ing, the U.S. government has set occupational noise exposure limits.
䉱 F I G U R E 1 4 . 6 Protect your hearing Continuous loud sounds can damage hearing, so

our ears may need protection, as shown here. Note in Table 14.2 that the intensity level of
lawn mowers is on the order of 90 dB.
14.4 SOUND PHENOMENA 503
TABLE 14.2 Sound Intensity Levels and Ear Damage Exposure Times
Time of Nonstop Exposure
Sound Decibels (dB) Examples That Can Cause Damage
Faint 30 Quiet library, whispering
Moderate 60 Normal conversation, sewing

machine
Very loud 80 Heavy traffic, noisy 10 hours

restaurant, screaming child
90 Lawn mower, motorcycle, Less than 8 hours
loud party
100 Chainsaw, subway train, Less than 2 hours
snowmobile
Extremely loud 110 Stereo headset at full blast, 30 minutes
rock concert
120 Dance clubs, car stereos, 15 minutes
action movies, some musical
toys
130 Jackhammer, loud computer Less than 15 minutes
games, loud sporting events
Painful 140 Boom stereos, gunshot blast, Only seconds (for
firecrackers example, hearing loss
can occur from a few
shots of a high-
powered gun if
protection is not worn)
Courtesy of The EAR Foundation.
DID YOU LEARN?

➥ Sound intensity expresses the rate of energy transfer (energy>time>area or
power>area).
➥ Doubling the sound intensity increases the intensity level by 3 dB.The decibel scale
is a logarithmic scale, not a linear scale.
➥ An intensity level difference of 10 dB corresponds to an intensity increase (or
decrease) of a factor of 10; a 20-dB difference to a factor of 100, and so on. In
general, the factor of intensity change is 10¢b, where b is in bels.
14.4 Sound Phenomena

➥ What is the difference between sound reflection, refraction, and diffraction?

➥ When does total constructive interference occur?
➥ When does total destructive interference occur?
REFLECTION, REFRACTION, AND DIFFRACTION

An echo is a familiar example of the reflection of sound—sound “bouncing” off a
surface. Sound refraction is less obvious than reflection, but you may have experi-
enced it on a calm summer evening, when it is possible to hear distant voices or
other sounds that ordinarily would not be audible. This effect is due to the refrac-
tion, or bending (change in direction), of the sound waves as they pass from one
region into a region where the air density is different. The effect is similar to what
would happen if the sound passed into another medium.
504 14 SOUND
Total
constructive
interference Warm
(two crests or
two troughs meet)
Cool
A
*
B 䉱 F I G U R E 1 4 . 7 Sound refraction Sound travels more slowly in the cool air near the
* water surface than in the upper, warmer air. As a result, the waves are refracted, or bent
downward. This bending increases the intensity of the sound at a distance where it other-
wise might not be heard.
Total
destructive
interference The required conditions for sound to be refracted downward are a layer of cooler
(a crest and air near the ground or water and a layer of warmer air above it, which provides a
a trough meet)
wave speed change. These conditions occur frequently over bodies of water, which
cool after sunset (䉱 Fig. 14.7). As a result of the cooling, the waves are refracted in
(a) an arc that may allow a distant person to receive an increased intensity of sound.
Another phenomenon is diffraction, described in Section 13.4. Sound may be dif-
A* fracted, or spread out, around corners or around an object. We usually think of
waves as traveling in straight lines. However, you can hear someone you cannot
see standing around a corner. This direction change is different from that of refrac-
␭AC LAC
tion, in which no obstacle causes the bending.
␭BC Reflection, refraction, and diffraction are described in a general sense here for
B*
sound. These phenomena are important considerations for light waves as well,
and will be discussed more fully in Chapters 22 and 24.
LBC C
In phase
INTERFERENCE
(b)
Like waves of any kind, sound waves interfere when they meet. Suppose that two
A* loudspeakers separated by some distance emit sound waves in phase at the same
frequency. If we consider the speakers to be point sources, then the waves will
spread out spherically and interfere (䉳 Fig. 14.8a). The lines from a particular
speaker represent wave crests (or condensations), and the troughs (or rarefac-
D Out of tions) lie in the intervening white areas.
phase
In particular regions of space, there will be constructive or destructive interfer-
B*
ence. But, if two waves meet in a region where they are exactly in phase (two
(c) crests or two troughs coincide), there will be total constructive interference
(Fig. 14.8b). Notice that the waves have the same motion at point C in the figure.
䉱 F I G U R E 1 4 . 8 Interference If, instead, the waves meet such that the crest of one coincides with the trough of
(a) Sound waves from two point the other (at point D), the two waves will cancel each other out (Fig. 14.8c). The
sources spread out and interfere.
(b) At points where the waves result will be total destructive interference. (See superposition in Section 13.4.)
arrive in phase (with a zero phase It is convenient to describe the path lengths traveled by the waves in terms of
difference), such as point C, con- wavelength 1l2 to determine whether they arrive in phase. Consider the waves
structive interference occurs. (c) At arriving at point C in Fig. 14.8b. The path lengths in this case are LAC = 4l and
points where the waves arrive com- LBC = 3l. The phase difference 1¢u2 is related to the path length difference 1¢L2
pletely out of phase (with a phase
difference of 180°), such as point D, by the simple relationship
destructive interference occurs. The
1¢L2
phase difference at a particular 2p (phase difference
point depends on the path lengths ¢u = (14.5)
l and path length difference)
the waves travel to reach that point.
14.4 SOUND PHENOMENA 505
Since 2p rad is equivalent, in angular terms, to a full wave cycle or wavelength,

multiplying the path length difference by 2p>l gives the phase difference in
radians. For the example illustrated in Fig. 14.8b,
1LAC - LBC2 = 14l - 3l2 = 2p rad

2p 2p
¢u =
l l
When ¢u = 2p rad, the waves are shifted by one wavelength. This is the same as
¢u = 0°, so the waves are in phase. Thus, the waves interfere constructively in the
region of point C, increasing the intensity, or loudness, of the sound detected
there.
From Eq. 14.5, it can be seen that the sound waves are in phase at any point
where the path length difference is zero or an integral multiple of the wavelength.
That is,
¢L = nl 1n = 0, 1, 2, 3, Á 2
(condition for
(14.6)
constructive interference)
A similar analysis of the situation in Fig. 14.8c, where LAD = 2 34 l and
LBD = 2 14 l, gives
A 2 4 l - 2 14 l B = p rad
2p 3
¢u =
l
or ¢u = 180°. At point D, the waves are completely out of phase, and destructive
interference occurs in this region.
Sound waves will be out of phase at any point where the path length difference
is an odd number of half-wavelengths 1l>22, or
¢L = ma b 1m = 1, 3, 5, Á 2
l (condition for
(14.7)
2 destructive interference)
At these points, a softer, or less intense, sound will be heard or detected. If the
amplitudes of the waves are exactly equal, the destructive interference is total and
no sound is heard.
Destructive interference of sound waves provides a way to reduce loud noises,
which can be distracting and cause hearing discomfort. The procedure is to have a
reflected wave or an introduced wave with a phase difference that cancels out the
original sound as much as possible. Ideally, this would be 180° out of phase with
the undesirable noise. A couple of such applications, automobile mufflers and
pilot headphones, were discussed in Section 13.4.
EXAMPLE 14.7 Pump Up the Volume: Sound Interference

At an open-air concert on a hot day (with an air temperature T H I N K I N G I T T H R O U G H . The sound waves from the speakers
of 25 °C), you sit 7.00 m and 9.10 m, respectively, from a pair will interfere. Is the interference, on which the intensity
of speakers, one at each side of the stage. A musician, warm- depends, constructive, destructive, or something in between?
ing up, plays a single 494-Hz tone. What do you hear in terms This depends on the path length difference, which can be
of intensity? (Consider the speakers to be point sources.) computed from the given distances.
SOLUTION.
Given: d1 = 7.00 m and d2 = 9.10 m Find: ¢L (path length difference in wavelength units
f = 494 Hz to determine interference)
T = 25 °C
The path length difference (2.10 m) between the waves arriv- of sound, v, at the given temperature is known. The speed v
ing at your location must be expressed in terms of the wave- can be found by using Eq. 14.1:
v = 1331 + 0.6TC2 m>s = 3331 + 0.612524 m>s = 346 m>s
length of the sound. To do this, we first need to know the
wavelength. Given the frequency, the wavelength can be
found from the relationship l = v>f, provided that the speed (continued on next page)
506 14 SOUND
The wavelength of the sound waves is then

v 346 m>s
l = = = 0.700 m
f 494 Hz
Thus, the distances in terms of wavelength are
d1 = 17.00 m2a b = 10.0l and d2 = 19.10 m2a b = 13.0l

l l
0.700 m 0.700 m
The path length difference in terms of wavelengths is
¢L = d2 - d1 = 13.0l - 10.0l = 3.0l
This is an integral number of wavelengths 1n = 32, so constructive interference occurs. The sounds of the two speakers reinforce
each other, and you hear an intense (loud) tone at 494 Hz.
F O L L O W - U P E X E R C I S E . Suppose that in this Example the tone traveled to a person sitting 7.00 m and 8.75 m, respectively, from
the two speakers. What would be the situation in that case?
Another interesting interference effect occurs when two tones of nearly the
same frequency 1f1 L f22 are sounded simultaneously. The ear senses pulsations
in loudness known as beats. The human ear can detect as many as seven beats per
second. A greater number of beats per second sounds “smooth” (continuous,
without any pulsations).
Suppose that two sinusoidal waves with the same amplitude, but slightly dif-
ferent frequencies, interfere (䉲 Fig. 14.9a). Figure 14.9b represents the resulting
sound wave. The amplitude of the combined wave varies sinusoidally, as shown
by the black curves (known as envelopes) that outline the wave.
What does this variation in amplitude mean in terms of what the listener per-
ceives? A listener will hear a pulsating sound (beats), as determined by the
envelopes. The maximum amplitude is 2A (at the point where the maxima of the
two original waves interfere constructively). Detailed mathematics shows that a
listener will hear the beats at a frequency called the beat frequency (fb), given by
fb = ƒ f1 - f2 ƒ (14.8)
The absolute value is taken because the frequency fb cannot be negative, even if
f2 7 f1. A negative beat frequency would be meaningless.
Beats can be produced when tuning forks of nearly the same frequency are vibrat-
䉲 F I G U R E 1 4 . 9 Beats Two trav- ing at the same time. For example, using forks with frequencies of 516 Hz and
eling waves of equal amplitude and 513 Hz, one can generate a beat frequency of fb = 516 Hz - 513 Hz = 3 Hz, and
slightly different frequencies inter- three beats are heard each second. Musicians tune two stringed instruments to the
fere and give rise to pulsating tones same note by adjusting the tensions in the strings until the beats disappear 1f1 = f22.
called beats. The beat frequency is
given by fb = ƒ f1 - f2 ƒ .
(a)
A f1
–A f2
Constructive Destructive Constructive Destructive Constructive

interference interference interference interference interference
2A
–2A
Resultant wave
fb = f1 – f2 (b)
14.5 THE DOPPLER EFFECT 507
DID YOU LEARN?

➥ Reflection is the “bouncing off” of sound from a surface; refraction is the “bending”
(change in direction) of sound when passing into a region of different wave speeds;
diffraction is the spreading out of sound around corners or objects.
➥ Total constructive interference occurs when two waves meet and are exactly in
phase (two crests or two troughs coincide).
➥ Total destructive interference occurs when two waves meet and are exactly out of
phase (a crest of one wave coincides with a trough of the other wave).
14.5 The Doppler Effect

➥ What gives rise to the Doppler effect?

➥ What does an aircraft sonic boom indicate?
➥ What meant by Mach number?
Standing along a highway, the pitch (the perceived frequency) of the sound of the
horn of a moving car or truck is heard to be higher as the vehicle approaches and
lower as it recedes. Variations in the frequency of the motor noise can also be
heard when a race car passes by in going around a track. A variation in the per-
ceived sound frequency due to the motion of the source is an example of the
Doppler effect. (The Austrian physicist Christian Doppler (1803–1853) first
described this effect.)
As 䉲 Fig. 14.10 shows, the sound waves emitted by a moving source tend to
bunch up in front of the source and spread out in back. The Doppler shift in fre-
quency can be found by assuming that the air is at rest in a reference frame such as
that depicted in 䉲 Fig. 14.11. The speed of sound in air is v, and the speed of the
moving source is vs. The frequency of the sound produced by the source is fs. In
one period, T = 1>fs , a wave crest moves a distance d = vT = l. (The sound
wave would travel this distance in still air in any case, regardless of whether the
source is moving.) But in one period, the source travels a distance ds = vs T before
emitting another wave crest. The distance between the successive wave crests is
thus shortened to a wavelength l¿ :
l¿ = d - ds = vT - vs T = 1v - vs2T =
v - vs
fs
Observer behind source 3 Observer in front of source

hears lower pitch hears higher pitch
(longer wavelength) (shorter wavelength)
4
1 2 3 4 5
䉱 F I G U R E 1 4 . 1 0 The Doppler effect for a moving source The sound waves bunch up in
front of a moving source—the whistle—giving a higher frequency there. The waves trail out
behind the source, giving a lower frequency there.
508 14 SOUND
䉴 F I G U R E 1 4 . 1 1 The Doppler d = vT = ␭
effect and wavelength Sound from
a moving car’s horn travels a dis-
tance d in a time T. During this time,
the car (the source) travels a dis- ␭′
tance ds before putting out a second
pulse, thereby shortening the
observed wavelength of the sound
in the approaching direction. ds = vsT
1 2 1
The frequency heard by the observer (fo) is related to the shortened wavelength by
fo = v>l¿ , and substituting l¿ gives
v v
fo = = ¢ ≤f
l¿ v - vs s
or
(source moving toward

1 a stationary observer
fo = § ¥ fs (14.9)
vs where vs = speed of source
1 -
v and v = speed of sound)
Since 1 - 1vs>v2 is less than 1, fo is greater than fs in this situation. For example,
suppose that the speed of the source is a tenth of the speed of sound; that is,
vs = v>10, or vs>v = 10
1
. Then, by Eq. 14.9, fo = 10
9 fs.
Similarly, when the source is moving away from the observer 1l¿ = d + ds2,
the observed frequency is given by
1 (source moving away

v
fo = ¢ ≤f = § ¥ fs from a stationary (14.10)
v + vs s vs
1 + observer)
v
Here, fo is less than fs. (Why?)

Combining Eqs. 14.9 and 14.10 yields a general equation for the observed fre-
quency with a moving source and a stationary observer:
- for source moving toward

v 1 stationary observer
fo = ¢ ≤f = § ¥ fs d (14.11)
v ⫾ vs s vs + for source moving away
1⫾ from stationary observer
v
As you might expect, the Doppler effect also occurs with a moving observer
and a stationary source, although this situation is a bit different. As the observer
moves toward the source, the distance between successive wave crests is the nor-
mal wavelength (or l = v>fs), but the measured wave speed is different. Relative
to the approaching observer, the sound from the stationary source has a wave
speed of v¿ = v + vo, where vo is the speed of the observer and v is the speed of
sound in still air. (The observer moving toward the source is moving in a direction
opposite that of the propagating waves and thus meets more wave crests in a
given time.)
With l = v>fs the observed frequency is then
v¿ v + vo
fo = = ¢ ≤ fs
l v
or
(observer moving toward a

vo stationary source where
fo = ¢ 1 + ≤f (14.12)
v s vo = speed of observer and
v = speed of sound)
Similarly, for an observer moving away from a stationary source, the perceived
wave speed is v¿ = v - vo and
v¿ v - vo
fo = = ¢ ≤ fs
l v
or
vo (observer moving away from a
fo = ¢ 1 - ≤f (14.13)
v s stationary source)
Equations 14.12 and 14.13 can be combined into a general equation for a moving
observer and a stationary source:
- for observer moving
v ⫾ vo vo toward stationary source
fo = ¢ ≤ fs = ¢ 1 ⫾ ≤ fs d (14.14)
v v + for observer moving away
from stationary source
You may find it difficult to remember whether a plus or minus sign is used in the general
equations for the Doppler effect. Let your experience help you. For the common case of a
stationary observer, the frequency of the sound increases when the source approaches,
so the denominator in Eq. 14.11 must be smaller than the numerator. Accordingly, in this
case you use the minus sign. When the source is receding, the frequency is lower. The
denominator in Eq. 14.11 must then be larger than the numerator, and you use the plus
sign. Similar reasoning will help you choose a plus or minus sign for the numerator in
Eq. 14.14. See Eq. 14.14a in footnote on the next page.
EXAMPLE 14.8 On the Road Again: The Doppler Effect

As a truck traveling at 96 km>h approaches and passes a per- T H I N K I N G I T T H R O U G H . This situation is an application of
son standing along the highway, the driver sounds the horn. the Doppler effect, Eq. 14.11, with a moving source and a sta-
If the horn has a frequency of 400 Hz, what are the frequen- tionary observer. In such problems, it is important to identify
cies of the sound waves heard by the person (a) as the truck the data correctly.
approaches and (b) after it has passed? (Assume that the
speed of sound is 346 m>s.)
SOLUTION. In Doppler situations, it is important to clearly list what is to be found.

Given: vs = 96 km>h = 27 m>s Find: (a) fo (observed frequency while truck is approaching)
fs = 400 Hz (b) fo (observed frequency while truck is moving away)
v = 346 m>s
(a) From Eq. 14.11 with a minus sign (source approaching sta- (b) A plus sign is used in Eq. 14.11 when the source is moving
tionary observer), away:
≤ fs = a b1400 Hz2 = 371 Hz

346 m>s
≤ fs = a b1400 Hz2 = 434 Hz
v 346 m>s v
fo = ¢ fo = ¢
v - vs 346 m>s - 27 m>s v + vs 346 m>s + 27 m>s
F O L L O W - U P E X E R C I S E . Suppose that the observer in this Example were initially moving toward and then past a stationary 400-Hz
source at a speed of 96 km>h. What would be the observed frequencies? (Would they differ from those for the moving source?)
510 14 SOUND
There are also cases in which both the source and the observer are moving,
either toward or away from one another. These will not be considered mathemati-
cally here, but will be conceptually in the next Conceptual Example.*
CONCEPTUAL EXAMPLE 14.9 It’s All Relative: Moving Source and Moving Observer
Suppose a sound source and an observer are moving away moves away from a stationary source, the observed frequency
from each other in opposite directions, each at half the speed is also lower (Eq. 14.13). With both source and observer mov-
of sound in air. In this case, the observer would (a) receive ing away from each other in opposite directions, the com-
sound with a frequency higher than the source frequency, bined effect would make the observed frequency even less, so
(b) receive sound with a frequency lower than the source fre- neither (a) nor (c) is the answer.
quency, (c) receive sound with the same frequency as the It would appear that (b) is the correct answer, but (d) must
source frequency, or (d) receive no sound from the source. logically eliminated for completeness. Remember that the
speed of sound relative to the air is constant. Therefore,
REASONING AND ANSWER. As we know, when a source (d) would be correct only if the observer is moving faster than the
moves away from a stationary observer, the observed fre- speed of sound relative to the air. Since the observer is moving
quency is lower (Eq. 14.10). Similarly, when an observer at only half the speed of sound, (b) is the correct answer.
F O L L O W - U P E X E R C I S E . In this Example, what would be the result if both the source and the observer were traveling in the same
direction with the same subsonic speed? (Subsonic, as opposed to supersonic, refers to a speed that is less than the speed of sound
in air.)
The Doppler effect also applies to light waves, although the equations describ-
ing the effect are different from those just given. When a distant light source such
as a star moves away from us, the frequency of the light we receive from it is low-
ered. That is, the light is shifted toward the red (long-wavelength) end of the spec-
trum, an effect known as a Doppler red shift. Similarly, the frequency of light from
an object approaching us is increased—the light is shifted toward the blue (short-
wavelength) end of the spectrum, producing a Doppler blue shift. The magnitude of
the shift is related to the speed of the source.
The Doppler shift of light from astronomical objects is very useful to astronomers.
The rotation of a planet, a star, or some other body can be established by looking at
the Doppler shifts of light from opposite sides of the object: because of the rotation,
one side is receding (and hence is red-shifted) and the other is approaching (and thus
is blue-shifted). Similarly, the Doppler shifts of light from stars in different regions of
our galaxy, the Milky Way, indicate that the galaxy is rotating.
You have been subjected to a practical application of the Doppler effect if you
have ever been caught speeding in your car by police radar, which uses reflected
radio waves. (Radar stands for radio detecting and ranging and is similar to under-
water sonar, which uses ultrasound.) If radio waves are reflected from a parked
car, the reflected waves return to the source with the same frequency. But for a car
that is moving toward a patrol car, the reflected waves have a higher frequency, or
are Doppler-shifted.
Actually, there is a double Doppler shift: In receiving the wave, the moving car
acts like a moving observer (the first Doppler shift), and in reflecting the wave, the
car acts like a moving source emitting a wave (the second Doppler shift). The
magnitudes of the shifts depend on the speed of the car. A computer quickly calcu-
lates this speed and displays it for the police officer.
For other important medical and weather applications of the Doppler effect, see
Insight 14.3, Doppler Applications: Blood Cells and Raindrops.
*In the case of both the observer and source moving,

v ⫾ vo
fo = ¢ ≤f (14.14a)
v ⫿ vs s
From experience, you should know there would be a frequency increase when the observer and source
approach each other, and vice versa. That is, the upper signs in the numerator and denominator apply if
the observer and source move toward each other, and the lower signs apply if they are moving away.
(See Exercise 54.)
SONIC BOOMS
Consider a jet plane that can travel at supersonic speeds. As the speed of a moving
source of sound approaches the speed of sound, the waves ahead of the source
come close together (䉲 Fig. 14.12a). When a plane is traveling at the speed of
sound, the waves can’t outrun it, and they pile up in front. At supersonic speeds,
the waves overlap. This overlapping of a large number of waves produces many
points of constructive interference, forming a large pressure ridge, or shock wave.
This kind of wave is sometimes called a bow wave because it is analogous to the
wave produced by the bow of a boat moving through water at a speed greater
than the speed of the water waves. Figure 14.12b shows the shock wave of a bullet
traveling at 500 m>s.
From aircraft traveling at supersonic speed, the shock wave trails out to the
sides and downward. When this pressure ridge passes over an observer on the
ground, the large concentration of energy produces what is known as a sonic
boom. There is really a double boom, because shock waves are formed at both
ends of the aircraft. Under certain conditions, the shock waves can break windows
and cause other damage to structures on the ground. (Sonic booms are no longer
heard as frequently as in the past. Pilots are now instructed to fly supersonically
only at high altitudes and away from populated areas.)
On a smaller scale, you have probably heard a “mini” sonic boom—the “crack”
of a whip. This must mean that the whip’s tip has somehow attained supersonic
speed. How does this happen? Whips generally taper down from the handle to
vs = 0
vs
vs < v
Subsonic
vs
vs = v
Mach 1
(b)
vs
vs > v
Supersonic
䉱 F I G U R E 1 4 . 1 2 Bow waves and sonic booms (a) When an
Tail Nose aircraft exceeds the speed of sound in air, vs , the sound waves
shock shock form a pressure ridge, or shock wave. As the trailing shock
wave wave wave passes over the ground, observers hear a sonic boom
(actually, two booms, because shock waves are formed at the
front and tail of the plane). (b) A bullet traveling at a speed of
500 m>s. Note the shock waves produced (and the turbulence
Atmospheric behind the bullet). The image was made by using interferome-
pressure try with polarized light and a pulsed laser, with an exposure
(a) time of 20 ns.
512 14 SOUND
Conical the tip, which may have several frayed strands. When the whip is given a flick of
shock wave the wrist, a wave pulse is sent down the length of the whip. Treating the whip
vs pulse as a string wave pulse, recall that the speed of the pulse depends inversely
on the linear mass density, which decreases toward the whip’s tip. Thus the speed
vt u of the pulse increases to the point that at the tip, it is greater than the speed of
sound. The “crack” is made by the air rushing back into the region of reduced
pressure created by the final flip of the whip’s tip, much as the sonic boom from a
supersonic jet trails behind the jet.
A common misconception is that a sonic boom is heard only when a plane
vst breaks the sound barrier. As an aircraft approaches the speed of sound, the pres-
sure ridge in front of it is essentially a barrier that must be overcome with extra
power. However, once supersonic speed is reached, the barrier is no longer there,
䉱 F I G U R E 1 4 . 1 3 Shock wave and the shock waves, continuously created, trail behind the plane, producing
cone and Mach number When the booms for everyone along its ground path.
speed of the source (vs) is greater Ideally, the sound waves produced by a supersonic aircraft form a cone-shaped
than the speed of sound in air (v), shock wave (䉳 Fig. 14.13). The waves travel outward with a speed v, and the speed
the interfering spherical sound
waves form a conical shock wave
of the source (plane) is vs. Note from the figure that the angle between a line tangent
that appears as a V-shaped pressure to the spherical waves and the line along which the plane is moving is given by
ridge when viewed in two dimen-
sions. The angle u is given by vt v 1
sin u = v>vs , and the inverse ratio sin u = = = (14.15)
vs>v is called the Mach number.
vs t vs M
INSIGHT 14.3 Doppler Applications: Blood Cells and Raindrops

BLOOD CELLS a less invasive alternative to other diagnostic procedures,
As learned in Insight 14.1, Ultrasound in Medicine, ultra- such as arteriography (X-ray pictures of an artery after the
sound provides a variety of uses in the medical field. Since injection of a dye).
the Doppler effect can be used to detect and provide infor- Another medical use of ultrasound is the echocardiogram,
mation on moving objects, it is used to examine blood flow which is an examination of the heart. On a monitor, this
in the major arteries and veins of the arms and legs ultrasonic technique can display the beating movements of
(Fig. 1a). In this application, the Doppler effect is used to the heart, and the physician can see the heart’s chambers,
measure the blood flow speed. Ultrasound reflects from red valves, and blood flow into and out of the organ (Fig. 1b).
blood cells with a change in frequency according to the
speed of the cells. The overall flow speed helps physicians RAINDROPS
diagnose such things as blood clots, arterial occlusion (clos- Radar has been used since the early 1940s to provide infor-
ing), and venous insufficiency. Ultrasound procedures offer mation about rainstorms and other forms of precipitation.
(a) (b)
F I G U R E 1 (a) Blood flow and blockage This Doppler ultrasound scan shows a deep vein thrombosis in a patient’s leg. The
thrombus (clot) blocking the vein is the dark area right of center. Blood flow in an adjacent artery (orange) is slowed due to the
clot. In extreme cases, a clot can break away and be carried to the lungs, where it can block an artery and cause a potentially fatal
pulmonary embolism (blockage of a blood vessel). (b) Echocardiogram This ultrasonic procedure can display the beating move-
ments of the heart, the heart chambers, valves, and blood flow as it makes it way in and out of the organ.
The inverse ratio of the speeds is called the Mach number (M), named after Ernst
Mach (1838–1916), an Austrian physicist who used it in studying supersonics, and is
given by
vs 1
M = = (14.16)
v sin u
If v equals vs , the plane is flying at the speed of sound, and the Mach number is 1
(that is, vs>v = 1). Therefore, a Mach number less than 1 indicates a subsonic
speed, and a Mach number greater than 1 indicates a supersonic speed. In the lat-
ter case, the Mach number tells the speed of the aircraft in terms of a multiple of
the speed of sound. A Mach number of 2, for instance, indicates a speed twice the
speed of sound. Note that since sin u … 1, no shock wave can exist unless M Ú 1.
DID YOU LEARN?

➥ The Doppler effect is caused by the relative motion of a sound source and observer,
giving rise to a variation in perceived frequency.
➥ Sonic booms are caused by a pressure ridge generated by supersonic aircraft—one
flying faster than the speed of sound.
➥ The Mach number (M) is given by the ratio of the speed of the sound source (vs)
and the speed of sound (v), so, for example, if M = vs>v = 2, then vs = 2v and the
source is traveling twice the speed of sound.
This information is obtained from the intensity of the 2 min with conventional radar. Doppler radar has saved
reflected signal. Such conventional radars can also detect the many lives with this increased warning time. The National
hooked (rotational) “signature” of a tornado, but only after Weather Service has a network of Doppler radars around the
the storm is well developed. United States, and Doppler radar scans are now common on
A major improvement in weather forecasting came about both TV weather forecasts and the Internet.
with the development of a radar system that could measure Doppler radars installed at major airports have another use:
the Doppler frequency shift in addition to the magnitude of to detect wind shears. Several airplane crashes and near-crashes
the echo signal reflected from precipitation (usually rain- have been attributed to downward wind bursts (also known as
drops). The Doppler shift is related to the velocity of the pre- microbursts or downbursts). Such strong downdrafts cause
cipitation blown by the wind. wind shears capable of forcing landing aircraft to crash. Wind
A Doppler-based radar system (Fig. 2a) can penetrate a bursts generally result from high-speed downdrafts in the tur-
storm and monitor its wind speeds. The direction of a storm’s bulence of thunderstorms, but they can also occur in clear air
wind-driven rain gives a wind “field” map of the affected when rain evaporates high above ground. Since Doppler radar
region. Such maps provide strong clues of developing torna- can detect the wind speed and the direction of raindrops in
does, so meteorologists can detect tornadoes much earlier clouds, as well as dust and other objects floating in the air, it can
than was ever before possible (Fig. 2b). With Doppler radar, provide an early warning against dangerous wind shear condi-
forecasters have been able to predict tornadoes as much as tions. Two or three radar sites are needed to detect motions in
20 min before they touch down, compared with just over two or three directions (dimensions), respectively.
F I G U R E 2 Doppler
radar (a) A Doppler radar
installation. (b) Doppler
radar depicts the precipi-
tation inside a thunder-
storm. A hook echo is a
signature of a possible
tornado.
(a) (b)
514 14 SOUND
14.6 Musical Instruments and Sound Characteristics

➥ In what sense are standing waves associated with musical instruments?

➥ What are the sensory effects of sound intensity, frequency, and waveform?
Musical instruments provide good examples of standing waves and boundary

conditions. On some stringed instruments, different notes are produced by using
finger pressure to vary the lengths of the strings (䉳 Fig. 14.14). As was learned in
Section 13.5, the natural frequencies of a stretched string (fixed at each end, as is
the case for the strings on an instrument) are fn = n1v>2L2, from Eq. 13.20. The
speed of the wave in the string is given by v = 1FT>m. Initially adjusting the ten-
sion in a string tunes it to a particular (fundamental) frequency. Then the effective
length of the string is varied by finger location and pressure.
Standing waves can be set up in air columns. You have probably done this in
䉱 F I G U R E 1 4 . 1 4 A shorter blowing across the open top of a soda bottle, producing an audible tone
vibrating string, a higher frequency (䉲 Fig. 14.15a)*. The blowing across the bottle excites the fundamental mode of the
Different notes are produced on
stringed instruments such as gui- column of air in the bottle. The frequency of the tone depends on the length of the
tars, violins, and cellos by placing a
finger on a string to change its effec- * A more complicated phenomenon, called Helmholtz resonance, occurs here, but for simplicity
tive, or vibrating, length. and standing waves, assume the bottle to be a circular cylinder. (See Conceptual Question 20 and
answer to the Follow-up Exercise.)
Antinode Node
Antinode Antinode
␭ ␭ ␭ ␭ ␭ ␭
L = —1 L = 2(—2) L = 3 (—3) L = —1 L = 3 (—3) L = 5 ( —5)
2 2 2 4 4 4
f1 2f1 3f1 f1 3f1 5f1
(a) Open organ pipe (b) Closed organ pipe
䉱 F I G U R E 1 4 . 1 5 Standing waves
(a) When air is blown across the open
top of a bottle, the air flow can cause
an audible tone. (b) Longitudinal
standing waves (illustrated here as
sinusoidal curves) are formed in
vibrating air columns in pipes. An
open pipe has antinodes at both open
ends. A closed pipe has node at the
closed end and an antinode at the
open end. (c) A modern pipe organ.
The pipes can be open or closed. (c)
14.6 MUSICAL INSTRUMENTS AND SOUND CHARACTERISTICS 515
air column. With more liquid in the bottle, the air column is shorter and the fre-
quency higher. There will be an antinode at the open end of the bottle and a node
at the liquid surface or the bottom of an empty bottle.
Standing waves are the basis of wind musical instruments. For example, con-
sider a pipe organ with fixed lengths of pipe, which may be open or closed (Fig.
14.15b). An open pipe is open at both ends, and a closed pipe is closed at one end
and open at the other (the end with the antinode). Analysis similar to that done in
Section 13.5 for a stretched string with the proper boundary conditions shows that
the natural frequencies of the pipes are
(natural frequencies
= na b = nf1 1n = 1, 2, 3, Á 2
v v
fn = for an open pipe –open (14.17)
ln 2L on both ends)
and
(natural frequencies
= ma b = mf1 1m = 1, 3, 5, Á 2
v v
fm = for a closed pipe – (14.18)
lm 4L closed on one end)
where v is the speed of sound in air. Note that the natural frequencies depend on the
length of the pipe. This is an important consideration in a pipe organ (Fig. 14.15c),
particularly in selecting the dominant or fundamental frequency. (The diameter of
the pipe is also a factor, but is not considered in this simple analysis.)
The same physical principles apply to wind and brass instruments. In all of
these, human breath is used to create standing waves in an open tube. Most such
instruments allow the player to vary the effective length of the tube and thus the
pitch produced—either with the help of slides or valves that vary the actual length
of tubing in which the air can resonate, as in most brasses, or by opening and clos-
ing holes in the tube, as in woodwinds (䉲 Fig. 14.16).
Recall from Section 13.5 that a musical note or tone is referenced to the funda-
mental vibrational frequency of an instrument. In musical terms, the first overtone
is the second harmonic, the second overtone is the third harmonic, and so on.
Note that for a closed organ pipe (Eq. 14.18), the even harmonics are missing.
Air Vibrating
in air Holes
All holes covered

L
First five holes covered Higher f

L
First three holes covered Even

L higher f
1
f⬀
L
(a) (b)
䉱 F I G U R E 1 4 . 1 6 Wind instruments (a) Wind instruments, such as clarinets, are essen-

tially open tubes. (b) The effective length of the air column, and hence the pitch of the
sound, is varied by opening and closing holes along the tube. The frequency f is inversely
proportional to the effective length L of the air column.
516 14 SOUND
EXAMPLE 14.10 Pipe Dreams: Fundamental Frequency

A particular open organ pipe has a length of 0.653 m. Taking T H I N K I N G I T T H R O U G H . The fundamental frequency 1n = 12
the speed of sound in air to be 345 m>s, what is the funda- of an open pipe is given directly by Eq. 14.17. Physically, there
mental frequency of this pipe? is a half-wavelength 1l>22 in the length of the pipe, so l = 2L.
SOLUTION.
Given: L = 0.653 m Find: f1 (fundamental frequency)

v = 345 m>s (speed of sound)
With n = 1 in Eq. 14.17,
v 345 m>s
f1 = = = 264 Hz
2L 210.653 m2
This frequency is middle C (C4).
F O L L O W - U P E X E R C I S E . A closed organ pipe has a fundamental frequency of 256 Hz. What would be the frequency of its first
overtone? Is this frequency audible?
Perceived sounds are described by terms whose meanings are similar to those
used to describe the physical properties of sound waves. Physically, a wave is gen-
erally characterized by intensity, frequency, and waveform (harmonics). The corre-
sponding terms used to describe the sensations of the ear are loudness, pitch, and
quality (or timbre). These general correlations are shown in 䉲 Table 14.3. However,
the correspondence is not perfect. The physical properties are objective and can be
measured directly. The sensory effects are subjective and vary from person to per-
son. (Think of temperature as measured by a thermometer and by the sense of
touch.)
Sound intensity and its measurement on the decibel scale were covered in
Section 14.3. Loudness is related to intensity, but the human ear responds differ-
ently to sounds of different frequencies. For example, two tones with the same
intensity (in watts per square meter) but different frequencies might be judged by
the ear to be different in loudness.
Frequency and pitch are often used synonymously, but again there is an
objective–subjective difference: If the same low-frequency tone is sounded at
two intensity levels, most people say that the more intense sound has a lower
pitch, or perceived frequency.
The curves in the graph of intensity level versus frequency shown in 䉴 Fig. 14.17
are called equal-loudness contours (or Fletcher–Munson curves, after the researchers
who generated them). These contours join points representing intensity–frequency
combinations that a person with average hearing judges to be equally loud. The top
curve shows that the decibel level of the threshold of pain (120 dB) does not vary a
great deal over the normal hearing range, regardless of the frequency of the sound.
In contrast, the threshold of hearing, represented by the lowest contour, varies
widely with frequency. For a tone with a frequency of 2000 Hz, the threshold of
hearing is 0 dB, but a 20-Hz tone would have to have an intensity level of over 70 dB
just to be heard (the extrapolated y-intercept of the lowest curve).
TABLE 14.3 General Correlation between Perceptual and Physical

Characteristics of Sound
Sensory Effect Physical Wave Property
Loudness Intensity
Pitch Frequency
Quality (timbre) Waveform (harmonics)
Threshold of pain
120
100
Intensity level (dB)

80
60
40
20
Threshold of
hearing
0
20 50 100 500 1000 5000
10,000
Frequency (Hz)
䉱 F I G U R E 1 4 . 1 7 Equal-loudness contours The curves indicate tones that are judged to

be equally loud, although they have different frequencies and intensity levels. For exam-
ple, on the lowest contour, a 1000-Hz tone at 0 dB sounds as loud as a 50-Hz tone at 40 dB.
Note that the frequency scale is logarithmic to compress the large frequency range.
It is interesting to note the dips (or minima) in the curves. The hearing curves
show a significant dip in the 2000–5000 Hz range, the ear being most sensitive
around 4000 Hz. A tone with a frequency of 4000 Hz can be heard at intensity levels
below 0 dB. The high sensitivity in the 2000–5000 Hz region is very important for
the understanding of speech. (Why?) Another dip in the curves, or region of sensi-
tivity, occurs at about 12 000 Hz.
The minima occur as a result of resonance in a closed cavity in the auditory
canal (similar to a closed pipe). The length of the cavity is such that it has a funda-
mental resonance frequency of about 4000 Hz, resulting in extra sensitivity. As in a
closed cavity, the next natural frequency is the third harmonic (see Eq. 14.18),
which is three times the fundamental frequency, or about 12 000 Hz.
EXAMPLE 14.11 The Human Ear Canal: Standing Waves

Consider the human ear canal to be a cylindrical tube of T H I N K I N G I T T H R O U G H . The auditory ear canal as described is
length 2.54 cm (1.0 in.; see Fig. 1 in Insight 14.2). What would essentially a closed pipe—open at one end (outer ear canal) and
be the lowest sound resonance frequency? Take the air tem- closed at the other (eardrum). The lowest-frequency resonance
perature in the canal to be 37 °C. standing wave that will fit in the pipe is L = l>4 (Fig. 14.15),
and l = 4L. Then, the frequency is given by f1 = v>l1 = v>4L
(Eq. 14.18), where v is the speed of sound in air.
SOLUTION.
Given: L = 2.54 cm = 0.0254 m Find: f1 (lowest resonance frequency)

T = 37 °C, (normal body temperature)
First finding the speed of sound at 37 °C, Compare with the curves in Fig. 14.17, and note the dip in the
v = 1331 + 0.6TC2 m>s = 3331 + 0.613724 m>s = 353 m>s
curves at about this frequency. How about the other dip just
above 10 kHz? Check out the next natural frequency for the
and ear canal, f3.
v 353 m>s
f1 = = = 3.47 * 103 Hz = 3.47 kHz
4L 410.0254 m2
F O L L O W - U P E X E R C I S E . Children have smaller ear canals than adults, on the order of 1.30 cm in length. What is the lowest fun-
damental frequency for a child’s ear canal? Use the same air temperature as in the Example. (Note: With growth the ear canal
lengthens, and it has been experimentally determined that the “adult” ear canal length and lowest fundamental frequency is
reached at about age 7.)
518 14 SOUND
Fundamental
frequency
Harmonics (overtones)
(b)
Complex
waveform
(a)
䉱 F I G U R E 1 4 . 1 8 Waveform and quality (a) The superposition of sounds of different fre-

quencies and amplitudes gives a complex waveform. The harmonics, or overtones, deter-
mine the quality of the sound. (b) The waveform of a violin tone is displayed on an
oscilloscope.
The quality of a tone is the characteristic that enables it to be distinguished

from another tone of basically the same intensity and frequency. Tone quality
depends on the waveform—specifically, the number of harmonics (overtones)
present and their relative intensities (䉱 Fig. 14.18). The tone of a voice depends in
large part on the vocal resonance cavities. One person can sing a tone with the
same basic frequency and intensity as another, but different combinations of over-
tones give the two voices different qualities.
The notes of a musical scale correspond to certain frequencies; as we saw in
Example 14.10, middle C (C4) has a frequency of 264 Hz. When a note is played on
an instrument, its assigned frequency is that of the first harmonic, which is the
fundamental frequency. (The second harmonic is the first overtone, the third har-
monic is the second overtone, and so on.) The fundamental frequency is dominant
over the accompanying overtones that determine the sound quality of the instru-
ment. Recall from Section 13.5 that the overtones that are produced depend on
how an instrument is played. Whether a violin string is plucked or bowed, for
example, can be discerned from the quality of identical notes.
DID YOU LEARN?

➥ In musical instruments, different standing waves produce different notes.
➥ The perception or sensory characteristics of sound are loudness (intensity), pitch
(frequency), and quality (waveform–harmonics).
PULLING IT TOGETHER A “Sound” Thermometer?

As you are standing in a doorway with the outside cold air on mental modes, a beat frequency of 2.00 Hz is produced. The
one side and the inside warm air on the other, you simultane- length of pipe A is 1.00 m and that of pipe B is 2.10 m. See
ously hear sound from two different organ pipes. Pipe A is 䉴 Fig. 14.19. (a) What is the frequency of the sound produced
outside in the cold air, and pipe B is inside in the room tem- by pipe B? (b) What are the two possible beat frequencies of
perature air 120 °C2. Pipe A is closed at one end, while pipe B the sound produced by pipe A? (c) What are the two possible
is open at both ends. When both are excited in their funda- temperatures of the air in the region of pipe A?
T H I N K I N G I T T H R O U G H . The concepts in this example include resonant frequencies Antinode

in open and closed pipes, beats, and the speed of sound as a function of tempera-
ture. (a) Fig 14.19 shows the standing wave in pipe B and gives the wavelength of
the wave. Since the speed of sound is known from the room temperature, the fre-
quency of pipe B can be found. (b) Beats tell only the amount by which pipe A’s fre-
quency differs from pipe B’s frequency, so there are two possible answers.
(c) Fig. 14.19 shows the standing wave in pipe A and gives its wavelength. From
that and the two possible frequencies in part (b), two possible speeds for sound can
be found. Those speeds can then be translated into temperatures because the speed
of sound depends on the temperature.
SOLUTION.
Given: pipe A (closed), L = 1.00 m Find: (a) fB (frequency of pipe B)

pipe B (open), L = 2.10 m (b) fA (two possible frequen- Node
2.10 m
beat frequency, fb = 2.00 Hz cies of pipe A)
pipe B is in room temperature (c) TC (two possible tempera-
air 120 °C2 tures for pipe A)
(a) Fig. 14.19 shows pipe B’s standing wave to be a half-wavelength. This is because it
is in its fundamental (lowest frequency, longest wavelength) mode. Therefore,
1.00 m
lB = 2LB = 4.20 m. The speed of sound in B’s locale is:
vB = 1331 + 0.6TC2 m>s = 331 + 0.6120 °C2 = 343 m>s
From this, the frequency of B’s sound can be found:
vB 343 m>s
fB = = = 81.7 Hz
lB 4.20 m Antinode
Pipe A Pipe B
(b) The beat frequency tells only that pipe A’s frequency differs from that of pipe B by
2.00 Hz; therefore, there are two choices: 䉱 F I G U R E 1 4 . 1 9 Closed and open
pipes; nodes and antinodes
fA = 81.7 Hz ⫾ 2.00 Hz The fundamental modes in pipe A
= 83.7 Hz or 79.7 Hz and pipe B.
(c) Fig 14.19 shows pipe A’s standing wave to be a quarter-wavelength. This is
because it is in its fundamental (lowest frequency, longest wavelength) mode. There-
fore, lA = 4LA = 4.00 m. Since there are two possible frequencies for pipe A, the
speed of sound in A’s locale also has two possibilities. These are,
vA = fA lA = 183.7 Hz214.00 m2 = 334.8 m>s
and
vA = fA lA = 179.7 Hz214.00 m2 = 318.8 m>s
The dependence of the speed of sound on air temperature is given by

v = 1331 + 0.6TC2 m>s. Solving for TC for the two possible air temperatures for pipe A
gives:
vA - 331 m>s 334.8 m>s - 331 m>s

TC = = = 6.3 °C
0.6 0.6
and
vA - 331 m>s 318.8 m>s - 331 m>s

TC = = = - 20.3 °C
0.6 0.6
Without any further information, that is the best that can be done. The latter temper-
ature is a bit cold for playing an organ outside, so the first temperature is probably
the correct one. (You should be able to show that 6.3 °C = 43.3 °F and
-20.3 °C = - 4.54 °F
520 14 SOUND
Sound intensity level (decibels)

■ The sound frequency spectrum is divided into infrasonic
1f 6 20 Hz2, audible 120 Hz 6 f 6 20 kHz2, and ultra- 180 Rocket launch
sonic 1f 7 20 kHz2 frequency regions.
180 dB
(Upper limit) 1 GHz

140 Jet plane takeoff
120 Pneumatic drill

Ultrasonic
110 Rock band with amplifiers 120 dB
100 Machine shop
20 kHz
90 Subway train
Frequency
80 Average factory
70 City traffic
Audible 60 Normal conversation
50 Average home
60 dB
40 Quiet library
20 Hz
Infrasonic
20 Soft whisper
0 Threshold of hearing
■ The speed of sound in a medium depends on the elasticity of
the medium and its density. In general, vsolids 7 vliquids 7 ■ Sound wave interference of two point sources depends on
vgases. phase difference as related to path length difference. Sound
Speed of sound in dry air: waves that arrive at a point in phase reinforce each other
(constructive interference); sound waves that arrive at a point
v = 1331 + 0.6TC2 m>s (14.1) out of phase cancel each other (destructive interference).
■ The intensity of a point source is inversely proportional to A*

the square of the distance from the source.
␭AC LAC
B* ␭BC
R
2R
A
3R
I 4A
9A LBC C
I/4 I ⬀ 12 In phase
Point source R
I/9
Phase difference (where ¢L is the path length difference):
1¢L2
2p
¢u = (14.5)
l
Intensity of a point source: Condition for constructive interference:

P I2 R1 2 ¢L = nl 1n = 0, 1, 2, 3, Á 2 (14.6)
I = 2
and = ¢ ≤ (14.2, 14.3)
4pR I1 R2 Condition for destructive interference:
■ The sound intensity level is a logarithmic function of the
¢L = ma b 1m = 1, 3, 5, Á 2
l
sound intensity and is expressed in decibels (dB). (14.7)
2
Intensity level (in decibels, dB):
Beat frequency:
I 2
b = 10 log where Io = 10 -12
W>m (14.4) fb = ƒ f1 - f2 ƒ (14.8)
Io
■ The Doppler effect depends on the velocities of the sound Natural frequencies of an open organ pipe—open on both
source and observer relative to still air. When the relative ends:
motion of the source and observer is toward each other, the
b = nf1 1n = 1, 2, 3, Á 2
observed pitch increases; when the relative motion of the v
fn = na (14.17)
source and observer is away from each other, the observed 2L
pitch decreases.
Antinode
Doppler effect:
Moving source, stationary observer
v 1
fo = ¢ ≤f = § ¥f (14.11)
v ⫾ vs s vs s L
1⫾
v
- for source moving toward stationary observer
b
+ for source moving away from stationary obsever
where vs = speed of source
Antinode
and v = speed of sound
␭ ␭ ␭
L = —1 L = 2 (—2) L = 3(—3)
2 2 2
Moving observer, stationary source f1 2f1 3f1
v ⫾ vo vo Open organ pipe

fo = ¢ ≤ fs = ¢ 1 ⫾ ≤ fs (14.14)
v v
+ for observer moving toward stationary source Natural frequencies of a closed organ pipe—closed on one
b end:
- for observer moving away from stationary source
where vo = speed of observer
b = mf1 1m = 1, 3, 5, Á 2
v
fm = ma (14.18)
and v = speed of sound 4L
Moving observer and moving source

Node
v ⫾ vo
fo = ¢ ≤f (14.14a)
v ⫿ vs s
Upper signs if moving toward each other
Lower signs if moving away from each other
Angle for conical shock wave:

vt v 1
sin u = = = (14.15)
vs t vs M
Antinode
Mach number:
␭ ␭ ␭
vs L = —1 L = 3 (—3) L = 5 ( —5)
1 4 4 4
M = = (14.16) f1 3f1 5f1
v sin u
Closed organ pipe
vs
vs > v
Supersonic
Tail Nose
shock shock
wave wave
Atmospheric
pressure
522 14 SOUND

MULTIPLE CHOICE
14.1 SOUND WAVES 9. The intensity of a sound wave is directly proportional to

AND the (a) amplitude, (b) frequency, (c) square of the ampli-
14.2 THE SPEED OF SOUND tude, (d) square of the frequency.
1. A sound wave with a frequency of 15 Hz is in what 10. A sound with an intensity level of 30 dB is how many
region of the sound spectrum: (a) audible, (b) infrasonic, times more intense than the threshold of hearing: (a)10,
(c) ultrasonic, or (d) supersonic? (b) 100, (c) 1000, or (d) 3000?
2. A sound wave in air (a) is longitudinal, (b) is transverse,

(c) has longitudinal and transverse components, (d) trav-
14.4 SOUND PHENOMENA
els faster than a sound wave through a liquid.
AND
3. The speed of sound is generally greatest in (a) solids, 14.5 THE DOPPLER EFFECT
(b) liquids, (c) gases, (d) a vacuum.
11. Constructive and destructive interference of sound
4. The speed of sound in air (a) is about 1>3 km>s, (b) is waves depends on (a) the speed of sound, (b) diffraction,
about 1>5 mi>s, (c) depends on temperature, (d) all of the (c) phase difference, (d) all of the preceding.
preceding. 12. Beats are the direct result of (a) interference, (b) refrac-
5. The speed of sound in water is about 4.5 times that in air. tion, (c) diffraction, (d) the Doppler effect.
A single-frequency sound source in air is fo . On penetrat- 13. Police radar makes use of (a) refraction, (b) the Doppler
ing water, the frequency of the sound will be (a) 4fo , effect, (c) interference, (d) sonic boom.
(b) fo>4, (c) fo .
14.6 MUSICAL INSTRUMENTS AND

14.3 SOUND INTENSITY AND SOUND SOUND CHARACTERISTICS
INTENSITY LEVEL
14. Given open and closed pipes of the same length, which
6. If the air temperature increases, would the sound inten- would have the lowest natural frequency: (a) the open
sity from a constant-output point source (a) increase, pipe, (b) the closed pipe, or (c) they both would have the
(b) decrease, or (c) remain unchanged? same low frequency?
7. The decibel scale is referenced to a standard intensity of 15. The human ear can hear tones best at (a) 1000 Hz,
(a) 1.0 W>m2, (b) 10-12 W>m2, (c) normal conversation, (b) 4000 Hz, (c) 6000 Hz, (d) all frequencies.
(d) the threshold of pain. 16. Equal loudness curves vary with sound (a) quality,
8. If the intensity level of a sound at 20 dB is increased to (b) harmonics, (c) waveform, (d) pitch.
40 dB, the intensity would increase by a factor of (a) 10, 17. The quality of sound depends on its (a) waveform,
(b) 20, (c) 40, (d) 100. (b) frequency, (c) speed, (d) intensity.
14.1 SOUND WAVES 14.3 SOUND INTENSITY AND SOUND

AND INTENSITY LEVEL
14.2 THE SPEED OF SOUND 6. What is the difference between sound intensity and
1. Suggest a possible explanation of why some flying sound intensity level?
insects produce buzzing sounds and some do not. 7. Where is the intensity greater and by what factor: (1) at a
2. Explain why sound travels faster in warmer air than in point a distance R from a power source P, or (2) at a
colder air. point a distance 2R from a power source of 2P? Explain.
3. Why does the speed of sound vary with temperature? 8. The Richter scale, used to measure the intensity level of
4. The speed of sound in air depends on temperature. earthquakes, is a logarithmic scale, as is the decibel scale.
What effect, if any, should humidity have? Why are such logarithmic scales used?
5. What is the difference between ultrasonic and 9. Can there be negative decibel levels, such as - 10 dB? If
supersonic? so, what would these mean?
EXERCISES 523
14.4 SOUND PHENOMENA 14.6 MUSICAL INSTRUMENTS AND

AND SOUND CHARACTERISTICS
14.5 THE DOPPLER EFFECT 16. (a) Why does it seem particularly quiet after a snowfall?
10. Do interference beats have anything to do with the (b) Why do empty rooms sound hollow? (c) Why do
“beat” of music? Explain. people’s voices sound fuller or richer when they sing in
the shower?
11. (a) Is there a Doppler effect if a sound source and an
17. The frets on a guitar finger board are spaced closer
observer are moving with the same velocity? (b) What
together the farther they are from the neck. Why is this?
would be the effect if a moving source accelerated
What would be the result if they were evenly spaced?
toward a stationary observer?
18. Is it possible for an open organ pipe and a closed organ
12. As a person walks between a pair of loudspeakers that pipe, each of the same length, to produce notes of the
produce tones of the same amplitude and frequency, he same frequency? Justify your answer.
hears a varying sound intensity. Explain. 19. How would an increase in air temperature affect the
13. How can Doppler radar used in weather forecasting frequencies of an organ pipe?
measure both the location and internal motions of a 20. When you blow across the mouth of an empty soda bot-
storm? tle, a particular tone is produced. If the bottle is filled to one-
third its height with water, how would the tone be affected?
14. A stationary sound source and a stationary observer are
Explain. How about if it were filled to one-half? (Consider
a fixed distance apart. However, the air between them is
only standing waves in the bottle.)*
moving toward the observer with a constant speed. How
is the frequency received by the observer affected? 21. A crystal wine glass is partially filled with water. A per-
Explain. son wets her finger and rubs it around the rim of the
glass, which produces sound. Why is this?
15. Can a jet pilot flying faster than the speed of sound hear 22. Why are there no even harmonics in a closed organ pipe?
sound? Explain.
*A more complicated phenomenon, called Helmholtz resonance,
occurs here, but for simplicity and standing waves, assume the bottle
to be a circular cylinder. (See the answer to the Follow-up Exercise.)
EXERCISES
14.1 SOUND WAVES 7. IE ● ● A tuning fork vibrates at a frequency of 256 Hz.

AND (a) When the air temperature increases, the wavelength
14.2 THE SPEED OF SOUND of the sound from the tuning fork (1) increases,
(2) remains the same, (3) decreases. Why? (b) If the tem-
1. ● What is the speed of sound in air at (a) 10 °C and perature rises from 0 °C to 20 °C, what is the change in
(b) 20 °C? the wavelength?
2. ● The speed of sound in air on a summer day is 350 m>s.
What is the air temperature? 8. ●● Particles approximately 3.0 * 10-2 cm in diameter

are to be scrubbed loose from machine parts in an aque-
3. ● Sonar is used to map the ocean floor. If an ultrasonic
ous ultrasonic cleaning bath. Above what frequency
signal is received 2.0 s after it is emitted, how deep is the
should the bath be operated to produce wavelengths of
ocean floor at that location?
this size and smaller?
4. ● What temperature change from 0 °C would increase
the speed of sound by 1.0 %? 9. ●● Medical ultrasound uses a frequency of around

5. ● The wave speed in a liquid is given by v = 1B>r,
20 MHz to diagnose human conditions and ailments.
where B is the bulk modulus of the liquid and r is its (a) If the speed of sound in tissue is 1500 m>s, what is the
density. Show that this equation is dimensionally smallest detectable object? (b) If the penetration depth is
correct. What about v = 1Y>r for a solid? (Y is Young’s about 200 wavelengths, how deep can this instrument
modulus.) penetrate?
6. ● ● A 0.75-m-long metal rod is dropped vertically onto a 10. ●● Brass is an alloy of copper and zinc. Does the addi-
marble floor. When the rod strikes the floor, it is deter- tion of zinc to copper cause an increase or decrease in the
mined electronically that the impact produces a 4-kHz speed of sound in brass rods compared to copper?
tone. What is the speed of sound in the rod? Explain.
524 14 SOUND
11. ●● The speed of sound in steel is about 4.50 km>s. 19. ● ● ● Sound propagating through air at 30 °C passes
A steel rail is struck with a hammer, and an observer through a vertical cold front into air that is 4.0 °C. If the
0.400 km away has one ear to the rail. (a) How much sound has a frequency of 2500 Hz, by what percentage
time will elapse from the time the sound is heard does its wavelength change in crossing the boundary?
through the rail until the time it is heard through the air?
Assume that the air temperature is 20 °C and that no 14.3 SOUND INTENSITY AND SOUND
wind is blowing. (b) How much time would elapse if the
INTENSITY LEVEL
wind were blowing toward the observer at 36.0 km>h
from where the rail was struck? 20. ● Calculate the intensity generated by a 1.0-W point
source of sound at a location (a) 3.0 m and (b) 6.0 m from it.
12. ●● A person holds a rifle horizontally and fires at a tar-
get. The bullet has a muzzle speed of 200 m>s, and the 21. IE ● (a) If the distance from a point sound source triples,
person hears the bullet strike the target 1.00 s after firing the sound intensity will be (1) 3, (2) 1>3, (3) 9, (4) 1>9
it. The air temperature is 72 °F. What is the distance to times the original value. Why? (b) By how much must
the target? the distance from a point source be increased to reduce
the sound intensity by half?
13. ●● A freshwater dolphin sends an ultrasonic sound to
locate a prey. If the echo off the prey is received by the 22. ● Assuming that the diameter of your eardrum is 1 cm
dolphin 0.12 s after being sent, how far is the prey from (see Exercise 16), what is the sound power received by
the dolphin? the eardrum at the threshold of (a) hearing and (b) pain?
14. ●● A submarine on the ocean surface receives a sonar 23. ● A middle C note (262 Hz) is sounded on a piano to
echo indicating an underwater object. The echo comes help tune a violin string. When the string is sounded,
back at an angle of 20° above the horizontal and the echo nine beats are heard in 3.0 s. (a) How much is the violin
took 2.32 s to get back to the submarine. What is the string off tune? (b) Should the string be tightened or
object’s depth? loosened to sound middle C?
24. ● Calculate the intensity level for (a) the threshold of
15. ●● The speed of sound in human tissue is on the order
hearing and (b) the threshold of pain.
of 1500 m>s. A 3.50-MHz probe is used for an ultrasonic
procedure. (a) If the effective physical depth of the ultra- 25. ● Find the intensity levels in decibels for sounds with
sound is 250 wavelengths, what is the physical depth in intensities of (a) 10-2 W>m2, (b) 10-6 W>m2, and
meters? (b) What is the time lapse for the ultrasound to (c) 10-15 W>m2.
make a round trip if reflected from an object at the effec- 26. ●● At Cape Canaveral, on blastoff a rocket produces an
tive depth? (c) The smallest detail capable of being intensity level of 160 dB as measured 10 m from the
detected is on the order of one wavelength of the ultra- rocket. What would be the intensity level at 100 m away?
sound. What would this be? (Assume no energy is lost due to reflections, etc.)
16. ●● The size of your eardrum (the tympanum; see Fig. 1 27. IE ● ● (a) If the power of a sound source doubles, the
in Insight 14.2, The Physiology and Physics of the Ear intensity level at a certain distance from the source
and Hearing) partially determines the upper frequency (1) increases, (2) exactly doubles, or (3) decreases. Why?
limit of your audible region, usually between 16 000 Hz (b) What are the intensity levels at a distance of 10 m
and 20 000 Hz. If the wavelength is on the order of from a 5.0-W and a 10-W source, respectively?
twice the diameter of the eardrum and the air tempera- 28. ●● The intensity levels of two people holding a conver-
ture is 20 °C, how wide is your eardrum? Is your sation are 60 dB and 70 dB, respectively. What is the
answer reasonable? intensity of the combined sounds?
17. IE ● ● ● On hiking up a mountain that has several over- 29. ●● A point source emits radiation in all directions at a
hanging cliffs, a climber drops a stone at the first cliff to rate of 7.5 kW. What is the intensity of the radiation
determine its height by measuring the time it takes to 5.0 m from the source?
hear the stone hit the ground. (a) At a second cliff that is
30. ●● Two sound sources have intensities of 10-9 W>m2
twice the height of the first, the measured time of the
and 10-6 W>m2, respectively. Which source is more
sound from the dropped stone is (1) less than double,
intense and by how many times more?
(2) double, or (3) more than double that of the first.
Why? (b) If the measured time is 4.8 s for the stone drop- 31. ●● Average speech has an intensity level of about 60 dB.
ping from the first cliff, and the air temperature is 20 °C, Assuming that 20 people all speak at 60 dB, what is the
how high is the cliff? (c) If the height of a third cliff is total sound intensity?
three times that of the first one, what would be the mea- 32. ●● A rock band (with loud speakers) has an average
sured time for a stone dropped from that cliff to reach intensity level of 110 dB at a distance of 15 m from the
the ground? band. Assuming the sound is radiated equally over a
18. ● ● ● A bat moving at 15.0 m>s emits a high-frequency
hemisphere in front of the band, what is the total power
sound as it approaches a wall that is 25.0 m away. output?
Assuming that the bat continues straight toward the 33. ●● A person has a hearing loss of 30 dB for a particular fre-
wall, how far away is it when it receives the echo? quency. What is the sound intensity that is heard at this
(Assume the air temperature in the cave to be 0 ºC.) frequency that has an intensity of the threshold of pain?
EXERCISES 525
TABLE 14.4 Takeoff and Landing Noise Levels for Some Common
Commercial Jet Aircraft* (See Exercise 34)
Aircraft Takeoff Noise (dB) Landing Noise (dB)
737 85.7–97.7 99.8–105.3

747 89.5–110.0 103.8–107.8
DC-10 98.4–103.0 103.8–106.6
L-1011 95.9–99.3 101.4–102.8
*Noise level readings are taken from 198 m (650 ft). The range depends on the aircraft model and
the type of engine used.
34. ●● Noise levels for some common aircraft are given in

䉱 Table 14.4. What are the lowest and highest intensities
for (a) takeoff and (b) landing for these planes? A B
35. IE ● ● If the distance to a sound source is halved, (a) will 150 m 200 m
the sound intensity level change by a factor of (1) 2,
(2) 1>2, (3) 4, (4) 1>4, or (5) none of the preceding? Why?
(b) What is the change in the sound intensity level?
36. ●● A compact speaker puts out 100 W of sound power.
(a) Neglecting losses to the air, at what distance would
the sound intensity be at the pain threshold? (b) Neglect- 300 m
ing losses to the air, at what distance would the sound
intensity be that of normal speech? Does your answer
seem reasonable? Explain.
37. ●● What is the intensity level of a 23-dB sound after
being amplified (a) ten thousand times, (b) a million
times, (c) a billion times? C D
38. ●● In a neighborhood challenge to see who can climb a

tree the fastest, you are ready to climb. Your friends 䉱 F I G U R E 1 4 . 2 0 A big bang See Exercise 42.
have surrounded you in a circle as a cheering section;
each individual alone would cause a sound intensity 43. ●● An office in an e-commerce company has fifty com-
level of 80 dB at your location. If the actual sound level puters, which generate a sound intensity level of 40 dB
at your location is 87 dB, how many people are rooting (from the keyboards). The office manager tries to cut the
for you? noise to half as loud by removing twenty-five comput-
ers. Does he achieve his goal? What is the intensity level
39. IE ● ● A dog’s bark has a sound intensity level of 40 dB. generated by twenty-five computers?
(a) If two of the same dogs were barking, the intensity 44. ● ● ● A 1000-Hz tone from a loudspeaker has an intensity
level is (1) less than 40 dB, (2) between 40 dB and 80 dB, level of 100 dB at a distance of 2.5 m. If the speaker is
(3) 80 dB. (b) What would be the intensity level? assumed to be a point source, how far from the speaker
40. ●● At a rock concert, the average sound intensity level will the sound have intensity levels (a) of 60 dB and
for a person in a front-row seat is 110 dB for a single (b) barely high enough to be heard?
band. If all the bands scheduled to play produce sound 45. ● ● ● During practice in a huddle, a quarterback shouts
of that same intensity, how many of them would have to the play in anticipation of crowd noise during the actual
play simultaneously for the sound level to be at or above game. To a receiver 0.750 m away from the quarterback
the threshold of pain? in the huddle, it seems as loud as the noise from a
screaming child. When they get into practice formation,
41. ●● At a distance of 12.0 m from a point source, the inten- the quarterback yells at twice the output power, yet the
sity level is measured to be 70 dB. At what distance from instructions seem only about as loud as normal conver-
the source will the intensity level be 40 dB? sation. Use typical values in Table 14.2 to estimate how
42. ●● At a Fourth of July celebration, a firecracker explodes far from the quarterback the receiver is in the formation.
(䉴 Fig. 14.20). Considering the firecracker to be a point 46. ● ● ● A bee produces a buzzing sound that is barely audi-
source, what are the intensities heard by observers at ble to a person 3.0 m away. How many bees would have
points B, C, and D, relative to that heard by the observer to be buzzing at that distance to produce a sound with
at A? an intensity level of 50 dB?
526 14 SOUND
14.4 SOUND PHENOMENA 58. ●● The half-angle of the conical shock wave formed by a
AND supersonic jet is 30°. What are (a) the Mach number of
14.5 THE DOPPLER EFFECT the aircraft and (b) the actual speed of the aircraft if the
air temperature is -20 °C?
47. ● A violinist and a pianist simultaneously sound notes
with frequencies of 436 Hz and 440 Hz, respectively. 59. ●● An observer is traveling between two identical
What beat frequency will the musicians hear? sources of sound (frequency 100 Hz). His speed is
10.0 m>s as he approaches one and recedes from the
48. IE ● A violinist tuning her instrument to a piano note of other. (a) What frequency tone does he hear from each
264 Hz detects three beats per second. (a) The frequency source? (b) How many beats per second does he hear?
of the violin could be (1) less than 264 Hz, (2) equal to Assume normal room temperature.
264 Hz, (3) greater than 264 Hz, (4) both (1) and (3). Why?
(b) What are the possible frequencies of the violin tone? 60. ● ● ● A bystander hears a siren vary in frequency from
476 Hz to 404 Hz as a fire truck approaches, passes by,

49. ● What is the frequency heard by a person driving and moves away on a straight street (䉲 Fig. 14.21). What
60 km>h directly toward a factory whistle 1f = 800 Hz2 is the speed of the truck? (Take the speed of sound in air
if the air temperature is 0 °C? to be 343 m>s.)
50. IE ● On a day with a temperature of 20 °C and no wind
blowing, the frequency heard by a moving person from a
500-Hz stationary siren is 520 Hz. (a) The person is 476 Hz 404 Hz
(1) moving toward, (2) moving away from, or (3) station-
ary relative to the siren. Explain. (b) What is the person’s
speed?
51. ●● While standing near a railroad crossing, you hear a
train horn. The frequency emitted by the horn is 400 Hz. 䉱 F I G U R E 1 4 . 2 1 The siren’s wail See Exercise 60.
If the train is traveling at 90.0 km>h and the air tempera-
ture is 25 °C, what is the frequency you hear (a) when
61. ● ● ● Bats emit sounds of frequencies around 35.0 kHz
the train is approaching and (b) after it has passed?
and use echolocation to find their prey. If a bat is moving
52. ●● Two identical strings on different cellos are tuned to with a speed of 12.0 m>s toward a hovering, stationary
the 440-Hz A note. The peg holding one of the strings insect, (a) what is the frequency received by the insect if
slips, so its tension is decreased by 1.5%. What is the beat the air temperature is 20 °C? (b) What frequency of the
frequency heard when the strings are then played reflected sound is heard by the bat? (c) If the insect were
together? initially moving directly away from the bat, would this
affect the frequencies? Explain.
53. ●● How fast, in kilometers per hour, must a sound
source be moving toward you to make the observed fre- 62. ● ● ● A supersonic jet flies directly overhead relative to
quency 5.0% greater than the true frequency? (Assume an observer, at an altitude of 2.0 km (䉲 Fig. 14.22). When
that the speed of sound is 340 m>s.) the observer hears the first sonic boom, the plane has
flown a horizontal distance of 2.5 km at a constant
54. IE ● ● You are driving east at 25.0 m>s as you notice an
speed. (a) What is the angle of the shock wave cone?
ambulance traveling west toward you at 35.0 m>s. The
(b) At what Mach number is the plane flying? (Assume
sound you detect from the sirens has a frequency of
that the speed of sound is at an average constant temper-
300 Hz. (a) Is the true frequency of the sirens (1) greater
ature of 15 °C.)
than 300 Hz, (2) less than 300 Hz, or (3) exactly 300 Hz?
(b) Determine the true frequency of the sirens. Assume
normal room temperature.
55. ●● The frequency of an ambulance siren is 700 Hz. What v
are the frequencies heard by a stationary pedestrian as
the ambulance approaches and moves away from her at u
a speed of 90.0 km>h? (Assume that the air temperature
is 20 °C.)
56. ●● A jet flies at a speed of Mach 2.0. What is the half- 2.0 km
angle of the conical shock wave formed by the aircraft?
Can you tell the speed of the shock wave?
57. IE ● ● A fighter jet flies at a speed of Mach 1.5. (a) If the
jet were to fly faster than Mach 1.5, the half-angle of the
conical shock wave would (1) increase, (2) remain the 䉱 F I G U R E 1 4 . 2 2 Faster than a
same, (3) decrease. Why? (b) What is the half-angle of speeding bullet See Exercise 62.
the conical shock wave formed by the jet plane at
Mach 1.5?
14.6 MUSICAL INSTRUMENTS AND 70. ●● An open organ pipe 0.750 m long has its first over-
SOUND CHARACTERISTICS tone at a frequency of 441 Hz. What is the temperature of
the air in the pipe?
63. ● The first three natural frequencies of an organ pipe are
126 Hz, 378 Hz, and 630 Hz. (a) Is the pipe an open or a 71. IE ● ● When all of its holes are closed, a flute is essen-
closed pipe? (b) Taking the speed of sound in air to be tially a tube that is open at both ends, with the length
340 m>s, find the length of the pipe. measured from the mouthpiece to the far end (as in
64. ● A closed organ pipe has a fundamental frequency of
Fig. 14.16b). If a hole is open, then the length of the tube
528 Hz (a C note) at 20 °C. What is the fundamental fre- is effectively measured from the mouthpiece to the hole.
quency of the pipe when the temperature is 0 °C? (a) Is the position at the mouthpiece (1) a node, (2) an
antinode, or (c) neither a node nor an antinode? Why?
65. ● The human ear canal is about 2.5 cm long. It is open at
(b) If the lowest fundamental frequency on a flute is
one end and closed at the other. (See Fig. 1 in
262 Hz, what is the minimum length of the flute at 20 °C?
Insight 14.2) (a) What is the fundamental frequency of
(c) If a note of frequency 440 Hz is to be played, which
the ear canal at 20 °C? (b) To what frequency is the ear
hole should be open? Express your answer as a distance
most sensitive? (c) If a person’s ear canal is longer than
from the hole to the mouthpiece.
2.5 cm, is the fundamental frequency higher or lower
than that in part (a)? Explain. 72. ● ● ● An organ pipe that is closed at one end is filled with
66. ● ● An organ pipe that is closed at one end has a length helium. The pipe has a fundamental frequency of 660 Hz
of 0.80 m. At 20 °C, what is the distance between a node in air at 20 °C. What is the pipe’s fundamental frequency
and an adjacent antinode for (a) the second harmonic with the helium in it?
and (b) the third harmonic?
73. ● ● ● An open organ pipe, in its fundamental mode, has a
67. ● ● An open organ pipe and an organ pipe that is closed
length of 50.0 cm. A second pipe, closed at one end, is
at one end both have lengths of 0.52 m at 20 °C. What is also in its fundamental mode. A beat frequency of
the fundamental frequency of each pipe? 2.00 Hz is heard. Determine the possible lengths of the
68. ● ● An open organ pipe is 0.50 m long. If the speed of closed pipe. Assume normal room temperature.
sound is 340 m>s, what are the pipe’s fundamental fre-
74. ● ● ● Bats typically give off an ultrahigh-frequency
quency and the frequencies of the first two overtones?
sound at about 50 000 Hz. If a bat is approaching a sta-
69. ● ● An organ pipe that is closed at one end is 1.10 m
tionary object at 18.0 m>s, what will be the reflected
long. It is oriented vertically and filled with carbon diox-
frequency it detects? Assume the air in the cave is at
ide gas (which is denser than air and thus will stay in the
5 °C. [Hint: You will need to apply the Doppler equa-
pipe). A tuning fork with a frequency of 60.0 Hz can be
tions twice. Why?]
used to set up a standing wave in the fundamental
mode. What is the speed of sound in carbon dioxide?
75. At a rock concert there are two main speakers, each 77. IE An open organ pipe with a length of 50.0 cm is oscillat-
putting out 500 W of sound power. You are 5.00 m from ing in its second-overtone or third-harmonic mode.
one and 10.0 m from the other. (a) What are the sound Assume the air to be at room temperature and the pipe to
intensities at your location due to each speaker and what be at rest in still air. A person moves toward this pipe at
is the total sound intensity? (b) What are the sound 2.00 m>s and, at the same time, away from a highly reflec-
intensity levels at your location due to each speaker and tive wall. (a) Will the observer hear beats: (1) yes, (2) no, or
what is the total sound intensity level? (c) About how (3) can’t tell from the data given? (b) Calculate the fre-
long can you sit there without suffering permanent hear- quency of sound emitted. (c) Calculate the beat frequency
ing damage? the observer would hear. [Hint: There are two frequencies,
one directly from the pipe and one from the wall.]
76. You hear sound from two organ pipes that are equidis-
tant from you. Pipe A is open at one end and closed at 78. IE A whale is swimming at a steady speed either
the other, while pipe B is open at both ends. When both directly at or directly away from an underwater cliff
are oscillating in their first-overtone mode, you hear a (you don’t know which). When the whale is 300 m from
beat frequency of 5.0 Hz. Assume normal room tempera- the cliff, it emits a sound and it hears the echo 0.399 s
ture. (a) If the length of pipe A is 1.00 m, calculate the later. (a) Which way is the whale traveling: (i) toward the
possible lengths of pipe B. (b) Assuming your shortest cliff or (ii) away from the cliff? Explain your reasoning.
length for pipe B, what would the beat frequency be (b) How fast is the whale traveling? (c) If the emitted
(assuming both are still in their first-overtone modes) on sound has a frequency of 12.1 kHz, by how much has the
a hot desert summer day with a temperature of 40 °C? frequency changed by the time the whale hears the echo?
528 14 SOUND
79. IE A pair of speakers are separated by 4.00 m. Speaker A 80. An unstretchable steel string is used to replace a broken
puts out a constant volume of sound at a total power of violin string. A length of 5.00 m of this string has a mass
36.0 W. Speaker B operates at 100 W. You are located at of 25.0 g. When in place, the new string will be 30.0 cm
3.00 m directly in front of speaker A with the line con- long and oscillate at 256 Hz in its fundamental mode.
necting you to A being perpendicular to the line joining (a) After it is in place, what tension must the new string
the speakers. be placed under? (b) Assuming normal room tempera-
Neglecting sound energy absorption by the air, (a) ture, what is the wavelength of the sound emitted by the
how do the sound intensities at your location compare: (i) new string in its fundamental mode? (c) If you wanted to
IB 7 IA, (ii) IB 6 IA, (iii) IB = IA, (iv) you can’t tell from decrease this sound wavelength by 5.00%, what would
the data given? (Hint: Draw a sketch of the arrangement. you do to the tension: increase or decrease it? Explain.
You do NOT need to calculate each intensity; use ratio (d) Determine the required tension in part (c).
reasoning.) (b) Compute each speaker’s intensity at your
location. Do your results confirm your answer to part (a)?
(c) Compute the intensity level of each speaker and the
total intensity level at your location.
Electric Charge, Forces,
15 and Fields
15.1 Electric charge (530)
■ types of charge
■ conservation of charge
■ charge-force law
15.2 Electrostatic charging (532)

■ conduction
■ friction
■ induction
■ polarization
15.3 Electric force (536)

■ Coulomb’s law
■ force and charged objects
PHYSICS FACTS
■
15.4 Electric field (540)
direction and magnitude
■ superposition principle
✦ Charles Augustin de Coulomb
(1736–1806) was a French scientist
and the discoverer of the force law
between charged objects. He also
F ew natural processes deliver
such an enormous amount of
energy in a fraction of a second as a
made contributions to the cleanup
■ dipoles and parallel plates
of the Parisian water supply, Earth lightning bolt. Yet most people
magnetism, and soils engineering.
have never experienced its power
✦ The Taser stun gun works by gen-
15.5 Conductors and electric erating electric charge separation, at close range; luckily, only a few
fields (548) thus applying an electric field to
the body, disrupting normal elec- hundred people per year are struck
■ interior fields
trical signals and causing tempo-
■ surface fields by lightning in the United States.
rary incapacity.
■ surface charge density
✦ The electric eel acts electrically in a It might surprise you to realize
similar way to that of a Taser. It
uses this for locating prey and
that you have almost certainly had
*15.6 Gauss’s law for electric
stunning them before eating. a similar experience, at least in a
✦ Home air purifiers employ electric-
fields: a qualitative physics context. Have you ever
ity to reduce dust, bacteria, and
approach (550)
other particulates in the air. Electric
force removes electrons from the
walked across a carpeted room and
pollutants, making them positively received a shock when reaching for
charged. Then they are attracted
to negative plates until manually a metallic doorknob? Although the
removed.
scale is dramatically different, the
physical process involved (static
530 15 ELECTRIC CHARGE, FORCES, AND FIELDS
electricity discharge) is much the same as being struck by lightning—

mini-lightning, in this case.
Electricity sometimes gives rise to dramatic effects such as sparking electrical
outlets or lightning strikes. We know that electricity can sometimes be dangerous,
but also that electricity can be “domesticated.” In the home or office, its usefulness
is taken for granted. Indeed, our dependence on electric energy becomes evident
only when the power goes off unexpectedly, providing a dramatic reminder of the
role that it plays in our daily lives. Yet less than a century ago there were no power
lines crossing the country, no electric lights or appliances—none of the electrical
applications that are all around us today.
Physicists now know that the electric force is related to the magnetic force
(see Chapter 20). Together they are called the “electromagnetic force,” which is
one of the four fundamental forces in nature. (The other three are gravity
[Section 7.5] and two types of short-range nuclear forces discussed in Chapters 29
and 30.) Our study begins with the electric force and its properties.
15.1 Electric Charge

Proton (+)
➥ Must a pair of isolated and oppositely charged point charges always attract one
another?
➥ If two isolated charged point charges repel one another, does that mean they are
both negatively charged?
➥ What will be the sign of net charge on a neutral atom if two of its electrons are
removed?
Electron (–) What is electricity? One simple answer is that it is a term describing phenomena
associated with the electricity in our homes. But fundamentally it involves the
(a) Hydrogen atom study of the interaction between electrically charged objects. To demonstrate this,
our study will begin with the simplest situation, electrostatics, when electrically
charged objects are static or always at rest.
Like mass, electric charge is a fundamental property of matter. Electric charge
is associated with particles that make up the atom: the electron and the proton.
The simplistic solar system model of the atom, as illustrated in 䉳 Fig. 15.1, likens its
structure to that of the planets orbiting the Sun. The electrons are viewed as orbit-
ing a nucleus, a core containing most of the atom’s mass in the form of protons and
electrically neutral particles called neutrons. As learned in Section 7.5, the cen-
tripetal force that keeps the planets in orbit about the Sun is supplied by gravity.
Similarly, the force that keeps the electrons in orbit around the nucleus is the elec-
Nucleus (+) trical force. However, there are important distinctions between gravitational and
(b) Beryllium atom
electrical forces.
One difference is that there is only one type of mass, and gravitational forces
䉱 F I G U R E 1 5 . 1 Simplistic model are always attractive. Electric charge, however, comes in two types, distinguished
of atoms The so-called solar system by the labels positive 1+2 and negative 1-2. Protons are designated as having a
model of (a) a hydrogen atom and positive charge, and electrons as having a negative charge. Different combinations
(b) a beryllium atom views the elec-
trons (negatively charged) as orbit-
of the two types of charge can produce either attractive or repulsive net electrical
ing the nucleus (positively charged), forces.
analogously to the planets orbiting The directions of the electric forces on isolated charged particles are given by
the Sun. The electronic structure of the following principle, called the law of charges or the charge–force law:
atoms is actually much more com-
plicated than this. Like charges repel, unlike charges attract.
15.1 ELECTRIC CHARGE 531
TABLE 15.1 Subatomic Particles and Their Electric Charge

Particle Electric Charge† Mass†
Electron - 1.602 * 10-19 C me = 9.109 * 10-31 kg + +

+ +
+ + Glass
Proton + 1.602 * 10 -19
C mp = 1.673 * 10 -27
kg
rods
Neutron 0 mn = 1.675 * 10 -27
kg
+ + + +
†
Even though the values are displayed to four significant figures, only two or three will generally
be used in our calculations.
That is, two negatively charged particles or two positively charged particles repel
each other, whereas particles with opposite charges attract each other (䉴 Fig. 15.2).
The repulsive and attractive forces are equal and opposite, and act on different objects,
in keeping with Newton’s third law (action–reaction, discussed in Section 4.3).
The charge on an electron and that on a proton are equal in magnitude, but
opposite in sign. The magnitude of the charge on an electron is abbreviated as e – –
– –
and is the fundamental unit of charge, because it is the smallest charge observed in – – Rubber
nature.* The SI unit of charge is the coulomb (C), named for the French rods
physicist>engineer Charles A. de Coulomb (1736–1806), who discovered a rela-
– – – –
tionship between electric force and charge (Section 15.3). The charges and masses
of the electron, proton, and neutron are given in 䉱 Table 15.1, where it can be seen
that e = 1.602 * 10-19 C. Our general symbol for charge will be q or Q. Thus the
charge on the electron can be expressed as qe = - e = - 1.602 * 10-19 C, and that
on the proton as qp = + e = + 1.602 * 10-19 C (usually expressed to two significant (a)
figures).
Other terms are frequently used when discussing charged objects. Saying that
an object has a net charge means that the object has an excess of either positive or
negative charges. (It is common, however, to ask about the “charge” of an object
when we really mean the net charge.) As will be seen in Section 15.2, excess charge
is most commonly produced by a transfer of electrons, not protons. (Protons are –
–
–
bound in the nucleus and, under most common situations, stay fixed.) For exam- –
–
ple, if an object has a (net) charge of +1.6 * 10-18 C, then it has had ten electrons –
removed from it because 10 * 1.6 * 10-19 C = 1.6 * 10-18 C. That is, the total
+ + + +
number of electrons on the object no longer completely cancels the positive charge
of all the protons—resulting in a net positive charge. On an atomic level, this situa-
tion means that some of the atoms that compose the object are deficient in elec-
trons. Such positively charged atoms are termed positive ions. Atoms with an (b)
excess of electrons are negative ions.
Since the charge of the electron is such a tiny fraction of a coulomb, an object hav- 䉱 F I G U R E 1 5 . 2 The charge–force
ing a net charge on the order of one coulomb is rarely seen in everyday situations. law, or law of charges (a) Like
Therefore, it is common to express amounts of charge using microcoulombs charges repel. (b) Unlike charges
attract.
(mC, 10-6 C), nanocoulombs (nC, 10-9 C), and picocoulombs (pC, 10-12 C).
Because the (net) electric charge on an object is caused by either a deficiency or
an excess of electrons, it must always be an integer multiple of the charge on an
electron. A plus sign or a minus sign will indicate whether the object has a defi-
ciency or an excess of electrons, respectively. Thus, the (net) charge of an object,
can be written as
q = ⫾ne (15.1)
SI unit of charge: coulomb 1C2
where n = 1, 2, 3, Á . Note that the net charge on any object is “quantized”; that
is, it can occur only in integral multiples of the charge on the electron (with the
appropriate sign).
*Protons, as well as neutrons and other particles, are now known to be made up of more fundamen-
tal particles called quarks, which carry charges of ⫾ 13 and ⫾ 23 of the electronic charge. There is experi-
mental evidence of the existence of quarks within the nucleus, but free quarks have not been detected.
Current theory implies that direct detection of quarks may, in principle, be impossible (Chapter 30).
In dealing with any electrical phenomena, another important principle (based

on experiment) is conservation of charge:
The net charge of an isolated system remains constant.
For example, suppose that a system consists initially of two electrically neutral
objects, and some electrons are transferred from one object to the other. The object
with the added electrons will then have a net negative charge, and the object with
the reduced number of electrons will have a net positive charge of equal magnitude.
(See Example 15.1.) But the net charge of the system remains zero. If the universe is
considered as a whole, conservation of charge means that the net charge of the uni-
verse is constant. Lastly, this principle applies to an isolated system even if it has a
net charge; that is, the net charge value stays the same and it need not be zero.
Note that this principle does not prohibit the creation or destruction of charged
particles. In fact, physicists have known for a long time that charged particles can
be created and destroyed on the atomic and nuclear levels. However, because of
charge conservation, charged particles must be created or destroyed only in pairs
with equal and opposite charges.
INTEGRATED EXAMPLE 15.1 On the Carpet: Conservation of Quantized Charge

When you shuffle across a carpeted floor on a dry day, the
Given: qc = + 12.15 nC2 ¢
10-9 C
carpet acquires a net positive charge (for details on this mech- ≤ Find: n (number of
1 nC
anism, see Section 15.2). (a) Will you have (1) a deficiency or transferred
(2) an excess of electrons? (b) If the charge the carpet acquired = + 2.15 * 10-9 C electrons)
has a magnitude of 2.15 nC, how many electrons were trans- qe = - 1.60 * 10-19 C
ferred? (from Table 15.1)
(A) CONCEPTUAL REASONING. (a) Since the carpet has a net The net charge on you is
positive charge, it must have lost electrons and you must
have gained them. Thus, your charge is negative, indicating q = - qc = - 2.15 * 10-9 C
an excess of electrons, and the correct answer is (2). Thus
(B) QUANTITATIVE REASONING AND SOLUTION. Because the q - 2.15 * 10-9 C
charge of one electron is known, the excess of electrons can be n = = = 1.34 * 1010 electrons
qe -1.60 * 10-19 C>electron
quantified. Express the charge in coulombs, and state what is
to be found. As can be seen, net charges, even in everyday situations, can
involve huge numbers of electrons (here, more than 13 bil-
lion), because the charge of any one electron is very small.
F O L L O W - U P E X E R C I S E . In this Example, if your mass is 80 kg, by what percentage has your mass increased due to the excess
electrons? (Answers to all Follow-Up Exercises are given in Appendix IV at the back of the book.)
DID YOU LEARN?

➥ Oppositely charged objects attract each other by the charge–force law.
➥ To repel, two charges must only have the same sign of charge.
➥ Removing electrons from a neutral object makes it positively charged.
15.2 Electrostatic Charging

➥ If a charged particle attracts a nearby object, must that object have a net charge?
➥ When rubbed with a cloth, a rubber rod acquires a net negative charge.What sign
charge did the cloth acquire?
➥ If a proton is brought near one end of a long metal rod (electrically neutral), what is
the sign of charge on the far end of the rod?
The existence of two types of electric charge along with the attractive and repul-
sive electrical forces can be easily demonstrated. Before learning how this is done,
let’s distinguish between electrical conductors and insulators. What distinguishes
15.2 ELECTROSTATIC CHARGING 533
these broad groups of substances is their ability to conduct, or transmit, Relative magnitude Material
of conductivity
electric charge. Some materials, particularly metals, are good conductors
108 CONDUCTORS
of electric charge. Others, such as glass, rubber, and most plastics, are Silver
insulators, or poor electrical conductors. A comparison of the relative Copper

magnitudes of the conductivities of some materials is given in 䉴 Fig. 15.3.
In conductors, the valence electrons of the atoms—that is, the electrons Aluminum
in the outermost atomic orbits—are loosely bound. As a result, they can be 107
easily removed from the atom and moved about in the conductor, or can Iron
leave the conductor altogether. That is, valence electrons are not perma-
nently bound to a particular atom. In insulators, however, even the loosest Mercury
bound electrons are too tightly bound to be easily removed from their
atoms. Thus, charge is not available to move through an insulator, nor is it
Carbon
readily removed from an insulator.
As Fig. 15.3 shows, there is a “middle” class of materials called 103
SEMICONDUCTORS
semiconductors. Their ability to conduct charge is intermediate between
that of insulators and conductors. The movement of electrons in semicon- Germanium
(Transistors)
ductors is more difficult to describe than the simple valence electron Silicon
approach used for insulators and conductors. In fact, the details of semicon- (Computer chips)
ductor properties can be understood only with the aid of quantum mechan- 10 –9
INSULATORS
ics, which is beyond the scope of this book. 10–10
Wood
However, it is interesting to note that the conductivity of semiconduc-
10 –12
tors can be adjusted by adding atomic impurities in varying concentra- Glass
tions. Beginning in the 1940s, scientists undertook research into the
properties of semiconductors to create applications for such materials. Sci- 10–15 Rubber
entists used semiconductors to create transistors, then solid-state circuits,
and, eventually, modern computer microchips. The microchip is one of the major
developments responsible for the high-speed computer technology of today. 䉱 F I G U R E 1 5 . 3 Conductors,
semiconductors, and insulators A
Now knowing the basics about conductors and insulators, let’s investigate a comparison of the relative magni-
way of determining the sign of the charge on an object. The electroscope is one of tudes of the electrical conductivities
the simplest devices used to determine electric charge (䉲 Fig. 15.4). In one of its of various materials.
most basic forms, it consists of a metal rod with a metallic bulb at one end. The rod
is attached to a solid, rectangular piece of metal that has an attached foil “leaf,”
usually made of gold or aluminum. This arrangement is insulated from its protec-
tive glass container by a nonconducting frame. When charged objects are brought
close to the bulb, electrons in the bulb are either attracted to or repelled by the
charged objects. For example, if a negatively charged rod is brought near the bulb,
electrons in the bulb are repelled, and the bulb is left with a positive charge. The
electrons are conducted down to the metal rectangle and its attached foil leaf,
which then will swing away, because of the like charges on the metal and leaf
(Fig. 15.4b). Similarly, if a positively charged rod is brought near the bulb, the leaf
also swings away from the metal. (Can you explain why?)
䉳 F I G U R E 1 5 . 4 The electroscope
–
An electroscope can be used to

–
+
–
determine whether an object is elec-

+
–
Negatively Positively trically charged. When a charged

Bulb +– +– ++
++ charged rod
––
–– charged rod object is brought near the bulb, the
leaf moves away from the metal
piece.
(a) Neutral electroscope has (b) Electrostatic forces

charges evenly distributed; cause leaf to move away.
leaf is vertical. (Only excess or net charge is shown.)
Notice that the net charge on the electroscope remains zero in these instances.
Because the device is isolated, only the distribution of charge is altered. However, it
is possible to give an electroscope (and other objects) a net charge by different
methods, all of which are said to involve electrostatic charging. Consider the fol-
lowing types of processes that produce electrostatic charging.
CHARGING BY FRICTION
In the frictional charging process, certain insulator materials are rubbed, typically
with cloth or fur, and they become electrically charged by a transfer of charge. For
example, if a hard rubber rod is rubbed with fur, the rubber will acquire a net neg-
ative charge; rubbing a glass rod with silk will give the glass a net positive charge.
This process is called charging by friction. The transfer of charge is due to the fric-
tional contact between the materials, and the amount of charge transferred
depends, as you might expect, on the nature of those materials.
Example 15.1 was an example of frictional charging in which a net charge was
picked up from the carpet. If you had reached for a metal object such as a door-
knob, you might have been “zapped” by a spark. As your hand
–
–
approaches the knob, it becomes positively charged, thus attract-

–
++ –
– ing the electrons from your hand. As the electrons travel, they
++ –
collide with, and excite, the atoms of the air, which give off light
as they de-excite (lose energy). This light is seen as the spark of
– “mini-lightning” between your hand and the knob.
–
CHARGING BY CONDUCTION (CONTACT)

Bringing a charged rod close to an electroscope will reveal that
the rod is charged, but it does not tell you the sign of the charge
on the rod. The sign can be determined, however, if the electro-
(a) Neutral electroscope is (b) Charges are scope is first given a known type of (net) charge. For example,
touched with negatively transferred to bulb; electrons can be transferred to the electroscope from a negatively
charged rod. electroscope has charged object as illustrated in 䉳 Fig. 15.5a. The electrons in the
net negative charge; rod repel one another, and some will transfer onto the electro-
leaf moves out.
scope. Notice that the leaf is now permanently diverged from the
metal. In this case, we say that the electroscope has been charged
–
+
–
by contact or by conduction (Fig. 15.5b). “Conduction” in this

+
–
–––– case refers to the flow of charge during the short period of time
–––
the electrons are transferred.
If a negatively charged rod is brought close to the now nega-
–– tively charged electroscope, the leaf will diverge even further as
more electrons are repelled down from the bulb (Fig. 15.5c). A pos-
itively charged rod will cause the leaf to collapse by attracting
electrons up to the bulb and away from the leaf area (Fig. 15.5d).
CHARGING BY INDUCTION
It might be asked whether it is possible to create an electroscope
(c) Negatively charged (d) Positively charged
rod repels electrons; rod attracts electrons;
that is positively charged using a negatively charged rubber rod
leaf moves further out. leaf collapses. (already charged by friction). The answer is yes. This can be
accomplished by charging by induction. Starting with an
䉱 F I G U R E 1 5 . 5 Charging by con-
duction (a) The electroscope is ini- uncharged electroscope, you touch the bulb with a finger, which
tially neutral (but the charges are grounds the electroscope—that is, provides a path by which electrons can escape
separated), as a charged rod touches from the bulb to the ground (䉴 Fig. 15.6). Then, when a negatively charged rod is
the bulb. (b) Charge is transferred to brought close to (but not touching) the bulb, the rod repels electrons from the bulb
the electroscope. (c) When a rod of through your finger and body and down into the Earth (hence the term ground).
the same charge is brought near the
bulb, the leaf moves farther apart. Removing your finger while the charged rod is kept nearby leaves the electroscope with
(d) When an oppositely charged rod a net positive charge. This is because when the rod is removed, the electrons that
is brought nearby, the leaf collapses. moved to the Earth have no way back because the return path is gone.
15.2 ELECTROSTATIC CHARGING 535
– 䉳 F I G U R E 1 5 . 6 Charging by
–
+++ + induction (a) Touching the bulb
– +++
with a finger provides a path to
–
– ground for charge transfer. The
– symbol e - stands for “electron.”
– + (b) When the finger is removed, the
–
electroscope is left with a net posi-
tive charge, opposite that of the rod.
e–
Ground
(a) Repelled by the nearby negatively (b) After removing the finger first,
charged rod, electrons are transferred then later the rod, the electroscope
to ground through hand. is left with a net positive charge.
CHARGE SEPARATION BY POLARIZATION

Charging by contact and charging by induction create a net charge through the
movement of charge to or from an object. However, charge can be moved within the
object while keeping its net charge zero. For example, the induction process described 䉲 F I G U R E 1 5 . 7 Polarization
previously initially causes polarization, or separation of positive and negative (a) When the balloons are charged
charge. If the object is not grounded, it will remain electrically neutral, but have by friction and placed in contact
with the wall, the wall is polarized.
regions of equal and opposite charge. In this situation, it is said that the object has That is, an opposite charge is
become polarized. On the molecular level, molecules polarized like this (called electric induced on the wall’s surface, to
dipoles—two poles, one of each sign) can be permanent; that is, they don’t need a which the balloons then stick by the
nearby charged object to retain charge separation. A good example of this is the water force of electrostatic attraction. The
molecule. Examples of both permanent and nonpermanent electric dipoles and electrons on the balloon do not
leave the balloon because its mater-
forces that can act on them are shown in 䉲 Fig. 15.7. Now you can understand why, ial (rubber) is a poor conductor. (b)
when you rub a balloon on your sweater, it can stick to the wall. The balloon is Some molecules, such as those of
charged by friction, and bringing it near the wall polarizes the wall. The opposite water, are polar by nature; that is,
sign charge on the wall’s nearest surface creates a net attractive force. they have permanently separated
Electrostatic charge can be annoying, as when static cling causes clothes and regions of positive and negative
charge. But even some molecules
papers to stick together, or even dangerous, as when electrostatic spark discharges that are not normally dipolar can be
start a fire or cause an explosion in the presence of a flammable gas. To discharge polarized temporarily by the pres-
electric charge, many large trucks have dangling metal chains in contact with the ence of a nearby charged object. The
ground. At gas stations, there are warnings to fill your gas cans while they are on electric force induces a separation of
the ground, not on the truck bed or car trunk surface (why?). charge and, consequently, tempo-
rary molecular dipoles. (c) A stream
of water bends toward a charged
Negative charge orients balloon. The negatively charged bal-
the molecular dipoles loon attracts the positive ends of the
+ –– + water molecules, causing the stream
+ –– + to bend.
–– +
+ + –– +
+ – – –
+ –– + + +
–– + ––
+ –– –– +
+
+
Permanently-dipolar water molecules
+ –
– + – –++– –+
+ – – +– +– +– + Nonpolar
– – –+ –+ +– molecule
+ –
– –+ –
+ –
– –
– – + – – ++
–– – ++ + – – –
– – ––
+ – – ++
Balloon Wall Induced molecular dipole

(a) (b) (c)
However, electrostatic forces can also be beneficial. For example, the air we
breathe is cleaner because of electrostatic precipitators used in smokestacks. In these
devices, electrical discharges cause the particles (by-products of fuel combustion) to
acquire a net charge. The charged particles can then be removed from the flue gases
by attracting them to electrically charged surfaces. On a smaller scale, electrostatic
air cleaners are available for use in the home (see the opening Physics Fact).
DID YOU LEARN?

➥ A polarized object can feel a net electric force, yet have a net charge of zero.
➥ An object that loses electrons becomes positively charged.
➥ In a metal, free electrons can move due to electric forces from external charges.
15.3 Electric Force

➥ What are the SI units of the proportionality constant k in the Coulomb force law?
➥ Two close electrons are released from rest and move away from each other with a
decreasing acceleration.Why?
➥ How does the electric force between two point charges vary with the distance
between them?
The directions of electric forces on interacting charges are given by the charge–force
law. However, what about their magnitudes? This was investigated by Coulomb, who
found that the magnitude of the electric force between two “point” (very small)
charges q1 and q2 depended directly on the product of the magnitude of the charges
and inversely on the square of the distance between them. That is, Fe r q1 q2>r2. (Here
q1 means the magnitude of q1, etc.) This relationship is mathematically similar to that
for the force of gravity between two point masses 1Fg r m1 m2>r22, see Section 7.5.
Like Cavendish’s measurements to determine the universal gravitational constant
G (Section 7.5), Coulomb’s measurements provided a constant of proportionality, k, so
that the electric force could be written in equation form. The magnitude of the electric
force between two point charges is given by Coulomb’s law:
kq1 q2 (point charges only,

Fe = (15.2)
r 2 q means charge magnitude)
䉲 F I G U R E 1 5 . 8 Coulomb’s law Here, r is the distance between the charges (䉲 Fig. 15.8a) and k a constant whose
(a) The mutual electrostatic forces
experimental value is
on two point charges are equal and
opposite. (b) For a configuration of
k = 8.988 * 109 N # m2>C2 L 9.00 * 109 N # m2>C2
two or more point charges, the force
on a particular charge is the vector
Equation 15.2 gives the force between any two charged particles, but in many
sum of the forces on it due to all the
other charges. (Note: In each of these instances, we are concerned with the forces between more than two charges. In this
situations, all of the charges are of situation, the net electric force on any particular charge is the vector sum of the
the same sign. How can we tell that forces on that charge due to all the other charges (Fig. 15.8b). For a review of vector
this is true? Can you tell their sign? addition, using electric forces, see the next two Examples. (See Section 3.1 and 3.2
What is the direction of the force on
for a general review of vectors, vector components, and vector addition.)
q2 due to q3?)
q2 kq 1q 3
F13 =
r2 r32
kq 1q 2 q1 q2 kq 1q 2
F12 = F21 = Fnet = F1 = F12 + F13
r2 r2 q1
r r3 kq 1q 2
q3 F12 =
r22
(a) (b)
15.3 ELECTRIC FORCE 537
CONCEPTUAL EXAMPLE 15.2 Free of Charge: Electric Forces

paper is greater than the repulsion 1F22 between the comb and
B
A rubber comb combed through dry hair can acquire a net
negative charge. That comb will then attract small pieces of the paper’s negative end. Therefore, after adding these two
uncharged paper. This would seem to violate Coulomb’s force forces vectorially, the net force on the paper points toward the
law. Since the paper has no net charge, you might expect there comb, and if it is light enough, the paper will accelerate in
to be no electric force on it. Which charging mechanism, that direction.
(a) conduction, (b) friction, or (c) polarization, explains this
– – – 䉳 FIGURE 15.9
phenomenon, and how does it explain it? – – –– Comb and paper
REASONING AND ANSWER. Because the comb doesn’t touch – – A neutral object can
the paper, the paper cannot be charged by either conduction – feel an electric force.
–––––––
or friction, because both of these require contact. Thus, the –––––––
–––––
answer must be (c). When the charged comb is near the paper, F1
the paper becomes polarized (䉴 Fig. 15.9). The key to under-
standing the attraction is to observe that the charged ends of + + +
+
the paper are not the same distance from the comb. The posi- – +
–
tive end of the paper is closer to the comb than the negative –– –
end. Since the electric force decreases with distance, the
attraction 1F12 between the comb and the positive end of the F2
B
FOLLOW-UP EXERCISE. Does the phenomenon described in this Example tell you the sign of the charge on the comb? Why or
why not?
EXAMPLE 15.3 Coulomb’s Law: Vector Addition Involving Trigonometry

(a) Two point charges of - 1.0 nC and +2.0 nC are separated magnitudes. Then it is just a matter of computing compo-
by a distance of 0.30 m (䉲 Fig. 15.10a). What is the electric nents. (a) For the two point charges, Coulomb’s law is used,
force on each particle? (b) A configuration of three charges is noting that the forces are attractive. (Why?) (b) Here compo-
shown in Fig. 15.10b. What is the net electric force on q3? nents must be used to vectorially add the two forces acting
on q3 due to q1 and q2. The angle u can be found from the
T H I N K I N G I T T H R O U G H . Adding electric forces is no differ- given distances. This angle is necessary to calculate the x
ent from adding any other type of force. The only difference and y force components. (See the Problem-Solving Hint
here is that Coulomb’s law is used to calculate the force immediately following this Example.)
y
y
q1 = +2.5 nC
(0, 0.30 m) r31
F32
q1 = –1.0 nC q2 = +2.0 nC q3 = +3.0 nC
θ θ
F12 F21 x x
θ (0.40 m, 0)
q3 θ Fnet = F3
0.30 m F31
r32
(0, –0.30 m)
(a) q2 = +2.5 nC
Vector diagram
(b)
䉱 FIGURE 15.10
SOLUTION. Listing the data and converting nanocoulombs to coulombs,
(a) q1 = - 11.0 nC2 ¢

10-9 C
≤ = - 1.0 * 10-9 C
B B
Given: Find: (a) F12 and F21
1 nC B
(b) F3
q2 = + 12.0 nC2 ¢
10-9 C
≤ = + 2.0 * 10-9 C
1 nC
(b) Data given in Figure 15.10b. Convert charges to coulombs as in part (a).
(a) Equation 15.2 gives the magnitude of the force acting on each charge using the charge magnitudes and distance between
them:
kq1 q2 19.00 * 109 N # m2>C2211.0 * 10-9 C212.0 * 10-9 C2
F12 = F21 =
10.30 m22
=
r2
= 0.20 * 10-6 N = 0.20 mN
Note that Coulomb’s law gives only the force’s magnitude. However, because the charges are of opposite sign, the forces must
be mutually attractive, as shown in Fig. 15.10a.
B B
(b) The forces F31 and F32 must be added vectorially, using trigonometry and components, to find the net force. Since all the
charges are positive, the forces are repulsive, as shown in the vector diagram in Fig. 15.10b. Since q1 = q2 and the charges are
B B
equidistant from q3, it follows that F31 and F32 have the same magnitude.
Also from the figure, it can be seen that r31 = r32 = 0.50 m. (Why?) With data from the figure, using Eq. 15.2:
kq2 q3 19.00 * 109 N # m2>C2212.5 * 10-9 C213.0 * 10-9 C2
F32 =
10.50 m22
=
r232
= 0.27 * 10-6 N = 0.27 mN
B B
Taking into account the directions of F31 and F32 , by symmetry the y-components cancel to produce zero net vertical force.
B
Thus, F3 (the net force on q3) acts horizontally along the positive x-axis and has a magnitude of F3 = F31x + F32x = 2 F31x because
F31 = F32. The angle u can be determined from the triangles; that is, u = tan-1 a b = 37°.
0.30 m
0.40 m
B
Thus F3 has a magnitude of
F3 = 2 F31x = 2 F32 cos u
= 210.27 mN2 cos 37° = 0.43 mN
and acts in the positive x-direction (to the right).
FOLLOW-UP EXERCISE. In part (b) of this Example, calculate the net force on q1.
The signs of the charges can be used explicitly in Eq. 15.2 with a positive value for F mean-
ing a repulsive force and a negative value an attractive force. However, such an approach is
not recommended, because this sign convention is useful only for one-dimensional forces,
that is, those that have only one component, as in Example 15.3a. When forces are two-
dimensional, thus requiring components, Eq. 15.2 should instead be used to calculate the
magnitude of the force, using only the magnitude of the charges (as in Example 15.3b). Then
the charge–force law determines the direction of the force between each pair of charges.
(Draw a sketch and put in the angles.) Lastly, use trigonometry to calculate each force’s
components and then combine them appropriately. This latter approach is recommended
and the one that will be used in this text.
The magnitudes of the charges in Example 15.3 are typical of the magnitudes of
static charges produced by frictional rubbing; that is, they are tiny. Thus, the forces
involved are very small by everyday standards, much smaller than any force we
have studied so far. However, on the atomic scale, even tiny forces can produce
huge accelerations, because the particles (such as electrons and protons) have
extremely small mass. Consider the answers in Example 15.4 compared with the
answers in Example 15.3.
EXAMPLE 15.4 Inside the Nucleus: Repulsive Electrostatic Forces

(a) What is the magnitude of the repulsive electrostatic force T H I N K I N G I T T H R O U G H . (a) Coulomb’s law must be applied
between two protons in a nucleus? Take the distance from to find the repulsive force. (b) To find the initial acceleration,
Newton’s second law 1Fnet = maB2 can be used.
B
center to center of these protons to be 3.00 * 10-15 m. (b) If
the protons were released from rest, how would the magni-
tude of their initial acceleration compare with that of the
acceleration due to gravity on the Earth’s surface, g?
15.3 ELECTRIC FORCE 539
SOLUTION. Listing the known quantities:

Given: r = 3.00 * 10-15 m Find: (a) Fe (magnitude of force)
a
q1 = q2 = + 1.60 * 10-19 C (from Table 15.1) (b) (ratio of acceleration to g)
mp = 1.67 * 10-27 kg (from Table 15.1) g
(a) Using Coulomb’s law (Eq. 15.2),

kq1 q2 19.00 * 109 N # m2>C2211.60 * 10-19 C211.60 * 10-19 C2
Fe = = 25.6 N
13.00 * 10-15 m2
=
r2 2
This force is much larger than that in the previous Example and is equivalent to the weight of an object with a mass of about
2.5 kg. Thus, with its small mass, we expect the proton to experience a huge acceleration.
(b) If it acted alone on a proton, this force would produce an That is, a L 1027g. The factor of 1027 is enormous. To help see
acceleration of how large it is, if a uranium atom were subject to this acceler-
ation, the net force required would be about the same as the
Fe 25.6 N
a = = = 1.53 * 1028 m>s2 weight of a polar bear (a thousand pounds or so)!
mp 1.67 * 10-27 kg Most atoms contain more than two protons in their nuclei.
Then, With these enormous repulsive forces, you would expect
nuclei to fly apart. Because this doesn’t generally happen,
a 1.53 * 1028 m>s2 there must be a stronger attractive force holding the nucleus
= = 1.56 * 1027
g 9.8 m>s2 together. This is called the nuclear (or strong) force, and will
be discussed in Chapters 29 and 30.
F O L L O W - U P E X E R C I S E . Suppose you could anchor an isolated proton to the ground and wished to delicately place a second one
directly above the first so that the second proton was in static equilibrium. How far apart would the protons be?
Although there is a striking similarity between the mathematical form of the

expressions for the electric and gravitational forces, there is a huge difference in
the relative strengths of the two forces, as is shown in Example 15.5.
EXAMPLE 15.5 Inside the Atom: Electric Force versus Gravitational Force
Determine the ratio of the electric force to the gravitational SOLUTION. The charges and masses of the particles are
force between a proton and an electron. In other words, how known (Table 15.1), as are the electrical constant k and the
many times larger than the gravitational force is the electric universal gravitational constant G.
force?
Given: qe = - 1.60 * 10-19 C Find: Fe
(ratio of forces)
T H I N K I N G I T T H R O U G H . The distance between the proton qp = + 1.60 * 10 C-19 Fg
and electron is not given. However, both the electrical force me = 9.11 * 10-31 kg
and the gravitational force vary as the inverse square of the mp = 1.67 * 10-27 kg
distance, so the distance will cancel out in a ratio. By using
Coulomb’s law and Newton’s law of gravitation (Section 7.5), The expressions for the forces are
the ratio can be determined if the charges, masses, and appro- kqe qp Gme mp
priate electric and gravitational constants are known. Fe = 2
and Fg =
r r2
Forming a ratio of magnitudes for comparison purposes (and to cancel r) gives

Fe kqe qp
=
Fg Gme mp
19.00 * 109 N # m2>C2211.60 * 10-19 C2
2
= 2.27 * 1039
16.67 * 10-11 N # m2>kg 2219.11 * 10-31 kg211.67 * 10-27 kg2
=
or
Fe = 12.27 * 10392Fg
The magnitude of the electrostatic force between a proton and an electron is more than 1039 times the magnitude of the gravita-
tional force. While a factor of 1039 is incomprehensible to most, it should be perfectly clear that because of this large value, the
gravitational force between charged particles can generally be neglected in our study of electrostatics.
F O L L O W - U P E X E R C I S E . With respect to this Example, show that gravity is even more negligible compared with the electric force
between two electrons. Explain why this is so.
DID YOU LEARN?

➥ In Coulomb’s law, k has SI units of N # m2>C2.
➥ As the distance between point charge increases, the force on each decreases.
➥ The force between two point charges varies inversely as the square of the distance
between them; that is, if the distance doubles, the force is reduced to one-fourth its
initial value.
15.4 Electric Field

➥ At a location in space, the direction of the electric field is the same as the direction of
the electric force that would act on a charge of what sign?
➥ Must there be an electric force at a given location in order for there to be an electric
field there?
➥ What does the spacing between electric field lines in a region of space tell you
about the electric field there?
The electric force, like the gravitational force, is an “action-at-a-distance” force.

Since the range of the electric force is infinite (Fe r 1>r 2 and approaches zero only
as r approaches infinity), a particular configuration of charges can have an effect
on an additional charge placed anywhere nearby.
The idea of a force acting across space was difficult for early investigators to
accept, and the modern concept of a force field, or simply a field, was introduced.
An electric field is envisioned as surrounding every arrangement of charges.
Thus, the electric field represents the physical effect of a particular configuration
of charges on the nearby space. The field represents what is different about the
nearby space because those charges are there. The concept treats charges as
interacting with the electric field created by other charges, not directly with the
charges “at a distance.” The main idea of the electric field concept is as follows:
A configuration of charges creates its electric field in the space nearby. When
another charge is placed in this field, the field will exert an electric force on that
charge. Thus:
Charge configurations can create electric fields, and these fields in turn can exert elec-
tric forces on other charges.
An electric field is actually a vector field (it has direction as well as magnitude). It
E
enables us to determine the force (including direction) exerted on a charge placed
⫹⫹ ⫹ ⫹ anywhere in the field region. However, the electric field is not that force. Instead, the
⫹ ⫹ ⫹⫹ q⫹ magnitude (or strength) of the field is defined as the electric force exerted per unit
⫹ ⫹ ⫹ ⫹ ⫹⫹ ⫹
⫹ ⫹ ⫹⫹ charge. Determining an electric field’s strength may be imagined using the follow-
⫺ ⫹
⫺ ⫺⫺
⫺⫺⫺ ⫹⫹ ing procedure. Place a very small charge (called a test charge) at the location of inter-
⫺ ⫺ ⫺ ⫹ ⫹
⫺ ⫺ ⫺ ⫹ ⫹⫹ est. Measure the force acting on that test charge and divide by the amount of its
⫺ ⫺⫺
⫺⫺⫺
⫺⫺ charge, thus determining the force that would be exerted per coulomb. Next imagine
removing the test charge. The force disappears (why?), but the field remains,
䉱 F I G U R E 1 5 . 1 1 Electric field because it is created by the nearby charges, which remain. When the electric field is
direction By convention, B
the direc- determined in many locations, we have a “map” of the electric field strength (mag-
tion of the electric field E is in the nitude) but no direction. Thus the “mapping” is incomplete.
direction of the force experienced by
an imaginary (positive) test charge.
Because the field direction is specified by the direction of the force on the test
To see the direction, ask which way charge, it depends on whether the test charge is chosen as positive or negative.
the test charge would accelerate if The convention is that a positive test charge 1q+2 is used for measuring electric field
released. Here the “system of direction (see 䉳 Fig. 15.11). That is:
charges” produces a (net) electric
field upward and to the right at the The electric field direction at any location is in the direction of the force experienced
location of the test charge. Can you by a positive test charge imagined to be placed at that location.
explain this direction by noting the
signs and locations of charges in Once the electric field’s magnitude and direction due to a charge configuration
this specific system? are known, you can ignore the “source” charge configuration and talk in terms of
15.4 ELECTRIC FIELD 541
the field it produces. This way of visualizing electric interactions between charges
often facilitates calculations.
B
To summarize mathematically, the electric field E at any location is defined as
kq
follows: E=
r2
B +
B
Fon q+
E = (15.3)
q+
SI unit of electric field: newton>coulomb 1N>C2

B
The direction of E is in the direction of the force on a small positive test charge at (a) Electric field vectors
that location.
For the special case of the field due to a single point charge, Coulomb’s law can be
The closer together
used. To determine the magnitude of the electric field due to a point charge at a dis-
the lines of force,
tance r from that point charge, use Eq. 15.3 to express the electric force. The result is: the stronger the
Fon q+ 1kqq+>r22 kq field.
E = = = 2
q+ q+ r
That is,
kq (magnitude of electric field +

E = (15.4)
r 2 due to point charge q)
(Notice that in deriving Eq. 15.4, q+ , canceled out. This must always happen, because
the field is produced by the other charges, not the test charge q+.)
Some electric field vectors in the vicinity of a positive charge are illustrated in
䉴 Fig. 15.12a. Note that their directions are away from the positive charge because a
(b) Electric field lines
positive test charge would feel a force in this direction. Notice also that the magni-
tude of the field (proportional to the arrow length) decreases with increasing r. 䉱 F I G U R E 1 5 . 1 2 Electric field
If there is more than one charge creating an electric field, then the total, or net, (a) The electric field points away
electric field at any point is found using the superposition principle for electric from a positive point charge, in the
direction a force would be exerted
fields, which can be stated as follows: on a small positive test charge. The
For a configuration of charges, the total, or net, electric field at any point is the vector field’s magnitude (proportional to
the vector lengths) decreases as the
sum of the electric fields due to the individual charges of the configuration.
distance from the charge increases,
This principle is demonstrated in the next two Examples, and a way to qualita- reflecting the inverse-square dis-
tance relationship characteristic of
tively determine the direction of the electric field when more than one charge is
the field produced by a point
involved is shown in the accompanying Learn by Drawing 15.1, Using the Super- charge. (b) In this simple case, the
position Principle to Determine the Electric Field Direction. vectors are easily connected to give
the electric field line pattern due to
a positive point charge.
using the superposition principle to

determine the electric field direction q1 +
= q2 = –2q1
To estimate the direction of the electric field at any point P,
draw the individual electric field vectors and add them, tak-
ing into account the relative field magnitudes if you can. In
B B
this situation, E1 is much smaller than E2 because of both E2 E
distance and charge factors. Can you explain why the vector
B
representing E2, if drawn accurately, would be about eight
B
times as long as that of E1? The final step to finding the elec- P
B
tric field E at P is to complete the vector addition (parallelo-
B B B E1
gram), that is, find E = E1 + E2.
EXAMPLE 15.6 Electric Fields in One Dimension: Zero Field by Superposition

Two point charges are placed on the x-axis as in 䉲 Fig. 15.13. T H I N K I N G I T T H R O U G H . Each point charge produces its own
Find all locations on the axis where the electric field is zero. field. By the superposition principle, the electric field is the vec-
tor sum of the two fields. Thus the question is asking for loca-
tions where these fields are equal and opposite, so as to cancel
Where is E = 0? and give no (total or net) electric field. Since both charges are
positive, their fields point to the right at all locations to the right
q1 = +1.5 mC q2 = +6.0 mC of q2. Therefore, the fields cannot cancel in that region. Similarly,
+ + x (m) to the left of q1, both fields point to the left and cannot cancel.
0 0.10 0.20 0.30 0.40 0.50 0.60 The only possibility of cancellation is in the area between the
charges. In that region, the two fields will cancel if their magni-
䉱 F I G U R E 1 5 . 1 3 Electric field in one dimension tudes are equal, because they are oppositely directed.
SOLUTION. Let us specify the location as a distance x from q1 (located at x = 0) and convert charges from microcoulombs to
coulombs as usual.
Given: d = 0.60 m (distance between charges) Find: x [the location(s) of zero E]
q1 = + 1.5 mC = + 1.5 * 10-6 C
q2 = + 6.0 mC = + 6.0 * 10-6 C
Setting the magnitudes of the individual fields equal and With q2>q1 = 4, taking the square root of both sides:
q2>q1
solving for x:
1 4 1 2
or
B 1d - x22 B 1d - x22
kq1 kq2 = = =
E1 = E2 or A x2 x d - x
1d - x22
=
x2
Solving, x = d>3 = 0.60 m>3 = 0.20 m. (Why not use the
Rearranging this expression and canceling the constant k negative square root? Try it and see.) The result being closer
yields to q1 makes sense physically. Because q2 is the larger charge,
1 1q2>q12 for the fields to be equal in magnitude, the location must be
closer to q1.
1d - x22
2
=
x
FOLLOW-UP EXERCISE. Repeat this Example, changing only the sign of the right-hand charge in Fig. 15.13.
INTEGRATED EXAMPLE 15.7 Electric Fields in Two Dimensions: Using Vector Components
and Superposition
䉲 Fig. 15.14a shows a configuration of three point charges. ( A ) C O N C E P T U A L R E A S O N I N G . The electric field points toward
(a) In what quadrant is the electric field at the origin: (1) the negative point charges and away from positive point charges.
B B B
first quadrant, (2) the second quadrant, or (3) the third quad- Therefore, E1 and E2 point in the positive x-direction and E3
rant? Explain your reasoning, using the superposition princi- points along the positive y-axis. Because the electric field is
ple. (b) Calculate the magnitude and direction of the electric the sum of these three fields, both of its components are posi-
B
field at the origin due to these charges. tive. Therefore, E is in the first quadrant (Fig. 15.14b). Thus,
the correct answer is (1).
䉴 FIGURE 15.14 y (m)

Finding the electric field
Vector addition requires the
use of components.
y
4.00 q3 = –1.50 mC
E3
Ey E
q2 = +2.00 mC q1 = –1.00 mC θ E1
x
x (m) 0 Ex
0 E2
–5.00 3.50
(a) (b)
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . The directions of the individual electric fields are shown in the sketch in part (a).
According to the superposition principle, these fields are added vectorially to find the electric field 1E = E1 + E2 + E32.
B B B B
Listing the data given and converting the charges into coulombs:
B
Given: q1 = - 1.00 mC = - 1.00 * 10-6 C Find: E (electric field at origin)
q2 = + 2.00 mC = + 2.00 * 10-6 C
q3 = - 1.50 mC = - 1.50 * 10-6 C
r1 = 3.50 m
r2 = 5.00 m
r3 = 4.00 m
B B B
From the sketch Ey is due to E3 , and Ex is the sum of the magnitudes of E1 and E2. The magnitudes of the three fields are deter-
mined from Eq. 15.4. These magnitudes are
kq1 19.00 * 109 N # m2>C2211.00 * 10-6 C2
E1 = 2 = = 7.35 * 102 N>C
r1 13.50 m22
kq2 19.00 * 109 N # m2>C2212.00 * 10-6 C2
E2 = = 7.20 * 102 N>C
15.00 m22
=
r22
kq3 19.00 * 109 N # m2>C2211.50 * 10-6 C2
E3 = = 8.44 * 102 N>C
14.00 m22
=
r23
The x- and y-components of the field are

Ex = E1 + E2 = + 7.35 * 102 N>C + 7.20 * 102 N>C = + 1.46 * 103 N>C
and
Ey = E3 = + 8.44 * 102 N>C
In component form,
E = Ex xN + Ey yN = 11.46 * 103 N>C2xN + 18.44 * 102 N>C2yN
B
You should be able to show that in magnitude–angle form this is

E = 1.69 * 103 N>C at u = 30.0° (u is in the first quadrant, relative to the + x-axis)
FOLLOW-UP EXERCISE. In this Example, suppose q1 was moved to the origin. Find the electric field at its former location.
ELECTRIC LINES OF FORCE

A convenient way of graphically representing the electric field is by use of electric
lines of force, or electric field lines. To start, consider the electric field vectors near
a positive point charge, as in Fig. 15.12a. In Fig. 15.12b these vectors have been
“connected.” This method constructs the electric field line pattern due to a positive
point charge. Notice that the field lines are closer together (their spacing
decreases) nearer the charge, because the field increases in strength. Also note that
at any location on a field line, the electric field direction is tangent to the line. (The
lines usually have arrows attached to them that indicate the general field direc-
tion.) It should be clear that electric field lines can’t cross. If they did, it would
mean that at the crossing spot there would be two directions for the force on a
charge placed there—a physically unreasonable result.
The general rules for sketching and interpreting electric field lines are as follows:
1. The closer together the field lines, the stronger the electric field.
2. At any point, the direction of the electric field is tangent to the field lines.
3. The electric field lines start at positive charges and end at negative charges.
4. The number of lines leaving or entering a charge is proportional to the mag-
nitude of that charge. (See the accompanying Learn by Drawing 15.2, Sketch-
ing Electric Lines of Force for Various Point Charges.)
5. Electric field lines can never cross.
These rules, along with the superposition principle, enable us to “map” the
pattern of electric lines of force (electric field line pattern) due to various charge
configurations, for example, the pattern due to an electric dipole in Example 15.8.
An electric dipole consists of two equal, but opposite, electric charges (or
“poles,” as they were known historically). Even though the net charge on the
LEARN BY DRAWING 15.2 dipole is zero, it creates an electric field because the charges are separated. If the
charges were at the same location, their fields would cancel everywhere.
sketching electric
lines of force for EXAMPLE 15.8 Constructing the Electric Field Pattern Due
various point charges to a Dipole
How many lines should be
Use the superposition principle and the electric field line rules to construct a typical elec-
drawn for -1 12 q, and what
tric field line due to an electric dipole.
should their direction be?
T H I N K I N G I T T H R O U G H . The construction involves vector addition of the individual
Sketching Electric Field Lines electric fields from the two opposite ends of the dipole.
SOLUTION.
Given: an electric dipole of two equal and opposite Find: a typical electric field line
charges separated by a distance d
+
An electric dipole is shown in 䉲 Fig. 15.15a. To keep track of the two fields, let’s label
B B
+q the positive charge q+ and the negative charge q- . Their individual fields, E + and E - ,
will be designated by the same subscripts.
Because electric fields (and field lines) start at positive charges, let’s begin at location
A, near charge q+ . Because this is much closer to q+ , it follows that E+ 7 E- . We know
B B
that E + will always point away from q+ and E - will always point toward q-. Putting these
– two facts together enables us to qualitatively draw the two fields at A. The parallelo-
– 21 q B
gram method determines their vector sum: the electric field at A, EA.
To map the electric field line, the general direction of the electric field at A points us
approximately to our next location, B. At B, there is a reduced magnitude (why?) and
B B
slight directional change for both E + and E -. You should now be able to see how the
fields at C and D are determined. Location D is special because it is on the perpendicu-
lar bisector of the dipole axis (the line that connects the two charges). The electric field
–
points downward anywhere on this line. You should be able to continue the construc-
–1 21 q tion at points E, F, and G.
Lastly, to construct the electric field line, start at the positive end of the dipole,
because the field lines leave that end. Because the electric field vectors are tangent to
the field lines, draw the line to fulfill this requirement. (You should be able to sketch in
the other lines and understand the complete dipole field pattern shown in Fig. 15.15b.)
E+
B E+
A
q+ + EA EB
E– E– C
E+
E– EC
+
D
d
E– E+ –
ED
E
EE
q– F
–
G
EF
EG
(a) (b)
䉱 F I G U R E 1 5 . 1 5 Mapping the electric field due to a dipole.

F O L L O W - U P E X E R C I S E . Using the techniques in this Example, construct the field lines
that start (a) just above the positive charge, (b) just below the negative charge, and
(c) just below the positive charge.
In addition to being used to learn about electric field sketching, dipoles are E
important in themselves, because they occur in nature. For example, electric
dipoles can serve as a model for important polarized molecules, such as the water
molecule. (See Fig. 15.7.) Also see Insight 15.2, Electric Fields in Law Enforcement
and Nature: Stun Guns and Electric Fish.
䉴 Figure 15.16a shows the use of the superposition principle to construct the elec-
tric field line pattern due to a large uniformly charged positive plate. Notice that the + + + + + + + + +
field points perpendicularly away from the plate on both sides. Figure 15.16b shows
the result if the plate is negatively charged, the only difference being the field direc-
tion. Putting these two together, the field between two closely spaced and oppositely
charged plates is found. The result is the pattern in Fig. 15.16c. Due to the cancellation
of the horizontal field components (as long as we stay away from the plate edges),
the electric field is uniform and points from the positive to negative. (Think of the
direction of the force acting on a positive test charge placed between the plates.) E
The derivation of the expression for the electric field magnitude between two (a)
closely spaced plates is beyond the scope of this text. However, the result is
4pkQ E
E = (electric field between parallel plates, not near edges) (15.5)
A
– – – – – – – – – – – – – – –
where Q is the magnitude of the total charge on one of the plates and A is the area
of one plate. Parallel plates are common in electronic applications. For example, in
E
Chapter 16 an important circuit element called a capacitor will be studied. In its
simplest form a capacitor is just a set of parallel plates. Capacitors play a crucial (b)
role in lifesaving devices such as heart defibrillators, as will be seen in Chapter 16.
Cloud-to-ground lightning can be approximated by closely spaced parallel plates +++ + + + + + + + ++ + + +
as in the next Example. (See also Insight 15.1, Lightning and Lightning Rods.)
E
– – – – – – – – – – – – – – –
EXAMPLE 15.9 Parallel Plates: Estimating the Charge on
(c)
Storm Clouds
䉱 F I G U R E 1 5 . 1 6 Electric field
The electric field (magnitude) E required to ionize moist air is about 1.0 * 106 N>C. due to very large parallel plates
When the field reaches this value, the least bound electrons are pulled off their mole- (a) Above a positively charged
cules (ionization of the molecules), which can lead to a lightning stroke. Assume that plate, the net electric field points
the value for E between the negatively charged lower cloud surface and the positively upward. Here, the horizontal com-
charged ground is 1.00% of this, or 1.0 * 104 N>C. (See Fig. 1a of Insight 15.1.) Take the ponents of the electric fields from
clouds to be squares 10 miles on each side. Estimate the magnitude of the total nega- various locations on Bthe plate can-
tive charge on the lower surface. cel. Below the plate, E points down-
ward. (b) For a negatively charged
T H I N K I N G I T T H R O U G H . The electric field is given, so Eq. 15.5 can be used to estimate plate, the electric field directions
Q. The cloud area A (one of the “plates”) must be expressed in square meters. (shown on both sides of the plate)
are reversed. (c) For two oppositely
SOLUTION. charged, closely spaced plates, there
is field cancellation outside the
Given: E = 1.0 * 104 N>C Find: Q (the magnitude of the charge on
plates resulting in almost no field
d = 10 mi L 1.6 * 104 m the lower cloud surface)
there. Between the plates the fields
Using A = d2 for the area of a square and solving Eq. 15.5 for the magnitude of the from the two plates add, resulting in
(approximately) a uniform field in
charge (the cloud surface is negative):
that region. (The field at the edges
11.0 * 104 N>C211.6 * 104 m2 of the plates is not shown.)
2
EA
Q = = 23 C
4p19.0 * 109 N # m2>C22
=
4pk
This expression is justified only if the distance between the clouds and the ground is
much less than their size. (Why?) Such an assumption is equivalent to assuming that
the 10-mile-long clouds are less than several miles from the Earth’s surface.
This amount of charge is huge compared with the frictional static charges developed
when shuffling on a carpet. However, because the cloud charge is spread out over a
large area, any one region of the cloud does not contain a lot of charge.
F O L L O W - U P E X E R C I S E . In this Example, (a) what is the direction of the electric field

between the cloud and the Earth? (b) How much charge would be required to ionize
moist air?
INSIGHT 15.1 Lightning and Lightning Rods

Although the violent release of electrical energy in the form of tial flow is near the ground. As it continues, electrons posi-
lightning is common, we have a lot to learn about its forma- tioned successively higher begin to migrate downward.
tion. It is known that during the development of a cumu- Hence, the path of electron flow is extended upward in a
lonimbus (storm) cloud, a separation of charge occurs. How return stroke. The surge of charge in the return stroke causes
the separation of charge takes place in a cloud is not fully the conductive path to be illuminated, producing the bright
understood, but it must be associated with the rapid vertical flash seen by the eye and recorded in time-exposure pho-
movement of air and moisture within storm clouds. Whatever tographs (Fig. 1b). Most lightning flashes have a duration of
the mechanism, the cloud acquires regions of different charge, less than 0.50 s. Usually, after the initial discharge, ioniza-
with the bottom usually negatively charged. tion again takes place along the original channel, and
As a result, an opposite charge is induced on the Earth’s surface another return stroke occurs. Typical lightning events have
(Fig. 1a). Eventually, lightning may reduce this charge difference three or four return strokes.
by ionizing the air, allowing a flow of charge between the cloud Ben Franklin is often said to have been the first to demon-
and the ground. However, air is a good insulator, so the electric strate the electrical nature of lightning. In 1750, he suggested
field must be quite strong for ionization to occur. (See Example an experiment using a metal rod on a tall building. How-
15.9 for a quantitative estimate of the charge on a cloud.) ever, a Frenchman named Thomas François d’Alibard was
Most lightning occurs entirely within a cloud (intracloud dis- the one who first set up an experiment using a rod during a
charges), where it cannot be seen. However, visible discharges thunderstorm (Fig. 1c). Franklin later performed a similar
do take place between clouds (cloud-to-cloud discharges) and experiment with a kite, also during a thunderstorm.
between a cloud and the Earth (cloud-to-ground discharges). A practical outcome of Franklin’s work was the lightning
Special high-speed camera photographs of cloud-to-ground rod, a pointed metal rod connected by a wire to a metal rod
discharges reveal a nearly invisible downward ionization path. driven into the Earth, or “grounded.” The elevated rod’s tip,
The lightning discharges in a series of jumps or steps and so is with its dense accumulation of induced positive charge and
called a stepped leader. As the leader nears the ground, positively large electric field (see Fig. 15.19b), intercepts the downward,
charged ions in the form of a streamer rise from trees, tall build- ionized stepped leader, discharging it harmlessly to the
ings, or the ground to meet it. ground before the leader reaches a structure or makes contact
When a streamer and a leader make contact, the electrons with an upward streamer. This prevents the formation of a
along the leader channel begin to flow downward. The ini- damaging electrical surge associated with a return stroke.
(a) (b) (c)

F I G U R E 1 Lightning and lightning rods (a) Cloud polarization induces a charge on the Earth’s surface. (b) When the elec-
tric field becomes large enough, an electrical discharge results, which we call lightning. (c) A lightning rod mounted atop a
tall structure provides a path to ground so as to prevent damage.
The electric field patterns for some other point charge configurations are shown in
䉴 Fig. 15.17. You should be able to see how they are sketched qualitatively. Note that
the electric field lines always begin on positive charges and end on negative ones (or
at infinity when there is no nearby negative charge). Choose the number of lines ema-
nating from or ending at a charge in proportion to the magnitude of that charge. (See
Learn by Drawing 15.2, Sketching Electric Lines of Force for Various Point Charges.)
䉳 F I G U R E 1 5 . 1 7 Electric fields
Electric fields for (a) like point
charges and (b) unequal like point
charges.
+ + – ––
(a) Like point charges (b) Unequal like point charges
DID YOU LEARN?

➥ The electric field direction is in the direction of the electric force that would be
experienced by a positive charge.
➥ For an electric force to exist, both an electric charge and field must be present.
➥ The spacing between electric field lines is inversely related to the magnitude of
the field.
INSIGHT 15.2 Electric Fields in Law Enforcement and Nature: Stun Guns and Electric Fish
Stun guns and electric fish have similar electric field proper- In electrogenic fish, the charge separation is accomplished
ties. Stun guns (there are several types, the most familiar by the electric organ (shown for the elephant nose fish in
being the hand-held Taser) generate a charge separation by Fig. 2b), which is a specialized stack of electroplates. Each elec-
using batteries and internal circuitry. This circuitry can pro- troplate is a disklike structure that is normally uncharged.
duce a large charge polarization—that is, equal and opposite When the brain sends a signal, the disks become polarized
charge on the electrodes. Figures 1a and 1b show a typical through a chemical process similar to that of nerve action, cre-
Taser. The charges on the electrodes oscillate in sign, but at ating the fish’s field.
any instant, the field is close to that of a dipole (Fig. 1c). Tasers Weakly electric fish are capable of producing electric fields
are used for subduing criminals, theoretically without perma- about the same strength as those produced by batteries. These
nent harm. A law enforcement officer, holding the grip, fields are good only for electrocommunication and
applies the electrodes to the body—say, the thigh. The electric electrolocation. Strongly electric fish produce fields hundreds
field disrupts the electrical signals in the nerves that control of times stronger and can kill prey by touching them simulta-
the large thigh muscle, rendering the muscle inoperative and neously with the oppositely charged areas. The electric eel
making the criminal more easily subdued. has thousands of electroplates stacked in the electric organ,
The phrase electric fish conjures an image of an electric eel which typically extends from behind its head well into its tail
(which is actually an eel-shaped fish). However, there are and may take up to 50% of its body length (Fig. 2c).
other fish that are “electric.” The electric eel and others such As an example of how these fields are used for electrolocation,
as the electric catfish are strongly electric fish. They can gener- consider the change to the elephant nose fish’s normal electric
ate large electric fields to stun prey but can also use the fields field pattern (Fig. 2b) when it approaches a small conducting
for location and communication. Weakly electric fish, such as object (Fig. 3). Notice that the field lines change to bend toward
the elephant nose (Fig. 2a), use their fields (Fig. 2b) for loca- the object; because the object is conducting, the field lines must
tion and communication only. Fish that actively produce elec- be oriented at right angles to its surface (Section 15.5).
tric fields are called electrogenic fish.
Internal circuitry ⫹
and batteries
E
Electrodes
⫺
Hand grip
Active electrodes
(a) (b) (c)
F I G U R E 1 The Taser stun gun (a) The exterior of a stun gun; notice the grip and two electrodes. (b) The interior of a stun
gun: the circuitry necessary to increase the electric field and charge separation to the strength required to disrupt nerve
communication. (c) A sketch of the electric field between the electrodes at some instant in time.
Head and vital organs
⫹
E
⫹
⫺ Electroplates ⫺
stacked into
electric organ
(a) (b) (c)

F I G U R E 2 Electric fish (a) The elephant nose fish is a weakly electric fish—one that uses its electric field only for electrolocation
and communication. (b) At an instant in time, the approximate electric field produced by the elephant nose fish’s electric organ,
located near its tail. (c) At an instant in time, the approximate electric field produced by an electric eel. The electric organ in the
eel is capable of producing fields that can kill and stun as well as locate and communicate.
FIGURE 3
Electrolocation The
field from an elephant
nose fish with a con-
ducting object nearby.
Note the change in spac-
ing of the field lines as
⫹ ⫺
This results in a stronger field at the part of the fish’s skin they enter the skin sur-
surface near the object. Skin sensors detect this increase and face due to the change in
field produced by the
send a signal to the brain to that effect. A nonconducting
nearby object. This
object, such as a rock, would have the opposite effect. Thus, change in field strength
electrolocation and electrocommunication are determined is picked up by sensory
by an interplay of the electric field and the sensory organs. organs on the skin,
However, the basic properties of electrostatic fields provide which send a signal to
us with the general idea of how such fish operate. the fish’s brain.
15.5 Conductors and Electric Fields

➥ What is the value of electric field inside a conductor under electrostatic conditions?
➥ What is the direction of the electric field at the surface of a conductor under electro-
static conditions?
➥ Where is the highest charge density located on the surface of a conductor?
The electric fields associated with charged conductors have several interesting
properties. By definition in electrostatics, the charges are at rest. Because conduc-
tors possess electrons that are free to move but don’t, the electrons must experi-
ence no electric force and thus no electric field. Hence we conclude that:
The electric field is zero inside a charged conductor.
Excess charges on a conductor tend to get as far away from each other as possi-
ble, because they are highly mobile. Then:
Any excess charge on an isolated conductor resides entirely on the surface of the
conductor.
Another property of static electric fields and conductors is that there cannot be
a tangential component of the field at the surface of the conductor. If this were not
true, charges would move along the surface, contrary to our assumption of a static
situation. Thus:
The electric field at the surface of a charged conductor is perpendicular to the surface.
15.5 CONDUCTORS AND ELECTRIC FIELDS 549
䉳 F I G U R E 1 5 . 1 8 Electric fields
and conductors (a) Under static
conditions, the electric field is zero
E inside a conductor. Any excess
No! charge resides on the conductor’s
E=0 E surface. For an irregularly shaped
+ +
+ + conductor, the excess charge accu-
+ +
mulates in the regions of highest
Conductor curvature (the sharpest points), as
shown. The electric field near the
surface is perpendicular to that sur-
face and strongest where the charge
(a) (b) is densest. (b) Under static condi-
tions, the electric field must not have
a component tangential to the con-
ductor’s surface.
Lastly, the excess charge on a conductor of irregular shape is most closely
packed where the surface is highly curved (at the sharpest points). Since the
charge is densest there, the electric field will also be the largest at these loca-
tions. That is:
F F
Excess charge tends to accumulate at sharp points, or locations of highest curvature, – –
on charged conductors. As a result, the electric field is greatest at such locations. FII FII
These properties are summarized in 䉱 Fig. 15.18. Note that they are true only for con-
ductors under static conditions. Electric fields can exist inside nonconducting materi-
als and inside conductors when conditions vary with time.
To understand why most of the charge accumulates in the highly curved surface F
regions, consider the forces acting between charges on the surface of the conductor.
(See 䉴 Fig. 15.19a.) Where the surface is fairly flat, these forces will be directed
–
nearly parallel to the surface. The charges will spread out until the parallel forces
–
from neighboring charges in opposite directions cancel out. At a sharp end, the
F
forces between charges will be directed more nearly perpendicular to the surface,
and so there will be little tendency for the charges to move parallel to the surface.
Therefore, one would expect highly curved regions of the surface to accumulate
the highest concentration of charge. (a)
An interesting situation occurs when there is a large concentration of charge on
a conductor with a sharp point (Fig. 15.19b). The electric field above the point may
be high enough to ionize air molecules (to pull or push electrons off the mole-
cules). The freed electrons are then further accelerated by the field and can cause +
++ ++
secondary ionizations by striking other molecules. This results in an “avalanche” + +
of electrons, visible as a spark discharge. More charge can be placed on a gently + +
curved conductor, such as a sphere, before a spark discharge will occur. The con-
centration of charge at the sharp point of a conductor is one reason for the effec- + +
tiveness of lightning rods. (See Insight 15.1, Lightning and Lightning Rods.)
For some law enforcement and biological applications of electric fields and con-
ductors, refer to Insight 15.2, Electric Fields in Law Enforcement and Nature: Stun (b)
Guns and Electric Fish.
䉱 F I G U R E 1 5 . 1 9 Concentration
As an illustration of an early experiment done on conductors with excess of charge on a curved surface (a) On
charge, consider the following Example. a flat surface, the repulsive forces
between excess charges are parallel
to the surface and tend to push the
charges apart. On a sharply curved
CONCEPTUAL EXAMPLE 15.10 The Classic Ice Pail Experiment surface, in contrast, these forces are
directed at an angle to the surface.
A positively charged rod is held inside an isolated metal container that has uncharged Their components parallel to the
electroscopes conductively attached to its inside surface and to its outside surface surface are smaller, allowing charge
to concentrate in such areas.
(䉲 Fig. 15.20). What will happen to the leaves of the electroscopes: (a) neither electro-
(b) Taken to the extreme, a sharply
scope’s leaf will show a deflection; (b) only the outside-connected electroscope’s leaf pointed metallic needle has a dense
will show a deflection; (c) only the inside-connected electroscope’s leaf will show a concentration of charge at the tip.
deflection; or (d) the leaves of both electroscopes will show deflections? (Justify your This produces a large electric field
answer.) in the region above the tip, which is
(continued on next page) the principle of the lightning rod.
䉴 F I G U R E 1 5 . 2 0 An ice
pail experiment.
Gaussian
surface
+ +
+ ? ?
Insulator
(a)
Gaussian REASONING AND ANSWER. The positively charged rod will attract negative charges,
surface causing the inside of the metal container to become negatively charged. The outside
electroscope will thus acquire a positive charge. Hence, both electroscopes will be
charged (though with opposite signs) and show deflections, so the answer is (d). This
experiment was performed by the nineteenth-century English physicist Michael Fara-
– day using ice pails, so this setup is often called Faraday’s ice pail experiment.
F O L L O W - U P E X E R C I S E . Suppose in this Example that the positively charged rod actu-
ally touched the metal container. How would the electroscopes react now?
(b)
DID YOU LEARN?
Gaussian ➥ In electrostatics, the electric field inside a conductor is zero.
surface ➥ Under electrostatic conditions the electric field at the surface of a conductor is
perpendicular to the surface everywhere.
––
➥ The highest density of charge on a conductor’s surface is where surface’s curvature
is the greatest.
(c)
*15.6 Gauss’s Law for Electric Fields: A Qualitative
Gaussian Approach
surface 1
Gaussian
➥ Can a Gaussian surface have charges inside it and yet have no net number of field
surface 3 lines penetrating it?
➥ If a Gaussian surface surrounds an object with a net charge, are the net number of
lines penetrating its surface related to the sign of the charge?
+
➥ If a Gaussian surface contains only two protons and then an electron is added into
the mix, what happens to the net number of field lines penetrating the surface?
One of the fundamental laws of electricity was discovered by Karl Friedrich Gauss
–
(1777–1855), a German mathematician. Using it for quantitative calculations
involves techniques beyond the scope of this book. However, a conceptual look at
Gaussian this law can teach us some interesting physics.
surface 2
Consider the single positive charge in 䉳 Fig. 15.21a. Now picture an imaginary
closed surface surrounding this charge. Such a surface is called a Gaussian surface.
Gaussian Let us designate electric field lines that pass through the surface outwardly as pos-
surface 4 itive and inward pointing ones as negative. If the lines of both types are counted
(d)
and totaled (that is, subtract the number of negative lines from the number of pos-
itive ones), the total is positive, because in this case there are only positive lines.
䉱 F I G U R E 1 5 . 2 1 Various Gauss-
ian surfaces and lines of force This result reflects the fact that there is a net number of outward- pointing electric
(a) Surrounding a single positive field lines through the surface. Similarly, for a negative charge (Fig. 15.21b), the
point charge, (b) surrounding a sin- count would yield a negative total, indicating a net number of inward-pointing
gle negative point charge, and lines passing through the surface. Note that these results would be true for any
(c) surrounding a larger negative
closed surface surrounding the charge, regardless of its shape or size. If the magni-
point charge. (d) Four different sur-
faces surrounding various parts of tude of the negative charge is doubled (Fig. 15.21c), the negative field line count
an electric dipole. would also double. (Why?)
*15.6 GAUSS’S LAW FOR ELECTRIC FIELDS: A QUALITATIVE APPROACH 551
Figure 15.21d shows a dipole with four different imaginary Gauss-

ian surfaces. Surface 1 encloses a net positive charge and therefore Gaussian
has a positive line count. Similarly, surface 2 has a negative line surface 1
count. The more interesting cases are surfaces 3 and 4. Note that both
include zero net charge—surface 3 because it includes no charges at
all and surface 4 because it includes equal and opposite charges.
Note that both surfaces 3 and 4 have a net line count of zero, correlat-
ing with no net charge enclosed.
Gaussian
These situations can be generalized (conceptually) to give us the
surface 2
underlying physical principle of Gauss’s law:*
The net number of electric field lines passing through an imaginary closed 䉱 F I G U R E 1 5 . 2 2 Water analogy
surface is proportional to the amount of net charge enclosed within that surface. to Gauss’s law A net outward flow
of water indicates a source of water
An everyday analogy illustrated in 䉴 Fig. 15.22 may help you understand this inside closed surface 1. A net
principle. If you surround a lawn sprinkler with an imaginary surface (surface 1), inward flow of water indicates a
water drain inside closed surface 2.
you find that there is a net flow of water out through that surface—because inside
is a “source” of water (disregarding the pipe’s bringing water into the sprinkler).
In an analogous way, a net outward-pointing electric field indicates the presence
of a net positive charge inside the surface, because positive charges are “sources”
of the electric field. Similarly, a puddle would form inside our imaginary surface 2
because there would be a net inward water flow through the surface. The follow-
ing Example illustrates the power of Gauss’s law in its qualitative form.
EXAMPLE 15.11 Charged Conductors Revisited: Gauss’s Law

A net charge Q is placed on a conductor of arbitrary shape Gaussian 䉳 FIGURE 15.23
(䉴 Fig. 15.23). Use the qualitative version of Gauss’s law to surface Gauss’s law: Excess
prove that all the charge must lie on the conductor’s surface Conductor charge on a conductor
under electrostatic conditions. surface See Conceptual
Example 15.11.
REASONING AND ANSWER. Because the situation is static
equilibrium, there can be no electric field inside the volume of
the conductor; otherwise, the almost-free electrons would
move around. Take a Gaussian surface that follows the shape E=0
of the conductor, but is just barely inside the actual surface.
Because there are no electric field lines inside the conductor,
there are also no electric field lines passing through our imag-
inary surface. Thus zero electric field lines penetrate the
Gaussian surface. But by Gauss’s law, the net number of field
lines is proportional to the amount of charge inside the sur- Because our surface can be as close to the conductor surface
face. Therefore, there must be no net charge within the as wanted, it follows that the excess charge, if it cannot be
surface. inside the volume of the conductor, must be on the surface.
F O L L O W - U P E X E R C I S E . In this Example, if the net charge on the conductor is negative, what is the sign of the net number of
lines through a Gaussian surface that completely encloses the conductor? Explain your reasoning.
DID YOU LEARN?

➥ The net number of field lines penetrating a Gaussian surface is proportional to the
net charge inside that surface.
➥ For a net positive enclosed charge, there is a positive (net outward) field line count
and a negative (net inward) count if that charge is negative.
➥ The net number of field lines is positive (meaning outward) if the net enclosed charge
is positive.
*Strictly speaking, this is Gauss’s law for electric fields. There is also a version of Gauss’s law for
magnetic fields, which will not be discussed.
PULLING IT TOGETHER Deflecting Electrons with Electric Fields

Two closely spaced horizontal parallel plates are charged electric field must be vertically upward, and the correct
equally but oppositely, creating a field of 1.15 * 103 N>C. answer is (3). (Recall that the field direction is the force direc-
When an electron is fired horizontally to the right into the tion on an imaginary positive charge, but an electron has a
field, it deflects downward. (a) What is the direction of the negative charge.)
field due to the plates: (1) horizontally to the right, (2) hori- QUANTITATIVE REASONING AND SOLUTION. Since the field is
zontally to the left, (3) vertically upward, or (4) vertically upward and constant, the force on the electron is constant and
downward? (b) If the deflection is 2.15 cm after the electron downward. Thus two-dimensional projectile motion equations
has traveled 3.50 cm horizontally, what was its initial speed? can be used, where the acceleration due to gravity, g, is
(Neglect air resistance and gravity.) replaced by the electron’s acceleration. To find this acceleration,
C O N C E P T U A L R E A S O N I N G . Since the electron deflects down- Newton’s second law is required. Note that the vertical deflec-
ward, it feels a downward electric force. Therefore, the plate’s tion is downward and will be designated by a minus sign.
Given: E = 1.15 * 103 N>C Find: vo (initial electron speed)

ƒ qe ƒ = e = 1.60 * 10-19 C (from Table 15.1)
me = 9.11 * 10-31 kg (from back inside cover)
x = 3.50 cm = 0.035 m (horizontal distance into the field)
y = - 2.15 cm = - 0.0215 m (vertical deflection in the field)
Let the origin of our x–y coordinate system be where the electron enters the field. Then, using the modified projectile equations in two
dimensions, we have x = vo t and y = - 12 at2, where t is the time the electron is in the field. The first equation can be solved for t.
ax 2
Then t can be substituted into the second, solving for the initial speed. You should be able to show that the result is vo = - .
A 2y
We also know the only force on the electron is the electric force; thus, its acceleration is
Fe eE 11.60 * 10-19 C211.15 * 103 N>C2
a = = = = 2.02 * 1014 m>s2
me me 9.11 * 10-31 kg
and the speed is
ax 2 12.02 * 1014 m>s 2210.035 m22
vo = - = - = 2.40 * 106 m>s
C 2y C 21- 0.0215 m2
■ The law of charges, or charge–force law, states that like ■ Insulators are materials that do not easily gain, lose, or con-
charges repel and opposite charges attract. duct electric charge.
■ Electrostatic charging involves processes that enable an
object to gain a net charge. Among these processes are
+ + – –
charging by friction, contact (conduction), and induction.
+ + – –
+ + Glass – – Rubber
rods rods
– – – –
+ + +
–
–
++
++
■ Charge conservation means that the net charge of an iso-

lated system remains constant.
■ Conductors are materials that conduct electric charge readily
because their atoms have one or more loosely bound electrons.
E
■ The electric polarization of an object involves creating sep-
arate and equal amounts of positive and negative charge in
different locations on that object.
–
+ + + + + + + + +
+
–
+
–
+
++ Negatively –– Positively
++ charged rod –– charged rod
■ Electric field lines are a visualization of the electric field.

The line spacing is inversely related to the field strength,
and tangents to the lines give the field direction.
The closer together
the lines of force,
■ Coulomb’s law expresses the magnitude of the force the stronger the
between two point charges: field.
kq1 q2
Fe = (two point charges) (15.2)
r2
where k L 9.00 * 109 N # m2>C2. +
kq 1q 2 q1 q2 kq 1q 2
F12 = F21 =
r2 r2
r
■ Under electrostatic conditions, the electric fields associated

■ The electric field is a vector field that describes how
with conductors have the following properties:
charges modify the space around them. It is defined as the
electric force per unit positive charge, or The electric field is zero inside a charged conductor.
B Excess charge on a conductor resides entirely on its surface.
B
Fon q+
E = (15.3) The electric field near the surface of a charged conductor is
q+ perpendicular to that surface.
E Excess charge on the surface of a conductor is most dense at
⫹⫹ ⫹ ⫹
locations of highest surface curvature.
⫹ ⫹ ⫹⫹ q⫹
⫹ ⫹ ⫹ ⫹ ⫹⫹ ⫹ The electric field near the surface of a charged conductor is
⫹ ⫹ ⫹⫹
⫺ ⫹ ⫹
⫺ ⫺ ⫺⫺ ⫹ greatest at locations of highest surface curvature.
⫺ ⫺⫺ ⫺ ⫺ ⫹⫹ ⫹
⫺ ⫺ ⫺ ⫹ ⫹
⫺ ⫺⫺
⫺⫺⫺⫺⫺
■ According to the superposition principle for electric

fields, the (net) electric field at any location due to a configu-
E=0
ration of charges is the vector sum of the individual electric
fields from individual charges of that configuration.

15.1 ELECTRIC CHARGE 15.2 ELECTROSTATIC CHARGING

1. A combination of two electrons and three protons would 4. A rubber rod is rubbed with fur. The fur is then quickly
have a net charge of (a) + 1, (b) - 1, (c) +1.6 * 10-19 C, brought near the bulb of an uncharged electroscope. The
(d) -1.6 * 10-19 C. sign of the charge on the leaves of the electroscope is
2. An electron is just above a fixed proton. The direction of (a) positive, (b) negative, (c) zero.
the force on the proton is (a) up, (b) down, (c) zero. 5. A stream of water is deflected toward a nearby electri-
3. In Multiple Choice Question 2, which one feels the cally charged object that is brought close to it. The sign
greater magnitude force: (a) the electron, (b) the proton, of the charge on the object (a) is positive, (b) is negative,
or (c) both feel the same? (c) is zero, (d) can’t be determined by the data given.
6. A balloon is charged and then clings to a wall. The sign 15.5 CONDUCTORS AND ELECTRIC
of the charge on the balloon (a) is positive, (b) is nega- FIELDS
tive, (c) is zero, (d) can’t be determined by the data
given. 16. In electrostatic equilibrium, is the electric field just below
the surface of a charged conductor (a) the same value as
the field just above the surface, (b) zero, (c) dependent
15.3 ELECTRIC FORCE on the amount of charge on the conductor, or (d) given
by kq>R2?
7. How does the magnitude of the electric force between
two point charges change as the distance between them 17. An uncharged thin metal slab is placed in an external
is increased? The force (a) decreases, (b) increases, electric field that points horizontally to the left. What is
(c) stays the same. the electric field inside the slab: (a) zero, (b) the same
8. Compared with the electric force, the gravitational force value as the original external field but oppositely
between two protons is (a) about the same, (b) somewhat directed, (c) less than the original external field value but
larger, (c) very much larger, (d) very much smaller. not zero, or (d) it depends on the magnitude of the exter-
9. If the distance between two charged particles is tripled, nal field?
what happens to the magnitude of the electric force each 18. The direction of the electric field at the surface of a
exerts on the other: (a) it stays the same, (b) it is reduced charged conductor under electrostatic conditions (a) is
to one-third its original value, or (c) it is reduced to one- parallel to the surface, (b) is perpendicular to the surface,
ninth its original value? (c) is at a 45° angle to the surface, or (d) depends on the
10. In Multiple Choice Question 9, if you wanted to change the charge on the conductor.
amount of charge on each of the particles by the same
amount so that the force between them went back to its
original value, what would you do: (a) increase each charge *15.6 GAUSS’S LAW FOR ELECTRIC
by three times, (b) increase each charge by nine times, or FIELDS: A QUALITATIVE APPROACH
(c) decrease each charge to one-third its original value?
19. A Gaussian surface surrounds an object with a net
charge of -5.0 mC. Which of the following is true:
(a) more electric field lines will point outward than
15.4 ELECTRIC FIELD
inward; (b) more electric field lines will point inward
11. How is the magnitude of the electric field due to a point than outward; (c) the net number of field lines through
charge reduced when the distance from that charge is the surface is zero; or (d) there must be only field lines
tripled: (a) It stays the same, (b) it is reduced to one-third passing inward through the surface?
of its original value, (c) it is reduced to one-ninth of its 20. What can you say about the net number of electric field
original value, or (d) it is reduced to one-twenty-seventh lines passing through a Gaussian surface located com-
of its original value? pletely within the region between a set of oppositely
12. The SI units of electric field are (a) C, (b) N>C, (c) N, (d) J. charged parallel plates: (a) the net number of field lines
13. At a point in space, an electric force acts vertically upward points outward; (b) the net number of field lines points
on an electron. The direction of the electric field at that inward. (c) the net number is zero; or (d) the net number
point is (a) down, (b) up, (c) zero, (d) undetermined. depends on the amount of charge on each plate.
14. Two electrons are placed on the (vertical) y-axis, one at 21. Two concentric spherical surfaces enclose a charged par-
y = + 20 cm and the other at y = - 20 cm. What is the ticle. The radius of the outer sphere is twice that of the
direction of the electric field at the location y = 0 cm, inner one. Which sphere will have more electric field
x = + 40 cm: (a) right, (b) left, (c) up, or (d) down? lines passing through its surface: (a) the larger one,
15. In the previous question, what is the field direction at (b) the smaller one, (c) both spheres would have the
the location y = + 40 cm, x = 0 cm: (a) right, (b) left, same number of field lines passing through them, or
(c) up, or (d) down? (d) the answer depends on the charge on the particle.
15.1 ELECTRIC CHARGE 15.2 ELECTROSTATIC CHARGING

1. If historically the charge on the electron were designated 4. Fuel trucks often have metal chains reaching from their
as positive and the charge on the proton as negative, frames to the ground. Why is this important?
what do you think would be the overall effect on the
physical universe as we know it? 5. Is there an overall gain or loss of electrons by an object
when it is electrically polarized? Explain.
2. An electrically neutral object can be given a net charge
by several means. Does this violate the law of conserva- 6. Explain carefully the steps you would use to create an
tion of charge? Explain. electroscope that is positively charged by induction.
3. If a neutral piece of metal becomes negatively charged, After you are done, suppose a charged object was
does its mass increase or decrease? What if it becomes brought near the top of the electroscope and the leaves
positively charged? collapsed. What is the sign of the charge on this object?
7. Suppose the negatively charged balloon in Figure 15.7c 17. (a) Could the electric field due to two identical negative
was replaced by a positively charged glass rod. Which charges ever be zero at some location(s) nearby? Explain.
way would the water stream bend now, if at all? Explain If your answer is yes, describe and sketch the situation.
your reasoning. (b) How would your answer change if the charges were
equal but oppositely charged? Explain.
15.3 ELECTRIC FORCE 18. A large square (finite size) flat plate is uniformly posi-
tively charged and in the horizontal plane. Determine
8. Two point charges initially exert an electric force of mag- the direction of the electric field at the following loca-
nitude F on one another. Suppose the charge of one was tions: (a) just above the center of the plate, (b) just off any
doubled and that of the other was tripled. What would edge of the plate and in the same horizontal plane as the
be the new force between them in terms of F? Explain plate, and (c) a vertical distance above any edge of the
your reasoning. plate on the same order of magnitude as the length of
9. Two nearby electrons would fly apart if released. How one side of the plate.
could you prevent this by placing a single charge in their 19. At a distance much larger than the dimensions of an
neighborhood? Explain clearly what the sign of the object having a net positive charge, what must its electric
charge and its location would have to be. field line pattern approximate? Explain. [Hint: When any
10. Two point charges are initially separated by a distance d. object is viewed from a large distance, it will appear geo-
Suppose the charge of one is increased by twenty seven metrically as what?]
times while the charge of the other is reduced to one-
third its initial value. What would their separation dis- 15.5 CONDUCTORS AND ELECTRIC
tance have to be changed to in order to keep the force
FIELDS
between them the same? (Your answer should be
expressed in terms of d.) Explain your reasoning. 20. Is it safe to stay inside a car during a lightning storm
11. A small charged object is placed and held just above the (䉲 Fig. 15.25)? Explain.
positive end of an electric dipole. The dipole starts to
accelerate downward when released. (a) What is the sign
of the charge on the object? (b) What would happen to
the dipole if this same charged object were held just
below the negative end of the dipole?
15.4 ELECTRIC FIELD

12. How is the relative magnitude of an electric field at dif-
ferent locations determined from an electric field vector
diagram? 䉱 F I G U R E 1 5 . 2 5 Safe inside a car? See
13. How can the relative magnitudes of an electric field at Conceptual Question 20.
different locations be determined from an electric field 21. Under electrostatic conditions, it is found that the excess
line diagram? charge on a conductor is uniformly spread over its sur-
14. Explain clearly why electric field lines can never cross. face. What is the shape of the surface?
15. A positive charge is inside an isolated metal spherical 22. Tall buildings have lightning rods to protect them from
shell, as shown in 䉲 Fig. 15.24. Describe the electric field lightning strikes. Explain why the rods are pointed in
in the following three regions: between the charge and shape and taller than the buildings.
the inside surface of the shell, inside the shell itself, and 23. Sketch the electric field line pattern that results when a
outside the outer shell surface. What is the sign of the metal slab is placed between a pair of closely spaced,
charge on the two shell surfaces? How would your equal (but oppositely charged) parallel plates. (Assume
answers change if the charge were negative? the slab has the same area as the plates and is oriented in
their plane, but does not touch them.)
䉳 FIGURE 15.24 A 24. Repeat the previous question, but this time insert a solid
point charge inside a thick small metal sphere into the middle of the plate region.
metal spherical shell See
+ *15.6 GAUSS’S LAW FOR ELECTRIC
FIELDS: A QUALITATIVE APPROACH
25. The same Gaussian surface is used to surround two
charged objects separately. The net number of field lines
penetrating the surface is the same in both cases, but the
Metal conductor lines are oppositely directed. What can you say about
the net charges on the two objects?
16. At a certain location, the electric field due to the excess 26. If a net number of electric field lines points outward
charge on the Earth’s surface points downward. What is from a Gaussian surface, does that necessarily mean
the sign of the charge on the Earth’s surface at that loca- there are no negative charges in the interior? Explain
tion? Explain. with an example.
EXERCISES
Integrated Exercises (IE s) are two-part exercises. The first part typically requires a conceptual answer choice based
15.1 ELECTRIC CHARGE 11. ● An electron and a proton are separated by 2.0 nm.
(a) What is the magnitude of the force on the electron?
1. ● What is the net charge of an object that has 1.0 million
(b) What is the net force on the system?
excess electrons?
12. IE ● Two charges originally separated by a certain dis-
2. ● In walking across a carpet, you acquire a net negative
tance are moved farther apart until the force between
charge of 50 mC. How many excess electrons do you have?
them has decreased by a factor of 10. (a) Is the new dis-
3. ● ● An alpha particle is the nucleus of a helium atom
tance (1) less than 10, (2) equal to 10, or (3) greater than
with no electrons. (a) What would be the charge on two 10 times the original distance? Why? (b) If the original
alpha particles? (b) How many electrons would you need distance was 30 cm, how far apart are the charges?
to add to make an alpha particle into a helium atom?
13. ● Two charges are brought together until they are
4. IE ● ● A glass rod rubbed with silk acquires a charge of 100 cm apart, causing the electric force between them to
+ 8.0 * 10-10 C. (a) Is the charge on the silk (1) positive, increase by a factor of exactly 5. What was their initial
(2) zero, or (3) negative? Why? (b) What is the charge on separation distance?
the silk, and how many electrons have been transferred
14. ● The distance between neighboring singly charged
to the silk? (c) How much mass has the glass rod gained
sodium and chlorine ions in crystals of table salt (NaCl)
or lost?
is 2.82 * 10-10 m. What is the attractive electric force
5. IE ● ● A rubber rod rubbed with fur acquires a charge of between the ions?
- 4.8 * 10-9 C. (a) Is the charge on the fur (1) positive,
15. ● ● Two charges, q1 and q2 , are located at the origin and
(2) zero, or (3) negative? Why? (b) What is the charge on the
at (0.50 m, 0), respectively. Where on the x-axis must a
fur, and how much mass is transferred to or from the rod?
third charge, q3 , of arbitrary sign be placed to be in elec-
(c) How much mass has the rubber rod lost or gained?
trostatic equilibrium if (a) q1 and q2 are like charges of
equal magnitude, (b) q1 and q2 are unlike charges of
15.2 ELECTROSTATIC CHARGING equal magnitude, and (c) q1 = + 3.0 mC and
6. ●An initially uncharged electroscope is polarized by q2 = - 7.0 mC?
bringing a negatively charged rubber rod near the bulb. 16. ● ● Two negative point charges are separated by 10.0 cm
If the bulb end of the electroscope acquires a net charge and feel a mutual repulsive force of 3.15 mN. The charge
of + 2.50 pC, how many electrons are on the leaf end? of one is three times that of the other. (a) How much
7. ● An initially neutral electrscope is charged by induc- charge does each have? (b) What would be the force if
tion by bringing near a positively charged object. If the total charge were instead equally distributed on both
3.22 * 108 electrons flow through the ground wire to point charges?
Earth and the ground wire is then removed, what is the 17. ● ● An electron is placed on a line connecting two fixed
net charge on the electrscope? point charges of equal charge but opposite sign. The dis-
tance between the charges is 30.0 cm and the charge of
15.3 ELECTRIC FORCE each is 4.50 pC. (a) Compute the force on the electron at
5.0-cm intervals starting 5.0 cm from the leftmost charge
8. IE ● An electron that is a certain distance from a proton is and ending 5.0 cm from the rightmost charge. (b) Plot
acted on by an electrical force. (a) If the electron were the net force versus electron location using your com-
moved twice that distance away from the proton, would puted values. From the plot, can you make an educated
the electrical force be (1) 2, (2) 21, (3) 4, or (4) 14 times the guess as to where the electron feels the least force?
original force? Why? (b) If the initial electric force is F, and
18. ● ● ● Three charges are located at the corners of an equi-
the electron were moved to one-third the original distance
lateral triangle, as depicted in 䉲 Fig. 15.26. What are the
toward the proton, what would be the new electrical force
magnitude and the direction of the force on q1?
in terms of F?
9. ● Two identical point charges are a fixed distance apart. q1 = +4.0 ␮C 䉳 FIGURE 15.26
By what factor would the magnitude of the electric force Charge triangle See
between them change if (a) one of their charges were Exercises 18, 31, and 32.
doubled and the other were halved, (b) both their
m
20
c
charges were halved, and (c) one charge were halved

cm
20
and the other were left unchanged?

10. ● In a certain organic molecule, the nuclei of two carbon 20 cm
atoms are separated by a distance of 0.25 nm. What is the
q2 = +4.0 ␮C q3 = −4.0 ␮C
magnitude of the electric force between them?
EXERCISES 557
19. ● ● ● Four charges are located at the corners of a square, as 27. ●● What would be the magnitude and the direction of an
illustrated in 䉲 Fig. 15.27. What are the magnitude and the electric field that would just support the weight of a pro-
direction of the force (a) on charge q2 and (b) on charge q4? ton near the surface of the Earth? What about an electron?
q1 = −10 ␮C q2 = −10 ␮C 28. IE ● ● Two charges, -3.0 mC and - 4.0 mC, are located at
( -0.50 m, 0) and (0.50 m, 0), respectively. There is a point
0.10 m on the x-axis between the two charges where the electric
field is zero. (a) Is that point (1) left of the origin, (2) at
0.10 m 0.10 m the origin, or (3) right of the origin? (b) Find the location
of the point where the electric field is zero.
0.10 m
29. ● ● Three charges, +2.5 mC, - 4.8 mC, and - 6.3 mC, are
q4 = +5.0 ␮C q3 = +5.0 ␮C located at ( -0.20 m, 0.15 m), (0.50 m, - 0.35 m), and
( -0.42 m, -0.32 m) respectively. What is the electric field
䉱 F I G U R E 1 5 . 2 7 Charge square. See Exercises 19, 33, and 37. at the origin?
30. ● ● Two charges of + 4.0 mC and + 9.0 mC are 30 cm
20. ●●● Two 0.10-g pith balls are suspended from the same apart. Where on the line joining the charges is the electric
point by threads 30 cm long. (Pith is a light insulating field zero?
material once used to make helmets worn in tropical cli-
31. ● ● What is the electric field at the center of the triangle
mates.) When the balls are given equal charges, they
in Fig. 15.26?
come to rest 18 cm apart, as shown in 䉲 Fig. 15.28. What
32. ● ● Compute the electric field at a point midway
is the magnitude of the charge on each ball? (Neglect the
mass of the thread.) between charges q1 and q2 in Fig. 15.26.
33. ● ● What is the electric field at the center of the square in
Fig. 15.27?
34. ● ● A particle with a mass of 2.0 * 10 kg and a charge of
-5
θ +2.0 mC is released in a (parallel plate) uniform horizontal

θ electric field of 12 N>C. (a) How far horizontally does the
θ θ particle travel in 0.50 s? (b) What is the horizontal compo-
T T nent of its velocity at that point? (c) If the plates are 5.0 cm
Fe Fe on each side, how much charge is on each?
q q
35. ● ● Two very large parallel plates are oppositely and uni-
formly charged. If the field between the plates is

1.7 * 106 N>C, (a) how dense is the charge on each plate
w w
(in mC>m2)? (b) How much total charge is on each plate
䉱 F I G U R E 1 5 . 2 8 Repelling pith balls See Exercise 20. if they are 15.0 cm on a side?
36. ● ● ● Two square, oppositely charged conducting plates
measure 20 cm on each side. The plates are close

together and parallel to each other. They each have a
15.4 ELECTRIC FIELD
total charge of +4.0 nC and - 4.0 nC, respectively.
21. IE ● (a) If the distance from a charge is doubled, is the (a) What is the electric field between the plates? (b) What
magnitude of the electric field (1) increased, force is exerted on an electron located between the
(2) decreased, or (3) the same compared to the initial plates? (c) What would be the electron’s acceleration if it
value? (b) If the original electric field due to a charge is were released from rest?
1.0 * 10-4 N>C, what is the magnitude of the new elec- 37. ● ● ● Compute the electric field at a point 4.0 cm from q2
tric field at twice the distance from the charge? along a line running toward q3 in Fig. 15.27.
22. ● An electron is acted on by an electric force of 38. ● ● ● Two equal and opposite point charges form a
3.2 * 10-14 N. What is the magnitude of the electric field dipole, as shown in 䉲 Fig. 15.29. (a) Add the electric fields
at the electron’s location? due to each end at point P, thus graphically determining
23. ● An electron is acted on by two electric forces, one of the direction of the field there. (b) Derive a symbolic
2.7 * 10-14 N acting upward and a second of expression for the magnitude of the electric field at point
3.8 * 10-14 N acting to the right. What is the magnitude P, in terms of k, q, d, and x. (c) If point P is very far away,
of the electric field at the electron’s location? use the exact result to show that E L kqd>x 3. (d) Why is it
24. ● What are the magnitude and direction of the electric field an inverse-cube falloff instead of inverse-square? Explain
at a point 0.75 cm away from a point charge of +2.0 pC? why it is not an inverse square falloff.
25. ● At what distance from a proton is the magnitude of its + +q
electric field 1.0 * 105 N>C?
26. IE ● ● Two fixed charges, -4.0 mC and -5.0 mC, are sep- P
arated by a certain distance. (a) Is the net electric field at d
a location halfway between the two charges (1) directed x
toward the - 4.0 mC charge, (2) zero, or (3) directed
toward the - 5.0 mC charge? Why? (b) If the charges are – –q
separated by 20 cm, calculate the magnitude of the net
electric field halfway between the charges. 䉱 F I G U R E 1 5 . 2 9 Electric dipole field See Exercise 38.
15.5 CONDUCTORS AND ELECTRIC 44. ●● An electrically neutral thin, square metal slab, measur-
FIELDS ing 5.00 cm on a side, is placed in a uniform external field
that is perpendicular to its square area. (a) If the top of the
39. IE ● A solid conducting sphere is surrounded by a thick, slab becomes negatively charged, what is the direction of
spherical conducting shell. Assume that a total charge the external field? (b) If the external field strength is
+ Q is placed at the center of the sphere and released. 1250 N>C, what are the direction and strength of the field
(a) After equilibrium is reached, the inner surface of the that is generated by the charges induced on the slab?
shell will have (1) negative, (2) zero, (3) positive charge. (c) What is the total charge on the negative side of the slab?
(b) In terms of Q, how much charge is on the interior of
the sphere? (c) The surface of the sphere? (d) The inner
surface of the shell? (e) The outer surface of the shell? *15.6 GAUSS’S LAW FOR ELECTRIC
40. ● In Exercise 39, what is the electric field direction (a) in FIELDS: A QUALITATIVE APPROACH
the interior of the solid sphere, (b) between the sphere and 45. ● Suppose a Gaussian surface encloses both a positive
the shell, (c) inside the shell, and (d) outside the shell? point charge that has six field lines leaving it and a nega-
41. ●● In Exercise 39, write expressions for the electric field tive point charge with twice the magnitude of charge of
magnitude (a) in the interior of the solid sphere, the positive one. What is the net number of field lines
(b) between the sphere and the shell, (c) inside the shell, passing through the Gaussian surface?
and (d) outside the shell. Your answer should be in 46. IE ● ● A Gaussian surface has sixteen field lines leaving
terms of Q, r (the distance from the center of the it when it surrounds a point charge of + 10.0 mC and
sphere), and k. seventy-five field lines entering it when it surrounds an
42. ●● A flat, triangular piece of metal with rounded corners unknown point charge. (a) The magnitude of the
has a net positive charge on it. Sketch the charge distrib- unknown charge is (1) greater than 10.0 mC, (2) equal to
ution on the surface and the electric field lines near the 10.0 mC, (3) less than 10.0 mC. Why? (b) What is the
surface of the metal (including their direction). unknown charge?
43. ●● Approximate a metal needle as a long cylinder with a 47. ●● If ten field lines leave a Gaussian surface when it com-
very pointed, but slightly rounded, end. Sketch the pletely surrounds the positive end of an electric dipole,
charge distribution and outside electric field lines if the what would the count be if the surface surrounded (a) just
needle has an excess of electrons on it. the other end? (b) What if it surrounded both ends?
48. On average, the electron and proton in a hydrogen atom 49. A negatively charged pith ball (mass 6.00 * 10-3 g,
are separated by a distance of 5.3 * 10-11 m charge -1.50 nC) is suspended vertically from a light
(䉲 Fig. 15.30). Assuming the orbit of the electron to be cir- nonconducting string of length 15.5 cm. This apparatus
cular, (a) what is the electric force on the electron? is then placed in a horizontal uniform electric field. After
(b) What is the electron’s orbital speed? (c) What is the being released, the pith ball comes to a stable position at
magnitude of the electron’s centripetal acceleration in an angle of 12.3° to the left of the vertical. (a) What is the
units of g? direction of the external electric field? (b) Determine the
magnitude of the electric field.
v 50. A positively charged particle with a charge of 9.35 pC is
suspended in equilibrium in the electric field between
e− two oppositely charged horizontal parallel plates. The
square plates each have a charge of 5.50 * 10-5 C, are
separated by 6.25 mm, and have an edge length of
11.0 cm. (a) Which plate must be positively charged?
p+ (b) Determine the mass of the particle.
r = 5.3 × 10−11 m 51. An electron starts from one plate of a charged closely
spaced (vertical) parallel plate arrangement with a veloc-
ity of 1.63 * 104 m>s to the right. Its speed on reaching
the other plate, 2.10 cm away, is 4.15 * 104 m>s. (a) What
type of charge is on each plate? (b) What is the direction
䉱 F I G U R E 1 5 . 3 0 Hydrogen atom See Exercise 48. of the electric field between the plates? (c) If the plates
are square with an edge length of 25.4 cm, determine the
charge on each.
52. An electron in a computer monitor enters midway positive end. Assuming an electric dipole is free to move
between two parallel oppositely charged plates, as and rotate and starts from rest, (a) use a sketch to show
B
shown in 䉲 Fig. 15.31. The initial speed of the electron is that if it is placed in a uniform external field E, it will
B
6.15 * 107 m>s and its vertical deflection (d) is 4.70 mm. B
begin to rotate so that p tries to “line up” with E.
(a) What is the magnitude of the electric field between (b) Show that the magnitude of the torque exerted on the
the plates? (b) Determine the magnitude of the surface dipole about its center is given by t = pE sin u, where u
B
charge density on the plates in C>m2. is the angle between p B
and E. (c) What is the net force on
this dipole? (d) For what angle(s) is the torque at its max-
– – – – – imum? What about at its minimum?
54. A proton is fired into a uniform electric field, opposite to
e−
1.0 cm the direction of the field. The proton’s speed upon enter-
d ⫽ 4.70 mm ing the field is 3.15 * 105 m>s, and it comes to rest
+ + + + + 5.25 cm after entering the field. (a) What is the electric
field strength? (b) What is the proton’s velocity when it
10 cm
is only 3.50 cm into the field? [Hint: There is more than
one answer. Why?]
䉱 F I G U R E 1 5 . 3 1 Electron in a computer monitor See
Exercise 52. 55. An electric dipole has charges of ⫾4.55 pC that are sepa-
rated by 6.00 cm. The dipole lies on the x-axis and its
center is at the origin. Located at y = + 4.00 cm is a point
53. For an electric dipole, the product qd is called the dipole charge carrying a charge of - 2.50 pC. (a) Determine the
moment and is given the symbol p. Here d is the distance net force on the dipole and its initial center of mass accel-
between poles and q is the magnitude of the charge on eration (including direction) if it has a total mass of
B
either end. The dipole moment vector p has a mgnitude 7.25 ng. (b) Determine the torque on the dipole about its
of qd and, by convention, points from the negative to the center of mass, including direction.
Electric Potential, Energy,
16 and Capacitance
16.1 Electric potential energy
and electric potential
difference (561)
■ voltage
16.2 Equipotential surfaces

and the electric field (568)
16.3 Capacitance (575)

■ charge and
energy storage
16.4 Dielectrics (579)

■ dielectric constant PHYSICS FACTS
16.5 Capacitors in series

and in parallel (582)
✦ The unit of electrical capacitance, the
farad, is named for the British scientist
Michael Faraday (1791–1867).
Although he had little formal educa-
T he young people in the chap-
ter-opening photo are clearly
experiencing an electrical effect as
tion, he discovered electromagnetic
■ equivalent capacitance induction, the principle behind elec- they are electrically charged to
tric power plants.
several thousand volts. Household
✦ In electrochemistry, an important
quantity of charge called the faraday. circuits operate at 120 volts and
The name honors Michael Faraday for
his experiments showing that 1 fara-
can give a potentially dangerous
day of charge deposits 1 mole of silver shock. Yet they don’t seem to be
on a negatively charged cathode.
✦ Count Alessandro Volta was born in having a problem. What’s going
Como, Italy in 1745. Because he did
on? You will find the explanation,
not speak until age 4, his family was
convinced he was mentally retarded. along with explanations of other
However, in 1778 he was the first to
isolate methane and later con-
electrical phenomena, in this and
structed the first electric battery. The the two following chapters. In this
unit of electromotive force, the volt
(V), is named in his honor. chapter, the concept of electric
✦ Electric eels can kill or stun prey by
potential will be introduced and its
producing voltages up to 650 volts.
Other electric fish, such as the properties and usefulness examined.
elephantnose, generate only about
1 volt, useful for location but not
Although this chapter concen-
killing. trates on the study of fundamental
16.1 ELECTRIC POTENTIAL ENERGY AND ELECTRIC POTENTIAL DIFFERENCE 561
electrical concepts such as voltage and capacitance, there are discussions involving
practical applications. For example, your dentist’s X-ray machine works by using high
voltage to accelerate electrons. Heart defibrillators use capacitors to temporarily
store the electrical energy required to stimulate the heart into its correct rhythm.
Capacitors are used to store the energy that triggers the flash unit in your camera.
The body’s nervous system, its communication network, is capable of sending thou-
sands of electrical voltage signals per second that shuttle back and forth along
“cables” called nerves. These signals are generated by chemical activity. The body
uses them to do many processes we take for granted, such as muscle movement,
thought processes, vision, and hearing. In the following chapters, practical uses of
electricity, such as electric appliances, computers, medical instruments, electrical
energy distribution systems, and household wiring, will be presented.
16.1 Electric Potential Energy and Electric Potential

Difference
➥ What is the difference between the terms electric potential energy difference and
electric potential difference?
➥ What is the SI unit for electric potential difference?
➥ When released in an electric field, negative electric charges tend to gain kinetic
energy as they experience what kind of change in electric potential?
In Chapter 15, electrical effects were analyzed in terms of electric field vectors and
electric field lines (of force). Recall that the study of mechanics in early chapters
also began with a vector approach, using Newton’s laws, free-body diagrams, and
forces (vectors). A search for a simpler approach to understanding mechanics led to
a scalar approach, in which scalar quantities such as work, kinetic energy, and
potential energy became extremely useful. Energy methods using these quantities
were employed to solve problems that can be much more difficult using the vector
(force) approach. It turns out to be extremely useful, both conceptually and for
problem solving, to extend energy methods to the study of electric fields.
ELECTRIC POTENTIAL ENERGY

To investigate electric potential energy, let’s start with one of the simplest electric
field patterns: the field between two large, oppositely charged parallel plates. As was
learned in Chapter 15, near the center of the plates the field is uniform in magnitude
and direction (䉲 Fig. 16.1a). Suppose a small positive charge q+ is moved at constant
+ + + + + B B
+
F Fg
h g= m
d E = qe
+
q+
mA
+
– – – – – A
(a) (b)
䉱 F I G U R E 1 6 . 1 Changes in potential energy in uniform electric and gravitational fields

(a) Moving a positive charge q+ against the electric field requires positive work and involves
an increase in electric potential energy. (b) Moving a mass m against the gravitational field
requires positive work and involves an increase in gravitational potential energy.
562 16 ELECTRIC POTENTIAL, ENERGY, AND CAPACITANCE
B
velocity against the electric field E in a straight line from the negative plate (A) to the
positive one (B). To do this an external force 1Fext2 with the same magnitude as the
B
electric force is required (why?), and so Fext = q+ E. The work done by this external
force is positive, because the force and displacement are in the same direction. Thus
the work done by the external force is Wext = Fext1cos 0°2d = q+ Ed.
Suppose the positive charge is now released from the positive plate. It will
accelerate toward the negative plate, gaining kinetic energy. This kinetic energy is a
result of the work done on the charge. At B, the initial energy is not kinetic, and
thus it must be some form of potential energy. The conclusion is that in moving
from A to B, the charge’s electric potential energy, Ue, has increased 1UB 7 UA2
by an amount equal to the external work done on it. So this change in electric
potential energy is
¢Ue = UB - UA = q+ Ed
The gravitational analogy to the parallel plate electric field is the gravita-
tional field near the Earth’s surface, where it is uniform. When an object is
raised a vertical distance h at constant velocity, the change in its potential
energy is positive 1UB 7 UA2 and equal to the work done by the external (lift-
ing) force. If this is done at constant velocity, then the external force must equal
the object’s weight, or Fext = w = mg (Fig. 16.1b). Thus the increase in gravita-
tional potential energy is,
¢Ug = UB - UA = Fext h = mgh
(Note: Different distance symbols (h and d) are used to distinguish between electri-
cal and gravitational situations, respectively.)
ELECTRIC POTENTIAL DIFFERENCE

Recall that defining the electric field as the electric force per unit positive test charge
eliminated the dependence on the test charge. Thus knowing the electric field, the
force on any charge placed in it could be determined from Fe = q+ E. Similarly, the
electric potential difference (¢V ) between any two points in space is defined as
the change in potential energy per unit positive test charge:
¢Ue
¢V = (electric potential difference) (16.1)
q+
SI unit of electric potential difference:
joule>coulomb (J>C) or volt (V)
The SI unit of electric potential difference is the joule per coulomb 1J>C2. This unit
¢V is independent of is named the volt (V) in honor of Alessandro Volta (1745–1827), an Italian scientist
the reference point who constructed the first battery (Section 17.1), and 1 V = 1 J>C. Potential differ-
ence is commonly called voltage, and the symbol for potential difference is rou-
VB = 300 V B VB´ = 1300 V tinely changed from ¢V to just V, as is done later in the chapter.
Notice a crucial point: Electric potential difference, although based on electric
potential energy difference, is not the same. Electric potential difference is
defined as electric potential energy difference per unit charge, and therefore does
not depend on the amount of charge moved. Like the electric field rather than
∆V = VB − VA ∆V´ = V´B − V´A the electric force, electric potential difference is more useful than electric poten-
= 200 V = 200 V tial energy difference. This is because, once ¢V is known, ¢Ue can then be deter-
mined for any amount of charge moved. To illustrate this idea, let’s calculate the
potential difference associated with the uniform field between two parallel
plates:
VA = 100 V A VA´ = 1100 V ¢Ue q+ Ed potential difference

¢V = = = Ed (16.2)
∆V = ∆V´ q+ q+ (parallel plates only)
Notice that the amount of charge moved, q+ , cancels out. Therefore the potential dif-
ference ¢V depends on only the characteristics of the charged plates—that is, the
field produced (E) and the separation (d). This result can be described as follows:
For a pair of oppositely charged parallel plates, the positively charged plate is at a
higher electric potential than the negatively charged plate by an amount equal to ¢V.
+
–
Notice that electric potential difference is defined without defining electric
potential itself (V). Although this may seem backwards, there is a good reason for
+
–
it. Of the two, electric potential difference is the physically meaningful quantity; + B + A
–
that is, it is the quantity actually measured. (Electric potential differences, or volt- Fext qp
ages, are measured with voltmeters; see Section 18.4.) The electric potential V, in
+
–
contrast, isn’t definable in an absolute way—it depends entirely on the choice of a +
–
reference point. This means that an arbitrary constant can be added to, or sub- +
tracted from, potential values (V), changing them. However, this has no affect on –
the physically meaningful quantity of potential difference. +
–
This idea was encountered during the study of potential energy associated with
d
springs and gravitation (Sections 5.2 and 5.4). Recall that only changes in potential
(a)
energies were important. Specific values of potential energy could be determined,
but only after the zero reference point was defined. For example, in the case of
gravity, gravitational potential energy is sometimes chosen to be zero at the +
Earth’s surface. However, it is just as correct (and sometimes more convenient) to –
B A
define the zero point to be located at an infinite distance from Earth (Section 7.5). +
–
These ideas also hold for electric potential energy and potential. The electric ++ v
potential may be chosen as zero at the negative plate of a pair of parallel plates. –
qp
However, it is sometimes convenient to locate the zero value at infinity, as will be +
–
seen in the case of a point charge. Either way, differences are unaffected. For a visu- +
alization of this, refer to Learn by Drawing 16.1, ¢V is Independent of the Refer- –
ence Point on page 562. This shows that with a certain choice of zero electric +
–
potential, point A is at a potential of + 100 V and B is at a potential of + 300 V. With +
–
a different zero reference choice, the potential at A might be +1100 V, in which
d
case, the electric potential at B would then be +1300 V. Regardless of the choice of
zero potential, B will always be 200 V higher in potential than A. (b)
Example 16.1 illustrates the relationship between electric potential energy and
electric potential.
+
–
+
EXAMPLE 16.1 Energy Methods in Moving a Proton: Potential –
B A
Energy versus Potential ++ + –
–
+
Imagine moving a proton from the negative plate to the positive plate of a parallel plate
arrangement (䉴 Fig. 16.2a). The plates are 1.50 cm apart, and the field is uniform with a mag- +
–
nitude of 1500 N>C. (a) What is the change in electric potential energy? (b) What is the elec-
++
tric potential difference (voltage) between the plates? (c) If the proton is released from rest at –
the positive plate (Fig. 16.2b), what speed will it have just before it hits the negative plate? + B
–
THINKING IT THROUGH. (a) The change in potential energy can be computed from the d
work required to move the proton. (b) The electric potential difference between the (c)
plates can then be found by dividing the work by the charge moved. (c) When the pro-
ton is released, its electric potential energy is converted into kinetic energy. Since the 䉱 F I G U R E 1 6 . 2 Accelerating a
proton’s mass is known, its speed can be calculated. charge (a) Moving a proton from
the negative to the positive plate
SOLUTION. The magnitude of the electric field, E, is given. Because a proton is increases the proton’s potential
involved, its mass and charge can be found in Table 15.1. energy. (See Example 16.1.)
(b) When it is released from the pos-
Given: E = 1500 N>C Find: (a) ¢Ue (potential energy change) itive plate, the proton accelerates
qp = + 1.60 * 10-19 C (b) ¢V (potential difference between toward the negative plate, gaining
mp = 1.67 * 10-27 kg plates) kinetic energy at the expense of
d = 1.50 cm = 1.50 * 10-2 m (c) v (speed of released proton just electric potential energy. (c) The
before it reaches negative plate) work done to move a proton
between any two points in an elec-
(continued on next page) tric field, such as A and B or A and
B¿ , is independent of the path.
(a) The electric potential energy increases, because positive work is done to move the
proton against the field, toward the positive plate:
¢Ue = qp Ed = 1+1.60 * 10-19 C211500 N>C211.50 * 10-2 m2
= + 3.60 * 10-18 J
(b) The potential difference, or voltage, is the potential energy change per unit charge
(defined by Eq. 16.1):
¢Ue + 3.60 * 10-18 J
¢V = = = + 22.5 V
qp +1.60 * 10-19 C
One would then say that the positive plate is 22.5 V higher in electric potential than the
negative one.
(c) The total energy of the proton is constant; therefore, ¢K + ¢Ue = 0. The proton has
no initial kinetic energy 1Ko = 02. Hence, ¢K = K - Ko = K. From this, the speed of
the proton can be calculated:
¢K = K = - ¢Ue
or
1 2
2 mp v = - ¢Ue
But on its return to the negative plate, the proton’s potential energy change is negative
(why?). Hence ¢Ue = - 3.60 * 10-18 J and its speed is
21- ¢Ue2 23-1 -3.60 * 10-18 J24
v = = = 6.57 * 104 m>s
C mp C 1.67 * 10-27 kg
Notice that even though the kinetic energy gained is very small, the proton acquires a
high speed because its mass is extremely small.
F O L L O W - U P E X E R C I S E . In this Example, what would be your answers if an alpha par-
ticle were moved instead of a proton? (An alpha particle is the nucleus of a helium
atom and has a charge of +2e and a mass approximately four times that of a proton.)
(Answers to all Follow-Up Exercises in Appendix VI in the back of the book.)
The principles in Example 16.1 can be used to show another interesting prop-
erty of electric potential energy (and potential) changes: Both are independent of the
path on which the charged particle is taken. Recall from Section 5.5 that this means
the electrostatic force is a conservative force, conserving total mechanical energy. As
shown in Fig. 16.2c, the work done in moving the proton from A to B is the same,
regardless of the route. The alternative wiggly paths from A to B and A to B¿
require the same work as do the straight-line paths. This is because movement at
right angles to the field requires no work. (Why?)
The gravitational analogy to Example 16.1 is that of raising an object in a uni-
form gravitational field. When the object is raised, gravitational potential energy
increases, because the force of gravity acts downward. However, with electricity
there are two types of charge, and the force between them can be repulsive or
attractive. At this point, the analogy to gravity breaks down.
To understand this failure of the analogy, consider how the discussion in
Example 16.1 would differ if an electron instead of a proton had been moved.
Because an electron is negatively charged, it would be attracted to plate B, and the
external force would have to be opposite the electron’s displacement (to prevent
the electron from accelerating). Thus in the case of moving an electron, the exter-
nal force would do negative work, thereby decreasing the electric potential energy.
Unlike the proton, the electron is attracted to the positive plate (the plate at the
higher electric potential). If allowed to move freely, electrons would “fall” (accel-
erate) toward regions of higher potential. Recall that the proton “fell” (acceler-
ated) toward the region of lower potential. Regardless, both the proton and
electron ended up losing electric potential energy and gaining kinetic energy.
Thus the behavior of charged particles in electric fields can be summarized in
“potential language” as follows:
Positive charges, when released in an electric field, accelerate toward regions of lower
electric potential.
Negative charges,when released in an electric field,accelerate toward regions of higher
electric potential.
Consider the following Example of a medical application involving the creation
of X-rays from fast-moving electrons, accelerated by large electric potential differ-
ences (voltages).
EXAMPLE 16.2 Creating X-Rays: Accelerating Electrons

Modern dental offices use X-ray machines for diagnosing hidden dental problems
(䉴 Fig. 16.3a). In essence, electrons are accelerated through electric potential differences
(voltages) of 25 000 V. When the electrons hit the positive plate, their kinetic energy is
converted into high-energy particles called X-ray photons (Fig. 16.3b). (Photons are par- (a)
ticles of light discussed in Chapter 27.) Suppose a single electron’s kinetic energy is dis-
tributed equally among five X-ray photons. How much energy would one photon
have?
THINKING IT THROUGH. From energy conservation, the kinetic energy gained by one X-rays
electron is equal in magnitude to the electric potential energy it loses. From the kinetic
energy lost by one electron, the energy of one X-ray photon can be calculated.
+ + + +
SOLUTION. The charge of an electron is known (from Table 15.1), and the accelerating
voltage is given.
Given: q = - 1.60 * 10-19 C Find: E (energy of one X-ray photon) −

+
¢V = 2.5 * 104 V v
The electron leaves the negatively charged plate and moves toward the region of high- −
est electric potential (“uphill”). Thus, the change in its electric potential energy is V
¢Ue = q¢V = 1- 1.60 * 10-19 C21 +2.5 * 104 V2 = - 4.0 * 10-15 J −
The gain in kinetic energy comes from this loss in electric potential energy. Because the −
electrons have no appreciable kinetic energy when they start, − − − −
K = ƒ ¢Ue ƒ = 4.0 * 10-15 J

Therefore, if equally shared, one photon will have an energy of
K (b)
E = = 8.0 * 10-16 J
5
䉱 F I G U R E 1 6 . 3 X-ray production
FOLLOW-UP EXERCISE. In this Example, use energy methods to determine the speed of (a) An illustration of a dental X-ray
one electron when it is halfway to the positive plate. machine. (b) A schematic diagram of
the X-ray production. See Example 16.2.
ELECTRIC POTENTIAL DIFFERENCE DUE TO A POINT

CHARGE A
In nonuniform electric fields, the potential difference between two points is deter-
rA
mined by applying the fundamental definition (Eq. 16.1). However, in this case
the field strength (and thus work done) varies, making the calculation beyond the I
HIGHER
scope of this text. The only nonuniform field which will be considered in any detail is LOWER
+
that due to a point charge (䉴 Fig. 16.4). For this situation, the potential difference (volt- POTENTIAL rB POTENTIAL
age) between two points at distances rA and rB from a point charge q is given by: II
B
kq kq electric potential difference VB > VA
¢V = - (16.3) ∆V = VB – VA
rB rA (point charge only)
is positive
In Fig. 16.4, the point charge is positive. Since point B is closer to the charge
than A, the potential difference is positive, that is, VB - VA 7 0, or VB 7 VA. Thus 䉱 F I G U R E 1 6 . 4 Electric field and
potential due to a point charge
B is at a higher potential than A. This is because changes in potential are deter- Electric potential increases as you
mined by visualizing the movement of a positive test charge. Here it takes positive move closer to a positive charge.
work to move such a charge from A to B. Thus, B is at a higher potential than A.
From this it can be seen that electric potential increases as one moves nearer to a
positive charge. Notice also (in Fig. 16.4) that the work done by taking path II is
the same as that for path I. Because the electric force is conservative, the potential
difference is also the same, regardless of path.
Consider what would happen if the central point charge were negative. In this
case, B would be at a lower potential than A because the work required to move a
positive test charge closer would be negative (why?).
Changes in electric potential thus follow these rules:
Electric potential increases when moving nearer to positive charges or farther from
negative charges
and
Electric potential decreases when moving farther from positive charges or nearer to
negative charges.
The electric potential at a very large distance from a point charge is usually chosen
to be zero (as was done for the gravitational case of a point mass in Chapter 7).
With this choice, the electric potential V at a distance r from a point charge is
kq electric potential
V = (point charge only, (16.4)
r
zero at infinity)
Even though this expression is for the electric potential, V, keep in mind that only elec-
tric potential differences 1¢V2 are important, as Integrated Example 16.3 illustrates.
INTEGRATED EXAMPLE 16.3 Describing the Hydrogen Atom: Potential Differences Near a Proton
According to the Bohr model of the hydrogen atom orbit compare: (1) the smaller orbit is at a higher potential,
(Chapter 27), the electron in orbit around the proton can exist (2) the larger is at a higher potential, or (3) they have the same
only in certain sized circular orbits. The smallest orbit has a potential? Explain your reasoning. (b) Verify your answer to
radius of 0.0529 nm, and the next largest has a radius of part (a) by calculating the values of the electric potential at the
0.212 nm. (a) How do the values of electric potential at each locations of the two orbits.
(A) CONCEPTUAL REASONING. The electron orbits in the electric field of a proton, whose charge is positive. Because electric poten-
tial increases with decreasing distance from a positive charge, the answer must be (1).
(B) QUANTITATIVE REASONING AND SOLUTION. The charge on the proton is known, so Eq. 16.4 can be used to find the potential
values. Listing the values,
Given: qp = + 1.60 * 10-19 C Find: V (the value of the electric
r1 = 0.0529 nm = 5.29 * 10-11 m potential for each orbit)
r2 = 0.212 nm = 2.12 * 10-10 m 1Note: 1 nm = 10-9 m2
Applying Equation 16.4 to the smaller orbit,

kqp 19.00 * 109 N # m2>C221 +1.60 * 10-19 C2
V1 = = = + 27.2 V
r1 5.29 * 10-11 m
and for the larger orbit, the result is
kqp 19.00 * 109 N # m2>C221 +1.60 * 10-19 C2
V2 = = = + 6.79 V
r2 2.12 * 10-10 m
F O L L O W - U P E X E R C I S E . In this Example, suppose the electron were moved from the smallest to the next orbit. (a) Has it moved
to a region of higher or lower electrical potential? (b) What would be the change in the electric potential energy of the atom?
ELECTRIC POTENTIAL ENERGY OF VARIOUS CHARGE

CONFIGURATIONS
In Chapter 7 the gravitational potential energy of systems of masses was consid-
ered in some detail. The expressions for electric force and gravitational force are
mathematically similar, and so therefore are those for potential energy, except that
charge takes the place of mass (remember that charge comes in two signs). r12 Very
distant
In the case of two masses, the mutual gravitational potential energy is + + +
q1 q2
negative, because the force is always attractive. For electric potential kq1q2
(U = 0)
energy, the result can be positive or negative, because the electric force can U12 =
r12
be repulsive or attractive.
(a)
For example, consider a positive point charge, q1, fixed in space. Suppose a
second positive charge q2 is brought toward it from a very large distance
(that is, let its initial location r : q ) to a distance r12 (䉴 Fig. 16.5a). In this
− q3
case, the work required is positive (why?). Therefore, this particular system
gains electric potential energy. The potential at a large distance 1Vq2 is, as is r13
usual for point charges and masses, chosen as zero. (The zero point is arbi-
trary.) Thus, from Eq. 16.3, the change in potential energy is r23
q1 +
kq1 kq1 q2
¢Ue = q2 ¢V = q21V1 - Vq2 = q2 ¢ - 0≤ = r12
r12 r12
+
Because Vq is chosen as zero, it follows that ¢Ue = U12 - Uq = U12. q2
With this choice of reference, the electric potential energy of any two- U = U12 + U23 + U13
charge system is
(b)
kq1 q2 mutual electric potential energy

U12 = (16.5) 䉱 F I G U R E 1 6 . 5 Mutual electric
r12 (two-charge system only)
potential energy of point charges
(a) If a positive charge is moved
Notice that for unlike charges the electric potential energy is negative, and for like from a large distance to a distance
charges the value is positive. So if the two charges are of the same sign, when r12 from another positive charge,
released, they will move apart, gaining kinetic energy as they lose potential there is an increase in potential
energy because positive work must
energy. Conversely, it would take positive work to increase the separation of two be done to bring the mutually
opposite charges, such as the proton and the electron, much like stretching a repelling charges closer. (b) For
spring. (See the Follow-Up Exercise in Integrated Example 16.3.) more than two charges, the system’s
Because energy is a scalar, for a configuration of any number of point charges (a electric potential energy is the sum
charge system), the total potential energy (U) is the algebraic sum of the mutual of the mutual potential energies of
each pair.
potential energies of all pairs of charges:
U = U12 + U23 + U13 + U14 Á (16.6)
Only the first three terms of Eq. 16.6 would be needed for the configuration shown
in Fig. 16.5b. Note that the signs of the charges keep things straight mathemati-
cally, as the biomolecular situation in Example 16.4 shows.
EXAMPLE 16.4 Molecule of Life: The Electric Potential Energy of a Water Molecule
The water molecule is the foundation of life as it is known. Many of its properties q1 = +5.20 × 10−20 C
(such as the reason it is a liquid on the Earth’s surface) are related to the fact that it is a + H
permanent polar molecule (see Section 15.4 on electric dipoles). A simple picture of the
water molecule, including the charges, is shown in 䉴 Fig. 16.6. The distance from each
hydrogen atom to the oxygen atom is 9.60 * 10-11 m, and the angle 1u2 between the
two hydrogen–oxygen bond directions is 104°. What is the total electrostatic energy of r13
the water molecule? q3 =
−10.4 × 10−20 C θ
O −−
r12
r23
䉴 F I G U R E 1 6 . 6 Electrostatic potential energy of
a water molecule The charges shown on the water
molecule are net average charges because the atoms
within the molecule share electrons. So the charges + H
on the ends of the water molecule can be smaller
q2 = +5.20 × 10−20 C
than the charge on the electron or proton.
T H I N K I N G I T T H R O U G H . The model of this molecule involves trigonometry. The total electrostatic potential energy is the
three charges. The charges are given, but the distance algebraic sum of the potential energies of the three pairs of
between the hydrogen atoms must be calculated using charges (that is, Eq. 16.6 will have three terms).
SOLUTION. The following data are taken from Fig. 16.6.

Given: q1 = q2 = + 5.20 * 10-20 C Find: U (total electrostatic potential
q3 = - 10.4 * 10-20 C of energy water molecule)
r13 = r23 = 9.60 * 10-11 m
u = 104°
1r12>22
= sina b . This can be solved for r12:
u
Notice that
r13 2
r12 = 2r13 asin b = 219.60 * 10-11 m21sin 52°2 = 1.51 * 10-10 m

u
2
Before determining the total potential energy of this system, let’s calculate each pair’s contribution separately. Note that
U13 = U23. (Why?) Applying Eq. 16.5,
kq1 q2 19.00 * 109 N # m2>C221 +5.20 * 10-20 C21 +5.20 * 10-20 C2
U12 = =
r12 1.51 * 10-10 m
= + 1.61 * 10 J
-19
and
kq2 q3 19.00 * 109 N # m2>C221 +5.20 * 10-20 C21 -10.4 * 10-20 C2
U13 = U23 = =
r23 9.60 * 10-11 m
= - 5.07 * 10-19 J
Thus the total electrostatic potential energy is
U = U12 + U13 + U23 = 1+1.61 * 10-19 J2 + 1- 5.07 * 10-19 J2 + 1- 5.07 * 10-19 J2
= - 8.53 * 10-19 J
The negative result indicates that the molecule requires positive work to break it apart. (That is, it must be pulled apart.)
F O L L O W - U P E X E R C I S E . Another common polar molecule is carbon monoxide (CO), a toxic gas commonly produced during
incomplete hydrocarbon fuel combustion. The carbon atom is, on average, positively charged and the oxygen atom is, on
average, negative. The distance between the carbon and oxygen atoms is 0.120 nm, and the total electrostatic potential energy of
this molecule is - 3.27 * 10-19 J. Determine the magnitude of the (average) charge on each end of the molecule.
DID YOU LEARN?

➥ Electric potential difference is electric potential energy difference per coulomb.
➥ Electric potential difference is expressed in volts (V), which is the name given to the
joule per coulomb.
➥ Negative charges gain kinetic energy by losing electric potential energy as they
move to regions of higher electric potential.
16.2 Equipotential Sur faces and the Electric Field

➥ What is meant by an equipotential surface in electrostatics?

➥ How does the spacing between equipotential surfaces relate to the electric field
strength in that region?
➥ How is the electric field direction in a region in space related to the change in poten-
tial in that region?
EQUIPOTENTIAL SURFACES
Suppose a positive charge is moved perpendicularly to an electric field (such as
path I of 䉴 Fig. 16.7a). As the charge moves from A to A¿ , no work is done by the elec-
tric field (why?). If no work is done, then the value of the potential energy does not
16.2 EQUIPOTENTIAL SURFACES AND THE ELECTRIC FIELD 569
change, so ¢UAA¿ = 0. From this, it can be concluded that these two points (A and
+
+
+
+
+
+
A¿ )—and all other points on path I—are at the same potential V; that is, I A
A
¢UAA¿ E
¢VAA¿ = VA¿ - VA = = 0 or VA = VA¿
q II
This result actually holds for all points on the plane parallel to the plates and
– – – – – – –
containing path I. A surface like this plane, on which the potential is constant, is
(a)
called an equipotential surface (or simply an equipotential). The word equipotential
means “same potential.” Note that, unlike this special case, equipotential surfaces,
+
+
+
+
+
+
are, in general, not planes.
B
Since no work is required to move a charge along an equipotential surface, it B
must be generally true that I E
Equipotential surfaces are always at right angles to the electric field. A II
Moreover, because the electric field is conservative, the work is the same whether
– – – – – – –
path I, path II, or any other path from A to A¿ is taken (Fig. 16.7a). As long as the
(b)
charge returns to the same equipotential surface from which it started, the work
done on it is zero and the value of the electric potential is the same.
B
If the positive charge is moved opposite to the direction of E (for example, path I
in Fig. 16.7b)—at right angles to the equipotentials—the electric potential energy,
and hence the electric potential, increases. (Why? Think of the sign of the work VB V A
required.) When B is reached, the charge is on a different equipotential surface— VA
one of a higher potential value than the surface that A is on. If, instead, the charge
had been moved from A to B¿ , the work would be the same as that in moving from
A to B. Hence, B and B¿ are on the same equipotential surface. For parallel plates,
the equipotentials are planes parallel to the plates (Fig. 16.7c).
(c)
To help understand the concept of an electric equipotential surface, consider a
gravitational analogy. If the gravitational potential energy is designated as zero at 䉱 F I G U R E 1 6 . 7 Construction of
ground level and an object is raised a height h = hB - hA (from A to B in 䉲 Fig. 16.8), equipotential surfaces between par-
then the work done by an external force is mgh and is positive. For horizontal move- allel plates (a) The work done in
ment, the potential energy does not change. This means that the dashed plane at moving a charge is zero as long as
you start and stop on the same
height hB is a gravitational equipotential surface—and so is the plane at hA, but it
equipotential surface. (Compare
has a lower gravitational potential value than the plane at hB. Therefore, surfaces of paths I and II.) (b) Once the charge
constant gravitational potential energy are planes parallel to the Earth’s surface. moves to a higher potential (for
Topographic maps, which display land contours by plotting lines of constant eleva- example, from point A to point B), it
tion (usually relative to sea level), are thus also maps of constant gravitational can stay on that new equipotential
surface by moving perpendicularly
potential (䉲 Fig. 16.9a, b). Note how the equipotentials near a point charge (Fig.
to the electric field (B to B¿ ). The
16.9c, d) are qualitatively similar to the gravitational contours due to a hill. change in potential is independent
It is useful to know how to sketch equipotential surfaces, because they are inti- of the path, since the same change
mately related to the electric field and to practical aspects such as voltage. Learn occurs whether path I or path II is
used. (Why?) (c) The actual equipo-
tential surfaces within the parallel
plates are planes parallel to those
plates. Two such plates are shown,
with VB 7 VA.
B B
UB = UB´ = mghB
g
h II I
hB
A m
UA = mghA < UB
hA
Ug = 0
䉱 F I G U R E 1 6 . 8 Gravitational potential energy analogy Raising an object in a uniform

gravitational field results in an increase in gravitational potential energy, and UB 7 UA. At
a given height, the object’s potential energy is constant as long Bas it remains on that (gravi-
B
tational) equipotential surface. Here, g points downward, like E in Fig. 16.7.
䉴 F I G U R E 1 6 . 9 Topographic Increasing
maps—a gravitational analogy to gravitational
equipotential surfaces (a) A sym- potential
metrical hill with slices at different 10 m
elevations. Each slice is a plane of 15 m
constant gravitational potential. 25 m h4
(b) A topographic map of the slices 20 m
in (a). The contours, where the 20 m h3 25 m
planes intersect the surface, repre- 15 m h2
sent increasingly larger values of U4U3 U2 U1
gravitational potential as one goes 10 m h1
up the hill. (c) The electric potential
V near a point charge q forms a sim-
ilar symmetrical hill. V is constant at
fixed distances from q. (d) Electrical
equipotentials around a point (a)
charge are spherical (in two dimen- (b)
sions they are circles) centered on Increasing
the charge. The closer the equipo- electric
tential to the positive charge, the potential
larger its electric potential.
V4
V3
V2 + V4 V3 V2 V1
V1
(c)
(d)
by Drawing 16.2, Graphical Relationship between Electric Field Lines and Equipo-
tentials, summarizes a qualitative method useful for sketching equipotential
surfaces if given an electric field line pattern. As this feature shows, the method is
+
also useful for the converse problem: sketching the electric field lines if the
E V1 equipotential surfaces are given. Can you see how these ideas were used to con-
struct the equipotentials of an electric dipole in 䉳 Fig. 16.10?
V2 To determine the mathematical relationship between the electric field (E) and
the electric potential (V), consider the special case of a uniform electric field
(䉲 Fig. 16.11). The potential difference 1¢V2 between any two equipotential planes
V3 (labeled V1 and V2 in the figure) can be calculated with the same technique used to
−
derive Eq. 16.2. The result is
V1 > V2 > V3
¢V = V3 - V1 = E¢x (16.7)
䉱 F I G U R E 1 6 . 1 0 Equipotentials 䉳 F I G U R E 1 6 . 1 1 Relationship
of an electric dipole Equipotentials between the potentialB change (¢V )
are perpendicular to electric field x
and the electric field (E) The
+
–
lines. V1 7 V2 because equipotential + electric field direction is that of
surface 1 is closer to the positive E +
– maximum decrease in potential, or
charge than is surface 2. To under-
x + opposite the direction of maximum
stand how equipotentials are con- –
increase in potential. (Here, this
structed, see Learn by Drawing 16.2. maximum is in the direction of the
+ +
+
–
solid blue arrow, not the angled
– x
+ ones; why?) The electric field mag-
+ nitude is given by the maximum
– rate at which the potential changes
+ over distance (usually in volts per
–
+
meter).
V1 V2 > V1 V3 > V2
V
graphical relationship between electric

field lines and equipotentials
E
Because it takes no work to move a charge along an
equipotential surface, such surfaces must be perpendicular D
to the electric field lines. Also, the electric field has a mag- C
nitude equal to the maximum rate of change in potential
with unit distance 1V>m2 and points in the direction in
B
A
which the potential decreases most rapidly. These facts can V1 > V2 > V3 > V4 > V5 > V6
be used to construct equipotentials if the field pattern is
known. The reverse is also true: Given the equipotentials, F I G U R E 2 Mapping the electric field from equipotentials
the electric field lines can be constructed. Furthermore, if Start at a convenient point, and trace a line that crosses each
the potential value (in volts) associated with each equipo- equipotential at a right angle. Repeat the process as often as
tential is known, the strength and direction of the field can needed to reveal the field pattern, adding arrows to indicate
be estimated from the rate at which the potential changes the direction of the field lines from high to low potential. In
going from one potential to the next, plan ahead so that
with distance (Eq. 16.8).
each succeeding equipotential is also crossed at right angles.
One example of each of these situations should provide a
better insight into the connection between equipotential
surfaces and their associated electric fields. Consider Fig. 1,
in which the electric field lines are given and the object is to Now suppose instead that the the equipotentials are
determine the shape of the equipotentials. Pick any point, given instead of the field lines (Fig. 2). The electric field
such as A, and begin moving at right angles to the field lines point in the direction of decreasing V and are perpen-
lines. Keep moving so as to maintain this perpendicular ori- dicular to the equipotential surfaces. Thus, to map the field,
entation to the lines. Between lines you may have to approx- start at any point, and move in such a way that your path
imate, but plan ahead to the next field line so it is crossed at intersects each equipotential surface at a right angle. The
a right angle. resulting field line is shown in Fig. 2, beginning at point A.
To find another equipotential, start at another point, such Starting at points B, C, and D provides additional field
as B, and proceed the same way. Sketch as many equipoten- lines that suggest the complete electric field pattern. One
tials as you need to map the area of interest. The figure need only add the arrows in the direction of decreasing
shows the result of sketching four equipotentials, from A (at potential to indicate field direction.
the highest potential—can you tell why?) to D (at the lowest B
Lastly, suppose an estimate of the magnitude of E at some
potential). point P (Fig. 3) is wanted. Assume further that the values of
the equipotentials 1.0 cm on either side of it are known
(shown). From this, it can be seen that the electric field
points roughly from A to B (why?) and its approximate
VA > VB > VC > VD magnitude is
11000 V - 950 V2
E = ` `
A ¢V
=
¢x max 2.0 * 10-2 m
B +
= 2.5 * 103 V>m
E
C
V > VA
D VA = 1000 V
A
F I G U R E 1 Sketching equipotentials from electric field lines 2.0 cm P

If you know the electric field pattern, pick a point in the VB = 950 V
region of interest and move so that your path is always per- B
pendicular to the next field line. Keep your path as smooth V< V
as possible, planning ahead so that each succeeding field B
line is also crossed at right angles. To map a surface with a
higher (or lower) potential, start at a different location in F I G U R E 3 Estimating the magnitude of the electric field
the electric field and repeat the process. Here, VA 7 VB, and The magnitude of the potential change per meter at any
so on. point gives the strength of the electric field at that point.
Thus, if you start on equipotential surface 1 and move perpendicularly away from it
and opposite to the electric field to equipotential surface 3, there is a potential increase
1¢V2 that depends on the electric field strength (E) and the distance 1¢x2.
For a given distance ¢x, this movement perpendicular to the equipotential sur-
faces and opposite the direction of the electric field yields the maximum possible
gain in potential. Think of taking one step of length ¢x in any direction, starting
from surface 1. The way to maximize the increase would be to step onto surface 3.
A step in any direction not perpendicular to surface 1 (for example, ending on sur-
face 2) yields a smaller increase in potential.
Notice that the direction of the maximum potential increase is the direction
B
opposite that of E. Thus, as a general rule:
B
The direction of the electric field E is that in which the electric potential decreases the
most rapidly, or equivalently, opposite the direction in which the electric potential
increases the most rapidly.
Then, at any location, the magnitude of the electric field is the maximum rate of
change of the potential with distance, or
E = ` `
¢V
(16.8)
¢x max
The unit of electric field is volts per meter 1V>m2. Previously, E was expressed in
newtons per coulomb (N>C; see Section 15.4). You should show, through dimen-
sional analysis, that 1 V>m = 1 N>C. A graphical interpretation of the relationship
B
between E and V is shown in Learn by Drawing 16.2.
In many practical situations, it is the potential difference (called voltage), rather
than the electric field, that is specified. For example, a D-cell flashlight battery has a
terminal voltage of 1.5 V, meaning that it can maintain a potential difference of 1.5 V
between its terminals. Most automotive batteries have a terminal voltage of about 12
V. Some of the common potential differences, or voltages, are listed in 䉲 Table 16.1.
Whether you know it or not, you live in an electric field near the Earth’s sur-
face. This field varies with weather conditions and, consequently, can be an indi-
cator of approaching storms. Example 16.5 applies the equipotential surface
concept to help in understanding the Earth’s electric field.
TABLE 16.1 Common Electric Potential Differences (Voltages)

Source Approximate Voltage ( ¢V)
Across nerve membranes 100 mV

Small-appliance batteries 1.5 to 9.0 V
Automotive batteries 12 V
Household outlets (United States) 110 to 120 V
Household outlets (Europe) 220 to 240 V
Automotive ignitions (spark plug firing) 10 000 V
Laboratory generators 25 000 V
High-voltage electric power delivery lines 300 kV or more
Cloud-to-Earth surface during thunderstorm 100 MV or more
EXAMPLE 16.5 The Earth’s Electric Field and Equipotential Surfaces: Electric Barometers?
Under normal atmospheric conditions, the Earth’s surface is elec- and in what direction does the electric potential decrease the
trically charged. This creates an approximately constant electric most rapidly? (b) How far apart are two equipotential surfaces
field of about 150 V>m pointing down near the surface. (a) Under that have a 1000-V difference between them? Which has a higher
these conditions, what is the shape of the equipotential surfaces, potential, the one farther from the Earth or the one closer?
T H I N K I N G I T T H R O U G H . (a) Near the Earth’s surface, the Eqs. 16.7 and 16.8 enables us to determine which way the
electric field is approximately uniform, so the equipotentials potential increases. (b) Equation 16.8 can then be used to
are similar to those of parallel plates. The discussion of determine how far apart the equipotential surfaces are.

Given: E = 150 V>m, downward Find: (a) shape of equipotential surfaces and direction of decrease in potential
¢V = 1000 V (b) ¢x (distance between equipotentials)
(a) Uniform electric fields are associated with plane equipo- ¢V 1000 V
tentials; in this case, the planes are parallel to the Earth’s sur- ¢x = = = 6.67 m
E 150 V>m
face. The electric field points downward. This is the direction
in which the potential decreases most rapidly. Because the potential decreases as we move downward (in
B
the direction of E), the higher potential is associated with the
(b) To determine the distance between the two equipotentials, surface that is 6.67 m farther from the ground.
think of moving vertically so that ¢V>¢x has its maximum
value. Solving Eq. 16.8 for ¢x yields
F O L L O W - U P E X E R C I S E . Re-examine this Example under storm conditions. During a lightning storm, the electric field can rise to
many times the normal value as well as reverse in direction. (a) Under these conditions, if the field is 900 V>m and points upward,
how far apart are two equipotential surfaces that differ by 2000 V? (b) Which surface is at a higher potential, the one closer to the
Earth or the one farther away? (c) Can you tell how far the two surfaces are from the ground? Why or why not?
Equipotential surfaces can be useful for describing the field near a charged con-
ductor, as Conceptual Example 16.6 shows.
CONCEPTUAL EXAMPLE 16.6 The Equipotential Surfaces Outside a Charged Conductor

A solid conductor with an excess positive charge is shown in associated with flat plates. While it might be tempting to pick
䉲 Fig. 16.12a. Which of the following best describes the shape answer (b), a quick look at the electric field near the surface of a
of the equipotential surfaces just outside the conductor’s sur- charged conductor (Section 15.5), in conjunction with Learn by
face: (a) flat planes, (b) spheres, or (c) approximately the Drawing 16.2, shows that the correct answer must be (c). To ver-
shape of the conductor’s surface? Explain your reasoning. ify that (c) is the correct answer, recall that near the surface the
electric field is perpendicular to that surface. Since the equipo-
REASONING AND ANSWER. Choice (a) can be eliminated tential surfaces are perpendicular to the field lines, they must
immediately, because flat (plane) equipotential surfaces are follow the contour of the conductor’s surface (Fig. 16.12b).
E E E
+++ +++ +++

++
+
++
+
++
+
+ + +
+ ++
+ ++
+ ++
+
+
+ + +
++
++
++
+ +++ + ++ + + ++ +
Equipotential
V1
(a) (b) V2 (c)
䉱 F I G U R E 1 6 . 1 2 Equipotential surfaces near a charged conductor
F O L L O W - U P E X E R C I S E . In this Example, (a) which of the two equipotentials (1 or 2) shown in Fig. 16.12c is at a higher potential?
(b) What is the approximate shape of the equipotential surfaces very far from this conductor? Explain your reasoning. [Hint: What
does the conductor look like when you are very far from it?]
THE ELECTRON-VOLT
The concept of electric potential provides a unit of energy that is particularly use-
ful in molecular, atomic, nuclear, and elementary particle physics. An energy unit
named the electron-volt (eV) is defined as the kinetic energy acquired by an elec-
tron (or proton) accelerated through a potential difference, or voltage, of exactly
1 V. The gain in kinetic energy is equal (but opposite) to the change in electric
potential energy. For an electron, its gain in kinetic energy in joules is:
¢K = - ¢Ue = - 1e¢V2 = - 1-1.60 * 10-19 C211.00 V2 = + 1.60 * 10-19 J
Since this is what is meant by 1 electron-volt, the conversion factor between the
electron volt and the joule (to three significant figures) is
1 eV = 1.60 * 10-19 J
The electron-volt is typical of energies on the atomic scale, so it is convenient to

express atomic energies in terms of electron-volts instead of joules. The energy of
any charged particle accelerated through any potential difference can be expressed
in electron-volts. For example, if an electron is accelerated through a potential dif-
ference of 1000 V, its gain in kinetic energy 1¢K2 is one thousand times that of a
1-eV electron, or
¢K = e¢V = 11 e211000 V2 = 1000 eV = 1 keV
The abbreviation keV stands for kiloelectron-volt.

The electron-volt is defined in terms of a particle with the minimum charge (the
electron or proton). However, the energy of a particle with any charge can also be
expressed in electron-volts. Thus, if a particle with a charge of +2e, such as an alpha
particle, were accelerated through a potential difference of 1000 volts, it would gain
a kinetic energy of ¢K = e¢V = 12e211000 V2 = 2000 eV = 2 keV. Note how easy
it is to compute the kinetic energy if it is done in electron-volts.
Occasionally, units larger than the electron-volt are needed. For example, in
nuclear and elementary particle physics, it is not uncommon to find particles with
energies of megaelectron-volts (MeV) and gigaelectron-volts (GeV); 1 MeV = 106 eV
and 1 GeV = 109 eV.*
In working problems, it is important to be aware that the electron-volt (eV) is
not an SI unit. Hence, when using energies in calculators, electron-volts must first
be converted into joules. For example, to determine the speed of an electron accel-
erated from rest through a potential difference of 10.0 V, first convert the kinetic
energy (10.0 eV) to joules:
K = 110.0 eV211.60 * 10-19 J>eV2 = 1.60 * 10-18 J
Continuing in the SI system, the mass of the electron must be in kilograms.

Then the speed is
v = 22K>m = 4211.60 * 10-18 J2>19.11 * 10-31 kg2 = 1.87 * 106 m>s
DID YOU LEARN?

➥ Equipotential surfaces are those surfaces on which all the points have the same
value for electric potential.
➥ The spacing between equipotential surfaces is inversely related to the electric field
strength.
➥ The electric field points in the direction of maximum decrease in electric
potential.
*At one time, a billion electron-volts was referred to as BeV, but this usage was abandoned because
confusion arose. In some countries, such as Great Britain and Germany, a billion means 1012 (which is
called a trillion in the United States).
16.3 CAPACITANCE 575
16.3 Capacitance
➥ How does the capacitance of a pair of oppositely charged conductors depend on

the amount of charge stored on one of the conductors?
➥ If the charge on a capacitor is doubled, how does the energy stored in it change?
➥ If a capacitor is charged by a 12-V battery, how much more charge is stored in it
compared to when it is charged by a 6-V battery?
A pair of parallel plates, if charged, stores electrical energy (䉴 Fig. 16.13). +Q –Q

Such an arrangement of conductors is an example of a capacitor. (Any + –
pair of conductors qualifies as a capacitor.) The energy storage occurs + –
+ – Q = +Q = CV
because it takes work to transfer the charge from one plate to the other. + –
Imagine that one electron is moved between a pair of initially + –
uncharged plates. Once that is done, transferring a second electron is + –
+ E –
more difficult, because it is not only repelled by the first electron on the + – V
negative plate, but also attracted by a double positive charge on the + –
positive plate. Separating the charges requires more and more work as + –
+ –
more and more charge accumulates on the plates. (This is analogous to
stretching a spring. The more you stretch it, the more work it takes to + –
stretch it further.) Battery
Metal
The work needed to charge parallel plates can be done quickly (usu- plates
ally in a few microseconds) by a battery. Although battery action won’t be
(a) Parallel plate capacitor
discussed until Chapter 17, all you need to know now is that a battery
removes electrons from the positive plate and transfers, or “pumps,”
them through a wire to the negative plate. In the process of doing work, C
the battery loses some of its internal chemical potential energy. Of pri- + –
mary interest here is the result: a separation of charge and the creation of
an electric field in the capacitor. The battery will continue to charge the
capacitor until the potential difference between the plates is equal to the
terminal voltage of the battery—for example, 12 V if you use a standard
automotive battery. When the capacitor is disconnected from the battery,
it becomes a storage “reservoir” of electrical energy. + –
For a capacitor, the charge Q on the plates is proportional to the volt-
age (electric potential difference) across the plates, or Q r V.* (Here, Q V
denotes the magnitude of the charge on either plate, not the net charge
(b) Schematic diagram
on the whole capacitor, which is zero.) This proportionality can be
made into an equation by using a constant, C, called capacitance: 䉱 F I G U R E 1 6 . 1 3 Capacitor and
circuit diagram (a) Two parallel
Q metal plates are charged by a battery
Q = CV or C = (16.9) that moves electrons from the posi-
V
tive plate to the negative one
SI unit of capacitance: coulomb per volt (C>V), or farad (F) through the wire. Work is done
while charging the capacitor, and
The coulomb per volt equals a farad; 1 C>V = 1 F. The farad is a large unit (see energy is stored in the electric field of
Example 16.7), so the microfarad 11 mF = 10-6 F2, the nanofarad 11 nF = 10-9 F2, the capacitor. (b) This diagram repre-
and the picofarad 11 pF = 10-12 F2 are commonly used. sents the charging situation shown
in part (a). It also shows the symbols
Capacitance represents the amount of charge stored per volt. When a capacitor commonly used for a battery (V) and
has a large capacitance, it holds a large amount of charge per volt compared with one a capacitor (C). The longer line of the
of smaller capacitance. If you connected the same battery to two different capacitors, battery symbol is the positive termi-
the one with the larger capacitance would store more charge and more energy. nal, and the shorter line represents
Capacitance depends only on the geometry (size, shape, and spacing) of the plates the negative terminal. The symbol
for a capacitor is similar, but the lines
(and the material between the plates, discussed in Section 16.5) and not the charge on are of equal length.
the plates. Consider the parallel plate capacitor, which has an electric field given by
Eq. 16.5:
4pkQ
E =
A
*At this point, will be used V to denote potential differences instead of ¢V. This is a common practice.
The voltage across the plates can be computed from Eq. 16.2:
4pkQd
V = Ed =
A
The capacitance of a parallel plate arrangement is then
= a b
Q 1 A
C = (parallel plates only) (16.10)
V 4pk d
It is common to replace the expression in the parentheses in Eq. 16.10 with a
single quantity called the permittivity of free space (Eo). The value of this con-
stant (to three significant figures) is
1 C2
eo = = 8.85 * 10-12 (permittivity of free space) (16.11)
4pk N # m2
eo is a quantity that describes the electrical properties of free space (vacuum), but its
value in air is only 0.05% larger. In our calculations, they will be taken to be the same.
It is common to rewrite Eq. 16.10 in terms of eo:
eo A
C = (parallel plates only) (16.12)
d
Let’s use Eq. 16.12 in the next Example to show just how unrealistically large an
air-filled capacitor with a capacitance of 1.0 F would be.
EXAMPLE 16.7 Parallel Plate Capacitors: How Large Is a Farad?

What would be the plate area of an air-filled 1.0-F parallel plate capacitor if the plate
separation were 1.0 mm? Would it be realistic to consider building such a capacitor?
T H I N K I N G I T T H R O U G H . The area can be calculated directly from Eq. 16.12. Remember
to keep all quantities in SI units, so that the answer will be in square meters. The vac-
uum value of eo for air can be used without creating a significant error.
Given: C = 1.0 F Find: A (area of one of the plates)
d = 1.0 mm = 1.0 * 10-3 m
Solving Eq. 16.12 for the area gives
Cd 11.0 F211.0 * 10-3 m2
A = = 1.1 * 108 m2
8.85 * 10-12 C2>1N # m22
=
eo
This is more than 100 km2 140 mi 22, that is, a square more than 10 km (6.2 mi) on a side.
It is unrealistic to build a parallel-plate capacitor that big; 1.0 F is therefore a very large
V value of capacitance. There are ways, however, to make compact high-capacity capaci-
tors (Section 16.4).
Voltage (in volts)
V+0
V = = V
2 2 F O L L O W - U P E X E R C I S E . In this Example, what would the plate spacing have to be if
you wanted the capacitor to have a plate area of 1 cm2? Compare your answer with a
= 1/C typical atomic diameter of 10-9 to 10-10 m. Is it feasible to build this capacitor?
pe
Slo
Q The expression for the energy stored in a capacitor can be obtained by graphical
Charge (in coulombs) analysis, since both Q and V vary during charging—for example, as the charge is
䉱 F I G U R E 1 6 . 1 4 Capacitor volt- separated by a battery. A plot of voltage versus charge for charging a capacitor is a
age versus charge A plot of voltage straight line with a slope of 1>C, because V = 11>C2Q (䉳 Fig. 16.14). The graph rep-
(V) versus charge (Q) for a capacitor resents the charging of an initially uncharged capacitor 1Vo = 02 to a final voltage
is a straight line with slope 1>C (V). The work done is equivalent to transferring the total charge, using an average
(because V = 11>C2Q). The average
voltage is V = 12 V, and the total
voltage V. Because the voltage varies linearly with charge, the average voltage is
work done is equivalent to transfer- half the final voltage V:
ring the charge through V. Thus,
UC = W = Q V = 12 QV, the area Vfinal + Vinitial V + 0 V
V = = =
under the curve (a triangle). 2 2 2
16.3 CAPACITANCE 577
The energy stored in the capacitor is equal to the work done by the battery. Since,
by definition, ¢V = V = W>Q it follows that this stored energy UC is given by
UC = W = QV = 12 QV
Because Q = CV, this result can be rewritten in several equivalent forms:
Q2
UC = 12 QV = = 12 CV2 (energy storage in a capacitor) (16.13)
2C
Typically, the form UC = 12 CV2 is the most practical, since the capacitance and the
voltage are usually the known quantities. A very important medical application of
the capacitor is the cardiac defibrillator, discussed in Example 16.8.
EXAMPLE 16.8 Capacitors to the Rescue: Energy Storage in a Cardiac Defibrillator

During a heart attack, the heart can beat in an erratic fashion,
called fibrillation. One way to get it back to normal rhythm is
to shock it with electrical energy supplied by a cardiac defibril-
lator (䉴 Fig. 16.15). About 300 J of energy is required to pro-
duce the desired effect. Typically, a defibrillator stores this
energy in a capacitor charged by a 5000-V power supply.
(a) What capacitance is required? (b) What is the charge on
the capacitor’s plates?
T H I N K I N G I T T H R O U G H . (a) To find the capacitance, solve for
C in Eq. 16.13. (b) The charge then follows from the definition
of capacitance (Eq. 16.9).
Given: UC = 300 J Find: (a) C (the capacitance)
V = 5000 V (b) Q (charge on capacitor)
(a) The most useful form of Eq. 16.13 is UC = 12 CV2. Solving
for C,
2UC 21300 J2
C = = 2.40 * 10-5 F = 24.0 mF
15000 V22
2
=
V 䉱 F I G U R E 1 6 . 1 5 Defibrillator A burst of electric current
(flow of charge) from a defibrillator may restore a normal
(b) The charge (magnitude) on either plate is then
heartbeat to people in cardiac fibrillation. Capacitors store the
Q = CV = 12.40 * 10-5 F215000 V2 = 0.120 C electrical energy on which the device depends.
F O L L O W - U P E X E R C I S E . For the capacitor in this Example, if the maximum allowable energy for any single defibrillation attempt
is 750 J, what is the maximum voltage that should be used?
Sometimes capacitors can be used to model real-life phenomena. For example,

a lightning storm can be considered to be the discharge of a negatively charged
cloud to the positively charged ground—in effect, a “cloud–ground” capacitor.
Another interesting application of electric potential treats nerve membranes as
cylindrical capacitors to help explain nerve signal transmission. (See Insight 16.1,
Electric Potential and Nerve Signal Transmission.)
DID YOU LEARN?

➥ The capacitance of a pair of conductors depends only on their geometry.
➥ Capacitor energy storage depends on the square of the capacitor’s charge.
➥ Capacitor charge depends linearly on its voltage.
INSIGHT 16.1 Electric Potential and Nerve Signal Transmission

The human body’s nervous system is responsible for the rior of the nerve cell. This ion selectivity gives rise to a polariza-
reception of external stimuli through our senses (such as tion of charge across the membrane. The exterior is positive
touch) as well as communication between the brain and our (with Na + trying to enter the region of lower concentration),
organs and muscles. If you touch something hot, nerves in attracting the negative proteins to the inner surface of the mem-
your hand detect the problem and send a signal to your brain; brane (Fig. 1b). Thus a cylindrical capacitor-like charge storage
your brain then sends the signal “Pull back!” through other system exists across an axon membrane at rest. The resting mem-
parts of the nervous system to your hand. But what are these brane potential (the voltage across the membrane) is defined as
signals, and how do they work? ¢V = Vin - Vout. Because the outside is positively charged, as
A typical nerve consists of a bundle of nerve cells called defined, the resting potential is a negative quantity; it ranges
neurons, much like individual telephone wires bundled into a from about - 40 to -90 mV (millivolts), with a typical value of
single cable. The structure of a typical neuron is shown in -70 mV in humans.
Fig 1a. The cell body, or soma, has long branchlike extensions Signal conduction occurs when the cell membrane receives
called dendrites, which receive the input signal. The soma is a stimulus from the dendrites. Only then does the membrane
responsible for processing the signal and transmitting it potential change, and this change is propagated down the
down a long extension called the axon. At the other end of the axon. The stimulus triggers Na + channels in the membrane
axon are projections with knobs called synaptic terminals. At (which are closed while resting, like a gate) to open and tem-
these knobs, the electrical signal is transmitted to another porarily allows sodium ions to enter the cell (Fig. 1b). These
neuron across a gap called the synapse. The human body con- positive ions are attracted to the negative charge layer on the
tains on the order of 100 billion neurons, and each neuron can interior and are driven by the difference in concentration. In
have several hundred synapses! Running the nervous system about 0.001 s, enough sodium ions have passed through the
costs the body about 25% of its energy intake. gated channel to cause a reversal of polarity, and the mem-
To understand the electrical nature of nerve signal transmis- brane potential rises, typically to +30 mV in humans. The
sion, let us focus on the axon. A vital component of the axon is its time sequence for this change in membrane potential is
cell membrane, which is typically about 10 nm thick and consists shown in Fig. 2.
of phospholipids (electrically polarized hydrocarbon molecules) When the difference in Na+ concentration causes the mem-
and embedded protein molecules (Fig. 1b). The membrane has brane voltage to become positive, the Na + channels close. A
proteins called ion channels, which form pores where large pro- chemical process involving proteins known as the Na>K
tein molecules regulate the flow of ions (primarily sodium) –ATPase molecular pump then re-establishes the resting poten-
across the membrane. The key to nerve signal transmission is the tial at about -70 mV by selectively transporting the excess
fact that these ion channels are selective: They allow only certain Na + back to the cell’s exterior.
types of ions to cross the membrane; others cannot. This temporary change in membrane potential (a total of
The fluid outside the axon, although electrically neutral, con- 100 mV, from - 70 mV to + 30 mV) is called the cell’s action
tains sodium ions 1Na+2 and chlorine ions 1Cl -2 in solution. In potential. The action potential is the signal that is transmitted
contrast, the axon’s internal fluid is rich in potassium ions 1K +2 down the axon. This “voltage wave” travels at speeds of 1 to
and negatively charged protein molecules. If it were not for the 100 m>s on its way to triggering another such pulse in the
selective nature of the cell membrane, the Na + concentration adjacent neuron. This speed, along with other factors such as
would be equal on both sides of the membrane. Under normal time delays in the synapse region, is responsible for typical
(or resting) conditions, it is difficult for Na+ to penetrate the inte- human reaction times totaling a few tenths of a second.
Axon Cross Section

Sodium channel Na+ +50
Dendrites
Membrane potential (mV)
opens + +30 mV
+ – –
–
+ + + +
Soma 0
–
+ + Action
∆V Cl – potential
– –
+ – + – +
+ –
Axon + + −50
Cell membrane –
+
– Cell interior −70

(phospholipids
– + + + – –
and protein + – + – + Membrane
Synaptic −100
molecules) – – resting
terminals 0 1 2 3 4
+ + – potential
+ Time (ms)
Synapse + + + +
– –
– + + + + Na+ Na+ Na+ channel
Next + channel channel closed,
neuron closed open Na/K–ATPase
Cell exterior
pump activated
(a) (b) F I G U R E 2 As the sodium channels open and

sodium ions rush to the cell’s interior, the mem-
F I G U R E 1 (a) The structure of a typical neuron. (b) An enlargement of brane potential changes quickly from its resting
the axon membrane, showing the membrane (about 10 nm thick) and the value of -70 mV to about + 30 mV. The resting
concentration of ions inside and outside the cell. potential is restored about 4 ms later.
16.4 DIELECTRICS 579
16.4 Dielectrics TABLE 16.2 Dielectric

LEARNING PATH QUESTIONS Constants for Some
➥ What does the dielectric constant depend upon? Materials
➥ Does the insertion of a dielectric sheet into a capacitor always increase its capacitance?
Dielectric
➥ Does the insertion of a dielectric sheet into a capacitor always result in an increase Material Constant (k)
in energy storage in that capacitor?
Vacuum 1.0000
In most capacitors, a sheet of insulating material, such as paper or plastic, lies Air 1.00059
between the plates. Such an insulating material, called a dielectric, serves several
Paper 3.7
purposes. For one, it keeps the plates from coming into contact. Contact would
allow the electrons to flow back onto the positive plate, neutralizing the charge on Polyethylene 2.3
the capacitor and the energy stored. A dielectric also allows flexible plates of Polystyrene 2.6
metallic foil to be rolled into a cylinder, giving the capacitor a more compact (more Teflon 2.1
practical) size. Finally, a dielectric increases the charge storage capacity of the
Glass (range) 3–7
capacitor and therefore, under the right conditions, the energy stored in the capac-
itor. This capability depends on the type of material and is characterized by the Pyrex glass 5.6
dielectric constant (K). Values of the dielectric constant for some common materi- Bakelite 4.9
als are given in 䉴 Table 16.2. Silicon oil 2.6
How a dielectric affects the electrical properties of a capacitor is illustrated in
B Water 80
䉲 Fig. 16.16. Here the air-filled capacitor is fully charged (creating a field E o) and then
disconnected from the battery, after which a dielectric is inserted (Fig. 16.16a). In the Strontium
titanate 233
dielectric material, work is done on molecular dipoles by the existing electric field,
aligning them with that field (Fig. 16.16b). (The molecular polarization may be per-
manent or temporarily induced by the electric field. In either case, the effect is the
same.) Work is also done on the dielectric sheet as a whole, because the charges on
the plates pull it into the region between the plates.
B
The result is that the dielectric creates a “reverse” electric field (Ed in Fig.
16.16c) that partially cancels the field between the plates. This means that the net
_ +
_ +
Dielectric
_ +
_
+
_ +
+Qo −Qo +Qo −Qo +Qo Eo −Qo +Qo −Qo

+
_
+ _ + _ + – + – + – + –
+ _ + _ + – + _ + _ + – + _ – + _ +
–
+ – + – + + – + –
_ + _ + _ _ +
+ – + – + – + –
+ – + – + + – + –
_ + _ + _ _ +
+ – + – + – + –
+ Eo – + – + + – + E –
+ – + _ + _ + – + _ – + _ +
–
+ – + – + + – + –
_ + _ + _ _ +
+ – + – + – + –
+
Vo = Eod
+ – + – + – + –
d Ed E < Eo
V = Ed
Charged capacitor Electric field Effect on electric field
Charged capacitor with dielectric inserted diagram of same and voltage
(a) (b) (c)
䉱 F I G U R E 1 6 . 1 6 The effects of a dielectric on an isolated capacitor (a) A dielectric material with ran-
domly oriented permanent molecular dipoles (or dipoles induced by the electric field) is inserted
between the plates of an isolated charged capacitor. As the dielectric is inserted, the capacitor tends to
pull it in, thus doing work on it. (Note the attractive forces between the plate charges and those induced
on the dielectric surfaces.) (b) When the material is in the capacitor’s B
electric field, the dipoles orient
themselves with the field, giving rise to an opposing electric field Ed. (c) The dipole field partially can-
cels the field due to the plate charges. The net effect is a decrease in both the electric field and the voltage.
Because the stored charge remains the same, the capacitance increases.
field 1E2 between the plates is reduced, and so is the voltage across the plates
B
(because V = Ed). The dielectric constant k of the material is defined as the ratio
of the voltage with the material in place (V) to the vacuum voltage (Vo). Because
V is proportional to E, this ratio is the same as the electric field ratio:
Vo Eo (only when the capacitor
k = = (16.14)
V E charge is constant)
Note that k is dimensionless and is greater than 1, because V 6 Vo. Equation
16.14 shows that the dielectric constant can be determined by measuring the two
voltages. (Voltmeters are discussed in detail in Chapter 18.) Because the battery
was disconnected and the capacitor isolated, the charge on the plates, Qo, is unaf-
fected. Because V = Vo>k, the value of the capacitance with the dielectric inserted
is larger than the vacuum value by a factor of k. In effect, the same amount of charge
is now stored at a lower voltage, and the result is an increase in capacitance. To
understand this mathematically, apply the definition of capacitance:
Q Qo Qo
= k¢ ≤
1Vo>k2
C = = or C = kCo (16.15)
V Vo
So inserting a dielectric into an isolated capacitor results in a larger capacitance.

But what about energy storage? Because there is no energy input (the battery is
disconnected) and the capacitor does work on the dielectric by pulling it into the
region between the plates, the stored energy drops by a factor of k (䉲 Fig. 16.17a), as
the following calculation shows:
Q2 Q2o 1Q 2o>2Co2 Uo
UC = = = = 6 Uo (battery disconnected)
2C 2kCo k k
䉴 F I G U R E 1 6 . 1 7 Dielectrics and
capacitance (a) A parallel plate + – + – + –
capacitor in air (no dielectric) is + – + – + –
charged by a battery to a charge Qo +Qo + – –Qo +Qo + – –Qo +Qo + – –Qo
and a voltage Vo (left). If the battery + – Disconnect + – + –
is disconnected and the potential + – battery + – + –
across the capacitor is measured by a + – + – + –
voltmeter, a reading of Vo is obtained
Vo
(center). But if a dielectric is now + –
Vo V < Vo
inserted between the capacitor V
plates, the voltage drops to V = Vo>k 2
3
4 2
3
4
(right), so the stored energy 1 0 5 1 0 5
decreases. (Can you estimate the
dielectric constant from the voltage Voltmeter
readings?) (b) A capacitor is charged
as in part (a), but the battery is left (a)
connected. When a dielectric is
inserted into the capacitor, the volt-
age is maintained at Vo. (Why?) + – + – Q > Qo
+ –
+ –
However, the charge on the plates +Qo + – –Qo +Q + –
– –Q
increases to Q = kQo. Therefore, + – + –
+ –
more energy is now stored in the + – + –
+ – + –
capacitor. In both cases, the capaci- + –
+ – + –
tance increases by a factor of k. + –
Vo Vo
+ – + –
(b)
16.4 DIELECTRICS 581
A different situation occurs, however, if the dielectric is inserted and the battery
remains connected. In this case, the voltage stays constant and the battery supplies
more charge to the capacitor—and therefore does work (Fig. 16.17b). Because the
battery does work, the energy stored in the capacitor increases. With the battery
remaining connected, the charge on the plates increases by a factor k, or Q = kQo.
Once again the capacitance increases, but now it is because more charge is stored
(a)
at the same voltage. From the definition of capacitance, the result is the same as
Eq. 16.15, because C = Q>V = kQo>Vo = k1Qo>Vo2 = kCo. Thus,
The effect of a dielectric is to increase the capacitance by a factor of k, regardless of

the conditions under which the dielectric is inserted.
In the case of a capacitor kept at constant voltage, its energy storage increases at
the expense of the battery. To see this, let’s calculate the energy with the dielectric
in place under these conditions:
UC = 12 CV2 = 12 kCo V2o = k A 12 Co V2o B = kUo 7 Uo (battery connected)
For a parallel plate capacitor with a dielectric, the capacitance is increased over
its (air) value in Eq. 16.12 by a factor of k: (b)
keo A 䉱 F I G U R E 1 6 . 1 8 Capacitor
C = kCo = (parallel plates only) (16.16) designs (a) The dielectric material
d between the capacitor plates enables
the plates to be constructed so that
This relationship is sometimes written as C = eA>d, where e = keo is called the they are close together, thus increas-
dielectric permittivity of the material, which is always greater than eo. (How do ing the capacitance. In addition, the
you know this?) plates can then be rolled up into a
compact, more practical capacitor.
A sketch of the inside of a typical cylindrical capacitor and an assortment of real (b) Capacitors (flat, brown circles
capacitors is shown in 䉴 Fig. 16.18. Changes in capacitance can be used to monitor and purple cylinders) among other
motion in our technological world, as Example 16.9 shows. circuit elements in a microcomputer.
EXAMPLE 16.9 The Capacitor as a Motion Detector?

Consider a capacitor filled with a flexible dielectric.
(䉴 Fig. 16.19). The capacitor is connected to a 12.0-V
battery and has a normal (uncompressed) plate separa-
tion of 3.00 mm and a plate area of 0.750 cm2. (a) What
is the required dielectric constant if the capacitance is
1.10 pF? (b) How much charge is stored on the plates +
under normal conditions? (c) How much charge flows Flexible d
V d′
onto the plates (that is, what is the change in their charge) dielectric
–
if they are compressed to a separation of 2.00 mm? (Battery still connected but
not shown)
THINKING IT THROUGH. (a) The capacitance of air-
filled plates can be found from Eq. 16.12, and then the
䉱 F I G U R E 1 6 . 1 9 Capacitors in use Capacitors can be used to con-
dielectric constant can be determined from Eq. 16.15. vert movement into electrical signals that can be measured and ana-
(b) The charge follows from Eq. 16.9. (c) The com- lyzed by computer. As the distance between the plates changes, so
pressed plate separation distance must be used to does the capacitance, which causes a change in the charge on the
recompute the capacitance. Then the new charge can be capacitor. Some early computer keyboards operated. In this way, as
found as in (b) and the charge difference determined. do other instruments such as seismographs (Chapter 13).
SOLUTION. The given data are as follows:

Given: V = 12.0 V Find: (a) k (dielectric constant)
d = 3.00 mm = 3.00 * 10-3 m (b) Q (initial capacitor charge)
A = 0.750 cm2 = 7.50 * 10-5 m2 (c) ¢Q (change in capacitor charge)
C = 1.10 pF = 1.10 * 10-12 F
d¿ = 2.00 mm = 2.00 * 10-3 m
(a) From Eq. 16.12, the capacitance if the plates were separated by air would be
eo A 18.85 * 10-12 C2>N # m2217.50 * 10-5 m22
Co = = = 2.21 * 10-13 F
d 3.00 * 10-3 m
Because the dielectric increases the capacitance, its value is
C 1.10 * 10-12 F
k = = = 4.98
Co 2.21 * 10-13 F
(b) The initial charge is then
Q = CV = 11.10 * 10-12 F2112.0 V2 = 1.32 * 10-11 C
(c) Under compressed conditions, the capacitance is
keo A 14.98218.85 * 10-12 C2>N # m2217.50 * 10-5 m222
C¿ = = = 1.65 * 10-12 F
d¿ 2.00 * 10-3 m
The voltage remains the same, thus Q¿ = C¿V = 11.65 * 10-12 F2112.0 V2 = 1.98 * 10-11 C. Because the capacitance was
increased by the compression, the charge increased by
¢Q = Q¿ - Q = 11.98 * 10-11 C2 - 11.32 * 10-11 C2 = + 6.60 * 10-12 C
As the key is depressed, a charge, whose magnitude is related to the displacement, flows onto the capacitor, providing a way of
measuring the movement electrically.
F O L L O W - U P E X E R C I S E . In this Example, suppose instead that the spacing between the plates were increased by 1.00 mm from
the normal value of 3.00 mm. Would charge flow onto or away from the capacitor? How much charge would this be?
DID YOU LEARN?

➥ The dielectric constant depends on the type of material.
➥ The insertion of a dielectric sheet into a capacitor always increases its capacitance.
➥ The insertion of a dielectric sheet into a charged capacitor may increase or
decrease its energy storage depending on whether it remains connected to a
power supply (voltage source) or not.
16.5 Capacitors in Series and in Parallel

➥ How does the equivalent capacitance of two capacitors in series compare to their
individual capacitance values?
➥ How does the equivalent capacitance of two capacitors in parallel compare to their
individual capacitance values?
➥ When two capacitors in series are connected to a battery, how does the charge on
each of them compare?
➥ When two capacitors in parallel are connected to a battery, how does the voltage
across each of them compare?
Capacitors can be connected in two basic ways: in series or in parallel. In series, the
capacitors are connected head to tail (䉴 Fig. 16.20a). When connected in parallel, all
the leads on one side of the capacitors have a common connection. (Think of all
the “tails” connected together and all the “heads” connected together; Fig. 16.20b.)
CAPACITORS IN SERIES
When capacitors are wired in series, the charge Q must be the same on all the plates:
Q = Q1 = Q2 = Q3 = Á
16.5 CAPACITORS IN SERIES AND IN PARALLEL 583
Q = Q1 = Q2 = Q3
V1 (Q's equal)
+Q1
C1 – Q1
+Q
+ +Q2 +
V C2 V2 V V Cs V = V1 + V2 + V3
– – Q2 –
–Q
+Q3
C3
– Q3 V3
1 1 1 1
= + +
Cs C1 C2 C3
(a) Capacitors in series
Q = Q1 + Q2 + Q3
(Q's not necessarily equal)
+Q1 +Q2 +Q3 +Q

+ +
V C1 C2 C3 V V Cp V
– –
– Q1 – Q2 – Q3 –Q
Cp = C1 + C2 + C3
(b) Capacitors in parallel
Q1 Q2 Q3
Qtotal
V V
Qtotal = Q1 + Q2 + Q3 +
(c) Capacitors in parallel
䉱 F I G U R E 1 6 . 2 0 Capacitors in series and in parallel (a) All capacitors connected in series

have the same charge, and the sum of the voltage drops is equal to the voltage of the bat-
tery. The total series capacitance is equivalent to the value of Cs. (b) When capacitors are
connected in parallel, the voltage drops across the capacitors are the same, and the total
charge is equal to the sum of the charges on the individual capacitors. The total parallel
capacitance is equivalent to the value of Cp. (c) In a parallel connection, thinking of the
plates makes it easier to see why the total charge is the sum of the individual charges. +Q A
–Q B
+ +Q C
V – D
–Q
+Q E
To see why this must be true, examine 䉴Fig. 16.21. Note that only plates A and F
–Q F
are actually connected to the battery. Because the plates labeled B and C are iso-
lated, the total charge on them must always be zero. So if the battery puts a
charge of +Q on plate A, then -Q is induced on B at the expense of plate C, 䉱 F I G U R E 1 6 . 2 1 Charges on
which acquires a charge of +Q. This charge in turn induces -Q on D, and so on capacitors in series Plates B and C
down the line. together had zero net charge to
Remember that the term “voltage drop” is just another name for “change in start. When the battery placed + Q
electrical potential energy per unit charge.” When all the series capacitor voltage on plate A, charge - Q was induced
on B; thus, C must have acquired
drops are added (see Fig 16.20a), their total must be the same as the voltage across +Q for the BC combination to
the battery terminals. Thus, in series, the sum of the individual voltage drops remain neutral. Continuing this
across all the capacitors is equal to the voltage of the source: way through the string, we see that
all the charges must be the same in
V = V1 + V2 + V3 + Á magnitude.
The equivalent series capacitance (Cs) is defined as the value of a single capac-
itor that could replace the series combination and store the same charge and
energy at the same voltage. Because the combination of capacitors stores a charge
of Q at a voltage of V, it follows that Cs = Q>V, or V = Q>Cs. However, the indi-
vidual voltages are related to the individual charges by V1 = Q>C1, V2 = Q>C2,
V3 = Q>C3, and so on.
Substituting these expressions into the voltage equation,
Q Q Q Q
= + + + Á
Cs C1 C2 C3
Canceling the common Q’s, the result is
1 1 1 1
= + + + Á (equivalent series capacitance) (16.17)
Cs C1 C2 C3
This means that Cs is always smaller than the smallest capacitance in the series
combination. For example, try Eq. 16.17 with C1 = 1.0 mF and C2 = 2.0 mF. You
should be able to show that Cs = 0.67 mF, which is less than 1.0 mF (the general
proof will be left to you). Physically, the reasoning goes like this. In series, all the
capacitors have the same charge, so the charge stored by this arrangement is
Q = Ci Vi (where the subscript i refers to any of the individual capacitors in the
string). Because Vi 6 V, the series arrangement stores less charge than any indi-
vidual capacitor connected by itself to the same battery.
It also makes sense that in series the smallest capacitance receives the largest
voltage. A small value of C means less charge stored per volt. In order for the
charge on all the capacitors to be the same, the smaller the value of capacitance,
the larger the fraction of the total voltage required 1Q = CV2.
CAPACITORS IN PARALLEL
With a parallel arrangement (Fig. 16.20b), the voltages across the capacitors are the
same (why?), and each individual voltage is equal to that of the battery:
V = V1 = V2 = V3 = Á
The total charge is the sum of the charges on each capacitor (Fig 16.20c):
Qtotal = Q1 + Q2 + Q3 + Á
The equivalent capacitance in parallel is expected to be larger than the largest

capacitance, because more charge per volt can be stored in this way than if any one
capacitor were connected to the battery by itself. The individual charges are given
by Q1 = C1 V, Q2 = C2 V, and so on. A capacitor with the equivalent parallel
capacitance (Cp) would hold this same total charge when connected to the battery,
so Cp = Qtotal>V, or Qtotal = Cp V. Substituting these expressions into the previous
equation gives
Cp V = C1 V + C2 V + C3 V + Á
and canceling the common V results in
Cp = C1 + C2 + C3 + Á (equivalent parallel capacitance) (16.18)
In the parallel case, the equivalent capacitance Cp is the sum of the individual
capacitances. In this case, the equivalent capacitance is larger than the largest indi-
vidual capacitance. Because capacitors in parallel have the same voltage, the
largest capacitance will store the most charge (and energy). For a comparison of
capacitors in series and in parallel, consider Example 16.10.
EXAMPLE 16.10 Charging Capacitors in Series and in Parallel

Given two capacitors, one with a capacitance of 2.50 mF and T H I N K I N G I T T H R O U G H . (a) Capacitors in series have the
the other with that of 5.00 mF, what are the charges on each same charge. From Eq. 16.17 the equivalent capacitance can
and the total charge stored if they are connected across a be found and then the charge on each capacitor can be deter-
12.0-V battery (a) in series and (b) in parallel? mined. (b) Capacitors in parallel have the same voltage; from
that, the charge on each can be easily determined because
their individual capacitances are known.

Given: C1 = 2.50 mF = 2.50 * 10-6 F Find: (a) Q (on each capacitor in series) and Qtotal (total charge)
C2 = 5.00 mF = 5.00 * 10-6 F (b) Q (on each capacitor in parallel) and Qtotal (total charge)
V = 12.0 V
(a) In series, the total (equivalent) capacitance is determined as follows:
1 1 1 3
= + =
Cs 2.50 * 10 F
-6
5.00 * 10 F
-6
5.00 * 10-6 F
so
Cs = 1.67 * 10-6 F
(Note: Cs is less than the smallest capacitance in the series, as expected.)
Because the charge on each capacitor is the same in series (and the same as the total), we have
Qtotal = Q1 = Q2 = Cs V = 11.67 * 10-6 F2112.0 V2 = 2.00 * 10-5 C
(b) Here, the parallel equivalent capacitance relationship is used:
Cp = C1 + C2 = 2.50 * 10-6 F + 5.00 * 10-6 F = 7.50 * 10-6 F
(This result is reasonable because it is greater than the largest individual value in the parallel arrangement.)
Therefore,
Qtotal = Cp V = 17.50 * 10-6 F2112.0 V2 = 9.00 * 10-5 C
In parallel, each capacitor has the full 12.0 V across it; hence,
Q1 = C1 V = 12.50 * 10-6 F2112.0 V2 = 3.00 * 10-5 C
Q2 = C2 V = 15.00 * 10-6 F2112.0 V2 = 6.00 * 10-5 C
As a final double check, notice that the total stored charge is equal to the sum of the charges on both capacitors.
F O L L O W - U P E X E R C I S E . In this Example, determine which combination, series or parallel, stores the most energy by calculating
how much is stored in each arrangement.
Capacitor arrangements generally can involve both series and parallel connec-
tions, as shown in Integrated Example 16.11. In this situation, the circuit is simpli-
fied, using the equivalent parallel and series capacitance expressions, until it results
in one single, overall equivalent capacitance. To find the results for each individual
capacitor, the steps are undone until the original arrangement is reached.
INTEGRATED EXAMPLE 16.11 Forward, Then Backward: A Series–Parallel Combination of Capacitors

Three capacitors are connected in a circuit as shown in (A) CONCEPTUAL REASONING. For part (a), consider that C1
䉲 Fig. 16.22a. (a) By looking at the capacitance values, how do and C2 are in parallel, hence they have the same voltage. Thus
the voltages across the various capacitors compare: answers (1) and (4) cannot be correct. Their capacitances (C1
(1) V3 7 V2 7 V1 , (2) V3 6 V2 = V1 , (3) V3 = V2 = V1 , or and C2) add to 0.30 mF, which is less than that of C3. Since the
(4) V3 = V2 7 V1? (b) By looking at the capacitance values, parallel combination (C1 and C2) is in series with C3, most of
how does the energy in C3 compare to the total energy stored the 12-V drop will occur across them. So the correct choice is
in C1 plus C2: (1) U3 7 U1 + 2 , (2) U3 = U1 + 2 , or (3) U3 6 U1 + 2? (2) V3 6 V2 = V1 . For part (b), because C1 and C2 (as a parallel
(c) Determine the voltages across each capacitor. (d) Deter- combination) are in series with C3, the parallel combination
mine the stored energy in each capacitor and use your results
to validate your choice in part (b). (continued on next page)
has the same (total) charge as does C3. In series, the lowest tion to a single equivalent capacitance. Two of the capacitors are
value of capacitance has the largest energy storage, thus the in parallel. Their single equivalent capacitance (Cp) is itself in
correct choice is (3) U3 6 U1 + 2 . series with the last capacitor—a fact that enables the total capac-
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . For part (c), itance to be found. Working backward will allow the voltage
the voltage across each capacitor could be found from V = Q>C across each capacitor to be found. For part (d), once the voltages
if the charge on each capacitor were known. The total charge on are known, individual energy storages can be calculated most
the capacitors is found by reducing the series–parallel combina- conveniently using UC = 12 CV2.
C3 Given: Values of capacitance and Find: (c) V1, V2, and V3

0.60 µ F battery voltage given in (voltages across each capacitor)
Figure 16.22a (d) U1, U2, and U3
V (energy stored in each capacitor)
C1 C2 (c) Starting with the parallel combination,
12 V
Cp = C1 + C2 = 0.10 mF + 0.20 mF = 0.30 mF
0.10 µ F 0.20 µ F Now the arrangement is partially reduced, as shown in Fig. 16.22b. Next, consider-
ing Cp in series with C3, the total, or equivalent, capacitance of the arrangement is
found as follows:
(a) 1 1 1 1 1 1 2 1
= + = + = + =
Cs C3 Cp 0.60 mF 0.30 mF 0.60 mF 0.60 mF 0.20 mF
Therefore,
V3 C3 Cs = 0.20 mF = 2.0 * 10-7 F
This is the equivalent capacitance of the arrangement when the circuit is reduced to
one equivalent capacitor in Fig. 16.22c. Treating the problem as one single capacitor,
the charge on that equivalent capacitor is:
V12 Cp Q = Cs V = 12.0 * 10-7 F2112 V2 = 2.4 * 10-6 C
This is the charge on C3 and Cp because they are in series. This fact can be used to
calculate the voltage across C3:
(b) Q 2.4 * 10-6 C
V3 =
= = 4.0 V
C3 6.0 * 10-7 F
The sum of the voltages across the capacitors equals the voltage across the battery
terminals. The voltages across C1 and C2 are the same because they are in parallel.
V
Because the voltage across C1 (or C2) plus the voltage across C3 equals the total volt-
Cs = Ctotal age (the battery voltage), V = V12 + V3 = 12 V. (See Fig. 16.22a.) Here, V12 repre-
sents the voltage across either C1 or C2. Thus
(c) V12 = V - V3 = 12 V - 4.0 V = 8.0 V
䉱 F I G U R E 1 6 . 2 2 Circuit reduction (d) The individual energies are found as follows:
U1 = 12 C1 V1 = 12 10.10 * 10-6 F218.0 V22 = 3.2 * 10-6 J = 3.2 mJ
When capacitances are combined, the 2
combination of capacitors is reduced to
U2 = 12 C2 V2 = 12 10.20 * 10-6 F218.0 V22 = 6.4 * 10-6 J = 6.4 mJ
a single equivalent capacitance. 2
and
U3 = 12 C3 V3 = 12 10.60 * 10-6 F214.0 V22 = 4.8 * 10-6 J = 4.8 mJ
2
The total energy stored in capacitors 1 and 2 is 9.6 mJ, which is greater than that
stored in capacitor 3.
F O L L O W - U P E X E R C I S E . In this Example, (a) what value of capacitance for capacitor 2 would make V3 = V1? (b) After the change
is made in part (a), what is the ratio of energy stored in capacitor 3 to that stored in capacitor 1?
DID YOU LEARN?

➥ The series equivalent capacitance is smaller than the smallest capacitance value in
the string.
➥ The parallel equivalent capacitance is larger than the largest capacitance value in
the arrangement.
➥ In series, all capacitors acquire the same charge.
➥ In parallel, all capacitors have the same voltage.
PULLING IT TOGETHER Capacitors, Batteries, Fields and Work

An air-filled capacitor consists of two circular metal plates T H I N K I N G I T T H R O U G H . (a) This is a straightforward calcula-
1.50 mm apart, each with a diameter of 5.16 cm. This capaci- tion using the expression for a parallel plate capacitor. (b) The
tor is connected to a 12.0-V car battery and is fully charged. charge is determined by knowing the plate voltage and the
(a) What is the capacitance of this arrangement? (b) How capacitance [part (a)]. The electric field is the rate of change of
much charge is on each plate and what is the magnitude of electric potential with distance between the plates. (c) Energy
the electric field between the plates? (c) Calculate the energy storage is most easily found using the voltage and capaci-
stored in this capacitor. (d) Suppose the capacitor is discon- tance. (d) When a new battery with a lower terminal voltage
nected from this battery and reconnected to a 3.00-V battery is used, the result will be less charge on the plates. From
instead. How much charge is moved between the plates and determining the new charge, the difference can be found.
in what direction? (e) How much work is required to move (e) The work done by the new battery will be equal to the
this charge? change in stored capacitor energy.
SOLUTION. The data are listed, and converted to meters.

Given: D = 5.16 cm = 5.16 * 10-2 m (plate diameter) Find: (a) C (capacitance)
d = 1.50 mm = 1.50 * 10-3 m (plate separation) (b) Q (charge on the capacitor) and E (electric field
V1 = 12.0 V (initial battery voltage) between its plates)
V2 = 3.00 V (final battery voltage) (c) UC1 (energy stored in the capacitor)
(d) ¢Q (charge moved and its direction)
(e) W (work to move the charge)
(a) The expression for the capacitance of a parallel plate arrangement yields
eo A eopR2
C = =
d d
5.16 * 10-2 m 2
18.85 * 10-12 C2>N # m22p ¢ ≤
2
= = 1.23 * 10-11 F
1.50 * 10-3 m
(b) The charge (magnitude) on each plate is
Q = CV1 = 11.23 * 10-11 F2112.0 V2 = 1.48 * 10-10 C
The electric field (magnitude) between the plates is constant and given by the rate of change of potential with distance or
V1 12.0 V
E = = = 8.00 * 103 V>m
d 1.50 * 10-3 m
(c) Using the voltage across the plates and the capacitance, the stored energy is
2
UC1 = 12 CV1
2 11.23
1
= * 10-11 F2112.0 V22 = 8.86 * 10-10 J
(d) A new (lower voltage) battery means less charge on the capacitor plates. Thus the charge on each plate must be reduced and
¢Q will be negative. The battery does this by moving some negative charge (electrons) to the positive plate, accomplishing the
reduction of charge on both plates at the same time. The new charge (Q¿ ) on the battery is Q¿ = CV2 = 11.23 * 10-11 F213.00 V2 =
0.370 * 10-10 C. Therefore the change in the charge on the capacitor is
¢Q = Q¿ - Q = 0.370 * 10-10 C - 1.48 * 10-10 C
= - 1.11 * 10-10 C
(e) Since the electrons are naturally attracted to the positive plate, and since the capacitor’s energy storage will be reduced, the
work done by the battery will be negative. (Recall that charging a capacitor requires positive work by the battery.) The new bat-
tery’s work is equal to the difference in stored energy in the capacitor, so the new stored energy must be found:
2
UC2 = 12 CV2
= 1
2 = 11.23 * 10-11 F213.00 V22 = 5.34 * 10-11 J
Thus the new battery’s work is
Wbatt = UC2 - UC1 = 0.534 * 10-10 J - 8.86 * 10-10 J
= - 8.31 * 10-10 J
■ The electric potential difference (or voltage) between two ■ The electric field is related to the rate of change of electric
potential with distance. The electric field 1E2 is in the direc-
B
points is the work done per unit positive charge between
those two points, or the change in electric potential energy tion of the most rapid decrease in electric potential (V). The
per unit positive charge. Expressed in equation form, this electric field magnitude (E) is the rate of change of the
relationship is potential with distance, or
¢Ue W
E = ` `
¢V = = (16.1) ¢V
q+ q+ (16.8)
¢x max
A
■ The electron-volt (eV) is the kinetic energy gained by an
rA
electron or a proton accelerated through a potential differ-
I ence of 1 volt.
HIGHER
LOWER
+ POTENTIAL
rB
POTENTIAL
II
■ A capacitor is any arrangement of two metallic plates.
B Capacitors store charge on their plates, and therefore elec-
tric energy.
■ Capacitance is a quantitative measure of how effective a
capacitor is in storing charge. It is the magnitude of the
■ Equipotential surfaces (surfaces of constant electric poten- charge stored on either plate per volt, or
tial, also called equipotentials) are surfaces on which a
charge has a constant electric potential energy. These sur- Q
faces are everywhere perpendicular to the electric field. Q = CV or C = (16.9)
V
+Q –Q
VB VA + –
VA + –
+ – Q = +Q = CV
+ –
+ –
+ –
+ E –
+ – V
■ The expression for the electric potential due to a point + –
+ –
charge (choosing V = 0 at r = q ) is + –
kq + –
V = (16.4) Battery
r Metal
plates
■ The electric potential energy for a pair of point charges is
(choosing U = 0 at r = q ) ■ The capacitance of a parallel plate capacitor (in air) is
kq1 q2
U12 = (16.5) eo A
r12 C = (16.12)
d
r12 Very
where eo = 8.85 * 10-12 C2>1N # m22 is called the

distant
+ + +
q1 q2
(U = 0)
kq1q2 permittivity of free space.
U12 =
r12
■ The energy stored in a capacitor depends on its capacitance
■ The electric potential energy of a configuration of more and the charge the capacitor stores (or, equivalently, the
than two point charges is the sum of point charge pair voltage across its plates). There are three equivalent expres-
terms from Eq. 16.5: sions for this energy:
Utotal = U12 + U23 + U13 + Á (16.6)
Q2
UC = 12 QV = = 12 CV2 (16.13)
− q3 2C
r13
■ A dielectric is a nonconducting material that increases
r23 capacitance.
q1 +
■ The dielectric constant K describes the effect of a dielectric
r12
on capacitance. A dielectric increases the capacitor’s capaci-
+ tance over its value with air between the plates by a factor of k
q2
U = U12 + U23 + U13
C = kCo (16.15)
■ Capacitors in series are equivalent to one capacitor, with a ■ Capacitors in parallel are equivalent to one capacitor, with a
capacitance called the equivalent series capacitance Cs. The capacitance called the equivalent parallel capacitance Cp.
equivalent series capacitance is In parallel, all the capacitors have the same voltage. The
1 1 1 1 equivalent parallel capacitance is
= + + + Á (16.17)
Cs C1 C2 C3 Cp = C1 + C2 + C3 + Á (16.18)
Q = Q1 = Q2 = Q3 Q = Q1 + Q2 + Q3
V1 (Q's equal) (Q's not necessarily equal)
+Q1
C1
– Q1
+Q +Q1 +Q2 +Q3 +Q
+ +Q2 + + +
V C2 V2 V V Cs V = V1 + V2 + V3 V C1 C2 C3 V V Cp V
– – Q2 – – –
–Q – Q1 – Q2 – Q3 –Q
+Q3
C3
– Q3
V3
1 = 1 + 1 + 1
Cp = C1 + C2 + C3
Cs C1 C2 C3

16.1 ELECTRIC POTENTIAL ENERGY AND (c) it stays the same, or (d) you can’t tell from the infor-
ELECTRIC POTENTIAL DIFFERENCE mation given?
1. The SI unit of electric potential difference is the (a) joule,
(b) newton per coulomb, (c) newton-meter, (d) joule per 16.2 EQUIPOTENTIAL SURFACES AND
coulomb. THE ELECTRIC FIELD
2. How does the electrostatic potential energy of a system 8. On an equipotential surface (a) the electric potential is
of two positive point charges change when the distance constant, (b) the electric field is zero, (c) the electric
between them is tripled: (a) it is reduced to one-third its potential is zero, (d) there must be equal amounts of neg-
original value, (b) it is reduced to one-ninth its original ative and positive charge.
value, (c) it is unchanged, or (d) it is increased to three
9. Equipotential surfaces (a) are parallel to the electric field,
times its original value?
(b) are perpendicular to the electric field, (c) can be at
3. An electron is moved from the positive plate to the nega- any angle with respect to the electric field.
tive plate of a charged parallel plate arrangement. How
10. An electron is moved from an equipotential surface at
does the sign of the change in the system’s electrostatic
+ 5.0 V to one at + 10.0 V. It is moving generally in a
potential energy compare to the sign of the change in
direction (a) parallel to the electric field, (b) opposite to
electrostatic potential the electron experiences: (a) both
the electric field, (c) you can’t tell how its direction com-
are positive, (b) the energy change is positive and the
pares to that of the electric field from the data given.
potential change is negative, (c) the energy change is
negative and the potential change is positive, or (d) both 11. As an electron is moved perpendicularly farther away
are negative? from a large uniformly charged plate, the system’s elec-
trostatic potential energy is observed to decrease. The
4. An isolated system consists of three point charges. Two
charge on the plate is (a) positive, (b) negative, (c) zero.
are negative and one is positive. What can you say about
the sign of this system’s electrostatic potential energy: 12. An electron is first moved perpendicularly away from a
(a) it is positive, (b) it is is negative, (c) it is zero, or large uniformly positively charged plate. Then after
(d) you can’t tell from the information given? removing the electron, the identical movement is repeated
with a proton. How do the electric potential differences
5. A small positive point charge is fixed at the origin, and experienced by each compare: (a) the electron’s is larger,
an electron is brought near it from a large distance away. (b) the proton’s is larger, or (c) they are the same?
The magnitude of the work taken to move the electron is
13. If the equipotential surfaces due to some charge distribu-
W. The change in the system’s electrostatic potential
tion are vertical planes, what can you say about the elec-
energy is (a) + W, (b) -W, (c) zero, (d) unrelated to W.
tric field direction in this region: (a) it is vertically
6. The spacing between two closely spaced oppositely upward, (b) it is vertically downward, (c) it is horizon-
charged parallel plates is decreased. What happens to tally to the left, (d) it is horizontally to the right, or
the electrostatic potential difference between the plates, (e) either (c) or (d) could be correct?
assuming they form an isolated system: (a) it increases, 14. A proton with an initial kinetic energy of 9.50 eV is fired
(b) it decreases, (c) it stays the same, or (d) you can’t tell directly at another proton whose location is fixed. When
from the information given? the moving proton has reached its point of closest
7. The amount of charge on each of two closely spaced approach, by how much has the electric potential energy
oppositely charged parallel plates is increased equally. of this two-particle system changed: (a) + 9.50 eV,
What happens to the electrostatic potential difference (b) -9.50 eV, (c) zero, or (d) it depends on the distance of
between the plates: (a) it increases, (b) it decreases, closest approach, so you can’t tell from the data given?
16.3 CAPACITANCE charge, (d) causes the plates to discharge because the
dielectric is a conductor.
15. The SI unit of capacitance is the farad, which is equiva-
lent to which of the following: (a) coulomb>volt, 23. A parallel plate capacitor is connected to a battery and
(b) joule, (c) volt, (d) coulomb, or (e) joule>coulomb? remains so. If a dielectric is then inserted between the
plates, (a) the capacitance decreases, (b) the voltage
16. How do the SI units of the permittivity of free space increases, (c) the voltage decreases, (d) the charge
compare with those of the Coulomb force constant k: increases.
(a) they are the same, (b) they are the inverse of one
another, or (c) they are unrelated? 24. A parallel plate capacitor is first connected to a battery
for a while and then disconnected from the battery. If a
17. A capacitor is first connected to a 6.0-V battery and then dielectric is then inserted between the plates, what hap-
disconnected and connected to a 12.0-V battery. How pens to the charge on its plates: (a) the charge decreases,
does its capacitance change: (a) it increases, (b) it (b) the charge increases, or (c) the charge stays the same?
decreases, or (c) it stays the same?
25. A parallel plate capacitor is connected to a battery and
18. A capacitor is first connected to a 6.0-V battery and then remains so. If a dielectric is then inserted between the
disconnected and connected to a 12.0-V battery. How plates, what happens to the electric field there: (a) it
does the charge on one of its plates change: (a) it decreases, (b) it increases, (c) it remains the same, (d) the
increases, (b) it decreases, or (c) it stays the same? field may increase, decrease, or not change depending
19. A capacitor is first connected to a 6.0-V battery and then on the dielectric constant?
disconnected and connected to a 12.0-V battery. By how 26. A parallel plate capacitor is first connected to a battery
much does the electric field strength between its plates for a while and then disconnected from the battery. If a
change: (a) it is two times greater, (b) it is four times dielectric is then inserted between the plates, what hap-
greater, or (c) it stays the same? pens to the electric field there: (a) it decreases, (b) it
20. The distance between the plates of a capacitor is cut in increases, (c) it remains the same, (d) the field may
half. By what factor does its capacitance change: (a) it is increase, decrease, or not change depending on the
cut in half, (b) it is reduced to one-fourth its original dielectric constant?
value, (c) it is doubled, or (d) it is quadrupled?
21. The area of the plates of a capacitor is reduced. How 16.5 CAPACITORS IN SERIES AND IN
should the distance between those plates be adjusted to PARALLEL
keep the capacitance constant: (a) increase it,
(b) decrease it, or (c) changing the distance cannot make 27. Capacitors in series must have the same (a) voltage,
up for the plate area change? (b) charge, (c) energy storage, (d) none of these.
28. Capacitors in parallel must have the same (a) voltage,
(b) charge, (c) energy storage, (d) none of these.
16.4 DIELECTRICS
29. Capacitors 1, 2, and 3 have the same capacitance value C.
22. Putting a dielectric into a charged parallel plate capaci- Capacitors 1 and 2 are in series and their combination is
tor that is not connected to a battery (a) decreases the in parallel with 3. What is their effective total capaci-
capacitance, (b) decreases the voltage, (c) increases the tance: (a) C, (b) 1.5C, (c) 3C, or (d) C>3?
16.1 ELECTRIC POTENTIAL ENERGY AND 6. If two locations are at the same electrical potential, how
ELECTRIC POTENTIAL DIFFERENCE much work does it take to move a charge from the first
location to the second? Explain.
1. What is the difference between (a) electrostatic potential
energy and electric potential and (b) electric potential
difference and voltage? 16.2 EQUIPOTENTIAL SURFACES AND
2. When a proton approaches another fixed proton, what THE ELECTRIC FIELD
happens to (a) the kinetic energy of the approaching pro-
7. Sketch the topographic map you would expect as you
ton, (b) the electric potential energy of the system, and
walk away from the ocean up a gently sloping uniform
(c) the total energy of the system?
beach. Label the gravitational equipotentials as to rela-
3. Using the language of electrical potential and energy tive height and potential value. Show how to predict,
(not forces), explain why positive charges speed up as from the map, which way a ball would accelerate if it
they approach negative charges. was initially rolled up the beach away from the water.
4. An electron is released in a region where the electric
potential decreases to the left. Which way will the elec- 8. Explain why two equipotential surfaces cannot intersect.
tron begin to move? Explain. 9. Suppose a charge starts at rest on an equipotential sur-
5. An electron is released in a region where the electric face, is moved off that surface, and is eventually
potential is constant. Which way will the electron accel- returned to the same surface at rest after a round trip.
erate? Explain. How much work did it take to do this? Explain.
EXERCISES 591
10. What geometrical shape are the equipotential surfaces factor does the electric field between the plates of the
between two charged parallel plates? capacitor change?
11. (a) What is the approximate shape of the equipotential sur-
faces inside the axon cell membrane? (See Insight 16.1
16.4 DIELECTRICS
Fig. 1.) (b) Under resting potential conditions, where inside
the membrane is the region of highest electric potential? 18. Give several reasons why a conductor would not be a
(c) What about during reversed polarity conditions? good choice as a dielectric for a capacitor.
12. Near a fixed positive point charge, if you move from one 19. A parallel plate capacitor is connected to a battery and
equipotential surface to another with a smaller radius, then disconnected. If a dielectric is inserted between the
(a) what happens to the value of the potential? (b) What plates, what happens to (a) the capacitance, (b) the volt-
was your general direction relative to the electric field? age across the capacitor’s plates, and (c) the electric field
13. (a) If a proton is accelerated from rest by a potential dif- between the plates?
ference of 1 million volts, how much kinetic energy does 20. Explain why the electric field between two parallel
it gain? (b) How would your answer to part (a) change if plates of a capacitor decreases when a dielectric is
the accelerated particle had twice the charge of the pro- inserted if the capacitor is not connected to a power sup-
ton (same sign) and four times the mass? ply, but remains the same when it is connected to a
14. (a) Can the electric field at a point be zero while there is power supply.
also a nonzero electric potential at that point? (b) Can the
electric potential at a point be zero while there is also a 16.5 CAPACITORS IN SERIES AND IN
nonzero electric field at that point? Explain. If your PARALLEL
answer to either part is yes, give an example.
21. Under what conditions would two capacitors in series
have the same voltage across them? What if they were in
16.3 CAPACITANCE parallel?
15. If the plates of an isolated parallel plate capacitor are 22. Under what conditions would two capacitors in parallel
moved farther apart from each other, does the energy have the same charge on them? What if they were in series?
storage increase, decrease, or remain the same? Explain. 23. If you are given two capacitors, how should you connect
16. If the potential difference across a capacitor is doubled, them to get (a) maximum equivalent capacitance and
what happens to (a) the charge on the capacitor and (b) minimum equivalent capacitance?
(b) the energy stored in the capacitor? 24. You have N (an even number Ú 2) identical capacitors,
17. A capacitor is connected to a 12-V battery. If the plate each with a capacitance of C. In terms of N and C, what
separation is tripled and the capacitor remains con- is their total effective capacitance (a) if they are all con-
nected to the battery, (a) by what factor does the charge nected in series? (b) If they are all connected in parallel?
on the capacitor change? (b) By what factor does the (c) If two halves (N>2 each) are connected in series and
energy stored in the capacitor change? (c) By what these two sets are connected in parallel?
EXERCISES
16.1 ELECTRIC POTENTIAL ENERGY AND 4. ● An electron is accelerated by a uniform electric field
ELECTRIC POTENTIAL DIFFERENCE 11000 V>m2 pointing vertically upward. Use Newton’s
laws to determine the electron’s velocity after it moves
1. ● A pair of parallel plates is charged by a 12-V battery. If
0.10 cm from rest.
the electric field between the plates is 1200 N>C, how far
apart are the plates? 5. ● (a) Repeat Exercise 4, but find the speed by using energy
2. ● A pair of parallel plates is charged by a 12-V battery.
methods. Find the direction in which the electron is mov-
How much work is required to move a particle with a ing by considering electric potential energy changes.
charge of - 4.0 mC from the positive to the negative (b) Does the electron gain or lose potential energy?
plate? 6. IE ● Consider two points at different distances from a
3. ● If it takes + 1.6 * 10-5 J to move a positively charged positive point charge. (a) The point closer to the charge is
particle between two charged parallel plates, (a) what is at a (1) higher, (2) equal, (3) lower potential than the
the charge on the particle if the plates are connected to a point farther away. Why? (b) How much different is the
6.0-V battery? (b) Was it moved from the negative to the electric potential 20 cm from a charge of 5.5 mC com-
positive plate or from the positive to the negative plate? pared to 40 cm from the same charge?
7. IE ● ● (a) At one-third the original distance from a posi- q1 = + 4.0 ␮C

tive point charge, by what factor is the electric potential
changed: (1) 1>3, (2) 3, (3) 1>9, or (4) 9? Why? (b) How far
from a +1.0-mC charge is a point with an electric poten-
cm
20
tial value of 10 kV? (c) How much of a change in poten-
cm
20
tial would occur if the point were moved to three times
that distance?
20 cm
8. IE ● ● According to the Bohr model of the hydrogen
atom (see Chapter 27), the electron can exist only in cir- q2 = +4.0 ␮C q3 = −4.0 ␮C
cular orbits of certain radii about a proton. (a) Will a
larger orbit have (1) a higher, (2) an equal, or (3) a lower 䉱 F I G U R E 1 6 . 2 4 A charge triangle See Exercises 15 and 17.
electric potential than a smaller orbit? Why? (b) Deter-
mine the potential difference between two orbits of radii 16. ●● Compute the energy necessary to bring together
0.21 nm and 0.48 nm. (from a very large distance) the charges in the configura-
9. ●● In Exercise 8, by how much does the potential energy tion shown in 䉲 Fig. 16.25.
of the atom change if the electron changes location (a) q1 = −10 ␮C q2 = −10 ␮C
from the lower to the higher orbit, (b) from the higher to
the lower orbit, and (c) from the larger orbit to a very 0.10 m
large distance?
0.10 m 0.10 m
10. ●● How much work is required to completely separate
two charges (each - 1.4 mC) and leave them at rest if they
were initially 8.0 mm apart? 0.10 m
11. ●● In Exercise 10, if the two charges are released at their q4 = +5.0 ␮C q3 = +5.0 ␮C
initial separation distance, how much kinetic energy
would each have when they are very distant from one 䉱 F I G U R E 1 6 . 2 5 A charge rectangle See Exercises 16 and 18.
another?
17. ● ● ● What is the value of the electric potential at (a) the
12. ●● It takes + 6.0 J of work to move two charges from a center of the triangle and (b) a point midway between q2
large distance apart to 1.0 cm from one another. If the and q3 in Fig. 16.24?
charges have the same magnitude, (a) how large is each 18. ● ● ● What is the value of electric potential at (a) the cen-
charge, and (b) what can you tell about their signs? ter of the square and (b) a point midway between q1 and
13. ●● A + 2.0-mC charge is initially 0.20 m from a fixed q4 in Fig. 16.25?
-5.0-mC charge and is then moved to a position 0.50 m 19. IE ● ● ● In a computer monitor, electrons are accelerated
from the fixed charge. (a) How much work is required to from rest through a potential difference in an “electron
move the charge? (b) Does the work depend on the path gun” arrangement (䉲 Fig. 16.26). (a) Should the left side
through which the charge is moved? of the gun be at (1) a higher, (2) an equal, or (3) a lower
potential than the right side? Why? (b) If the potential
14. ●● An electron is moved from point A to point B and
difference in the gun is 5.0 kV, what is the “muzzle
then to point C along two legs of an equilateral triangle
speed” of the electrons emerging from the gun? (c) If the
with sides of length 0.25 m (䉲 Fig. 16.23). If the horizontal
gun is directed at a screen 25 cm away, how long do the
electric field is 15 V>m, (a) what is the magnitude of the
electrons take to reach the screen?
work required? (b) What is the potential difference
between points A and C? (c) Which point is at a higher
potential? 䉳 F I G U R E 1 6 . 2 6 Electron
speed See Exercise 19.
Electron gun
B
10 kV
0.25 m 0.25 m E = 15 V/m
35 cm
C 0.25 m A 16.2 EQUIPOTENTIAL SURFACES AND

THE ELECTRIC FIELD
䉱 F I G U R E 1 6 . 2 3 Work and energy See 20. A uniform electric field of 10 kV>m points vertically
●
Exercise 14. upward. How far apart are the equipotential planes that
differ by 100 V?
15. ●● Compute the energy necessary to bring together 21. ● In Exercise 20, if the ground is designated as zero
(from a very large distance) the charges in the configura- potential, how far above the ground is the equipotential
tion shown in 䉴 Fig. 16.24. surface corresponding to 7.0 kV?
EXERCISES 593
22. ● Determine the potential 2.5 mm from the negative 33. ● ● ● Consider a point midway between the two large
plate of a pair of parallel plates separated by 20.0 mm charged plates in Fig. 16.27. Compute the change in electric
and connected to a 24-V battery. potential if from there you moved (a) 1.0 mm toward the
23. ● Relative to the positive plate in Exercise 22, where is positive plate, (b) 1.0 mm toward the negative plate, and
the point with a potential of 14 V? (c) 1.0 mm parallel to the plates. What do your answers tell
24. ● If the radius of the equipotential surface of a point
you about the direction of the electric field in that region?
charge is 10.5 m and is at a potential of + 2.20 kV (com- 34. ● ● ● Repeat Exercise 33 if the plates are instead con-
pared to zero at infinity), what are the magnitude and nected to a 24-V battery. Also determine the electric field
sign of the point charge? (direction and magnitude) at the midway point between
25. IE ● (a) The equipotential surfaces in the neighborhood of the plates. Compare your answers to Exercise 33 and
a positive point charge are spheres. Which sphere is asso- comment on the source of any differences.
ciated with the higher electric potential: (1) the smaller
one, (2) the larger one, or (3) they are associated with the
same potential? (b) Calculate the amount of work (in 16.3 CAPACITANCE
electron-volts) it would take to move an electron from
35. ● How much charge flows through a 12-V battery when
12.6 m to 14.3 m away from a +3.50-mC point charge.
a 2.0-mF capacitor is connected across its terminals?
26. ● The potential difference between the cloud and
ground in a typical lightning discharge may be up to 36. ● A parallel plate capacitor has a plate area of 0.525 m2
100 MV (million volts). What is the gain in kinetic energy and a plate separation of 2.15 mm. What is its capacitance?
of an electron accelerated through this potential differ-
37. ● What plate separation is required for a parallel plate
ence? Give your answer in both electron-volts and
capacitor to have a capacitance of 9.00 nF if the plate area
joules. (Assume that there are no collisions.)
is 0.425 m2?
27. ● In a typical Van de Graaff linear accelerator, protons are
accelerated through a potential difference of 20 MV. What is 38. IE ● (a) For a parallel plate capacitor with a fixed plate
their kinetic energy if they started from rest? Give your separation distance, a larger plate area results in (1) a
answer in (a) eV, (b) keV, (c) MeV, (d) GeV, and (e) joules. larger capacitance value, (2) an unchanged capacitance
value, (3) a smaller capacitance value. (b) A 2.50-nF par-
28. ● In Exercise 27, how do your answers change if a doubly
charged 1+ 2e2 alpha particle is accelerated instead? (An

allel plate capacitor has a plate area of 0.514 m2. If the
alpha particle consists of two neutrons and two protons.) plate area is doubled, what is the new capacitance value?
29. ● ● In Exercises 27 and 28, compute the speed of the pro- 39. ●● A 12.0-V battery remains connected to a parallel plate
ton and alpha particle after being accelerated. capacitor with a plate area of 0.224 m2 and a plate sepa-
30. ● ● Calculate the voltage required to accelerate a beam of
ration of 5.24 mm. (a) What is the charge on the capaci-
protons initially at rest, and calculate their speed if they tor? (b) How much energy is stored in the capacitor?
have a kinetic energy of (a) 3.5 eV, (b) 4.1 keV, and (c) What is the electric field between its plates?
(c) 8.0 * 10-16 J. 40. ●● If the plate separation of the capacitor in Exercise 39
31. ● ● Repeat the calculation in Exercise 30 for electrons changed to 10.48 mm after the capacitor is disconnected
instead of protons. from the battery, how do your answers change?
32. ● ● ● Two large parallel plates are separated by 3.0 cm and
41. ●● Current state-of-the-art capacitors are capable of
connected to a 12-V battery. Starting at the negative plate
storing many times the energy of older ones. Such a
and moving 1.0 cm toward the positive plate at a 45° angle
capacitor, with a capacitance of 1.0 F, is able to light a
(䉲 Fig. 16.27), (a) what value of potential would be
small 0.50-W bulb at steady full power for 5.0 s before it
reached, assuming the negative plate were defined as zero
quits. What is the terminal voltage of the battery that
potential? (b) Repeat part (a) except move 1.0 cm directly
charged the capacitor?
toward the positive plate. Explain why your answers to
(a) and (b) are different. (c) After the movement in (b), 42. ● ● ● A 1.50-F capacitor is connected to a 12.0-V battery
suppose you moved 0.50 cm parallel to the plane of the for a long time, and then is disconnected. The capacitor
plates. What would be the electric potential value then? briefly runs a 1.00-W toy motor for 2.00 s. After this time,
(a) by how much has the energy stored in the capacitor
decreased? (b) What is the voltage across the plates?
– + (c) How much charge is stored on the capacitor? (d) How
– 0.50 cm +
– + much longer could the capacitor run the motor, assum-
cm
– + ing the motor ran at full power until the end?

0
– +
1.
45° + 43. ●●● Two parallel plates have a capacitance value of

–
– + 0.17 mF when they are 1.5 mm apart. They are connected
3.0 cm permanently to a 100-V power supply. If you pull the
plates out to a distance of 4.5 mm, (a) what is the electric
– + 12 V field between them? (b) By how much has the capaci-
tor’s charge changed? (c) By how much has its energy
storage changed? (d) Repeat these calculations assuming
䉱 F I G U R E 1 6 . 2 7 Reaching our potential the power supply is disconnected before you pull the
See Exercises 32 and 33. plates further apart.
16.4 DIELECTRICS uncharged capacitors, one with a capacitance of 0.75 mF

and the other with that of 0.30 mF are connected in series
44. ● A capacitor has a capacitance of 50 pF, which increases
to a 12-V battery. Then the capacitors are disconnected,
to 150 pF when a dielectric material is between its plates.
discharged, and reconnected to the same battery in paral-
What is the dielectric constant of the material?
lel. Calculate the energy loss of the battery in both cases.
45. ● A 50-pF capacitor is immersed in silicone oil, which 52. ● ● For the arrangement of three capacitors in 䉲 Fig. 16.28,
has a dielectric constant of 2.6. When the capacitor is what value of C1 will give a total equivalent capacitance
connected to a 24-V battery, (a) what will be the charge of 1.7 mF?
on the capacitor? (b) How much energy is stored in the
capacitor?
46. ●● The dielectric of a parallel plate capacitor is to be a
slab of glass that completely fills the volume between the V C1 C3 C2
plates. The area of each plate is 0.50 m2. (a) What thick-
ness should the glass have if the capacitance is to be
0.10 mF? (b) What is the charge on the capacitor if it is 6.0 V 0.30 µ F 0.20 µ F
connected to a 12-V battery? (c) How much more energy
is stored in this capacitor compared to an identical one
without the dielectric insert?
47. IE ● ● ● A parallel plate capacitor has a capacitance of 䉱 F I G U R E 1 6 . 2 8 A capacitor triad See Exercises 52 and 56.
1.5 mF with air between the plates. The capacitor is con-
nected to a 12-V battery and charged. The battery is then 53. IE ● ● (a) Three capacitors of equal capacitance are con-
removed. When a dielectric is placed between the plates, nected in parallel to a battery, and together they acquire
a potential difference of 5.0 V is measured across the a certain total charge Q from that battery. Will the charge
plates. (a) What is the dielectric constant of the material? on each capacitor be (1) Q, (2) 3Q, or (3) Q>3? (b) Three
(b) What happened to the energy storage in the capaci- capacitors of 0.25 mF each are connected in parallel to a
tor: (1) it increased, (2) it decreased, or (3) it stayed the 12-V battery. What is the charge on each capacitor?
same? (c) By how much did the energy storage of this (c) How much total charge was acquired from the battery?
capacitor change when the dielectric was inserted? 54. IE ● ● (a) If you are given three identical capacitors, you
48. IE ● ● ● An air-filled parallel plate capacitor has rectan- can obtain (1) three, (2) five, (3) seven different capaci-
gular plates with dimensions of 6.0 cm * 8.0 cm. It is tance values. (b) If the three capacitors each have a
connected to a 12-V battery. While the battery remains capacitance of 1.0 mF, what are the different values of
connected, a sheet of 1.5-mm-thick Teflon (dielectric con- equivalent capacitance?
stant of 2.1) is inserted and completely fills the space 55. ● ● What are the maximum and minimum equivalent
between the plates. (a) While the dielectric was being capacitances that can be obtained by combinations of
inserted, (a) charge flowed onto the capacitor, (2) charge three capacitors of 1.5 mF, 2.0 mF, and 3.0 mF?
flowed off the capacitor, (3) no charge flowed. (b) Deter- 56. ● ● ● If the capacitance of C1 is 0.10 mF, (a) what is the
mine the change in the charge storage of this capacitor charge on each of the capacitors in the circuit in Fig. 16.28?
because of the dielectric insertion. (c) Determine the (b) How much energy is stored in each capacitor?
change in energy storage in this capacitor because of the
57. ● ● ● Four capacitors are connected in a circuit as illus-
dielectric insertion. (d) By how much was the battery’s
trated in 䉲 Fig. 16.29. Find the charge on, the voltage dif-
stored energy changed?
ference across, and the energy stored for each of the
capacitors.
16.5 CAPACITORS IN SERIES AND IN
PARALLEL
C1 C2
49. ● What is the equivalent capacitance of two capacitors
with capacitances of 0.40 mF and 0.60 mF when they are
connected (a) in series and (b) in parallel? V 0.40 µ F 0.40 µ F
50. ● Two identical capacitors are connected in series and
their equivalent capacitance is 1.0 mF. What is each one’s C3 C4
12 V
capacitance value? Repeat the calculation if, instead,
they were connected in parallel.
0.20 µ F 0.60 µ F
51. IE ● ● (a) Two capacitors can be connected to a battery in
either a series or parallel combination. The parallel combi-
nation will require (1) more, (2) equal, (3) less energy from
a battery than the series combination. Why? (b) Two 䉱 F I G U R E 1 6 . 2 9 Double parallel in series See Exercise 57.
58. IE A tiny dust particle in the form of a long thin needle electric field (including direction) in the membrane
has charges of 7.14 pC on its ends. The length of the under these conditions. (c) Estimate the force on that
particle is 3.75 mm. (a) Which location is at a higher first sodium ion. (d) What is the electric field (including
potential: (1) 7.65 mm above the positive end, (2) 5.15 mm direction) under normal conditions when the voltage
above the positive end, or (3) both locations are at the across the membrane is -70 mV?
same potential? (b) Compute the potential at the two 64. In Exercise 63, assume that the inside and outside surfaces
points in part (a). (c) Use your answer from part (b) to of the axon membrane act like a parallel plate capacitor
determine the work needed to move an electron from the with an area of 1.1 * 10-9 m2. (a) Estimate the capaci-
near point to the far point. tance of an axon’s membrane, assuming it is filled with
59. A vacuum tube has a vertical height of 50.0 cm. An elec- lipids with a dielectric constant of 3.0. (b) How much
tron leaves from the top at a speed of 3.2 * 106 m>s charge would be on each surface under resting potential
downward and is subjected to a “typical” Earth field of conditions? (c) How much electrostatic energy would be
150 V>m downward. (a) Use energy methods to deter- stored in this axon under resting potential conditions?
mine whether it reaches the bottom surface of the tube. 65. Two parallel plates, 9.25 cm on a side, are separated by
(b) If it does, with what speed does it hit? If not, how 5.12 mm. (See 䉲 Fig. 16.30a.) (a) Determine their capaci-
close does it come to the bottom surface? (c) How does tance if the volume from one plate to midplane is filled
the gravitational force on the electron compare to the with a material of dielectric constant 2.55 and the rest is
electric force on it, both in magnitude and in direction? filled with a different material of dielectric constant 4.10.
60. A helium atom with one electron already removed (a (b) If these plates are connected to a 12-V battery, what is
positive helium ion) consists of a single orbiting electron the electric field strength in each dielectric region? [Hint:
and a nucleus of two protons. The electron is in its mini- Do you see two capacitors in series?]
mum orbital radius of 0.027 nm. (a) What is the potential
energy of the system? (b) What is the centripetal acceler- A d
ation of the electron? (c) What is the total energy of the
system? (d) What is the minimum energy required to ␬1 2.55
“ionize” this atom, in other words, to cause the electron ␬2 4.10
to leave completely?
61. Suppose that the three capacitors in Figure 16.22 have the (a)
following values: C1 = 0.15 mF, C2 = 0.25 mF, and
C3 = 0.30 mF. (a) What is the equivalent capacitance of
this arrangement? (b) How much charge will be drawn A d
from the battery? (c) What is the voltage across each capac-
itor? (d) What is the energy storage in each capacitor? ␬1 2.55 ␬2 4.10
62. IE Two very large horizontal parallel plates are sepa-
rated by 1.50 cm. An electron is to be suspended at rest (b)
in midair between them. (a) The top plate should be at
(1) a higher potential, (2) an equal potential, (3) a lower 䉱 F I G U R E 1 6 . 3 0 Double-stuffed capacitor See Exercise 65.
potential compared with the bottom plate. Explain.
(b) What voltage across the plates is required? (c) Does 66. Repeat Exercise 65 except fill the volume from one edge
the electron have to be positioned midway between the to the middle with the same two materials. (See Figure
plates, or is any location between the plates just as good? 16.30b.) (Do you see two capacitors in parallel?)
63. (Before attempting this one, see Insight 16.1, Electric 67. A capacitor 15.70 mF2 is connected in a series arrange-
Potential and Nerve Signal Transmission and Learn by ment with a second capacitor 12.30 mF2 and a 12-V bat-
B
Drawing 16.2 on graphical relationships between E and tery. (a) How much charge is stored on each capacitor?
V.) Suppose an (axon) cell membrane is experiencing the (b) What is the voltage drop across each capacitor? The
end of a stimulus event and the voltage across the cell battery is then removed, leaving the two capacitors iso-
membrane is instantaneous at 30 mV. Assume the mem- lated. (c) If the smaller capacitor’s capacitance is now
brane is 10 nm thick. At this point the Na>K-ATPase doubled, by how much does the charge on each and the
molecular pump starts to move the excess Na+ ions back voltage across each change?
to the exterior. (a) How much work does it take for the 68. Repeat Exercise 67 assuming, instead, that the capacitors
pump to move the first sodium ion? (b) Estimate the are, instead, connected in parallel.
Electric Current and
17 Resistance
17.1 Batteries and
direct current (597)
■ emf
■ terminal voltage
17.2 Current and

drift velocity (600)
■ electric current
■ the ampere defined
17.3 Resistance and

Ohm’s law (602)
■ resisitivity
■ ohmic resistances
PHYSICS FACTS
17.4
■
Electric power (609)
appliances and efficiency
✦ André Marie Ampère (1775–1836), a
French physicist/mathematician, was
known for his work with electric cur-
rents. The SI unit of current, the ampere,
I f you were asked to think of elec-
tricity and its uses, many favorable
images would probably come to
was named in his honor.
mind, including such diverse applica-
✦ In a metal wire, the electric energy trav-
els close to speed of light, much, much tions as lamps, television sets, and
faster than the charge carriers them-
selves. The latter travel at only several
computers. You might also think of
millimeters per second. some unfavorable images, such as
✦ The SI unit of electrical resistance,
the ohm ( Æ ), is named after Georg
lightning, a shock, or sparks from an
Simon Ohm (1789–1854), a German overloaded electric outlet.
mathematician and physicist. A
quantity called electrical conductiv- Common to all of these images is
ity, proportional to the inverse of
resistance, is named, appropriately
the concept of electric energy. For
enough, the mho—his last name electric appliances, energy is supplied
spelled backwards.
✦ Generating voltages of up to
by electric current in wires; for light-
600 volts, electric eels and rays can, ning or a spark, it is conducted
for brief times, discharge as much
as 1 ampere of current through through the air. In general, the light,
flesh. The energy is delivered at
600 J/s, or about 0.75 hp.
heat, or mechanical energy is simply
electric energy converted to a differ-
ent form. In the chapter-opening
17.1 BATTERIES AND DIRECT CURRENT 597
photograph, for example, the light given off by the spark is actually emitted by air
molecules excited by electrons moving from one wire to the other.
In this chapter, the fundamental principles governing electric circuits are our
primary concern. These principles will enable us to answer questions such as: What
is electric current and how does it travel? What causes an electric current to move
through an appliance when it is switched on? Why does the electric current cause
the filament in a bulb to glow brightly, but not affect the connecting wires in the
same way? Electrical principles can be applied to gain an understanding of a wide
range of phenomena, from the operation of household appliances to the workings
of Nature’s spectacular fireworks—lightning.
17.1 Batteries and Direct Current

➥ On a battery, which electrode is at a higher electric potential?

➥ Under what conditions are the emf and terminal voltage of a battery approximately
equal?
➥ Does a set of identical batteries produce the maximum possible emf when con-
nected in series or parallel?
After studying electric force and energy in Chapters 15 and 16, you can probably
guess what is required to produce an electric current, or a flow of charge. Here are some Electron Electron
flow flow
analogies to help. Water naturally flows downhill, from higher to lower gravitational
potential energy—that is, because there is a difference in gravitational potential energy.
Heat flows naturally because of temperature differences. In electricity, a flow of electric
charge is caused by an electric potential difference—which is called “voltage.” V
In solid conductors, particularly metals, some of the outer electrons of atoms are
relatively free to move. (In liquid conductors and charged gases called plasmas, posi-
A Anode B Cathode
tive and negative ions as well as electrons can move.) Energy is required to move – –
–
electric charge. Electric energy is generated through the conversion of other forms of – Electrolyte –
– B+
energy, giving rise to a potential difference, or voltage. Any device that can produce – + –
B –
and maintain a potential difference is called by the general name of a power supply. –
A+ – B+ – + B+
+ + – B
– A+ B B – +
–
BATTERY ACTION A+ – B
+ +
B B B+ B+
A+ A+
One common type of power supply is the battery. A battery converts stored
chemical potential energy into electrical energy. The Italian scientist Alessandro Membrane
Volta constructed one of the first practical batteries. A simple battery consists of
䉱 F I G U R E 1 7 . 1 Battery action in
two unlike metal electrodes in an electrolyte, a solution that conducts electric charge. a chemical battery or cell Chemical
With the appropriate electrodes and electrolyte, a potential difference develops processes involving an electrolyte
across the electrodes as a result of chemical action (䉴 Fig. 17.1). and two unlike metal electrodes
When a complete circuit is formed, for example, by connecting a lightbulb and cause ions of both metals to dissolve
wires (Fig. 17.1), electrons from the more negative electrode (B) will move through into the solution at different rates.
Thus, one electrode (the cathode)
the wire and bulb to the less negative electrode (A).* The result is a flow of electrons becomes more negatively charged
in the wire. As electrons move through the bulb’s filament, colliding with and trans- than the other (the anode). The
ferring energy to its atoms (typically tungsten), the filament reaches a sufficient tem- anode is at a higher potential than
perature to give off a glow of visible light. Since electrons move to regions of higher the cathode. By convention, the
potential, electrode A is at a higher potential than B. Thus the battery action has cre- anode is designated the positive ter-
minal and the cathode the negative
ated a potential difference ( ¢V or simply V) across its terminals. Electrode A is the
anode and is labeled with a plus 1 +2 sign. Electrode B is the cathode and is labeled
terminal. This potential difference
(V) can cause a current, or a flow of
as negative 1 - 2. It is easy to keep track of this sign convention because the nega- charge (electrons), in the wire. The
tively charged electrons will move through the wire from B 1 - 2 to A 1 +2. positive ions migrate as shown. (A
membrane is necessary to prevent
*As will be seen shortly, a complete circuit is any complete loop consisting of wires and electrical mixing of the two types of ion;
devices (such as batteries and lightbulbs). why?)
598 17 ELECTRIC CURRENT AND RESISTANCE
For our study of circuits, a battery will be pictured as a “black box” that main-
tains a constant potential difference across its terminals. Inserted into a circuit, a bat-
tery can do work on, and transfer energy to, electrons in the wire (at the expense of
its own internal chemical energy), which in turn can deliver that energy to external
circuit elements. In these elements, the energy is converted into other forms, such as
mechanical motion (as in electric fans), heat (as in immersion heaters), and light (as
in flashlights). Other sources of voltage (that is, other types of power supplies), such
as generators and photocells, will be considered later.
To help better visualize the role of a battery, consider the gravitational analogy in
䉳 Fig. 17.2. A gasoline-fueled pump (analogous to the battery) does work on the water
as it lifts it. The increase in the water’s gravitational potential energy comes at the
expense of the chemical potential energy of the gasoline molecules. The water then
䉱 F I G U R E 1 7 . 2 Gravitational returns to the pump by flowing down the trough (analogous to the wire) into the pond.
analogy to a battery and lightbulb A
On the way down, the water does work on the wheel, resulting in rotational kinetic
gasoline-powered pump lifts water
from the pond, increasing the energy, analogous to the electrons transferring energy to an appliance such as a fan.
potential energy of the water. As the
water flows downhill, it transfers BATTERY EMF AND TERMINAL VOLTAGE
energy to (or does work on) a water-
wheel, causing the wheel to spin. The potential difference across the terminals of a battery when it is not connected to a
This action is analogous to the circuit is called the battery’s electromotive force (emf), symbolized by e. The name is
delivery of energy to a fan.
misleading, because emf is not a force, but a potential difference, or voltage. To avoid
confusion with force, the electromotive force is called just emf. Thus a battery’s emf
represents the work done by the battery per coulomb of charge that passes through it.
If a battery does 1 joule of work on 1 coulomb of charge, then its emf is 1 joule per
coulomb 11 J>C2, or 1 volt (1 V).
The emf actually represents the maximum potential difference across the termi-
nals (䉲 Fig. 17.3a). Under practical circumstances, when a battery is in a circuit and
charge flows, the voltage across the terminals is always slightly less than the emf. This
“operating voltage” (V) of a battery (the battery symbol is the pair of unequal-length
parallel lines in Fig. 17.3b) is called its terminal voltage. Because batteries in actual
operation are of the most interest, it is the terminal voltage that is important.
Under many conditions, the emf and terminal voltage are essentially the same. Any
difference is due to the battery’s internal resistance (r), shown explicitly in the circuit
diagram in Fig. 17.3b. (Resistance, defined in Section 17.3, is a quantitative measure of
the opposition to charge flow.) Internal resistances are typically small, so the terminal
voltage of a battery is essentially the same as the emf, that is, V L e. However, when a
battery supplies a large current or when its internal resistance is high (as in older bat-
teries), the terminal voltage may drop appreciably below the emf. This occurs because
it takes some voltage just to produce a current in the internal resistance itself. Mathe-
matically, the terminal voltage is related to the emf, current, and internal resistance by
V = e - Ir, where I is the electric current (Section 17.2) in the battery.
For example, some cars have a battery “voltage readout.” Upon startup, the 12-V
battery’s voltage typically reads only 10 V (this value is normal). Because of the large
䉴 F I G U R E 1 7 . 3 Electromotive
force (emf) and terminal voltage
(a) The emf 1e2 of a battery is the
maximum potential difference
across its terminals. This maximum R
occurs when the battery is not con-
nected to an external circuit.
(b) Because of internal resistance (r), V Electron V Electron
the terminal voltage V when the + + flow r flow
– – + –
battery is in operation is less than
the emf e. Here, R is the resistance
of the lightbulb. Internal
resistance
Circuit diagram
(a) Electromotive (b) Terminal voltage

force (emf)
17.1 BATTERIES AND DIRECT CURRENT 599
+ – V = V1 + V2 + V3 V1
+ – V = V1 = V2 = V3 䉳 F I G U R E 1 7 . 4 Batteries in series
V1 and in parallel (a) When batteries
V V
+ – + – are connected in series, their voltages
V2
V2 add, and the voltage across the resis-
+ – V3 + – tance R is the sum of the voltages.
V3 (b) When batteries of the same volt-
age are connected in parallel, the
voltage across the resistance is the
same, as if only a single battery were
present. In this case, each battery
supplies part of the total current.
+
– + + +
+ V = V1 + V2 + V3 R V = V1 = V2 = V3 R
– – – –
+
–
Circuit diagram Circuit diagram
(a) Batteries in series (b) Batteries in parallel

(equal voltages)
current required at startup, the Ir term (here it has a value of 2 V) reduces the emf by
about 2 V to the measured terminal voltage of 12 V - 2 V = 10 V. When the engine
is running and supplying most of the electric energy to run the car’s functions, the
current required from the battery is essentially zero and the battery readout rises back
to normal voltage levels (that is, close to its emf value). Thus, the terminal voltage, and
not the emf, is a true indication of the state of the battery. Unless otherwise specified,
negligible internal resistance will be assumed, so that V L e, an ideal battery.
There exists a wide variety of batteries. One of the most common is the 12-V auto- electric circuit
mobile battery, consisting of six 2-V cells connected in series.* That is, the positive ter-
minal of each cell is connected to the negative terminal of the next (see the three cells symbols and circuits
in 䉱 Fig. 17.4a). When batteries or cells are connected in this fashion, the voltages add.
– +
If cells are connected in parallel, the positive terminals are connected to each Battery
other, as are the negative ones (Fig. 17.4b). When identical batteries are connected
this way, the potential difference or terminal voltage is the same for all of them Resistor
and equal to the voltage of any one of them.† However, each supplies only a frac-
tion of the current to the circuit. For example, if there are three batteries with equal Capacitor
voltages connected in parallel, each supplies one-third of the total current. A par-
Wire
allel connection of two batteries is the main method for “jump-starting” a car. For
such a start, the weak (high-r) battery is connected in parallel to a normal (low-r)
Two
battery, which delivers most of the current to start the car. or unconnected
wires
CIRCUIT DIAGRAMS AND SYMBOLS
Two
To help analyze and visualize circuits, it is common to draw circuit diagrams that are connected
schematic representations of the wires, batteries, appliances, and so on. Each element wires
in the circuit is represented by its own symbol in such a diagram. As in Fig. 17.3b and
Fig. 17.4, the battery symbol is two parallel lines, the longer one representing the posi- Open
tive terminal and the shorter one the negative terminal. Any element (such as a light- switch
bulb or appliance) that opposes the flow of charge is represented by the symbol
Closed
. (Electrical resistance is defined in Section 17.3; here we merely introduce the switch
symbol.) Connecting wires are unbroken lines and are assumed, unless stated other-
wise, to have negligible resistance. Where lines cross, it is assumed that they do not
contact one another, unless they have a dot at their intersection. Lastly, switches are
shown as “drawbridges,” capable of going up (to open the circuit and stop the cur-
rent) and down (closed to complete the circuit and allow current). These symbols, Putting it
along with that of the capacitor (from Chapter 16), are summarized in Learn by Draw- together
ing 17.1, Electric Circuit Symbols and Circuits. The use of these symbols and circuit
diagrams to understand circuits conceptually is demonstrated in the next Example. + A complete
– circuit
*Chemical energy is converted to electrical energy in a chemical cell. The term battery generally
refers to a collection, or “battery,” of cells.
†
If the battery voltages differ then the higher voltage battery will supply, at its expense, energy to
the lower voltage battery.
CONCEPTUAL EXAMPLE 17.1 Asleep at the Switch?

䉴 Fig. 17.5 shows a circuit diagram that represents two identi- A 䉳 F I G U R E 1 7 . 5 What
cal batteries (each with a terminal voltage V) connected in happens to the voltage?
parallel to a lightbulb (represented by a resistor). Because it is
assumed that the wires have no resistance, before switch S1 is V V
opened, the voltage across the lightbulb equals V (that is,
VAB = V). What happens to the voltage across the lightbulb VAB
when S1 is opened: (a) the voltage remains the same (V) as
that before the switch was opened, (b) the voltage drops to S1 S2
V>2, because only one battery is now connected to the bulb,
or (c) the voltage drops to zero?
B
REASONING AND ANSWER. It might be tempting to choose
answer (b), because there is now just one battery. But look
again. The remaining battery is still connected to the light- means that the answer cannot be (b), because the remaining
bulb. This means that there must be some voltage across the battery itself will maintain a voltage of V across the bulb.
lightbulb, so the answer certainly cannot be (c). But it also Hence, the answer is (a).
FOLLOW-UP EXERCISE. In this Example, what would the correct answer be if, in addition to S1, switch S2 were also opened?
Explain your answer and reasoning. (Answers to all Follow-Up Exercises are given in Appendix VI at the back of the book.)
DID YOU LEARN?

➥ On a battery, the anode ( + ) is at a higher electric potential than the cathode ( - ).
➥ A small internal resistance means that the battery’s emf and terminal voltage are
almost the same.
➥ Battery terminal voltages add to their maximum possible value when connected in
series.
17.2 Current and Drift Velocity

➥ How is the SI unit of current, the ampere, related to the coulomb, the SI unit of
charge?
➥ What is the order of magnitude of the drift velocity in a typical current-carrying
metal wire?
➥ What is the order of magnitude for the speed at which electric energy travels
through metal wires?
R
S From the previous discussion, it should be clear that sustaining an electric current
requires a voltage source and a complete circuit—the name given to a continuous
conducting path. Most practical circuits include a switch to “open” or “close” the
circuit. An open switch eliminates the continuous part of the path, thereby stop-
Electron
flow ping the flow of charge in the wires. (This is situation is called an open circuit.
When the switch is closed, the circuit is closed.)
I Conventional
current
V ELECTRIC CURRENT
+ – Because it is the electrons that move in any circuit’s wires, the charge flow is away
from the negative terminal of the battery.* Historically, however, circuit analysis
Battery has been done in terms of conventional current. The conventional current’s direc-
tion is that in which positive charges would flow, that is, opposite the actual elec-
䉱 F I G U R E 1 7 . 6 Conventional tron flow (䉳 Fig. 17.6).
current For historical reasons, cir- The battery is said to deliver current to a circuit or a component of that circuit (a
cuit analysis is usually done with
conventional current. Conventional circuit element). Alternatively, it is sometimes said that a circuit (or its compo-
current is in the direction in which nents) draws current from the battery. The current then returns to the battery.
positive charges would flow, or
opposite to the electron flow. *Some situations do exist in which a positive charge flow is responsible for the current—for exam-
ple, in semiconductors.
17.2 CURRENT AND DRIFT VELOCITY 601
A battery can produce a current in only one direction. One-directional charge flow
is called direct current (dc). (Note that if the current changes direction and>or mag-
nitude, it is alternating current. This type of current is studied in detail in Chapter 21.)
Quantitatively, the electric current (I) is the time rate of flow of net charge. Ini-
tially our primary concern is with steady charge flow. In this case, if a net charge q
䉱 F I G U R E 1 7 . 7 Electric current
passes through a cross-sectional area in a time interval t (䉴 Fig. 17.7), the electric
Electric current (I) in a wire is
current associated with that charge flow is defined as defined as the rate at which the net
q charge (q) passes through the wire’s
I = (electric current) (17.1) cross-sectional area: I = q>t. The
t units of I are amperes (A), or amps
SI unit of current: coulomb per second 1C>s2 or ampere (A)
for short.
The coulomb per second is designated the ampere (A), in honor of the French
physicist André Ampère (1775–1836), an early investigator of electrical and mag-
netic phenomena. In everyday usage, the ampere is commonly shortened to amp.
Thus, a current of 10 A is read as “ten amps.” Small currents are expressed in
milliamperes (mA, or 10-3A), microamperes (mA, or 10-6 A), or nanoamperes (nA, or
10-9A). These units are usually shortened to milliamps, microamps, and nanoamps,
respectively. In a typical household circuit, it is not unusual for the wires to carry
several amps of current. To understand the relationship between charge and cur-
rent, consider Example 17.2.
EXAMPLE 17.2 Counting Electrons: Current and Charge

Suppose there is a steady current of 0.50 A in a flashlight bulb T H I N K I N G I T T H R O U G H . The current and time elapsed are
lasting for 2.0 min. How much charge passes through the bulb given; therefore, the definition of current (Eq. 17.1) allows the
during this time? How many electrons does this represent? calculation of the charge q. Since each electron carries a charge
of magnitude 1.6 * 10-19 C, q can be converted into the num-
ber of electrons.
SOLUTION. Listing the data given and converting the time into seconds:
Given: I = 0.50 A Find: q (amount of charge)
t = 2.0 min = 1.2 * 102 s n (number of electrons)
By Eq. 17.1, I = q>t, so the magnitude of the charge is
q = It = 10.50 A211.2 * 102 s2 = 10.50 C>s211.2 * 102 s2 = 60 C
Solving for the number of electrons (n) from q = ne,
q 60 C
n = = = 3.8 * 1020 electrons
e 1.6 * 10-19 C>electron
(That’s a lot of electrons!)

F O L L O W - U P E X E R C I S E . Many sensitive laboratory instruments can measure currents in the nanoamp range or smaller. How
long, in years, would it take for 1.0 C of charge to flow past a given point in a wire that carries a current of 1.0 nA?
DRIFT VELOCITY, ELECTRON FLOW, AND ELECTRIC

ENERGY TRANSMISSION
Although charge flow is sometimes described as analogous to water flow, electric
charge traveling in a conductor does not move in the same way that water flows in
a pipe. In the absence of a potential difference across the ends of a metal wire, the
free electrons move in random directions at high speeds, colliding many times per
second with the metal atoms. As a result, there is no net flow of charge, thus there
is no current under these conditions.
However, when a potential difference (voltage) is applied across the wire (such
as by a battery), an electric field appears in the wire in one direction. A flow of
electrons then begins opposite that direction (why opposite?). This does not mean
the electrons move directly from one end of the wire to the other. They still move
in all directions as they collide with the atoms of the conductor, but there is now a
Drift velocity vd very small added component (in one direction) to their velocities. The result is that
their velocities are now directed, on average, more toward the positive terminal of
the battery than away (䉳 Fig. 17.8).
This net electron flow is characterized by an average velocity called the drift
velocity. The drift velocity is much smaller than the random (thermal) velocities
e–
of the electrons themselves. Typically the magnitude of the drift velocity is on the
order of 1 mm>s. At that speed, it would take an electron about 17 min to travel
E
1 m along a wire. Yet a lamp comes on almost instantaneously when the switch is
closed (completing the circuit), and electronic signals carrying telephone conver-
䉱 F I G U R E 1 7 . 8 Drift velocity
Because of collisions with the atoms sations travel almost instantaneously over miles of wire. How can that be?
of the conductor, electron motion is Evidently, something must be moving faster than the “drifting” electrons.
random. However, when the con- Indeed, this something is the electric field. When a potential difference is applied,
ductor is connected, for example, to the associated electric field in the conductor travels at a speed close to that of light
a battery to form a complete circuit, (in the material, roughly 108 m>s). Thus the electric field influences the electrons
there is a small net motion in the
direction opposite the electric field throughout the conductor almost instantaneously. This means that the current starts
[toward the high-potential (posi- everywhere in the circuit essentially simultaneously. You don’t have to wait for
tive) terminal, or anode]. The speed electrons to “get there” from a distant place (say, near the switch). Thus in the light
and direction of this net motion are bulb, the electrons that are already in its filament begin to move almost immedi-
the drift velocity of the electrons. ately, delivering energy and creating light with no noticeable delay.
This effect is analogous to toppling a row of standing dominos. When you tip a
domino at one end, that signal (or energy) is transmitted rapidly down the row.
Very quickly, at the other end, the last domino topples (and delivers energy). Note
that the domino delivering the signal or energy is not the one you pushed. It was
the energy, not the dominos, that traveled down the row.
DID YOU LEARN?

➥ The ampere is defined as a coulomb per second.
➥ Drift velocities in wires are typically a few millimeters per second.
➥ Electric energy in metal wires travels at speeds on the order of the speed of light (in
metal), or about 108 m>s.
17.3 Resistance and Ohm’s Law

➥ How is resistance defined in terms of voltage and current?

➥ What is special about an ohmic resistor?
➥ How does the resistance of a cylindrical object depend on its length and
cross-sectional area?
➥ How does resistance change with temperature for most common materials?
If a voltage (potential difference) is placed across any conducting material in a com-

plete circuit, what factors determine the resulting current? As might be expected,
usually the greater the voltage, the greater the current. However, another factor also
influences current. Just as internal friction (viscosity; see Section 9.5) affects fluid
flow in pipes, the resistance of the material will affect the flow of charge. Any object
that offers significant resistance to electrical current is called a resistor and is repre-
sented by the zigzag symbol (Section 17.1). This symbol is used to represent all types
of resistors, from the cylindrical color-coded ones on printed circuit boards to electrical
devices and appliances such as hair dryers and lightbulbs (䉳 Fig. 17.9).
䉱 F I G U R E 1 7 . 9 Resistors in use But how is resistance quantified? For example, if a large voltage applied across an
A printed circuit board, typically object produces only a small current, then it is clear that the object has a high electri-
used in computers, includes resis- cal resistance. Keeping this notion in mind, the electrical resistance of any object is
tors of different values. The large, defined as the ratio of the voltage applied across it to the resulting current in it, or
striped cylinders are resistors; their
four-band color code indicates their
resistance in ohms. V
R = (electrical resistance) (17.2a)
I
SI unit of resistance: volt per ampere 1V>A2, or ohm 1Æ2

17.3 RESISTANCE AND OHM’S LAW 603
The units of resistance are volts per ampere 1V>A2, called the ohm (æ) in honor R
of the German physicist Georg Ohm (1789–1854), who investigated the relation-
ship between current and voltage. Large values of resistance are expressed as
kilohms 1kÆ2 and megohms 1MÆ2. A schematic circuit diagram showing how, in R= V
I
principle, resistance is determined is illustrated in 䉴 Fig. 17.10a. (Chapter 18
includes a detailed study of the instruments used to measure electrical currents I
and voltages, called ammeters and voltmeters, respectively.)
For some materials, the resistance may be constant over a range of voltages. A
resistor that exhibits constant resistance is said to obey Ohm’s law, or to be ohmic. + –
The law was named after Ohm, who found that many materials, particularly met-
als, possessed this property. A plot of voltage (V) versus current (I) for a material
with an ohmic resistance gives a straight line with a slope equal to its resistance R V
(Fig. 17.10b). A common and practical form of Ohm’s law is (a)
V
V = IR (Ohm’s law) (17.2b)
1or I r V, only when R = constant2
Voltage
I
Ohm’s law is not a fundamental law in the same sense as, for example, the law V/
=
of conservation of energy. There is no “law” that states that materials must have R
e =
constant resistance. Indeed, many advances in electronics are based on materials op
Sl
such as semiconductors, which have nonlinear (nonohmic) voltage–current
relationships. I
Unless specified otherwise, resistances will be assumed ohmic. Always Current
remember, however, that many materials are nonohmic. For instance, the resis- (b)
tance of tungsten filaments in lightbulbs increases with temperature, being
larger at their operating temperature than at room temperature. Example 17.3 䉱 F I G U R E 1 7 . 1 0 Resistance and
Ohm’s law (a) In principle, any
shows how the resistance of the human body can make the difference between object’s electrical resistance can be
life and death. determined by dividing the voltage
across it by the resulting current
through it. (b) If the object obeys
Ohm’s law (meaning constant resis-
EXAMPLE 17.3 Danger in the House: Human Resistance tance), a plot of voltage versus cur-
rent is a straight line with a slope
Any room in the house that is exposed to water and electrical voltage can present haz- equal to R, the element’s resistance.
ards. (See the discussion of electrical safety in Section 18.5.) For example, suppose a
person steps out of a shower and inadvertently touches an exposed 120-V wire (per-
haps a frayed cord on a hair dryer) with a finger thus creating a complete circuit though
the body to ground. The human body, when wet, can have an electrical resistance as
low as 300 Æ . Using this value, estimate the current in that person’s body.
T H I N K I N G I T T H R O U G H . The wire is at an electrical potential of 120 V above the floor,
which is “ground” and taken to be at 0 V. Therefore the voltage (or potential difference)
across the body is 120 V. To determine the current, Eq. 17.2, the definition of resistance
can be used.
Given: V = 120 V Find: I (current in the body)
R = 300 Æ
From Eq. 17.2,
V 120 V
I = = = 0.400 A = 400 mA
R 300 Æ
While this is a small current by everyday standards, it is a large current for the human
body. A current over 10 mA can cause severe muscle contractions, and currents on the
order of 100 mA can stop the heart. So this current is potentially deadly. (See Chapter 18,
Insight 18.2, Electricity and Personal Safety, Table 1.)
F O L L O W - U P E X E R C I S E . When the human body is dry, its resistance (over its length)
can be as high as 100 kÆ . Under these conditions, what voltage would produce a cur-
rent of 1.0 mA (the value that a person can barely feel)?
FACTORS THAT INFLUENCE RESISTANCE

Length On the atomic level, resistance arises when electrons collide with the atoms that
make up a material. Thus, resistance depends on the type of material of which an
Material
object is composed. However, geometrical factors also influence resistance. Basi-
cally the resistance of an object of uniform cross-section, such as a length of wire,
depends on four properties: (1) the type of material, (2) its length, (3) cross-
Temperature Cross-sectional
area sectional area, and (4) temperature (䉳 Fig. 17.11).
As might be expected, the resistance (R) of an object (such as a piece of wire) is
䉱 F I G U R E 1 7 . 1 1 Resistance factors inversely proportional to its cross-sectional area (A), and directly proportional to its
Factors directly affecting the electrical length (L); that is, R r L>A. For example, a uniform metal wire 4.0 m long offers
resistance of a cylindrical conductor twice as much resistance as a similar wire 2.0 m long, but a wire with a cross-
are the type of material it is made of,
its length (L), its cross-sectional area sectional area of 0.50 mm2 has only half the resistance of one with an area of
(A), and its temperature (T). 0.25 mm2. These geometrical resistance conditions are analogous to those for liq-
uid flow in a pipe. The longer the pipe, the greater its resistance (drag), but the
larger its cross-sectional area, the more liquid it can carry per second. To see an
interesting use of the dependence of resistance on length and area by living organ-
isms, see Insight 17.1, The “Bio-Generation” of High Voltage.
INSIGHT 17.1 The “Bio-Generation”of High Voltage

As discussed in Section 17.1, two different metals in acid can Anterior
generate a constant separation of charge (a voltage) and thus can
produce electric current. However, living organisms can also
create voltages by a process sometimes called “bio-generation.”
Electric eels (see Chapter 15, Insight 15.2, Electric Fields in

Law Enforcement and Nature: Stun Guns and Electric Fish ), Protein/lipid
membranes V1 0
in particular, can generate 600 V, more than enough voltage to
kill humans. But how do they accomplish this? It turns out
that the process has similarities both to regular “dry cells”
and to nerve signal transmission.
Eels have three organs related to their electrical activities. The
Sachs’ organ generates low-voltage pulsations for navigation. Posterior
The other two, named the Hunter organ and the Main organ, are
(a)
the sources of the high voltage (Fig. 1). In these organs, cells
called electrocytes, or electroplates, are arranged in a stack. Each
cell has a flat, disklike shape. The electroplate stack is a series
connection similar to that in a car battery, in which there are six

cells at 2.0 V each, producing a total of 12 V. In an eel, each elec- V1 0
troplate is capable of producing only about 0.15 V, but four or
five thousand in series can add up to a voltage of 600 V! The elec-
troplates are electrically similar to muscle cells in that they, like
muscle cells, receive nerve impulses by synaptic connection. Vtotal 0

Sachs‘ organ

Main and Hunter‘s organs

(b)
Adult length 艐 2m
F I G U R E 2 (a) A single, resting electroplate One of the thou-
F I G U R E 1 Anatomy of an electric eel Eighty percent of an elec- sands of electroplates in the eel’s electric organs has, under
tric eel’s body is devoted to voltage generation. Most of that por- resting conditions, equal amounts of positive charge at its top
tion contains the two organs (Main and Hunter’s) responsible for and bottom, resulting in no voltage. (b) Resting electroplates
the high voltage associated with killing of prey. The Sachs’ organ in series Several thousand electroplates in series under rest-
produces a lower pulsating voltage used for navigation. ing conditions have a total voltage of zero.
However, these nerve impulses do not cause movement. attached to the bottom. From what you know about resistance
Instead, they trigger voltage generation by the following mecha- (see the discussion of Eq. 17.3 and R r L>A), it should be
nism. clear that both the reduction in area and increase in length of
Each electroplate has the same structure. The top and bot- the neurons serve to increase neuron resistance compared
tom membranes behave similarly to nerve membranes (see with those attached to more distant electroplates. Increased
Chapter 16, Insight 16.1, Electric Potential and Nerve Signal resistance means that the action potential travels slower
Transmission). Under resting conditions, the Na + ions cannot through the closer neurons, thus enabling the closer electro-
penetrate the membrane. To equilibrate their concentrations plates to receive their signal at the same time as the more dis-
on both sides, the ions reside near the outside surface. This, in tant ones—a very interesting and practical use of physics
turn, attracts the (interior) negatively charged proteins to the (from the eel’s perspective, not the prey’s).
interior surface. As a result, the interior is at a potential of 0.08
V lower than the outside. Therefore, under resting conditions,

the outside top (toward the head, or anterior) surface and the
outside bottom (posterior) surface of all the electroplates are
positive (one is shown in Fig. 2a) and exhibit no voltage
1¢V1 = 02. Hence under resting conditions a series stack has V1 艐 0.15V
no voltage 1¢Vtotal = g ¢Vi = 02 from top to bottom (Fig. 2b).
However, when an eel locates prey, the eel’s brain sends a
signal along a neuron to only the bottom membrane of each elec-

troplate (one cell is shown in Fig. 3a). A chemical

(acetylcholine) diffuses across the synapse onto the membrane, Na
briefly opening the ion channels and allowing in Na+ . For a
few milliseconds the lower membrane polarity is reversed, creating (a)
a voltage across one cell of ¢V1 L 0.15 V. The whole stack
does this simultaneously, causing a large voltage across the Signal
from Anterior
ends of the stack ( ¢Vtotal L 4000 ¢V1 = 600 V; see Fig. 3b).
When the eel touches the prey with the stack ends, the result- brain

ing current pulse through the prey (on the order of 0.5 A)
delivers enough energy to kill or at least stun.
An interesting biological “wiring” arrangement enables all
electroplates to be triggered simultaneously—a requirement
crucial for generation of the maximum voltage. Since each
electroplate is a different distance from the brain, the action

potential traveling down the neurons must be carefully
timed. To do this, the neurons attached to the top of the stack
(closest to the brain) are longer and thinner than those Vtotal 艐 600V

F I G U R E 3 (a) An electroplate in action Upon location of prey, a signal
is sent from the eel’s brain to each electroplate along a neuron attached
only to the bottom of the plate. This triggers a brief opening of the ion
channel allowing Na+ ions to the interior, temporarily reversing the
polarity of the lower membrane. This creates a temporary electric poten-
tial difference (voltage) between the top and bottom membranes. Each
electroplate voltage is typically a few tenths of a volt. (b) A series stack of
electroplates in action When each electroplate in the stack is triggered
into action by the lower neuron signal, this results in a large voltage
between the top and bottom of the stack, typically on the order of 600 V.
This large voltage enables the eel to deliver a pulse of current on the Posterior
order of a few tenths of an ampere through the prey. The energy
deposited in the prey is usually enough to stun or kill it. (b)
RESISTIVITY
The resistance of an object is partly determined by its material’s atomic properties,
quantitatively described by that material’s resistivity (R). The resistance of an
object with a uniform cross-section is given by
b
L
R = ra (uniform cross-section only) (17.3)
A
SI unit of resistivity: ohm-meter1Æ # m2
TABLE 17.1 Resistivities (at 20 °C) and Temperature Coefficients of Resistivity for Various Materials*
r1Æ # m2 a11>°C2 r1Æ # m2 a11>°C2
Conductors Semiconductors
Aluminum 2.82 * 10 -8
4.29 * 10 -3
Carbon 3.6 * 10-5 - 5.0 * 10-4
Copper 1.70 * 10-8 6.80 * 10-3 Germanium 4.6 * 10-1 - 5.0 * 10-2
2
Iron 10 * 10 -8
6.51 * 10 -3
Silicon 2.5 * 10 - 7.0 * 10-2
Mercury 98.4 * 10-8 0.89 * 10-3
Nichrome 100 * 10-8 0.40 * 10-3
(alloy of nickel
and chromium)
Insulators
Nickel 7.8 * 10 -8
6.0 * 10 -3
Glass 1012
Platinum 10 * 10-8 3.93 * 10-3 Rubber 1015
Silver 1.59 * 10 -8
4.1 * 10 -3
Wood 1010
Tungsten 5.6 * 10-8 4.5 * 10-3
*Values for semiconductors are general ones, and resistivities for insulators are typical orders of magnitude.
The units of resistivity 1r2 are ohm-meters 1Æ # m2. (You should show this.) Thus,
from knowing its resistivity (the material type) and using Eq. 17.3, the resistance
of any constant-area object can be calculated, as long as its length and cross-
sectional area are known.
The values of the resistivities of some conductors, semiconductors, and insula-
tors are given in 䉱 Table 17.1. The values strictly apply only at 20 °C, because resis-
tivity generally depends upon temperature. Most common wires are composed of
copper or aluminum with cross-sectional areas on the order of 10-6 m2 or 1 mm2.
You should be able to show that the resistance of a 1.5-m-long copper wire with
this area is on the order of 0.025 Æ 125 mÆ2. This explains why wire resistances
are neglected in circuits—their values are much less than most household devices.
An interesting and potentially important medical application involves the mea-
surement of human body resistance and its relationship to body fat. (See Insight
17.2, Bioelectrical Impedance Analysis (BIA).) To get a feeling for the magnitudes
of these quantities in living tissue, consider Example 17.4.
EXAMPLE 17.4 Electric Eels: Cooking with Bio-Electricity?

Suppose an electric eel touches the head and tail of a long SOLUTION. Listing the data:
approximately cylindrically shaped fish, and applies a voltage of Given: L = 20 cm = 0.20 m Find: r (resistivity)
600 V across it. (See Insight 17.1.) If a current of 0.80 A results
d = 2.0 cm = 2.0 * 10-2 m
(likely killing the fish), estimate the average resistivity of the
V = 600 V
fish’s flesh, assuming it is 20 cm long and 2.0 cm in diameter.
I = 0.80 A
T H I N K I N G I T T H R O U G H . With cylindrical geometry, the fish’s The cross-sectional area of the fish is
length and cross-sectional area can be found from its dimen- 2
d 2 p12.0 * 10-2 m2
sions. From the voltage and current, its resistance can be deter- A = pr = pa b =
2
= 3.1 * 10-4 m2
mined. Lastly, its resistivity can be estimated using Eq. 17.3. 2 4
The fish’s overall resistance is R = V>I = 600 V>0.80 A = 7.5 * 102 Æ . The average resistivity of the flesh can be found using
Eq. 17.3:
RA 17.5 * 102 Æ213.1 * 10-4 m22

r = = = 1.2 Æ # m, or about 120 Æ # cm
L 0.20 m
Comparing this to the values in Table 17.1, the fish’s flesh is much more resistive than metals (as expected). Its value is on the
order of the resistivities of human tissues, such as cardiac muscle’s value of about 175 Æ # cm.
F O L L O W - U P E X E R C I S E . Suppose for its next meal, the eel in this Example chooses a different species of fish. The next fish has
twice the average resistivity, half the length, and half the diameter of the first fish. What current would be expected in this fish if
the eel applied 400 V across its body?
For many materials, especially metals, the temperature dependence of resistiv-

ity is nearly linear if the temperature does not vary too far from room tempera-
ture. Under these conditions, the resistivity at a temperature T, after a temperature
change ¢T = T - To, is given by
r = ro11 + a¢T2
(temperature variation
(17.4)
of resistivity)
where a is a constant (over only a certain temperature range) called the

temperature coefficient of resistivity and ro is a reference resistivity at To (usu-
ally 20 °C). Equation 17.4 can be rewritten as
¢r = roa¢T (17.5)
where ¢r = r - ro is the change in resistivity that occurs when the temperature

changes by ¢T. The ratio ¢r>ro is dimensionless, so a has units of inverse degrees
Celsius, written as 1>°C or °C-1. Physically, a represents the fractional change in
resistivity 1¢r>ro2 per degree Celsius. The temperature coefficients of resistivity for
some materials are listed in Table 17.1. These coefficients will be assumed constant
over normal temperature ranges. Notice that for semiconductors and insulators the
coefficients are given only as orders of magnitude and are usually not constant.
Since resistance is directly proportional to resistivity (Eq. 17.3), an object’s resis-
tance has the same dependence on temperature as its resistivity (Eqs. 17.3 and
17.4). Hence the resistance of an object varies with temperature as:
R = Ro11 + a¢T2 or
(temperature variation
¢R = Ro a¢T (17.6)
of resistance)
INSIGHT 17.2 Bioelectrical Impedance Analysis (BIA)

Traditional methods for estimating body fat percentages In actuality, the voltage alternates in polarity at a frequency of
involve the use of buoyancy tanks or calipers to pinch the 50 kHz, because this frequency is known not to trigger electri-
flesh. However, in recent years, electrical resistance experi- cally excitable tissues, such as nerves and cardiac muscle.
ments have been designed to measure the body fat of the From what has been presented in this chapter (for example,
human body.* In theory, these measurements—termed Eq. 17.3), you should be able to understand some of the fac-
bioelectrical impedance analysis (BIA)—have the potential to tors involved in interpreting the results of human resistance
determine, with more accuracy than traditional methods, a measurements. The measured resistance is the total resis-
patient’s total water content, fat-free mass, and body fat tance. However, the current travels not through a uniform
(adipose tissue). material but rather through the arm, trunk, and leg. Not only
The principle of BIA is based on the water content of the does each of these body parts have a different fat-to-muscle
human body. Water in the human body is a relatively good con- ratio, which affects resistivity 1r2, but also they differ widely
ductor of electric current, due to the presence of ions such as in length (L) and cross-sectional area (A). The arm and the leg,
potassium 1K +2 and sodium 1Na+2. Because muscle tissue usually dominated by muscle and with a small cross-sectional
holds more water per kilogram than fat, it is a better conductor area, offer the most resistance. The trunk, which usually con-
than fat. Thus for a given voltage, the difference in currents tains a relatively high percentage of fat and has a large cross-
should be a good indicator of the fat-to-muscle percentage. sectional area, has low resistance.
In practice, one electrode of a low-voltage power supply is By subjecting BIAs to statistical analysis, researchers hope
connected to the wrist and the other to the opposite ankle to understand how the wide range of physical and genetic
during a BIA test. The current is kept below 1 mA for safety, parameters present in humans affects resistance measure-
with typical currents being about 800 mA. The subject cannot ments. Among these parameters are height, weight, body
feel this small current. Typical resistance values are about type, and ethnicity. Once the correlations are understood, BIA
250 Æ . From Ohm’s law, the required voltage is may well become a valuable medical tool in routine physicals
V = IR = 18 * 10-4 A21250 Æ2 = 0.200 V, or about 200 mV. and in the diagnosis of various diseases.
*Technically, this technique measures the body’s impedance, which includes effects of capacitance and magnetic effects as well as resistance.
(See Chapter 21.) However, these contributions are about 10% of the total. Hence, the word resistance is used here.
Here, ¢R = R - Ro, the change in resistance relative to its reference value Ro, is
usually taken as 20 °C. The variation of resistance with temperature provides a
means of measuring temperature in the form of an electrical resistance thermometer,
as illustrated in Example 17.5.
EXAMPLE 17.5 An Electrical Thermometer: Variation of Resistance with Temperature

A platinum wire has a resistance of 0.50 Æ at 0 °C. It is placed T H I N K I N G I T T H R O U G H . Using the temperature coefficient of
in a water bath, where its resistance rises to a final value of resistivity for platinum from Table 17.1, ¢T can be found
0.60 Æ . What is the temperature of the bath? from Eq. 17.6 and added to 0 °C, the initial temperature, to
find the temperature of the bath.
SOLUTION.
Given: To = 0 °C Find: T (temperature of the bath)
Ro = 0.50 Æ
R = 0.60 Æ
a = 3.93 * 10-3>°C (Table 17.1)
The ratio ¢R>Ro is the fractional change in the initial resistance Ro (at 0 °C). Solving Eq. 17.6 for ¢T, using the given values:
¢R R - Ro 0.60 Æ - 0.50 Æ
= 51 °C
13.93 * 10-3>°C210.50 Æ2
¢T = = =
aRo aRo
Thus, the bath is at T = To + ¢T = 0 °C + 51 °C = 51 °C.

FOLLOW-UP EXERCISE. In this Example, if the material had been copper with Ro = 0.50 Æ , rather than platinum, what would
its resistance be at 51 °C? From this, you should be able to explain which material makes the more “sensitive” thermometer, one
with a high temperature coefficient of resistivity or one with a low value.
SUPERCONDUCTIVITY
Although carbon and other semiconductors have negative temperature coeffi-
cients of resistivity, many materials, including most metals, have positive temper-
ature coefficients. This means that their resistances decrease as the temperature
decreases. You might wonder how far electrical resistance can be reduced by low-
ering the temperature. In certain cases, the resistance can reach zero—not just
close to zero, but, as accurately as can be measured, exactly zero. This phenome-
non is called superconductivity (first discovered in 1911 by Heike Kamerlingh
Onnes, a Dutch physicist). Currently the required temperatures are about 100 K or
below. Thus, at present, practical usage is mainly restricted to high-tech laboratory
apparatus, medical research and industrial equipment.
However, superconductivity does have the potential for important new everyday
applications, especially if materials can be found whose superconducting tempera-
ture is near room temperature. Among the applications are superconducting mag-
nets (already in use in labs and small-scale naval propulsion units). In the absence of
resistance, high currents and very high magnetic fields are possible (Chapter 19).
Used in motors or engines, superconducting electromagnets would be more effi-
cient, providing more power for the same energy input. Superconductors might also
be used as electrical transmission lines with no resistive losses. Some envision
superfast superconducting computer memories. The absence of electrical resistance
opens almost endless possibilities. You are likely to hear much more about super-
conductor applications in the future as new materials are developed.
DID YOU LEARN?

➥ The resistance of an object is defined as the ratio of the voltage across it to the
resulting current in it.
➥ Ohmic resistors have constant resistance values.
➥ The resistance of a cylindrical object increases with length and decreases with
cross-sectional area.
➥ For most materials, the resistance increases approximately linearly with temperature.
17.4 ELECTRIC POWER 609
17.4 Electric Power

➥ For a given voltage, how does the power dissipated in a resistor depend on its
resistance?
➥ Which appliance has a higher value of electrical resistance: a small fan or a hair dryer?
➥ How is the industrial unit of electrical energy, the kilowatt-hour, related to the joule?
When a sustained current exists in a circuit, the electrons are given energy by the
voltage source, such as a battery. As these charge carriers pass through circuit
components, they collide with the atoms of the material (that is, they encounter
resistance) and lose energy. The energy transferred in the collisions can result in an
increase in the temperature of the components. In this way, electrical energy can
be transformed, at least partially, into thermal energy.
However, electric energy can also be converted into other forms of energy such
as light (as in lightbulbs) and mechanical motion (as in electric drills). According
to conservation of energy, whatever forms the energy may take, the total energy
delivered to the charge carriers by the battery must be completely transferred to the
circuit elements (neglecting losses in the wires). That is, on return to the voltage
source or battery, a charge carrier loses all the energy it gained from that source,
and the process is repeated.
The energy gained by an amount of charge q from a voltage source (voltage V) is
qV. [A quick unit check gives the correct result, since qV Q 1 C 21J> C 2 = J.] Over a
time interval t, the rate at which energy is delivered may not be constant. Thus the
average rate of energy delivery, called the average electric power 1P2 is given by
W qV
P = = I q carries
t t qV of energy
In the special case when the current and voltage are steady with P = IV
time (as with a battery), then the average power is the same as the V = 12 V R = 2.0 Ω = (6.0 C/s)(12 J/C)
power at all times. For steady (dc) currents, I = q>t (Eq. 17.1). Thus, = 72 J/s = 72 W
under these conditions the preceding equation can be rewritten as:
I = 6.0 A
P = IV (dc electric power) (17.7a) (a)
Recall from Section 5.6 that the SI unit of power is the watt (W). The
ampere (I) times the volt (V) is the joule per second 1J>s2, or watt (W). m = 12 kg/bucket
(You should show this.) 1 bucket
every 6.0 s
A visual mechanical analogy to help explain Eq. 17.7a is given in
䉴 Fig. 17.12. The figure depicts a simple electric circuit as a system for
transferring energy, analogous to a conveyor belt delivery system.

Because R = V>I, power can be written in three alternate (but Delivery rate
equivalent) forms: = (1/6 bucket/s)(12 kg/bucket)
= 2.0 kg/s
V2 motor (b)
P = IV = = I 2R (electric power) (17.7b)
R 䉱 F I G U R E 1 7 . 1 2 Electric power analogy Electric
circuits can be thought of as energy delivery systems
much like a conveyor belt. (a) Imagine the current as
JOULE HEAT being made up of consecutive segments of charge
q = 1.0 C, each carrying qV = 12 J of energy sup-
The thermal power expended in a current-carrying resistor is plied by the battery. The current is I = V>R = 6.0 A,
referred to as joule heat, or I2R losses (pronounced “I squared R” or 6.0 C>s. Then the power (or energy delivery rate)
losses). In many instances (such as in electrical transmission lines), to the resistor is 16.0 C>s2112 J>C2 = 72 J>s = 72 W.
joule heating is an undesirable side effect. However, in other situa- (b) The conveyor belt comprises a series of buckets,
tions, the conversion of electrical energy to thermal energy is the each carrying 12 kg of sand (analogous to the energy
carried by each charge q), arriving at a rate of one
main purpose. Heating applications include the heating elements bucket every 6.0 s (analogous to the current I). The
(burners) of electric stoves, hair dryers, immersion heaters, and delivery rate in kg>s is analogous to the power in J>s
toasters. in part (a).
䉴 F I G U R E 1 7 . 1 3 Power ratings
(a) Lightbulbs are rated in watts.
Operated at 120 V, this 60-W bulb
converts 60 J of electric energy into
heat and light each second.
(b) Appliance ratings list either volt-
age and power or voltage and cur-
rent. From either, the current, power,
and effective resistance can be found.
Here, one appliance is rated at 120 V
and 18 W and the other at 120 V and
300 mA. Can you compute the cur-
rent and resistance for the former
and the power and resistance for the
latter when in normal operation?
(a) (b)
Electric lightbulbs are rated in watts (power)—for example, 60 W (䉱 Fig. 17.13a).

Incandescent lamps are relatively inefficient as light sources. Typically, less than
2% of the electrical energy is converted to visible light; most of the energy pro-
duced is invisible infrared radiation and heat. Recently, in the interest of energy
conservation, many people have switched to compact fluorescent (CF) light bulbs,
which typically produce the same amount of light at only one-fourth the power.
(For more on other aspects of energy conservation and electrical appliances, see
Example 17.8.)
Electrical appliances are tagged or stamped with their power ratings. Either the
voltage and power requirements or the voltage and current requirements are
given (Fig. 17.13b). In either case, the current, power, and effective resistance can
be calculated. Typical power requirements of some household appliances are
given in 䉲 Table 17.2. Even though most common appliances specify a nominal
operating voltage of 120 V, it should be noted that household voltage can vary
from 110 V to 120 V and still be considered in the “normal” range. Take a look at
Example 17.6 for more insight on everyday appliances and their operation.
TABLE 17.2 Typical Power and Current Requirements for Various Household Appliances (120 V)
Appliance Power Current Appliance Power Current
Air conditioner, room 1500 W 12.5 A Heater, portable 1500 W 12.5 A

Air-conditioning, central 5000 W 41.7 A* Microwave oven 900 W 5.2 A
Blender 800 W 6.7 A Radio–cassette player 14 W 0.12 A
Clothes dryer 6000 W 50 A* Refrigerator, frost-free 500 W 4.2 A
Clothes washer 840 W 7.0 A Stove, top burners 6000 W 50.0 A*
Coffeemaker 1625 W 13.5 A Stove, oven 4500 W 37.5 A*
Dishwasher 1200 W 10.0 A Television, color 100 W 0.83 A
Electric blanket 180 W 1.5 A Toaster 950 W 7.9 A
Hair dryer 1200 W 10.0 A Water heater 4500 W 37.5 A*
*A high-power appliance such as this is typically wired to a 240-V house supply to reduce the current to half these values (Section 18.5).
EXAMPLE 17.6 A Potentially Dangerous Repair: Don’t Do It Yourself!

A hair dryer is rated at 1200 W for a 115-V operating voltage. The current will increase, because the voltage is the same
Its uniform wire filament breaks near one end, and the owner 1V2 = V12. Thus V2 = I2 R2 = V1 = I1 R1. Now the new current
repairs it by removing the section near the break and simply can be expressed in terms of the original current:
reconnecting it. The resulting filament is 10.0% shorter than
≤ I = 11.112I1
its original length. What will be the heater’s power output R1 R1
I2 = ¢ ≤ I1 = ¢
after this “repair”? R2 0.900R1 1
T H I N K I N G I T T H R O U G H . The wire always operates at 115 V. Hence, the current after the repair is about 11% larger than
Thus shortening the wire, which decreases its resistance, will before.
result in a larger current. With this increase in current, one The original power is P1 = I1 V = 1200 W. The power after
would expect the power output to go up. the repair is P2 = I2 V (note that the voltages have no sub-
SOLUTION. Let’s use a subscript 1 to indicate “before break- scripts because they remained the same and will cancel out).
age” and a subscript 2 to mean “after the repair.” Listing the Forming a ratio gives
given values: P2 I2 V I2
= = = 1.11
Given: P1 = 1200 W Find: P2 (power output P1 I1 V I1
V1 = V2 = 115 V after the repair)
This can be solved for P2 as follows:
L2 = 0.900L1
After the repair, the wire has 90.0% of its original resistance, P2 = 1.11P1 = 1.1111200 W2 = 1330 W
because a wire’s resistance is directly proportional to its
length (see Eq. 17.3). To show the reduction to 90% explicitly, The power output of the dryer has been increased by more
let’s express the resistance after repair 1R2 = rL2>A2 in terms than 120 W. Do not perform such a repair job, as the resulting
of the original resistance 1R1 = rL1>A2: power output may exceed the manufacturer’s requirements and
cause the dryer to melt or, worse, start a fire!
L2 0.900L1 L1
R2 = r = r = 0.900 ¢ r ≤ = 0.900 R1
A A A
as expected.
FOLLOW-UP EXERCISE. In this Example, determine the (a) initial and final resistances and (b) initial and final currents.
People often complain about their electric bills, but what are we actually paying
for? What is being sold is electric energy, usually measured in units of the
kilowatt-hour (kWh). Power is the rate at which work is done P = W>t, or
W = Pt so work has units of watt-seconds 1power * time2. Converting this unit
to the larger unit of kilowatt-hours (kWh), it can be seen that the kilowatt-hour is a
unit of work (or energy), equivalent to 3.6 million joules, because, with W = Pt
1 kWh = 11000 W213600 s2 = 11000 J>s213600 s2 = 3.6 * 106 J
Thus, consumers pay the “power” company for electrical energy used to do work
with their appliances.
The cost of electric energy varies with location and with time. Currently in the
United States, this cost ranges from a low of several cents (per kilowatt-hour) to several
times that value. In the late 1990’s electric energy rates have been deregulated. Coupled
with increasing demand (without a corresponding increase in supply), deregulation
has given rise to skyrocketing rates in some areas of the country. Do you know the
price of electricity in your locality? Check an electric bill to find out, especially if you
are in one of the areas affected by dramatically rising rates. Let’s look at a comparison
of the electric energy costs to run some typical appliances in Integrated Example 17.7.
INTEGRATED EXAMPLE 17.7 A Modern Appliance Dilemma: Computing or Eating

(a) Consider two appliances that operate at the same voltage. computer is on 15 h>day every day and the higher-powered
Appliance A has a higher power rating than appliance B. How broiler is only used for a half hour twice a week, estimate the
does the resistance of A compare with that of B: is it (1) larger, electrical cost that each generates per month. Your calculations
(2) smaller, or (3) the same? (b) A computer system includes a should show that the lower power appliance actually costs
color monitor with a power requirement of 200 W, whereas a more to run per month. Explain why. Assume 15 cents per kWh
countertop broiler>toaster oven is rated at 1500 W. What is the and 30 days and 4.3 weeks per month.
resistance of each if both are designed to run at 120 V? (c) If the (continued on next page)
( A ) C O N C E P T U A L R E A S O N I N G . Power depends on current (b) The monitor current is (from Eq. 17.7)
and voltage. Because the two appliances operate at the same
Pm 200 W
voltage, they can’t carry the same current and still have differ- Im = = = 1.67 A
ent power requirements. Therefore, answer (3) cannot be cor- V 120 V
rect. Because both appliances operate at the same voltage, the and that in the broiler>toaster is
one with higher power (A) must carry more current. For
appliance A to carry more current at the same voltage as B, it Pb 1500 W
Ib = = = 12.5 A
must have less resistance than B. Therefore, the correct V 120 V
answer is (2); A has less resistance than B.
Thus the resistances are
( B ) A N D ( C ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . For
both appliances in part (b) the definition of resistance V 120 V
(R = V>I, Eq. 17.2) can be applied. However, this requires the Rm = = = 71.9 Æ
Im 1.67 A
current to be determined first, which can be done because the
power ratings are known [see Eq. 17.7 1P = IV2]. In part (c), and
the energy used depends on not only the rate (power) of V 120 V
usage but also the time of usage. This will account for differ- Rb = = = 9.60 Æ
Ib 12.5 A
ent energies and costs. Listing the data, using the subscript m
for monitor and b for broiler>toaster: Because all operate at the same voltage, an appliance’s resis-
tance is inversely related to its power requirement.
Given: Pm = 200 W Find: R (resistance of each
Pb = 1500 W appliance)
V = 120 V
(c) The time of operation of each appliance (per month) is needed. This is determined as follows (employing the same sub-
scripts as in part (b):
tm = a 15 b a 30 b = 450 h and tb = a 0.5 b a 2 b a4.3 b = 4.3 h
h d h d wk
d mo d wk mo
Thus the energy and its cost to run the monitor are
Em = 450 h * 200 W = 90 * 103 Wh = 90 kWh,
or a monthly cost of
190 kWh21$0.15>kWh2 = $13.50
and for the broiler they are
Eb = 4.3 h * 1500 W = 6.45 * 103 Wh = 6.45 kWh,
or a monthly cost of
16.45 kWh21$0.15>kWh2 = $0.98.
Thus the higher power (rate) associated with the broiler is more than offset by its much shorter time of usage, resulting in the
monitor costing about 13–14 times more per month.
F O L L O W - U P E X E R C I S E . An immersion heater is a common appliance in most college dorms, useful for heating water for tea, cof-
fee, or soup. Assuming that 100% of the heat produced goes into the water, (a) what must be the heater’s resistance (operating at
120 V) to heat a cup of water (mass 250 g) from room temperature 120 °C2 to boiling in 3.00 min? (b) How much will this usage
increase your monthly electric bill if you prepare two cups of water per day this way?
ELECTRICAL EFFICIENCY AND NATURAL RESOURCES

About 25% of the electric energy generated in the
䉴 F I G U R E 1 7 . 1 4 All lit up A
satellite image of the Americas at United States is used for lighting (䉳 Fig. 17.14).
night. Can you identify the major This percentage is roughly equivalent to the
population centers in the United output of 100 electric-generating (power)
States and elsewhere? The red spots plants. Refrigerators consume about 7% of the
across part of South America indi-
electric energy produced in the United States
cate large-scale burning of vegeta-
tion. The small yellow spot in (the output of about twenty-eight such plants).
Central America shows burning gas This huge (and growing) consumption of
flares at oil production sites. At the electric energy has prompted the federal gov-
top right edge, you can just glimpse ernment and many state governments to set
a few of the white city lights of
minimum efficiency limits for such appliances
Europe. This image was recorded
by a visible–infrared system. as refrigerators, freezers, air conditioners, and
water heaters (䉴 Fig. 17.15). Also, more efficient
compact fluorescent lighting (䉴 Fig. 17.16) has
been put into more common use.
䉳 F I G U R E 1 7 . 1 5 Energy guide
Consumers are made aware of the
efficiencies of appliances in terms of
the average yearly cost of their
operation. Sometimes the yearly
cost is given for different kilowatt-
hour (kWh) rates, which vary
around the United States.
䉱 F I G U R E 1 7 . 1 6 Compact fluorescents Can

you tell from the label how many times more
efficient this bulb is than an old-fashioned
incandescent bulb with the same visible light
output?
The result of these measures has been significant energy savings as new, more
efficient appliances gradually replace inefficient models. Energy saved translates
directly into savings of fuels and other natural resources, as well as a reduction in
environmental hazards such as pollution and also global warming. To see what
kind of results can be achieved from applying just one energy efficiency standard,
consider Examples 17.8 and 17.9.
EXAMPLE 17.8 Electric Energy Cost: The Price of Coolness

If the motor of a frost-free refrigerator runs 15% of the time, T H I N K I N G I T T H R O U G H . From the power and the time the
how much does it cost to operate per month (to the nearest motor is on per day, the electrical energy the refrigerator
cent) if the power company charges 11¢ per kilowatt-hour? requires daily can be determined. That can be used to project
(Assume 30 days in a month.) to a 30-day month.
SOLUTION. Practical energy amounts are usually written in kilowatt-hours because the joule is a relatively tiny unit. Listing the
data:
Given: P = 500 W (Table 17.2) Find: operating cost per month
Cost = $0.11>kWh
The refrigerator motor operates 15% of the time, so in one day it runs t = 10.152124 h2 = 3.60 h. Because P = E>t, the electrical
energy required per day is
E = Pt = 1500 W213.60 h>day2 = 1.80 * 103 Wh>day = 1.80 kWh>day
So the cost (per day) is
a ba b = $0.20>day
1.80 kWh $0.11
day kWh
or about 20¢ per day. For a 30-day month, the cost would be
$ 0.20 30 day
a ba b L $ 6>month
day month
F O L L O W - U P E X E R C I S E . How long would you have to leave a 60-W lightbulb on to use the same amount of electrical energy that
the refrigerator motor in this Example uses each hour it is on?
Using the results from this Example, let’s see what can be saved by increasing
the efficiency of the refrigerator as in Example 17.9.
EXAMPLE 17.9 What Can Be Saved: Increasing Electrical

Efficiency
Many modern power plants produce electric energy at a rate of about 1.0 GW (109 watts).
Estimate how many fewer such power plants the United States would need if all its
households switched from the 500-W refrigerators of Example 17.8 to more-efficient
400-W refrigerators. (Assume that there are about 80 million homes in the United States
with an average of 1.2 refrigerators operating per home.)
THINKING IT THROUGH. The results from Example 17.8 can be used to calculate the
overall effect.
SOLUTION.
Given: Plant output rate = 1.0 GW = 1.0 * 106 kW Find: how many fewer
Energy requirement, power plants would
500-watt model = 1.80 kWh>day (Example 17.8) be required after
Number of homes = 80 * 106 switching to more-
Number of refrigerators per home = 1.2 efficient refrigerators
For the entire country, the energy usage per day with the less-efficient refrigerators is
1.80 kWh>day 1.2 refrigerators

a b180 * 106 homes2a b = 1.8 * 108
kWh
refrig home day
The more-efficient refrigerators would save 20% of this 1100 W>500 W = 0.202, or
0.211.8 * 108 kWh>day2 = 3.6 * 107 kWh>day.
Per day, the electric energy production of a 1.0-GW power plant is
1kWh>plant2
a1.0 * 106 b a24 b = 2.4 * 107
kW h
plant day day
Thus the number of plants could be reduced by
3.6 * 107 kWh>day

L 1.5 plants
2.4 * 107 kWh>1plant-day2
Note that this saving results from a change in a single appliance. Imagine what could be
done if all appliances, including lighting, were made more efficient. Developing and
using more efficient electrical appliances is one way to avoid having to build new elec-
tric energy generating plants and dealing with the rising cost associated with nonre-
newable fuels.
F O L L O W - U P E X E R C I S E . Electric and gas water heaters are often said to be equally
efficient—typically, about 95%. In reality, while gas water heaters are capable of 95%
efficiency, it might be more accurate to describe electric water heaters as only about
30% efficient, even though approximately 95% of the electrical energy they require is
transferred to the water in the form of heat. Explain. [Hint: What is the source of
energy for an electric water heater? Compare this to the energy delivery of natural
gas. Recall the discussions of electrical generation in Section 12.4 and Carnot effi-
ciency in Section 12.5.]
DID YOU LEARN?

➥ At a constant voltage, power dissipation of an appliance or instrument is inversely
related to its resistance.
➥ At the same operating voltage, lower power appliances have more resistance.
➥ The kWh is equal to 3.6 * 106 J or 3.6 MJ.
PULLING IT TOGETHER Designing an Immersion Heater

A 120-V electrical immersion heater (coil) is to be designed T H I N K I N G I T T H R O U G H . The power output of the heater is
that will heat 400 mL of water for tea in 3.00 min, starting related to the applied voltage across it and its resistance. The
from a water temperature of 20.0 ºC and heating it 90.0 ºC. power output can be determined from time and the total
The heater’s wire is made of Nichrome and the total wire (heat) energy needed to heat the water and cup. After the
length is 70.0 cm. If the heater is to be 50.0% efficient and the resistance is found, knowing the resistivity and length of the
glass cup has a mass of 150 g, what must the diameter of the (cylindrical) wire should enable the calculation of the wire’s
wire be? Treat the 120-V as a dc voltage and assume that the cross-sectional area, and then its diameter.
wire is ohmic and the cup’s temperature is the same as the
water that is in it. Neglect the variation in resistivity with
temperature of the Nichrome.
SOLUTION. The data are listed, and converted to SI units.

Given: V = 120 V Find: d (wire diameter)
L = 70.0 cm = 0.700 m
r = 100 * 10-8 Æ # m (from Table 17.1)
Ti = 20.0 °C
Tf = 90.0 °C
¢T = 70.0 °C
t = 3.00 min = 180 s
e = 50.0% = 0.500 (efficiency)
Vw = 400 mL = 0.400 L
cw = 4186 J>1kg # °C2 (Table 11.1)
mg = 150 g = 0.150 kg
cg = 840 J>1kg # °C2 (Table 11.1)
The total (heat) energy needed, Q, is the sum of that needed by the water and the cup. The mass of water is needed. This is easily
found from its density of 1 g>cm3, resulting in a mass of 400 g, or 0.400 kg. Thus
Q = mwc w ¢T + mg cg ¢T
= 10.400 kg214186 J>1kg # °C22170.0 °C2 + 10.150 kg21840 J>1kg # °C22170.0 °C2
= 1.17 * 105 J + 0.0882 * 105 J = 1.26 * 105 J
This is 50.0% of the actual energy E expended by the heater, that is, Q = 0.500E or
Q 1.26 * 105 J
E = = = 2.52 * 105 J
0.500 0.500
The heater power is
E 2.52 * 105 J
P = = = 1.40 * 103 W
t 180 s
V2
Now the heater resistance can be found as follows, since P = :
R
V2 1120 V22
R = = = 10.3 Æ
P 1.40 * 103 W
L
But R is related to the wire by R = r ; therefore
A
= 1100 * 10-8 Æ # m2a b = 6.80 * 10-8 m2

L 0.700 m
A = r
R 10.3 Æ
pd2
Finally, the diameter can be found from this circular area, since A = , and thus
4
4A 416.80 * 10-8 m22

d = = = 2.94 * 10-4 m L 0.3 mm
Ap C p
■ A battery produces an electromotive force (emf), or volt- ■ The electrical resistance (R) of an object is the voltage across
age, across its terminals. The high-voltage terminal is the the object divided by the current in it, or
anode ( ⴙ), and the low-voltage one is the cathode (ⴚ).
V
R = or V = IR (17.2)
I
Electron Electron
flow flow
The units of resistance are the ohm (æ), or the volt per
ampere.
R
V
A Anode B
Cathode R= V
– – I
–
– Electrolyte –
– B+
–
– B+ – I
–
A+ – B+
– + B+
+ + – B
– A+ B B – +
–
A+ – B + –
A+ A+ B+ B+ B+ B+
V
■ Electromotive force (emf e) is measured in volts and repre-
sents the number of joules of energy that a battery (power ■ A circuit element obeys Ohm’s law if it exhibits constant
supply) gives to 1 coulomb of charge passing through it; electrical resistance. Ohm’s law is commonly written as
that is, 1 J>C = 1 V. V = IR, where R is constant.
V
+ – Voltage
I
V/
=
R
e=
op
Sl
I
Current
■ Electric current (I) is the rate of charge flow. Its direction is ■ The resistance of an object depends on the resistivity (R) of
assumed to be that of the conventional current, which is the the material (based on its atomic properties and possibly its
direction in which positive charge actually flows or appears temperaturre), the cross-sectional area A, and the length L.
to flow. In metals, because the charge flow is electrons, the For objects of uniform cross-section,
conventional current direction is thus opposite the direction
b
of electron flow. Current is measured in amperes L
11 A = 1 C>s2 and defined as
R = ra (17.3)
A
q
I = (17.1)
t
Length
R
S
Material
Electron
flow Temperature Cross-sectional
area
I Conventional
current
V ■ Electric power (P) is the rate at which work is done by a bat-
+ – tery (power supply), or the rate at which energy is trans-
ferred to a circuit element. The power delivered to a circuit
Battery
element depends on the element’s resistance, the current in
■ For an electric current to exist in a circuit, it must be a it, and the voltage across it. Electrical power can be written
complete circuit—that is, a circuit (set of circuit elements in three equivalent ways:
and wires) that connects both terminals of a battery or V2
power supply with no break. P = IV = = I 2R (17.7b)
R

In this chapter, assume that all batteries have negligible internal resistance unless otherwise indicated.
17.1 BATTERIES AND DIRECT CURRENT 17.3 RESISTANCE AND OHM S LAW
1. When a battery with a significant internal resistance is 13. The ohm is just another name for the (a) volt per ampere,
part of a complete circuit, the voltage across its terminals (b) ampere per volt, (c) watt, (d) volt.
is its (a) emf, (b) terminal voltage, (c) power output, 14. Two ohmic resistors are connected to the electrodes of a
(d) all of the preceding. 12-V battery one at a time. The current in resistor A is
2. As a battery ages, assuming it is always connected into twice that in B. What can you say about their resistance
the same complete circuit, its (a) emf increases, (b) emf values: (a) RA = 2RB, (b) RA = RB, (c) RA = RB>2, or
decreases, (c) terminal voltage increases, (d) terminal (d) none of the preceding?
voltage decreases. 15. An ohmic resistor is placed across the terminals of two
3. When four 1.5-V batteries are connected, the output volt- different batteries one at a time. When the resistor is con-
age of the combination is measured as 1.5 V. These bat- nected to battery A, the resulting current is three times the
teries therefore are connected (a) in series, (b) in parallel, current compared to when it is attached to battery B.
(c) as a pair in series connected in parallel to the other What can you say about the battery voltages: (a) VA = 3VB
pair in series, (d) you can’t tell the connection from the , (b) VA = VB, (c) VB = 3VA, or (d) none of the preceding?
data given. 16. If you double the voltage across a resistor while at the
4. When helping someone whose car has a “dead” battery, same time cutting its resistance to one-third its original
how should your car’s battery be connected in relation value, what happens to the current in the resistor: (a) it
to the dead battery: (a) in series, (b) in parallel, or doubles, (b) it triples, (c) it increases by six times, or
(c) either in series or in parallel would work fine? (d) you can’t tell from the data given?
5. When several 1.5-V batteries are connected in series, the 17. Both the length and diameter of cylindrical resistor are
overall output voltage of the combination is measured to doubled. What happens to the resistor’s resistance: (a) it
be 12 V. How many batteries are needed to achieve this doubles, (b) it is cut in half, (c) it is reduced to one-fourth
voltage: (a) two, (b) ten, (c) eight, or (d) six? its initial value, or (d) none of these?
6. To move 3.0 C of charge from one electrode to the other, 18. Two wires are identical except that one is aluminum and
a 12-V battery must do how much work: (a) 12 V, (b) 12 J, the other is copper. Which one’s resistance will increase
(c) 3.0 J, (d) 36 W, or (e) 36 J? more rapidly as they are heated: (a) the aluminum wire,
7. In a circuit diagram, a battery is represented by (a) two (b) the copper wire, (c) both would increase at the same
parallel equal-length lines, (b) a straight line in the direc- rate, or (d) you can’t tell?
tionof the wires, (c) two unequal-length lines, (d) a wig-
gly jagged symbol.
17.4 ELECTRIC POWER
19. The electric power unit, the watt, is equivalent to what
combination of SI units: (a) A2 # Æ , (b) J>s, (c) V 2> Æ , or
17.2 CURRENT AND DRIFT VELOCITY
8. In which of these situations does more charge flow past (d) all of the preceding?
a given point on a wire: when the wire has a current of 20. If the voltage across an ohmic resistor is doubled, the
(a) 2.0 A for 1.0 min, (b) 4.0 A for 0.5 min, (c) 1.0 A for power expended in the resistor (a) increases by a factor
2.0 min, or (d) all are the same? of 2, (b) increases by a factor of 4, (c) decreases by half, or
9. Which of these situations involves the least current: a (d) none of the preceding.
wire that has (a) 1.5 C passing a given point in 1.5 min, 21. If the current through an ohmic resistor is halved, the
(b) 3.0 C passing a given point in 1.0 min, or (c) 0.5 C power expended in the resistor (a) increases by a factor
passing a given point in 0.10 min? of 2, (b) increases by a factor of 4, (c) decreases by half,
10. In a dental X-ray machine, the accelerated electrons (d) decreases by a factor of 4.
move to the east. The conventional current in the 22. A cylindrical resistor dissipates thermal energy at a cer-
machine is in what direction: (a) east, (b) west, or (c) you tain power rate, P, when connected to a battery. It is dis-
can’t tell from the data given? connected and cut in half lengthwise. One of the halves
11. In a current-carrying metal wire, the drift velocity of the is then reconnected across the same battery. The new
electrons is on the order of (a) the speed of light, (b) the power rate for the shortened resistor is (a) P, (b) 2P,
speed of sound, (c) a few millimeters per second. (c) P>2, (d) P>4.
12. When a light switch that controls a single light bulb is 23. Two wires are of the same length and thickness, but one
thrown to the “on” position, the electric energy gets to is aluminum and the other is copper. Both are connected,
the light bulb at a speed on the order of (a) the speed of one at a time, to the terminals of the same battery. Which
light, (b) the speed of sound, (c) a few millimeters per one has a higher power output: (a) the aluminum wire,
second, or (d) you can’t tell from the data given since it (b) the copper wire, (c) they have the same power out-
depends on the power output of the bulb. put, or (d) you can’t tell?
17.1 BATTERIES AND DIRECT CURRENT will the current in the wire be affected? (b) How will the
current be affected if, instead, the new wire has the same
1. Explain why electrode A (in the battery design shown in
length as the old one but half the diameter? In both
Fig. 17.1) is labeled with a plus sign when it has an
cases, explain your reasoning.
excess of electrons, which carry a negative charge.
12. A real battery always has some internal resistance r
2. Why does the battery design shown in Fig. 17.1 require a that increases with the battery’s age (䉲 Fig. 17.17).
chemical membrane? Explain why, in a complete circuit connection, this
3. The manufacturer’s rating of a battery is 12 V. Does this results in a drop of the terminal voltage V of the battery
mean that the battery will necessarily measure 12 V with time.
across its terminals when it is placed in a complete cir-
cuit? Explain. R
4. Sketch the following complete circuits, using the symbols
shown in Learn by Drawing 17.1: (a) two ideal 6.0-V bat-
teries in series wired to a capacitor followed by a resis-
tor; (b) two ideal 12.0-V batteries in parallel, connected S Electron
+ – flow
as a unit to two resistors in series with one another; (c) a V
nonideal battery (one with internal resistance) wired to
r +–
two capacitors that are in parallel with each other, fol-
lowed by two resistors in series with one another.
(Battery) − V = Ir
䉱 F I G U R E 1 7 . 1 7 Emf and terminal voltage See Conceptual

17.2 CURRENT AND DRIFT VELOCITY Question 12 and Exercise 15.
5. In the circuit shown in Fig. 17.4a, what is the direction of
(a) the electron flow in the resistor, (b) the conventional 13. If you cut a long wire in half, what would you have to do
current in the resistor, and (c) the conventional current in to its diameter in order to keep its resistance constant?
the battery? Explain your reasoning.
6. The drift speed of electrons in a complete circuit, is typi-
cally a few millimeters per second. Yet a lamp 3.0 m
17.4 ELECTRIC POWER
away from a light switch turns on instantaneously when
you flip the switch. Explain this apparent paradox. 14. Assuming that the resistance of your hair dryer obeys
7. In the battery design shown in Fig. 17.1, how does the Ohm’s law, what would happen to its power output if
direction of current inside the battery compare to that in you plugged it directly into a 240-V outlet in Europe if it
the wire connecting its two electrodes? is designed to be used in the 120-V outlets of the United
States?
8. To move charges in metallic wires, an internal electric
field must exist in the wire (see, for example, Fig. 17.8). 15. Most lightbulb filaments are made of tungsten and are
In such a wire, are the electric field and the electric cur- about the same length. What then must be different
rent in the same or opposite directions? Explain. about the tungsten filament in a 60-W bulb compared
with that in a 40-W bulb?
16. Which one gives off more power (as joule heating) when
17.3 RESISTANCE AND OHM S LAW connected across the terminals of a 12-V battery: a 5.0-Æ
resistor or a 10-Æ resistor? Explain your reasoning.
9. If the voltage (V) were plotted (vertically) on the same 17. From the electric power relationship P = I 2R, it would
graph versus current (I) for two ohmic conductors with appear that when increasing the value of an appliance’s
different resistances, how could you tell which is less resistance, its power output would also increase. Yet a
resistive? 60-W light bulb has less resistance than its lower power
10. Filaments in lightbulbs usually fail just after the bulbs 40-W counterpart. Explain.
are turned on rather than when they have already been 18. Explain clearly why, in terms of electric energy
on for a while. Why? generation from nonrenewable fuels, an electric appli-
11. A wire is connected across a steady voltage source. (a) If ance can be, at most, only about 33% efficient. [Hint: See
that wire is replaced with one of the same material that is Chapter 12 on thermal cycle efficiencies and modern
twice as long and has twice the cross-sectional area, how power plants.]
EXERCISES 619
EXERCISES
17.1 BATTERIES AND DIRECT CURRENT to pass that location if the current in the wire is
doubled?
Assume all batteries are ideal unless told otherwise.
12. ● ● ● Car batteries are often rated in “ampere-hours” or
1. ● (a) Three 1.5-V batteries are connected in series. What is A # h. (a) Show that the A # h has units of charge and that
the total voltage of the combination? (b) What would be 1 A # h = 3600 C. (b) A fully charged, heavy-duty battery is
the total voltage if the cells were connected in parallel? rated at 100 A # h and can deliver a current of 5.0 A steadily
2. ● What is the voltage across six 1.5-V batteries when until depleted. What is the maximum time this battery can
they are connected (a) in series, (b) in parallel, (c) three in deliver that current, assuming it isn’t being recharged?
parallel with one another and this combination wired in (c) How much charge will the battery deliver in this time?
series with the remaining three? 13. IE ● ● ● Imagine that some protons are moving to the left
3. ● Two 6.0-V batteries and one 12-V battery are con- at the same time that some electrons are moving to the
nected in series. (a) What is the voltage across the whole right past the same location. (a) Will the net current be
arrangement? (b) What arrangement of these three bat- (1) to the right, (2) to the left, (3) zero, or (4) none of the
teries would give a total voltage of 12 V? preceding? (b) In 4.5 s, 6.7 C of electrons flow to the right
at the same time that 8.3 C of protons flow to the left.
4. ●● Given three batteries with voltages of 1.0 V, 3.0 V, and
What are the direction and magnitude of the current due
12 V, what are the minimum and maximum voltages that
to the protons? (c) What are the direction and magnitude
could be achieved by connecting them in series?
of the current due to the electrons? (d) What are the
5. IE ● ● You are given four AA batteries that are rated at direction and magnitude of the total current?
1.5 V each. The batteries are grouped in pairs. In 14. ● ● ● In a proton linear accelerator, a 9.5-mA proton cur-
arrangement A, the two batteries in each pair are in rent hits a target. (a) How many protons hit the target
series, and then the pairs are connected in parallel. In each second? (b) How much energy is delivered to the
arrangement B, the two batteries in each pair are in par- target each second if each proton has a kinetic energy of
allel, and then the pairs are connected in series. (a) Com- 20 MeV and loses all its energy in the target? (c) If the
pared with arrangement B, will arrangement A have target is a 1.00-kg block of copper, at what rate will its
(1) a higher, (2) the same, or (3) a lower total voltage? temperature increase if it is not cooled?
(b) What are the total voltages for each arrangement?
17.3 RESISTANCE AND OHM S LAW*

17.2 CURRENT AND DRIFT VELOCITY 15. ● A battery labeled 12.0 V supplies 1.90 A to a 6.00-Æ
6. ● How long does it take for a charge of 3.50 C to pass resistor (Fig. 17.17). (a) What is the terminal voltage of
through the cross-sectional area of a wire that is carrying the battery? (b) What is its internal resistance?
a current of 0.57 A? 16. ● How much current is drawn from an ideal 12-V battery
7. ● A net charge of 30 C passes through the cross-sectional (no significant internal resistance) when a 15-Æ resistor is
area of a wire in 2.0 min. What is the current in the wire? connected across its terminals?
17. ● What terminal voltage must an ideal battery (no sig-
8. ● (a) How long would it take for a net charge of 2.5 C to
nificant internal resistance) have to produce a 0.50-A cur-
pass a location in a wire if it is to carry a steady current
rent through a 2.0-Æ resistor?
of 5.0 mA? (b) If the wire is actually connected directly to
18. ● What is the emf of a battery with a 0.15-Æ internal
the two electrodes of a battery and the battery does 25 J
of work on the charge during this time, what is the ter- resistance if the battery delivers 1.5 A to an externally
minal voltage of the battery? connected 5.0-Æ resistor?
19. IE ● Some states allow the use of aluminum wire in
9. ● A small toy car draws a 0.50-mA current from a 3.0-V
houses in place of copper. (a) If you wanted the resistance
NiCd (nickel–cadmium) battery. In 10 min of operation,
of your aluminum wire to be the same as that of copper
(a) how much charge flows through the toy car, and
(assuming the same lengths), would the aluminum wire
(b) how much energy is lost by the battery?
have to have (1) a greater diameter than, (2) a smaller
10. ●A car’s starter motor draws 50 A from the car’s battery diameter than, or (3) the same diameter as the copper
during startup. If the startup time is 1.5 s, how many wire? (b) Calculate the ratio of the thickness of aluminum
electrons pass a given location in the circuit during that to that of copper needed to make their resistances equal.
time?
11. ● ● A net charge of 20 C passes a location in a wire in *Assume that the temperature coefficients of resistivity given in
1.25 min. How long does it take for a net 30-C charge Table 17.1 apply over large temperature ranges.
20. ● During a research experiment on the conduction of same as, or (3) less than that before the stretch? (b) A
current in the human body, a medical technician attaches 1.0-m length of copper wire with a 2.0-mm diameter is
one electrode to the wrist and a second to the shoulder of stretched out; its length increases by 25% while its cross-
a patient. If 100 mV is applied across the two electrodes sectional area decreases, but remains uniform. Compute
and the resulting current is 12.5 mA, what is the overall the resistance ratio (final to initial).
resistance of the patient’s arm? 32. ● ● ● 䉲 Figure 17.18 shows data on the dependence of the
21. ● A 0.60-m-long copper wire has a diameter of 0.10 cm. current through a resistor on the voltage across that
What is the resistance of the wire? resistor. (a) Is the resistor ohmic? Explain your reason-
22. ● ● A material is formed into a long rod with a square ing. (b) What is the value of its resistance? (c) Use the
cross-section 0.50 cm on each side. When a 100-V voltage is data to predict what voltage would be needed to pro-
applied across a 20-m length of the rod, a 5.0-A current is duce a 4.0-A current in the resistor.
carried. (a) What is the resistivity of the material? (b) Is the V (V)
material a conductor, an insulator, or a semiconductor?
23. ● ● Two copper wires have equal cross-sectional areas and 40
lengths of 2.0 m and 0.50 m, respectively. (a) What is the
ratio of the current in the shorter wire to that in the longer 30
one if they are connected to the same power supply? (b) If 20
you wanted the two wires to carry the same current, what
would the ratio of their cross-sectional areas have to be? 10
(Give your answer as a ratio of longer to shorter.) 0 I (A)
5.0 10 15 20
24. IE ● ● Two copper wires have equal lengths, but the
diameter of one is three times that of the other. (a) The 䉱 F I G U R E 1 7 . 1 8 An ohmic resistor? See Exercise 32.
resistance of the thinner wire is (1) 3, (2) 13 , (3) 9, (4) 19
times that of the resistance of the thicker wire. (b) If the 33. ● ● ● At 20 °C, a silicon rod of uniform cross-section is
thicker wire has a resistance of 1.0 Æ , what is the resis- connected to a battery with a terminal voltage of 6.0 V
tance of the thinner wire? and a 0.50-A current results. The temperature of the rod
25. ● ● The wire in a heating element of an electric stove is then increased to 25 °C. (a) What is its new resistance?
burner has a 0.75-m effective length and a 2.0 * 10-6 -m2 (b) How much current does it carry? (c) If you wanted to
cross-sectional area. (a) If the wire is made of iron and cut the current from its room temperature value of 0.50 A
operates at 380 °C, what is its operating resistance? to 0.40 A, at what temperature would the rod have to be?
(b) What is its resistance when the stove is “off”? 34. IE ● ● ● A platinum wire is connected to a battery. (a) If the
26. ● ● (a) What is the percentage variation of the resistivity of temperature increases, will the current in the wire
copper over the temperature range from room tempera- (1) increase, (2) remain the same, or (3) decrease? Why?
ture (20 °C) to 100 °C? (b) Assume that a copper wire’s (b) An electrical resistance thermometer is made of plat-
resistance changes due to only resistivity changes over this inum wire that has a 5.0-Æ resistance at 20 °C. The wire is
temperature range. Further assume that it is connected to connected to a 1.5-V battery. When the thermometer is
the same power supply. By what percentage would its cur- heated to 2020 °C, by how much does the current change?
rent change? Would it be an increase or decrease? (c) By how much does the wire’s joule heating rate change?
27. ● ● A copper wire has a 25-mÆ resistance at 20 °C. When
the wire is carrying a current, heat produced by the cur- 17.4 ELECTRIC POWER
rent causes the temperature of the wire to increase by
27 °C. (a) What is the change in the wire’s resistance? (b) If 35. ● A digital video disk (DVD) player is rated at 100 W at
its original current was 10.0 mA, what is its final current? 120 V. What is its resistance?
36. ● A freezer of resistance 10 Æ is connected to a 110-V
28. ● ● When a resistor is connected to a 12-V source, it
draws a 185-mA current. The same resistor connected to source. What is the power delivered when this freezer is on?
a 90-V source draws a 1.25-A current. (a) Is the resistor 37. ● The current in a refrigerator with a resistance of 12 Æ
ohmic? Justify your answer mathematically. (b) What is is 13 A (when the refrigerator is on). What is the power
the rate of Joule heating in this resistor in both cases? delivered to the refrigerator?
● Show that the quantity volts squared per ohm 1V > Æ2
2
29. ● ● A particular application requires a 20-m length of 38.
aluminum wire to have a 0.25-mÆ resistance at 20 °C. has SI units of power.
(a) What is the wire’s diameter? (b) What would its resis- 39. ● An electric water heater is designed to produce 50 kW
tance be if its length was halved and it was then placed of heat when it is connected to a 240-V source. What is its
in an ice water bath? resistance?
30. ● ● (a) If the resistance of the wire in Exercise 29 cannot 40. ● ● If the heater in Exercise 39 is 90% efficient, how long
vary by more than ;5.0% from its value at 20 °C, to what would it take to heat 50 gal of water from 20 °C to 80 °C?
operating temperature range should it be restricted? 41. IE ● ● An ohmic resistor in a circuit is designed to oper-
(b) What would be the operating range if the wire were, ate at 120 V. (a) If you connect the resistor to a 60-V
instead, made of copper? power source, will the resistor dissipate heat at (1) 2,
31. IE ● ● ● As a wire is stretched out so that its length (2) 4, (3) 21 , or (4) 14 times the designed power? Why? (b) If
increases, its cross-sectional area decreases, while the the designed power is 90 W at 120 V, but the resistor is
total volume of the wire remains constant. (a) Will the connected to a 30-V power supply, what is the power
resistance after the stretch be (1) greater than, (2) the delivered to the resistor?
42. ●● An electric toy with a resistance of 2.50 Æ is operated short time, the temperature of the coil increases by
by a 3.00-V battery. (a) What current does the toy draw? 150 °C because of joule heating. (a) Will the dissipated
(b) Assuming that the battery delivers a steady current for power (1) increase, (2) remain the same, or (3) decrease?
its lifetime of 4.00 h, how much charge passed through the Why? (b) What is the corresponding change in the
toy? (c) How much energy was delivered to the toy? power? (c) What is the percentage change in its current?
43. ● ● A welding machine draws 18 A of current at 240 V. 51. ● ● A 20-Æ resistor is connected to four 1.5-V batteries.
(a) What is its power rating? (b) What is its resistance? What is the joule heat loss per minute in the resistor if the
(c) When it is inadvertently connected to a 120 V outlet, batteries are connected (a) in series and (b) in parallel?
the current in it is 10 A. Is the machine’s resistance 52. ● ● A 5.5-kW water heater operates at 240 V. (a) Should
ohmic? Prove your answer. the heater circuit have a 20-A or a 30-A circuit breaker?
44. ● ● On average, an electric water heater operates for (A circuit breaker is a safety device that opens the circuit
2.0 h each day. (a) If the cost of electricity is $0.15>kWh, at its rated current.) (b) Assuming 85% efficiency, how
what is the cost of operating the heater during a 30-day long will the heater take to heat the water in a 55-gal
month? (b) What is the resistance of a typical water tank from 20 °C to 80 °C?
heater? [Hint: See Table 17.2.] 53. ● ● A student uses an immersion heater to heat 0.30 kg of
45. ● ● (a) What is the resistance of an immersion-type heat- water from 20 °C to 80 °C for tea. (a) If the heater is 75%
ing coil if it is to generate 15 kJ of heat per minute when efficient and takes 2.5 min to heat the water. what is its
it is connected to a 120-V source? (b) What would the resistance? (b) How much current is in the heater?
coil’s resistance have to be if instead 10 kJ of heat per (Assume 120-V household voltage.)
minute was desired? 54. ● ● An ohmic appliance is rated at 100 W when it is con-
46. ● ● A 200-W computer power supply is on 10 h per day. nected to a 120-V source. If the power company cuts the
(a) If the cost of electricity is $0.15>kWh, what is the voltage by 5.0% to conserve energy, what is (a) the cur-
energy cost (to the nearest dollar) if this computer is rent in the appliance and (b) the power consumed by the
used like this for a year (365 days)? (b) If this power sup- appliance after the voltage drop?
ply is replaced by a more efficient 100-W version and it 55. ● ● A lightbulb’s output is 60 W when it operates at
costs $75, how long will it take the decreased operating 120 V. If the voltage is cut in half and the power dropped
cost to pay for this power supply? to 20 W during a brownout, what is the ratio of the
47. ● ● A 120-V air conditioner unit draws 15 A of current. If bulb’s resistance at full power to its resistance during the
it operates for 20 min, (a) how much energy in kilowatt- brownout?
hours does it use in that time? (b) If the cost of electricity 56. ● ● To empty a flooded basement, a water pump must
is $0.15>kWh, what is the cost (to the nearest penny) of do work (lift the water) at a rate of 2.00 kW. If the pump
operating the unit for 20 min? (c) If the air conditioner is wired to a 240-V source and is 84% efficient, (a) how
initially cost $450 and it is operated, on average, 4 h per much current does it draw and (b) what is its resistance?
day, how long does it take before the operating costs
57. ● ● ● (a) Find the individual monthly (30-day) electric
equal the price?
energy costs (to the nearest dollar) for each of the follow-
48. ● ● Two resistors, 100 Æ and 25 kÆ , are rated for a maxi-
ing household appliances if the utility rate is $0.12>kWh:
mum power output of 1.5 W and 0.25 W, respectively. central air conditioning that runs 30% of the time; a
(a) What is the maximum voltage that can be safely blender that is used 0.50 h>month; a dishwasher that is
applied to each resistor? (b) What is the maximum cur- used 8.0 h>month; a microwave oven that is used
rent that each resistor can have? 15 min>day; the motor of a frost-free refrigerator that
49. ● ● A wire 5.0 m long and 3.0 mm in diameter has a resis- runs 15% of the time; a stove (burners plus oven) that is
tance of 100 Æ . A 15-V potential difference is applied used a total of 10 h>month; and a color television that is
across the wire. Find (a) the current in the wire, (b) the operated 120 h>month. (b) Determine the percentage
resistivity of its material, and (c) the rate at which heat is that each appliance contributes to the total monthly elec-
being produced in the wire. tric cost. (c) What is the resistance of each appliance and
50. IE ● ● When connected to a voltage source, a coil of current in each appliance when they are operating? (Use
tungsten wire initially dissipates 500 W of power. In a the information given in Table 17.2.)
58. IE A piece of carbon and a piece of copper have the to that of carbon at the raised temperature. (c) Assuming
same resistance at room temperature. (a) If the tempera- they have the same voltage across them, calculate the
ture of each piece is increased by 10.0 °C, will the copper ratio of the current in the copper compared to the carbon
piece have (1) a higher resistance than, (2) the same resis- at the raised temperature. (d) Repeat part (c) for the ratio
tance as, or (3) a lower resistance than the carbon piece? of the power delivered.
Why? (b) Calculate the ratio of the resistance of copper
59. Two pieces of aluminum and copper wire are identical in lethal. (b) Calculate the ratio of electric current in the
length and diameter. At some temperature, one of the wires (assumed ohmic) at 500 kV to when the wires
wires will have the same resistance that the other has at operate at 120 V. (c) Calculate the ratio of heating loss in
20 °C. (a) What is that temperature? (Hint: There may be a given length of wire (assumed ohmic) carrying current
more than one temperature.) (b) If the two wires are con- at 500 kV to when it operates at 120 V.
nected in series, compute the ratio of the total resistance 67. In a country setting it is common to see a hawk sitting on
initially to the total resistance at the temperatures in part a single high-voltage electric power line searching for a
(a). (c) Repeat part (b) if the wires were instead con- roadkill meal (䉲 Fig. 17.19). To understand why this bird
nected in parallel. isn’t electrocuted, let’s do a ballpark estimate of the volt-
60. A battery delivers 2.54 A to a resistor rated at 4.52 Æ . age between her feet. Assume dc conditions in a power
When it is connected to a 2.21-Æ resistor, it delivers 4.98 A. line that is 1.0 km long, has a resistance of 30 Æ , and is at
Determine the battery’s (a) internal resistance (assumed an electric potential of 250 kV above the other wire (the
constant), (b) emf, and (c) terminal voltage (in both cases). one the bird is not on), which is at ground or zero volts.
61. An external resistor is connected to a battery with a vari- (a) If the wires are carrying energy at the rate of 100 MW,
able emf but constant internal resistance of 0.200 Æ. At an what is the current in them? (b) Assuming the bird’s feet
emf of 3.00 V, the resistor draws a current of 0.500 A, and at are 15 cm apart, what is the resistance of that segment of
6.00 V, the resistor draws a current of 1.50 A. (a) Is the exter- the hot wire? (c) What is the voltage difference between
nal resistor ohmic? Prove your answer. (b) Determine the the bird’s feet? Comment on the size of your answer and
value of the external resistance under the two different con- whether you think this might be dangerous. (d) What is
ditions. (c) In both cases, determine the ratio of Joule heat- the voltage difference between her feet if she places one
ing rates in the external resistor to that in the battery. on the ground wire while continuing to hold onto the
62. An electric eel delivers a current of 0.75 A to a small hot wire? Comment on the size of your answer and
pencil-thin prey 15 cm long. If the eel’s “bio-battery” whether you think this might be dangerous.
was charged to 500 V, and it was constant for 20 ms
before dropping to zero, estimate (a) the resistance of the 䉳 F I G U R E 1 7 . 1 9 Bird
fish, (b) the energy delivered to the fish, and (c) the aver- on a wire See Exercise 67.
age electric field (magnitude) in the fish’s flesh.
63. Most modern TVs have an “instant warm-up” feature.
Even though the set appears to be off, it is “off” only in
that there is no picture and audio. To provide a “quick
on” feature, the TV’s electronics are kept ready. This
takes about 10 W of electric power, constantly. Under
these conditions, what is (a) the current in the TV and
(b) the resistance of the TV? (c) Assuming that there is
one TV with this feature for every two households, esti-
mate how many electric power plants this feature takes
to run just in the United States.
64. A computer CD-ROM drive that operates on 120 V is rated
at 40 W when it is operating. (a) How much current does 68. A cylindrical resistor is made of carbon and is 10.0 cm
the drive draw? (b) What is the drive’s resistance? (c) How long with a diameter of 1.00 cm. Assuming the resistor is
much energy (in kWh) does this drive use per month kept at room temperature, (a) what is its resistance? (b) If
assuming it operates 15 min per day? (d) Estimate the elec- it is then connected to a 12.0-V battery, how many elec-
tric energy bill per month, assuming 15 cents per kWh. trons pass through one end of the resistor in 1 min? (c) In
65. The tungsten filament of an incandescent lamp has a that 1 min, how much stored energy did the battery lose?
resistance of 200 Æ at room temperature. (a) What would (d) Repeat these calculations for the resistor if it is sliced
the resistance be at an operating temperature of 1600 °C? in half lengthwise.
(b) Assuming it is plugged into a 120-V outlet, what is its 69. A high-voltage power supply (10.0 kV) is encased in a
power output when just starting up? (c) By how much metal box and rests on a metal table, which is grounded.
does its power output change from the room tempera- Between the power supply’s box and the table is a
ture value when it is at its operating temperature? square sheet of rubber, 2.54 cm thick and 30.0 cm on a
66. A common sight in our modern world is high-voltage side. (a) What is the resistance of the rubber mat?
lines carrying electric energy over long distances from (b) Suppose the output of the power supply touches its
power plants to populated areas. The delivery voltage of metal framed box, which touches the rubber mat uni-
these lines is typically 500 kV, whereas by the time the formly over its whole area. How much current flows
energy reaches our households it is down to 120 V (see through the rubber.? (c) If the rubber mat were to be
Chapter 20 for how this is done). (a) Explain clearly why replaced by a wooden square, what would the square’s
electric power is delivered over long distances at high thickness have to be to maintain the same protection as
voltages when high voltages are known to be potentially the rubber? Assume the areas stay the same.
18 Basic Electric Circuits
18.1 Resistances in series,
parallel, and series–parallel
combinations (624)
■ equivalent resistances
18.2 Multiloop circuits and

Kirchhoff ’s rules (631)
■ junction rule
■ loop rule
■ circuits analysis
18.3 RC circuits (637)

■ charging and discharging
capacitors
■ time constant
PHYSICS FACTS
■
18.4 Ammeters and
voltmeters (640)
electrical measurements:
✦ More than one hair dryer cannot
be used on the same household
circuit without tripping a (15A) cir-
cuit breaker. If two are used at the
M etallic wires are usually
thought of as the “connec-
tors” between elements in a circuit.
design and usage same time, two separate circuits
are needed.
However, wires are not the only
✦ Less than 10 mA of current through conductors of electricity, as the
the human body can trigger mus-
18.5 Household circuits and cle paralysis. If a person touches
chapter-opening photo shows.
electrical safety (644) exposed wiring and cannot then let Because the bulb is lit, the circuit
go, death could result if the current
passes through a vital organ. must be complete (the power sup-
✦ Specialized pacemaker cells ply is not shown). It can be con-
located in a small region of the
heart trigger your heartbeat. Their cluded, therefore, that the “lead” in
electrical signals travel across the
heart in about 50 ms. These cells
a pencil (actually a form of carbon
can be influenced by the body’s called graphite) conducts electric-
nervous system, so the rate at
which they trigger the heart to ity. The same must be true for the
beat can vary dramatically— from
a calm 60 beats per minute when
liquid in the beaker—in this case, a
asleep to more than 100 beats per solution of water and ordinary
minute during physical exertion.
table salt.
Electric circuits are of many
kinds and can be designed for
624 18 BASIC ELECTRIC CIRCUITS
many specific purposes, from boiling water to lighting a Christmas tree to restart-
ing a heart. Circuits containing “liquid” conductors (as in the photo) have practical
applications in the laboratory and industry; for example, they can be used to syn-
thesize or purify chemical substances and to electroplate metals. (Electroplating
means to chemically attach metals to surfaces using electrical techniques, such as
in making silver plate.)
Building on the principles discussed in Chapters 15, 16, and 17, this chapter
emphasizes the analysis of electric circuits and their applications. Circuit analysis
most often deals with voltage, current, and power requirements. A circuit may be
analyzed theoretically before being assembled. The analysis might show that the
circuit would not function properly as designed or that there could be a safety
problem (such as overheating due to joule heat). To help in this chapter’s analysis,
circuit diagrams will be used to visualize and understand circuit functions. (A few
of these diagrams were included in Chapter 17.)
Our circuit analysis begins by looking at some of the ways that resistive ele-
ments, such as lightbulbs, can be connected.
18.1 Resistances in Series, Parallel, and Series–Parallel

Combinations
➥ Two resistors of differing resistance are in series and connected to a battery to form
a complete circuit.Which resistor has more current?
➥ Two resistors of differing resistance are in series and connected to a battery to form
a complete circuit.Which resistor has more voltage across it?
➥ Two resistors of differing resistance are in parallel and connected to a battery to
form a complete circuit.Which resistor has more electric power dissipation?
The resistance symbol can represent any type of circuit element, such as a
toaster. Here all elements will be considered as ohmic (of constant resistance)
unless otherwise stated. In addition, wire resistance will be neglected, unless
otherwise stated.
RESISTORS IN SERIES
In analyzing a circuit, because voltage represents energy per unit charge, to con-
serve energy, the sum of the voltages around a complete circuit loop is zero. Remember
that voltage means “change in electrical potential,” so voltage gains and losses are
represented by + and - signs, respectively. For the circuit in 䉴 Fig. 18.1a, by con-
servation of energy (per coulomb), the individual voltages (Vi , where i = 1, 2, and
3) across the resistors add to equal the voltage (V) across the battery terminals.
Each resistor in series must carry the same current (I) because charge can’t “pile
up” or “leak out” at any location in the circuit. The result of summing the voltage
gains and losses, is V - g Vi = 0. Since each resistor’s voltage is related to its
resistance by Vi = IRi , this can be substituted into the previous equation, and the
result is
V - g 1IRi2 = 0 or V = g 1IRi2 (18.1)
The elements in Fig. 18.1a are connected in series, or head to tail. When resistors
are in series, the current must be the same through all the resistors, as required by the
18.1 RESISTANCES IN SERIES, PARALLEL, AND SERIES–PARALLEL COMBINATIONS 625
V1
I
R1 V1 = IR1 I
V2
+ V– V R2 V2 = IR2 V Rs
V3
R3 V3 = IR3
Rs = R1 + R2 + R3
V = V 1 + V2 + V3
(a) (b)
䉱 F I G U R E 1 8 . 1 Resistors in series (a) When resistors (representing the resistances of

lightbulbs here) are in series, the current in each is the same. g Vi, the sum of the voltage
drops across the resistors, is equal to V, the battery voltage. (b) The equivalent resistance Rs
of the resistors in series is the sum of the resistances.
conservation of charge. If this were not true, then charge would build up or disap-
pear at particular locations in the circuit, which does not happen. 䉴 Figure 18.2 shows
a situation analogous to the electric circuit in Fig. 18.1: the flow of water (“current”)
over a smooth streambed (“wires”) punctuated by a series of rapids (“resistance”).
Labeling the common current in the resistors as I, Eq. 18.1 can
be written explicitly for three series resistors (such as in Fig.
18.1a):
V = V1 + V2 + V3 ∆Ug1
= IR1 + IR2 + IR3 = I1R1 + R2 + R32

Equivalent series resistance (Rs) is defined as the resistance ∆Ugtotal ∆Ug2
value of a single resistor that could replace the actual resistors
and yet maintain the same current. This means that V = IRs , or
∆Ug3
Rs = V>I. Hence, three resistors in series have an equivalent
resistance given by
V ∆Ugtotal = ∆Ug1 + ∆Ug2 + ∆Ug3
Rs = = R1 + R2 + R3
I
䉱 F I G U R E 1 8 . 2 Water flow anal-
That is, the equivalent resistance of three resistors in series is the sum of the three ogy to resistors in series Even
individual resistances. Thus the three resistors in Fig. 18.1a could be replaced by a though, in general, a different
single resistor of resistance Rs (Fig. 18.1b) without affecting the current. For exam- amount of gravitational potential
ple, if each resistor in Fig. 18.1a had a value of 10 Æ , then Rs would be 30 Æ . energy (per kilogram) is lost as the
water flows down each set of
This result for three resistors can be extended to any number of resistors in series: rapids, the current of water is the
(equivalent series same everywhere. The total loss of
Rs = R1 + R2 + R3 + Á = g Ri (18.2) gravitational potential energy (per
resistance) kilogram) is the sum of the losses.
(To make this a “complete” water
Note that the equivalent series resistance is larger than the resistance of the largest resistor circuit, some external agent, such as
in the series. a pump, would need to continu-
Series connections are not common in some circuits, such as house wiring, ously do work on the water by
because there are two major disadvantages. The first is clear considering what hap- returning it to the top of the hill,
pens if one of the bulbs in Fig. 18.1a burns out (or is simply turned off). In this case, restoring its original gravitational
potential energy.)
all the bulbs would go out, because the circuit would no longer be complete, or con-
tinuous. In this situation, the circuit is said to be open. An open circuit is said to have
an infinite equivalent resistance, because the current in it is zero, even though the
battery voltage is not.
䉴 F I G U R E 1 8 . 3 Resistors in paral- V = V 1 = V2 = V3
lel (a) When resistors are connected I
in parallel, the voltage drop across
each resistor is the same. The cur- + V – I1 I2 I3
1 2 3
rent from the battery divides (gener-
ally unequally) among the resistors.
(b) The equivalent resistance, Rp , of I
resistors in parallel is given by a rec-
iprocal relationship.
V Rp
I
I = I 1 + I2 + I3
I1 I3
I2
V V = V1 = V2 = V3
R1 R2 R3 1 1 1 1
= + +
Rp R1 R2 R3
(a) (b)
The second disadvantage of series connections is that each resistor operates at

less than the battery voltage (V). Consider what would happen in Fig. 18.1a if a
fourth resistor were added. The voltage across each of the original bulbs (and the
current) would decrease, resulting in reduced power delivered to all bulbs. That
is, the bulbs would not glow at their rated light output. Clearly, this situation
would not be acceptable in a household setting. Compare these disadvantages to
the parallel circuits that follow.
RESISTORS IN PARALLEL
Resistors can also be connected in parallel (䉱 Fig. 18.3a). In this case, all the resis-
tors have common connections—that is, all the leads on one side of the resistors
are attached together and then to one terminal of the battery. The remaining leads
are attached together and then to the other terminal. When resistors are connected in
parallel to a source of emf, the voltage drop across each resistor must be the same. It may
not surprise you to learn that household circuits are wired in parallel. (See Section
䉲 F I G U R E 1 8 . 4 Analogies for
resistors in parallel (a) When a road 18.5.) When wired in parallel, each appliance operates at full voltage, and turning
forks, the total number of cars enter- one appliance off or on does not affect the others.
ing the two branches each minute is Unlike resistors in series, the current in a parallel circuit divides into the different
equal to the number of cars arriving paths (Fig. 18.3a). This occurs at any junction (a location where several wires come
at the fork each minute. Movement
together), much as traffic divides or merges together when it reaches a junction in the
of charge into and then out of a
junction can be considered in the road (䉲 Fig. 18.4a). Thus by charge conservation, the total current out of the battery
same way. (b) When water flows must be the same as the sum of the separate currents. Specifically, for three resistors
from a dam, the gravitational poten- in parallel, I = I1 + I2 + I3. Notice that if the resistances are equal, the current will
tial energy lost (per kilogram of divide so that each resistor has the same current. However, in general, the resistances
water) in falling to the stream below
are not equal and the current will divide among the resistors in inverse proportion to
is the same regardless of the path.
This is analogous to voltages across their resistances. This means that the largest current will take the path of least resis-
parallel resistors. tance. Remember, however, that no one resistor carries the total current.
∆Ug
(a) (b)
The equivalent parallel resistance (Rp) is the value of a single resistor that
could replace all the resistors and maintain the same current. Thus, Rp = V>I, or
I = V>Rp. In addition, the voltage drop (V) must be the same across each resistor.
To visualize this, imagine two separate water paths, each leading from the top of a
dam to the bottom. The water loses the same amount of gravitational potential
energy per gallon (analogous to V) regardless of the path (Fig. 18.4b). For electric-
ity, a given amount of charge loses the same amount of electrical potential energy,
regardless of which parallel resistor it passes through.
The current in each resistor is thus given by Ii = V>Ri. (The subscript i repre-
sents any of the resistors: 1, 2, or 3.) Substituting for each current,
V V V
I = I1 + I2 + I3 = + +
R1 R2 R3
Therefore,
V 1 1 1 1
= V¢ ≤ = V¢ + + ≤
Rp Rp R1 R2 R3
By equating the expressions in the parentheses, it can be seen that Rp is related to
the individual resistances by a reciprocal equation
1 1 1 1
= + +
Rp R1 R2 R3
This result is actually true for any number of resistors in parallel:
1 1 1 1 1 (equivalent parallel
= + + + Á = g¢ ≤ (18.3)
Rp R1 R2 R3 Ri resistance)
For only two resistors in parallel Eq. 18.3 can be solved for Rp:
R1R2 (two parallel

Rp = (18.3a)
R1 + R2 resistors)
Note that Eq. 18.3 gives 1>Rp, not Rp. At the end of the calculation, the reciprocal must be
taken to find Rp. Unit analysis will show that the units are not ohms until inverted.
As usual, carrying units along with calculations makes errors of this type less likely
to occur.
Note that the equivalent resistance of resistors in parallel is always less than the smallest
resistance in the arrangement. Two parallel resistors, for example, of resistances
6.0 Æ and 12.0 Æ , are equivalent to a single one with a resistance of 4.0 Æ (you
should show this). But why does this occur?
To see that this result makes physical sense, consider a 12-V battery in a circuit
with a single 6.0-Æ resistor. The current in the circuit is 2.0 A 1I = V>R2. Now
imagine connecting a 12.0-Æ resistor in parallel to the 6.0-Æ resistor. The current
through the 6.0-Æ resistor will be unaffected—it will remain at 2.0 A. (Why?)
However, the new resistor will have a current of 1.0 A (using I = V>R again). Thus
the total current in the circuit is 1.0 A + 2.0 A = 3.0 A. Now look at the overall
result. When the second resistor is attached in parallel, the total current delivered
by the battery increases. Since the voltage did not increase, the equivalent resis-
tance of the circuit must have decreased (below its initial value of 6.0 Æ ) when the
12-Æ resistor was attached. In other words, every time an extra parallel path is
added, the result is more total current at the same voltage. Thus when adding more
resistors in parallel, the circuit’s equivalent resistance is always decreased. (Con-
versely, removing parallel resistors increases the equivalent parallel resistance.)
Notice that this argument does not depend on the value of the added resistor. All
that matters is that another path with some resistance is added. (Try this using a 2-Æ
or a 2-MÆ resistor in place of the 12-Æ resistor. A decrease in equivalent resistance
always happens. However, the value of the equivalent resistance will be different.)
To see how these parallel and series connection calculations work, consider
Example 18.1.
EXAMPLE 18.1 Connections Count: Resistors in Series and in Parallel

What is the equivalent resistance of three resistors (1.0 Æ , (c) From the equivalent series resistance and the battery volt-
2.0 Æ , and 3.0 Æ ) when connected (a) in series (Fig. 18.1a) age the total current is:
and (b) in parallel (Fig. 18.3a)? (c) How much total current
V 12 V
will be delivered by a 12-V battery in each of these arrange- I = = = 2.0 A
ments? (d) How much current will be in each resistor and Rs 6.0 Æ
what is the voltage drop across each resistor in each of these For the parallel arrangement, the total current is:
arrangements?
V 12 V
THINKING IT THROUGH. To find the equivalent resistances for I = = = 22 A
parts (a) and (b), apply Eqs. 18.2 and 18.3, respectively. For the Rp 0.55 Æ
series arrangement in part (c), the current supplied by the battery Note that the current for the parallel combination is much
is the total current and can be determined by treating the battery larger than that for the series combination. (Why?)
as if it were connected to a single resistor—the series equivalent
(d) The voltage drop across each resistor in series can be cal-
resistance. For the parallel arrangement, the total current can be
culated; note that the current in each is already known since it
determined by doing the same with the parallel equivalent resis-
is the same as the total current (2.0 A):
tance. (d) When in series, the total current is the same as the cur-
rent in all the resistors. From the current and resistance values, V1 = IR1 = 12.0 A211.0 Æ2 = 2.0 V
V2 = IR2 = 12.0 A212.0 Æ2 = 4.0 V
voltage drops can be calculated. When in parallel, each resistor
has the has the same voltage across it; hence the individual cur-
rents can be found by using the resistance values. V3 = IR3 = 12.0 A213.0 Æ2 = 6.0 V
SOLUTION. Listing the data: Notice that to ensure that the current in each series resistor is
the same, it must be that in series, the larger resistors require
Given: R1 = 1.0 Æ Find: (a) Rs (series resistance)
more voltage. As a check, note that the sum of the resistor volt-
R2 = 2.0 Æ (b) Rp (parallel resistance) age drops 1V1 + V2 + V32 equals the battery voltage.
R3 = 3.0 Æ (c) I (total current for each case) Now the current through each resistor in parallel can be
V = 12 V (d) current in and voltage across determined, because each has a voltage of 12 V across it.
each resistor (for each case) Therefore,
(a) The equivalent series resistance is
V 12 V
Rs = R1 + R2 + R3 = 1.0 Æ + 2.0 Æ + 3.0 Æ = 6.0 Æ I1 = = = 12 A
R1 1.0 Æ
The result is larger than the largest resistance, as expected. V 12 V
I2 = = = 6.0 A
(b) The equivalent parallel resistance is found as follows: R2 2.0 Æ
1 1 1 1 1 1 1 V 12 V
= + + = + + I3 = = = 4.0 A
Rp R1 R2 R3 1.0 Æ 2.0 Æ 3.0 Æ R3 3.0 Æ
6.0 3.0 2.0 11 As a check, note that the sum of the currents is equal to the
= + + =
6.0 Æ 6.0 Æ 6.0 Æ 6.0 Æ current through the battery.
and finally inverting, As can be seen, for resistors in parallel, the resistor with the
smallest resistance gets most of the total current because resis-
6.0 Æ
Rp = = 0.55 Æ tors in parallel experience the same voltage. (Note that for
11 parallel arrangements the least resistance never has all the
which is less than the least resistance, also as expected. current, just the largest.)
F O L L O W - U P E X E R C I S E . (a) Calculate the power delivered to each resistor for both arrangements in this Example. (b) What gen-
eralizations can you make? For instance, which resistor gets the most power in series? In parallel? (c) For each arrangement, ver-
ify the total power delivered to all the resistors is equal the power output of the battery. (Answers to all Follow-Up Exercises are
As a practical wiring application, consider strings of Christmas tree lights. In the

past, such strings had large bulbs connected in series. When one bulb burned out, all
the others in the string went out, leaving you to hunt for the faulty bulb. Now, in
newer strings that have smaller bulbs, one or more bulbs may burn out, but the
others remain lit. Does this mean that the bulbs are now wired in parallel? No; parallel
wiring would give a small resistance and large current, which could be dangerous.
Instead, an insulated jumper, or “shunt,” is wired in parallel with each bulb’s Filament
filament (䉴 Fig. 18.5). In normal operation, the shunt is insulated from the filament
Shunt
wires and does not carry current. When the filament breaks as the bulb “burns
out,” there is momentarily an open circuit, and for a short instance no current in the Glass bead
string. Thus, the voltage across the open circuit at the broken filament will be the
full 120-V household voltage. This voltage causes sparking that burns off
the shunt’s insulating material. Now the shunt is in electrical contact with the
other filament wires, again completing the circuit, and the rest of the lights in the
string continue to glow. (The shunt, a wire with little resistance, is indicated by the
small resistance symbol in the circuit diagram of Fig. 18.5. Under normal opera-
tion, there is a gap—the insulation—between the shunt and the filament wire.) To
understand what happens to the remaining bulbs in a string with a burnt-out
bulb, consider Conceptual Example 18.2.
CONCEPTUAL EXAMPLE 18.2 Oh, Tannenbaum! Christmas Tree

Lights Burning Brightly Shunt
Consider a string of Christmas tree lights composed of bulbs with jumper shunts, as in Filament
Fig. 18.5b. If the filament of one bulb burns out and the shunt completes the circuit, will
the other bulbs each (a) glow a little more brightly, (b) glow a little more dimly, or (c) be
unaffected?
REASONING AND ANSWER. If one bulb filament burns out and its shunt completes the
circuit, there will be less total resistance in the circuit, because the shunt’s resistance is
much less than the filament’s resistance. (Note that the filaments of the good bulbs and
the shunt of the burnt-out bulb are in series, so their resistances add.)
With less total resistance, there will be more current in the circuit, and the remaining 䉱 F I G U R E 1 8 . 5 Shunt-wired
good bulbs will glow a little brighter because the light output of a bulb is directly related to Christmas tree lights A shunt, or
the power delivered to that bulb. (Recall that electrical power is related to the current by “jumper,” in parallel with the bulb
filament reestablishes a complete
P = I 2R.) So the answer is (a). For example, suppose the string initially has eighteen iden-
circuit when one of the filaments
tical bulbs. Because the total voltage across the string is 120 V, the voltage drop across any
bulb is 1120 V2>18 = 6.7 V. If one bulb is out (and shunted), the voltage across each of the
burns out (lower right bulb). With-
out the shunt, if one were to burn
remaining lighted bulbs becomes 1120 V2>17 = 7.1 V. This increased voltage causes the out, all the bulbs would go out.
current to increase. Both increases contribute to more power delivered to each bulb, and
brighter lights (recall the alternative expression for electric power, P = IV).
F O L L O W - U P E X E R C I S E . In this Example, using a brand new string of bulbs, if one bulb
was removed, what would be the voltage across (a) its empty socket and (b) any of the
remaining bulbs? Explain.
SERIES–PARALLEL RESISTOR COMBINATIONS

Resistors may be connected in a circuit in a variety of series–parallel combina-
tions. As shown in 䉲 Fig. 18.6, circuits with only one voltage source can sometimes
be reduced to a single equivalent loop, containing just the voltage source and one
equivalent resistance, by applying the series and parallel results.
A procedure for analyzing such combination circuits (that is, for determining
voltage, current, and power for each circuit element) is as follows:
1. Determine which groups of resistors are in series and which are in parallel,
and reduce all groups to equivalent resistances, using Eqs. 18.2 and 18.3.
2. Reduce the circuit further by treating the separate equivalent resistances
(from Step 1) as individual resistors. Proceed until you get to a single loop
with one total (overall or equivalent) resistance value.
3. Find the current delivered to the reduced circuit using I = V>Rtotal.
4. Expand the reduced circuit back to the actual circuit by reversing the reduc-
tion steps, one at a time. Use the current of the reduced circuit to find the cur-
rents and voltages for the resistors in each step.
To see this procedure in use, consider Example 18.3.
R1 = R1 Rp = R1 Rs =
1 1
I 6.00 Ω R4 = R3R4 Rp + R5 =
R3 =
1
6.00 Ω 2.00 Ω R3 + R4 4.00 Ω
V= R2 = = 1.50 Ω
24.0 V V V R2 =
4.00 Ω R2
4.00 Ω
R5 = R5 =
2.50 Ω 2.50 Ω
(a) (b) (c)
R1 =
6.00 Ω
R2Rs
1
V Rp = V 䉱 F I G U R E 1 8 . 6 Series–parallel
2 R2 + Rs Rtotal = R1 + Rp
2
= 2.00 Ω
1
= 8.00 Ω combinations and circuit reduction
The process of reducing series com-
binations and parallel combinations
to equivalent resistances reduces the
circuit with one voltage source to a
single loop with a single equivalent
(d) (e) resistance. (See Example 18.3.)
EXAMPLE 18.3 Series–Parallel Combination of Resistors: Same Voltage or Same Current?

What are the voltages across and the currents in each of the Then, R2 and Rs1 are in parallel and can be reduced (again
resistors R1 through R5 in Fig. 18.6a? using Eq. 18.3) to Rp2 (Fig. 18.6d):
1 1 1 1 1 2
THINKING IT THROUGH. Applying the steps described previ- = + = + =
ously, it is important to identify parallel and series combina- Rp2 R2 Rs1 4.00 Æ 4.00 Æ 4.00 Æ
tions before starting. It should be clear that R3 is in parallel Thus Rp2 is given by
with R4 (written R3 ‘ R4). This parallel combination is itself in
series with R5. Furthermore, the 1R3 ‘ R42 + R5 leg is in paral- Rp2 = 2.00 Æ
lel with R2. Lastly, this parallel combination is in series with This operation leaves two resistances (R1 and Rp2) in series.
R1. Combining the resistors step by step should enable the Hence the total equivalent resistance (Rtotal) of the circuit is
determination of the total equivalent circuit resistance (Fig. 18.6e):
(Step 2). From that value, the total current can be calculated.
Then, working backward, the current in, and voltage across, Rtotal = R1 + Rp2 = 6.00 Æ + 2.00 Æ = 8.00 Æ
each resistor can be found. Therefore, the battery delivers a current of
SOLUTION. To avoid rounding errors, the results will be car- V 24.0 V
ried to three significant figures. I = = = 3.00 A
Rtotal 8.00 Æ
Given: Values in Fig. 18.6a Find: Current and voltage Now let’s work backward and “rebuild” the actual circuit.
for each resistor Note that the battery current is the same as the current
(Fig. 18.6a) through R1 and Rp2 , because they are in series. (In Fig. 18.6d,
The parallel combination at the right-hand side of the circuit I = I1 = 3.00 A and I = Ip2 = 3.00 A.) Therefore, the voltages
diagram can be reduced to the equivalent resistance Rp1 (see across these resistors are
Fig. 18.6b), using Eq. 18.3: V1 = I1 R1 = 13.00 A216.00 Æ2 = 18.0 V
1 1 1 1 1 4 and
= + = + =
Vp2 = Ip2 Rp2 = 13.00 A212.00 Æ2 = 6.00 V
Rp1 R3 R4 6.00 Æ 2.00 Æ 6.00 Æ
Thus Rp1 is
Because Rp2 is made up of R2 and Rs1 (Fig. 18.6c and d), there
Rp1 = 1.50 Æ must be a 6.00-V drop across both of these resistors. So the
This reduction leaves a series combination of Rp1 and R5 along current in each can be determined as follows:
that side, which is reduced to Rs1, using Eq. 18.2 (Fig. 18.6c): V2 6.00 V
I2 = = = 1.50 A
Rs1 = R p1 + R5 = 1.50 Æ + 2.50 Æ = 4.00 Æ R2 4.00 Æ
18.2 MULTILOOP CIRCUITS AND KIRCHHOFF’S RULES 631
and With these voltages and known resistances, the last two cur-
Vs1 6.00 V rents, I3 and I4 , are
Is1 = = = 1.50 A
Rs1 4.00 Æ V3 2.25 V
I3 = = = 0.375 A
Next, notice that Is1 is also the current in Rp1 and R5 , because R3 6.00 Æ
they are in series. (In Fig. 18.6b, Is1 = Ip1 = I5 = 1.50 A.) and
The resistors’ individual voltages are therefore
Vp1 = Is1 Rp1 = 11.50 A211.50 Æ2 = 2.25 V
V4 2.25 V
I4 = = = 1.13 A
R4 2.00 Æ
and
V5 = Is1 R5 = 11.50 A212.50 Æ2 = 3.75 V The current 1Is12 is expected to divide at the R3-R4 junction.
Thus, a double-check is available: I3 + I4 does, in fact, equal
(As a check, note that the voltages do, in fact, add to 6.00 V.) Is1 , within rounding errors.
Finally, the voltage across R3 and R4 is the same as Vp1
(why?), and
Vp1 = V3 = V4 = 2.25 V
F O L L O W - U P E X E R C I S E . In this Example, verify that the total power delivered to all of the resistors is the same as the power out-
put of the battery. Why must this be true?
DID YOU LEARN?

➥ In series, all circuit elements carry the same current.
➥ For resistors in series, the resistor with the most resistance has the largest voltage
across it.
➥ For resistors in parallel, the resistor with the least resistance dissipates the most R2
power.
R1 V2
R4 R5
R3
18.2 Multiloop Circuits and Kirchhoff ’s Rules V1
➥ Kirchhoff’s junction rule amounts to applying conservation of what quantity?

➥ Kirchhoff’s loop rule amounts to applying conservation of what quantity? (a)
➥ If a a resistor is traversed in the direction opposite to its current direction, what is the
sign of the voltage across it? A
I1 I3 I2
Rs
Series–parallel circuits with a single voltage source can be reduced to a single loop, as
seen in Example 18.3. However, circuits may contain several loops, each one having I1 V2
several voltage sources, resistances, or both. In many cases, resistors may not be in
V1 Rp
series or in parallel. As an example of this situation, a multiloop circuit, which does
R3
not lend itself to the methods of Section 18.1, is shown in 䉴 Fig. 18.7a. Even though
some groups of resistors may be replaced by their equivalent resistances (Fig. 18.7b), I3
I1 I2
this circuit can be reduced only so far by using parallel and series procedures.
Analyzing these types of circuits requires a more general approach—that is, the B
application of Kirchhoff’s rules.* These rules embody conservation of charge and (b)
energy. (Although not stated specifically, Kirchhoff’s rules were applied to the 䉱 F I G U R E 1 8 . 7 Multiloop circuit
parallel and series arrangements in Section 18.1.) First, it is useful to introduce In general, a circuit that contains volt-
some terminology that will help us describe more complex circuits: age sources in more than one loop can-
not be completely reduced by series
■ A point where three or more wires are joined is called a junction or node—for and parallel methods alone. However,
example, point A in Fig. 18.7b. some reductions within each loop may
be possible, such as from part (a) to
■ A path connecting two junctions is called a branch. A branch may contain one part (b). At a junction the current
or more circuit elements, and there may be more than two branches between divides or comes together, as at junc-
two junctions. tions A and B in part (b), respectively.
Any path between two junctions is
called a branch. In part (b), there are
three branches—that is, there are three
*Gustav Robert Kirchhoff (1824–1887) was a German scientist who made important contributions different ways to get from junction A
to electrical circuit theory and light spectroscopy. to junction B.
KIRCHHOFF’S JUNCTION THEOREM

Kirchhoff’s first rule, or junction theorem, states that the algebraic sum of the
currents at any junction is zero:
(sum of currents
g Ii = 0 (18.4)
at a junction)
This means that the sum of the currents entering a junction (taken as positive) and
the currents leaving the junction (taken as negative) is zero. This rule is just a state-
ment of charge conservation—no charge can pile up at a junction (why?). For the
junction at point A in Fig. 18.7b, for example, using the sign conventions the alge-
braic sum of the currents is
I1 - I2 - I3 = 0
or equivalently
I1 = I2 + I3
current in = current out
(This rule was applied in analyzing parallel resistances in Section 18.1.)
Sometimes it is not evident whether a particular current is directed into or out of a junc-
tion just from looking at a circuit diagram. In this case, a direction is simply assumed.
Then the currents are calculated, without worry about their directions. If some of the
assumed directions turn out to be opposite to the actual directions, then negative
answers for these currents will result. This outcome means that the directions of these
V>0 currents are opposite to the directions initially chosen (or guessed).
– +
KIRCHHOFF’S LOOP THEOREM
V<0 Kirchhoff’s second rule, or loop theorem, states that the algebraic sum of the
Across battery potential differences (voltages) across all of the elements of any closed loop is zero:
(sum of voltages
(a) g Vi = 0 (18.5)
around a closed loop)
V<0 This expression means that the sum of the voltage rises (an increase in potential)
equals the sum of the voltage drops (a decrease in potential) around a closed loop,
R which must be true if energy is conserved. (This rule was used in analyzing series
I
resistances in Section 18.1.)
Notice that traversing a circuit loop in different directions will yield either a
V>0
voltage rise or a voltage drop across each circuit element. Thus, it is important to
Across resistor establish a sign convention for voltages. The sign conventions used in this book
are illustrated in 䉳 Fig. 18.8. The voltage across a battery is taken as positive (a
(b)
voltage rise) if it is traversed from the negative terminal to the positive (Fig.
䉱 F I G U R E 1 8 . 8 Sign convention 18.8a) and negative if it is traversed from positive to negative. (Note that the
for Kirchhoff’s loop theorem (a) The direction of the current through the battery has nothing to do with the sign of the
battery voltage is taken as positive if battery voltage. The sign of this voltage depends only on the direction chosen to
it is traversed from the negative to cross the battery.)
the positive terminal. It is assigned a The voltage across a resistor is taken as negative (a decrease) if the resistor is
negative value if traversed from the
positive to the negative. (b) The traversed in the same direction as the assigned current, in essence going “down-
voltage across a resistor is taken as hill” potential-wise (Fig. 18.8b). The voltage will be positive (an increase in poten-
negative if the resistor is traversed tial) if the resistor is traversed in the direction opposite to the current. Used
in the direction of the assigned cur- together, these sign conventions allow the summation of the voltages around a
rent (“downstream”). It is taken as closed loop, regardless of the direction chosen to do that sum. It should be clear
positive if the resistance is traversed
in the direction opposite that of the that Eq. 18.5 is the same in either case. To see this, note that reversing the chosen
assigned branch current loop direction simply amounts to multiplying Eq. 18.5 (from the original direc-
(“upstream”). tion) by -1. This operation, of course, does not change the equation or the physics.
In applying Kirchhoff’s loop theorem, the sign of a voltage across a resistor is deter-
mined by the direction of the current in that resistor. However, there can be situations
in which the current direction is not obvious. How do you handle the voltage signs in
such cases? The answer is simple: After assuming a direction for the current, follow
the voltage sign convention based on this assumed direction. This guarantees that the
signs of all the voltage drops are mathematically consistent. Thus, if it turns out that
the actual current direction is opposite your choice, the voltage drops will automati-
cally reflect that.
A graphical interpretation of Kirchhoff’s loop theorem is presented in Learn by

Drawing 18.1, Kirchhoff Plots: A Graphical Inerpretation of Kirchhoff’s Loop Theo-
rem. Integrated Example 18.4, in which a simple parallel circuit is reexamined using
Kirchhoff’s rules, shows that our previous series–parallel considerations were con- I
sistent with these circuit rules. In this Example, take care to notice how important it J
is draw a correct circuit diagram—it can guide you as to how to proceed. I2 I1
+
V R2 R1
INTEGRATED EXAMPLE 18.4 A Simple Circuit: Overkill for –
Kirchhoff’s Rules?
R3
Two resistors R1 and R2 are connected in parallel. This combination is in series with a
third resistor R3, which has the largest resistance of the three. A battery completes the
circuit, with one electrode connected to the beginning and the other to the end of this (a)
network. (a) Which resistor will carry the most current: (1) R1, (2) R2, or (3) R3? Explain.
(b) In the actual circuit, assume R1 = 6.0 Æ , R2 = 3.0 Æ , R3 = 10.0 Æ , and the bat- R1 =
6.0 Ω
tery’s terminal voltage is 12.0 V. Apply Kirchhoff’s rules to determine the current in
each resistor and the voltage across each resistor.
(A) CONCEPTUAL REASONING. It is best to first look at a schematic circuit diagram based
on the word description of the network (䉴 Fig. 18.9). It might appear that the resistor with V=
the least resistance would carry the most current. But be careful; this holds only if all the 12 V
resistors are in parallel. This is not true here. The two parallel resistors each carry only a
portion of the total current. However, because the total of their two currents is in R3, that
resistor carries the total, and therefore the most, current. Thus, the correct answer is (3). R3 =
(B) QUANTITATIVE REASONING AND SOLUTION. 10.0 Ω
Given: R1 = 6.0 Æ Find: I1, I2, and I3 (current in each resistor) and
R2 = 3.0 Æ V1, V2, and V3 (voltage across each resistor) (b)
R3 = 10.0 Æ
V = 12.0 V
There are three unknown currents: the total current (I) and the currents in each of the
R2 =
parallel resistors (I1 and I2). Since there is only one battery, the current must be clock- 3.0 Ω
wise (shown in the figure). Applying Kirchhoff’s junction theorem to the first junction V=
(J in Fig. 18.9a) 12 V
g Ii = 0 or I - I1 - I2 = 0 (1)
Using the loop theorem in the clockwise direction in Fig. 18.9b, the battery is crossed
R3 =
from the negative to the positive terminal and then R1 and R3 are traversed to complete 10.0 Ω
the loop. The resulting equation (showing the voltage signs explicitly) is
g Vi = 0 or + V + 1- I1R12 + 1 -IR32 = 0 (2)
(c)
A third equation can be obtained by applying the loop theorem but this time going
through R2 instead of R1 (Fig. 18.9c). This yields 䉱 F I G U R E 1 8 . 9 Sketching circuit
g Vi = 0 or + V + 1 -I2 R22 + 1 -IR32 = 0 (3) diagrams and Kirchhoff’s rules
(a) The circuit diagram resulting
Putting in the battery voltage (in volts) and resistances (in ohms) and rearranging gives from the written description in Inte-
three equations and three unknowns (the currents): grated Example 18.4. (b) and (c) The
(continued on next page) two loops used in the analysis of
Integrated Example 18.4.
I = I1 + I2 (1a)
12 - 6I1 - 10I = 0 or 6 - 3I1 - 5I = 0 (2a)
12 - 3I2 - 10I = 0 (3a)

Adding Eqs. (2a) and (3a) yields 18 - 31I1 + I22 - 15I = 0. However, from Eq. (1a),
I = I1 + I2. Therefore, this equation becomes
18 - 3I - 15I = 0 or 18I = 18
and solving for the total current yields I = 1.00 A.
Eqs. (3a) and (1a) can then be solved for the remaining currents:
2 1
I2 = 3 A and I1 = 3 A
These answers are consistent with our circuit diagram reasoning in part (a).
Because the currents are now known, the voltages can be obtained from Ohm’s law,
V = IR. Thus,
V1 = I1R1 = A 13 A B 16.0 Æ2 = 2.0 V
V2 = I2R2 = A 23 A B 13.0 Æ2 = 2.0 V
V3 = I3R3 = 11.0 A2110.0 Æ2 = 10.0 V
Take a quick look at the results to see if they are reasonable. As expected, the voltage
drops across the parallel resistors are equal. Because of that, two-thirds of the total cur-
rent is in the resistor with the least resistance. Also, the total voltage across the network
is 12.0 V, as it must be.
F O L L O W - U P E X E R C I S E . (a) In this Example, predict what will happen to each of the
R2 = 9.0 Ω currents if R2 is increased. Explain your reasoning. (b) Rework part (b) of this Example,
I2 changing R2 to 8.0 Æ , and see if your reasoning is correct.
I2 Note that answers in Example 18.4 were in amperes and volts because amps, volts, and
Loop 2
ohms were used consistently throughout. By staying within this SI system (that is,
I2 expressing electrical quantities in volts, amps and ohms), carrying units is not necessary;
I3 the answers will automatically be in these units. (Of course, it is always a good idea to
check units.)
V2 = R3 =
I1 2.0 Ω
12 V
I1 APPLICATION OF KIRCHHOFF’S RULES
R1 = Loop 1
6.0 Ω As indicated by its whimsical title, Integrated Example 18.4 could have been
worked as a relatively simple series–parallel combination. However, more compli-
cated, multiloop circuits (where resistors are neither in parallel nor in series) require
a more structured approach using basic principles. In this book, the following gen-
V1 = 6.0 V
eral steps will be used when applying these principles (Kirchhoff’s rules):
1. Assign a current and direction of current for each branch in the circuit. This
Loop 3 assignment is done most conveniently at junctions.
䉱 F I G U R E 1 8 . 1 0 Application of 2. Indicate the loops and the directions in which they are to be traversed
Kirchhoff’s rules To analyze a circuit (䉳 Fig. 18.10). Every branch must be in at least one loop.
such as the one shown for Example
3. Apply Kirchhoff’s first rule (junction rule) at each junction that gives a
18.5, assign a current and its direc-
tion for each branch in the circuit unique equation. (This step gives a set of equations that includes all currents,
(most conveniently done at junc- but there may be redundant equations from two different junctions.)
tions). Identify each loop and the 4. Traverse the number of loops necessary to include all branches. In traversing
direction of traversal. Then write
current equations for each indepen- a loop, apply Kirchhoff’s second rule, the loop theorem (using V = IR for
dent junction (using Kirchhoff’s each resistor), and write the equations, using the proper sign conventions.
junction theorem). Also write volt-
age equations for as many loops as If this procedure is applied properly, Steps 3 and 4 give a set of N equations if there
needed to include every branch are N unknown currents. These equations may then be solved for the currents. If
(using Kirchhoff’s loop theorem). Be more loops are traversed than necessary, one (or more) redundant loop equation(s)
careful to observe sign conventions. might appear. Only the number of loops that includes each branch once is needed.
This procedure may seem complicated, but it’s generally straightforward, as

Integrated Example 18.5 shows.
INTEGRATED EXAMPLE 18.5 Branch Currents: Using Kirchhoff’s Rules and Energy Conservation
Consider the circuit diagrammed in Fig. 18.10. (a) What can you After substituting in the values and rearranging
say about the magnitude of the electric potential change between 9I2 - 2I3 = 12 (3a)
the two nodes if the path taken is through R2 compared to the
magnitude if instead the path went through R3 and battery #2: Equations (1), (2a), and (3a) form a set of three equations with
(1) the change is greater if the path goes through R2, (2) the three unknowns. The currents can be found in many ways.
change is greater if the path goes through the battery, (3) the One is to substitute Eq. (1) into Eq. (2a) and eliminate I1:
change is the same for both paths? Explain. (b) Find the current 31I2 + I32 + I3 = - 3
in each branch and determine the two changes in part (a).
(c) Determine the power dissipated in each resistor and compare After rearranging and dividing by 3, this simplifies to
the total to that rate at which energy is lost (or gained) by the two I2 = - 1 - 43 I3 (4)
batteries.
Then, substituting Eq. (4) into Eq. (3a) eliminates I2:
(A) CONCEPTUAL REASONING. The electric potential difference
between the nodes must be the same. If not, the nodes would 9 A -1 - 4
3 I3 B - 2I3 = 12
have different potential values, which cannot be the case. Hence
the correct answer is (3), the change is the same for both paths. Finishing the algebra and solving for I3,
(B) AND (C) QUANTITATIVE REASONING AND SOLUTION. - 14I3 = 21 or I3 = - 1.5 A
The solution is begun by assigning current directions (“best
guesses”) in each loop, and then using the junction theorem The minus sign means the wrong direction was assumed for I3.
and the loop theorem (twice—once for each inner loop) to Putting the value of I3 into Eq. (4) gives I2:
generate three equations, because there are three currents. I2 = - 1- 43 1- 1.5 A2 = 1.0 A
From the currents and resistance/battery values, potential
changes can be determined. Lastly, these quantities can be Then, from Eq. (1),
used to find power levels in each of the circuit elements. I1 = I2 + I3 = 1.0 A - 1.5 A = - 0.5 A
Given: Values in Find: (b) I1, I2, and I3 (current in each of Again, the sign means the direction of I1 was wrong.
Fig. 18.10 the three branches) and The electric potential change, from the left to the right
V (voltage drop between the node through battery #2, is determined by the sign conven-
two nodes via two different tions. The current through R3 is actually to the left, so its
paths) potential change is positive; however, the battery is traversed
(c) P1, P2, and P3 (power dissi- from anode to cathode, and this change is negative:
pated in each resistor) and V = - 12 V + ƒ I3 ƒ R3 = - 12 V + ƒ - 1.5 A ƒ 12.0 Æ2 = - 9.0 V
Pb1 and Pb2 (rate of energy lost
The electric potential change, going from the left node to the
or gained by each battery)
right node through R2, is negative, because the path is taken
(b) The chosen current directions and loop traversal direc- in the direction of the current. Its value is
V = - I2R2 = - 11.0 A219.0 Æ2 = - 9.0 V
tions are shown in the figure. (Remember, these directions are
not unique; choose them, and check the final current signs to see
if they were correct.) There is a current in every branch, and in agreement with the conceptual part (a).
every branch is in at least one loop. (Some branches are in more (c) The resistors all dissipate energy as follows:
than one loop, which is acceptable.) PR1 = I 21 R1 = 10.50 A2216.0 Æ2 = 1.5 W
Applying Kirchhoff’s first rule at the left-hand junction
PR2 = I 22 R2 = 11.0 A2219.0 Æ2 = 9.0 W
I1 - I2 - I3 = 0 or I1 = I2 + I3 (1)
Going around loop 1 as in Fig. 18.10 and applying Kirch- PR3 = I 23 R3 = 11.50 A2212.0 Æ2 = 4.5 W
hoff’s loop theorem with the sign conventions gives for a total dissipation rate of 15 W.
g Vi = + V1 + 1- I1 R12 + 1- V22 + 1 -I3 R32 = 0 (2) The batteries can gain or lose energy, depending upon the
Putting in the numerical values yields direction of their current. Battery #2 is losing energy because
its current leaves its anode. Thus
+ 6 - 6I1 - 12 - 2I3 = 0
Pb2 = I3 Vb2 = 11.5 A2112 V2 = 18 W (a loss)
Rearranging this equation and dividing both sides by 2
3I1 + I3 = - 3 Battery #1 increasing its stored energy (being “recharged”)
because its current enters the anode.
For convenience, units are omitted (they are all in amps and
ohms, and are self-consistent). Pb1 = I1 Vb1 = 1-0.5 A216.0 V2 = - 3.0 W (a gain)
For loop 2, the loop theorem yields
The end energy rate (power) result shows conservation of energy,
g Vi = + V2 + 1 - I2 R22 + 1 +I3 R32 = 0 (3) since the net battery rate equals the total resistor loss rate of 15 W.
FOLLOW-UP EXERCISE. Find the currents in this Example by using the junction theorem and loops 3 and 1 instead of loops 1 and 2.
LEARN BY DRAWING 18.1 For elaborate circuits, this graphical method may prove to
be too complicated for practical use. Nevertheless, it is
Kirchhoff plots: a graphical interpretation always good to keep this concept in mind, as it illustrates
the fundamental physics behind the loop theorem.
of Kirchhoff ’s loop theorem As an example of the power of this method, consider the cir-
cuit in Fig. 1: a battery with internal resistance r wired to a sin-
The equation form of Kirchhoff’s loop theorem has a geometri-
gle external resistor R. The direction of the current is from the
cal visualization that may help you develop better insight into
anode to the cathode through the external resistor. The poten-
its meaning. This graphical approach allows the visualization
tial of the battery’s cathode is chosen as zero. Starting there and
of how the potential changes in a circuit, either to anticipate the
traversing the circuit in the direction of the current, there is a
results of mathematical analysis or to qualitatively confirm the
rise in potential from the battery cathode to the anode. From
results. (Don’t forget that a complete analysis usually also
there the potential remains constant as the current travels
includes the junction theorem—see Example 18.5.)
through the wires to the external resistor. That is, no significant
The idea is to make a three-dimensional plot based on the
voltage drop should be indicated along connecting wires (why?).
circuit diagram. The wires and elements of the circuit form
At the resistor, there must be a drop in potential. How-
the basis for the x–y plane, or the diagram’s “floor.” Plotted
ever, it must not drop to zero, because there must be some
perpendicularly to this plane, along the z-axis, is the electric
potential difference left to produce current in the internal
potential (V), with an appropriate choice for zero. Such a
resistance. Thus it can be reasoned visually why the termi-
diagram is called a Kirchhoff plot (Fig. 1).
nal voltage of the battery, V, must be less than its emf (the
The rules for constructing a Kirchhoff plot are simple:
rise between a and b).
Start at a known potential value, and go around a complete
Figure 2 shows two resistors in series, and that combination
loop, finishing where you started. Because you come back
in parallel with a third resistor. For simplicity, all three resis-
to the same location, the sum of all the rises in potential
tors have the same resistance (R) and the battery’s internal
(positive voltages) must be balanced by the sum of the
resistance is assumed to be zero. Starting at point a, there is a
drops (negative voltages). This requirement is the geometri-
rise in potential corresponding to the battery voltage. Then, as
cal expression of energy conservation, embodied mathemat-
the loop is traced, it takes a route through the single resistor, so
ically by Kirchhoff’s loop theorem.
there must be one drop in potential equal in magnitude to e.
Thus, if the potential increases (say, in traversing a battery
Following the loop that includes the two resistors, each
from negative terminal to positive terminal), draw a rise in the
must have half the total drop (why?). So each will carry half
z-direction. In this instance, the rise represents the terminal
the current of the single resistor. Recall that in parallel cir-
voltage of the battery. Similarly, if the potential decreases (for
cuits, the largest resistance carries the least current. Notice
example, in traversing a resistor in the direction of the cur-
how nicely the geometrical approach helps develop intu-
rent), make sure that the potential drops. If possible, try to
ition and allows anticipation of quantitative results.
draw the rises and drops (the voltages) to scale.
As an exercise, try redrawing Fig. 2 if, instead, the series
resistors had resistance values of R and 2R. Which of these
two now has the largest voltage? How do the currents in the
Potential resistors compare to the previous situation? Lastly, analyze
the circuit mathematically, to see whether your expectations
are confirmed.
ε IR
Potential
V
ε I c
R d ε
b
f
ε a e
r e Ir I R
V = terminal voltage = ε – Ir < ε c
F I G U R E 1 Kirchhoff plots: A graphical problem-solving b R d R g
strategy The schematic of the circuit is laid out in the x–y
plane, and the electric potential is plotted perpendicularly ε a
along the z-axis. Usually, the zero of the potential is taken to
be the negative terminal of the battery. A direction for cur-
rent is assigned, and the value of the potential is plotted
around the circuit, following the rules for gains and losses. F I G U R E 2 Kirchhoff plot of a more complex circuit
This particular plot shows a rise in potential when the bat- Imagine how the plot would change if you were to vary the
tery is traversed from cathode to anode, followed by a drop values of the three resistors. Then analyze the circuit mathe-
in potential across the external resistor, and a smaller drop matically to see whether your plot allowed you to anticipate
in potential across the battery’s internal resistance. the voltages and currents.
18.3 RC CIRCUITS 637
DID YOU LEARN?

➥ Kirchhoff’s junction theorem is equivalent to conservation of charge.
➥ Kirchhoff’s loop theorem is equivalent to conservation of energy.
➥ Current in a resistor is in the direction of decreasing electric potential; going
opposite the direction of the current means an increase in electric potential.
I⫽0
18.3 RC Circuits
R
Q⫽0
➥ How does the charging time constant depend on the resistance in an RC circuit?
➥ How does the discharging time constant depend on the capacitance in an RC circuit? C
➥ After one time constant has elapsed for a discharging capacitor, what percentage of
the initial value is the capacitor’s voltage? Vo
S
Until now, only circuits that have constant currents have been considered. In some
direct-current (dc) circuits, the current can vary with time while maintaining a con- (a)
stant direction. Such is the case in RC circuits, which consist of resistors, capaci-
tors, and power supplies in various combinations.
R
CHARGING A CAPACITOR THROUGH A RESISTOR I +Q
++
The charging of an uncharged capacitor by a battery is depicted in 䉴 Fig. 18.11. C ––
After the switch is closed, even though there is a gap (the capacitor plates), charge
Vo –Q
must flow (that is, there must a current in the circuit) while the capacitor is charging.
The maximum charge (designated as Qo) that the capacitor can attain depends on S
its capacitance (C) and the battery voltage (Vo). To determine Qo and understand
(b)
how the circuit current and capacitor charge vary with time, consider the following.
At t = 0, there is no charge on the capacitor and thus no voltage across it. By Kirch-
hoff’s loop theorem, this means that the full battery voltage must appear across the
resistor, resulting in an initial (maximum) current given by Io = Vo>R. As charge on R
the capacitor increases, so must the voltage across its plates, thereby reducing the I0 +Qo
resistor’s voltage and current. Eventually, when the capacitor is charged to its maxi- ++ ++
C –– ––
mum, the current becomes zero. At this time, the resistor’s voltage is zero and the
capacitor’s voltage is Vo. Because of the relationship between the charge on a capaci- Vo –Qo
tor and its voltage [Chapter 16, Eq. (19)], the maximum capacitor charge is Qo = CVo. S
(This time sequence is depicted in Fig. 18.11.)
The resistance value is one of two factors that determines how quickly the capaci- (c)
tor is charged, because the larger its value, the greater the resistance to charge flow.
䉱 F I G U R E 1 8 . 1 1 Charging a
The capacitance is the other factor that influences the charging speed—it takes capacitor in a series RC circuit
longer to charge a larger capacitor. Analysis of this type of circuit requires mathe- (a) Initially there is no current and
matics beyond the scope of this book. However, it can be shown that as a capacitor is no charge on the capacitor.
charged, the voltage across it increases exponentially with time according to (b) When the switch is closed, there
is a current in the circuit until the
VC = Vo31 - e -t>1RC24
(charging capacitor capacitor is charged to its maximum
(18.6) value. The rate of charging depends
voltage in an RC circuit)
on the circuit’s time constant,
where e has an approximate value of 2.718. (Recall that the irrational number e is t1 =RC2. (c) For times much larger
the base of the system of natural logarithms.*) A graph of VC versus t is shown in than t, the current is very
䉲 Fig. 18.12a. As expected, VC starts at zero and approaches Vo, the capacitor’s close to zero, and the capacitor is
said to be fully charged.
maximum voltage, after a “long” time.
A graph of I versus t is shown in Fig. 18.12b. The current in the circuit varies
with time according to
I = Io e -t>1RC2 (18.7)
The current decreases exponentially with time and has its largest value initially, as
expected.
*For a review of exponential functions, see Appendix I.

VC According to Eq. 18.6, it would take an infinite time for the capacitor to become
fully charged. However, in practice, most capacitors become close to completely
Voltage
charged in relatively short times. It is therefore customary to use a special value to

express the “charging time.” This value, called the time constant (T), is
Vo
t = RC (time constant for RC circuits) (18.8)
0.63V o
(You should be able to show that RC has units of seconds.) After an elapsed time of
one time constant, that is, t = t = RC, the voltage across a charging capacitor has
t risen to 63% of the maximum possible. This can be seen by evaluating VC (Eq.
τ = RC 18.6), replacing t with t1=RC2.
Time
(a) VC = Vo11 - e -t>t2 = Vo11 - e -12
L Vo a1 - b = 0.63Vo
I 1
Current
2.718
Because Q r VC, the capacitor also has 63% of its maximum possible charge after
Io
one time constant. You should be able to show that after one time constant, the
current has dropped to 37% of its initial (maximum) value, Io.
After a time equal to two time constants has elapsed 1t = 2t = 2RC2, the
capacitor is charged to more than 86% of its maximum value; at t = 3t = 3RC, the
0.37I o
capacitor is charged to 95% of its maximum value; and so on. (Make sure you
t know how these results were obtained.) As a general rule of thumb, a capacitor is
τ = RC
considered to be “fully charged” after “several time constants” have elapsed.
Time
(b)
DISCHARGING A CAPACITOR THROUGH A RESISTOR
䉱 F I G U R E 1 8 . 1 2 Capacitor
charging in a series RC circuit (a) In a 䉲Figure 18.13a shows a capacitor being discharged through a resistor. In this case,
series RC circuit, as the capacitor the voltage across the capacitor decreases exponentially with time, as does the cur-
charges, the voltage across it rent. The expression for the decay of the capacitor’s voltage (from its initial maxi-
increases nonlinearly, reaching 63% mum voltage of Vo) is
of its maximum voltage (Vo) in one
time constant, t. (b) The current in (discharging capacitor
this circuit is initially a maximum VC = Vo e -t>1RC2 = Vo e -t>t (18.9)
voltage in an RC circuit)
1Io = Vo>R2 and decays exponen-
tially, falling to 37% of its initial After one time constant, the capacitor voltage is at 37% of its original value
value in one time constant, t. (Fig. 18.13b). The current in the circuit decays exponentially also, following
Eq 18.7. For example, the capacitor in a heart defibrillator will discharge its
stored energy (as a flow of charge or current) to the heart (resistance R) in a
S VC
I
Vo
++ ++
VC R
––––C
0.37Vo
t
τ = RC
Q = CVC Time
(a) (b)
䉱 F I G U R E 1 8 . 1 3 Capacitor discharging in a series RC circuit (a) The capacitor is initially

fully charged. When the switch is closed, current appears in the circuit as the capacitor begins
to discharge. (b) In this case, the voltage across the capacitor (and the current in the circuit)
decays exponentially with time, falling to 37% of its initial value in one time constant, t.
18.3 RC CIRCUITS 639
discharge time (several time constants) of about 0.1 s. RC circuits are also an
integral part of cardiac pacemakers, which charge a capacitor, transfer the
energy to the heart, and repeat this at a rate determined by the time constant.
For details on these interesting and important instruments, refer to Insight 18.1,
Applications of RC Circuits to Cardiac Medicine. As a practical application, con-
sider the use of RC circuits in cameras in Example 18.6.
INSIGHT 18.1 Applications of RC Circuits to Cardiac Medicine

The normal human heart beats between 60 and 70 times per If needed, a second charge can be applied quickly after of the
minute, with each beat delivering about 70 mL of blood—about first. To handle this, the capacitor should be able to recharge
a gallon per minute. Your heart is essentially a pump com- in about 5 s (Fig. 1c). Thus the charging time constant should
posed of specialized muscle cells. The cells are triggered to beat be on the order of 1 s. Thus a charging resistor (of resistance
when they receive electrical signals. These signals are sent by Rc) should have a resistance value of several hundred kilo-
pacemaker cells located in the upper chambers of the heart. hms, since Rc = t> C L 105 Æ = 100 kÆ .
During a heart attack or after an electrical shock, the heart In some forms of heart disease, the heart beats irregularly
may go into an unregulated beat pattern. If left untreated, this due to problems with the pacemaker cells. The heart can be
condition could be fatal in minutes. Fortunately, it is possible triggered to beat in correct (or sinous) rhythm by implanting a
to return the heart to its normal pattern by passing an electri- cardiac pacemaker. This unit is the size of a book of matches,
cal current through it. The instrument that does this is called a powered by a long-life battery, and inserted surgically near
cardiac defibrillator. The main component of a defibrillator is, the location of the pacemaker cells.
in essence, a capacitor charged to a high voltage.* Most pacemakers are controlled by a sophisticated trigger-
Several hundred joules of electric energy are needed to ing circuit that allows the pacemaker to send signals to the
restart a heart. The high-voltage and low-voltage plates of the heart only if needed (“on demand” pacemakers; see Figs. 2a
capacitor are attached to the patient’s skin by two “paddles” and 2b). The triggering circuit sends a signal to the pacemaker
placed just above the two sides of the heart (Fig. 1a and to “fire” only if the heart does not beat. If the heart is beating
Fig. 1b). When a switch is thrown, charge flows through the normally, the capacitor switch is in the “full charge” position,
heart, transferring the capacitor’s energy to the heart. waiting for the signal to fire (discharge).
The discharge through the heart is essentially that of an RC For our purposes, the pacemaker can be represented by an
circuit. Typically the capacitor has a capacitance in the 10 mF RC circuit. The capacitor (typically 10 mF) is kept charged by
range and is charged by voltages on the order of thousands of the battery, and must be ready to release its energy at a rapid
volts. The resistance of a heart (Rh) is typically about 1000 Æ , rate in the worst-case scenario of the pacemaker cells not
giving a discharging time constant RC on the order of 10 ms. operating at all. Typically, the resistance of the heart muscle is
Thus the capacitor is essentially fully discharged after 50 ms. about 100 Æ , meaning the pacemaker discharge time constant
is t L 1 ms. Thus it is effectively fully discharged in 5 ms.
Typically, the capacitor takes about 10 ms. Since it takes
*Because portable batteries aren’t capable of high voltages, the about 5 ms to discharge, it has about 9 ms to recharge, mean-
charging uses a phenomenon called electromagnetic induction, which ing a recharge time constant of about 2 ms. This requires a
will be studied in Chapter 20. recharge resistor Rc to be about 200 Æ (Fig. 2c).
I Rc
⫹⫹ ⫹⫹
High V
source C
⫺⫺ ⫺⫺
I
(a) (b) (c)
F I G U R E 1 Restart the heart! (a) Paddles are placed externally to either side of the heart, and energy from a charged capacitor
passes through it, hopefully triggering it into a normal beating pattern. (b) This shows the schematic diagram for correct defibril-
lator use. The discharge is that of an RC circuit. (c) Recharging the defibrillator’s capacitor, getting it ready to go again, through a
(charging) resistor Rc L 105 Æ .
Vc
Rc S
Rn
t (ms)
Vc C Rh 0 5 10 15
Discharge Recharge
through heart capacitor
(S to right) (S to left)
(a) (b) (c)
F I G U R E 2 Cardiac pacemaker (a) The typical pacemaker (shown as a capacitor in a box) is implanted on or near the heart surface,
with its leads attached to the heart muscle (resistance Rn). (The capacitor’s charging circuit is not shown.) Other leads (not shown)
receive signals from the heart to determine whether the pacemaker needs to “fire.” (b) The sensing circuit determines the position of
the capacitor’s “switch.” If the heart is not beating, the sensing circuit flips the switch to the right, initiating energy discharge through
the heart muscle. If the heart is beating properly, the sensing circuit sets the switch to the left, keeping the capacitor fully charged.
(c) If the pacemaker is in operation, one full cycle requires about 15 ms. About 5 ms is for discharge through the heart muscle, and 10
ms to recharge the capacitor. The recharge is accomplished using a long-life battery, Vc .
EXAMPLE 18.6 RC Circuits in Cameras: Flash Photography Is as Easy as Falling Off a Log(arithm)
In many cameras, the built-in flash gets its energy from that Putting the data into Eq. 18.6, VC = Vo 11 - e -t>t2, the result is
stored in a capacitor. The capacitor is charged using long-life 7.20 = 9.0011 - e -5.00>t2
batteries with voltages of typically 9.00 V. Once the bulb is
fired, the capacitor must recharge quickly through an internal Rearranging this equation yields e -5.00>t = 0.20, and the reci-
RC circuit. If the capacitor has a value of 0.100 F, what must procal of this expression (to make the exponent positive) is
the resistance be so that the capacitor is charged to 80% of its e 5.00>t = 5.00
maximum charge (the minimum charge to fire the bulb again)
To solve for the time constant, recall that if e a = b, then a is the
in 5.00 s?
natural logarithm (ln) of b. Thus, in this Example, 5.00>t is the
T H I N K I N G I T T H R O U G H . After one time constant, the capaci- natural logarithm of 5.00. A calculator quickly shows that
tor will be charged to 63% of its maximum voltage and ln 5.00 = 1.61. Therefore
charge. Because the capacitor needs 80%, the time constant 5.00
must be less than 5.00 s. Eq. 18.6 can be used (along with a cal- = ln 5.00 = 1.61
t
culator) to determine the time constant. From that, the
or
required value of resistance can be determined.
5.00
t = RC = = 3.11 s
SOLUTION. The data given include the final voltage across 1.61
the capacitor, VC, which is 80% of the battery’s voltage, which Solving for R yields
means that Q is 80% of the maximum charge.
3.11 s 3.11 s
R = = = 31.1 Æ
Given: C = 0.100 F Find: R (the resistance C 0.100 F
VB = Vo = 9.00 V required so the The time constant is less than 5.00 s, because achieving 80% of
VC = 0.80Vo = 7.20 V capacitor is 80% the maximum voltage requires a time interval longer than one
t = 5.00 s charged in 5.00 s) time constant.
F O L L O W - U P E X E R C I S E . (a) In this Example, how does the energy stored in the capacitor (after 5.00 s) compare with the maxi-
mum energy storage? Explain why it isn’t 80%. (b) If you waited 10.00 s to charge the capacitor, what would its voltage be? Why
isn’t it twice the voltage that exists across the capacitor after 5.00 s?
An interesting application of an RC circuit is diagrammed in 䉴 Fig. 18.14a. This

circuit is called a blinker circuit (or a neon tube relaxation oscillator). The resistor and
capacitor are initially wired in series, and then a miniature neon tube is connected
in parallel with the capacitor.
18.4 AMMETERS AND VOLTMETERS 641
䉳 F I G U R E 1 8 . 1 4 Blinker circuit
V (a) When a neon tube is connected
R across the capacitor in a series RC
circuit that has the proper voltage
Tube voltage
C
V Vb source, the voltage across the tube
will oscillate with time. As a result,
Neon Vm the tube periodically flashes or blinks.
tube
(b) A graph of tube voltage versus
(90−120 V) t time shows the voltage oscillating
Time between Vb, the “breakdown” volt-
age, and Vm, the “maintaining” volt-
(a) (b) age. See text for detailed discussion.
When the circuit is closed, the voltage across the capacitor (and the neon tube)
rises from 0 to Vb, which is the breakdown voltage of the neon gas in the tube (about 80
V). At that voltage, the gas becomes ionized (that is, electrons are freed from atoms,
creating positive and negative charges that are free to move). Thus the gas begins to
conduct electricity, and the tube lights. When the tube is in this conducting state, the
capacitor discharges through it, and the voltage across the neon tube falls rapidly
(Figure 18.14b). When the tube’s voltage drops below Vm , a state called its
maintaining voltage, the ionization in the tube cannot be sustained, and it stops con-
ducting, thus going dark. Next the capacitor begins charging, the tube voltage rises
from Vm to Vb , and the cycle repeats continually, causing the tube to blink on and off.
DID YOU LEARN?

➥ The time constant for an RC circuit varies directly with the resistance.
➥ The time constant for an RC circuit varies directly with the capacitance.
➥ After one time constant, the voltage across a discharging capacitor is about 37% of
its initial value.
18.4 Ammeters and Voltmeters

➥ How should a voltmeter be connected to measure the voltage drop across a

resistor? I
➥ How should an ammeter be connected to measure the current in a 1 2 3 4 Permanent
0
resistor? magnet
➥ To be as accurate as possible, how much resistance should an ammeter
have?
B
As the names imply, an ammeter measures current through circuit ele- N S
ments and a voltmeter measures voltages across circuit elements. The Wire Cylindrical
coil A iron core
basic component common to both of these meters is a galvanometer
(䉴 Fig. 18.15a). The galvanometer operates on magnetic principles covered (a)
in Chapter 19. In this chapter, it will be treated simply as a circuit element
with an internal resistance r (typically about 50 Æ ) whose needle deflec- r
tion is proportional to the current in it (Fig. 18.15b). A G B
(b)
THE AMMETER
A galvanometer measures current, but because of its small resistance, it can mea- 䉱 F I G U R E 1 8 . 1 5 The galvanome-
ter (a) A galvanometer is a current-
sure only currents in the microampere range without burning out its wires. How-
sensitive device whose needle
ever, a galvanometer can be used to construct an ammeter to measure larger deflection is proportional to the cur-
currents. To do this, a small shunt resistor (resistance Rs) is employed in parallel rent in its coil. (b) The circuit sym-
with a galvanometer. The job of the shunt resistor (or “shunt” for short) is to take bol for a galvanometer is a circle
most of the current (䉲 Fig. 18.16). This requires the shunt to have much less resis- containing a G. The internal resis-
tance than the galvanometer 1Rs = r2. Example 18.7 illustrates how the resistance tance (r) of the meter is indicated
explicitly as r.
of the shunt is determined in the design of an ammeter, while also providing
another application of Kirchhoff’s laws to circuit analysis.
䉴 F I G U R E 1 8 . 1 6 A dc ammeter Ammeter A
Here, R is the resistance of the resis-
tor whose current is being mea-
sured. (a) A galvanometer in
G
parallel with a shunt resistor (Rs)
creates an ammeter capable of mea- Ig r
suring various ranges of current,
depending on the value of Rs. I I I I
(b) The circuit symbol for an amme- J A
ter is a circle with an A inside it. (See R R
Is Shunt
Example 18.7 for a detailed discus- resistor
sion of ammeter design.) (b)
Rs
(a)
EXAMPLE 18.7 Ammeter Design Using Kirchhoff’s Rules: Choosing a Shunt

Suppose a galvanometer can safely carry a maximum coil cur- T H I N K I N G I T T H R O U G H . The galvanometer can carry only a
rent of only 200 mA (called its full-scale sensitivity) and has a small current, so most of the current has to be diverted
coil resistance of 50 Æ . If it is to be used in an ammeter to through the shunt. Thus, the shunt resistance must be much
measure currents up to 3.0 A (at full scale), what is the less than the galvanometer’s internal resistance. Because the
required shunt resistance? (See Fig. 18.16a.) shunt and coil resistance are in parallel, they have the same
voltage across them. This reasoning along with Kirchhoff’s
laws should enable the determination of Rs.

Given: Ig = 200 mA = 2.00 * 10-4 A Find: Rs (shunt resistance)
r = 50 Æ
Imax = 3.0 A
The voltages across the galvanometer and shunt are equal. Using the subscripts “g” for galvanometer and “s” for shunt
Vg = Vs or Ig r = Is Rs
Using Kirchhoff’s junction rule at J, the current I in the external circuit splits into two currents: I = Ig + Is , or Is = I - Ig . Substi-
tuting this into the previous equation, the result is
Ig r = 1I - Ig2Rs
Thus the shunt’s resistance Rs can be found as follows.
Ig r
Rs =
Imax - Ig
12.00 * 10-4 A2150 Æ2
=
3.0 A - 2.00 * 10-4 A
= 3.3 * 10-3 Æ = 3.3 mÆ
As expected, the shunt resistance is much smaller than the coil’s. This allows most of the current (2.9998 A at full scale) to pass
through the shunt. This ammeter then read currents linearly up to 3.0 A. For example, if a current of 1.5 A were to flow into the
ammeter, there would be 100 mA (half the maximum) in the coil, which would show a half-scale reading, or 1.5 A.
F O L L O W - U P E X E R C I S E . In this Example, if a shunt resistance of 1.0 mÆ had been used instead, what would be the full-scale reading
(maximum current reading) of the ammeter?
THE VOLTMETER
A voltmeter that is capable of reading voltages higher than the microvolt range
(anything higher than this value would burn out the galvanometer alone) is con-
structed by connecting a large multiplier resistor in series with a galvanometer
(䉴 Fig. 18.17). Because the voltmeter has a large resistance, due to the multiplier
18.4 AMMETERS AND VOLTMETERS 643
Voltmeter V 䉳 F I G U R E 1 8 . 1 7 A dc voltmeter
Here, R is the resistance of the resis-
tor whose voltage is being mea-
Rm r sured. (a) A galvanometer in series
G with a multiplier resistor (Rm) is a
voltmeter capable of measuring var-
Ig << I Multiplier ious ranges of voltage, depending
I ≈ IR resistor on the value of Rm. (b) The circuit
Ig V
symbol for a voltmeter is a circle
IR R with a V inside it. (See Example 18.8
I I R
I I for a detailed discussion of volt-
V meter design.)
(a) (b)
resistor, it draws little current from the circuit element when it is properly connected
in parallel across that element. However, the current in the voltmeter is proportional
to the voltage across the circuit element. Thus, the voltmeter can be calibrated in
volts. To better understand this configuration, consider Example 18.8.
EXAMPLE 18.8 Voltmeter Design: Using Kirchhoff’s Rules to Choose a Multiplier Resistor
Suppose that the galvanometer in Example 18.7 is to be used The voltages across the galvanometer and multiplier resistors
instead in a voltmeter with a full-scale reading of 3.0 V. What are given by, respectively,
is the required multiplier resistance?
Vg = Ig r and Vm = Ig Rm
THINKING IT THROUGH. To turn a galvanometer into a volt-
Combining these three equations,
meter, a reduction in current is needed, and this is accom-
plished by adding a large “multiplier resistor” in series with V = Vg + Vm = Ig r + Ig Rm
the galvanometer. All the data necessary to calculate the mul-
tiplier resistance are given here and in Example 18.7. Solving for the resistance of the multiplier,
V - Ig r
SOLUTION. Listing the data, Rm =
Ig
Given: Ig = 200 mA Find: Rm (multiplier
= 2.00 * 10-4 A resistance) 3.0 V - 12.00 * 10-4 A2150 Æ2
=
(from Example 18.7) 2.00 * 10-4 A
r = 50 Æ
= 1.5 * 104 Æ = 15 kÆ
(from Example 18.7)
Vmax = 3.0 V Notice that the second term in the numerator (Igr) is negligi-
The resistances of the galvanometer and multiplier are in ble compared with the full-scale reading of 3.0 V. Thus, to a
series. This combination is itself in parallel with the external good approximation, Rm L V>Ig , or V r Ig , thus proving
circuit element (resistance R). Therefore, the voltage across that the measured voltage is, in fact, proportional to the cur-
the external circuit element is the sum of the voltages across rent in the galvanometer.
the galvanometer and multiplier (Fig. 18.17):
V = Vg + Vm
F O L L O W - U P E X E R C I S E . The voltmeter in this Example is used to measure the voltage of a single resistor connected to a battery.
A current of 1.00 A flows through the ohmic resistor (R = 2.00 Æ ) before the voltmeter is connected. Assuming that the battery
voltage does not change when the voltmeter is connected, determine the current in the galvanometer, the current in the resistor,
and the reading after the connection.
For versatility, ammeters and voltmeters may be constructed with a multirange

feature. This is accomplished by providing a choice of shunt or multiplier resistors
(䉲 Fig. 18.18a and 18.18b). Combinations of these meters are manufactured and
sold as multimeters, which are capable of measuring voltage, current, and often
resistance. Electronic digital multimeters are now commonplace (Fig. 18.18c). In
place of mechanical galvanometers, these use electronic circuits to analyze digital
signals and calculate voltages, currents, and resistances, which are then displayed.
Rm1 Rm2Rm3Rm4
Rs1 Multiplier
resistors
Rs2
Switch
Rs3
Shunt
Meter resistors Meter
I Switch
terminals terminals
V
(a) Multirange ammeter (b) Multirange voltmeter (c)
䉱 F I G U R E 1 8 . 1 8 Multirange meters (a) An ammeter or (b) a voltmeter can measure different ranges of current and
voltage by switching among different shunt or multiplier resistors, respectively. (Instead of a switch, there may be an
exterior terminal for each range.) (c) Both functions can be combined in a multimeter, shown here on the left measuring
the voltage across a lightbulb. (How can you tell it is not measuring the current?)
DID YOU LEARN?

➥ A voltmeter should be connected in parallel with a circuit element to measure its
voltage.
➥ An ammeter should be inserted in series with the circuit element whose current is
to be measured.
➥ An accurate ammeter should have a very small resistance to avoid affecting the
current to be measured.
䉲 F I G U R E 1 8 . 1 9 Household
wiring schematic A 120-V circuit is 18.5 Household Circuits and Electrical Safety
obtained by connecting either of the
“hot” lines to the ground line. A LEARNING PATH QUESTIONS
voltage of 240 V (for appliances that ➥ Why are common household circuits wired in parallel?
require a lot of power such as elec- ➥ What is the purpose of a circuit breaker in a household circuit?
tric stoves) can be obtained by con-
➥ How does grounding the case of an appliance make it safer?
necting the two “hot” lines of
opposite polarity. (Note: For clarity,
the dedicated ground wire [the Although household circuits use alternating current, which has not yet been dis-
third line that takes the rounded cussed, their operation (and many practical applications) can be understood using
prong] is not shown.) the same circuit principles just studied.
For example, would you expect the elements in a
household circuit (lamps, appliances, and so on) to be in
series or parallel? From the discussion of Christmas tree
lights (Section 18.1), it should be apparent that they must
be in parallel. When a bulb in a lamp in your kitchen
burns out, other appliances on that circuit, such as the
coffee maker, must continue to work. Moreover, house-
hold appliances and lamps are generally designed to
operate at approximately 120 V. If appliances were in
(Refrigerators
run on 120 V)
series, they would each operate at only a fraction of 120 V.
Electrical power is supplied to a house by a three-
(Electric stoves wire system (䉳 Fig. 18.19). There is an average differ-
run on 240 V) ence in potential of 240 V between the two “hot,” or
Circuit high-potential, wires. Each of these “hot” wires has an
breaker
average 120-V difference in potential with respect to
the ground. The two “hot” wires are always main-
+120 V tained at opposite polarities. The third wire is
∆V = 120 V grounded at the point where the wires enter the house,
Circuit
0V ∆V = 240 V usually by a metal rod driven into the ground. This
breaker
Ground ∆V = 120 V wire is defined to be at zero potential and is called the
–120 V ground, or neutral, wire.
18.5 HOUSEHOLD CIRCUITS AND ELECTRICAL SAFETY 645
The 120 V needed by most appliances is obtained by connecting them between

the ground and either high-potential wire. The result is the same in either case,
because ¢V = 120 V - 0 V = 120 V or ¢V = 0 V - 1- 120 V2 = 120 V. (See
Fuse strip
Fig. 18.19.) Note that even though the ground wire is at zero potential, it is a current-
carrying wire, because it is part of the complete circuit. High-power appliances such
as air conditioners, ovens, and water heaters typically operate at the higher voltage Current
of 240 V to reduce current and thus heating losses. The 240 V is obtained by connect- path
ing such appliances between the two hot wires: ¢V = 120 V - 1-120 V2 = 240 V.
The current needed for the operation of an appliance may be given on a rating tag. If (a)
not, it can usually be determined from the power rating on the tag (since I = P>V).
For example, a stereo rated at 180 W would require a current of 1.50 A. ( I = P>V =
180 W>120 V = 1.50 A.)
There are limitations on the total power of the appliances in a circuit because of
a limitation on the total current in the wires of that circuit. Specifically, joule heat-
ing (or I2R loss) of the wires must be carefully considered. Remember that the
more elements that are in parallel, the smaller their equivalent resistance. Thus
adding appliances (usually by flipping them “on”) increases the total current. Real
wires have resistance and can be subject to significant joule heating if the current
is large. Therefore, by adding too many appliances, it is possible to overload a
household circuit and produce too much heat in the wires. This could melt insula-
tion and perhaps even start a fire. (b)
This potential overloading is prevented by limiting the current. Two types of
devices are commonly employed as limiters: fuses and circuit breakers. Fuses are 䉱 F I G U R E 1 8 . 2 0 Fuses (a) A
still seen in some older homes (䉴 Fig. 18.20). Inside the fuse is a metal strip that fuse contains a metallic strip that
melts when the current exceeds a
melts when the current is larger than the rated value (typically 15 A for a 120-V cir- rated value. This action opens the
cuit). The melting of the strip opens the circuit, and the current drops to zero. An circuit and prevents overheating.
Edison-base fuse has threads like those on the base of a lightbulb type-s fuses have (b) Edison-base fuses (left) have
threads specific to their current ratings. (See Fig. 18.20b.) threads similar to lightbulbs. The
Circuit breakers are used exclusively in newer homes. One type (䉲 Fig. 18.21) threads are identical in this type of
fuse; thus, fuses with different
uses a bimetallic strip (see Chapter 10). As the current in the strip increases, the ampere ratings can be inter-
strip becomes warmer and bends. At the rated value of current, the strip will bend changed—something that is not
sufficiently to open the circuit. The strip then cools, so the breaker can be reset. desirable. (Why?) Type-S fuses
However, a blown fuse or a tripped circuit breaker indicates that the circuit is (right) have different threads for dif-
attempting to draw too much current! Find and correct the problem before replacing ferent ratings and thus cannot be
interchanged.
the fuse or resetting the circuit breaker. Also, under no circumstances should the
blown fuse be even temporarily replaced by one with a higher current rating
(why?). If a fuse with the correct current rating is not available, for safety purposes
it is better to leave that circuit open (unless it controls items needed for emergency
or crucial for living) until the correct fuse is found.
Switches, fuses, and circuit breakers are placed in the “hot” (high-potential)
side of the circuit. They would, of course, also work if placed in the grounded side.
To see why they aren’t in the grounded side, consider the following. If they were
there, even if the switch were open, the fuse blown, or the breaker tripped, the
appliances would all remain connected to a high voltage—which could be danger-
ous if a person made electrical contact with the high voltage (䉲 Fig. 18.22a).
Latch Electrical
Bimetallic Thermal Circuit
strip contacts trip broken
Current in Current out After current overload

(a)
䉱 F I G U R E 1 8 . 2 1 Circuit breakers (a) A diagram of a thermal trip element.

With increased current and joule heating, the element bends until it opens the cir-
cuit at some preset current value. Trip elements using magnetic principles also
exist. (b) A typical bank of household circuit breakers. (b)
䉴 F I G U R E 1 8 . 2 2 Electrical safety Hot side Hot side

(a) Switches and fuses or circuit (high potential) Motor (high potential)
breakers should always be wired in
the hot side of the line, not in the
grounded side as shown. If these
elements are wired in the grounded Fuse Fuse
side, the line (and potentially the blown
metallic case of the appliance)
remains at a high voltage even
when the fuse is blown or a switch Electrical
is open. (b) Even if the fuse or cir- contact with case
Ground Ground
cuit breaker is wired in the hot side,
a potentially dangerous situation (a) (b)
exists. If an internal wire comes in
contact with the metal casing of an
appliance or power tool, a person
Even with fuses or circuit breakers wired correctly into the “hot” side of the cir-
touching the casing, which is at high
voltage, can get a shock. To prevent cuit, there is still a possibility of receiving an electrical shock from a defective
this possibility, a third dedicated appliance that has a metal casing, such as a hand drill. For example, if a wire
ground line runs from the case to comes loose inside, it could make contact with the casing, which would then be at a
ground (see Fig. 18.23). high voltage (Fig. 18.22b). If a person’s body touched the wire, it could then pro-
vide a path for current to ground, and would thus receive a shock. For a discus-
Hot side Motor
sion of the effects of electric shock, see Insight 18.2, Electricity and Personal Safety.
(high potential) To prevent a shock, a third, dedicated grounding wire is usually added to the
circuit, which grounds the metal casing of appliances such as power tools
(䉳 Fig. 18.23). This wire provides a path of very low resistance, bypassing the tool,
Dedicated
and does not normally carry current. If a hot wire comes in contact with the cas-
grounding ing, the circuit is completed through this grounding wire. Then the fuse is blown
or the circuit breaker tripped. If you touched the wire, most of the current would
travel through the ground wire rather than you. Most likely you would not be
Ground harmed. Remember, however, that the breaker, if reset, will continue to trip unless
you find the source of the problem and fix it.
䉱 F I G U R E 1 8 . 2 3 Dedicated
On three-prong grounded plugs, the large, round prong connects with the
grounding For safety, a third wire is
connected from an appliance or grounding wire. Adapters can be used between a three-prong plug and a two-
power tool to ground. This dedi- prong socket. Such an adapter has a grounding lug or grounding wire
cated grounding wire normally car- (䉲 Fig. 18.24a) that should be fastened to the receptacle box by the plate-fastening
ries no current (as opposed to the screw. The receptacle box is itself attached to the grounding wire. If the adapter
grounded wire of the circuit). If the
lug or wire is not connected, the system is left unprotected, which defeats the pur-
hot wire should come in contact
with the metal case, the current will pose of the dedicated grounding safety feature.
follow the ground wire (the path of You may have noticed another type of plug, a two-prong plug that fits in the
least resistance) rather than go socket in only one orientation, because one prong is wider than the other, as is one
through the body of the operator of the slits of the receptacle (Fig. 18.24b). This type is called a polarized plug.
holding the case. The plug used for
Polarizing in the electrical circuit sense is a method of identifying the hot and
this is shown in Fig. 18.24.
grounded sides of the line so that particular connections can be made.
Such polarized plugs and sockets are now a common safety feature. Wall recep-
tacles are wired so that the small slit connects to the hot side and the large slit con-
nects to the neutral, or ground, side. Having the hot side identified in this way
makes two safeguards possible. First, the manufacturer of an electrical appliance
䉴 F I G U R E 1 8 . 2 4 Plugging into
ground (a) To accommodate the
dedicated ground wire (Fig. 18.23),
a three-prong plug is used. The
adapter shown enables a three-
prong plug to be used in a two-
prong socket. The grounding lug
(loop) on the adapter should be con-
nected to the plate-fastening screw
on the grounded receptacle box—
otherwise, this safety feature is lost.
(b) A polarized plug. The differently
sized prongs permit prewired iden-
tification of the high and ground
sides of the line. See text for details. (a) (b)
18.5 HOUSEHOLD CIRCUITS AND ELECTRICAL SAFETY 647
can design it so that the switch is always in the hot side of the line. Thus, all of the
wiring of the appliance beyond the switch is safely neutral when the switch is
open and the appliance is off. Moreover, the casing of an appliance is connected by
the manufacturer to the grounded side by means of a polarized plug. Should a hot
wire inside the appliance come loose and contact the metal casing, the effect
would be similar to that with a dedicated grounding system. The hot side of the
line would be shorted to the ground, which would blow a fuse or trip a circuit
breaker. Once again, you would be spared.
Another type of electrical safety device, the ground fault circuit interrupter, or
GFCI, is discussed in Chapter 20.
DID YOU LEARN?

➥ Household circuits are wired in parallel so that each appliance operates at the same
voltage and is independently switchable.
➥ A circuit breaker’s main purpose is to limit the total current in the wire supplying a
household circuit, thereby preventing overloading, heating, and fires.
➥ By grounding the case of an appliance, any accidental case current will go through
the ground wire and not through you.
INSIGHT 18.2 Electricity and Personal Safety

Safety precautions are necessary to prevent injuries when TABLE 1 Effects of Electric Current on the
people work with electrical devices or wiring. Electrical con- Human Body*
ductors (wires) are coated with insulating materials so they
can be handled safely. However, if a person comes in contact Current (approximate) Effect
with a charged conductor, a difference in potential could exist
across part of the person’s body. A bird can sit on a high- 2.0 mA (0.002 A) Mild shock or heating
voltage line without any problem because both of its feet are 10 mA (0.01 A) Paralysis of motor muscles
at the same potential—thus there is no difference in potential to
20 mA (0.02 A) Paralysis of chest muscles, causing
generate a current in the bird. But if a person carrying an alu-
respiratory arrest; fatal in a few
minum (conducting) ladder touches it to a bare electrical line,
minutes
a difference in potential exists between the line and the
ground. Thus the ladder and person become part of a current- 100 mA (0.1 A) Ventricular fibrillation, preventing
carrying circuit. coordination of the heart’s beating;
The extent of personal injury in such cases depends on the fatal in a few seconds
current in the body and on its path through the body. If the body 1000 mA (1 A) Serious burns; fatal almost instantly
as a whole is subjected to a voltage (say from shoulder to foot), *The effect of a given amount of current depends on a variety of
the current in the body is I = V>Rbody and clearly depends on conditions. This table gives only general and relative descriptions
the body’s resistance. which can vary. If the skin is dry, the total and assumes a circuit path that includes the upper chest.
body resistance can be as high as 0.50 MÆ 10.50 * 106 Æ2 or
more. For a voltage of 120 V, the current would be only about
one-quarter of a milliamp, since worse. Some of the possible effects of this type of path are
V 120 V also given in Table 1.
I = = = 0.24 * 10-3 A = 0.24 mA Injury results because the current interferes with muscle
Rbody 0.50 * 106 Æ function and/or causes burns. Muscle function is regulated
This current is almost too weak to be felt (see 䉴 Table 1). But if the by electrical nerve impulses, which can be influenced by
skin is wet, then Rbody can drop to as low as 5.0 kÆ and the cur- external currents. Muscle reaction and pain can occur from a
rent would then be 24 mA (you should show this), a value which current of just a few milliamperes. At about 10 mA, muscle
could potentially be dangerous. (See Table 1 again.) paralysis can prevent a person from releasing the conductor.
A basic precaution is to avoid contact with any exposed At about 20 mA, contraction of the chest muscles occurs,
electrical conductor that might cause a voltage across any which can cause impairment or stoppage of breathing. Death
part of your body. The physical damage resulting from such can occur in a few minutes. At 100 mA, rapid uncoordinated
contact depends on the current path through the body. If movements of the heart muscles (called ventricular fibrillation)
that path is from a finger to the thumb on one hand, a large prevent proper heart pumping action and can be fatal in sec-
current probably would result in only a burn. However, if onds. Working safely with electricity requires a knowledge of
the path is from hand to hand through the chest (and there- fundamental electrical principles and common sense. Electric-
fore likely through the heart), the effect could be much ity must be treated with respect.
PULLING IT TOGETHER Circuit Sketching, Parallel and Series Combinations and More!
A complete DC circuit consists of a 24.0-V battery and three ing. (a) The sketch is straightforward when following the cir-
ohmic resistors. The resistors have resistances of 2.00 Æ, 6.00 Æ cuit-sketching rules in Chapter 18. The equivalent resistance is
and 12.0 Æ. The resistor with the smallest resistance is wired to the series combination of the single and parallel-combination.
the positive terminal of the battery and the other two follow it The battery output power is determined from its voltage and
in parallel with each other. the total current. (b) Working back to the actual circuit from the
(a) Sketch the circuit. Find its equivalent resistance and the equivalent single resistor enables the determination of the cur-
total power delivered by the battery. (b) What are the current, rent and voltage for each resistor. Their product is the power.
voltage and power for each resistor? (c) Which one of the fol- (c) Adding more resistance to a parallel-combination
lowing circuit modifications would result in the largest increase increases the combination’s equivalent resistance, thus both
in the output power of the battery: (1) changing the 12.0-Æ choices (1) and (2) result in more total circuit resistance and
resistor to one with a higher resistance, (2) changing the 6.00-Æ therefore a decrease in battery power. Choice (3) results in a
resistor to one with a higher resistance, or (3) connecting some resistance lower than the original 2.00-Æ value, thus decreas-
unknown resistor in parallel to the 2.00-Æ resistor? Explain ing the total circuit resistance and increasing the power out-
your reasoning. (d) Using your answer to part (c), determine put of the battery. Thus the correct choice is (3), add any
the changed/added resistor value required to increase the bat- resistor in parallel to the 2.00 Æ, one. (d) The new battery
tery output power by 10%. power implies a new equivalent circuit resistance. From that,
one can work backward to determine the resistor to be con-
T H I N K I N G I T T H R O U G H . This example brings together the con-
nected to the 2.00-Æ one.
cepts of circuit sketching, equivalent resistance and Joule heat-

Given: V = 24.0 V Find: (a) sketch the circuit; find Req (circuit equivalent resistance)
R1 = 2.00 Æ and Pb (battery power)
R2 = 6.00 Æ (b) currents, voltages and power for each resistor
R3 = 12.0 Æ (c) the best way to increase battery power output
(d) R4 (new resistor to increase battery power output by 10%).
(a) The circuit diagram is shown in 䉲 Fig. 18.25 . The equiva- (b) The total current is the current in R1 (why?), hence
lent resistance of the two parallel resistors 1Rp2 is determined V1 = I R1 = 8.00 V and P1 = I V1 = 32.0 W
as follows:
The remainder of the voltage drop 124.0 V - 8.00 V = 16.0 V2
1 1 1 1 1 3 is the voltage across the other two resistors (why?) thus for R2,
= + = + =
Rp R2 R3 6.00 Æ 12.0 Æ 12.0 Æ V2 16.0 V
I2 = = = 2.67 A
therefore, Rp = 4.00 Æ. R2 6.00 Æ
This combination is in series with the 2.00-Æ resistor, for a and
circuit equivalent resistance of P2 = I2 V2 = 42.7 W
Req = R1 + Rp = 2.00 Æ + 4.00 Æ = 6.00 Æ and for R3
V3 16.0 V
The total current (I) will be as if this resistance were con- I3 = = = 1.33 A
nected to the battery; therefore R3 12.0 Æ
and
V 24.0 V P3 = I3V3 = 21.3 W
I = = = 4.00 A
Req 6.00 Æ
Notice that the total of the currents after the first junction
and the battery power output is agrees with the incoming current and the battery power output
agrees with the total power of the three resistors.
Pb = I V = 14.00 A2124.0 V2 = 96.0 W
(c) As discussed in the Thinking It Through section, only
choice (3) results in a reduction in total circuit resistance, thus
increasing the power output of the battery.
(d) To find the resistor to be added in parallel to the 2.00-Æ
I R1 = 2.00 Ω resistor realize that the battery’s power output is to be raised
to Pbœ = 1.10Pb = 106 W. Based on the same voltage, the new
I2 I3 current 1I œ2 is
R2 = 6.00 Ω R3 = 12.0 Ω Pbœ 106 W
Iœ = = = 4.40 A
V = 24.0 V V 24.0 V
This is no longer the current through R1 because this resistor
I will have a parallel “partner” whose resistance is to be deter-
mined. But it is known that new total current is the current in
䉱 FIGURE 18.25 the original parallel combination (equivalent resistance of
4.00 Æ ).Thus the new voltage across this combination is therefore,

Vpœ = I œRp = 14.40 A214.00 Æ2 = 17.6 V. This means that
1 1 1
6.40 V1 = 24.0 V - 17.6 V2 appears across the new parallel = - ,
combination whose equivalent resistance 1Rx2 is R4 Rx R1
or
6.40 V
Rx = = 1.45 Æ 1 1 1
4.40 A = - = 0.190 Æ -1
The new resistor 1R42 can be found because
R4 1.45 Æ 2.00 Æ
hence
1 1 1
+ = ; R4 = 5.27 Æ
R1 R4 Rx
■ When resistors are wired in series, the current through each ■ Kirchhoff’s loop theorem states that in traversing a com-
of them is the same. The equivalent series resistance of plete circuit loop, the algebraic sum of the voltage gains and
resistors in series is losses is zero, or the sum of the voltage gains equals the sum
of the voltage losses (conservation of energy in an electric
Rs = R1 + R2 + R3 + Á= g Ri (18.2)
circuit). In terms of voltages, this can be written as
V1
(sum of voltage around
I g Vi = 0 (18.5)
a closed loop)
R1 V1 = IR1
V2
Potential
+ V– V R2 V2 = IR2
V3
IR
R3 V3 = IR3 ε
V
ε I c
V = V1 + V2 + V3 R d
b
f
■ When resistors are wired in, the voltage across each of them ε a
is the same. The equivalent parallel resistance is r e Ir
1 1 1 1 1 ■ The time constant (T) for an RC circuit is a characteristic

= + + + Á = g (18.3) time by which we measure the capacitor’s charging and dis-
Rp R1 R2 R3 Ri
charging rate. t is given by
V = V1 = V2 = V3 t = RC (18.8)
I
+ V –
■ An ammeter is a device for measuring current; it consists of
I1 I2 I3
1 2 3 a galvanometer and a shunt resistor in parallel. Ammeters
are connected in series with the circuit element carrying the
current to be measured, and have very little resistance.
I I
A
I R
I = I1 + I2 + I3
I1
I2
I3 ■ A voltmeter is a device for measuring voltage; it consists of
V
R1 R2 R3
V = V1 = V2 = V3 a galvanometer and a multiplier resistor wired in series.
Voltmeters are connected in parallel, with the circuit ele-
ment experiencing the voltage to be measured, and have
■ Kirchhoff’s junction theorem states that the total current large resistance.
into any junction equals the total current out of that junc- V
tion (conservation of electric charge).
g Ii = 0 1sum of currents at a junction2
R
I I
(18.4)

18.1 RESISTANCES IN SERIES, about the sign of the change in electric potential (the
PARALLEL, AND SERIES—PARALLEL voltage): (a) it is negative, (b) it is positive, (c) it is zero,
COMBINATIONS or (d) you can’t tell from the data given?
11. By our sign conventions, if a battery is traversed in the
1. Which of the following quantities must be the same for
actual direction of the current in it, what can you say
resistors in series: (a) voltage, (b) current, (c) power, or
about the sign of the change in electric potential (the
(d) energy?
battery’s terminal voltage): (a) it is negative, (b) it is
2. Which of the following quantities must be the same for positive, (c) it is zero, or (d) you can’t tell from the data
resistors in parallel: (a) voltage, (b) current, (c) power, or given?
(d) energy?
12. You are given a multiloop circuit with one battery that
3. Two resistors (A and B) are connected in series to a 12-V has a terminal voltage of 12 V. After leaving the positive
battery. Resistor A ends up with 9 V across it. Which terminal of the battery, a short wire takes you to a junc-
resistor has the least resistance: (a) A, (b) B, (c) both have tion where the current (and circuit) splits into three
the same, or (d) you can’t tell from the data given? wires, each containing two resistors. Later on, the three
4. Two resistors (A and B) are connected in parallel to a arms of the circuit rejoin and then are connected to the
12-V battery. Resistor A ends up with 2.0 A of current negative terminal of the battery.
and the total current in the battery is 3.0 A. Which resis- As you traverse each wire/resistor arm separately
tor has the most resistance: (a) A, (b) B, (c) both have the leading to the final junction, what can you say about the
same, or (d) you can’t tell from the data given? sum of the voltages across the two resistors in each wire:
5. Two resistors (one with a resistance of 2.0 Æ and the (a) they all total +12 V, (b) they all total -12 V, (c) they
other with that of 6.0 Æ ) are connected in parallel to a all total less than 12 V in magnitude.
battery. Which one produces the most joule heating:
(a) the 2.0-Æ resistor, (b) the 6.0-Æ resistor; (c) both pro-
duce the same, or (d) you can’t tell from the data given?
18.3 RC CIRCUITS
6. Two resistors (one with a resistance of 2.0 Æ and the 13. A fully charged capacitor stores 2.5 mJ of electric energy.
other with that of 6.0 Æ ) are connected in series to a bat- Then it has a resistor connected across its oppositely
tery. Which one produces the most joule heating: (a) the charged plates. What can you say about the total heat
2.0-Æ resistor, (b) the 6.0-Æ resistor; (c) both produce the generated in the resistor (ignore wire resistance): (a) it is
same, or (d) you can’t tell from the data given? greater than 2.5 mJ, (b) it is equal to 2.5 mJ, (c) it is less
7. Resistors A and B are wired in parallel, and that combi- than 2.5 mJ.
nation is in turn connected in series to resistor C. The 14. As a capacitor discharges through a resistor, the voltage
whole network is then connected to a battery. Which of across the resistor is a maximum (a) at the beginning of
the following statements is true: (a) the current in C must the process, (b) near the middle of the process, (c) at the
be less than the current in either A or B; (b) the current in end of the process, (d) after one time constant.
C must be more than the current in either A or B; (c) the 15. When a capacitor discharges through a resistor, the cur-
current in C must equal the sum of the currents in A and rent in the circuit is a minimum (a) at the beginning of
B; or (d) the current in C must exceed the sum of the cur- the process, (b) near the middle of the process, (c) at the
rents in A and B. end of the process, (d) after one time constant.
8. For the circuit in Question 7, which of the following state- 16. A charged capacitor discharges through a resistor (call
ments is true: (a) the voltage across C must be less than this discharging situation #1). If the value of the resistor
that across either A or B; (b) the voltage across C must be is doubled and the identically charged capacitor is
more than the voltage across either A or B; (c) the voltage allowed to discharge again (situation #2), how do the
across C must equal the sum of the voltages across A and time constants compare: (a) t1 = 2t2, (b) t1 = t2,
B; or (d) the voltages across A and B must be equal? (c) t1 = 12 t2 ?
17. An uncharged capacitor is charged by a battery through
18.2 MULTILOOP CIRCUITS AND a resistor (call this charging situation #1). The capacitor
KIRCHHOFF’S RULES is then completely discharged and recharged (using a
different battery but the same resistor) to twice the final
9. You have a multiloop circuit with one battery. After leav- charge as in situation #1 (call this charging situation #2).
ing the battery, the current encounters a junction into two How do the charging time constants compare:
wires. One wire carries 1.5 A and the other 1.0 A. What is (a) t1 = 2t2, (b) t1 = t2, (c) t1 = 12 t2?
the current in the battery: (a) 2.5 A, (b) 1.5 A, (c) 1.0 A,
(d) 5.0 A, or (e) it can’t be determined from the given data?
10. By our sign conventions, if a resistor is traversed in the
direction opposite of the current in it, what can you say
18.4 AMMETERS AND VOLTMETERS 18.5 HOUSEHOLD CIRCUITS AND

18. To accurately measure the voltage across a 1-kÆ resistor, ELECTRICAL SAFETY
a voltmeter should have a resistance that is (a) much 22. The ground wire in household wiring (a) is a current-
larger than 1 kÆ , (b) much smaller than 1 kÆ , (c) about carrying wire, (b) is at a voltage of 240 V from one of the
the same as 1 kÆ , (d) as close to zero as possible. “hot” wires, (c) carries no current, (d) none of the
19. To accurately measure the current in a 1.0-kÆ resistor, an preceding.
ammeter should have a resistance that is (a) much larger 23. A dedicated grounding wire (a) is the basis for the polar-
than 1.0 kÆ , (b) much smaller than 1.0 kÆ , (c) about the ized plug, (b) is necessary for a circuit breaker, (c) nor-
same as 1.0 kÆ , (d) as large as possible. mally carries no current, (d) none of the preceding.
20. To correctly measure the voltage across a circuit element, 24. The circuit breaker in a normal 120-V household circuit
a voltmeter should be connected (a) in series with it, limits the total current in its circuit to (a) about 15 to
(b) in parallel with it, (c) between the high potential side 20 A, (b) about 1 A, (c) about 120 V, (d) about 100 A.
of the element and ground, (d) in none of the preceding. 25. Two identical appliances are connected to the same
21. Which of the following enables an ammeter to make a household circuit and the circuit breaker trips. Which of
correct measurement of current in a circuit element the following is true (there may be more than one correct
(there may be more than one correct answer:) (a) connect answer): (a) if the breaker is reset and only one appliance
the ammeter in series with the element and just before it; is connected, the breaker will not trip; (b) connecting only
(b) connect the ammeter in series with the element and one appliance on this circuit might not trip the breaker
just after it; (c) connect the ammeter in parallel with the again; (c) each appliance had too low a resistance to work
element; (d) connect the ammeter between the high with a normal 120-V circuit breaker; and/or (d) the appli-
potential side of the element and ground; and/or ances together had a total resistance that was too low to
(e) none of these is the correct way to use an ammeter. work with a normal 120-V circuit breaker.
18.1 RESISTANCES IN SERIES, largest current, (b) the largest voltage, and (c) the largest
PARALLEL, AND SERIES–PARALLEL power output?
COMBINATIONS 8. Three resistors have values of 5.0 Æ , 2.0 Æ , and 1.0 Æ .
1. Are the voltage drops across resistors in series generally The first one is followed in series by the last two wired in
the same? If not, under what circumstance(s) could they parallel. When this arrangement is connected to a bat-
be the same? tery, which resistor has (a) the largest current, (b) the
largest voltage, and (c) the largest power output?
2. Are the joule heating rates for resistors in series gener-
ally the same? If not, under what circumstance(s) could
they be the same? 18.2 MULTILOOP CIRCUITS AND
3. Are the currents in resistors in parallel generally the KIRCHHOFF’S RULES
same? If not, under what circumstance(s) could they be 9. Must current always leave from the positive terminal of
the same? a battery that is in a complete circuit? Explain. If not,
4. Are the joule heating rates in resistors in parallel gener- give an example in which the current can enter at the
ally the same? If not, under what circumstance(s) could positive terminal.
they be the same? 10. Use Kirchhoff’s junction theorem to explain why the
5. If a large resistor and a small resistor are connected in total equivalent resistance of a circuit is reduced, not
series, will the value of the effective resistance be closer increased, by connecting a second resistor in parallel to
to that of the large resistance or that of the small one? another resistor.
What if they are connected in parallel? 11. Use Kirchhoff’s loop theorem to explain why a 60-W
6. Lightbulbs are labeled with their power output. For lightbulb produces more light than one rated at 100 W
example, when a lightbulb is labeled 60 W, it is assumed when they are connected in series to a 120-V source.
that the bulb is connected to a 120-V source. Suppose [Hint: Recall that the power ratings are meaningful only
you have two bulbs. A 60-W bulb is followed by a 40-W at 120 V.]
bulb in series to a 120-V source. Which one glows 12. Use both of Kirchhoff’s theorems to explain why a 60-W
brighter? Why? What happens if you switch the order of lightbulb produces less light than one rated at 100 W when
the bulbs? Are either of them at full power rating? [Hint: they are connected in parallel to a 120-V source. [Hint:
Consider their relative resistance values.] Recall that the power ratings are meaningful only at 120 V.]
7. Three identical resistors are connected to a battery. Two 13. Use Kirchhoff’s loop theorem to explain why, in a series
are wired in parallel, and that combination is followed in connection, the largest resistance has the greatest voltage
series by the third resistor. Which resistor has (a) the drop across it.
18.3 RC CIRCUITS you want to measure the voltage across just one of them
with just one measurement. (c) Three resistors are wired
14. An alternative way to describe the discharge/charge
in series and you want to measure the total voltage across
time of an RC circuit is to use a time interval called the
them. (d) Three resistors are wired in series and you want
half-life, which is defined as the time for the capacitor to
to measure the voltage across just one of them.
lose half its initial charge. Based on this definition, is the
time constant longer or shorter than the half-life?
Explain your reasoning. 18.5 HOUSEHOLD CIRCUITS AND
15. Is the time it takes to charge a capacitor in an RC circuit ELECTRICAL SAFETY
to 25% of its maximum value longer or shorter than one
time constant? Is the time it takes to discharge a capaci- 22. In terms of electrical safety, explain clearly what is
tor to 25% of its initial charge longer or shorter than one wrong with the circuit in 䉲 Fig. 18.26, and why.
time constant? Explain your answers.
16. Use Kirchhoff’s loop theorem to explain why the current
Motor
in an RC circuit that is discharging a capacitor decreases
as time goes on. Use the loop theorem to explain why the 120 V
current in a charging RC circuit also decreases with time. S
[Hint: The loop theorem will tell you about the voltage
across the resistor, which is directly related to the current
in the circuit.]
䉱 F I G U R E 1 8 . 2 6 A safety problem? (The purple circular
element represents a fuse or circuit breaker.) See Conceptual
18.4 AMMETERS AND VOLTMETERS Question 22.
17. (a) What would happen if an ammeter were connected 23. The severity of bodily injury from electrocution depends
in parallel with a current-carrying circuit element? on the magnitude of the current and its path, yet you
(b) What would happen if a voltmeter were connected in commonly see signs that warn “Danger: High Voltage”
series with a current-carrying circuit element? (䉲 Fig. 18.27). Shouldn’t such signs be changed to refer to
18. Explain clearly, using Kirchhoff’s laws, why the resis- high current? Explain.
tance of an ideal voltmeter is infinite.
19. If designed properly, a good ammeter should have a
very small resistance. Why? Explain clearly, using Kirch-
hoff’s laws.
20. Draw the circuit diagrams indicating the correct placement
for the ammeter in the following situations. (Use a circle
with an “A” in it to represent the ammeter.) (a) Three resis-
tors are wired in parallel and you want to measure the
total current through all of them with just one measure-
ment. (b) Three resistors are wired in parallel and you 䉱 F I G U R E 1 8 . 2 7 Danger—high voltage Shouldn’t the sign
read “high current” instead of “high voltage”? See Conceptual
want to measure the current of just one of them with just
Question 23.
one measurement. (c) Three resistors are wired in series
and you want to measure the total current through all of 24. Explain why it is safe for birds to perch with both feet on
them. (d) Three resistors are wired in series and you want the same high-voltage wire, even if the insulation is
to measure the current through just one of them. worn through.
21. Draw the circuit diagrams indicating the correct place- 25. After a collision with a power pole, you are trapped in
ment for the voltmeter in the following situations. (Use a your car, with a high-voltage line (with frayed insulation)
circle with an “V” in it to represent the voltmeter.) in contact with the hood of the car. If you must get out
(a) Three resistors are wired in parallel and you want to before help arrives, is it safer to step out of the car one foot
measure the total voltage across all of them with just one at a time or to jump with both feet leaving the car at the
measurement. (b) Three resistors are wired in parallel and same time? Explain your reasoning.
EXERCISES
Assume that all resistors are ohmic unless otherwise stated.
EXERCISES 653
18.1 RESISTANCES IN SERIES, PARALLEL, 12. ●● Find the current in and voltage across the 10-Æ resis-
AND SERIES–PARALLEL COMBINATIONS tor shown in 䉲 Fig. 18.30.
R1 = 10 Ω
1. ● Three resistors that have values of 10 Æ , 20 Æ , and 䉳 FIGURE 18.30
R3 = 5.0 Ω
30 Æ are to be connected. (a) How should you connect Current and voltage
R2 = 2.0 Ω
them to get the maximum equivalent resistance, and drop of a resistor See
what is this maximum value? (b) How should you con- Exercises 12 and 22.
V = 10 V
nect them to get the minimum equivalent resistance, and
what is this minimum value?
2. ● Two identical resistors (each with resistance R) are 13. ● ● For the circuit shown in 䉲 Fig. 18.31, find (a) the cur-
connected together in series and then this combination is rent in each resistor, (b) the voltage across each resistor,
wired in parallel to a 20-Æ resistor. If the total equivalent and (c) the total power delivered.
resistance is 10 Æ , what is the value of R? R2 = 20 Ω
䉳 FIGURE 18.31
3. ● Two identical resistors (R) are connected in parallel
Circuit reduction See
and then wired in series to a 40-Æ resistor. If the total Exercises 13 and 23.
equivalent resistance is 55 Æ , what is the value of R? V = 20 V R3 = 20 Ω
4. IE ● (a) In how many different ways can three 4.0-Æ R1 = 20 Ω
resistors, be wired: (1) three, (2) five, or (3) seven?

(b) Sketch the different ways you found in part (a) and 14. ●● Suppose that the resistor arrangement in Fig. 18.28 is
determine the equivalent resistance for each. connected to a 12-V battery. What will be (a) the current
5. ● Three resistors with values of 5.0 Æ , 10 Æ , and 15 Æ in each resistor, (b) the voltage drop across each resistor,
are connected in series in a circuit with a 9.0-V battery. and (c) the total power delivered?
(a) What is the total equivalent resistance? (b) What is 15. ● ● Suppose, in Exercise 14, that another 2.0-Æ resistor is
the current in each resistor? (c) At what rate is energy in series with one in the lower branch. (a) Redraw the
delivered to the 15-Æ resistor? circuit and predict how the currents in the three different
6. ● Three resistors with values 1.0-Æ , 2.0-Æ , and 4.0-Æ are arms will change (increase, decrease, or stay the same)
connected in parallel in a circuit with a 6.0-V battery. compared to those in the original circuit. (b) Calculate
What are (a) the total equivalent resistance, (b) the volt- the new currents in each arm of the new circuit.
age across each resistor, and (c) the power delivered to 16. ● ● ● The terminals of a 6.0-V battery are connected to
the 4.0-Æ resistor? points A and B in Fig. 18.29. (a) How much current is in
7. IE ● ● A length of wire with a resistance R is cut into two each resistor? (b) How much power is delivered to each?
equal-length segments. These segments are then twisted (c) Compare the sum of the individual powers with the
together to form a conductor half as long as the original power delivered to the equivalent resistance for the circuit.
wire. (a) The resistance of the shortened conductor is (1) 17. ● ● ● Lightbulbs with the power ratings (expressed in
R>4, (2) R>2, (3) R. Explain your reasoning. (b) If the resis- watts) given in 䉲 Fig. 18.32 are connected in a circuit as
tance of the original wire is 27 mÆ and the wire is, instead, shown. (a) What current does the voltage source deliver
cut into three equal segments and then twisted together, to the circuit? (b) Find the power delivered to each bulb.
what is the resistance of the shortened conductor? (Take the bulbs’ resistances to be the same as at their nor-
8. ● ● You are given four 5.00-Æ resistors. (a) Show how to mal operating voltage.)
connect all the resistors so as to produce an effective
total resistance of 3.75 Æ . (b) If this network were then
60 W 䉳 F I G U R E 1 8 . 3 2 Watt’s up?
See Exercise 17.
connected to a 12-V battery, determine the current in and 100 W
voltage across each resistor.
9. ● ● Two 8.0-Æ resistors are connected in parallel, as are two 15 W 40 W
120 V
4.0-Æ resistors. These two combinations are then connected
in series in a circuit with a 12-V battery. What is the current
in each resistor and the voltage across each resistor? 18. ● ● ● Two resistors R1 and R2 are in series with a 7.0-V bat-
10. ● ● What is the equivalent resistance of the resistors in tery. If R1 has a resistance of 2.0 Æ and R2 receives energy
䉲 Fig. 18.28? at the rate of 6.0 W, what is (are) the value(s) for the cir-
R1 = 2.0 Ω
䉳 FIGURE 18.28 cuit’s current(s)? (There may be more than one answer.)
Series–parallel combi- 19. ● ● ● For the circuit in 䉲 Fig. 18.33, find (a) the current in
R2 = 2.0 Ω nation See Exercises each resistor, (b) the voltage across each resistor, (c) the
R4 = 2.0 Ω
10 and 14. power delivered to each resistor, and (d) the total power
R3 = 2.0 Ω delivered by the battery.
11. IE ● ● What is the equivalent resistance between points 䉳 F I G U R E 1 8 . 3 3 Resistors

V = 10 V and currents See Exercise 19.
A and B in 䉲 Fig. 18.29?
R1 = 10 Ω R2 = 5.0 Ω
R1 = 6.0 Ω
䉳 FIGURE 18.29 R3 = 10 Ω
A B
R4 =
Series–parallel combination
10 Ω R2 = 4.0 Ω See Exercises 11 and 16.
R5 = 20 Ω R4 = 5.0 Ω
R3 = 6.0 Ω
20. ● ● ● (a) Determine the equivalent resistance of the circuit 28. ●●● Find the currents in the circuit branches in 䉲Fig 18.38.
in 䉲 Fig. 18.34 Find (b) the current in each resistor, (c) the
voltage across each resistor, and (d) the total power 䉳 FIGURE 18.38
V 1 = 20 V V 2 = 10 V R2 = How many loops?
delivered to the circuit. 4.0 Ω
R 5 = 2.0 Ω R1 = See Exercise 28.
䉳 FIGURE 18.34 5.0 Ω
R3 =
10 Ω 2.0 Ω R 4 = 2.0 Ω 6.0 Ω
Power dissipation
6.0 Ω
V = 24 V 4.0 Ω See Exercise 20. R 6 = 2.0 Ω
12 Ω
29. ●●● For the multiloop circuit shown in 䉲 Fig. 18.39, what
10 Ω is the current in each branch?
5.0 Ω
R 1 = 2.0 Ω R 2 = 4.0 Ω
䉳 FIGURE 18.39
18.2 MULTILOOP CIRCUITS AND Triple-loop circuit
KIRCHHOFF’S RULES V 3 = 6.0 V See Exercise 29.
V 1 = 6.0 V R 3 = 6.0 Ω
21. ● (a) For the circuit showin in Fig. 18.10, traverse loop 3 R 5 = 10 Ω
opposite to the direction shown, and demonstrate that the R 4 = 8.0 Ω
V 2 = 12 V
resulting equation is the same as that obtained if you R 6 = 12 Ω
had followed the direction of the arrows. (b) Repeat the
procedure in part (a) by traversing loops 1 and 2 in the
direction opposite that taken in the text, and demon- 18.3 RC CIRCUITS
strate that equations equivalent to those in Example 18.5 30. ● A capacitor in a single-loop RC circuit is charged to
are obtained. 63% of its final voltage in 1.5 s. Find (a) the time constant
22. ● ● Use Kirchhoff’s loop theorem to find the current in for the circuit and (b) the percentage of the circuit’s final
each resistor in Fig. 18.30. voltage after 3.5 s.
23. ● ● Apply Kirchhoff’s rules to the circuit in Fig. 18.31 to 31. ● In Fig. 18.11b, the switch is closed at t = 0, and the
find the current in each resistor. capacitor begins to charge. What is the voltage across the
24. IE ● ● Two batteries with terminal voltages of 10 V and resistor and across the capacitor, expressed as fractions
4 V are connected with their positive terminals together. of Vo (to two significant figures), (a) just after the switch
A 12-Æ resistor is wired between their negative termi- is closed, (b) after two time constants have elapsed, and
nals. (a) The current in the resistor is (1) 0 A, (2) between (c) after many time constants have elapsed?
0 A and 1.0 A, (3) greater than 1.0 A. Explain your choice. 32. IE ● In a flashing neon sign display, a certain time con-
(b) Use Kirchhoff’s loop theorem to find the current in stant is desired. (a) To increase this time constant, you
the circuit and the power delivered to the resistor. should (1) increase the resistance, (2) decrease the resis-
(c) Compare this result with the power output of each tance, (3) eliminate the resistor. Why? (b) If a 2.0 s time
battery. Do both batteries lose stored energy? Explain. constant is to be tripled and you have a 1.0-mF capacitor,
25. ● ● Using Kirchhoff’s rules, find the current in each by how much should the resistance change?
resistor in 䉲 Fig. 18.35. 33. ● ● How many time constants will it take for a charged
R 1 = 10 Ω capacitor to be discharged to one-fourth of its initial

䉳 FIGURE 18.35
stored energy?
Single-loop circuit
V 1 = 20 V V 2 = 10 V See Exercise 25. 34. ● ● A 1.00-mF capacitor, initially charged to 12 V, dis-
charges when it is connected in series with a resistor.

(a) What resistance is necessary to cause the capacitor to
R 2 = 20 Ω
have only 37% of its initial charge 1.50 s after starting?
26. ● ● Apply Kirchhoff’s rules to the circuit in 䉲 Fig. 18.36, (b) What is the voltage across the capacitor at t = 3t if
and find (a) the current in each resistor and (b) the rate at the capacitor is instead charged by the same battery
which energy is being delivered to the 8.0-Æ resistor. through the same resistor?
35. ● ● A 3.00-mF capacitor, initially charged to 24 V, dis-
䉳 FIGURE 18.36 charges when it is connected in series with a resistor.
R 1 = 4.0 Ω
A loop in a loop See (a) How much energy does this capacitor store when
V 1 = 12 V R 3 = 2.0 Ω
R 2 = 6.0 Ω
Exercise 26. fully charged? (b) What is the capacitor’s voltage when it
has only half of its maximum energy? Is it 12 V? Why or
R 5 = 2.0 Ω V 2 = 6.0 V
R 4 = 8.0 Ω why not? (c) What resistance is necessary to cause the
capacitor to have only 50% of its energy left after 0.50 s
27. ● ● ● Find the current in each resistor in the circuit shown
of discharge? (d) What is the current in the resistor at
in 䉲 Fig. 18.37. this time?
36. ● ● A series RC circuit with C = 40 mF and R = 6.0 Æ has
䉳 FIGURE 18.37 a 24-V source in it. With the capacitor initially uncharged,
R 2 = 4.0 Ω
Double-loop circuit an open switch in the circuit is closed. (a) What is the volt-
R 1 = 4.0 Ω V 2 = 5.0 V See Exercise 27. age across the resistor immediately afterward? (b) What is
V 3 = 5.0 V
V 1 = 10 V the voltage across the capacitor at that time? (c) What is
R 3 = 4.0 Ω
the current in the resistor at that time?
EXERCISES 655
37. ●● (a) For the circuit in Exercise 36, after the switch has simple ohmic resistor. Suppose that the ammeter is con-
been closed for t = 4t, what is the charge on the capaci- nected in series with the resistor (which is connected to
tor? (b) After a long time has passed, what are the volt- an ideal power source with voltage V) and the voltmeter
ages across the capacitor and the resistor? is placed across the resistor only. (a) Sketch this circuit
38. ● ● ● A series RC circuit consisting of a 5.0-MÆ resistor (with instruments connected) and use it to explain why
and a 0.40-mF capacitor is connected to a 12-V battery. If V
the correct resistance is not given by R = , where V is
the capacitor is initially uncharged, (a) what is the I
change in voltage across it between t = 2t and t = 4t? the voltmeter reading and I is the ammeter reading.
(b) By how much does the capacitor’s stored energy (b) Show that the actual resistance of the element is
change in the same time interval? larger than the result in part (a) and is instead given by
39. ● ● ● A 3.0-MÆ resistor is connected in series with an ini- V
I - 1V>RV2
R = , where RV is the resistance of the
tially uncharged 0.28-mF capacitor. This arrangement is
then connected across four 1.5-V batteries (also in series). voltmeter. (c) Show that the result in part (b) reduces to
(a) What is the maximum current in the circuit and when V
R = for an ideal voltmeter.
does it occur? (b) What percentage of the maximum cur- I
rent is in the circuit after 4.0 s? (c) What is the maximum 47. ● ● ● In Exercise 46, suppose instead that the ammeter is
charge on the capacitor and when does it occur? connected in series with the resistor and that the volt-
(d) What percentage of the maximum charge is on the meter is placed across both the ammeter and the resistor.
capacitor after 4.0 s? (e) How much energy is stored in (a) Sketch this circuit (with instruments connected) and
the capacitor after one time constant has elapsed? use it to explain why the correct resistance is not given
V
by R = , where V is the voltmeter reading and I is the
18.4 AMMETERS AND VOLTMETERS I
ammeter reading. (b) Show that the actual resistance of
40. IE ● A galvanometer with a full-scale sensitivity of the element is smaller than the result in part (a) and is
2000 mA has a coil resistance of 100 Æ . It is to be used in instead given by R = 1V>I2 - RA, where RA is the resis-
an ammeter with a full-scale reading of 30 A. (a) Should tance of the ammeter. (c) Show that the result in part (b)
you use (1) a shunt resistor, (2) a zero resistor, or (3) a V
multiplier resistor? Why? (b) What is the necessary resis- reduces to R = for an ideal ammeter.
I
tance for your answer choice in part (a)?
41. IE ● The galvanometer in Exercise 40 is to be used in a
voltmeter with a full-scale reading of 15 V. (a) Should
you use (1) a shunt resistor, (2) a zero resistor, or (3) a 18.5 HOUSEHOLD CIRCUITS AND
multiplier resistor? Why? (b) What is the required resis- ELECTRICAL SAFETY
tance for your answer choice in part (a)? 48. ● Suppose you are using a drill that is incorrectly wired
42. ● A galvanometer with a full-scale sensitivity of 600 mA as in Fig. 18.22a, and you make electrical contact with an
and a coil resistance of 50 Æ is to be used to build an ungrounded metal case. (a) Explain why this is a danger-
ammeter designed to read 5.0 A at full scale. What is the ous situation for you. (b) Estimate the current in you,
required shunt resistance? assuming an overall body resistance of 300 Æ between
43. ● A galvanometer has a coil resistance of 20 Æ . A current your hand and feet. Would this be dangerous? [Hint:
of 200 mA deflects the needle through ten divisions at Check Insight 18.2.]
full scale. What resistance is needed to convert the gal- 49. ● In Exercise 48, suppose instead the case had been
vanometer to a full-scale 10-V voltmeter? properly wired and grounded as shown in Figure 18.23.
44. ● ● An ammeter has a resistance of 1.0 mÆ . Sketch the cir- If the grounding wire had a total resistance of 0.10 Æ .
cuit diagram and find (a) the current in the ammeter and what is the ratio of the current in you to the current in
(b) the voltage drop across a 10-Æ resistor that is in series the ground wire, assuming that the fuse/circuit breaker
with a 6.0-V ideal battery when the ammeter is properly does not “trip.” (b) Show that there is enough current to
connected to that 10-Æ resistor. (Express your answer to “trip” it, thus safely opening the circuit, reducing the
five significant figures to show how the current differs current to zero (almost immediately), and probably sav-
from 0.60 A and the voltage differs from 6.0 V, which are ing you from serious injury.
the expected values when no ammeter is in place.) 50. ●● One day your electric stove does not turn on. You
45. ● ● A voltmeter has a resistance of 30 kÆ . (a) Sketch the decide to check the 240-V outlet to see if it is the prob-
circuit diagram and find the current in a 10-Æ resistor lem. You use two voltmeter probes inserted into the out-
that is in series with a 6.0 V ideal battery when the volt- let slots, but because of cramped conditions, you
meter is properly connected across that 10-Æ resistor. accidentally touch the metal part of both probes, one
(b) Find the voltage across the 10-Æ resistor under the with each hand. (a) How much current is in you during
same conditions. (Express your answer to five significant the time you are touching the probes, assuming that the
figures to show how the current differs from 0.60 A and outlet was actually operating properly and there is a
the voltage differs from 6.0 V, which are the expected resistance of 100 Æ between your hands? (b) Is this
values when no voltmeter is in place.) enough current to be dangerous to you? (c) Is there
46. ● ● ● In principle, when used together, an ammeter and enough current to “trip” the circuit breaker and save the
voltmeter allow for the measurement of the resistance of day? Comment on the relative sizes of your answers to
any circuit element. Let’s assume that that element is a parts (b) and (c).
51. Find the (a) current in, (b) voltage across, and (c) power gen- 55. A battery has three cells connected in series, each with
erated for each resistor shown in the circuit in 䉲 Fig. 18.40. an internal resistance of 0.020 Æ and an emf of 1.50 V.
This battery is connected to a 10.0-Æ resistor. (a) Deter-
䉳 FIGURE 18.40
R 1 = 2.0 Ω mine the voltage across the resistor. (b) How much
Kirchhoff’s rules See
V 1 = 6.0 V Exercise 51. current is in each cell? (c) What is the rate at which heat
V 2 = 6.0 V R 2 = 4.0 Ω is generated in the battery and how does it compare to
the Joule heating rate in the external resistor?
R 3 = 8.0 Ω 56. A 10.0-mF capacitor in a heart defibrillator unit is charged

V 3 = 6.0 V
fully by a 10 000-V power supply. Each capacitor plate is
connected to the chest of a patient by wires and flat “pad-
dles,” one on either side of the heart. The energy stored in
52. Four resistors are connected in a circuit with a 110-V source,
the capacitor is delivered through an RC circuit, where R is
as shown in 䉲 Fig. 18.41. (a) What is the current in each
the resistance of the body between the two paddles. Data
resistor? (b) How much power is delivered to each resistor?
indicate that it takes 75.1 ms for the voltage to drop to 20.0
䉳 FIGURE 18.41 V. (a) Find the time constant.
R2 = Joule heat losses (b) Determine the resistance, R. (c) How much time does it
V = 110 V 25 Ω R 4 = 25 Ω
See Exercise 52. take for the capacitor to lose 90% of its stored energy?
R3 =
R 1 = 100 Ω (d) If the paddles are left in place for many time constants,
50 Ω
how much energy is delivered to the chest/heart area of
53. Nine resistors, each of value R, are connected in a the patient?
“ladder” fashion, as shown in 䉲 Fig. 18.42. (a) What is the 57. During an operation, one of the electrical instruments in
effective resistance of this network between points A and use has its metal case shorted to the 120-V “hot” wire
B? (b) If R = 10 Æ and a 12.0-V battery is connected from that powers it. The attending physician, who is isolated
point A to point B, how much current is in each resistor? from ground because of rubber-soled shoes, inadver-
R R R tently touches the case with his elbow, while simultane-
A
䉳 F I G U R E 1 8 . 4 2 A resis- ously touching the patient’s chest with his opposite
tance ladder See Exercise 53.
hand. The patient, lying on a metal table, is well
R R R
grounded. (a) Draw a schematic circuit diagram of this
B potentially dangerous complete circuit.
R R R (b) If the patient’s head-to-ground resistance is 2200 Æ ,
54. 䉲 Fig. 18.43 shows a schematic circuit of an instrument called what is the minimum resistance for the physician so
a potentiometer, which is a device for determining very accu- that they both feel, at most, a “mild shock”? [Hint: Use
rate emf values of power supplies. It consists of three batter- your diagram to determine how the “resistors” are
ies, an ammeter, several resistors, and a uniform wire that connected.]
can be “tapped” for a specific fraction of its total resistance. 58. An air-filled parallel plate capacitor is being used in an
eo is the emf of a working battery, e1 designates a battery electrical circuits laboratory. The plates are separated by
with a precisely known emf, and e2 designates a battery 1.50 mm and each has a diameter of 15.0 cm. (a) What is
whose emf is unknown. The switch S is thrown toward bat- the capacitance of this plate arrangement? (b) The capac-
tery 1, and the point T (for “tapped”) is moved along the itor is then connected in series to a 100-Æ resistor and a
resistor until the ammeter reads zero. 100-V DC power supply. How long does it take to charge
Let’s call the resistance of this arrangement R1. This pro- the capacitor to 80% of its maximum charge, and what is
cedure is repeated with the switch thrown toward battery 2, that maximum charge? (c) How long does it take to
and the point T is moved to T ¿ until the ammeter again charge the capacitor to 80% of its maximum stored
reads zero. Let’s designate the resistance of this arrange- energy, and what is the maximum stored energy? (d) Are
ment as R2. Show that the unknown emf can be determined the times for part (c) and (d) the same? Explain.
R2
from the following relationship: e2 = e. 59. An air-filled parallel plate capacitor consists of square
R1 1
plates 10.0 cm on a side separated by 2.25 mm. (a) What
Ᏹo
䉳 F I G U R E 1 8 . 4 3 The is the capacitance of this arrangement? (b) The capacitor
potentiometer See Exer- is then connected in series to a 500-Æ resistor and a 25-V
R2 cise 54. DC power supply. What is the time constant of this cir-
R1
cuit? (c) What plate spacing would double the time con-
T T'
stant? (d) To double the time constant by changing the
S area, what would the new length of each side be? (e) To
double the time constant by inserting a dielectric mater-
A
Ᏹ1 Ᏹ2
ial to completely fill the space between plates, what
dielectric constant would the material have to have?
19 Magnetism
19.1 Permanent magnets,
magnetic poles, and magnetic
field direction (658)
■ the pole-force law
■ magnetic field mapping
19.2 Magnetic field strength

and magnetic force (660)
■ the tesla unit
■ right-hand (force) rule
19.3 Applications: charged

particles in magnetic Stator Guidance
magnet
fields (664)
■ mass spectrometer Support Guidance
magnet rail
PHYSICS FACTS
W
19.4 Magnetic forces on
current-carrying wires (667) ✦ The SI unit of current, the coulomb hen thinking about mag-
per second, or ampere, is actually
■ right-hand (force) rule
defined in terms of the magnetic
netism, most people tend
■ torque an current loops field it creates and the magnetic to envision an attraction because it
force that field can exert on
another current. is well known that certain things
19.5 Applications: current- ✦ Nikola Tesla (1856–1943) was a Ser-
can be picked up with a magnet.
carrying wires in magnetic bian-American researcher known
fields (671) for the Tesla coil, which is capable of You have probably encountered
producing high voltages discharges
■ dc motors
and is a common sight at science magnetic latches that hold cabinet
fairs. The SI unit of magnetic field,
the tesla, was named in his honor.
doors shut, or used magnets to
19.6 Electromagnetism: currents
as a magnetic field source (673)
When Westinghouse secured the stick notes to the refrigerator. It is
patent rights to Tesla’s alternating-
■ wires, loops, and solenoids current, designs of electric energy less likely that one thinks of repul-
generation and transmission has
become the primary means of deliv- sion. Yet repulsive magnetic forces
ering electric energy throughout
19.7 Magnetic materials (678) the world.
exist—and they can be just as use-
■ ferromagnets
✦ Pierre Curie, a French scientist, ful as attractive ones.
and magnetic permeability (1859–1906) was a pioneer in widely
varying areas ranging from magnet- In this regard, the chapter-
ism to radioactivity. He discovered
opening photo shows an interest-
*19.8 Geomagnetism: the that ferromagnetic substances lost
earth’s magnetic field (682) their magnetic behavior above a ing example. At first glance, the
certain temperature, now known as
the Earth’s magnetic poles
■
the Curie temperature. vehicle looks like an ordinary train.
■ the aurora
658 19 MAGNETISM
But where are its wheels? In fact, it isn’t a conventional train at all, but a high-speed,
magnetically levitated one. It doesn’t physically touch the rails when moving;
rather, it “floats” above them, supported by repulsive forces produced by powerful
magnets. The advantages are obvious: with no wheels, there is no rolling friction
and no bearings to lubricate—in fact, there are few moving parts of any kind.
Where do these magnetic forces come from? For centuries, the properties of mag-
nets were attributed to the supernatural. The original “natural” magnets were called
lodestone found in the ancient Greek province of Magnesia. Magnetism is now associ-
ated with electricity, because physicists discovered that both are actually different
aspects of a single force: the electromagnetic force. Electromagnetism is used in motors,
generators, radios, and many other familiar applications. In the future, the development
of high-temperature superconductors (Section 17.3) may open the way for the practical
application of many electromagnetic devices now found only in the laboratory.
Although electricity and magnetism are actually just different manifestations
of the same force, it is instructive to consider them individually and then put them
together, so to speak, as electromagnetism. This chapter and the next will investi-
gate magnetism and its intimate relationship to electricity.
19.1 Permanent Magnets, Magnetic Poles, and Magnetic

Field Direction
➥ What must be the magnetic polarity of the end of a bar magnet that is attracted to
the north end of another bar magnet?
➥ Just above the end of a vertical bar magnet, a compass points downward.What is
the magnetic polarity of that end?
➥ At a specific location, how is the spacing between magnetic field lines related to the
field strength?
One of the features of a common bar magnet is that it has two “centers” of force,
called poles, near each end (䉳 Fig. 19.1). To avoid confusion with the plus–minus
designation used for electric charge, these poles are instead labeled as north (N)
and south (S). This terminology stems from the early use of the magnetic compass
to determine direction. The north pole of a magnetic compass needle was histori-
cally defined as the north-seeking pole—that is, the end that points north on the
䉱 F I G U R E 1 9 . 1 Bar magnet The Earth. The other end of the compass needle was labeled as south, or a south pole.
iron filings (acting as little compass By using two bar magnets, the nature of the forces acting between magnetic
needles) indicate the poles, or cen- poles can be determined. Each pole of a bar magnet is attracted to the opposite
ters of force, of a common bar mag- pole of the other magnet and repelled by the same pole of the other magnet. Thus
net. The small red compass’s
direction determines the magnetic
we have the pole–force law, or law of poles (䉲 Fig. 19.2):
field direction and thus the mag- Like magnetic poles repel each other, and unlike magnetic poles attract each other.
netic polarity of each end of the bar
magnet. (See Fig. 19.3.)
䉴 F I G U R E 1 9 . 2 The pole–force
law, or law of poles Like poles (N–N
or S–S) repel, and unlike poles N S N
(N–S) attract. N S S
S N S
Like poles repel Unlike poles attract

19.1 PERMANENT MAGNETS, MAGNETIC POLES, AND MAGNETIC FIELD DIRECTION 659
A sometimes confusing result of this historical definition occurs because the north
pole of a compass needle is attracted to the Earth’s north polar region (that is,
geographic north). Thus that area must be, magnetically speaking, a south magnetic
pole. Because of this directional convention, the Earth’s south magnetic pole is in
the general vicinity of its north geographic pole. (See Section 19.8 for more details
on the geophysics of the Earth’s magnetic field.)
Two opposite magnetic poles, such as those of a bar magnet, form a magnetic
dipole. At first glance, a bar magnet’s field might appear to be the magnetic analog
of the electric dipole. There are, however, fundamental differences between the
two. For example, permanent magnets always have two poles occurring together,
never one by itself. You might think that breaking a bar magnet in half would yield
two isolated poles. However, the resulting pieces of the magnet always turn out to
be two shorter magnets, each with its own set of north and south poles. While a single
isolated magnetic pole (a magnetic monopole) could exist in theory, it has yet to be
found experimentally.
The fact that there is no magnetic analog to electric charge provides a strong hint
about the differences between electric and magnetic fields. For example, the source
of magnetism is electric charge, just like the electric field. However, as will be seen in
Sections 19.6 and 19.7, magnetic fields are produced by electric charges only when
they are in motion, such as electric currents in circuits and orbiting (or spinning)
atomic electrons. The latter is actually the source of the bar magnet’s field.
MAGNETIC FIELD DIRECTION

The historical approach to analyzing a bar magnet’s field was to try to express the
magnetic force between poles in a mathematical form similar to Coulomb’s law for
the electric force (Section 15.3). In fact, Coulomb developed such a law, using mag-
netic pole strengths in place of electric charge. However, this approach is rarely used,
because it does not fit our modern understanding—that is, single magnetic poles do
not exist. Instead, the modern description uses the concept of the magnetic field.
Recall that electric charges produce electric fields, which can be represented by
electric field lines. The electric field (vector) is defined as the force per unit charge at
any location in space, or E = Fe>qo. Similarly, magnetic interactions can be described
B B
B
in terms of the magnetic field, a vector quantity represented by the symbol B. Just as
electric fields exist near electric charges, magnetic fields occur near permanent mag-
nets. The magnetic field pattern surrounding a magnet can be made visible by sprin-
kling iron filings on paper or glass covering it (Fig. 19.1). Because of the magnetic field,
the iron filings become magnetized into little magnets (basically compass needles)
B
and, behaving like compasses would line up in the direction of B.
Because the magnetic field is a vector field, both magnitude (or “strength”) and
direction must be specified. The direction of a magnetic field (usually called a “B
field”) is defined in terms of a compass that has been calibrated (for direction)
using the Earth’s magnetic field:
B
The direction of a magnetic field B at any location is the direction that the north pole
of a compass would point if placed at that location.
This provides a method for mapping a magnetic field by moving a small compass
to various locations in the field. At any location, the compass needle will line up in
the direction of the B field that exists there. If the compass is then moved in the
direction in which its needle (the north end) points, the path of the needle traces
out a magnetic field line, as illustrated in 䉲 Fig. 19.3a. Because the north end of a
compass points away from the north pole of a bar magnet, the field lines of a bar
magnet point away from that pole and point toward its south pole.
In summary, the rules that govern the interpretation of the magnetic field lines
are very similar to those that apply to electric field lines:
The closer together (that is, the denser) the B field lines, the stronger the magnetic
field. At any location, the direction of the magnetic field is tangent to the field line, or,
equivalently, the direction that the north end of a compass points.
660 19 MAGNETISM
䉳 F I G U R E 1 9 . 3 Magnetic
fields (a) Magnetic field lines can
Compass be traced and outlined by using
iron filings or a compass, as
shown in the case of the mag-
netic field due to a bar magnet.
W
S
N The filings behave like tiny com-
N
E
passes and line up with the field.
The closer together the field
P B lines, the stronger the magnetic
field. (b) Iron filing pattern for
the magnetic field between
unlike poles; the field lines con-
S
verge. (c) Iron filing pattern for
the magnetic field between like
poles; the field lines diverge.
(a)
Notice the concentration of iron filings in the pole regions (Figs. 19.3b and
19.3c). This indicates closely spaced field lines and therefore a relatively strong
magnetic field compared with other locations. As an example of field direction,
observe that just outside the middle of the magnet the field is downward, tangent
B
to the field line at point P in the sketch in Fig. 19.3a. The magnitude of B is defined
in terms of the magnetic force exerted on a moving electric charge, which is
discussed in the following section.
DID YOU LEARN?

➥ The south end of a bar magnet is attracted to north end of another bar magnet.
➥ If a compass points towards a bar magnet’s end, that end is a south magnetic pole.
➥ The spacing between magnetic field lines and the field strength are inversely
related.
(b) 19.2 Magnetic Field Strength and Magnetic Force

➥ What must be the direction of the velocity of a charged particle if it experiences no
force while in a magnetic field?
➥ What is the orientation between a charged particle’s velocity and the magnetic field
if the particle is to experience the maximum magnetic force?
➥ If a charged particle moves perpendicularly to a uniform magnetic field, what is the
shape of its trajectory and how does that shape depend on the particle’s speed?
Experiments indicate that the magnetic force on a particle depends partly on that
particle’s electric charge. That is, there is a connection between electrical properties
of objects and how they respond to magnetic fields. The study of these interactions
is called electromagnetism. Consider the following electromagnetic interaction.
Suppose a positively charged particle is moving at a constant velocity as it enters a
uniform magnetic field. For simplicity, it is assumed that the particle’s velocity is
perpendicular to the magnetic field. (A fairly uniform B field exists between the
poles of a “horseshoe” magnet, as shown in 䉴 Fig. 19.4a.) When the charged parti-
cle enters the field, it experiences a magnetic force and is deflected into an upward
(c) curved path, which is actually an arc segment of a circular path (if the B field is
uniform), as shown in Fig. 19.4b.
From the study of circular motion (Section 7.3), for a particle to move in a circu-
lar arc, a centripetal force must act on that particle. Recall that this centripetal
19.2 MAGNETIC FIELD STRENGTH AND MAGNETIC FORCE 661
(“center-seeking”) force is perpendicular to the particle’s velocity. But what pro-

vides this force here? No electric field is present. The gravitational force, besides
being too weak to cause such a deflection, would deflect the particle into a down- N
ward parabolic arc, not an upward circular one. Evidently, the force is a magnetic
one, due to the interaction of the moving charge and the magnetic field. Thus a
magnetic field can exert a force on a moving charged particle. B
From detailed measurements, the magnitude of this force is found to be propor-
tional to the particle’s charge and its speed. When the particle’s velocity 1v
B
2 is per-
pendicular to the magnetic field 1B2, the magnitude of the field, or the field
B
S
strength B, is defined as
(a)
F B
(valid only when v
B = B (19.1)
qv is perpendicular to B)
v
SI unit of magnetic field: newtons per ampere-meter 3N>1A # m2, or tesla 1T24 +
S
B
Physically, the magnitude of B represents the magnetic force exerted on a
charged particle per unit charge (coulomb) and per unit speed 1m>s2. From this defin-
N
ition, the units of B must be N>1C # m>s2 or N>1A # m2, because 1 A = 1 C>s. This
combination of units is named the tesla (T) after Nikola Tesla (1856–1943), an early B
researcher in magnetism, and 1 T = 1 N>1A # m2.
q +
Most everyday magnetic field strengths, such as those from permanent mag- v
nets, are much smaller than 1 T. In such situations, it is common to express mag-
netic field strengths in milliteslas 11 mT = 10-3 T2 or microteslas 11 mT = 10-6 T2.
(b)
A non-SI unit commonly used by geologists and geophysicists, called the gauss
(G), is defined as one ten-thousandth of a tesla 11 G = 10-4 T = 0.1 mT2. For 䉱 F I G U R E 1 9 . 4 Force on a mov-
example, the Earth’s magnetic field is on the order of several tenths of a gauss, or ing charged particle (a) A horseshoe
several hundredths of a millitesla. On the other hand, conventional laboratory magnet, created by bending a per-
manent bar magnet, produces a
magnets can produce fields as high as 3 T, and superconducting magnets up to 25 fairly uniform field between its
T or higher. poles. (b) When a charged particle
Thus, if the magnetic field strength is known, the magnitude of the magnetic enters a magnetic field, the particle
force F on any charged particle moving at any speed can be found by rearranging is acted on by a force whose direc-
Eq. 19.1.* tion is obvious by the deflection of
the particle from its original path.
B
(valid only if v
F = qvB B (19.2)
is perpendicular to B)
In the more general case, a particle’s velocity may not be perpendicular to the
field. Then the magnitude of the force depends on the sine of the angle 1u2
between the velocity vector and the magnetic field vector. In general, the magni-
tude of the magnetic force is
(magnetic force on
F = qvB sin u (19.3)
a charged particle)
and B are parallel 1u = 0°2 or

B B
This means that the magnetic force is zero when v
oppositely directed 1u = 180°2, because sin 0° = sin 180° = 0. The force has maxi-
mum value when these two vectors are perpendicular. With u = 90° 1sin 90° = 12,
this maximum value is F = qvB sin 90° = qvB.
THE RIGHT-HAND FORCE RULE FOR MOVING CHARGES

The direction of the magnetic force on any moving charged particle is determined
by the orientation of the particle’s velocity relative to the magnetic field. Experi-
ment shows that the magnetic force direction is given by the right-hand force rule
(䉲 Fig. 19.5a):
*Strictly speaking, the speeds must be considerably less than the speed of light to avoid relativistic
complications (Chapter 26).
662 19 MAGNETISM
F F F
F
v v
+ +
v + v
+
u u
u B
B B
B
(a) (b) (c) (d)
䉱 F I G U R E 1 9 . 5 Several equivalent right-hand rules for magnetic

B
force direction (a) When the fingers of the right hand B
are pointed in
B
the direction of v and then curled toward the direction of B, the extended thumb B
points in the direction of the force F on a positive
B
charge. (b) The magnetic force is always perpendicular to the plane formed by B and v and thus is always perpendicular to the
B
direction of the particle’s motion.
B
(c) When the extended forefinger of the right hand B
points in the direction of v and the middle fin-
ger points in the direction of B, the extendedB
right thumb points in the direction of F on a positive charge. (d) When the fingers
B
of the
B
right hand are pointed in the direction of B and the thumb in the direction of v , the palm points in the direction of the force F on a
positive charge. (Regardless of the rule you employ, remember to continue to use your right hand but, at the end, reverse the force
direction for a negative charge.)
When the fingers of the right hand are pointed in the direction of a charged particle’s
B
velocity vB and then curled (through the smallest angle) toward the vector B, the
B
extended thumb points in the direction of the magnetic force F that acts on a positive
charge. If the particle has a negative charge, the magnetic force is in the direction
opposite to that of the thumb.
It is sometimes convenient to imagine the fingers of the right hand as physically
B B B B
turning or rotating the vector v into B until v and B are aligned, much like rotating
a right-hand screw thread. Several common physically equivalent alternatives are
shown in Figs. 19.5c and 19.5d.
B B
Notice that the magnetic force is always perpendicular to the plane formed by v and B
(see Fig. 19.5b). Because the force is perpendicular to the particle’s direction of
motion 1v B
2, it cannot do any work on the particle. (This follows from the definition of
work in Section 5.1, with a right angle between the force and displacement,
W = Fd cos 90° = 0.) Therefore, a magnetic field does not change the speed (that is,
kinetic energy) of a particle—only its direction.
For negative charges, start by assuming the charge is positive. Next, determine
the force direction, using the right-hand force rule. Lastly, reverse this direction to
find the actual force direction on the negative charge. To see how this rule is
applied to both charge signs, consider the following Conceptual Example.
CONCEPTUAL EXAMPLE 19.1 Even “Lefties” Use the Right-Hand Rule

In a linear particle accelerator, a beam of protons travels horizontal. If it were, it would deflect the protons down or up.
zontally northward. To deflect the protons eastward with a Thus, choices (b) and (d) can be eliminated. Use the right-
B
uniform magnetic field, which direction should the field hand force rule (assuming a positive charge) to see if B could
point: (a) vertically downward, (b) west, (c) vertically be downward (answer a). You should verify (make a sketch)
upward, or (d) south? that for a downward magnetic field, the force would be to the
west. Hence, the answer must be (c). The magnetic field must
REASONING AND ANSWER. Because the force is perpendicu- point upward to deflect the protons to the east. This answer
B B
lar to the plane formed by v and B, the field cannot be hori- should be verified using the right-hand force rule.
F O L L O W - U P E X E R C I S E . What direction would the particles in this Example deflect if they were electrons moving horizontally
southward in the same B field? (Answers to all Follow-Up Exercises are given in Appendix VI at the back of the book.)
Note: B fields that point into the plane of the page are designated by .
B fields that point out of the plane are indicated by •. Visualize these symbols as
the feathered end and the tip of an arrow, respectively.
19.2 MAGNETIC FIELD STRENGTH AND MAGNETIC FORCE 663
According to the previous discussion, charged particles traveling in uniform y

magnetic fields have circular arcs as their trajectory. To see this in more detail, con-
sider Example 19.2 carefully.
EXAMPLE 19.2 Going Around in Circles: Force on a Moving Charge B

− v
A particle with a charge of - 5.0 * 10-4 C and a mass of 2.0 * 10-9 kg moves at
1.0 * 103 m>s in the + x-direction. It enters a uniform magnetic field of 0.20 T that points
in the + y-direction (see 䉴Fig. 19.6a). (a) Which way will the particle deflect as it enters x
the field? (b) What is the magnitude of the force on the particle when it is in the field?
z
(c) What is the radius of the circular arc that the particle will travel while in the field?
T H I N K I N G I T T H R O U G H . The initial deflection is in the direction of the initial magnetic
(a) Side view
force. Since the particle is negative, care must be taken in applying the right-hand force
rule. A circular arc is expected, as the magnetic force is always perpendicular to the parti- x
cle’s velocity. The magnitude of the magnetic force on a single charge is given by Eq. 19.3.
This force is the only significant force on the electron; therefore, it is also the net force. From v
this, Newton’s second law should enable the determination of the circular orbit radius. F
SOLUTION. Listing the given data:
Given: q = - 5.0 * 10-4 C Find: (a) initial deflection direction
v = 1.0 * 103 m>s 1 +x-direction2 (b) F (magnitude of initial v
m = 2.0 * 10-9 kg magnetic force) F
B = 0.20 T 1 + y-direction2 (c) r (radius of the orbit)
F
(a) By the right-hand force rule, the force on a positive charge would be in the −
v
+ z-direction (see direction of palm, Fig 19.6a). Because the charge is negative, the force
B (out of page)
is actually opposite this; thus, the particle will begin to deflect in the -z-direction. z
(b) The force’s magnitude can be determined from Eq. 19.3. Because only the magni- (b) Top view
tude is of interest, the sign of q can be dropped.
䉱 F I G U R E 1 9 . 6 Path of a charged
F = qvB sin u
particle in a magnetic field (a) A
= 15.0 * 10-4 C211.0 * 103 m>s210.20 T21sin 90°2 = 0.10 N charged particle entering a uniform
magnetic field will be deflected—in
(c) Because the magnetic force is the only force acting on the particle, it is also the net the - z direction by the right-hand
force (Fig. 19.6b). This net force points toward the circle’s center and is centripetal force. rule, because the charge is negative.
Therefore, in describing circular motion, Newton’s second law becomes (b) In the field, the force is always
B
Fc = maB c perpendicular to the particle’s
velocity. The particle moves in a cir-
Now substitute the magnetic force (from Eq. 19.2, because u = 90°) for the net force and cular path if the field is constant and
the expression for centripetal acceleration (a c = v2>r; see Section 7.3) to obtain: the particle enters the field perpen-
dicularly to its direction. (See
mv2 mv Example 19.2.)
qvB = or r =
r qB
Finally, inserting the numerical values,
mv 12.0 * 10-9 kg211.0 * 103 m>s2
r = = 2.0 * 10-2 m = 2.0 cm
15.0 * 10-4 C210.20 T2
=
qB
(You should check that the SI unit for the combination of mv>qB is the meter.)
F O L L O W - U P E X E R C I S E . In this Example, if the particle had been a proton traveling ini-
tially in the + z-direction, (a) in what direction would it be initially deflected? (b) If the
radius of its circular path were 5.0 cm and its speed were 1.0 * 105 m>s, what would be
the magnetic field strength?
DID YOU LEARN?

➥ If the velocity of a charged particle is either in the direction of, or opposite to, a
magnetic field, it will experience no magnetic force.
➥ For maximum magnetic force, the velocity of a charged particle should be
perpendicular to the field.
➥ If a charged particle moves at a right angles to a uniform magnetic field, its
trajectory is a circle whose diameter is linearly related to the particle speed.
664 19 MAGNETISM
19.3 Applications: Charged Par ticles in Magnetic Fields

➥ In a mass spectrometer that uses a constant B field, which ion’s path has the larger
diameter: the carbon dioxide or the nitrogen ion?
➥ In a mass spectrometer that uses a constant B field, which ion’s path has twice the
diameter that a proton would have assuming all have the same speeds: He+ or He2++?
A charged particle moving in a magnetic field can experience a magnetic force. This
force deflects the particle by an amount that may depend on its mass, charge, and
velocity (speed and direction), as well as the field strength. Let’s take a look at the
important role the magnetic force plays in some common appliances, machines, and
䉱 F I G U R E 1 9 . 7 Cathode-ray tube
(CRT) The motion of the deflected instruments.
beam traces a pattern on a fluores-
cent screen.
THE CATHODE-RAY TUBE (CRT): OSCILLOSCOPE
Magnetic deflecting coils
SCREENS, TELEVISION SETS, AND COMPUTER
Cathode (⫺) MONITORS*
Horizontal Vertical
Anode (⫹) The cathode-ray tube (CRT) is a vacuum tube that, until recently,
was commonly used as a display screen for laboratory instruments
such as the oscilloscope (䉳 Fig. 19.7). The basic operation of the oscillo-
scope and the old-fashioned television picture tube is similar, and is
shown in 䉳 Fig. 19.8. Electrons are emitted from a hot metallic fila-
ment and accelerated by a voltage applied between the cathode 1- 2
Electron
gun
and the anode 1+ 2 in an “electron gun” arrangement. In one kind of
Beam of design, these tubes use current-carrying coils to produce a magnetic
electrons field (Section 19.6), which in turn controls the deflection of the elec-
tron beam. As the field strength is quickly varied, the electron beam
Fluorescent scans the fluorescent screen in a fraction of a second. When the elec-
screen trons hit the fluorescent material, they cause the atoms to emit light
䉱 F I G U R E 1 9 . 8 Television tube
(Section 27.4).
An old-type television picture tube
is a cathode-ray (electron) tube, or
CRT. The electrons are accelerated THE VELOCITY SELECTOR AND THE MASS SPECTROMETER
between the cathode and anode and Have you ever thought about how the mass of an atom or molecule is measured?
are then deflected to the proper
location on a fluorescent screen by
Electric and magnetic fields provide a way in a mass spectrometer (“mass spec”
magnetic fields produced by current- for short). Mass spectrometers perform many functions in modern laboratories.
carrying coils. For example, they can be used to track short-lived molecules in studies of the bio-
chemistry of living organisms. They can also determine the structure of large
organic molecules and analyze the composition of complex mixtures, such as a
sample of smog-laden air. In criminal cases, forensic chemists use the mass spec-
trometer to identify traces of materials, such as a streak of paint from a car acci-
dent. In other fields such as archaeology and paleontology, these instruments can
be used to separate atoms to establish the age of ancient rocks and human arti-
facts. In modern hospitals, mass spectrometers are essential for measuring and
maintaining the proper balance of gaseous medications, such as anesthetic gases
administered during an operation.
In actuality, it is the masses of ions, or charged molecules, that are measured in a
mass spectrometer.† Ions with a known charge 1+q2 are produced by removing
electrons from atoms and molecules. At this point, the ions in the beam would
have a distribution of speeds, rather than a single speed. If these entered the
*Vacuum tube displays, for the most part, have been replaced by flat-screen displays that employ
materials such as liquid crystals (LCD), especially in TVs and computer monitors. These do not use
magnetic forces in their operation.
†
Recall that removing electrons from or adding electrons to an atom or molecule produces an ion.
However, an ion’s mass is negligibly different from that of its neutral atom, because the electron’s mass
is very small compared to the masses of the protons and neutrons in the atomic nuclei.
19.3 APPLICATIONS: CHARGED PARTICLES IN MAGNETIC FIELDS 665
Velocity selector B2 (out of page) 䉳 F I G U R E 1 9 . 9 Principle of the

d mass spectrometer Ions pass
+ + + + through the velocity selector; only
Slits
those with a particular velocity
E 1v = E>B12 then enter a magnetic
q + field (B2). These ions are deflected,
v with the radius of the circular path
depending on the mass and charge
of the ion. Paths of two different
– – – – r radii indicate that the beam contains
ions of two different masses (assum-
B1 ing that they have the same charge).
(into page)
Detector
Top view
mass spectrometer, then ions with different speeds would take different paths in
the mass spectrometer. Thus, before the ions enter the spectrometer, a specific
ion velocity must be selected for analysis. This can be accomplished by using a
velocity selector. This instrument consists of an electric field and a magnetic field
at right angles.
This arrangement allows particles traveling only at a unique velocity to pass
through undeflected. To see this, consider a positive ion approaching the crossed-
field arrangement at right angles to both fields. The electric field produces a
downward force 1Fe = qE2, and the magnetic field produces an upward force
1Fm = qvB12. (You should verify each force’s direction in 䉱 Fig. 19.9.)
If the beam is not to be deflected, the net force on each ion must be zero. In
other words, these two forces must cancel. Therefore, they must be equal in mag-
nitude and oppositely directed. Equating the two force magnitudes,
Fe = Fm or qE = qvB1
which can be solved for the “selected” speed:
E
v = (19.4a)
B1
If the plates are parallel, the electric field between them is given by E = V>d,
where V is the voltage across the plates and d is the distance between them. Thus a
more practical version of the previous equation is
V (speed selected by
v = (19.4b)
B1 d a velocity selector)
The desired speed is usually selected by varying the plate voltage V.

Once through the velocity selector, the beam passes through a slit into another
magnetic field 1B22 that is perpendicular to the beam direction. At this point, the
B
ions are bent into a circular arc. The analysis is identical to that in Example 19.2.
Therefore,
v2
Fc = ma c or qvB2 = m
r
Using Eq. 19.4, the mass of the particle is given by
qdB1 B2 (mass determined with
m = ¢ ≤r (19.5)
V a mass spectrometer)
666 19 MAGNETISM
The quantity inside the parentheses is a constant (assuming all ions have the same
charge). Hence, the larger the mass of an ion, the larger the radius of its circular
path. Two paths of different radii are shown in Fig. 19.9. This indicates that the
beam actually contains ions of two different masses. If the radius is measured (say,
by recording the position where the ions hit a detector), the ion mass can be deter-
mined using Eq. 19.5.
In a mass spectrometer of slightly different design, the detector is kept at a fixed
position. This design employs a time-varying magnetic field (B2), and a computer
records and stores the detector reading as a function of time. In this design, m is
proportional to B2. To see this, rewrite Eq. 19.5 as m = 1qdB1 r>V2B2. Because the
quantity in the parentheses is a constant, then m r B2. Thus as B2 is varied, the
detector data in connection with a high-speed computer enable us to determine
䉱 F I G U R E 1 9 . 1 0 Mass spectrom- the masses and relative number (that is, the percentage) of ions of each mass.
eter Display of a mass spectrome-
Regardless of design, the result—called a mass spectrum (the number of ions
ter, with the number of molecules
plotted vertically and the molecular plotted against their mass)—is typically displayed on an oscilloscope or computer
mass horizontally. Such patterns, screen and digitized for storage and analysis (䉳 Fig. 19.10). Consider the following
mass spectra, help determine the Example of a mass spectrometer arrangement.
composition and structure of mole-
cules. The mass spectrometer can
also be used to identify tiny
amounts of a molecule in a complex EXAMPLE 19.3 The Mass of a Molecule: A Mass Spectrometer
mixture.
One electron is removed from a methane molecule before it enters the mass spectrome-
ter in Fig. 19.9. After passing through the velocity selector, the ion has a speed of
1.00 * 103 m>s. It then enters the main magnetic field region, in which the field
strength is 6.70 * 10-3 T. From there it follows a circular path and lands 5.00 cm from
the field entrance. Determine the mass of this molecule. (Neglect the mass of the elec-
tron that is removed.)
T H I N K I N G I T T H R O U G H . The centripetal force is provided by the magnetic force on the

ion. Since one electron has been removed, the ion’s charge is + e. Because the velocity
and magnetic field are at right angles, the magnetic force is given by Eq. 19.2. By apply-
ing Newton’s second law to circular motion, the molecule’s mass can be determined.
SOLUTION. First, list the given data.

Given: q = + e = 1.60 * 10-19 C Find: m (mass of a methane molecule)
r = 15.00 cm2>2 = 0.0250 m
B2 = 6.70 * 10-3 T
v = 1.00 * 103 m>s
The centripetal force on the ion (Fc = mv2>r) is provided by the magnetic force
(Fm = qvB2):
mv2
= qvB2
r
Solving this equation for m and putting in the numerical values:
qB2 r 11.60 * 10-19 C216.70 * 10-3 T210.0250 m2
m = = = 2.68 * 10-26 kg
v 1.00 * 103 m>s
F O L L O W - U P E X E R C I S E . In this Example, if the magnetic field between the velocity

selector’s parallel plates, which are 10.0 mm apart, is 5.00 * 10-2 T, what voltage must
be applied to the plates?
SILENT PROPULSION: MAGNETOHYDRODYNAMICS

In a search for quiet and efficient methods of propulsion at sea, engineers have
invented a system based on magnetohydrodynamics—the study of the interactions of
moving fluids and magnetic fields. This method of propulsion relies on the magnetic
force and does not require the moving parts, such as motors, bearings, and shafts,
that are common to most ships and submarines. To better avoid detection, this
“silent-running” feature is of particular importance to modern submarine design.
19.4 MAGNETIC FORCES ON CURRENT-CARRYING WIRES 667
Basically, seawater enters the front of the unit and is accelerated and expelled at
high speeds out the rear (䉴 Fig. 19.11). A superconducting electromagnet (see
Section 19.7) is used to produce a large magnetic field. At the same time, an elec-
tric generator produces a large dc voltage, sending a current through the seawater.
[Seawater is a good conductor because it has a large concentration of sodium
1Na +2 and chlorine 1Cl -2 ions. Fig. 19.11 shows only what happens to the Na + B
ions; you should show that the Cl - ions are also pushed rearward. ] The magnetic
F
force on the electric current pushes the water backward, expelling it as a jet of
water. By Newton’s third law, a reaction force acts forward on the submarine, –
Ejected
enabling it to accelerate silently. + sea-
E +
+ water
DID YOU LEARN? +
➥ In a constant B-field mass spectrometer, the diameter of the ion’s path depends
linearly on its mass. Incoming
➥ In a constant B field mass spectrometer, the diameter of the ion’s path is inversely seawater
related to its charge.
䉱 F I G U R E 1 9 . 1 1 Propulsion via
magnetohydrodynamics In magne-
19.4 Magnetic Forces on Current-Carrying Wires tohydrodynamic propulsion, sea-
water has an electric current passed
LEARNING PATH QUESTIONS through it by a dc voltage. A mag-
➥ A long, straight, current-carrying wire is placed in a uniform magnetic field that netic field exerts a force on the cur-
rent, pushing the water out of the
points vertically upward.The wire has no magnetic force on it. In what direction(s)
submarine or boat. The reaction
must the current point? force pushes the vessel in the oppo-
➥ A long, straight, current-carrying wire is in a magnetic field and has the maximum site direction (shown here only for
magnetic force exerted on it.What is the direction of the wire’s current relative to the positive ions).
that of the B field?
➥ A rectangular current-carrying wire loop is in the horizontal plane. If it is placed in
a uniform magnetic field and is to have no torque on it, which way should that
field point?
A charged particle moving in a magnetic field will generally experience a

magnetic force. Because an electric current is composed of moving F F
charges, it should be expected that a current-carrying wire, when placed B
in a magnetic field, would experience such a force as well. The sum of the
individual magnetic forces on the charges that make up the current gives I I
the magnetic force on the wire. q v q v Current
Recall that the direction of the “conventional current” assumes that elec-
tric current in a wire is due to the motion of positive charges, as depicted in
䉴Fig. 19.12.* In the orientation depicted, the magnetic force is a maximum,
L = vt
because u = 90°. In a time t, any one charge qi would move on the average a
length L = vt, where v is the average drift speed (the same for all moving 䉱 F I G U R E 1 9 . 1 2 Force on a wire
charges). All the moving charges (total charge = g qi) in this length of wire are segment Magnetic fields exert
acted on by a magnetic force in the same direction. Thus the total force on this length forces on wires that carry currents,
because electric current is composed
of wire can be determined by summing Eq. 19.2 over all the charges. Assuming the
of moving charged particles. The
field to be uniform, both the field (B) and the drift velocity (v) can be factored out of maximum magnetic force on a
the sum, yielding, current-carrying wire is shown,
F = g 1qi vB2 = 1g qi2vB

because the angle between the
charge velocity and the field is 90°.
Substituting L>t for v and rearranging gives

g qi
F = 1g qi2a b B = ¢
L
≤ LB
t t
*Use the right-hand force rule to convince yourself that electrons traveling to the left would give the
same direction of magnetic force.
668 19 MAGNETISM
䉴 F I G U R E 1 9 . 1 3 A right-hand F
force rule for current-carrying wires F
The direction of the force is given I
I B B
by pointing the fingers of the right
B
hand in the direction of the con- B
ventional current BI and then curl- N S N S
ing them toward B. The extended I
thumb points in the direction of F.
B
F
The force is (a) upward and then F
(b) downward if the current is
reversed. I
(a) (b)
But g qi>t is, by definition, the current in the wire (I). Therefore, this result can be
rewritten in terms of the current as
(valid only if current and
F = ILB (19.6)
magnetic field are perpendicular)
This result is the maximum force on the wire. If the current makes an angle u with
respect to the field direction, then the force on the wire will be less. In general, the
force on a length of current-carrying wire in a uniform magnetic field is
(magnetic force on a
F = ILB sin u (19.7)
curent-carrying wire)
As is expected, if the current is parallel to or directly opposite to the field, then the
F force on the wire is zero.
The direction of the magnetic force on a current-carrying wire is also given by a
right-hand rule. As was the case for individual charged particles, there are several
equivalent versions of the right-hand force rule for a current-carrying wire, the
I most common being:
When the fingers of the right hand are pointed in the direction of the (conventional)
B
current I and then curled toward the vector B, the extended thumb points in the
direction of the magnetic force on the wire.
B
This version is illustarted in 䉱 Figs. 19.13a and 19.13b. An equivalent alternative
䉱 F I G U R E 1 9 . 1 4 An alternative
right-hand force rule When the fin-
technique is shown in 䉳 Fig. 19.14.
gers of the right hand are extendedBin When the fingers of the right hand are extended in the direction of the magnetic field B,
B
the direction of the magnetic field B

and the thumb is pointed in the direc-
and the thumb points in the direction of the (conventional) current I carried by the wire,
tion of the conventional current I, the palm of the right hand points in the direction of the magnetic force on the wire.
then the palm points in the direction
B
of F. You should check that this gives Both techniques give the same direction, because they are extensions of the right-
the same direction as the technique hand force rules for individual charges. To see how current-carrying wires can
used in Fig. 19.13. interact magnetically with the Earth’s field, consider Integrated Example 19.4.
INTEGRATED EXAMPLE 19.4 Magnetic Forces on Wires at the Equator

Because a current-carrying wire is acted on by a magnetic ( A ) C O N C E P T U A L R E A S O N I N G . The magnetic force direction
force, it would seem possible to suspend such a wire at rest must be upward (i.e. away from the Earth’s center), because
above the ground using the Earth’s magnetic field. (a) Assum- gravity acts downward (䉴 Fig. 19.15). The Earth’s magnetic
ing this could be done, consider long, straight wire located at field at the equator is parallel to the ground and points north.
the equator. What would the current direction have to be to Because the magnetic force is perpendicular to both the cur-
perform such a feat: (1) up, (2) down, (3) east, or (4) west? rent and the field, the current cannot be up or down, which
Draw a sketch to show your reasoning. (b) Calculate the cur- eliminates the first two choices. To decide between east and
rent required to suspend the wire, assuming that the Earth’s west, simply choose one and see if it works (or doesn’t work).
magnetic field is 0.40 G (gauss) at the equator and the wire is Suppose the current is to the west. The right-hand force rule
1.0 m long with a mass of 30 g. shows that the magnetic force acts downward. Because this is
19.4 MAGNETIC FORCES ON CURRENT-CARRYING WIRES 669
incorrect, the only correct answer is (3), east. You should ver- The current and the field are at right angles to each other;
ify that this is correct by using the force right-hand rule hence, the magnetic force is given by Eq. 19.6, and from that,
directly. the current can be determined.
Listing the data and converting to SI units
䉳 FIGURE 19.15 Given: m = 30 g = 3.0 * 10-2 kg Find: I (current
Defying gravity by B = 10.40 G2110-4 T>G2 required to
using a magnetic field? = 4.0 * 10-5 T suspend the
B Near the Earth’s equa-
L = 1.0 m wire)
tor it is theoretically
F
possible to cancel the The wire’s weight is w = mg = 13.0 * 10-2 kg219.8 m>s22
pull of gravity with an = 0.29 N. With the wire suspended at rest, this must equal the
w upward magnetic force
Equator magnitude of the magnetic force, to give a net force of zero on
on a wire. the wire. Thus
w = ILB
Solving for the required current gives
w 0.29 N
I = = 7.4 * 103 A
11.0 m214.0 * 10-5 T2
=
LB
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . The mass of (As always, verify that the units are consistent.) This is a huge
the wire is known, and therefore its weight can be calculated. current, so suspending the wire in this manner is probably
This must be equal and opposite to the magnetic force. not a practical idea.
F O L L O W - U P E X E R C I S E . (a) Using the right-hand force rule, show that the idea of suspending a wire, as in this Example, could not
work at either the south or north magnetic poles of the Earth. (b) In this Example, what would the wire’s mass have to be for it to be
suspended when carrying a more reasonable current of 10 A? Does this seem like a reasonable mass for a 1-m length of wire?
F Axis of rotation
TORQUE ON A CURRENT-CARRYING LOOP F
Another important use of magnetism is to exert forces and torques on current- I

carrying loops, such as the rectangular loop in a uniform field shown in N L
䉴 Fig. 19.16a. (The wires connecting the loop to a voltage source that provides the
S
current are not shown.) Suppose the loop is free to rotate about an axis passing B
F
through opposite sides, as shown. There is no net force or torque due to the forces F
acting on its pivoted sides (the sides through which the rotation axis passes). The
forces on these sides are equal and opposite and in the plane of the loop. Therefore, (a)
they produce no net torque or force. However, the equal and opposite forces on the
two sides of the loop parallel to the axis of rotation, while not exerting a net force, do F
create a net torque (see Section 8.2).
To see how this works, consider Fig. 19.16b, which is a side view of Fig.19.16a.
The magnitude of the magnetic force F on each of the nonpivoted sides (length L) u
r⬜ B
is given by F = ILB. The torque produced by this force (Section 8.2) is t = r⬜ F, u
w/
where r⬜ is the perpendicular distance (lever arm) from the axis of rotation to the 2
line of action of the force. From Fig. 19.16b, r⬜ = 12 w sin u, where w is the width of
the loop and u is the angle between the normal to the loop’s plane and the direc- F
tion of the field. The net torque t is due to the torques from both forces and is their Line of action
sum, or twice one of them. Thus the net torque on the loop is given by
t = 2r⬜ F = 2 A 12 w sin u B F = wF sin u
(b) Side view (pivot side)
䉱 F I G U R E 1 9 . 1 6 Force and
= w1ILB2 sin u torque on a current-carrying pivoted
loop (a) A current-carrying rectan-
Then, because wL is just the area (A) of the loop, the torque on a single pivoted, gular loop oriented in a magnetic
current-carrying loop can be rewritten as field as shown is acted on by a force
on each of its sides. Only the forces
t = IAB sin u (torque on one current-carrying loop) (19.8) on the sides parallel to the axis of
rotation produce a torque that causes
the loop to rotate. (b) A side view
(Although derived for a rectangular loop, Eq. 19.8 is valid for a flat loop of any shows the geometry for determining
shape and area.) the torque. (See text for details.)
670 19 MAGNETISM
m A coil is composed of N individual loops wired in series (where N = 2, 3, Á ).

Thus, for a coil, the torque is N times that on one loop (because each loop carries
the same current). Therefore the torque on a coil is
(torque on current–carrying
t = NIAB sin u (19.9)
I coil of N loops)
B
It is convenient to define the magnitude of a coil’s magnetic moment vector, m , as
m = NIA (magnitude of a coil's magnetic moment) (19.10)
2 2
(a) (SI units of magnetic moment: ampere # meter or A # m )
B
The direction of the magnetic moment vector m is determined by circling the fin-
gers of the right hand in the direction of the (conventional) current in the coil. The
B B
F thumb then points in the direction of m . Thus the vector m is always perpendicu-
m 90° lar to the plane of the coil (䉳 Fig. 19.17a). Equation 19.10 can now be rewritten in
B
terms of the magnetic moment:
F t = mB sin u (19.11)
Normal
The magnetic torque tends to align the magnetic moment vector 1m B
2 with the
(Maximum torque) magnetic field direction. To see this, notice that a loop or coil in a magnetic field is
subject to a torque until sin u = 0 (that is, u = 0°), at which point the forces pro-
ducing the torque are parallel to the plane of the loop (see Fig. 19.17b). This situa-
(b) tion exists when the plane of the loop is perpendicular to the field. If the loop is
started from rest with its magnetic moment making some angle with the field, the
loop will undergo an angular acceleration that will rotate it toward the zero angle
F position.
Rotational inertia will carry it through this equilibrium position (zero angle,
Fig. 19.17c) and on to the other side. On that side, the torque will then begin to
slow the loop, eventually stopping it, and then act to reaccelerate it back toward
equilibrium. In other words, the torque on the loop is restoring and tends to cause
m
B the magnetic moment to oscillate about the field direction, much like a compass
90° needle as it settles down to point north.
EXAMPLE 19.5 Magnetic Torque: Doing the Twist?

F A laboratory technician makes a circular coil out of 100 loops of thin copper wire with a
resistance of 0.50 Æ . The coil diameter is 10 cm and the coil is connected to a 6.0-V bat-
(Zero torque) tery. (a) Determine the magnetic moment (magnitude) of the coil. (b) Determine the
maximum torque (magnitude) on the coil if it were placed between the pole faces of a
magnet where the magnetic field strength was 0.40 T.
(c)
T H I N K I N G I T T H R O U G H . The magnetic moment includes not only the number of loops
䉱 F I G U R E 1 9 . 1 7 Magnetic and the area of the coil, but also the current in the wires. Ohm’s law can be used to find
moment of a current-carrying loop the current. The maximum torque occurs when the angle between the magnetic
(a) A right-hand rule determines the
moment vector and the B field is 90°, as given by Eq. 19.11.
direction of the loop’s magnetic
B
moment vector m . The fingers wrap SOLUTION. Listing the given data with the radius of the circle expressed in SI units:
around the loop in the direction of Given: N = 100 loops Find: (a) m (coil magnetic moment)
the current, and the thumb gives the r = d>2 = 5.0 cm = 5.0 * 10-2 m (b) t (maximum torque on the coil)
B
direction of m . (b) Condition of max-
R = 0.50 Æ
imum torque. (c) Condition of zero
torque. If the loop is free to rotate, V = 6.0 V
the magnetic moment vector of the (a) The magnetic moment is given by Eq. 19.10, so the area and current are needed:
A = pr2 = 13.14215.0 * 10-2 m2 = 7.9 * 10-3 m2
loop will tend to align with the direc- 2
tion of the external magnetic field.
and
V 6.0 V
I = = = 12 A
R 0.50 Æ
Therefore the magnetic moment magnitude is
m = NIA = 11002112 A217.9 * 10-3 m22 = 9.5 A # m2
19.5 APPLICATIONS: CURRENT-CARRYING WIRES IN MAGNETIC FIELDS 671
(b) The magnitude of the maximum torque (using u = 90° in Eq. 19.11) is:
t = mB sin u = 19.5 A # m2210.40 T21sin 90°2 = 3.8 m # N
F O L L O W - U P E X E R C I S E . In this Example, (a) show that if the coil were rotated so that its
magnetic moment vector was at 45° from the B field direction, the torque would not be half
the maximum torque. (b) At what angle would the torque be half the maximum torque?
DID YOU LEARN?

➥ For there to be no magnetic force on a wire, its current must be either in the
direction of the field or opposite that direction.
➥ The maximum magnetic force on a current-carrying wire occurs when the current
and B field are at right angles.
➥ No torque is exerted on a current-carrying loop if the loop’s magnetic moment and
the field direction are either the same or opposite.
19.5 Applications: Current-Carrying Wires in Magnetic Zero current With current
Fields Permanent
LEARNING PATH QUESTIONS magnet
➥ In a dc motor, how do the directions of the magnetic field and of the plane of the
␾
coil compare when the torque on the coil is at a maximum?
➥ What is the function of the split-ring commutator in a dc motor?
➥ In a galvanometer, how is the needle deflection related to the current in its coil?
With the principles of electromagnetic interactions learned so far, the operation of

some familiar applications and instruments can be understood. Consider some of N
S
the examples that follow.
Coil
Cylindrical
THE GALVANOMETER: THE FOUNDATION OF THE AMMETER iron core
AND VOLTMETER Spiral spring
Recall that ammeters and voltmeters use the galvanometer as the heart of their (a)
design.* Now you can understand how the galvanometer works. As 䉴 Fig. 19.18a
shows, a galvanometer consists of a coil of wire loops on an iron core that pivots F Pivot
between the pole faces of a permanent magnet. When a current is in the coil, a
torque is exerted on the coil. A small spring supplies a countertorque, and when
N S
the two torques cancel (equilibrium) a pointer indicates a deflection angle f that is
proportional to the coil’s current.
A problem arises if the galvanometer’s magnetic field is not shaped correctly. If F
the coil rotated from its position of maximum torque 1u = 90°2, the torque would Iron core
lessen, and the pointer deflection f would not then be proportional to the current.
(b)
This problem is avoided by making the pole faces curved and by wrapping the
coil on a cylindrical iron core. The core tends to concentrate the field lines such 䉱 FIGURE 19.18
B
that B is always perpendicular to the nonpivoted side of the coil (Fig. 19.18b). The galvanometer (a) The deflec-
With this design, the deflection angle is proportional to the current through the tion 1f2 of the needle from its zero-
galvanometer 1f r I2, as required. current position is proportional to
the current in the coil. A galvanome-
ter can therefore detect and measure
currents. (b) A magnet with curved
THE DC MOTOR pole faces is used so that the field
In general, an electric motor is a device that converts electrical energy into mechanical lines are always perpendicular to
energy. Such a conversion actually occurs during the movement of a galvanometer the core surface and the torque does
not vary with f.
needle. However, a galvanometer is not considered a motor, because a practical dc
motor must have continuous rotation for continuous energy output.
*Even though most voltmeters and ammeters are now digital, it is useful to understand how the
mechanical versions use magnetic forces to make electrical measurements.
672 19 MAGNETISM
Rotation axis
I
F
Current reverses,
producing
unstable
N equilibrium F
I
F F B
F S F
– F
+ F
F
Contact
I brush F F
F
Split-ring (1) (2) (3) (4) (5)
commutator
(a) (b) Side view of loop, showing clockwise loop rotation sequence
䉱 F I G U R E 1 9 . 1 9 A dc motor (a) A split-ring commutator reverses the polarity and cur-

rent each half-cycle, so the coil rotates continuously. (b) An end view shows the forces on
the coil and its orientation during a half-cycle. [For simplicity, a single loop is depicted, but
the coil actually has many (N) loops.] Notice the current reversal (shown by the dot and
cross notation) between situations (3) and (4).
Because of the restoring nature of the magnetic torque on a pivoted, current-car-

rying coil in a magnetic field, it will not be able to make a complete revolution (see
Fig. 19.16b for an example). To provide continuous rotation of such a coil, the design
must be altered. Somehow the current must be reversed every half-turn so that the
torque-producing forces are reversed, thus producing continual acceleration in the
same direction. This is typically done by using a split-ring commutator, which is an
arrangement of two metal half-rings insulated from each other (䉱 Fig. 19.19a).
Here the ends of the coil’s wire are fixed to the half-rings so they rotate together.
The current is supplied to the coil through the commutator by means of contact
brushes. Then, with one half-ring electrically positive and the other negative, the coil
and ring will start to rotate. When they have gone through half a rotation, the half-
rings come in contact with the opposite brushes. Because their polarity is now
reversed, the current in the coil also reverses. In turn, this action reverses the direc-
tions of the magnetic forces, keeping the torque in the same direction (clockwise in
Fig. 19.19b). Even though the torque is zero at the equilibrium position, the coil is in
unstable equilibrium and has enough rotational momentum to continue through
the equilibrium point, whereupon the torque takes over and angularly accelerates
the coil through the next half-cycle. The process repeats in continuous operation. In
a real motor, the rotating shaft is called the armature.
THE ELECTRONIC BALANCE

Traditional laboratory balances measure mass by balancing the weight of an
unknown mass against a known one. Digital electronic balances (䉴 Fig. 19.20a)
work on a different principle. In one type of design, there is still a suspended
beam with a pan on one end that holds the object to be weighed, but no known
mass is needed. Instead, the balancing downward force is supplied by current-car-
rying coils of wire in the field of a permanent magnet (Fig. 19.20b). The coils move
up and down in the cylindrical gap of the magnet, and the downward force is pro-
portional to the current in the coils. The weight of the object in the pan is deter-
mined from the coil current, which produces a force just sufficient to balance the
beam. From the weight, the balance determines the object’s mass using the local
value for g and the relationship between weight and mass 1m = w>g2.
The current required to produce balance is controlled automatically by photosen-
sors and an electronic feedback loop. When the beam is balanced and horizontal, a
knife-edge obstruction cuts off part of the light from a source that falls on a photo-
19.6 ELECTROMAGNETISM: CURRENTS AS A MAGNETIC FIELD SOURCE 673
䉳 F I G U R E 1 9 . 2 0 Electronic
balance (a) A digital electronic
balance. (b) Diagram of the princi-
ple of an electronic balance. The
Unknown balance force is supplied by electro-
mass magnetism.
Knife edge
Electric
eye
Coils
g
Magnet Digital Amplifier

ammeter
N
(a) (b)
W E
sensitive “electric eye,” the resistance of which depends on the amount of light
S
falling on it. This resistance controls the current that an amplifier sends through the
coil. For example, if the beam tilts so that the knife edge rises and more light strikes
the eye, the current in the coil is increased to counterbalance the tilting. In this man-
ner, the beam is electronically maintained in nearly horizontal equilibrium. Lastly,
the current that keeps the beam in the horizontal position is read out on a digital Switch
ammeter calibrated in mass units (usually grams or milligrams) instead of amperes. open
DID YOU LEARN? + –

➥ For a current-carrying coil to experience maximum torque in a magnetic field, the
magnetic field must be parallel to its plane.
➥ In a dc motor, the split-ring commutator reverses the coil current every half rotation
so that the magnetic torque on the coil continues to rotate it in the same direction.
➥ In a galvanometer, the needle deflection is linearly related to the coil current.
(a) No current
19.6 Electromagnetism: Currents as a Magnetic Field Source

➥ How does the magnetic field strength B vary with the perpendicular distance from a
long, straight, current-carrying wire? Switch
➥ When looking down on the circular plane of a current-carrying loop of wire, the cur- closed
rent is clockwise, what is the B field direction at the loop’s center?
➥ Two parallel wires carry currents. When is the force between them attractive? I + –
Repulsive?
Electric and magnetic phenomena, although apparently quite different, are actu-
ally closely and fundamentally related. As has been seen, the magnetic force on a
particle in a magnetic field depends on the particle’s electric charge. But what is (b) Current
the source of the magnetic field? Danish physicist Hans Christian Oersted discov-
ered the answer in 1820, when he found that electric currents produce magnetic fields. 䉱 F I G U R E 1 9 . 2 1 Electric current
His studies marked the beginnings of the discipline called electromagnetism, and magnetic field (a) With no cur-
rent in the wire, the compass needle
which involves the relationship between electric currents and magnetic fields.
points north. (b) With a current in
In particular, Oersted first noted that an electric current could produce a deflec- the wire, the compass needle is
tion of a compass needle. This property can be demonstrated with an arrangement deflected, indicating the presence of
such as that shown in 䉴 Fig. 19.21. When the circuit is open and there is no current, an additional magnetic field super-
the compass needle points, as usual, in the northerly direction. However, when imposed on that of the Earth. In this
case, the strength of the additional
the switch is closed and there is current in the circuit, the compass needle points in
field is roughly equal in magnitude
a different direction, indicating that an additional magnetic field (due to the cur- to that of the Earth. How can
rent) must be affecting the needle. you tell?
674 19 MAGNETISM
I I
4 B
Fingers
1 d 3
Wire curl in
B circular
Compass Thumb sense 2
points in of field
direction
of current
4
I out
B of page
3
Magnetic 1
field
B
2
View from above
(a)
(b)
䉱 F I G U R E 1 9 . 2 2 Magnetic field around a long, straight, current-carrying wire (a) The

field lines form concentric circles around the wire, as revealed by the pattern of iron filings.
(b) The circular sense of the field lines is given by the right-hand source rule, and the mag-
netic field vector is tangent to the circular field line at any point.
Developing expressions for the magnetic field created by various configura-

tions of current-carrying wires requires mathematics beyond the scope of this
book. Here we present the results for the magnetic fields for several common cur-
rent configurations.
MAGNETIC FIELD NEAR A LONG, STRAIGHT, CURRENT-

CARRYING WIRE
At a perpendicular distance d from a long, straight wire carrying a current I
B
(䉱 Fig. 19.22), the magnitude of B is given by
mo I (magnetic field due

B = (19.12)
2pd to a long, straight wire)
where mo = 4p * 10-7 T # m>A is a proportionality constant called the magnetic

permeability of free space. For long, straight wires, the field lines are closed cir-
B
cles centered on the wire (Fig. 19.22a). Note in Fig. 19.22b that the direction of B
due to a current in a long, straight wire is given by a right-hand source rule:
If a long, straight, current-carrying wire is grasped with the right hand and the
extended thumb points in the direction of the current (I), then the curled fingers indi-
cate the circular sense of the magnetic field lines.
EXAMPLE 19.6 Common Fields: Magnetic Field from a Current-Carrying Wire

The maximum household current in a wire is about 15 A. T H I N K I N G I T T H R O U G H . Eq. 19.12 allows for the magnetic
Assume that this current exists in a long, straight, horizontal field determination for a long, straight wire. The direction of
wire in a west-to-east direction (䉴 Fig. 19.23). What are the the field is given by the right-hand source rule.
magnitude and direction of the magnetic field the current
produces 1.0 cm directly below the wire?
Up SOLUTION.
B
Given: I = 15 A Find: B (magnitude
N d = 1.0 cm = 0.010 m and direction)
From Eq. 19.12, the magnitude of the field 1.0 cm below the
W wire is
14p * 10-7 T # m>A2115 A2
E
I mo I
d B = = = 3.0 * 10-4 T
2pd 2p10.010 m2
B
S By the source rule (Fig. 19.23), the field direction directly
below the wire is north. (Note fingers of the right hand, below
Down the wire, point north.)
䉱 F I G U R E 1 9 . 2 3 Magnetic field Finding the magnitude F O L L O W - U P E X E R C I S E . (a) In this Example, what is the field
and direction of the magnetic field produced by a straight direction 5.0 cm above the wire? (b) What current is needed to
current-carrying wire using one version of the right-hand produce a magnetic field at this new location with one-half
source rule (see Fig. 19.22b). the strength of the field in the Example?
MAGNETIC FIELD AT THE CENTER OF A CIRCULAR

CURRENT-CARRYING WIRE LOOP
At the center of a circular coil of wire consisting of N loops, each of radius r and
each carrying the same current I (䉲 Fig. 19.24a shows one such loop), the magni-
B
tude of B is given by
mo NI (magnetic field at center

B = (19.13a)
2r of circular coil of N loops)
In this case (and all circular current arrangements, such as solenoids, discussed
next), it is convenient to determine the magnetic field direction using the right-
hand source rule that is slightly different from (but equivalent to) the one for
straight wires (see Fig. 19.24b):
If a circular loop of current-carrying wires is grasped with the right hand so the fingers
are curled in the direction of the current, the magnetic field direction inside the circular
area formed by the loop is the direction in which the extended thumb points.
In all cases, the magnetic field lines form closed loops, the direction of which is
B
determined by the right-hand source rule. Recall however, that the direction of B
is tangent to the field line, and therefore depends on the location (Fig. 19.24c).
Notice that the overall field pattern of the loop is geometrically similar to that of a
bar magnet (shown overlaid and shadowy in Fig. 19.24c). More about this later.
B I
r
r P
N S B
B
(a) (b) (c) (d)
䉱 F I G U R E 1 9 . 2 4 Magnetic field due to a circular current-carrying loop (a) Pattern of iron filings for a current-carrying
loop. Note that the magnetic field at the loop’s center is perpendicular to the loop’s plane. (b) The direction of the field in
the area enclosed by the loop is given by a right-hand source rule. With B
the fingers wrapped around the loop in the direc-
tion of the conventional current, the thumb indicates the direction of B in the plane of the loop. (c) The overall magnetic
field of a current-carrying circular loop is similar to that of a bar magnet. (d) The magnetic field on the central axis of a
current-carrying loop.
676 19 MAGNETISM
(a) (b)
䉱 F I G U R E 1 9 . 2 5 Magnetic field
of a solenoid (a) The magnetic field More generally, on the central axis of a circular coil of wire consisting of N loops
of a current-carrying solenoid is
(Fig. 19.24d shows this location P), each of radius r and each carrying a current I,
fairly uniform near the central axis B
of the solenoid, as seen in this pat- the magnitude of B varies with the distance x from the loops’ center according to
tern of iron filings. (b) The direction
of the field in the interior can be mo NIr 2 (magnetic field on central axis
B = (19.13b)
21r + x 2
determined by applying the right- 2 2 3>2 of circular coil of N loops)
hand source rule to any of the loops.
Notice its resemblance to the field
near a bar magnet.
MAGNETIC FIELD IN A CURRENT-CARRYING SOLENOID
A solenoid is constructed by winding a long wire into a coil, or helix, with many cir-
cular loops, as shown in 䉱 Fig. 19.25. If the solenoid’s radius is small compared
with its length (L), the interior magnetic field is parallel to the solenoid’s longitu-
dinal axis and constant in magnitude. Notice how the solenoid’s external field
(Fig. 19.25) resembles that of a permanent bar magnet.
As usual, the direction of the interior field is given by the right-hand source
rule for circular geometry. If the solenoid has N turns and each carries a current I,
the magnitude of the magnetic field near its center is given by
mo NI (magnetic field near

B = (19.14)
L the center of a solenoid)
Notice that the solenoid field depends on how closely packed it is (note the N>L)
or how densely the turns are wound. Thus to quantify this, the linear turn density n is
defined as n = N>L. The units of n are turns per meter (in SI terms, this is just m-1).
Eq. 19.14 is sometimes rewritten in terms of the turn density as B = mo nI.
To see why the solenoid might be well suited for magnetic applications requir-
ing a large and fairly uniform magnetic field, consider Example 19.7.
EXAMPLE 19.7 Wire versus Solenoid: Concentrating the Magnetic Field

A solenoid is 0.30 m long with 300 turns and carries a current SOLUTION.
of 15.0 A. (a) What is the magnitude of the magnetic field near Given: I = 15.0 A Find: (a) B (magnitude of mag-
the center of this solenoid? (b) Compare your result with the N = 300 turns netic field near the
field near the single wire in Example 19.6, which carries the L = 0.30 m solenoid center)
same current, and comment. (b) compare the answer in
part (a) with that from a
T H I N K I N G I T T H R O U G H . The field strength depends on the
long straight wire in
number of turns (N), the solenoid length (L), and the current
Example 19.6
(I). This is a direct application of Eq. 19.14.
(a) From Eq. 19.14,

mo NI 14p * 10-7 T # m>A213002115.0 A2
B = = = 6p * 10-3 T L 18.8 mT
L 0.30 m
(b) This field is more than sixty times larger than the field produced by the wire in Example 19.6. Winding many loops close
together in a helix fashion increases the field while enabling the use of the same current. This is because the solenoid’s field is
the vector sum of the fields from 300 loops, and the individual magnetic field directions are all approximately the same, result-
ing in a larger net field for the same current.
F O L L O W - U P E X E R C I S E . In this Example, if the current were reduced to 1.0 A and the solenoid shortened to 10 cm, how many
turns would be needed to create the same magnetic field?
Integrated Example 19.8 uses most of the aspects of electromagnetism discussed

so far: the production of magnetic fields by electric currents and the resulting forces
those fields can exert on other electric currents. Study the example carefully, espe-
cially the use of the appropriate both types of right-hand rules (source and force).
INTEGRATED EXAMPLE 19.8 Attraction or Repulsion? Magnetic

Fields of, and Forces on, Parallel Wires
Two long, parallel wires carry currents in the same direction, as illustrated in 䉴 Fig. 19.26a. d
(a) Is the magnetic force between these wires (1) attractive or (2) repulsive? Make a sketch
to show how you obtained your result. (b) Wire 1 carries a current of 5.0 A and the current I1 I2
in wire 2 is 10 A. Both have a length of 50 cm, and they are separated by 3.0 mm. Deter-
mine the magnitude of the magnetic field created by each wire at the other wire’s loca-
tion. (c) Determine the magnetic force that each wire exerts on the other.
(A) CONCEPTUAL REASONING. Choose one wire at a time and determine the direction of
the magnetic field it produces at the location of the other wire using the source right-hand
B B
rule. In Fig. 19.26b, the fields from both wires (B1 and B2), at the location of the other wire, are 1 2
B
shown. (Notice that B2 is represented by a longer arrow. Why?)
(a)
Next, use the right-hand force rule on each wire to determine the direction of the
force on it. The result, shown in Fig. 19.26c, is an attractive force on each wire. So (1) is
the correct answer. (In keeping with Newton’s third law, these forces are represented by d
arrows of the same length and oppositely directed. The magnitudes should come out
the same in part (c).) I1 I2
B1
(B) AND (C) QUANTITATIVE REASONING AND SOLUTION. To find the magnitude of the
magnetic field produced by each wire, Eq. 19.12 should be used. Because both fields are
at right angles to the currents in the wires, the magnitude of the force on each wire is a
maximum and is given by ILB. Be careful to use the appropriate field and current, that
is, watch the subscripts carefully. B2
The symbols are shown in Fig. 19.26. Listing the data and converting to SI units: 1 2
Given: I1 = 5.0 A Find: (b) B1 and B2 (the magnitude of the (b)
I2 = 10 A magnetic fields due to each wire)
d = 3.0 mm = 3.0 * 10-3 m (c) F1 and F2 (the magnitude of the
L = 50 cm = 0.50 m magnetic forces on each wire) d
(b) The magnitude of the magnetic field due to wire 1 at the location of wire 2 is I1
14p * 10-7 T # m>A215.0 A2
I2
mo I1 B1
B1 = = = 3.3 * 10-4 T F2 F1
2pd 2p13.0 * 10-3 m2
The magnitude of the magnetic field due to wire 2 at the location of wire 1 is
mo I2 14p * 10-7 T # m>A2110 A2
B2 = = = 6.6 * 10-4 T B2
2pd 2p13.0 * 10-3 m2 1 2
(c) The magnitude of the magnetic force on wire 1 due to the field created by wire 2 is (c)
F1 = I1 LB2 = 15.0 A210.50 m216.6 * 10-4 T2 = 1.7 * 10-3 N
䉱 F I G U R E 1 9 . 2 6 Mutual interac-
The magnitude of the magnetic force on wire 2 due to the field created by wire 1 is tion of parallel current-carrying wires
F2 = I2 LB1 = 110 A210.50 m213.3 * 10-4 T2 = 1.7 * 10-3 N (a) Two parallel wires carry current
in the same direction. (b) Wire 1 cre-
As expected, the forces are the same magnitude, in keeping with Newton’s third law. ates a magnetic field at the site of
(Wire 2 is in a weaker field than wire 1, so how can it experience the same force as wire 1?) wire 2 and vice versa. (c) The wires
F O L L O W - U P E X E R C I S E . (a) In this Example, determine the force direction if the current exert equal but oppoosite (attrac-
in either one of the wires is reversed. (b) If the magnitude of the force on the wires is to tive) forces on each other. (See Inte-
grated Example 19.8 for details.)
be kept the same as in the Example, but the current in each is tripled, how far apart must
the wires be placed?
678 19 MAGNETISM
The magnetic force between parallel wires such as in Example 19.8 provides the
modern basis for the definition of the ampere. The National Institute of Standards
and Technology (NIST) defines the ampere as follows.
One ampere is a current that, if maintained in each of two long parallel wires sepa-
rated by a distance of exactly 1 m in free space, produces a magnetic force between
the wires of exactly 2 * 10-7 N per meter of wire.
This definition was chosen as a universal standard, in part because it is easier to

measure forces than to count electrons (that is, measure charge) over time.
DID YOU LEARN?

➥ The magnetic field strength varies inversely with the perpendicular distance from
long straight current-carrying wire.
➥ For a current-carrying wire loop, the direction of the magnetic field at its center is
determined by the thumb of your right hand after placing your fingers in the
current’s direction.
➥ The force between two parallel wires with currents in the same direction is
attractive. If the currents are opposite each other, the wires repel.
19.7 Magnetic Materials

➥ What characteristic of magnetic domains produces an unmagnetized iron bar magnet?

➥ In a given magnetic domain, what can be said about the electron spin directions?
➥ The magnetic field near the center of an air-filled solenoid is 0.75 mT.With an iron core,
the field strength is 0.75 T.What is the value of the iron’s magnetic permeability?
Why is it that some materials are magnetic or easily magnetized and others are
not? How can a bar magnet create a magnetic field when it carries no obvious cur-
rent? To answer these questions, let’s start with some basics. It is known that an
electric current produces a magnetic field. If the magnetic fields of a bar magnet
and a long solenoid are compared (see Figs. 19.1 and 19.25), it seems logical that
the magnetic field of the bar magnet might be due to internal currents. Perhaps
these “invisible” currents are due to electrons orbiting the atomic nucleus or elec-
tron spin. However, detailed analysis of atomic structure shows that the net mag-
netic field produced by orbital motion is usually zero or very small.
What then is the source of the magnetism produced by magnetic materials?
Modern atomic quantum theory tells us that the permanent type of magnetism,
like that exhibited by an iron bar magnet, is produced by electron spin. Classical
physics likens a spinning electron to the Earth rotating on its axis. However, this
mechanical analog is not actually the case. Electron spin is a quantum mechanical
effect with no direct classical analog. Nonetheless, the picture of spinning elec-
trons creating magnetic fields is useful for qualitative thinking and reasoning. In
effect, each “spinning” electron produces a field similar to a current loop (Fig.
19.24c). This pattern, resembling that from a small bar magnet, enables us to treat
electrons, magnetically speaking, as tiny compass needles.
In multielectron atoms, the electrons usually are arranged in pairs with their
spins oppositely aligned (that is, one with “spin up” and one with “spin down,” in
chemistry parlance). In this case, their magnetic fields will effectively cancel, and
the material cannot be magnetic. Aluminum is such a material.
However, in certain materials, known as ferromagnetic materials, the fields
due to electron spins in individual atoms do not cancel. Thus each atom possesses
a magnetic moment. There is a strong interaction between these neighboring
moments that leads to the formation of regions called magnetic domains. In a
given domain, the electron spin moments are aligned in approximately the same
direction, producing a relatively strong (net) magnetic field within that domain.
Not many ferromagnetic materials occur naturally. The most common are iron,
19.7 MAGNETIC MATERIALS 679
Domains more
closely aligned
with field
Growth at expense
of other domains
(a) No external magnetic field (b) With external magnetic field (c) Resulting bar magnet
䉱 F I G U R E 1 9 . 2 7 Magnetic domains (a) With no external magnetic field, the magnetic

domains of a ferromagnetic material are randomly oriented and the material is unmagnetized.
(b) In an external magnetic field, domains with orientations parallel to the field may grow at the Switch open
expense of other domains, and the orientations of some domains may become more aligned
with the field. (c) As a result, the material becomes magnetized, or exhibits magnetic properties.
Iron filings
nickel, and cobalt. Gadolinium and certain manufactured alloys, such as

neodymium and other rare earth alloys, are also ferromagnetic. Switch closed
In an unmagnetized ferromagnetic material, the domains are randomly oriented
I
and there is no net magnetization (䉱 Fig. 19.27a). But when a ferromagnetic material
B
(such as an iron bar) is placed in an external magnetic field, the domains change
their orientation and size (Fig. 19.27b). Remember the picture of the electron as a (a)
small compass; the electrons begin to “line up” in the external field. As the external
field and the iron bar begin to interact, the iron exhibits the following two effects:
1. Domain boundaries change, and the domains with magnetic orientations in
the direction of the external field grow at the expense of the others.
N
2. The magnetic orientation of some domains may change slightly so as to be
more aligned with the field.
B
Upon removal of the external fields, the iron domains remain more or less aligned
in the original external field direction, thus creating an overall “permanent” mag- S
netic field of their own. Iron sliver
Now you can also understand why an unmagnetized piece of iron is attracted N
to a magnet and why iron filings line up with a magnetic field. Essentially, the (b)
pieces of iron become induced magnets (Fig. 19.27c). Some uses of permanent
magnets and magnetic forces in modern medicine are discussed in Insight 19.1,
The Magnetic Force in Future Medicine.
ELECTROMAGNETS AND MAGNETIC PERMEABILITY

Ferromagnetic materials are used to make electromagnets, usually by wrapping
a wire around an iron core (䉴 Fig. 19.28a). The current in the coil creates a mag-
netic field in the iron, which in turn creates its own field that is typically many
times larger than the coil’s field. The magnetic field itself can be turned on and
off at will by turning the current on and off. When the current is on, it induces
magnetism in ferromagnetic materials (for example, the iron sliver in Fig.
19.28b) and, if the forces are large enough, it can be used to pick up large
(c)
amounts of scrap iron (Fig. 19.28c).
The iron used in an electromagnet is called soft iron. In this type of iron, when 䉱 F I G U R E 1 9 . 2 8 Electromagnet
the external field is removed, the magnetic domains become unaligned and the (a) (top) With no current in the cir-
iron reverts to its demagnetized state. The adjective soft refers not to mechanical cuit, there is no magnetic force.
(bottom) However, with a current in
hardness, but to magnetic properties.
the coil, there is a magnetic field and
When an electromagnet is on (lower drawing in Fig. 19.28a), the iron core is the iron core becomes magnetized.
magnetized and adds to the field of the solenoid. The total field is given by (b) Detail of the lower end of the
electromagnet in part (a). The sliver
mNI (magnetic field at center of iron is attracted to the end of the
B = (19.15) electromagnet. (c) An electromagnet
L of the iron core solenoid)
picking up scrap metal.
680 19 MAGNETISM
INSIGHT 19.1 The Magnetic Force in Future Medicine

From ancient times to the present, humans have sought heal- held in place by external magnetic fields. One application could
ing power in magnetism. Claims that magnetism can heal be the treatment of diabetic foot ulcers—open lesions that are
bunions, eliminate tennis elbow, and cure cancer have been difficult to heal due to diabetes-related circulation problems. The
made, but none have ever been substantiated. However, there wound would be covered by thin, strong magnets in a bandage.
are many magnetic applications in modern medicine, such as An injection of microspheres filled with slow-release drugs, such
the well-known magnetic resonance imaging (MRI) system as an antibiotic, would follow. The magnets would attract the
(see Chapter 28). microspheres to the ulcer site and hold them in place. As the
Many new ideas are on the horizon as well. For example, cer- microspheres would break down over the course of several
tain types of bacteria can create nanometer-sized permanent weeks, they would release the drugs slowly, helping the body
magnets inside themselves (see Insight 19.2, Magnetism in repair the wound. Microspheres filled with radioactive material
Nature). Scientists have proposed raising these bacteria and har- could potentially treat liver, lung, brain, and other deep tumors.
vesting their tiny magnets, which are just small enough to fit Another experimental therapy uses magnetically induced
through a hypodermic needle. These magnets could then be heating (hyperthermic techniques) to target breast cancer. This
attached to drug molecules. By placing a magnetic field near the therapy could be especially important in destroying the
site of interest, the molecules could be attracted and held there. smaller tumors now being found by modern imaging tech-
Using the magnetic force to hold the drug molecules in place niques. For these tumors, fluid magnetite (Fe3O4) would be
would increase their effectiveness and reduce side effects that injected directly into the tumor. For tumors larger than a few
can occur when drugs circulate into other parts of the body. cubic millimeters, the iron particles could be delivered via the
A major problem with this proposal is the need to develop circulatory system if they were first attached to biomolecules
techniques for extracting the tiny bacterial magnets and grow- that specifically target cancer cells. Regardless of the method,
ing them in large enough quantities. Alternative proposals the treatment would be the same.
include creating nano-sized unmagnetized pieces of iron by In the presence of an external alternating magnetic field,
chemical means, attaching them to the drug molecules, and the iron particles would heat up due to induced currents (see
moving them around with magnetic fields in a microscopic Chapter 20 on electromagnetic induction). A rise in tempera-
version of iron filings. Both proposals present dangers, such ture of only a few degrees Celsius above normal body tem-
as the drug molecules attracting each other, thus clumping perature has been shown to kill cancer cells. In theory, this
and potentially blocking blood flow. heating would occur locally only at the tumor sites and be
Instead, perhaps, tiny magnetic microspheres could be filled minimally invasive. Initial experiments have had positive
with drugs or radioactive material and steered to the site and results, and the outlook is promising.
Notice that this equation is identical to that for the magnetic field of an air core
solenoid (Eq. 19.14), except that it contains m instead of mo, the permeability of free
space. Here, M represents the magnetic permeability of the core material, not free
space. The role permeability plays in magnetism is similar to that of the permittiv-
ity e in electricity (Chapter 16). For magnetic materials, the magnetic permeability
is defined in terms of its value in free space; thus,
m = km mo (19.16)
where km is called the relative permeability (dimensionless) and is the magnetic
analog of the dielectric constant k.
The value of km for a vacuum is equal to unity. (Why?) For ferromagnetic mate-
rials, the total magnetic field far exceeds that from the wire wrapping. Thus it fol-
lows that for ferromagnetic materials, m W mo and km W 1. A core of a
ferromagnetic material with a large permeability in an electromagnet can enhance
its field thousands of times compared with an air core. In other words, ferromag-
netic materials have values of km on the order of thousands. To see the effect of fer-
romagnetic materials, refer to Example 19.9.
EXAMPLE 19.9 Magnetic Advantage: Using Ferromagnetic Materials

A laboratory solenoid with 200 turns in a length of 30 cm is lim- T H I N K I N G I T T H R O U G H . The field strength depends upon the
ited to carrying a maximum current of 2.0 A. The scientists in the number of turns (N), the solenoid length (L), the current (I),
lab need an interior magnetic field strength of at least 2.0 T and and the permeability of the core material 1m2. This is a direct
are debating whether they need to employ a ferromagnetic core. application of Eqs. 19.14 and 19.15.
(a) Is their field possible if no material fills its core? (b) If not,
determine the minimum magnetic permeability of the ferromag-
netic material that would comprise the core.
19.7 MAGNETIC MATERIALS 681
SOLUTION.
Given: Imax = 2.0 A Find: (a) B (is 2.0 T possible with no core material?)
N = 200 turns (b) m (magnetic permeability required to attain B = 2.0 T)
L = 0.30 m
(a) From Eq. 19.14, without any core material, the interior field clearly would not be large enough, since:
mo NI 14p * 10-7 T # m>A21200212.0 A2
B = = = 1.7 * 10-3 T = 1.7 mT
L 0.30 m
(b) The required field is about 2.0 T>1.7 * 10-3 T or about 1200 times as strong as the answer to part (a). Thus, because B r m if
everything else is constant, attaining a value of 2.0 T requires a permeability of m Ú 1200 mo or m Ú 1.5 * 10-3 T # m>A.
F O L L O W - U P E X E R C I S E . In this Example, if the scientists found a way for the solenoid to handle up to 5.0 A, what would be the
minimum relative permeability?
An electromagnet’s magnetic field strength depends on its current. Large cur-

rents produce large fields, but this is accompanied by much greater joule heating
(I2R losses) in the wires, which could require water cooling. The problem can be
alleviated by using superconducting wire. Because superconducting materials
have zero resistance, there is no joule heating. This technology currently exists in
laboratories. For commercial use, superconducting magnets are just being
designed and implemented but as yet are largely impractical because of the
energy required for cooling. Remember that superconductors need to be at a cer-
tain low temperature to exist in their superconducting state. If (and when)
near–room temperature superconductors are found, high-strength magnetic fields
will become economical and commonplace in many yet-to-be invented appliances
and equipment.
The type of iron that retains some of its magnetism after being in an external
magnetic field is called hard iron and is used to make so-called permanent mag-
nets. You may have noticed that a paper clip or a screwdriver blade becomes
slightly magnetized after being near a magnet. Permanent magnets are produced
by heating pieces of some ferromagnetic material in an oven and then cooling
them in a strong magnetic field to get the maximum effect. In permanent magnets,
the domains remain largely aligned when the external field is removed.
A permanent magnet is not truly permanent, because its magnetism can be
destroyed. Hitting such a magnet with a hard object or dropping it on the floor
can cause a loss of some domain alignment, reducing the magnet’s magnetic
field. Another loss of magnetism is caused by increasing the random (thermal)
motions of atoms. This also tends to disrupt domain alignment. For example,
one of the worst things you can do to a credit card’s magnetic strip is to leave
the card in a car on a hot day. The increased thermal motion of the electron
spins can partially destroy the magnetic pattern on the strip. Above a certain
critical temperature, called the Curie temperature (or Curie point), domain
coupling is destroyed by these increased thermal oscillations, and a ferromag-
netic material loses its “permanent“ magnetism. This effect was discovered by
the French physicist Pierre Curie* (1859–1906). The Curie temperature for iron
is 770 °C.
Ferromagnetic domain alignment plays an important role in geology and geo-
physics. For example, it is well known that when cooled, lava flows that initially
contain iron above its Curie temperature can retain some magnetism due to the
Earth’s field as it existed when the lava cooled below the Curie temperature and
solidified. Measuring the strength and orientation of older lava flows at various
locations has enabled geophysicists to map the changes in the Earth’s magnetic
field and polarity over time.
Some of the first evidence in support of plate tectonic motion came from mea-
suring the direction of the magnetic polarity of seafloor samples containing iron.
*Pierre Curie was the husband of Madame Curie, a famous name in radioactivity research. See
Chapter 29.
682 19 MAGNETISM
The seafloor near the mid-Atlantic ridge, for example, is composed of lava flows
from underwater volcanoes. These solidified flows were found to exhibit perma-
nent magnetism, but the polarity varied with time (older samples are farther out
from the ridge) as the Earth’s magnetic polarity changed.
DID YOU LEARN?

➥ When magnetic domains are randomly aligned, no net magnetic field is produced.
➥ In a given magnetic domain, the electron spins are, on average, in the same
direction.
➥ A material’s magnetic permeability is the ratio of the magnetic field when it is in
place to the field’s value when it is not in place.
*19.8 Geomagnetism: The Ear th’s Magnetic Field

➥ Which direction would a compass point if it were near the Earth’s south geographic pole?
➥ Why does a compass generally not point to true geographic north?
➥ At what latitudes are cosmic rays most efficiently diverted by the Earth’s
magnetic field?
The magnetic field of the Earth was used for centuries before people had any clues
about its origin. In ancient times, navigators used lodestones or magnetized nee-
dles to locate north. Some other forms of life, including certain bacteria and hom-
ing pigeons, also use the Earth’s magnetic field for navigation. (See Insight 19.2,
Magnetism in Nature.)
An early study of magnetism was carried out by the English scientist Sir
William Gilbert in about 1600. In investigating the magnetic field of a specially cut
lodestone (the name for naturally occurring magnetized rocks) that simulated that
of the Earth, he concluded that the Earth as a whole acts as a magnet. Gilbert
thought that a large body of permanently magnetized material within the Earth
might produce its field.
Rotation axis In fact, the Earth’s external magnetic field, or the geomagnetic field, does have a
Magnetic configuration similar to that which would be produced by a large interior bar
north Geographic magnet with the south pole of the magnet pointing north (䉳 Fig. 19.29). The magni-
north tude of the horizontal component of the Earth’s magnetic field at the magnetic
equator is on the order of 10-5 T (about 0.4 G), and the vertical component at the
geomagnetic poles is about 10-4 T (roughly, 1 G). It has been calculated that for a
S B ferromagnetic material of maximum magnetization to produce this field, it would
have to occupy only about 0.01% of the Earth’s volume.
The idea of a ferromagnet of this size within the Earth may not seem unreason-
able at first, but this cannot be a correct model. It is known that the interior tem-
N
peratures deep inside the Earth are well above the Curie temperatures of iron and
nickel, the ferromagnetic materials believed to be the most abundant in the Earth’s
interior. For example, the Curie temperature for iron is attained at a depth of only
100 km below the Earth’s surface. The temperatures are even higher at greater
depths. So the existence of a permanent internal Earth magnet is not possible.
Knowing that electric currents produce magnetic fields has led scientists to
䉱 F I G U R E 1 9 . 2 9 Geomagnetic speculate that the Earth’s field is associated with motions in the liquid outer core,
field The Earth’s magnetic field is
which, in turn, may be connected in some way with the Earth’s rotation. It is
similar to that of a bar magnet.
However, a permanent solid mag- known, for example, that Jupiter, a planet that is largely gaseous and rotates very
net could not exist within the Earth rapidly, has a magnetic field much larger than that of the Earth. Mercury and
because of the high temperatures Venus have very weak magnetic fields; these planets are more like Earth and
there. The Earth’s magnetic field is rotate relatively slowly.
believed to be associated with
Several theoretical models have been proposed to explain the Earth’s magnetic
motions in the liquid outer core
deep within the planet. field. For example, it has been suggested that the field arises from currents associ-
ated with thermal convection cycles in the liquid outer core caused by heat from
the inner core. But the details of this mechanism are still not clear.
*19.8 GEOMAGNETISM: THE EARTH’S MAGNETIC FIELD 683
Notice that the axis of the Earth’s magnetic field does not coincide with the
planet’s rotational axis, which defines the geographic poles. Hence, the Earth’s
(south) magnetic pole and the geographic North Pole do not coincide (see Fig. 19.29).
INSIGHT 19.2 Magnetism in Nature

For centuries, humans have relied on compasses to provide even when its direction changed. It became apparent that
directional information (Fig. 19.29). Studies indicate that cer- members of this species act like biological magnetic dipoles, or
tain organisms seem to have their own built-in directional biological compasses. Once aligned with the field, they migrate
sensors. For example, some species of bacteria are along magnetic field lines by moving their flagella (whiplike
magnetotactic—that is, able to sense the presence and direction appendages), as shown in Fig. 2.
of the Earth’s magnetic field. What makes these bacteria act as living compasses? Even
In the 1980s, experiments were performed on bacteria com- among known magnetotactic species, “new” bacteria (formed
monly found in bogs, swamps, and ponds.* In a laboratory by cell division) do not initially have this magnetotactic sense.
magnetic field, when a droplet of muddy water was viewed However, if they live in a solution containing a minimum
under a microscope, one species of bacteria always aligned and concentration of iron, they are able to synthesize a chain of
migrated in the direction of the field (Fig. 1)—just as these bac- small magnetic particles (Fig. 2). Oddly enough, these inter-
teria do in their natural environment with the Earth’s field. nal compasses have the same chemical composition as the
Furthermore, when these bacteria died and therefore could no original slivers of naturally occurring ore used as compasses
longer migrate, they maintained their alignment with the field by ancient sailors: magnetite (chemical symbol Fe3O4). The
individual particles in the chain are approximately 50 nm
*See, for example, R. P. Blakemore and R. B. Frankel, “Magnetic across, and the chain of a mature bacterium typically contains
Navigation in Bacteria,” Scientific American, December 1981. We are about twenty such particles, each of which is a single mag-
indebted to Professor Frankel for several interesting discussions of netic domain.
this topic. In essence, these bacteria are passively steered by their
internal compasses. But why is it biologically important for
these bacteria to follow the Earth’s magnetic field? A piece of
F I G U R E 1 Magnetotactic bac-
teria in migration Bacteria in a the puzzle was found while investigators were studying the
drop of muddy water, as same species from Southern Hemisphere waters. These bacte-
viewed under a microscope, ria migrate opposite the direction of the Earth’s field, unlike
align along the direction of the their Northern Hemisphere counterparts.
applied magnetic field (north to Recall that in the Northern Hemisphere the Earth’s mag-
the left) and accumulate at the netic field inclines downward and that the reverse is true in
edge. When the field is the Southern Hemisphere. This discovery led scientists to
reversed, so is the direction of believe that the bacteria are using the field direction for sur-
migration. vival. Oxygen is toxic to them, so they are most likely to sur-
vive in the muddy, oxygen-poor depths of their bog, swamp,
F I G U R E 2 A magnetotactic or pond, and following the Earth’s magnetic field direction
bacterium A freshwater mag- enables them to head that way (Fig. 3). This directional sense
netotactic bacterium, shown in also aids them near the equator. There it does not direct them
an electron micrograph. Two downward but instead keeps them at a constant depth, thus
whiplike appendages, or fla- avoiding an upward migration to the deadly oxygen-rich sur-
gella, are clearly visible, along face waters.
with a chain of magnetite Evidence of magnetic field navigation has been found not
particles.
only in bacteria, but also in such diverse organisms as bees,
butterflies, homing pigeons, and dolphins.
BEarth BEarth
BEarth
(a) Northern Hemisphere (b) Southern Hemisphere (c) Equator
F I G U R E 3 Survival of the fittest? (a) In the Northern Hemisphere, where the Earth’s magnetic field inclines downward, mag-
netotactic bacteria follow the field to the oxygen-poor depths. (b) In the Southern Hemisphere, the Earth’s field is inclined
upward, but the bacteria migrate opposite the field and therefore are also able to head for deep waters, like their Northern
Hemisphere cousins. (c) Near the equator, the bacteria move parallel to the water surface and thus are kept away from shallow,
oxygen-rich (and hence toxic to them) waters.
684 19 MAGNETISM
N
u Magnetic The magnetic pole is about a thousand kilometers south of the geographic
north North Pole (true north). The Earth’s north magnetic pole is displaced even
more from its south geographic pole, meaning that the magnetic axis does
not even pass through the center of the Earth.
W E
A compass indicates the direction of magnetic north, not “true,” or geo-
graphic, north. The angular difference in these two directions is called the
magnetic declination (䉳 Fig. 19.30). The magnetic declination varies with
S location. Knowing these variations has historically been particularly
(a) important for the accurate navigation of airplanes and ships, as you can
imagine. Most recently, with the advent of super-accurate GPSs (Global
Geographic
Positioning Systems), high-tech travelers no longer have to depend on
North Pole compasses.
The Earth’s magnetic field exhibits a variety of fluctuations with time.
The permanent magnetism created in iron-rich rocks as they cooled in the
True Earth’s field has provided much evidence of these fluctuations over long
north
time scales. For example, the Earth’s magnetic poles have switched polar-
Earth's
magnetic ity at various times in the past, most recently about 700 000 years ago.
"South" Pole During a period of reversed polarity, the south magnetic pole is near the
15° south geographic pole—the opposite of today’s polarity. The mechanism
Magnetic
north for this periodic magnetic polarity reversal is not clearly understood and
scientists are still investigating it.
On a much shorter time scale, the magnetic poles tend to “wander,” or
change location. For example, the Earth’s south magnetic pole (in the
north polar region) has recently been moving about 1° in latitude
15°W (roughly 110 km or 70 mi) per decade. Currently it is in the Arctic ocean
15°E moving towards Siberia. This long-term polar drift means that the mag-
10°W netic declination map (Fig. 19.30b) varies with time and must be updated
periodically.
10°E On a still shorter time scale, there are sometimes dramatic daily shifts
5°E 0° 5°W
of magnetic pole location by as much as 80 km (50 mi), followed by a
(b) return to the starting position. These shifts are thought to be caused by
charged particles from the Sun that reach the Earth’s upper atmosphere
䉱 F I G U R E 1 9 . 3 0 Magnetic decli- and set up currents that change the planet’s overall magnetic field.
nation (a) The angular difference Charged particles from the Sun entering the Earth’s magnetic field give rise
between magnetic north and “true,”
to another phenomenon. A charged particle that enters a uniform magnetic
or geographic, north is called the
magnetic declination. (b) The mag- field at an angle that is not perpendicular to the field spirals in a helix
netic declination varies with loca- (䉴Fig. 19.31a). This is because the component of the particle’s velocity parallel
tion and time. The map shows to the field does not change. (Recall that a magnetic field acts only on the
isogonic lines (lines with the same perpendicular component of the velocity.) The motions of charged particles in a
magnetic declination) for the conti-
nonuniform field are quite complex. However, for a bulging field such as that
nental United States. For locations
on the 0° line, magnetic north is in depicted in Fig. 19.31b, the particles spiral back and forth as though in a
the same direction as true (geo- “magnetic bottle.”
graphic) north. On either side of this An analogous phenomenon occurs in the Earth’s magnetic field, giving rise to
line, a compass has an easterly or regions with concentrations of charged particles. Two large donut-shaped regions
westerly variation. For example, on
at altitudes of several thousand kilometers are called the Van Allen radiation belts
a 15°E line, a compass has an east-
erly declination of 15°. (Magnetic (Fig. 19.31c). In the lower Van Allen belt, light emissions called auroras occur—the
north is 15° east of true north.) aurora borealis, or northern lights, in the Northern Hemisphere and the aurora
australis, or southern lights, in the Southern Hemisphere. These eerie, flickering
lights are most commonly observed in the Earth’s polar regions, but have been
seen at lower latitudes (䉴 Fig. 19.32).
An aurora is created when charged solar particles become trapped in the
Earth’s magnetic field. Maximum aurora activity occurs after a solar disturbance,
such as a solar flare—a violent magnetic storm on the Sun that spews out enor-
mous quantities of charged particles. Trapped in the Earth’s magnetic field, these
particles are guided toward the polar regions, where they excite or ionize oxygen
and nitrogen atoms in the atmosphere. When the excited atoms return to their nor-
mal state and the ions regain their normal number of electrons, light is emitted
(Section 27.4), producing the glow of the aurora.
*19.8 GEOMAGNETISM: THE EARTH’S MAGNETIC FIELD 685
y
B
+
B B Protons
+q
z
x
Electrons
(a) (b)
(c)
䉱 F I G U R E 1 9 . 3 1 Magnetic confinement (a) A charged particle entering a uniform mag-
netic field at an angle other than 90° moves in a spiraling path. (b) In a nonuniform,
bulging magnetic field, particles spiral back and forth as though confined in a magnetic
bottle. (c) Charged particles are trapped in the Earth’s magnetic field, and the regions
where they are concentrated are called Van Allen belts.
DID YOU LEARN?

➥ At the Earth’s polar regions, compasses point perpendicular to the ground with their
directions determined by the magnetic polarity of that region.
➥ A compass points to the magnetic pole which does not usually coincide with the
geographic pole.
➥ The most efficient latitudes for cosmic ray deflection by the Earth’s magnetic field are
where the particles enter at right angles to the field, that is, in the equatorial region.
PULLING IT TOGETHER Electric Currents, Magnetic Fields, Torques,

and Checking of SI Units
A tightly wrapped cylindrical solenoid has a 5.00-cm diameter and a 25.0-cm length and
consists of 1000 circular wire windings with a total resistance of 2.00 Æ . It forms a complete
circuit with a 12.0-V battery. Near the center of this solenoid there is a small circular current-
carrying coil (radius 0.600 cm) that has a current of 1.25 A and 25 concentric loops. The coil
is oriented such that the solenoid’s magnetic field is parallel to its area. (a) How much cur- 䉱 F I G U R E 1 9 . 3 2 Aurora borealis:
rent does this solenoid have? (b) What is the magnetic field strength near the center of this the northern lights This spectacular
solenoid? (c) How much torque is exerted on the small coil by the solenoid’s field? display is caused by energetic solar
particles trapped in the Earth’s
T H I N K I N G I T T H R O U G H . This exercise includes solenoids, Ohm’s law, and torques on magnetic field. The particles excite
coils. (a) The relationship between voltage, current, and resistance from Eq. 17.2a enables or ionize air atoms; on de-excitation
determination of the solenoid current. Note that since it is the only element connected to (or recombination) of the atoms,
the battery, the battery voltage is the same as that across the solenoid. (b) The solenoid’s light is emitted.
magnetic field depends on its total number of turns, length, and current. Thus the field
can be calculated directly from the data. (c) From the description, the angle between the
small coil’s magnetic moment and the solenoid’s field must be 90°. (How do you know
this?) The torque exerted on the coil depends on this angle, the solenoid’s magnetic field,
which is known from part (b), and the coil’s magnetic moment. This moment depends on
the coil current and area, both of which are known.
SOLUTION. Listing the data given and converting to SI units where needed,
Given: D = 5.00 cm Find: (a) Is (current in solenoid)
L = 25.0 cm = 0.250 m (b) B (solenoid’s magnetic
Rs = 2.00 Æ field strength)
Ns = 1000 turns (c) t (torque exerted on
Vs = 12.0 V small coil)
r = 0.600 cm = 6.00 * 10-3 m
Ic = 1.25 A
Nc = 25 loops (continued on next page)
686 19 MAGNETISM
(a) The solenoid’s current is determined by the voltage across it and its resistance;
Vs 12.0 V
Is = = = 6.00 A
Rs 2.00 Æ
(b) The solenoid’s magnetic field can then be found from the solenoid field expression:
11000 turns216.00 A2
= 14p * 10-7 T # m>A2
Ns Is
B = mo = 3.02 * 10-2 T
L 0.250 m
(c) The torque on a current-carrying coil of N loops in a uniform magnetic field is
given by Eq. 19.11. The coil’s area is a circle, and therefore its magnetic moment m
(magnitude) is
m = Nc Ic A = Nc Ic 1pr22
= 125 loops211.25 A231p216.00 * 10-3 m2 4 = 3.53 * 10-3 A # m2
2
Since the coil’s area is not penetrated by the solenoid’s magnetic field, the normal to the
coil’s area must be perpendicular 190°2 to the solenoid’s magnetic field. Therefore in
Eq. 19.11, the torque on the coil is a maximum. The magnitude of this torque is
t = mB sin u
= 13.53 * 10-3A # m2213.02 * 10-2 T21sin 90°2
= 1.07 * 10-4 T # A # m2
Let’s check to see if this combination is equivalent to the SI unit of torque. Since
1 T = 1 # , 1 T # A # m2 = a 1 # b11 A # m22 = 1 m # N, which is the SI unit of
N N
A m A m
torque (Section 8.2).
■ The pole–force law, or law of poles: opposite magnetic ■ A series of N current-carrying circular loops, each with a
poles attract and like poles repel. plane area A and carrying a current I, can experience a
magnetic torque when placed in a magnetic field. The mag-
nitude of the torque on such an arrangement is
N S N
S
N
N
S
S
S t = NIAB sin u (19.9)
F Axis of rotation
Like poles repel Unlike poles attract
The magnetic field 1B2 has SI units of the tesla (T), where
F
B
■
I
1 T = 1 N>1A # m2. Magnetic fields can exert forces on mov- N L
S
ing charged particles and electric currents. The magnitude B
of the magnetic force on a charged particle is F
F
F = qvB sin u (19.3) ■ The magnitude of the magnetic field produced by a long,
straight wire is
The magnitude of the magnetic force on a current-carrying
mo I
wire is B = (19.12)
2pd
F = ILB sin u (19.7)
■ Right-hand force rules determine the direction of a
magnetic force on moving charged particles and current-
carrying wires.
v + I
where mo = 4p * 10-7 T # m>A is the magnetic permeability
u
B of free space. For long, straight wires, the field lines are
B closed circles centered on the wire.
■ The magnitude of the magnetic field produced by an arrange- ■ Right-hand source rules are used to determine the direction
ment of N circular current-carrying loops of radius r is of the magnetic field from various current configurations.
I
mo NI (magnetic field at the center I

B = (19.13a)
2r of circular coil of N loops) Fingers
4 B
1 d 3
curl in
B circular
2 Thumb sense 2
mo NIr (magnetic field on central axis points in of field
B = (19.13b) direction
21r + x 2
of current
2 2 3>2 of circular coil of N loops) 4
1
B
2
r View from above
B
■ In ferromagnetic materials, the electron spins align, creat-
I ing domains. When an external field is applied, the effect is
to increase the size of those domains that already point in
the direction of the field at the expense of the others. When
the external magnetic field is removed, a permanent mag-
■ The magnitude of the magnetic field produced near the center net remains.
of the interior of a solenoid with N windings and a length L is
mo NI
B = (19.14) (a) No external magnetic field
L Domains more
closely aligned
with field
B Growth at expense
of other domains
I
(b) With external magnetic field
(c) Resulting bar magnet

19.1 PERMANENT MAGNETS, MAGNETIC 19.2 MAGNETIC FIELD STRENGTH AND

POLES, AND MAGNETIC FIELD MAGNETIC FORCE
DIRECTION 6. A proton moves vertically upward and perpendicular to
1. When the ends of two bar magnets are near each other, a uniform magnetic field. It deflects to the left as you
they are found to attract. The ends must be (a) one north, watch it. What is the magnetic field direction: (a) directly
the other south, (b) both north, (c) both south, (d) either away from you, (b) directly toward you, (c) to the right,
(a) or (b). or (d) to the left?
2. A compass is placed just off the end of a permanent bar 7. An electron is moving horizontally to the east in a uniform
magnet, and points away from that end of the magnet. It magnetic field that is vertical. It is found to deflect north.
can be concluded that this end of the magnet (a) acts as a What direction is the magnetic field: (a) up, (b) down, or
north magnetic pole, (b) acts as a south magnetic pole, (c) the direction can’t be determined from the given data?
(c) you can’t conclude anything about the magnetic 8. If a negatively charged particle were moving upward
properties of the permanent magnet. along the right edge of this page, which way should a
3. If you look directly at the south pole of a bar magnet, its magnetic field (perpendicular to the plane of the paper)
magnetic field points (a) to the right, (b) to the left, be oriented so that the particle would initially be
(c) away from you, (d) toward you. deflected to the left: (a) out of the page, (b) in the plane
4. Which way would a compass point if placed midway of the page, or (c) into the page?
between the ends of the two bar magnets shown in Fig. 9. An electron passes through a magnetic field without being
19.3b: (a) up, (b) down, (c) left, or (d) right? deflected. What can you conclude about the angle between
5. Which way would a compass point if placed just to the the magnetic field direction and that of the electron’s veloc-
right of the midway point between the ends of the two ity, assuming that no other forces act: (a) they could be in
bar magnets shown in Fig. 19.3c: (a) up, (b) down, the same direction, (b) they could be perpendicular, (c)
(c) left, or (d) right? they could be opposite, or (d) both (a) and (c) are possible?
688 19 MAGNETISM
19.3 APPLICATIONS: CHARGED 18. You are looking directly into one end of a long solenoid.
PARTICLES IN MAGNETIC FIELDS The magnetic field at its center points directly away from
you. What is the direction of the current in the solenoid, as
10. In a mass spectrometer two ions with identical charge viewed by you: (a) clockwise, (b) counterclockwise,
and speed are accelerated into two different semicircular (c) directly toward you, or (d) directly away from you?
arcs. Ion A’s arc has a radius of 25.0 cm and ion B’s arc
has a radius of 50.0 cm. What can you say about their rel- 19. A current-carrying loop of wire is in the plane of this
ative masses: (a) mA = mB, (b) mA = 2mB, (c) mA = 12 mB, paper. Outside the loop, its magnetic field points into the
or (d) you can’t say anything given just this data? paper. What is the direction of the current in the loop?
(a) clockwise, (b) counterclockwise, or (c) you can’t tell
11. In a mass spectrometer two ions with identical mass and
from the data given.
speed are accelerated into two different semicircular
arcs. Ion A’s arc has a radius of 25.0 cm and ion B’s arc 20. Consider a current-carrying circular loop of wire. On its
has a radius of 50.0 cm. What can you say about their net central axis (that is, the line perpendicular to the area of
charges: (a) qA = qB, (b) qA = 2qB, (c) qA = 12 qB, or the loop and passing through its center), which location
(d) you can’t say anything given just this data? has the least magnetic field strength: (a) the center of the
12. In the velocity selector shown in Fig. 19.9, which way loop, (b) 10 cm above the center of the loop, or (c) 20 cm
will an ion be deflected if its velocity is less than E>B1: above the center of the loop?
(a) up, (b) down, or (c) there will be no deflection?
19.7 MAGNETIC MATERIALS

19.4 MAGNETIC FORCES ON CURRENT-
21. The main source of magnetism in magnetic materials is
CARRYING WIRES (a) electron orbits, (b) electron spin, (c) magnetic poles,
AND (d) nuclear properties.
19.5 APPLICATIONS: CURRENT-
22. When a ferromagnetic material is placed in an external
CARRYING WIRES IN MAGNETIC FIELDS
magnetic field, (a) the domain orientation may change,
13. A long, straight, horizontal wire located on the equator (b) the domain boundaries may change, (c) new domains
carries a current directed toward the west. What is the are created, (d) both (a) and (b).
direction of the force on the wire due to the Earth’s mag- 23. Heating a permanent magnet can significantly reduce the
netic field: (a) east, (b) west, (c) south, or (d) downward? magnetic field it produces. The reduction in field strength
14. A long, straight, horizontal wire located on the equator by this method is mainly due to (a) domain orientation
carries a current. In what direction should the current be if change, (b) domain boundary change, (c) increase thermal
the purpose is to balance the wire’s weight with the mag- motion of electron spin directions, (d) all of these.
netic force on it: (a) east, (b) west, (c) south, or (d) upward?
15. You are looking horizontally due west directly at the cir-
cular plane of a current-carrying coil. The coil is in a uni- *19.8 GEOMAGNETISM: THE EARTH’S
form vertically upward magnetic field. When released, MAGNETIC FIELD
the top of the coil starts to rotate away from you as the
24. The Earth’s magnetic field (a) has poles that coincide with
bottom rotates toward you. Which direction is the cur-
the geographic poles, (b) only exists at the poles, (c) reverses
rent in the coil: (a) clockwise, (b) counterclockwise, or
polarity every few hundred years, (d) none of these.
(c) you can’t tell from the data given?
16. For the current-carrying loop shown in Fig. 19.16, the maxi- 25. Aurora (see Fig. 19.32) (a) occur only in the Northern Hemi-
mum torque occurs for what value of the angle u: (a) 0°, sphere, (b) are related to the lower Van Allen belt, (c) occur
(b) 90°, (c) 180°, or (d) the torque is the same for all angles? because of Earth’s magnetic pole reversals, (d) happen pre-
dominantly when there are no solar disturbances.
26. If the direction of your compass pointed straight up,
19.6 ELECTROMAGNETISM: CURRENTS where would you be: (a) near the Earth’s north geo-
AS A MAGNETIC FIELD SOURCE graphic pole, (b) near the equator, or (c) near the Earth’s
17. A long, straight wire is parallel to the ground and carries south geographic pole?
a steady current to the west. At a point directly below 27. If a proton was orbiting above the Earth’s equator in the
the wire, what is the direction of the magnetic field the Van Allen belt, which way would it have to be moving:
wire produces: (a) north, (b) east, (c) south, or (d) west? (a) to the west, (b) to the east, or (c) either direction?
19.1 PERMANENT MAGNETS, MAGNETIC that magnetic field lines must leave from the north pole
POLES, AND MAGNETIC FIELD DIRECTION of a permanent bar magnet and enter its south pole.
1. Given two identical iron bars, one of which is a perma- 3. (a) As you start in the very middle of Fig. 19.3b and
nent magnet and the other unmagnetized, how could move horizontally to the right, what happens to the
you tell which is which by using only the two bars? magnetic field spacing as indicated by the iron filing pat-
2. The direction of any magnetic field is taken to be in the tern? What does this imply in terms of the magnetic field
direction that a compass points. Explain why this means strength? (b) What is the direction of the magnetic field
in this region of the figure? Can you tell the direction 19.3 APPLICATIONS: CHARGED
from the iron filing pattern alone? Explain. PARTICLES IN MAGNETIC FIELDS
9. Explain how a nearby magnet can distort the display of
19.2 MAGNETIC FIELD STRENGTH AND an older computer monitor or television picture tube
MAGNETIC FORCE that is based on CRT technology. [Hint: See Fig. 19.8 for a
working diagram of these instruments.]
4. A proton and an electron are moving at the same veloc-
ity perpendicularly to a constant magnetic field. (a) How 10. The enlarged circular inset in Fig. 19.11 shows how the
do the magnitudes and directions of the magnetic forces positive sodium 1Na+2 ions in seawater are accelerated
on them compare? (b) What about the magnitudes of out the rear of the submarine to provide a propulsive
their accelerations? force. But what about the negative chlorine 1Cl -2 ions in
5. If a charged particle moves in a straight line and there the seawater? Because they have charge of the opposite
are no other forces on it except possibly from a magnetic sign, aren’t they accelerated forward, resulting in a net
field, can you say with certainty that no magnetic field is force of zero on the submarine? Explain.
present? Explain. 11. Explain clearly why the speed selected in a velocity
6. Three particles enter the same uniform magnetic field as selector setup such as in Fig. 19.9 does not depend on the
shown in 䉲 Fig. 19.33a. Particles 1 and 3 have equal charges of the ions passing through.
speeds and charges of the same magnitude. What can 12. Redraw the charged particle path in the apparatus dia-
you say about (a) the charges of the particles and grammed in Fig. 19.9 if the ions were negatively charged
(b) their masses? instead of positively charged.
13. (a) Redraw the charged particle path in the apparatus dia-
v B 䉳 FIGURE 19.33 grammed in Fig. 19.9 if the electric field of the velocity selec-
1
v
Charges in motion See tor were reduced in magnitude. (b) Redraw the charged
Conceptual Questions 6 particle path in the apparatus diagrammed in Fig. 19.9 if the
and 7. magnetic field of the velocity selector were reduced in mag-
nitude. In both cases, explain your reasoning.
v
2
19.4 MAGNETIC FORCES ON CURRENT-

CARRYING WIRES
v AND
3 +
v 19.5 APPLICATIONS: CURRENT-
(a) (b) CARRYING WIRES IN MAGNETIC FIELDS
7. You want to deflect a positively charged particle in an
S-shaped path, as shown in Fig. 19.33b, using only mag- 14. Two straight wires are parallel to each other and carry
netic fields. (a) Explain how this could be done by using different currents in the same direction. Do they attract
magnetic fields perpendicular to the plane of the page. or repel each other? How do the magnitudes of these
(b) How does the emerging particle’s kinetic energy forces on each wire compare?
compare with the particle’s initial kinetic energy? 15. Predict what should happen to the length of a metal spring
8. A magnetic field can be used to determine the sign of when a large current passes through it. [Hint: Consider the
charge carriers in a current-carrying wire. Consider a direction of the current in the neighboring spring coils.]
wide conducting strip in a magnetic field oriented as
16. (a) How would you orient a square current loop in a uni-
shown in 䉲 Fig. 19.34.
form magnetic field so that there is no torque on the
The charge carriers are deflected by the magnetic force
loop? (b) How would the orientation change to maxi-
and accumulate on one side of the strip, giving rise to a
mize the torque on the loop? (c) In each case, is there a
measurable voltage across it. (This phenomenon is known
net magnetic force on this loop? Explain.
as the Hall effect.) If the sign of the charge carriers is
unknown (they are either positive charges moving as 17. Explain the operation of the doorbell and door chimes
indicated by the arrows in the figure or negative charges illustrated in 䉲 Fig. 19.35.
moving in the opposite direction), how does polarity or
sign of the measured voltage allow the sign of the charge
Tone bar Tone bar
to be determined? Assume that only one type of charge Bell Solenoid 䉳 FIGURE 19.35
carrier is responsible for the current. core Electromagnetic
B applications Both
Contacts Hammer (a) a doorbell and
q v q v Spring (b) door chimes uti-
+ lize electromagnets.
See Conceptual
q v q v Armature Push
q v Contacts button Question 17.
Spring
䉱 F I G U R E 1 9 . 3 4 The Hall effect See Conceptual Question 8 (a) (b)

690 19 MAGNETISM
18. In a long, straight, current-carrying wire, the electrons 25. Two circular wire loops are coplanar (that is, their areas
are moving to the west. If the wire is in a uniform mag- are in the same plane) and have a common center. The
netic field pointing upward, what is the direction of the outer loop carries a current of 10 A in the clockwise
force on the wire? Answer this from two different view- direction. To create a zero magnetic field at the center of
points: that of the force on the electrons, and then that of the loops, what should be the direction of the current in
the force on the wire considering conventional current the inner loop? Should its current be 10 A, larger than
direction. Are your answers the same? 10 A, or smaller than 10 A?
19. (a) Show that the SI unit for magnetic moment multi-
plied by the SI unit for magnetic field yields the SI unit
for torque. (b) If you are looking down onto the area of a 19.7 MAGNETIC MATERIALS
current-carrying loop of wire and the current is counter- 26. If you are looking down on the orbital plane of the elec-
clockwise, what is the direction of the loop’s magnetic tron in a hydrogen atom and the electron orbits counter-
moment? clockwise, what is the direction of the magnetic field the
20. Suppose a long, straight, current-carrying wire had a electron produces at the proton?
current from west to east. If it were immersed in a verti- 27. What is the purpose of the iron core often used at the
cally upward magnetic field that was stronger on its center of a solenoid?
west side than on its east side, what would be the initial
motion of the wire if released from rest? Explain. 28. Discuss several ways that the magnetic field of a perma-
nent magnet can be destroyed or reversed.
29. A lava flow from the Kilauea Iki volcano on the “Big
19.6 ELECTROMAGNETISM: CURRENTS Island” of Hawaii cools below its Curie temperature as it
moves and finally solidifies. Describe the direction of the
AS A MAGNETIC FIELD SOURCE
remnant magnetism in this lava flow. [Hint: Take a look
21. A circular current-carrying loop is lying flat on a table. A at the direction of the Earth’s magnetic field there.]
calibrated compass, when placed at the center of the
loop, points downward. If you look straight down on the
*19.8 GEOMAGNETISM: THE EARTH’S
loop, what is the direction of the current? Explain your
reasoning.
MAGNETIC FIELD
22. If you doubled your distance from a long current- 30. Determine the direction of the force due to the Earth’s
carrying wire, what changes would need to be made to magnetic field on an electron near the equator when the
the current to keep the magnetic field strength the same electron’s velocity is directed (a) due south, (b) north-
as at the nearer position but reversed direction? west, and (c) upward.
23. Given two solenoids, one with 100 turns and the other 31. In a relatively short time, geologically speaking, data
with 200 turns. If both carry the same current, will the indicate that the Earth’s magnetic field direction will
one with more turns necessarily produce a stronger reverse. After that, what would be the polarity of the
magnetic field at its center? Explain. magnetic pole near the Earth’s geographic North Pole?
24. To minimize the effects of the magnetic field when 32. Based on Fig. 19.30, approximately how far off (in angle
needed, the wires carrying current to equipment or and direction) would your compass direction be from
appliances are placed close together. Explain how this geographic north if you were located in (a) Phoenix,
works to reduce the magnetic field created by the cur- (b) Chicago, and (c) New Orleans?
rent in the wire.
EXERCISES
19.1 PERMANENT MAGNETS, MAGNETIC in and oriented horizontally so that its north end is closest
POLES, AND MAGNETIC FIELD DIRECTION to #1 and 2.5 cm directly to the right of its north pole. What
is the magnetic force on the north end of magnet #1 now?
1. ● A permanent bar magnet (#1) is vertically oriented so
that its north end is below its south end. The north end of 2. ● Two identical bar magnets of negligible width are
this magnet feels an upward magnetic force of 1.5 mN located in the x-y plane. Magnet #1 lies on the x-axis
from an identical vertically oriented magnet (#2) that has and its north end is at x = + 1.0 cm, while its south end
one end located 2.5 cm directly below the north end of #1. is at x = + 5.0 cm. Magnet #2 lies on the y-axis and its
(a) Make a sketch showing the orientation of the second north end is at y = + 1.0 cm, while its south end is at
magnet’s poles. (b) A third identical magnet (#3) is brought y = + 5.0 cm. (a) In what direction would a compass
EXERCISES 691
point if it were located at the origin? (b) Repeat part (a) other angle(s) would the magnetic force on it be the
for the situation where magnet #1 is reversed in polarity. same? Would the direction be the same? Explain.
[Hint: Make a sketch of the two magnets and their indi- 11. ● ● ● A beam of protons exits from a particle accelerator
vidual fields at the origin.] due east at a speed of 3.0 * 105 m>s. The protons then
3. ● ● Two bar very narrow magnets are located in the x-y enter a uniform magnetic field of magnitude 0.50 T that
plane. Magnet #1 lies on the x-axis and its north end is at is oriented at 37° above the horizontal relative to the
x = + 1.0 cm, while its south end is at x = + 5.0 cm. beam direction.
Magnet #2 lies on the y-axis and its north end is at (a) What is the initial acceleration of a proton as it enters
y = + 1.0 cm, while its south end is at y = + 5.0 cm. the field? (b) What if the magnetic field were angled at 37°
Magnet #2 produces a magnetic field that is only one- below the horizontal instead? (c) If the beam were made of
half the magnitude of magnet #1. (a) In what direction electrons traveling at the same speed rather than protons
would a compass point if it were located at the origin? and the field were angled upward at 37°, would there be
(b) Repeat part (a) for the situation where magnet #1 is any difference in the force on the electrons compared to the
reversed in polarity. protons? Explain. (d) In part (c), what would be the ratio of
the acceleration of an electron to that of a proton?
19.2 MAGNETIC FIELD STRENGTH AND

MAGNETIC FORCE 19.3 APPLICATIONS: CHARGED
PARTICLES IN MAGNETIC FIELDS
4. ● A positive charge moves horizontally to the right
across this page and enters a magnetic field directed ver- 12. ● An ionized deuteron (a bound proton–neutron system
tically downward in the plane of the page. (a) What is with a net + e charge) passes through a velocity selector
the direction of the magnetic force on the charge: (1) into whose perpendicular magnetic and electric fields have
the page, (2) out of the page, (3) downward in the plane magnitudes of 40 mT and 8.0 kV>m, respectively. Find
of the page, or (4) upward in the plane of the page? the speed of the ion.
Explain. (b) If the charge is 0.25 C, its speed is 13. ● In a velocity selector, the uniform magnetic field of 1.5 T
2.0 * 102 m>s, and it is acted on by a force of 20 N, what is produced by a large magnet. Two parallel plates with a
is the magnetic field strength? separation of 1.5 cm produce the perpendicular electric
5. ● A charge of 0.050 C moves vertically in a field of 0.080 T field. What voltage should be applied across the plates so
that is oriented 45° from the vertical. What speed must the that (a) a singly charged ion traveling at 8.0 * 104 m>s will
charge have such that the force acting on it is 10 N? pass through undeflected and (b) a doubly charged ion
6. ● A charge of 0.250 C moves vertically in a field of 0.500 T traveling at the same speed will pass through undeflected?
that is oriented some angle from the vertical. If the charge’s 14. ● A charged particle travels undeflected through per-
speed is 2.0 * 102 m>s, what field angle(s) will ensure that pendicular electric and magnetic fields whose magni-
the force acting on the charge is 5.0 N? tudes are 3000 N>C and 30 mT, respectively. Find the
7. ● ● A beam of protons is accelerated to a speed of speed of the particle if it is (a) a proton and (b) an alpha
5.0 * 106 m>s in a particle accelerator and emerges hori- particle. (An alpha particle is a helium nucleus—a posi-
zontally from the accelerator into a uniform magnetic tive ion with a double positive charge of + 2e.)
field. What magnetic field (give its direction and magni- 15. ● ● In an experimental technique for treating deep
tude) oriented perpendicularly to the velocity of the pro- tumors, unstable positively charged pions (p+ , elemen-
ton would cancel the force of gravity and keep the beam tary particles with a mass of 2.25 * 10-28 kg) penetrate
moving exactly horizontally? the flesh and disintegrate at the tumor site, releasing
8. ● An electron travels in the + x-direction in a magnetic energy to kill cancer cells. If pions with a kinetic energy
field and is acted on by a magnetic force in the of 10 keV are required and if a velocity selector with an
- y-direction. (a) In which of the following directions electric field strength of 2.0 * 103 V>m is used, what
could the magnetic field be oriented: (1) -x, (2) +y, must be the magnetic field strength?
(3) +z, or (4) -z? Explain. (b) If the electron speed is 16. ● ● In a mass spectrometer, a singly charged ion having a
3.0 * 106 m>s and the magnitude of the force is particular velocity is selected by using a magnetic field
5.0 * 10-19 N, what is the magnetic field strength? of 0.10 T perpendicular to an electric field of
9. ● An electron travels at a speed of 2.0 * 10 m>s
4 1.0 * 103 V>m. A magnetic field of this same magnitude
through a uniform magnetic field whose magnitude is is then used to deflect the ion, which moves in a circular
1.2 * 10-3 T. What is the magnitude of the magnetic path with a radius of 1.2 cm. What is the mass of the ion?
force on the electron if its velocity and the magnetic field 17. ● ● In a mass spectrometer, a doubly charged ion having a
(a) are perpendicular, (b) make an angle of 45°, (c) are particular velocity is selected by using a magnetic field of
parallel, and (d) are exactly opposite? 100 mT perpendicular to an electric field of 1.0 k V>m. This
10. ● ● (a) What angle(s) does a particle’s velocity have to same magnetic field is then used to deflect the ion in a cir-
make with the magnetic field direction for the particle to cular path with a radius of 15 mm. Find (a) the mass of the
be subjected to half the maximum possible magnetic ion and (b) the kinetic energy of the ion. (c) Does the kinetic
force, Fmax? (b) Express the magnetic force on a charged energy of the ion increase in the circular path? Explain.
particle in terms of Fmax if the angle between its velocity 18. ● ● ● In a mass spectrometer, a beam of protons enters a
and the magnetic field direction is (i) 10°, (ii) 80°, and magnetic field. Some protons make exactly a one-quarter
(iii) 100°. (c) If the particle’s velocity makes an angle of circular arc of radius 0.50 m. If the field is always
50° with respect to the magnetic field direction, at what perpendicular to the proton’s velocity, (a) what is the
692 19 MAGNETISM
field’s magnitude if exiting protons have a kinetic 24. ●●A straight current-carrying wire 25 cm long is ori-
energy of 10 keV? (b) How long does it take the proton to ented at right angles to a uniform horizontal magnetic
complete the quarter circle? (c) Find the net force field of 0.30 T pointing in the -x-direction. (a) Along
(magnitude) on a proton while it is in the field. which of the x-y-z axes would the current direction have
to be to cause the wire to be subject to a force of
(a) 0.050 N in the +y-direction, (b) 0.025 N in the
+ z-direction, and (c) 0.020 N in the + x-direction?
19.4 MAGNETIC FORCES ON CURRENT- 25. ● ● A wire carries a current of 10 A in the + x-direction.
CARRYING WIRES (a) Find the force per unit length on the wire if it is in a
AND magnetic field that has components of Bx = 0.020 T,
By = 0.040 T, and Bz = 0 T. (b) Find the force per unit
19.5 APPLICATIONS: CURRENT-
length on the wire if only the field’s x-component is
CARRYING WIRES IN MAGNETIC FIELDS
changed to Bx = 0.050 T. (c) Find the force per unit
19. ● (a) Use a right-hand force rule to find the direction of the length on the wire if only the field’s y-component is
current in the wires shown in 䉲 Fig. 19.36. In each case, the changed to By = - 0.050 T.
magnetic force direction is shown. (b) If in each case the 26. ● ● A nearly horizontal dc power line in the midlatitudes
wire is a straight segment 15 cm long carrying a current of of North America carries a current of 1000 A directly
5.5 A, and is in a B-field whose strength is 1.0 mT, eastward. If the Earth’s magnetic field at the location of
determine the magnitude of the magnetic force. the power line is northward with a magnitude of
5.0 * 10-5 T at an angle of 45° below the horizontal,
what are the magnitude and direction of the magnetic
F F force on a 15-m section of the line?
F 27. ● ● A wire is bent as shown in 䉲 Fig. 19.37 and placed in a
B B magnetic field with a magnitude of 1.0 T in the indicated
B direction. Find the net force on the whole wire if
x = 50 cm and it carries a current of 5.0 A in the direc-
(i) (ii) (iii) tion shown.
B 䉳 FIGURE 19.37
F Current-carrying wire in a
x
F=0 I
3x I magnetic field See
B B Exercise 27.
Bent wire
(iv) (v)
28. IE ● ● ● A loop of current-carrying wire is in a 1.6-T mag-
䉱 F I G U R E 1 9 . 3 6 The right-hand force rule See Exercise 19. netic field. (a) For the magnetic torque on the loop to be
at a maximum, should the plane of the coil be (1) paral-
lel, (2) perpendicular, or (3) at a 45° angle to the magnetic
20. ● A straight, horizontal segment of wire carries a current field? Explain. (b) If the loop is rectangular with dimen-
in the +x-direction in a magnetic field that is directed in sions 20 cm by 30 cm and carries a current of 1.5 A, what
the - z-direction. (a) Is the magnetic force on the wire is the magnitude of the magnetic moment of the loop,
directed in the (1) -x-, (2) + z-, (3) + y-, or and what is the maximum torque? (c) What would be the
(4) - y-direction? Explain. (b) If the wire is 1.0 m long angle(s) between the magnetic moment vector and the
and carries a current of 5.0 A and the magnitude of the magnetic field direction if the loop felt only 20% of its
magnetic field is 0.30 T, what is the magnitude of the maximum torque?
force on the wire? 29. ● ● ● A rectangular wire loop with a cross-sectional area
21. ● A 2.0-m length of straight wire carries a current of of 0.20 m2 carries a current of 0.25 A. The loop is free to
20 A in a uniform magnetic field of 50 mT whose direc- rotate about an axis that is perpendicular to a uniform
tion is at an angle of 37° from the direction of the current. magnetic field with strength 0.30 T. The plane of the loop
Find the force on the wire. is at an angle of 30° to the direction of the magnetic field.
(a) What is the magnitude of the torque on the loop?
22. ●● A horizontal magnetic field of 1.0 * 10-4 T is at an (b) How would you change the magnetic field to double
angle of 30° to the direction of the current in a straight, hor- the magnitude of the torque in part (a)? (c) How could
izontal wire 75 cm long. If the wire carries a current of 15 A, you change only the current to double the torque in part
(a) what is the magnitude of the force on the wire? (b) What (a)? (d) If you wanted to double the torque by changing
angle(s) would be required for the force to be half the value only the loop area, what would the new area have to be?
found in part (a), assuming nothing else is changed? (e) Could you double the torque in part (a) by changing
23. ●● A wire carries a current of 10 A in the + x-direction in only the angle?
a uniform magnetic field of 0.40 T. Find the magnitude of
the force per unit length and the direction of the force on
19.6 ELECTROMAGNETISM: CURRENTS
the wire if the magnetic field is (a) in the + x-direction,
(b) in the + y-direction, (c) in the +z-direction, (d) in the
AS A MAGNETIC FIELD SOURCE
-y-direction, (e) in the - z-direction, and (f) at an angle 30. ● The magnetic field at the center of a 50-turn coil of
of 45° above the + x-axis and in the x-y plane. radius 15 cm is 0.80 mT. Find the current in the coil.
EXERCISES 693
31. ● In Exercise 30, if you wanted to double the field is (1) toward or (2) away from the observer. (b) If the diam-
strength while keeping the current and turn count the eter of the loop is 12 cm and the current is 1.8 A, what is the
same, what would the coil area have to be? magnitude of the magnetic field at the center of the loop?
32. ● (a) Show that the right-hand side of Eq. 19.13b gives 43. ● ● A circular loop of wire with a radius of 5.0 cm carries
the correct SI units for magnetic field. (b) Show that a current of 1.0 A. Another circular loop of wire is con-
Eq. 19.13 b reduces to Eq. 19.13a when the central axis centric with (that is, has a common center with) the first
location is at the center of the loop. and has a radius of 10 cm. The magnetic field at the cen-
33. ● The magnetic field 7.5 cm directly above the center of ter of the loops is double what the field would be from
a 25-turn coil of radius 15 cm is 0.80 mT. Find the current the first one alone, but oppositely directed. What is the
in the coil. current in the second loop?
34. ● A long, straight wire carries a current of 2.5 A. Find 44. ● ● A current-carrying solenoid is 10 cm long and is
the magnitude of the magnetic field 25 cm from the wire. wound with 1000 turns of wire. It produces a magnetic
35. ● In a physics lab, a student discovers that the magni-
field of 4.0 * 10-4 T at the solenoid’s center. (a) How
tude of the magnetic field at a certain distance from a long would you make the solenoid in order to produce a
long wire is 4.0 mT. If the wire carries a current of 5.0 A, field of 6.0 * 10-4 T at its center? (b) Adjusting only the
what is the distance of the magnetic field from the wire? windings, what number would be needed to produce a
field of 8.0 * 10-4 T at the center? (c) What current in the
36. ● A solenoid is 0.20 m long and consists of 100 turns of
solenoid would be needed to produce a field of
wire. At its center, the solenoid produces a magnetic field
9.0 * 10-4 T but in the opposite direction?
with a strength of 1.5 mT. Find the current in the coil.
45. ● ● A solenoid is wound with 200 turns per centimeter.
37. ● ● Two long, parallel wires carry currents of 8.0 A and
An outer layer of insulated wire with 180 turns per cen-
2.0 A (䉲 Fig. 19.38). (a) What is the magnitude of the mag-
timeter is wound over the solenoid’s first layer of wire.
netic field midway between the wires? (b) Where on a
When the solenoid is operating, the inner coil carries a
line perpendicular to and joining the wires is the mag-
current of 10 A and the outer coil carries a current of 15 A
netic field zero?
in the direction opposite to that of the current in the
12 cm
䉳 FIGURE 19.38 inner coil (䉲 Fig. 19.40). (a) What is the direction of the
I 1 = 8.0 A I 2 = 2.0 A Parallel current- magnetic field at the center for this configuration?
carrying wires See (b) What is the magnitude of the magnetic field at the
Exercises 37, 40 and 56. center of the doubly wound solenoid?
Outer Inner
A 9.0 cm 䉳 FIGURE 19.40
Wire 1 Double it up? See
Wire 2 Exercise 45.
38. ●● Two long, parallel wires separated by 50 cm each carry
15 A 10 A
currents of 4.0 A in a horizontal direction. Find the mag-
netic field midway between the wires if the currents are
46. ●● A set of jumper cables used to start a car from
(a) in the same direction and (b) in opposite directions.
another car’s battery is connected to the terminals of
39. ● ● Two long, parallel wires separated by 0.20 m carry both batteries. If 15 A of current exists in the cables dur-
equal currents of 1.5 A in the same direction. Find the ing the starting procedure and the cables are parallel and
magnitude and direction of the magnetic field 0.15 m 15 cm apart, (a) do the wires repel or attract? Explain.
away from each wire on the side opposite the other wire (b) What is the magnetic field strength that each wire
(䉲 Fig. 19.39). produces at the location of the other? (c) What is the
0.20 m force per unit length on the cables? Include an explana-
䉳 F I G U R E 1 9 . 3 9 Magnetic tion of whether the wires repel or attract one another.
field summation See Exercise 39.
47. ● ● Two long, straight, parallel wires carry the same current
I = 1.5 A
in the same direction. (a) Use both the right-hand source
I = 1.5 A
and force rules to determine whether the forces on the wires
are (1) attractive or (2) repulsive. (b) If the wires are 24 cm
B= ? B= ?
apart and experience a force per unit length of 24 mN>m,
0.15 m 0.15 m determine the current in each wire. (c) What is the magnetic
field strength midway between the two wires?
48. ● ● ● Four wires running through the corners of a square
with sides of length a, as shown in 䉲 Fig. 19.41, carry
40. ●●In Fig. 19.38, find the magnetic field (magnitude and equal currents I. Calculate the magnetic field at the cen-
direction) at point A, which is located 9.0 cm away from ter of the square in terms of I and a.
wire 2 on a line perpendicular to the line joining the wires. a 䉳 F I G U R E 1 9 . 4 1 Current-
41. ● ● A coil of four circular loops of radius 5.0 cm carries a carrying wires in a square array
I I
current of 2.0 A clockwise, as viewed from above the coil’s See Exercise 48.
plane. What is the magnetic field at the center of the coil?
a
42. IE ● ● A circular loop of wire in the horizontal plane car-
ries a counterclockwise current, as viewed from above. I I
(a) Use the right-hand source rule to determine whether
the direction of the magnetic field at the center of the loop
694 19 MAGNETISM
19.7 MAGNETIC MATERIALS *19.8 GEOMAGNETISM: THE EARTH’S

49. ●● A 50-cm-long solenoid has 100 turns of wire and car- MAGNETIC FIELD
ries a current of 0.95 A. It has a ferromagnetic core com- 51. ●● (a) Exiting from a small linear accelerator near Wash-
pletely filling its interior where the field is 0.71 T. ington D.C., a proton is moving parallel to the ground.
Determine the (a) magnetic permeability and (b) relative What should the direction(s) of its velocity be in order to
magnetic permeability of the material. maximize the upward component of the Earth’s magnetic
50. ● ● ● A circular wire coil consists of 100 turns and is force on it? (b) Using a value of 0.05 mT for the value of
wound tightly around a very long iron cylinder with a the horizontal component of the Earth’s magnetic field
radius of 2.5 cm and a relative permeability of 2200. The at this location, what is the upward acceleration of the
loop has a current of 7.5 A in it. Determine the magnetic proton as it exits the accelerator at a speed of 2000 m>s?
field strength produced by the coil (a) at the center of the 52. ● ● ● A crude model of the Earth’s magnetic field consists
coil and (b) at a location on the central axis of the iron of a circular loop of current with a radius of 500 km with
cylinder 5.0 cm above the center of the circular coil. its center coincident with that of the Earth’s. Assuming
the plane of the loop to be approximately in the equatorial
plane, and using a value of 0.10 mT for the magnitude of
the Earth’s field at the poles, (a) estimate the current in the
theoretical loop. (b) What would be the magnetic moment
(magnitude and direction) of this supposed loop?
53. A particle with charge q and mass m moves in a horizon- rium. (It “floats.”) (b) If the lower wire has a linear mass
tal plane at right angles to a uniform vertical magnetic density of 1.5 * 10-3 kg>m and the wires carry the same
field B. (a) What are the period T and frequency f of the current, what should be the current?
particle’s circular motion in terms of q, B, and m? (This
? ?
frequency is called the cyclotron frequency.) You result –I I
should verify that the time for one orbit for any charged
particle in a uniform magnetic field is independent of its
I
speed and radius. (b) Compute the path radius and the
cyclotron frequency if the particle is an electron with a 䉱 F I G U R E 1 9 . 4 2 Magnetic suspension The bottom wire is
speed of 1.0 * 105 m>s traveling in a region where the magnetically attracted to the top (rigidly fixed) wire. See
field strength is 0.10 mT. Exercise 57.
54. What is the (a) “current” due to the electron orbiting in a
circular path about the proton in a hydrogen atom?
(b) What magnetic field strength does this “electron cur- 58. A beam of protons is accelerated easterly from rest
rent” create at the proton location? (c) If the electron is through a potential difference of 3.0 kV. It enters a region
orbiting clockwise, as viewed from above its orbital where there exists an upward pointing uniform electric
plane, what is the direction of this field? Take the orbital field. This field is created by two parallel plates separated
radius to be 0.0529 nm. [Hint: Find the electron’s period by 10 cm with a potential difference of 250 V across them.
by considering the centripetal force.] (a) What is the speed of the protons as they enter the
55. Two long, straight, parallel wires 10 cm apart carry cur- electric field? (b) Find the magnitude and direction rela-
rents in opposite directions. (a) Use the right-hand source tive to the velocity of the magnetic field (perpendicular
and force rules to determine whether the forces on the B
to E) needed so the beam passes undeflected through the
wires are (1) attractive or (2) repulsive. Show your reason- plates. (c) What happens to the protons if the magnetic
ing. (b) If the wires carry equal currents of 3.0 A, what is field is greater than the value found in part (b)?
the magnetic field magnitude that each produces at the 59. A cylindrical solenoid 10 cm long has 3000 turns of wire
other’s location? (c) Use the result of part (b) to determine and carries a current of 5.0 A. A second solenoid, consist-
the magnitude of the force per unit length they exert on ing of 2000 turns of wire and the same length as the first
each other. solenoid, surrounds it and is concentric (shares a common
56. In Figure 19.38, (a) what is the direction of the magnetic central axis) with it. The outer coil carries a current of 10 A
field produced by wire 1 at the location of wire 2? in the same direction as the current in inner one.
(b) What about midway between the wires? (c) What is (a) Find the magnetic field near their common center.
the force (including direction) per unit length on wire 1? (b) What current in the second solenoid (magnitude and
57. A long wire is placed 2.0 cm directly below a rigidly relative direction) would make the net field strength at
mounted second wire (䉴 Fig. 19.42). (a) Use the right- the center twice that of the first solenoid alone? (c) What
hand source and force rules to determine whether the current in the second solenoid (magnitude and relative
currents in the wires should be in (1) the same or (2) the direction) would result in zero net magnetic field near
opposite direction so that the lower wire is in equilib- their common center?
60. A proton enters a uniform magnetic field that is at a right between the inner and outer one, (3) only outside the
angle to its velocity. The field strength is 0.80 T and the larger one, or (4) inside the smaller one and outside the
proton follows a circular path with a radius of 4.6 cm. larger one? (b) The larger one is a 200-turn coil of wire
What are (a) the magnitude of its linear momentum and with a radius of 9.50 cm and carries a current of 11.5 A.
(b) its kinetic energy? (c) If its speed were doubled, what The second one is a 100-turn coil with a radius of
would then be the radius, momentum, and kinetic 2.50 cm. Determine the current in the inner coil so the
energy? magnetic field at their common center is zero. Neglect
61. Exiting a linear accelerator, a narrow horizontal beam of the Earth’s field.
protons travels due north. If 1.75 * 1013 protons pass a 65. Consider the following arrangement (called Helmholtz
given point per second, (a) determine the magnetic field coils) of two identical current-carrying coils. They are
direction and strength at a location of 2.40 m east of the “stacked” vertically with their centers on a common ver-
beam. (b) Does it seem likely this would demagnetize tical axis and their areas arranged horizontally, as shown
the encoded magnetic strip on, for example, an ATM in 䉲 Fig. 19.43. Assume each has 100 loops of wire, carries
card? [Hint: The ATM card “lives” safely in the Earth’s 7.5 A of current (same direction), and has a radius of
magnetic environment.] 10 cm. Their centers are separated by 10 cm.
62. A 200-turn circular coil of wire has a radius of 10.0 cm Determine the magnetic field strength at (a) the cen-
and a total resistance of 0.115 Æ . At its center the mag- ter of the each coil, (b) midway between the two coils,
netic field strength is 7.45 mT. (a) Determine the voltage and (c) 10 cm above or below the centers of the coils.
of the power supply creating the current in the coil. This arrangement is experimentally useful for producing
(b) What would be the field strength at a point 4.5 cm its largest magnetic field near the midway point between
directly above or below the center of the coil? the two coils. Plot the field strength as a function of loca-
63. A 100-turn circular coil of wire has a radius of 20.0 cm tion using your results. Does your graph indicate that
and carries a current of 0.400 A. The normal to the coil the maximum strength occurs near the midpoint? (It can
area points due east. A compass, when placed at the cen- also be shown that the field there is approximately uni-
ter of the coil, does not point east, but instead makes an form.)
angle of 60° north of east. Using this data, determine
(a) the magnitude of the horizontal component of the
Earth’s field at that location and (b) the magnitude of the 䉳 FIGURE 19.43
Earth’s field at that location if it makes an angle of 55° r I
Helmholtz coils. Two identi-
below the horizontal. cal coils stacked vertically.
64. A circular coil of current-carrying wire has the normal to See Exercise 65.
d
its area pointing upward. A second smaller concentric
circular coil carries a current in the opposite direction. r I
(a) Where, in the plane of these coils, could the magnetic
field be zero: (1) only inside the smaller one, (2) only
Electromagnetic
20 Induction and Waves
20.1 Induced emf: Faraday’s
law and Lenz’s law (697)
■ Faraday’s law
■ Lenz’s law
20.2 Electric generators and

back emf (705)
■ ac generators
20.3 Transformers and power

transmission (710)
■ step-up and step-down
transformers
20.4 Electromagnetic
waves (716) PHYSICS FACTS
■
■ em spectrum
radiation pressure
✦ Nikola Tesla (1856–1943), the
Serbian-American scientist–inventor
whose last name is the SI unit of
magnetic field strength, invented ac
A s was seen in Chapter 19, an
electric current produces a
magnetic field. But the relationship
dynamos, transformers, and motors.
He sold the patent rights to these to
between electricity and magnetism
George Westinghouse which led to does not stop there. In this chapter,
the first large-scale electric generator
at Niagara Falls. To prove the safety it will be shown that under the
of electric energy to a skeptical pub-
lic, Tesla gave exhibitions in which
right conditions, a magnetic field
he would light lamps by allowing can produce an electric field and
electricity to flow through his body.
✦ Radio waves, radar, visible light, and
electric current. How is this done?
X-rays are all electromagnetic waves, Chapter 19 considered only
better known as light. The only differ-
ence is their frequency and wave- constant magnetic fields. No cur-
length. In a vacuum, they all travel at
c 13.00 * 108 m>s2.
rent is produced in a loop of wire
✦ The Scottish physicist James Clerk that is stationary in a constant mag-
Maxwell (1831–1879) fully devel-
oped and integrated the equations
netic field. However, if the mag-
of electricity and magnetism. This netic field changes with time, or if
set became known as Maxwell’s
equation. From them he predicted the wire loop moves into or out of,
the value of c which was one of the
great achievements of nineteenth-
or is rotated in, the field, a current is
century physics. produced in the wire.
20.1 INDUCED EMF: FARADAY’S LAW AND LENZ’S LAW 697
The uses of this interrelationship of electricity and magnetism are many. One
example happens during the playing of a videotape, which is actually a magnetic
tape that has information encoded on it as variations in its magnetism. These varia-
tions can be used to produce electrical currents, which, in turn, are amplified and
the signal sent for replay to the television set. Similar processes are involved when
information is stored on or retrieved from a magnetic disk in your computer.
Now, early in the twenty-first century, a call for “alternative” and “renewable”
sources of electric energy is getting louder. At “wind farms” such as that in the
chapter-opening photo, one of the oldest and simplest energy sources on Earth—
wind—is used to generate “clean” electric energy. The windmill generators con-
vert some of the air’s kinetic energy into electric energy. But how does this last step
take place? Regardless of the source of the energy—the burning of oil, coal, or gas;
a nuclear reactor wind, waves, or falling water—the actual conversion to electric
energy is accomplished by means of magnetic fields and electromagnetic induc-
tion. This chapter not only examines the underlying principles that make such con-
version possible, but also discusses several practical applications. Moreover, it wil
be seen that the creation and propagation of electromagnetic radiation are inti-
mately related to electromagnetic induction.
20.1 Induced emf : Faraday’s Law and Lenz’s Law

➥ For a given coil orientation, how are the magnetic flux through the coil and the coil
area related? 䉲 F I G U R E 2 0 . 1 Electromagnetic
induction (a) When there is no rela-
➥ How can an induced emf be created in a coil without changing its area or the mag- tive motion between the magnet
netic field? and the wire loop, the number of
➥ Does the induced current in a wire coil depend on the magnitude of the magnetic field lines through the loop (in this
flux through the coil? case, seven) is constant, and the gal-
vanometer shows no deflection.
Recall from Section 17.1 that the term emf stands for electromotive force, which is a (b) Moving the magnet toward the
voltage or electric potential difference capable of creating an electric current. It can be loop increases the number of field
lines passing through the loop (now
observed experimentally that a magnet held stationary near a conducting wire loop twelve), and an induced current is
does not induce an emf (and therefore produces no current) in that loop (䉲 Fig. 20.1a). detected. (c) Moving the magnet
If the magnet is moved toward the loop, however, as shown in Fig. 20.1b, the deflec- away from the loop decreases the
tion of the galvanometer needle indicates that current exists in the loop, but only dur- number of field lines passing
ing the motion. Furthermore, if the magnet is moved away from the loop, as shown in through the loop (to five). The
induced current is now in the oppo-
Fig. 20.1c, the galvanometer needle is deflected in the opposite direction, which indi- site direction. (Note the needle
cates a reversal of the current’s direction, but, again, only during the motion. deflection.)
v v
I
N S N N
v=0 B I
B B
Stronger Weaker
B B B
I
– o + – o + – o +
I
Loop head on Loop head on

(a) No motion between magnet and loop (b) Magnet is moved toward loop (c) Magnet is moved away from loop
698 20 ELECTROMAGNETIC INDUCTION AND WAVES
Deflections of the galvanometer needle, indicating the presence of induced cur-

B rents, also occur if the loop is moved toward or away from the stationary magnet.
The effect thus depends on relative motion of the loop and magnet. The magnitude
Loop
head on of the induced current depends on the speed of that motion. However, experimen-
o
v tally, there is a noteworthy exception. If a loop is moved (but not rotated) in a
– +
uniform magnetic field, as shown in 䉳 Fig. 20.2, no current is induced. This situa-
tion will be discussed later in this section.
Yet another way to induce a current in a stationary wire loop is to vary the cur-
rent in another, nearby loop. When the switch in the battery-powered circuit in
䉱 F I G U R E 2 0 . 2 Relative motion 䉳 Fig. 20.3a is closed, the current in the loop on the right goes from zero to some con-
and no induction When a loop is
moved parallel to a uniform mag- stant value in a short time. Only during the buildup time does the magnetic field
netic field, there is no change in the caused by the current in this loop increase in the region of the loop on the left. Dur-
number of field lines passing ing the buildup, the galvanometer needle deflects, indicating current in the left loop.
through the loop, and there is no When the current in the right loop attains its steady value, the field it produces
induced current. becomes constant, and the current in the left loop drops to zero. Similarly, when the
switch in the right loop is opened (Fig. 20.3b), its current and field decrease to zero,
Induced Increasing and the galvanometer deflects in the opposite direction, indicating a reversal in
B I
current current and direction of the current induced in the left loop. The important fact to note is that
field induced current in a loop occurs only when the magnetic field through that loop changes.
In Fig. 20.1, moving the magnet changed the magnetic environment in a loop,
causing an induced emf that, in turn, caused an induced current. For the case of
two stationary loops (Fig. 20.3), a changing current in the right loop produced a
– o + I + changing magnetic environment in the left loop, thereby inducing an emf and a
current in the left loop.* There is a convenient way of summarizing what is hap-
pening in both Fig. 20.1 and Fig. 20.3: To induce currents in a loop or complete cir-
cuit, utilizing a process called electromagnetic induction, all that matters is
Galvanometer Switch whether the magnetic field through the loop or circuit is changing.
just closed
Detailed experiments on electromagnetic induction were done independently
(a)
by Michael Faraday in England and Joseph Henry in the United States around
Induced I Decreasing 1830. Faraday found that the important factor in electromagnetic induction was
current B current and the time rate of change of the number of magnetic field lines passing through the
field loop or circuit area. That is, he discovered that
An induced emf is produced in a loop or complete circuit whenever the number of
magnetic field lines passing through the plane of the loop or circuit is changing with time.
– o + I +
MAGNETIC FLUX
Because of Faraday’s discovery, determining induced emf requires that the num-
Switch
ber of field lines through the loop somehow be quantified. Consider a loop of wire
just opened in a uniform magnetic field (䉴 Fig. 20.4a). The number of field lines through the
(b) loop depends on the loop’s area, its orientation relative to the field, and the
strength of that field. To describe the loop’s orientation, the concept of an area
䉱 F I G U R E 2 0 . 3 Mutual induction
vector 1A2 is employed. Its direction is normal to the loop’s plane, and its magni-
B
tude is equal to the loop area. The angle between the magnetic field 1B2 and the
(a) When the switch is closing in the B
area vector 1A2, u, is a measure of their relative orientation. For example, in

right-loop circuit, the buildup of B
current produces a changing mag-
netic field in the other loop, induc- Fig. 20.4a, u = 0°, meaning that the vectors are in the same direction, or, alterna-
ing a current in it. (b) When the tively, that the area plane is perpendicular to the field.
switch is opened, the magnetic field
For the case of a magnetic field that does not vary over the area, the number of
collapses, and the magnetic field in
the left loop decreases. The induced magnetic field lines passing through a particular area (the area inside a loop in this
current in this loop is then in the case) is proportional to the magnetic flux (≥), which is defined as
opposite direction. The induced cur- magnetic flux
rents occur only when the magnetic £ = BA cos u (20.1)
(in a constant magnetic field)
field passing through a loop
changes and vanish when the field SI unit of magnetic flux: tesla-meter squared 1T # m22, or weber (Wb)†
reaches a constant value.
*The term mutual induction is used to describe the situation in which emfs and currents are induced
between two (or more) loops.
†
Wilhelm Eduard Weber (1804–1891), a German physicist, was noted for his work in magnetism
and electricity, particularly terrestrial magnetism. The unit name weber was introduced as the SI unit of
magnetic flux in 1935.
B Axis of Side view

rotation
B
A A
A θ = 0° θ = 180° θ
A θ = 90° A
(a) (b) Φ = +BA (c) Φ = –BA (d) Φ = 0 (e) Φ = BA cos θ
䉱 F I G U R E 2 0 . 4 Magnetic flux (a) Magnetic flux 1£2 is a measure of theBnumber of field

lines passing through an area (A). The area can be represented by a vector A perpendicular
to the plane of the area. (b) When the plane of a loop is perpendicular to the field and
u = 0°, then £ = £ max = + BA. (c) When u = 180°, the magnetic flux has the same magni-
tude, but is opposite in direction: £ = - £ max = - BA. (d) When u = 90°, then £ = 0.
(e) As the loop’s plane is changed from being perpendicular to the field to being more par-
allel to the field, less area is available to the field lines, and therefore the flux decreases. In
general, £ = BA cos u.
Since the SI unit of magnetic field is the tesla, the magnetic flux has SI units of
T # m2. This combination is sometimes called the weber, defined as 1 Wb = 1 T # m2.
u
The orientation of the loop with respect to the magnetic field affects the number of B
field lines passing through it, and this factor is accounted for by the cosine term in A cos u
Eq. 20.1. Let us consider several possible orientations: u Side view of
If B and A are parallel 1u = 0°2, then the magnetic flux is positive and has a
B B loop area A
■
maximum value of £ max = BA cos 0° = + BA. The maximum possible number = B (A cos u)
of magnetic field lines pass through the loop in this orientation (Fig. 20.4b).
(a)
If B and A are oppositely directed 1u = 180°2, then the magnitude of the mag-
B B
■
netic flux is a maximum again, but of opposite sign:
£ 180° = BA cos 180° = - BA = - £ max (Fig. 20.4c). B cos u
B B
■ If B and A are perpendicular, then there are no field lines passing through the
u
plane of the loop, and the flux is zero: £ 90° = BA cos 90° = 0 (Fig. 20.4d).
B
■ For situations at intermediate angles, the flux is less than the maximum value,
but nonzero (Fig. 20.4e). A cos u can be interpreted as the effective area of the Side view of loop
loop perpendicular to the field lines (䉴 Fig. 20.5a). Alternatively, B cos u can be area A
viewed as the perpendicular component of the field through the full area of the
loop, A, as shown in Fig. 20.5b. Thus, Eq. 20.1 can be interpreted as either = (B cos u) A
£ = 1B cos u2A or £ = B1A cos u2. Either way, the answer is the same.
(b)
FARADAY’S LAW OF INDUCTION AND LENZ’S LAW 䉱 F I G U R E 2 0 . 5 Magnetic flux

From quantitative experiments, Faraday determined that the emf 1e2 induced in a
through a loop: An alternative inter-
pretation Instead of defining the
coil (a coil, by definition, consists of a series connection of N individual loops) flux 1£2 (a) in terms of the magnetic
depends on the time rate of change of the magnetic field lines through all the field magnitude (B) passing through
loops, or the time rate of change of the magnetic flux through all the loops (total flux). a reduced area 1A cos u2, we can
This dependence, known as Faraday’s law of induction, is expressed mathemati- define it (b) in terms of the perpen-
dicular component of the magnetic
cally as field 1B cos u2 passing through A.
¢1N£2 Either way, £ is a measure of the
¢£ number of field lines passing
e = - = -N (Faraday’s law for induced emf) (20.2)
¢t ¢t through A and is given by
£ = BA cos u (Eq. 20.1).
where ¢£ is the change in flux through one loop. In a coil consisting of N loops,
the total change in flux is N¢£ . Note that the induced emf in Eq. 20.2 is an aver-
age value over the time interval ¢t (why?).
The minus sign is included in Eq. 20.2 to give an indication of the direction of the
induced emf, which has not been mentioned until now. The Russian physicist
Heinrich Lenz (1804–1865) discovered the law that governs the direction of the
induced emf. Lenz’s law is stated as follows:
Induced External
External I Induced
field increases
field I
(B2 > B1)
B1
+x Induced -x
+x
B
Induced I
(a) (b)
䉱 F I G U R E 2 0 . 6 Finding the direction of the induced current (a) An external magnetic

field is shown increasing to the right. The induced current creates its own magnetic field to
try to counteract the flux change that is occurring. (b) The (induced) current right-hand
Bar Small coil (source) rule determines the direction of the induced current. Here the direction of the
magnet induced field must be to the left. With the thumb of the right hand pointing left, the fingers
give the induced current direction.
S An induced emf in a wire loop or coil has a direction such that the current it creates
v
produces its own magnetic field that opposes the change in magnetic flux through
that loop or coil.
X X X X X X X X X
X X X X X X X X X This means that the magnetic field due to the induced current is in such a direction to
X X X X X X X X X tend to keep the flux through the loop from changing. For example, if the flux
X X X X X X X X X increases in the + x-direction, the magnetic field due to the induced current will be
X X X X X X X X X in the - x-direction (䉱 Fig. 20.6a). This effect tends to cancel the increase in the flux,
X X X X X X X X X
X X X X X X X X X or oppose the change. Essentially, the magnetic field due to the induced current tries
to maintain the existing magnetic flux. This effect is sometimes called “electro-
(a) magnetic inertia,” by analogy to the tendency of objects to resist changes in their
velocity. In the long run, the induced current cannot prevent the magnetic flux
X X X X X
from changing. However, during the time that the flux is changing, the induced
X X X X magnetic field will oppose that change.
X X X X X The direction of the induced current is given by the induced-current
X X X X right-hand rule:
X X X X X
X X X X With the thumb of the right hand pointing in the direction of the induced field, the
X X X X X fingers curl in the direction of the induced current.
(b) (See Fig. 20.6b and Integrated Example 20.1.) This rule is a version of the right-
Induced I
hand rule used to find the direction of a magnetic field produced by a current
(Chapter 19). Here it is used in reverse. Typically, the induced field direction is
X X X X X B from known (for example, -x in Fig. 20.6b) and the direction of the current that pro-
X
X X X induced duces it is to be determined. An application of Lenz’s law is illustrated in Inte-
X XXXXX X current tries
grated Example 20.1.
X XXX X to counteract
X XXXXX X the flux
X X X X reduction INTEGRATED EXAMPLE 20.1 Lenz’s Law and Induced Currents
X X X X X in (b)
(c) (a) The south end of a bar magnet is pulled far away from a small wire coil. (See
䉳 Fig. 20.7a.) Looking from behind the coil toward the south end of the magnet
䉱 F I G U R E 2 0 . 7 Using a bar mag- (Fig. 20.7b), what is the direction of the induced current: (1) counterclockwise, (2) clock-
net to induce currents (a) The south wise, or (3) there is no induced current? (b) Suppose that the magnetic field over the
end of a bar magnet is pulled away area of the coil is initially constant at 40.0 mT, the coil’s radius is 2.0 mm, and there are
from a wire loop. (b) The view from 100 loops in the coil. Determine the magnitude of the average induced emf in the coil if
the right of the loop shows the mag- the bar magnet is removed in 0.750 s.
netic field pointing away from the
( A ) C O N C E P T U A L R E A S O N I N G . There is initially magnetic flux into the plane of the coil
observer, or into the page, and
decreasing. (c) To counteract this (Fig. 20.7b), and later, when the magnet is far away from the coil, there is no flux; and
loss of flux into the page, current is the flux has changed. Therefore, there must be an induced emf, so answer (3) cannot be
induced in the clockwise direction, correct. As the bar magnet is pulled away, the field weakens but maintains the same
so as to provide its own field into the direction. The induced emf will produce an (induced) current that, in turn, will produce
page. See Integrated Example 20.1.
a magnetic field into the page so as to try to prevent this decrease in flux. Therefore, the
induced emf and current are in the clockwise direction, as found using the induced cur-
rent right-hand rule (Fig. 20.7c) and the correct answer is (2), clockwise.
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . This Example is a straightforward
application of Eq. 20.2. The initial flux is the maximum possible and the final flux is
zero (why?). The data are listed and converted to SI units:
Given: Bi = 40.0 mT = 0.0400 T Find: e (magnitude of average
Bf = 0 induced emf)
r = 2.00 mm = 2.00 * 10-3 m
N = 100 loops
¢t = 0.750 s
To find the initial magnetic flux through one loop of the coil, use Eq. 20.1 with an angle of
2
u = 0°. (Why?) The area is A = pr2 = p12.00 * 10-3 m2 = 1.26 * 10-5 m2. Therefore,
the initial flux, £ i, through one loop is positive (why?) and given by
£ i = Bi A cos u = (0.0400 T)11.26 * 10-5 m22 cos 0° = 5.03 * 10-7 T # m2
Because the final flux is zero, ¢ £ = £ f - £ i = 0 - £ i = - £ i. Therefore, the magni-
tude of the average induced emf is
35.03 * 10-7 1T # m22>loop4
= 1100 loops2
ƒ ¢£ ƒ
10.750 s2
ƒeƒ = N = 6.70 * 10-5 V
¢t
F O L L O W - U P E X E R C I S E . In this Example, (a) in which direction is the induced current if

instead a north magnetic pole approaches the coil quickly? Explain. (b) In this Example,
what would be the average induced current if the coil had a total resistance of 0.200 Æ ?
(Answers to all Follow-Up Exercises are in Appendix VI at the back of the book.)
Lenz’s law incorporates the principle of energy conservation. Consider a situa-

Increasing magnitude of B
tion in which a wire loop has an increasing magnetic flux through its area. Con-
trary to Lenz’s law, suppose instead that the magnetic field from the induced I
current added to the flux instead of keeping it at its original value. This increased
flux would then lead to an even greater induced current. In turn, this greater N
induced current would produce a still greater magnetic flux, which in turn would
give a greater induced current, and so on. Such a something-for-nothing energy v
situation would violate conservation of energy.
(a)
To understand the direction of the induced emf in a loop in terms of forces, con-
sider the case of the moving magnet (for example, Fig. 20.1b). Recall that a current- F I F
carrying loop creates its own magnetic field similar to that of a bar magnet. (See
Figs. 19.3 and 19.25.) The induced current sets up a magnetic field in the loop, and
that loop acts like a bar magnet with a polarity that will oppose the motion of the N N
real bar magnet (䉴 Fig. 20.8). You should be able to show that if the bar magnet is
v
pulled away from the loop, the loop exerts a magnetic attraction to try to keep the
magnet from leaving—electromagnetic inertia in action. (b)
When the expression for the magnetic flux 1£2 given by Eq. 20.1 is substituted
into Eq. 20.2, the result is 䉱 F I G U R E 2 0 . 8 Lenz’s law in
terms of forces (a) If the north end
N¢1BA cos u2 of a bar magnet is moved rapidly
¢£
e = -N = - (20.3) toward a wire loop, current is
¢t ¢t induced in the direction shown.
(b) While the induced current exists,
From this expression, it can be seen that an induced emf results if: the loop then acts like a bar magnet
with its “north end” close to the
1. the magnitude of the magnetic field changes, north end of the real bar magnet.
Thus there is a magnetic repulsion.
2. the loop area changes, and>or This is an alternative way of viewing
Lenz’s law: Induce a current so as to
3. the orientation between the loop area and the field direction changes.
try to keep the flux from changing—
in this case, to try to keep the bar
In situation (1), a flux change is created by a time-varying field, such as that magnet away from the loop and
from a time-varying current in a nearby circuit or that created by moving a mag- maintain its initial value of flux,
net near a coil, as in Fig. 20.1 (or by moving the coil near the magnet). zero.
In situation (2), a flux change results because of a varying loop area. This situa-
tion might occur if a loop had an adjustable circumference (such as the loop
around an inflatable balloon).
Finally, in situation (3), a change in flux can result from a change in orientation of
the loop. This can occur when a coil is rotated in a magnetic field. The change in the
number of field lines through a single loop is evident in the sequential views in
Fig. 20.4. Rotating a coil in a field is a common way of inducing an emf and will be
considered on its own in Section 20.2. The emfs that result from changing the field
strength and loop area are analyzed in the next two Examples. Also, see Insight 20.1,
Electromagnetic Induction at Work: Flashlights and Antiterrorism for ways in
which electromagnetic induction helps make our everyday lives safer and easier.
INSIGHT 20.1 Electromagnetic Induction at Work: Flashlights and Antiterrorism

Electromagnetic induction is used in our daily lives, in most cient light-emitting diode (LED). The resulting beam of light lasts
cases without our realizing it. One example is a flashlight that for several minutes before the flashlight needs to be reshaken.
works without a battery (Fig. 1a). As the flashlight is shaken, a This device could, at the least, play an important backup role to
strong permanent magnet in it oscillates through induction coils, the more traditional flashlights that rely on batteries.
inducing an oscillating emf and current. To charge a capacitor In the field of air travel safety, induction is used to prevent
(and thus store electric energy in it), the ac current must be dangerous metallic objects (such as knives and guns) from
rectified into dc current. (Rectification is the name of the process being carried onto airplanes. As a passenger walks through the
that converts ac into dc current.) The schematic of this flashlight arch of an airport metal detector (see Fig. 2), a series of large
is shown in Fig. 1b. Here a solid-state rectifier circuit (triangular “spiked” currents is periodically delivered to a coil (solenoid) in
symbol) acts as a “one-way current valve.” As shown, only one of the nonmagnetic sides of the archway.
clockwise dc current goes to the capacitor and charges it. After In the most common system, called PI (for pulsed induction),
about a minute of shaking, the capacitor is fully charged. When these current spikes occur hundreds of times per second. As the
the switch S is thrown, the capacitor discharges through an effi- current rises and falls, a changing magnetic field is created in the
passenger. If the passenger is carrying nothing metallic, there
will be no significant induced current and therefore no induced
magnetic field. However, if the passenger has a metal object, a
current will be induced in that object, which in turn will produce
its own (induced) magnetic field that can be sensed by the emit-
ting coil as a “magnetic echo.” Sophisticated electronics measure
the echo-induced emf and trigger a warning light to suggest that
further inspection of that passenger is warranted.
(a)
ac induced dc current
current Rectifier LED
S
I
I
+++ +++
N S N S
C––– –––
Coil
(b)
Oscillating
magnet
F I G U R E 1 A batteryless flashlight (a) A photo of a rela-
tively new type of flashlight that produces light using electric
energy generated by shaking (induction). (b) A schematic
diagram of the flashlight shown in part (a). As the flashlight
is shaken, its internal permanent magnet passes through a
coil, inducing a current. This current alternates in direction F I G U R E 2 Screening at the airport As passengers walk
(why?) and thus needs to be turned into dc (“rectified”) through the arch, they are subjected to a series of magnetic
before it can charge a capacitor. Once the capacitor is fully field pulses. If they have a metal object on their person, the
charged, it can be used to create a current through a light- currents induced in that object create their own magnetic field
emitting diode (LED), which in turn gives off light, typically “echo” that, when detected, gives the safety inspectors reason
for several minutes. to check the passenger more closely.
CONCEPTUAL EXAMPLE 20.2 Fields in the Fields: Electromagnetic Induction

In rural areas where electric power lines carry electricity to REASONING AND ANSWER. Magnetic field lines from long
big cities, it is possible to generate small electric currents by wires are circular. (See Fig. 19.23.) By the source right-hand rule,
means of induction in a conducting loop. The overhead the magnetic field direction at ground level is parallel to the
power lines carry alternating currents that periodically Earth’s surface and alternates in direction. The orientation
reverse direction sixty times per second. How would you ori- choices are shown in Fig. 20.9b. Neither answer (a) nor (c) can
ent the plane of the loop to maximize the induced current if be correct, because in these orientations there would never be
the power lines run north to south: (a) parallel to the Earth’s any magnetic flux passing through the loop. In the situation in
surface, (b) perpendicular to the Earth’s surface in the this Example, the flux would be constant and there would be no
north–south direction, or (c) perpendicular to the Earth’s sur- induced emf. Hence, the answer is (b). If the loop is oriented
face in the east–west direction? (See 䉲 Fig. 20.9a.) perpendicular to the Earth’s surface with its plane in the
north–south direction, the flux through it would vary from zero
to its maximum value and back sixty times per second, and this
would maximize the induced emf and current in the loop.
I
b
I c
N
B a
W 䉳 F I G U R E 2 0 . 9 Induced emfs below power
E lines (a) If current-carrying wires run in the
N S north–south direction, then directly below the
alternating current produces a magnetic field
that oscillates between pointing east and west.
(b) These are the three choices for loop orienta-
(a) (b) tion in Conceptual Example 20.2.
F O L L O W - U P E X E R C I S E . Suggest possible ways of increasing the induced current in this Example by changing only properties of
the loop and not those of the overhead wires.
EXAMPLE 20.3 Induced Currents: A Potential Hazard to Equipment?

Electrical instruments can be damaged or destroyed if they netic flux change can be determined from Eq. 20.1 with u = 0°
are in a rapidly changing magnetic field. This can occur if an and u = 180°. The average induced emf can then be calcu-
instrument is located near an electromagnet operating under lated from Eq. 20.2. (b) Once the emf is determined, the
ac conditions; the electromagnet’s external field could pro- induced current can be calculated from I = e>R.
duce a changing flux within a nearby instrument. If the
induced currents are large enough, they could damage the
instrument. Consider a computer speaker that is near such an Oscillating 䉳 FIGURE 20.10
electromagnet (䉴 Fig. 20.10). Suppose an electromagnet B field Instrument hazard?
exposes the speaker to a maximum magnetic field of 1.00 mT The coil of a speaker is
that reverses direction every 1>120 s. close to an alternating
Assume that the speaker’s coil consists of 100 circular current electromagnet.
loops (each with a radius of 3.00 cm) and has a total resistance The changing flux in
of 1.00 Æ . According to the manufacturer of the speaker, the the coil produces an
average current in the coil should not exceed 25.0 mA. (a) Cal- induced emf and, thus,
an induced current that
culate the magnitude of the average induced emf in the coil
depends on the resis-
during the 1>120-s interval. (b) Is the induced current likely to tance of the coil.
damage the speaker coil?
T H I N K I N G I T T H R O U G H . (a) The flux goes from a (maximum) Speaker coil
positive to a (maximum) negative value in 1>120 s. The mag- ac electromagnet
SOLUTION. Listing the data and converting to SI units,

Given: Bi = + 1.00 mT = + 1.00 * 10-3 T ( + pointing one way) Find: (a) e (magnitude of average induced emf)
Bf = - 1.00 mT = - 1.00 * 10-3 T (pointing the opposite way) (b) I (magnitude of average induced current)
¢t = 1>120 s = 8.33 * 10-3 s
N = 100 loops
R = 1.00 Æ
r = 3.00 cm = 3.00 * 10-2 m
2
(a) The circular loop area is A = pr2 = p13.00 * 10-2 m2 = 2.83 * 10-3 m2. Thus the initial flux through one loop is (see Eq. 20.1):
£ i = Bi A cos u = 11.00 * 10-3 T212.83 * 10-3 m2>loop21cos 0°2 = 2.83 * 10-6 T # m2>loop
Because the final flux is the negative of this, the change in flux through one loop is
¢ £ = £ f - £ i = - £ i - £ i = - 2£ i = - 5.66 * 10-6 T # m2>loop
Therefore, the magnitude of the average induced emf is (using Eq. 20.2)
5.66 * 10-6 A T # m2>loop B
= 1100 loops2 B
ƒ ¢£ ƒ
e = N R = 6.79 * 10-2 V
¢t 8.33 * 10-3 s
(b) This voltage is small by everyday standards, but keep in mind that the speaker coil’s resistance is also small. To determine
the average induced current in the coil, use the relationship between voltage, resistance, and current:
e 6.79 * 10-2 V
I = = = 6.79 * 10-2 A = 67.9 mA
R 1.00 Æ
This value exceeds the allowed average speaker current of 25.0 mA and therefore the speaker coil is possibly subject to damage.
FOLLOW-UP EXERCISE. In this Example, if the speaker coil were moved farther from the electromagnet, it could reach a point
where the induced average current would be below the “dangerous” level of 25.0 mA. Determine the maximum magnetic field
strength at this point.
As a special case, emfs and currents can be induced in conductors as they are
v
moved through a magnetic field. In this situation, the induced emf is called a
motional emf. To see how this works, consider the situation in 䉳 Fig. 20.11a. As the
bar moves upward, the circuit area increases by ¢A = L¢x (Fig. 20.11a.) At con-
∆A stant speed, the distance traveled by the bar in a time ¢t is ¢x = v¢t. Therefore,
∆x = v∆t ¢A = Lv¢t. The angle between the magnetic field and the normal to the area 1u2
Uniform
external B field
is always 0°. However, the area is changing, so the flux varies. However,
L £ = BA cos 0° = BA; hence ¢£ = B¢A, or ¢£ = BLv¢t. Therefore, from Fara-
day’s law, the magnitude of this “motional” (induced) emf, e, is ƒ e ƒ = ƒ ¢£>¢t ƒ =
2 BLv¢t> ¢t = BLv. This is the fundamental idea behind electric energy generation:
R 1 Move a conductor in a magnetic field, and convert the work done on it into electri-
cal energy. To see some of the details, consider the following Integrated Example.
(a)
INTEGRATED EXAMPLE 20.4 The Essence of Electric Energy

Generation: Mechanical Work into
v Electrical Current
Consider the situation in Fig. 20.11a but with the bar moving down instead. An external
force does work as the movable bar moves, and this work is converted to electrical
energy. Because the “circuit” (wires, resistor, and bar) is in a magnetic field, the flux
Induced through it changes with time, inducing a current. (a) What is the direction of the induced
B field Induced I current in the resistor: (1) from 1 to 2 or (2) from 2 to 1? (b) If the bar is 20 cm long and is
pulled at a steady speed of 10 cm>s, what is the induced current if the resistor has a value
2 of 5.0 Æ and the circuit is in a uniform magnetic field of 0.25 T?
R 1 ( A ) C O N C E P T U A L R E A S O N I N G . In Fig. 20.11a, the magnetic flux points left and is decreas-
ing. According to Lenz’s law, the field due to the induced current must then be to the left to
make up for the reduced flux. Using the induced-current right-hand rule, we find that the
(b) direction of the induced current is from 2 to 1 (Fig. 20.11b), and the correct answer is (2).
䉱 F I G U R E 2 0 . 1 1 Motional emf
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . The flux change is due to an area
(a) As the metal rod is pulled on the
metal frame, the area of the rectan- change as the bar is pushed downward. The analysis for motional emfs has been done
gular loop varies with time. A cur- in the preceding text. Lastly, once the motional emf has been found, the induced current
rent is induced in the loop as a can be determined using Ohm’s law.
result of the changing flux. (b) To Listing the data and converting to SI units:
counteract the increase of flux to the
left, an induced current creates its Given: B = 0.25 T Find: I (induced
own magnetic field to the right. See L = 20 cm = 0.20 m current in the
Integrated Example 20.4. v = 10 cm>s = 0.10 m>s resistor
R = 5.0 Æ
20.2 ELECTRIC GENERATORS AND BACK EMF 705
In the preceding text, it was shown that the magnitude of the induced emf e is given by
BLv, so
ƒ e ƒ = BLv = 10.25 T210.20 m210.10 m>s2 = 5.0 * 10-3 V
Hence the induced current is
e 5.0 * 10-3 V
I = = = 1.0 * 10-3 A
R 5.0 Æ
Clearly this arrangement isn’t a practical way to generate large amounts of electrical energy.
Here the power dissipated in the resistor is only 5.0 * 10-6 W. (You should verify this.)
F O L L O W - U P E X E R C I S E . In this Example, if the field were increased by three times and the
bar’s width changed to 45 cm, what would be the required bar speed to induce a current of
0.10 A?
DID YOU LEARN?

➥ The magnetic flux through a coil is directly proportional to the area of that coil.
➥ An induced emf can be produced in a coil that is rotating in a magnetic field, thus
changing the flux through the coil.
➥ The induced current in a coil depends on how rapidly the magnetic flux changes
with time in that coil.
20.2 Electric Generators and Back emf

➥ How is the generation of electric energy accomplished in a generator?

➥ Why does an electric generator produce only alternating (ac) voltage outputs? B
➥ What is the origin of the “back emf”in an electric motor?
S
One way to induce an emf in a loop is through a change in the loop’s orientation in its
magnetic field (Fig. 20.4). This is the operational principle behind electric generators. N
Brushes
ELECTRIC GENERATORS
Slip rings
An electric generator is a device that converts mechanical energy into electrical
energy. Basically, the function of a generator is the reverse of that of a motor. ac
Recall that a battery supplies direct current (dc). That is, the voltage polarity (and voltmeter
therefore the current direction) do not change. However, most generators produce (a)
alternating current (ac), named because the polarity of the voltage (and therefore the
current direction) change periodically. Thus, the electric energy used in homes and
industry is delivered in the form of alternating voltage and current. (See Chapter 21
for analysis of ac circuits and Chapter 18 for household wiring diagrams.) One cycle
An ac generator is sometimes called an alternator particularly in automobiles. The
Voltage
elements of a simple ac generator are shown in 䉳 Fig. 20.12. A wire loop called an
armature is mechanically rotated in a magnetic field by some external means, such as
water flow or steam hitting turbine blades. The rotation of the blades in turn causes a Time
rotation of the loop. This results in a change in the loop’s magnetic flux and an
ac voltage
induced emf in the loop. The ends of the loop are connected to an external circuit by
means of slip rings and brushes. In this case, the induced currents will be delivered to
(b)
that circuit. In practice, generators have many loops, or windings, on their armatures.
When the loop is rotated at a constant angular speed 1v2, the angle 1u2 between the 䉱 F I G U R E 2 0 . 1 2 A simple ac gen-
magnetic field vector and the area vector of the loop changes with time: u = vt erator (a) The rotation of a wire loop
(assuming that u = 0° at t = 0). As a result, the number of field lines through the loop in a magnetic field produces (b) a volt-
age output whose polarity reverses
changes with time, causing an induced emf. From Eq. 20.1, the flux (for one loop) with each half-cycle. This alternating
varies as voltage is picked up by a brush>slip
£ = BA cos u = BA cos vt ring arrangement as shown.
From this it can be seen that the induced emf will also vary with time. For a rotat-
ing coil of N loops, Faraday’s law yields
¢£ ¢1cos vt2
e = -N = - NBA ¢ ≤
¢t ¢t
Here, B and A have been removed from the time rate of change, because they are
constant. By using methods beyond the scope of this book, it can be shown that
the induced emf expression can be rewritten as
e = 1NBAv2 sin vt
Notice that the product of terms, NBAv, represents the magnitude of the maxi-
mum emf, which occurs whenever sin vt = 1. If NBAv is called eo, the maxi-
mum value of the emf, then the previous equation can be rewritten compactly as
e = eo sin vt (20.4)
Because the sine function varies between 1, the polarity of the emf changes with
time (䉳 Fig. 20.13). Note that the emf has its maximum value eo when u = 90° or
Side view of loop
u = 270°. That is, at the instants when the plane of the loop is par-
(sequential series of loop rotation) allel to the field, and the magnetic flux is zero, the emf will be at its
largest (magnitude). The change in flux is greatest at these angles,
B because although the flux is momentarily zero, it is changing
rapidly due to a sign change. Near the angles that produce the
flux’s largest value (u = 0° and u = 180°), the flux is approxi-
+ eo mately constant and thus the induced emf is zero at those angles.
Because the induced current is produced by this alternating
0° 90° 180° 270° 360°
induced emf, the current also changes direction periodically. In
everyday applications, it is common to refer to the frequency (f)
– eo of the armature [in hertz (Hz) or rotations per second], rather
than the angular frequency (v). Because they are related by
䉱 F I G U R E 2 0 . 1 3 An ac generator v = 2pf, Eq. 20.4 can be rewritten as
e = eo sin 12pft2 (alternator emf)
output A graph of the sinusoidal
output of a generator, with a side (20.5)
view of the corresponding loop ori-
The ac frequency in the United States and most of the western hemisphere is
entations during a cycle, showing the
flux variation with time. Note that 60 Hz. A frequency of 50 Hz is common in Europe and other areas.
the emf is a maximum when the flux Keep in mind that Eqs. 20.4 and 20.5 give the instantaneous value of the emf and
changes most rapidly, as it passes that e varies between +eo and - eo over half of an armature rotational period (1>120
through zero and changes in sign. of a second in the United States). For practical ac electrical circuits, time-averaged
values for ac voltage and current are more important. This concept will be devel-
oped in Chapter 21. To see how various factors influence the generator’s output,
examine Example 20.5 closely. Also, see Insight 20.2, Electromagnetic Induction at
Play: Hobbies and Transportation for ways in which electromagnetic induction
makes for an interesting hobby, and also is used to generate the electric energy
needed to power hybrid automobiles for more fuel-efficient transportation.
EXAMPLE 20.5 An ac Generator: Renewable Electric Energy

A farmer decides to use a waterfall to create a small hydroelec- The generator’s maximum (or peak) emf is given by
tric power plant for his farm. He builds a coil consisting of 1500 eo = NBAv. Because v = 2pf and, for a circle, A = pr2, this
circular loops of wire with a radius of 20 cm, which rotates on can be rewritten as
the generator’s armature at 60 Hz in a magnetic field. To gener-
ate an rms (“average”) voltage of 120 V, he needs to generate a eo = NB1pr2212pf2 = 2p2NBr2f
maximum emf of 170 V (this concept will be discussed in more
detail in Chapter 21). What is the magnitude of the generator’s Solving for B,
magnetic field necessary for this to happen?
eo 170 V
B = = 2.4 * 10-3 T
2p211500210.20 m22160 Hz2
THINKING IT THROUGH. The magnetic field can be deter- =
2p2Nr 2f
mined from the expression for eo.
SOLUTION. F O L L O W - U P E X E R C I S E . In this Example, suppose that the
Given: eo = 170 V Find: B (magnitude of the farmer wanted to generate an emf with an rms value of 240 V,
N = 1500 loops magnetic field) which requires a maximum emf of 340 V. If he chose to do so
r = 20 cm = 0.20 m by changing the size of the coil, what would the new radius
f = 60 Hz have to be?
INSIGHT 20.2 Electromagnetic Induction at Play: Hobbies and Transportation

F I G U R E 1 A two-coil helping power the car, the hybrid engine also supplies electric
metal detector Note both energy (through induction in a generator) to batteries and an
the transmitter (larger electric motor, which in turn supply power to the wheels. In
outer) and receiver this way, more work can be extracted from a gallon of gaso-
(smaller inner) coils. (See line than in a conventional engine.
text for description.)
A cutaway of a typical hybrid car is shown in Fig. 2a.
Hybrid cars currently come in two basic designs: parallel and
series. In the parallel hybrid arrangement (Fig. 2b), the gasoline
engine is connected to the wheels via a standard transmis-
sion. However, it also turns a generator that, through induc-
tion, creates and supplies electric energy to charge the
batteries and>or to operate the electric motor. Sophisticated
power electronics monitor the charge on the batteries and
divert current to where it is needed. The electric motor is con-
nected to the wheels through its own separate transmission,
hence the name parallel hybrid—the gasoline and electric
motor work together, in parallel. Fully hybrid models are capa-
ble of moving the car with either engine alone (for maximum
fuel economy—say, while cruising on a freeway) or both
Electromagnetic induction plays an important part in our leisure simultaneously (when more power is needed—say, while
and transportation activities. For example, hobbyists use metal accelerating onto a freeway).
detectors to hunt for metallic “buried treasure.” A common Alternatively the engines>motors can be connected in
design consists of two coils of wire at the end of a shaft used for series—in the series hybrid automobile. Here the electric motor
sweeping just above the ground (see Fig. 1). At the handheld end is what actually powers the wheels (Fig. 2c). The job of the
are electronics for displaying information about any detected gasoline engine is to supply electric energy (through induc-
items. The outer, or transmitter, coil contains a current oscillating tion in its generator) to the batteries and electric motor. If the
at several thousand hertz, creating an ever-changing magnetic batteries are fully charged and the motor is running well, the
field in the ground below it. (Usually it can penetrate a foot or gasoline engine can idle down or power off. With frequent
more below the surface, depending on soil type and condition.) accelerations, when the electric motor is called on for a high-
If no metallic objects are within range of this oscillating power output, the batteries may drain quickly. Under these
field, then no significant currents will be induced. Therefore, no conditions, the power electronics direct the gasoline engine to
induced magnetic field “echo” will be detected by the inner, or begin generating electric energy to recharge the batteries.
receiver, coil. However, if a metallic object is present, the current Regardless of the design, the object is the same: higher effi-
induced in it will create a magnetic echo (field) that the receiver ciency—that is, more miles per gallon. Hybrids, unlike purely
will detect as an induced emf and current. By means of sophis- electric cars, are never “plugged in”; they derive all their
ticated computer software that evaluates the strength of the energy from burning gasoline. However, they are much more
induced signal, the object’s depth and chemical makeup can be efficient, and thus considerably less polluting, than conven-
estimated. tional cars. Some recent car models employ hybrid engines that
The increasing price of gasoline has many drivers turning are capable of more horsepower than their gasoline counter-
to gas-electric hybrid automobiles, in which the gasoline engine parts. For these reasons, the hybrid engine is increasingly likely
is considerably smaller than conventional ones. In addition to to be the engine of choice for many drivers in the near future.
Integrated power electronics Battery pack Fuel tank

Engine
Electric motor Gasoline Transmission
Generator
Internal
combustion
engine Electric
Batteries Transmission
motor
(b)
Final drive Fuel tank

Gasoline Engine Generator Electric
Transmission Transmission
motor
& automated
Transmission clutch
Batteries
& automated clutch Final drive
(a) (c)
F I G U R E 2 Hybrid automobiles (a) A cutaway of a typical modern hybrid vehicle. (b) A schematic of the main systems in a
parallel hybrid. (c) A schematic of the main systems in a series hybrid. (See text for description.)
䉴 F I G U R E 2 0 . 1 4 Electrical
generation (a) Turbines such as
those depicted here generate electric
energy. (b) Gravitational potential
energy of water, here trapped
behind the Glen Canyon dam on the
Colorado River in Arizona, is con-
verted into electric energy. (c) Wind
kinetic energy is converted into
electric energy by turning the blades
on these windmills in an enormous
“wind farm” in the windy San Gor-
gonio pass east of Los Angeles, Cali-
fornia. The combined output of this
“farm” is equivalent to a medium-
size nuclear power plant. (d) An
artist’s conception of one possible
(a) (b)
wave generator design. Here the
magnet is held fixed to the ocean
bottom and the coil is attached to a
buoy that oscillates as the waves
pass by. An emf is thus generated in
the moving coil as the magnetic flux
through it changes with time.
(c) (d)
In most large-scale ac generators (power plants), the armature is actually stationary,

and magnets revolve about it. The revolving magnetic field produces a time-varying
flux through the coils of the armature and thus an ac output. A turbine supplies the
mechanical energy required to spin the magnets in the generator (䉱 Fig. 20.14a). Tur-
bines are typically powered by steam generated from the heat of combustion of fossil
fuels or by heat generated from nuclear fission (see Chapters 29 and 30).
More recently, as supplies of fossil fuels dwindle, demand for such fuels rises,
and concerns about their emissions arise relative to global climate change, there has
been increased interest in electric generation using renewable energy sources.
Hydroelectricity generates electric energy through the use of falling water to rotate a
turbine, as is shown in Fig. 20.14b.
Fig. 20.14c shows a huge “wind farm” in the San Gorgonio Pass region east of Los
Angeles, California. Intense winds driven by extreme desert heating are created by
the narrowing of the pass. The winds drive the blades which turn the turbines to gen-
erate electric energy from the kinetic energy of the wind.
Figure 20.14d shows an artist’s sketch of one of several proposed ways of using
wave motion to generate electric energy. As the waves go by, the coils oscillate about
the fixed magnets, inducing electric current, thus converting the wave energy into
electric energy. Keep in mind, however, that in terms of physics, the only basic dif-
ference between the various types of electric energy generation is the source of the
energy that turns the turbines. In all cases, some other form of energy is converted
into electric energy.
BACK EMF
Although their main job is to convert electric energy into mechanical energy,
motors also generate (induced) emfs at the same time. Like a generator, a motor
has a rotating armature in a magnetic field. For motors, the induced emf is called a
back emf (or counter emf), , because its direction is opposite that of the line volt-
age and tends to reduce the current in the armature coils.
If V is the line voltage, then the net voltage driving the motor is less than V
(because the line voltage and the back emf are of opposite polarity). Therefore, the
net voltage is Vnet = V - eb. If the motor’s armature has a resistance of R, the cur-
rent the motor draws while in operation is I = Vnet>R = 1V - eb2>R or, solving
for the back emf,
eb = V - IR (back emf of a motor) (20.6)
where V is the line voltage.

The back emf of a motor depends on the rotational speed of the armature and
Armature coils
increases from zero to some maximum value as the armature goes from rest to its
normal operating speed. On startup, the back emf is zero (why?). Therefore the
starting current is a maximum (Eq. 20.6 with eb = 0). Ordinarily, a motor turns R = 8.0 Ω
something, such as a drill bit; that is, it has a mechanical load. Without a load, the I
V = 120 V Ᏹb = 100 V
armature speed will increase until the back emf almost equals the line voltage. The
+ – – +
result is a small current in the coils, just enough to overcome friction and joule
heat losses. Under normal load conditions, the back emf is less than the line volt-
age. The larger the load, the slower the motor rotates and the smaller the back emf.
Driving Battery representation
If a motor is overloaded and turns very slowly, the back emf may be reduced so source of back emf induced
much that the current becomes very large (note that Vnet increases as eb decreases) in armature coils
and may burn out the coils. The back emf plays a vital role in the regulation of a
motor’s operation by limiting the current in it. 䉱 F I G U R E 2 0 . 1 5 Back emf The
back emf in the armature of a dc
Schematically, a back emf in a dc motor circuit can be represented as an motor can be represented as a bat-
“induced battery” with polarity opposite that of the driving voltage (䉴 Fig. 20.15). tery with a polarity opposite that of
To see how the back emf affects the current in a motor, consider the next Example. the driving voltage.
EXAMPLE 20.6 Getting up to Speed: Back emf in a dc Motor

A dc motor has windings that have a resistance of 8.00 Æ and (a) From Eq. 20.6, the current in the windings is
operates at a line voltage of 120 V. With a normal load, there is
a back emf of 100 V when the motor reaches full speed. (See V 120 V
Is = = = 15.0 A
Fig. 20.15.) Determine (a) the starting current drawn by the R 8.00 Æ
motor and (b) the armature current at operating speed under
(b) When the motor is at full speed, the back emf is 100 V;
a normal load.
thus, the current is less.
T H I N K I N G I T T H R O U G H . (a) The only difference between
startup and full speed is that there is no back emf at startup. V - eb 120 V - 100 V
I = = = 2.50 A
The net voltage and resistance determine the current, so R 8.00 Æ
Eq. 20.6 can be applied. (b) At operating speed, the back emf
increases and is opposite in polarity to the line voltage. With little or no back emf, the starting current is relatively
Equation 20.6 can again be used to determine the current. large. When a big motor, such as that of a central air-
conditioning unit, starts up, the lights in the building might
SOLUTION. Listing the data as usual: momentarily dim, because of the large starting current that
Given: R = 8.00 Æ Find: (a) Is (starting current) the motor draws. In some designs, resistors are temporarily
V = 120 V (b) I (operating current) connected in series with a motor’s coil to protect the windings
eb = 100 V from burning out as a result of large starting currents.
F O L L O W - U P E X E R C I S E . In this Example, (a) how much energy is required to bring the motor to operating speed if it takes 10 s
and the back emf averages 50 V during that time? (b) Compare this amount with the amount of energy required to keep the
motor running for 10 s once it reaches its operating conditions.
Because motors and generators are opposites, so to speak, and a back emf
develops in a motor, you may be wondering whether a back force develops in a
generator. The answer is yes. When an operating generator is not connected to an
external circuit, no current exists, and therefore there is no magnetic force on the
armature coils. However, when the generator delivers energy to an external circuit
and current is in the coils, the magnetic force on the armature coils produces a
countertorque that opposes the rotation of the armature. As more current is drawn,
the countertorque increases and a greater driving force is needed to turn the arma-
ture. Therefore, the higher the generator’s current output, the greater the energy
expended (that is, fuel consumed) in overcoming the countertorque.
DID YOU LEARN?

➥ Rotating a generator’s coil in a magnetic field induces an emf in the coil.
➥ The coil of a generator rotates, and thus the flux rate of change varies with time;
thus, the generator output also varies with time.
➥ In a motor, electric energy is used to rotate a coil in a magnetic field. In accordance
with Faraday’s and Lenz’s laws, this produces an emf in the reverse direction (called
a back emf ) in an attempt to impede the rotation.
20.3 Transformers and Power Transmission

➥ What fundamental law of electromagnetism is transformer action based on?

➥ How do the primary windings compare to the secondary windings in a step-up
transformer?
➥ Why are step-up transformers used before electric energy is transmitted over the
wires?
Electric energy is transmitted by power lines over long distances. It is desirable to

minimize I2R losses (joule heat) that can occur in these transmission lines. Because
the resistance of a line is fixed, reducing I2R losses means reducing current. How-
ever, the power output of a generator is determined by its outputs of current and
voltage 1P = IV2, and for a fixed voltage, such as 120 V, a reduction in current
would mean a reduced power output. It might appear that there is no way to
reduce the current while maintaining the power level. Fortunately, electromag-
netic induction enables the reduction of power transmission losses by increasing
voltage while simultaneously reducing current in such a way that the delivered
Iron core
power is essentially unchanged. This is done using a device called a transformer.
ac
source A simple transformer consists of two coils of insulated wire wound on the same
iron core (䉳 Fig. 20.16a). When ac voltage is applied to the input coil, or primary coil,
the alternating current produces an alternating magnetic flux concentrated in the
iron core, without any significant leakage of flux outside the core. Under these con-
ditions, the same changing flux also passes through the output coil, or secondary coil,
Primary Secondary inducing an alternating voltage and current in it. (Note that it is common in trans-
coil coil former design to refer to emfs as “voltages,” as was done in Chapter 18.)
(a) Step-up transformer: The ratio of the induced voltage in the secondary coil to that of the voltage in
high-voltage the primary coil depends on the ratio of the numbers of turns in the two coils. By
(low-current) output Faraday’s law, the induced voltage in the secondary coil is
ac ¢£
source Vs = - Ns
¢t
where Ns is the number of turns in the secondary coil. The changing flux in the
primary coil produces a back emf of
Primary Secondary
¢£
Vp = - Np
coil coil ¢t
(b) Step-down transformer: where Np is the number of turns in the primary coil. If the resistance of the pri-
low-voltage mary coil is neglected, this back emf is equal in magnitude to the external voltage
(high-current) output
applied to the primary coil (why?). Forming a ratio of output voltage (secondary)
䉱 F I G U R E 2 0 . 1 6 Transformers to input voltage (primary) yields
(a) A step-up transformer has more Vs -Ns1¢£>¢t2
-Np1¢£>¢t2
turns in the secondary coil than in the =
primary coil. (b) A step-down trans- Vp
former has more turns in the primary
coil than in the secondary coil. or
20.3 TRANSFORMERS AND POWER TRANSMISSION 711
Vs Ns
= (voltage ratio in a transformer) (20.7)
Vp Np
If the transformer is 100% efficient (that is, there are no energy losses), then the
power input is equal to the power output 1Pp = Ps2. Using the expression for elec-
tric power, P = IV, 100% efficiency can therefore be rewritten as
Ip Vp = Is Vs (20.8)
Although some energy is always lost to joule heat, this equation is a good approxi-
mation, as a well-designed transformer will have an efficiency greater than 95%.
(The details of transformer energy losses will be discussed shortly.) Assuming this
ideal case, from Eq. 20.8, the transformer currents and voltages are related to the
turn ratio by
Ip Vs Ns
= = (20.9)
Is Vp Np
To summarize the transformer action in terms of voltage and current output,
Ns
Vs = ¢ ≤V (20.10a)
Np p
and
Np
Is = ¢ ≤ Ip (20.10b)
Ns
If the secondary coil has more windings than the primary coil does (that is,
Ns>Np 7 1), as in Fig. 20.16a, the voltage is “stepped up,” because Vs 7 Vp. This is
called a step-up transformer. Notice that because of this there is less current in the
secondary than in the primary (Np>Ns 6 1 and Is 6 Ip).
If the secondary coil has fewer turns than the primary does, we have a step-
down transformer (Fig. 20.16b). In the usual transformer language, this means that
the voltage is “stepped down,” and the current, therefore, is increased. Depending
on the design details, a step-up transformer may be used as a step-down trans-
former by simply reversing output and input connections.
INTEGRATED EXAMPLE 20.7 Transformer Orientation: Step-Up or Step-Down Configuration?

An ideal 600-W transformer has 50 turns on its primary coil The secondary voltage can be found using Eq. 20.10a with a
and 100 turns on its secondary coil. (a) Is this transformer (1) a turn ratio of 2, because Ns = 2Np
step-up or (2) a step-down arrangement? (b) If the primary
≤ V = 1221120 V2 = 240 V
Ns
coil is connected to a 120-V source, what are the output volt- Vs = ¢
age and current of this transformer? Np p
( A ) C O N C E P T U A L R E A S O N I N G . Step-up and step-down refer to If the transformer is ideal, then the input power equals the
what happens to the voltage, not the current. Because the output power. On the primary side, the input power is
voltage is proportional to the number of turns, in this case the Pp = Ip Vp = 600 W, so the input current must be
secondary voltage is greater than the primary voltage. Thus 600 W 600 W
the correct answer is (1), a step-up transformer. Ip = = = 5.00 A
Vp 120 V
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . The output
Because the voltage is stepped up by a factor of two, the out-
voltage can be determined from Eq. 20.10a once the turn
put current should be stepped down by a factor of two. From
ratio is established. From the power, the current can be
Eq. 20.10b,
determined.
Np
≤ Ip = a b15.00 A2 = 2.50 A
Given: Np = 50 Find: Vs and Is (secondary volt- 1
Is = ¢
Ns = 100 age and current) Ns 2
Vp = 120 V
F O L L O W - U P E X E R C I S E . (a) When a European visitor (the average ac voltages are 240 V in Europe) visits the United States, what
type of transformer should be used to enable her hair dryer to work properly? Explain. (b) For a 1500-W hair dryer (assumed
ohmic), what would be the transformer’s input current in the United States, assuming it to be ideal?
The preceding relationships strictly apply only to ideal (or “lossless”) transform-
Pivot ers; actual transformers have electric energy losses during the “transformation.”
That is, some electric energy is converted into other types. Well-designed transform-
ers generally are designed to have such losses of less than 5%. Essentially, then,
S v there is no such thing as an ideal transformer.
I Many factors combine to determine how close a real transformer comes to per-
F
forming like an ideal one. First, there is flux leakage; that is, not all of the flux
F passes through the secondary coil. In some transformer designs, one of the insu-
I v N lated coils is wound directly on top of the other (interlocking) rather than having
two separate coils. This configuration helps minimize flux leakage while reducing
B transformer size.
Second, the ac current in the primary means there is a changing magnetic flux
(a) through those coils. In turn, this gives rise to an induced emf in the primary. This
is called self-induction. By Lenz’s law, the self-induced emf will oppose the change
in current and thus limit the primary current (this is an effect similar to that of the
S back emf in a motor).
FN–N A third reason that transformers are less than ideal is joule heating (I2R losses)
S v B
F due to the resistance of the wires. Usually this loss is small because the wires have
N
FS–S little resistance.
N Lastly, consider the effect of induction on the core material itself. To increase
magnetic flux, the core is made of a highly permeable material (such as iron), but
(b) such materials are also good conductors. The changing magnetic flux in the core
induces emfs there, which in turn create eddy (or “swirling”) currents in the core
Pivot material. These eddy currents can cause energy loss between the primary and sec-
ondary by heating the core (I2R losses again).
S To reduce the loss of energy due to eddy currents, transformer cores are made
of thin sheets of material (usually iron) laminated with an insulating glue between
them. The insulating layers between the sheets break up the eddy currents or con-
N
fine them to the thin sheets, greatly reducing energy loss.
The effects of eddy currents can be demonstrated by allowing a plate made of a
conductive, but nonmagnetic, metal, such as aluminum, to swing through a mag-
netic field (䉳 Fig. 20.17a). As it enters or leaves the field, induced eddy currents are
set up in the plate because the magnetic flux through its area is changing. By
Lenz’s law, eddy currents are induced in such a direction as to oppose the flux
change.
(c) When the plate enters the field (the position of the left-hand plate in
Fig. 20.17a), a counterclockwise current is induced. (You should apply Lenz’s law
䉱 F I G U R E 2 0 . 1 7 Eddy currents to show this.) The induced current produces its own magnetic field, which means
(a) Eddy currents are induced in a
that, in effect, the plate has a north magnetic pole near the permanent magnet’s
metal plate moving in a magnetic
field. The induced currents oppose north pole and a south magnetic pole near the permanent magnet’s south pole
the change in flux. These currents (Fig. 20.17b). Two repulsive magnetic forces act on the plate. The effect of the net
then experience a retarding mag- force is to slow the plate down as it enters the field. The plate’s eddy currents are
netic force to oppose the motion reversed in direction as it leaves the field, producing a net attractive magnetic
first into, then out of, the field
force, thus tending to slow the plate from leaving the field. In both cases, the
region. To see this, note that the cur-
rents reverse direction as the plate induced emfs act to slow the plate’s motion.
leaves the field. (b) An overhead The reduction of eddy currents (similar to how the laminated layers in a trans-
view as the plate swings toward the former work) can be demonstrated by using a plate with slits cut into it
field from
B
the left. The retarding (Fig. 20.17c). When this plate swings between the magnet’s poles, it swings rela-
force F (to slow it from entering the
tively freely, because the eddy currents are greatly reduced by the air gaps (slits).
field) results fromBthe two repulsive
B
forces (FN - N and FS - S) acting Consequently, the magnetic force on the plate is also reduced.
between magnetic poles. This is Eddy currents can actually have practical uses in some applications. For exam-
because the side of the plate closest ple, their damping effect has been applied in the braking systems of rapid transit
to the north pole of the permanent railcars. When an electromagnet (housed in the car) is turned on, it applies a mag-
magnet acts as a north pole, and the
netic field to a rail. The repulsive force due to the induced eddy currents in the rail
other side acts as a south pole. (c) If
the plate has slits, the eddy currents, acts as a braking force (䉴 Fig. 20.18). As the car slows, the eddy currents in the rail
and thus magnetic forces, are drasti- decrease, allowing a smooth braking action.
cally reduced and the plate will
swing more freely.
20.3 TRANSFORMERS AND POWER TRANSMISSION 713
Subway 䉳 F I G U R E 2 0 . 1 8 Electromagnetic
v car braking and mass transit When
braking, a train energizes an electro-
magnet onboard. This electromagnet
straddles a long metal rail. The
F S F
S induced currents in the rail produce
mutually repulsive forces 1F2
B
N N
between the rail and the train,
thereby slowing the train.
POWER TRANSMISSION AND TRANSFORMERS

For power transmission over long distances, transformers provide a way to
increase the voltage and reduce the current of an electric generator, thus cutting 24000 V
down the joule heating (I2 R) losses in the transmission wires that are carrying cur-
rent. A schematic diagram of an ac power distribution system is shown in Generator
䉴 Fig. 20.19. The voltage output of the generator is stepped up, reducing the cur-
rent. The energy is transmitted over long distances to an area substation near the
consumers. There, the voltage is stepped down, increasing the current. There are
further step-downs at distributing substations and utility poles before the electric- Step
ity is supplied to homes and businesses at the normal voltage and current. up
The following Example illustrates the benefits of being able to step up the volt-
age (and step down the current) for electrical power transmission.
230000 V
EXAMPLE 20.8 Cutting Your Losses: Power Transmission Station
at High Voltage
A small hydroelectric power plant produces energy in the form of electric current at
10 A and a voltage of 440 V. The voltage is stepped up to 4400 V (by an ideal trans-
former) for transmission over 40 km of power line, which has a total resistance of 20 Æ . Step
(a) What percentage of the original energy would have been lost in transmission if the down
voltage had not been stepped up? (b) What percentage of the original energy is actually
lost when the voltage is stepped up?
T H I N K I N G I T T H R O U G H . (a) The power output can be computed from P = IV and
Area
compared with the power lost in the wire, P = I 2R. (b) Equations 20.10a and 20.10b
substation
should be used to determine the stepped-up voltage and stepped-down currents, respec- 100000 V
tively. Then the calculation is repeated, and the results are compared with those of part (a).
SOLUTION.
Given: Ip = 10 A Find: (a) percentage energy loss without voltage step-up
Vp = 440 V (b) percentage energy loss with voltage step-up Step
Vs = 4400 V down
R = 20 Æ
(a) The power output by the generator is Distributing
P = Ip Vp = 110 A21440 V2 = 4400 W substation 20000 V
The rate of energy loss of the wire (joules per second, or watts) in transmitting a current
of 10 A is very high, because
Ploss = I 2R = 110 A22120 Æ2 = 2000 W
Step
Thus, the percentage of the produced energy lost to joule heat in the wires is nearly 50% down
because
Ploss 2000 W
% loss = * 100% = * 100% = 45% User 120–240 V
P 4400 W
(b) When the voltage is stepped up to 4400 V, this allows for transmission of energy at a 䉱 F I G U R E 2 0 . 1 9 Power transmis-
current that is reduced by a factor of 10 from its value in part (a). Thus the secondary sion A diagram of a typical electri-
current is cal power distribution system.
Vp
Is = ¢ ≤ Ip = a b110 A2 = 1.0 A
440 V
Vs 4400 V
The power (loss) is thus reduced by a factor of 100, because it varies as the square of
the current:
Ploss = I 2R = 11.0 A22120 Æ2 = 20 W
Therefore, the percentage of power lost is also reduced by a factor of 100 to a much
more acceptable level:
Ploss 20 W
% loss = * 100% = * 100% = 0.45%
P 4400 W
F O L L O W - U P E X E R C I S E . Some heavy-duty electrical appliances, such as water pumps,
can be wired to 240 V or 120 V. Their power rating is the same regardless of the voltage
at which they run. (a) Explain the efficiency advantage of operating such appliances at
the higher voltage. (b) For a 1.00-hp pump (746 W), estimate the ratio of the power lost
in the wires at 240 V to the power lost at 120 V (assuming that all resistances are ohmic
and the connecting wires are the same).
DID YOU LEARN?

➥ Transformer action is based on Faraday’s law of induction.
➥ There are more turns in the secondary windings than in the primary windings of a
step-up transformer
➥ Using a transformer, electric power is transmitted at very high voltage and low
current; this means much lower losses to joule heating in the wires.
20.4 Electromagnetic Waves

➥ What is the relationship between the electric and magnetic fields in an electromag-
netic wave?
➥ What is “radiation pressure”?
➥ How are the different regions of the electromagnetic spectrum categorized?
Electromagnetic waves (or electromagnetic radiation) were considered as a means

of heat transfer in Section 11.4. The production and characteristics of electromag-
netic radiation can now be understood, because these waves are composed of elec-
tric and magnetic fields.
Scottish physicist James Clerk Maxwell (1831–1879) is credited with first bring-
ing together, or unifying, electric and magnetic phenomena. Using mathematics
beyond the scope of this book, he took the equations that governed each field and
predicted the existence of electromagnetic waves. In fact, he went further and cal-
culated their speed in a vacuum, and his prediction agreed with the experiment.
Because of these contributions, the set of equations is known as Maxwell’s equa-
tions, although they were, for the most part, developed by others (for example,
Faraday’s law of induction).
Essentially, Maxwell showed how the electric field and the magnetic field could
be thought of as a single electromagnetic field. The apparently separate fields are
symmetrically related in the sense that either one can create the other under the
proper conditions. This symmetry is evident by looking at the equations (not
shown). A qualitative summary of the results is sufficient:
A time-varying magnetic field produces a time-varying electric field.
A time-varying electric field produces a time-varying magnetic field.
The first statement restates, in field language, our observations in Section 20.1: A
changing magnetic flux gives rise to an induced emf. The second statement
(which will not be studied in detail) is crucial to the self-propagating characteristic
of electromagnetic waves. Together, these two phenomena enable these waves to
travel through a vacuum, whereas all other waves, such as string waves, require a
supporting medium.
20.4 ELECTROMAGNETIC WAVES 715
c 䉳 F I G U R E 2 0 . 2 0 Source of elec-
Electric field
or Magnetic field tromagnetic waves Electromagnetic
waves are produced, fundamen-
l tally, by accelerating electric
Oscillator
charges. (a) Here charges (electrons)
in a metal antenna are driven by an
c oscillating voltage source. As the
c antenna polarity and current direc-
tion periodically change, alternating
Antenna B electric and magnetic fields propa-
E gate outward. The electric and mag-
netic fields are perpendicular to the
(a) c (b) direction of wave propagation.
Thus, electromagnetic waves are
transverse waves. (b) At large dis-
tances from the source, the initially
According to Maxwell’s theory, accelerating electric charges, such as an oscillating curved wavefronts become planar.
electron, produce electromagnetic waves. The electron in question could, for exam-
ple, be one of the many electrons in the metal antenna of a radio transmitter, driven
by an electrical (voltage) oscillator at a frequency of 106 Hz (1 MHz). As each electron
oscillates, it continually accelerates and decelerates and thus radiates an electromag-
netic wave (䉱 Fig. 20.20a). The driven oscillations of many electrons produce time-
varying electric and magnetic fields in the vicinity of the antenna. The electric field,
shown in red in Fig. 20.20a, is in the plane of the page and continually changes direc-
tion, as does the magnetic field (shown in blue and pointing into and out of the page).
Both the electric and the magnetic fields carry energy and propagate outward
at the speed of light. This speed is symbolized by the letter c. To three significant
figures, c = 3.00 * 108 m>s. At large distances from the source, these electromag-
netic waves become plane waves. (Figure 20.20b shows a wave at an instant in
time.) Here, the electric field 1E2 is perpendicular to the magnetic field 1B2, and
B B
B B
each varies sinusoidally with time. Both E and B are perpendicular to the direction
of wave propagation. Thus, electromagnetic waves are transverse waves, with the
fields oscillating perpendicularly to the direction of propagation. According to
Maxwell’s theory, as one field changes, it creates the other. This process, repeated
again and again, gives rise to the traveling electromagnetic wave we call light. An
important result of all this is as follows:
In a vacuum, all electromagnetic waves, regardless of frequency or wavelength, travel
at the same speed, c = 3.00 * 108 m>s.
For everyday distances, the time delay due to the speed of light can usually be
neglected. However, for interplanetary trips, this delay can be a problem. Consider
the following Example.
EXAMPLE 20.9 Long-Distance Guidance: The Speed of Electromagnetic Waves in a Vacuum

The first successful Mars landings were the Viking probes in T H I N K I N G I T T H R O U G H . This situation calls for a time–
1976. They sent radio and TV signals (both are electromag- distance calculation. The planets are farthest apart when they
netic waves) back to the Earth. How much longer would it are on opposite sides of the Sun and separated by a distance
have taken for a signal to reach us when Mars was farthest of dM + dE. (This arrangement requires signals to be sent
from the Earth than when it was closest to us? The average through the Sun, which, of course, is not possible. However, it
distances of Mars and the Earth from the Sun are 229 million does serve to determine the upper limit on transmission
km (dM) and 150 million km (dE) respectively. Assume that times.) The planets are closest when they are aligned on the
both planets have circular orbits, and use the average dis- same side of the Sun. In this case, their separation distance is
tances as the radii of the circles. at a minimum value of dM - dE. (Draw a diagram to help
visualize this.) Because the speed of electromagnetic waves in
a vacuum is known, the times can be found from t = d>c.
SOLUTION. Listing the data and converting the distances to meters:
Given: dM = 229 * 106 km = 2.29 * 1011 m Find: ¢t (difference in time for light to travel the
dE = 150 * 106 km = 1.50 * 1011 m longest and shortest distances)
Radio and TV waves travel at speed c. The longest travel time tL is

3.79 * 1011 m
= 1.26 * 103 s 1or 21.1 min2
dM + dE
tL = =
c 3.00 * 108 m>s
For the shortest distance, the shortest travel time tS is
7.90 * 1010 m
= 2.63 * 102 s 1or 4.39 min2
dM - dE
tS = =
c 3.00 * 108 m>s
The difference in the times is thus ¢t = tL - tS = 1.00 * 103 s (or 16.7 min).
F O L L O W - U P E X E R C I S E . Assume that a Rover Martian vehicle is heading for a collision with a rock 2.0 m ahead of it. When it is at
that distance, the vehicle sends a picture of the rock to controllers on Earth. If Mars is at the closest point to the Earth, what is the
maximum speed that the Rover could have and still avoid a collision? Assume that the video signal from the Rover reaches the
Earth and that the signal for it to stop is sent back immediately.
RADIATION PRESSURE
y An electromagnetic wave carries energy. Consequently, it can do work and can exert a
force on a material it strikes. Consider light striking an electron at rest on a surface
(䉳 Fig. 20.21). The electric field of the wave exerts a force on the electron, giving it a
E
downward velocity 1v B
2, as shown in the figure. Because a charged particle moving in a
magnetic field experiences a force, there is a magnetic force on the electron, due to the
e– magnetic field component of the light wave. By the force right-hand rule, this force is in
x B
the direction that the wave is propagating (Fig. 20.21a). Therefore, because the electro-
F z
magnetic wave will produce the same force on many electrons in a material, it exerts a
v
force on the surface as a whole, and that force is in the direction in which it is traveling.
The radiation force per area is called radiation pressure. Radiation pressure is neg-
ligibly small for most everyday situations, but it can be important in atmospheric and
astronomical phenomena, as well as in atomic and nuclear physics, where masses are
䉱 F I G U R E 2 0 . 2 1 Radiation pres- small and there is no friction. For example, radiation pressure plays a key role in deter-
sure The electric field of an electro- mining the direction in which the tail of a comet points. Sunlight delivers energy to the
magnetic wave that strikes a surface
acts on an electron, giving it a veloc- comet’s “head,” which consists of ice and dust. Some of this material evaporates as the
ity. The magnetic field then exerts a comet nears the Sun, and the evaporated gases are pushed away from the Sun by radi-
force on that moving charge in the ation pressure. Thus, the tail generally points away from the Sun, no matter whether
direction of propagation of the inci- the comet is approaching or leaving the Sun’s vicinity.
dent light. (Verify this direction, Another potential use of radiation pressure from sunlight is to propel interplane-
using the magnetic right-hand force
rule.) tary “sailing” satellites outward from the Sun toward the outer planets in an ever
enlarging, spiraling orbit (䉲 Fig. 20.22a). To create enough force, given the extremely
low pressure of the sunlight, the sails would have to be very large in area, and the
satellite would have to have as little mass as possible. The payoff is that no fuel
(except for small amounts for course corrections) would be needed once the satellite
was launched. Consider the following Conceptual Example, which concerns radia-
tion pressure and space travel.
䉴 F I G U R E 2 0 . 2 2 Sailing the
solar system (a) A space probe
launched from the Earth (E) F F pi = p and pf = 0
equipped with a large sail would be
acted on by radiation pressure from
sunlight (the Sun is at S). This cost- ∴∆ p = p
free force would cause the satellite E
Dark sail
to spiral outward. With proper plan-
ning, the craft could get to outer F
S F
planets with little or no extra fuel.
Note the reduction in force with dis- pf = – p
tance. (b) Is it better for the sail to be 2F
dark or shiny? See Conceptual
Example 20.10 and review momen- F pi = p
tum conservation. ∴∆ p = 2p
(a)
Shiny sail
(b)
CONCEPTUAL EXAMPLE 20.10 Sailing the Sea of Space: Radiation Pressure in Action
Consider the design of a relatively light spacecraft with a tic collision (such as a putty wad sticking to a door), and the
huge “sail” to be used as an interplanetary probe. It would be sail would acquire all of the momentum 1p B
2 originally pos-
designed to use the pressure of sunlight to propel it to the sessed by the radiation.
outer planets using little or no power of its own. To get the However, if the radiation is reflected, the situation is analo-
maximum propulsive force, what kind of surface should the gous to a completely elastic collision, like a Superball bounc-
sail have: (a) shiny and reflective, (b) dark and absorptive, or ing off a wall (see Section 6.1). Because the momentum of the
(c) surface characteristics would not matter? radiation after the collision would be equal to its original
momentum in magnitude, but opposite in direction, its
B B
REASONING AND ANSWER. At first glance, you might think momentum would thus be reversed (from p to - p ). To con-
that the answer is (c). However, as we have seen, radiation is serve momentum, the momentum transferred to the shiny
capable of exerting force and can transfer momentum to sail would be twice as great as that for the dark sail. Because
whatever it strikes. The interaction between the radiation and force is the rate of change of momentum, reflective sails
the sail can be described in terms of conservation of momen- would experience, on average, twice as much force as absorp-
tum, as shown in Fig. 20.22b. (See Section 6.3.) If the radiation tive ones. So the answer is (a).
is absorbed, the situation is analogous to a completely inelas-
F O L L O W - U P E X E R C I S E . (a) Would the sail in this Example provide less or more acceleration as the interplanetary sailing ship
moves farther from the Sun? (b) Explain how a change in sail area could counteract this change.
TYPES OF ELECTROMAGNETIC WAVES

Electromagnetic waves are classi-
fied by the range of frequencies or Microwaves Ultraviolet light Gamma rays
wavelengths they encompass.
Recall from Chapter 13 that fre-
quency and wavelength are Radio, TV waves Infrared light X-rays
inversely related by the traveling
104 106 108 1010 1012 1014 1016 1018
wave relationship l = c>f, where
the speed of light, c, has been substi-
Frequency in Hz
tuted for the general wave speed v.
The higher the frequency, the
Visible light
shorter the wavelength, and vice Red Orange Yellow Green Blue Violet
versa. The electromagnetic spec-
trum is continuous, so the limits of
the various types of radiation are 700 600 550 500 400
approximate (䉴 Fig. 20.23). 䉲 Table (4.3 × 1014 Hz) Wavelength in nm (7.5 × 1014 Hz)
20.1 lists these ranges for the gen-
eral types of electromagnetic waves. 䉱 F I G U R E 2 0 . 2 3 The electro-
magnetic spectrum The spectrum of
frequencies or wavelengths is
Power Waves Electromagnetic waves with a frequency of 60 Hz result from
divided into various regions, or
alternating currents in electric power lines. These power waves have a wave- ranges. The visible light region is a
length of 5.0 * 106 m, or 5000 km (more than 3000 mi). Waves of such low fre- very small part of the total spec-
quency are of little practical use. They may occasionally produce a so-called 60-Hz trum. For visible light, wavelengths
hum on your stereo or introduce, via induction, unwanted electrical noise in deli- are usually expressed in nanometers
11 nm = 10-9 m2. (The relative sizes
cate instruments. More serious concerns have been expressed about the possible
of the wavelengths at the top of the
effects of these waves on health. Some early research tended to suggest that very figure are not to scale.)
low-frequency fields may have potentially harmful biological effects on cells and
tissues. However, recent surveys indicate that this is not the case.
Radio and TV Waves Radio and TV waves are generally in the frequency range
from 500 kHz to about 1000 MHz. The AM (amplitude-modulated) band runs from
530 to 1710 kHz (1.71 MHz). Higher frequencies, up to 54 MHz, are used for
“shortwave” bands. TV bands range from 54 MHz to 890 MHz. The FM (frequency-
modulated) radio band runs from 88 to 108 MHz, which lies in a gap between chan-
nels 6 and 7 of the range of TV bands. Cellular phones use radio waves to transmit
TABLE 20.1 Classification of Electromagnetic Waves

Approximate Frequency Approximate Wavelength
Type of Wave Range (Hz) Range (m) Some Typical Sources
Power waves 60 5.0 * 106 Electric currents

Radio waves—AM 0.53 * 106 - 1.7 * 106 570 - 186 Electric circuits>antennae
6 6
Radio waves—FM 88 * 10 - 108 * 10 3.4 - 2.8 Electric circuits>antennae
6 6
TV 54 * 10 - 890 * 10 5.6 - 0.34 Electric circuits>antennae
9 11
Microwaves 10 - 10 10 -1
- 10 -3
Special vacuum tubes
Infrared radiation 1011 - 1014 10-3 - 10-7 Warm and hot bodies, stars
14 14
Visible light 4.0 * 10 - 7.0 * 10 10 -7
The Sun and other stars, lamps
14 17
Ultraviolet radiation 10 - 10 10 -7
- 10 -10
Very hot bodies, stars, special lamps
17 19
X-rays 10 - 10 10 -10
- 10 -12
High-speed electron collisions, atomic processes
Gamma rays Above 1019 Below 10-12 Nuclear reactions, nuclear decay processes
voice communication in the ultrahigh-frequency (UHF) band, with frequencies

similar to those used for TV channels 13 and higher.
Early global communications used the “shortwave” bands, as do amateur
(ham) radio operators today. But how are the normally straight-line radio waves
transmitted around the curvature of the Earth? This feat is accomplished by reflec-
tion off ionic layers in the upper atmosphere. Energetic particles from the Sun ion-
ize gas molecules, giving rise to several ion layers. Certain of these layers reflect
radio waves. By “bouncing” radio waves off these layers, radio transmissions can
be sent beyond the horizon, to any region of the Earth.
Such reflections requires the ionic layers to have fairly uniform density. When,
from time to time, a solar disturbance produces a larger-than-average shower of
energetic charged particles that upsets this uniformity, a communications “black-
out” can occur as the radio waves are scattered in various directions rather than
reflected in straight lines. To avoid such disruptions, global communications have,
in the past, relied largely on transoceanic cables. More recently, an orbiting net-
work of communications satellites, which can provide line-of-sight transmission
to any point on the globe, has been constructed.
Microwaves Microwaves, with frequencies in the gigahertz (GHz) range, are

produced by special vacuum tubes (called klystrons and magnetrons). Microwaves
are commonly used in communications, microware ovens, and radar applications.
In addition to its roles in navigation and guidance, radar provides the basis for the
speed guns used to time such things as baseball pitches and motorists through the
use of the Doppler effect (see Section 14.5). When the waves are reflected off an
object of interest, the magnitude and sign of the frequency shift allow determina-
tion of the object’s velocity.
Infrared (IR) Radiation The infrared (IR) region of the electromagnetic spec-
trum lies adjacent to the low-frequency, or long-wavelength, end of the visible
spectrum. A warm body emits IR radiation, which depends on that body’s tem-
perature. (See Chapter 27.) An object at or near room temperature emits radiation
in the far IR region. (“Far” means relative to the visible region.)
Recall from Section 11.4 that IR radiation is sometimes referred to as “heat
rays.” This is because water molecules, which are present in most materials and
which possess electrical permanent polarization, readily absorb electromagnetic
radiation at frequencies in the IR wavelength region. When they do, their random
thermal motion is increased—and the molecules “heat up,” as do their surround-
ings. IR lamps are used in therapeutic applications, such as easing pain in strained
muscles, and to keep food warm. IR is also associated with maintaining the
Earth’s temperature through the greenhouse effect. In this effect, incoming visible
light (which passes relatively easily through the atmosphere) is absorbed by the
Earth’s surface and reradiated as IR radiation. This IR radiation is in turn trapped
by “greenhouse gases,” such as carbon dioxide and water vapor, which are
opaque to IR radiation. Its name comes from the actual glass-enclosed green-
house, where the glass rather than atmospheric gases traps the IR energy.
Visible Light The region of visible light occupies only a small portion of the
electromagnetic spectrum and covers a frequency range from about 4 * 1014 Hz
to about 7 * 1014 Hz. In terms of wavelengths, the range is from about 700 to 400
nm (Fig. 20.23). Recall that 1 nanometer 1nm2 = 10-9 m. Only the radiation in this
region activates the receptors on the retina of human eyes. Visible light emitted or
reflected from objects provides us with visual information about our world. Visi-
ble light and optics will be discussed in Chapters 22 to 25.
It is interesting to note that not all animals are sensitive to the same range of
wavelengths. For example, snakes can visually detect infrared radiation, and the
visible range of many insects extends well into the ultraviolet range. The sensitiv-
ity range of the human eye conforms closely to the spectrum of wavelengths emit-
ted by the Sun. The human eye’s maximum sensitivity is in the same yellow-green
region where the Sun’s energy output is at its maximum (wavelengths of about
550 nm).
Ultraviolet (UV) Radiation The Sun’s spectrum has a small component of

ultraviolet (UV) light, whose frequency range lies beyond the violet end of the vis-
ible region. UV is also produced artificially by special lamps and very hot objects.
In addition to causing tanning of the skin, UV radiation can cause sunburn and>or
skin cancer if exposure to it is too high.
Upon its arrival at the Earth, most of the Sun’s UV emission is absorbed in the
ozone (O3) layer in the atmosphere, at an altitude of about 30 to 50 km (about 20 to
30 mi). Because the ozone layer plays a protective role, there is concern about its
depletion due to chlorofluorocarbon gases (such as Freon, once commonly used in
refrigerators) that drift upward and react with the ozone.
Most UV radiation is absorbed by ordinary glass. Therefore, you cannot get
much of a tan through glass windows. Sunglasses are labeled to indicate which
UV protection standards they meet in shielding the eyes from this potentially
harmful radiation. Certain types of high-tech glass (called “photogray” or “transi-
Heating
tion” glass) darken when exposed to UV. These materials are used to create sun- filament
glasses that darken when exposed to sunlight. Of course, these sunglasses aren’t
Cathode
very useful while driving a car. (Why?) Welders wear special glass goggles to pro-
tect their eyes from large amounts of UV produced by the arcs of welding torches. Accelerated
electrons
Similarly, it is important to shield your eyes from sunlamps or snow-covered sur-
faces. The UV component of sunlight reflected from snow-covered surfaces can
produce snowblindness in unprotected eyes. X-rays
X-Rays Beyond the ultraviolet region of the electromagnetic spectrum is the Target
important X-ray region. We are familiar with X-rays primarily through medical Anode
applications. X-rays were discovered accidentally in 1895 by the German physicist +
High-voltage
Wilhelm Roentgen (1845–1923) when he noted the glow of a piece of fluorescent source
paper, evidently caused by some mysterious radiation coming from a cathode ray
tube. Because of the apparent mystery involved, this radiation was named x-radia-
tion, or X-rays for short. 䉱 F I G U R E 2 0 . 2 4 The X-ray tube
The basic elements of an X-ray tube are shown in 䉴 Fig. 20.24. An accelerating Electrons accelerated through a
voltage, typically several thousand volts, is applied across the electrodes in a large voltage strike a target elec-
sealed, evacuated tube. Electrons emitted from the heated negative electrode trode. There they slow down and
interact with the electrons of the tar-
(cathode) are accelerated toward the positive electrode (anode). When striking the get material. Energy is emitted in
anode and thus decelerate, some of their kinetic-energy is converted to electro- the form of X-rays during this
magnetic energy in the form of X-rays. “braking” (deceleration) process.
A similar process takes place in color TV picture tubes, which use high voltages
and electron beams. When the high-speed electrons hit the screen, they can emit
X-rays. Fortunately, most tube televisions still in operation (that is, those that have
not been replaced by LCD or plasma technology) have the shielding necessary to
protect viewers from exposure to this radiation. In the early days of color televi-
sion, this was not always the case—hence the warning that came with the set: “Do
not sit too close to the screen.”
As will be discussed in Chapter 27, the energy carried by electromagnetic radia-
tion depends on its frequency. High-frequency X-rays have very high energies and
can cause cancer, skin burns, and other harmful effects. However, at low intensities,
(a) X-rays can be used with relative safety to view the internal structure of the human
body and other opaque objects.* X-rays can pass through materials that are opaque to
other types of radiation. The denser the material, the greater its absorption of X-rays
and the less intense the transmitted radiation will be. For example, as X-rays pass
through the human body, many more are absorbed or scattered by bone than by tis-
sue. If the transmitted radiation is directed onto a photographic plate or film, the
exposed areas show variations in intensity—a picture of internal structures.
The combination of computers and modern X-ray machines permits the forma-
tion of three-dimensional images by means of a technique called computerized
tomography, or CT (䉳 Fig. 20.25).
(b) Gamma Rays The electromagnetic waves of the uppermost frequency range of
the known electromagnetic spectrum are called gamma rays (g-rays). This high-
䉱 F I G U R E 2 0 . 2 5 CT scan In an frequency radiation is produced in nuclear reactions, in particle accelerators,
ordinary X-ray image, the entire and in certain types of nuclear decay (radioactivity). Gamma rays will be dis-
thickness of the body is projected
onto the film and internal structures cussed in more detail in Chapter 29.
often overlap, making details hard
to distinguish. In CT—computer- DID YOU LEARN?
ized tomography (from the Greek ➥ A changing magnetic field creates a changing electric field, which creates a
tomo, meaning “slice,” and graph, changing magnetic field, and so on.
meaning “picture”)—X-ray beams ➥ All electromagnetic waves can exert a force or pressure on an object with which
scan a slice of the body. (a) The they interact; this pressure is known as radiation pressure.
transmitted radiation is recorded by
➥ The different regions of the electromagnetic spectrum are categorized by either
a series of detectors and processed
by a computer. Using information wavelength or frequency, which are inversely related.
from multiple slices, the computer
constructs a three-dimensional
*Many health scientists believe that there is no safe “threshold” level for X-rays or other energetic
image. Any single slice can be dis-
radiation—that is, no level of exposure that is completely risk-free—and that some of the dangerous
played for further study. (b) CT
effects are cumulative over a lifetime. People should therefore avoid unnecessary medical X-rays or
image of a brain with a benign
any other unwarranted exposure to “hard” radiation (Chapter 29). However, when properly used, X-
tumor.
rays can be an extremely useful diagnostic tool capable of saving lives.
PULLING IT TOGETHER Newton Meets Faraday

A metal rod and frame are immersed in a uniform field velocity. Draw the free-body diagram for this situation and
(B = 0.120 T) similar to the setup in Fig. 20.11. The wires and determine the terminal speed.
rod have negligible resistance and the resistor R has a resis-
tance of 1.50 Æ . The arrangement is placed on a flat, noncon- T H I N K I N G I T T H R O U G H . (a) Part (a) involves no induced cur-
ducting tabletop with the magnetic field pointing toward rents or magnetic forces because the rod is initially at rest and the
the floor. The rod (mass 60.0 g and length 1.50 m) is attached magnetic flux through its “circuit” is not changing. Summing the
by a string to a 10.0-g mass dangling over the side. The rest of the forces to find the net force will enable the determina-
frame is anchored to the table top and the metal rod moves tion of the initial acceleration. (b) The net force must be decreas-
frictionlessly on the rails. The system is released from rest ing, probably due to induction. As the rod begins to move, there
at t = 0. will be a changing magnetic flux through its circuit and hence a
(a) Draw the rod and dangling mass free body diagrams at current. A current in a magnetic field feels a force. It is likely that
t = 0 and determine the rod’s initial acceleration. (b) As time this force will offset the pull of the string on the rod. (c) As the rod
goes on, the rod’s acceleration decreases. Explain this by accelerates and its speed increases, the induced current in it
redrawing the free body diagrams when there is motion, grows, resulting in an ever-increasing magnetic force on the wire.
applying Newton’s laws, as well as Faraday’s and Lenz’s At some speed, the magnetic force will cancel out the tension in
laws. (c) Eventually the rod will move at a constant (terminal) the string, creating a net force of zero and constant velocity.
SOLUTION.
Given: B = 0.120 T Find: (a) ao (rod’s initial acceleration)

L = 1.50 m (b) Explain the decrease in acceleration with time
m = 60.0 g = 0.0600 kg (c) vt (terminal velocity)
M = 10.0 g = 0.0100 kg
R = 1.50 Æ
(a) The forces on the rod are the normal forces (two, one from N
each rail, combined into one force, N), the pull of gravity, and
the tension in the string. The dangling mass has only two
forces, the tension and its weight. Thus the initial free-body dia- T
gram look like 䉴 Fig. 20.26, from a side view:
Summing the forces vertically on the dangling mass, choos-
ing + down, gives T
g Fy = W - T w
= Mg - T = Mao
Summing the forces horizontally on the dangling rod, choosing
+ to the right, gives W
g Fx = T = mao 䉱 F I G U R E 2 0 . 2 6 (a) The initial free-body diagrams for

the rod and the dangling mass.
Solving for their common initial acceleration gives
0.0100 kg v
ao = c dg = c d 9.80 m>s2 = 1.40 m>s2
M
m + M 0.060 kg + 0.010 kg
B I
(b) Looking from above, the field is pointed away from us and the
flux is increasing in that direction as the rod begins to move. Thus
there will be an induced current counterclockwise to create its
own flux to oppose this change. See 䉴 Fig. 20.27.
The induced current interacts with the magnetic field and then
causes a magnetic force Fmag opposite the rod’s velocity. (You
䉱 F I G U R E 2 0 . 2 7 (b) As the rod is pulled right, the
should be able to show this using a right-hand.) The free-body
changing magnetic flux induces a counterclockwise cur-
diagram thus looks like Fig. 20.26 with the addition of a
backward magnetic force on the rod. N
The net force on the rod in the horizontal direction is reduced
by the magnetic force and therefore its acceleration is less than at
the start. As the rod continues to accelerate, its speed increases, as Fmag T’
do the induced current and force on it. Thus as time goes on, the
net force decreases, resulting in a decreasing acceleration.
(c) At some time, the backward magnetic force will cancel T’
the tension force. Thus the net force on the rod (and dangling
mass) will become zero and the system will move at its con- w
stant terminal velocity. This is shown in 䉴 Fig. 20.28. If the dan-
gling mass does not accelerate the tension in the string (T¿)
must increase to cancel the weight (W), hence T¿ = Mg. Simi- W
larly, on the rod, the net horizontal force is zero therefore
Fmag = T¿ . Combining these results, 䉱 F I G U R E 2 0 . 2 8 (c) The free body diagrams for the rod
Fmag = Mg = 10.0100 kg219.80 m>s22 = 0.0980 N and the dangling mass under terminal velocity conditions.
However, the magnetic force on the wire is given by
Fmag = ILB. Thus under terminal conditions: Kirchhoff’s loop rule, when summed around the circuit,
shows this: g Vloop = + e - IR = 0. Hence
Fmag = ILB = 0.098 N
or e = IR = 10.544 A211.50 Æ2 = 0.816 V
0.0980 N 0.0980 N However, for a motional emf the emf is related to the rod
11.50 m210.120 T2
I = = = 0.544 A
LB speed by e = BLv. Thus:
This enables the determination of the voltage across the resis- e 0.816 V
11.50 m210.120 T2
vt = = = 4.53 m>s
tor, because that voltage is the induced emf in the circuit. BL
■ Magnetic flux 1£2 is a measure of the number of magnetic ■ An ac generator converts mechanical energy into electrical
field lines that pass through an area. For a single wire loop energy. The generator’s emf as a function of time is
of area A, it is defined as e = eo sin vt (20.4)
£ = BA cos u (20.1) where eo is the maximum emf.
where B is the magnetic field strength (assumed constant), A
is the loop area, and u is the angle between the direction of the
magnetic field and the normal to the area’s plane. B
B Axis of N
rotation
Brushes
Slip rings
A
ac
voltmeter
■ A transformer is a device that changes the voltage supplied

to it by means of induction. The voltage applied to the
■ Faraday’s law of induction relates the induced emf e in a
input, or primary (p), side of the transformer is changed
loop (or coil composed of N loops in series) to the time rate
into the output, or secondary (s), voltage. The current and
of change of the magnetic flux through that loop (or coil).
voltage relationships for a transformer are
¢£
e = -N (20.2) Ns
¢t Vs = ¢ ≤V (20.10a)
Np p
where ¢£ is the change in flux through one loop and there
are N total loops. Np
Is = ¢ ≤ Ip (20.10b)
Ns
Increasing magnitude of B
I ac Iron core
source
N
v
Primary Secondary
coil coil
■ An electromagnetic wave (light) consists of time-varying

■ Lenz’s law states that when a change in magnetic flux electric and magnetic fields that propagate at a speed of
induces an emf in a coil, loop, or circuit, the resulting, or c = 13.00 * 108 m>s2 in a vacuum. The different types of
induced, current direction is such as to create a magnetic electromagnetic radiation (such as UV, radio waves, and
field to oppose the change in flux. visible light) differ in frequency and wavelength.
l
F
I F
c
N N
B
v E

20.1 INDUCED EMF: FARADAY’S LAW strength of the magnetic field, (c) the orientation of the
AND LENZ’S LAW loop with respect to a fixed field direction, or (d) the
direction of the field relative to a fixed loop?
1. Which of the following is an SI unit of magnetic flux
3. For a current to be induced in a wire loop, (a) there must
(there may be more than one correct answer): (a) Wb,
be a large magnetic flux through the loop, (b) the loop’s
(b) T # m2, (c) T # m>A, or (d) T?
plane must be parallel to the magnetic field, (c) the
2. The magnetic flux through a loop can change due to a loop’s plane must be perpendicular to the magnetic
change in which of the following (there may be more field, (d) the magnetic flux through the loop must vary
than one correct answer): (a) the area of the coil, (b) the with time.
4. Identical single loops A and B are oriented so they ini- most rapidly, (b) not changing, (c) at its maximum value,
tially have the maximum amount of flux in a magnetic (d) it does not depend on the flux in any way.
field. Loop A is then quickly rotated so that its normal is 11. The back emf of an electric motor depends on which of the
perpendicular to the magnetic field, and in the same following (there may be more than one correct answer): (a)
time, B is rotated so its normal makes an angle of 45° the line voltage, (b) the current in the motor, (c) the arma-
with the field. How do their induced emfs compare: ture’s rotational speed, or (d) none of the preceding?
(a) they are the same; (b) A’s is larger than B’s; (c) B’s is
larger than A’s; or (d) you can’t tell the relative emf mag-
nitudes from the data given? 20.3 TRANSFORMERS AND POWER
5. Identical single loops A and B are oriented so they have TRANSMISSION
the maximum amount of flux when placed in a magnetic 12. A transformer at the local substation of the delivery sys-
field. Both loops maintain their orientation relative to the tem just before your house has (a) more windings in the
field, but in the same amount of time A is moved to a primary coil, (b) more windings in the secondary coil,
region of stronger field, while B is moved to a region of (c) the same number of windings in the primary and sec-
weaker field. How do their induced emfs compare: ondary coils.
(a) they are the same; (b) A’s is larger than B’s; (c) B’s is
larger than A’s; or (d) you can’t tell the relative emf mag- 13. The output power delivered by a realistic step-down
nitudes from the data given? transformer is (a) greater than the input power, (b) less
than the input power, (c) the same as the input power.
6. A bar magnet is thrust toward the center of a circular
metallic loop. The magnet approaches perpendicularly 14. A transformer located just outside a power plant before
with its length perpendicular to the coil’s plane. As the the energy is delivered over the wires would have
bar recedes from your view and approaches the coil, a (a) more windings in the primary coil than in the sec-
clockwise current is induced in the loop. What polarity is ondary, (b) more windings in the secondary coil than in
that end of the bar magnet nearest the coil: (a) north, the primary, (c) the same number of windings in the pri-
(b) south, (c) you can’t tell from the data given? mary and secondary coils.
7. The north end of a bar magnet is quickly pulled away 15. A transformer located just outside a power plant before
from the center of a circular metallic loop. The magnet’s the energy is delivered over the wires would have
length is always perpendicular to the coil’s plane. As the (a) more current in the primary coil than in the secondary,
south end of the magnet approaches you, what would be (b) more current in the secondary coil than in the primary,
the induced current direction in the coil: (a) clockwise, (b) (c) the same current in the primary and secondary coils.
counterclockwise, (c) the induced current would be zero,
or (d) you can’t tell from the data given?
20.4 ELECTROMAGNETIC WAVES
16. Relative to the blue end of the visible spectrum, the yel-
20.2 ELECTRIC GENERATORS AND BACK low and green regions have (a) higher frequencies,
EMF (b) longer wavelengths, (c) shorter wavelengths, (d) both
(a) and (c).
8. Increasing only the coil area in an ac generator would
result in (a) an increase in the frequency of rotation, (b) a 17. Which of the following electromagnetic waves has the
decrease in the maximum induced emf, (c) an increase in lowest frequency: (a) UV, (b) IR, (c) X-ray, or
the maximum induced emf, (d) no change in the genera- (d) microwave?
tor output. 18. Which of the following electromagnetic waves travels
9. In an ac generator, the maximum emf output occurs slowest in a vacuum: (a) green light, (b) infrared light,
when the magnetic flux through the coil is (a) zero, (c) gamma rays, (d) radiowaves, or (e) they all have the
(b) maximum, (c) not changing, (d) it does not depend same speed?
on the flux in any way. 19. If the frequency of an orange source of light was halved,
10. In an ac generator, the maximum emf output occurs what kind of light would it then put out: (a) red, (b) blue,
when the magnetic flux through the coil is (a) changing (c) violet, (d) UV, (e) X-ray, or (f) IR?
20.1 INDUCED EMF: FARADAY’S LAW 䉳 F I G U R E 2 0 . 2 9 A time-

AND LENZ’S LAW Galvanometer varying magnetic field What
1. A bar magnet is dropped through a coil of wire as shown
0 will the galvanometer measure?
N See Conceptual Question 1.
in 䉴 Fig. 20.29. (a) Describe what is observed on the gal- – +
S
vanometer by sketching a graph of induced emf versus t.
(b) Does the magnet fall freely? Explain.
2. In Fig. 20.1b, what would be the direction of the induced
current in the loop if the south pole of the magnet were
approaching instead of the north pole?
3. In Fig. 20.7a, how could you move the coil so as to pre- 11. In a dc motor, if the armature is jammed or turns very
vent any current from being induced in it? Explain. slowly under a heavy load, the coils in the motor may
4. Two identical strong magnets are dropped simultane- burn out. Explain why this can happen.
ously by two students into two vertical tubes of the same 12. If you wanted to make a more compact ac generator (oper-
dimensions (䉲 Fig. 20.30). One tube is made of copper, ating at the same frequency) by reducing the area of the
and the other is made of plastic. From which tube will coils, how would could you compensate by changing its
the magnet emerge first? Why? other physical characteristics in order to maintain the same
emf output? Explain how each characteristic would change
䉳 F I G U R E 2 0 . 3 0 Free fall? (larger, smaller, etc.) and why the change compensates.
20.3 TRANSFORMERS AND POWER

Magnets TRANSMISSION
Copper
Plastic 13. Explain why large-scale electric energy delivery systems
operate at such high voltages when such voltages can be
5. By using the appropriate SI units, show that the units on dangerous.
both sides of Eq. 20.2 (Faraday’s law) are the same. 14. For an emergency project in his automotive workshop, a
6. If a fixed-area metal loop is kept entirely within a uni- mechanic needs a step-down transformer, but has quick
form magnetic field and a current is induced in it, what access to only step-up transformers in the shop. Show
must its motion be? Explain. how he might be able to use a step-up transformer as a
step-down one.
7. A circular metal loop is kept entirely within a magnetic
field, but moved to a region of higher field strength 15. A metallic plate is in the plane of this paper. A uniform
while not rotated. What could you do to its diameter to magnetic field is perpendicular to the plane of the paper
prevent an induced current in it? Explain. and creates a magnetic flux in the plate. Give the direc-
tion of the eddy currents in the plate (clockwise, counter-
8. (a) In Fig. 20.10, what would happen to the direction of clockwise, or zero) for each of the following cases.
the induced current if the metal rod were moving down- Assume you are looking down onto the paper’s plane.
ward instead? (b) What would happen to the direction of (a) The field points away from you and is decreasing.
the induced magnetic field? Is the induced magnetic (b) The field points toward you and is increasing. (c) The
field necessarily opposite to the external field direction? field points toward you and is decreasing. (d) The field
[Hint: Compare your answer to part (b) with the direc- points away from you and does not change with time.
tion in the figure.]
20.4 ELECTROMAGNETIC WAVES

20.2 ELECTRIC GENERATORS AND BACK 16. (a) An antenna has been connected to a car battery.
EMF Under these conditions, will the antenna emit electro-
magnetic radiation? Why or why not? Explain.
9. (a) Explain why the maximum emf produced by an ac
(b) Repeat part (a) for the time when the battery is dis-
generator occurs when the flux through its armature coil
connected and the current in the antenna drops to zero.
is zero. (b) Explain why the emf produced by an ac gen-
erator is zero when the flux through its armature coil is 17. On a cloudy summer day, you work outside and feel
at its maximum. cool, yet that evening you find that you are sunburned.
Explain how this is possible.
10. A student has a bright idea for a generator that appar-
ently generates electric energy without a corresponding 18. Radiation exerts pressure on surfaces on which it falls
loss of energy somewhere else. His suggested arrange- (radiation pressure). (a) Will this pressure be greater on a
ment is shown in 䉲 Fig. 20.31. The magnet is initially shiny surface or a dark surface? (b) For a given type of
pulled down and released. When the magnet is attached surface, will the pressure be greater using a bright source
to a spring, the student thinks that there should be a con- compared to a source emitting the same color but fainter?
tinuous electrical output as the magnet oscillates. What (c) For a given surface, would you expect the force of radi-
is wrong with this idea? [Hint: Check the total energy ation pressure on it to vary with distance from a light
stored in the spring–magnet system and the forces on source? For all parts, clearly explain your reasoning.
the magnet as it oscillates.] 19. (a) When police radar waves bounce off an oncoming car
and are received by the transmitting radar “gun,” they
䉳 FIGURE 20.31 have a different frequency than the emitted waves.
Inventive genius? See Explain. (b) Is the frequency in part (b) higher or lower
Conceptual Question 10. than the original frequency? What about wavelength?
What about wave speed? [Hint: Remember the Doppler
effect.]
N
S
EXERCISES 725
EXERCISES
20.1 INDUCED EMF: FARADAY’S LAW netic field perpendicular to the plane of the coil. The
AND LENZ’S LAW radius of the coil is 10 cm, and the initial strength of the
magnetic field is 1.5 T. Assuming that the strength of the
1. ● What should be the diameter of a circular wire loop if field decreased with time, (a) what is the final strength of
it is to have a magnetic field of 0.15 T oriented perpen- the field? (b) If the field strength had, instead, increased,
dicular to its area which produces a magnetic flux of what would its final value have been? (c) Explain a
1.2 * 10-2 T # m2? method whereby you could, in principle, tell whether
2
2. ● A circular loop with an area of 0.015 m is in a uniform the field was increasing or decreasing in magnitude.
magnetic field of 0.30 T. What is the flux through the 11. IE ● ● A boy is traveling due north at a constant speed
loop’s plane if it is (a) parallel to the field, (b) at an angle while carrying a metal rod. The rod’s length is oriented
of 37° to the field, and (c) perpendicular to the field? in the east–west direction and is parallel to the ground.
3. ● A circular loop (radius of 20 cm) is in a uniform mag- (a) There will be no induced emf when the rod is (1) at
netic field of 0.15 T. What angle(s) between the normal to the equator, (2) near the Earth’s magnetic poles,
the plane of the loop and the field would result in a flux (3) somewhere between the equator and the poles. Why?
with a magnitude of 1.4 * 10-2 T # m2? (b) Assume that the Earth’s magnetic field is
4. ● The plane of a conductive loop with an area of 1.0 * 10-4 T near the North Pole and 1.0 * 10-5 T near
0.020 m2 is perpendicular to a uniform magnetic field of the equator. If the boy runs with a speed of 1.3 m>s
0.30 T. If the field drops to zero in 0.0045 s, what is the northward near each location, and the rod is 1.5 m long,
magnitude of the average emf induced in the loop? calculate the induced emf in the rod in each location.
5. ● An ideal solenoid with a current of 1.5 A has a radius 12. ● ● A metal airplane with a wingspan of 30 m flies hori-
of 3.0 cm and a turn density of 250 turns>m. (a) What is zontally along a north–south route in the northern hemi-
the magnetic flux (due to its own field) through only one sphere at a constant speed of 320 km>h in a region where
of its loops at its center? (b) What current would be the vertical component of the Earth’s magnetic field is
required to double the flux value in part (a)? 5.0 * 10-5 T. (a) What is the magnitude of the induced
6. ● ● A uniform magnetic field is at right angles to the
emf between the tips of its wings? (b) If the easternmost
plane of a wire loop. If the field decreases by 0.20 T in wing tip is negatively charged, is the plane flying due
1.0 * 10-3 s and the magnitude of the average emf north or due south? Explain.
13. ● ● Suppose that the metal rod in Fig. 20.11 is 20 cm long
induced in the loop is 80 V, (a) what is the area of the
loop? (b) What would be the value of the average and is moving at a speed of 10 m>s in a magnetic field of
induced emf if the field change was the same but took 0.30 T and that the metal frame is covered with an insulat-
twice as long to decrease? (c) What would be the value of ing material. Find (a) the magnitude of the induced emf
the average induced emf if the field decrease was twice across the rod and (b) the current in the rod. (c) Repeat
as much and it also took twice as long to change? these calculations if the wire were not covered and the
total resistance of the circuit (rod plus frame) were 0.15 Æ .
7. ● ● (a) A square loop of wire with sides of length 40 cm is
14. ● ● ● The flux through a loop of wire changes uniformly
in a uniform magnetic field perpendicular to its area. If
the field’s strength is initially 100 mT and it decays to from +40 Wb to -20 Wb in 1.5 ms. (a) What is the signifi-
zero in 0.010 s, what is the magnitude of the average emf cance of the negative number attached to the final flux
induced in the loop? (b) What would be the average emf value? (b) What is the average induced emf in the loop?
if the sides of the loop were only 20 cm? (c) If you wanted to double the average induced emf by
changing only the time, what would the new time interval
8. ● ● The magnetic flux through one loop of wire is
be? (d) If you wanted to double the average induced emf
reduced from 0.35 Wb to 0.15 Wb in 0.20 s. The average
by changing only the final flux value, what would it be?
induced current in the coil is 10 A. (a) Can you deter-
15. ● ● ● A fixed coil of wire with 10 turns and an area of
mine the area of the loop from the data given? Explain.
0.055 m2 is placed in a perpendicular magnetic field. This
(b) Find the resistance of the wire.
field oscillates in direction and magnitude at a frequency of
9. ● ● When the magnetic flux through a single loop of wire
10 Hz and has a maximum value of 0.12 T, (a) What is the
increases by 30 T # m2, an average current of 40 A is average emf induced in the coil during the time it takes for
induced in the wire. Assuming that the wire has a resis- the field to go from its maximum value in one direction to
tance of 2.5 Æ , (a) over what period of time did the flux its maximum value in the other direction? (b) Repeat part
increase? (b) If the current had been only 20 A, how long (a) for a time interval of one complete cycle. (c) At what
would the flux increase have taken? time(s) during a complete magnetic field cycle would you
10. ● ● In 0.20 s, a coil of wire with 50 loops experiences an expect the induced emf to have its maximum magnitude?
average induced emf of 9.0 V due to a changing mag- What about its minimum value? Explain both answers.
20.2 ELECTRIC GENERATORS AND BACK (a) when running at its operating speed, (b) when running
EMF at half its final rotational speed, and (c) when starting up?
25. ● ● ● A 240-V dc motor has an armature whose resistance
16. ● A hospital emergency room ac generator operates at a
frequency of 60 Hz. If the output voltage is at a maxi- is 1.50 Æ . When running at its operating speed, it draws
mum value (magnitude) at t = 0, when is it next (a) a a current of 16.0 A. (a) What is the back emf of the motor
maximum (magnitude), (b) zero, and (c) at its initial when it is operating normally? (b) What is the starting
value (direction and magnitude)? current? (Assume that there is no additional resistance in
the circuit.) (c) What series resistance would be required
17. ● A student makes a simple ac generator by using a single
to limit the starting current to 25 A?
square wire loop 10 cm on a side. The loop is then rotated
at a frequency of 60 Hz in a magnetic field of 0.015 T.
(a) What is the maximum emf output? (b) If she wanted to 20.3 TRANSFORMERS AND POWER
make the maximum emf output ten times larger by adding TRANSMISSION
loops, how many should she use in total? 26. IE ● The secondary coil of an ideal transformer has
18. ● ● A simple ac generator consists of a coil with 10 turns 450 turns, and the primary coil has 75 turns. (a) Is this
(each turn has an area of 50 cm2). The coil rotates in a transformer a (1) step-up or (2) step-down transformer?
uniform magnetic field of 350 mT with a frequency of Explain your choice. (b) What is the ratio of the current
60 Hz. (a) Write an expression in the form of Eq. 20.5 for in the primary coil to the current in the secondary coil?
the generator’s emf variation with time. (b) Compute the (c) What is the ratio of the voltage across the primary coil
maximum emf. to the voltage in the secondary coil?
19. ● ● An ac generator operates at a rotational frequency of
27. ● An ideal transformer steps 8.0 V up to 2000 V, and the
60 Hz and produces a maximum emf of 100 V. Assume 4000-turn secondary coil carries 2.0 A. (a) Find the num-
that its output at t = 0 is zero. What is the instantaneous ber of turns in the primary coil. (b) Find the current in
emf (a) at t = 1>240 s? (b) at t = 1>120 s? (c) at t? the primary coil.
(d) How much time elapses between successive 0-volt
28. ● The primary coil of an ideal transformer has 720 turns,
outputs? (e) What maximum emf would this generator
and the secondary coil has 180 turns. If the primary coil
produce if it were operated, instead, at 120 Hz?
carries 15 A at a voltage of 120 V, what are (a) the voltage
20. ● ● The armature of a simple ac generator has 20 circular
and (b) the output current of the secondary coil?
loops of wire, each with a radius of 10 cm. It is rotated
with a frequency of 60 Hz in a uniform magnetic field of 29. ●● The transformer in the power supply for a computer’s
800 mT. (a) What is the maximum emf induced in the 500-GB external hard drive changes a 120-V input voltage
loops? (b) How often is this value attained? (c) If the time (from a regular house line) to the 5.0-V output voltage that
period in part (b) were cut in half, what would be the is required to power the drive. (a) Find the ratio of the
new maximum emf value? number of turns in the primary coil to the number of turns
in the secondary coil. (b) If the drive is rated at 10 W when
21. ● ● The armature of an ac generator has 100 turns. Each
running and the transformer is ideal, what is the current in
turn is a rectangular loop measuring 8.0 cm by 12 cm. The
the primary and secondary when the drive is in operation?
generator has a sinusoidal voltage output with an ampli-
tude of 24 V. (a) If the magnetic field of the generator is 30. ●● The primary coil of an ideal transformer is connected
250 mT, with what frequency does the armature turn? to a 120-V source and draws 1.0 A. The secondary coil
(b) If the magnetic field was doubled and the frequency has 800 turns and supplies an output current of 10 A to
cut in half, what would be the amplitude of the output? run an electrical device. (a) What is the voltage across the
secondary coil? (b) How many turns are in the primary
22. IE ● ● (a) To increase the output of an ac generator, a stu-
coil? (c) If the maximum power allowed by the device
dent has the choice of doubling either the generator’s mag-
(before it is destroyed) is 240 W, what is the maximum
netic field or its frequency. To maximize the increase in
input current to this transformer?
emf output, (1) he should double the magnetic field, (2) he
should double the frequency, (3) it does not matter which 31. ● An ideal transformer has 840 turns in its primary coil
one he doubles. Explain. (b) Two students display their ac and 120 turns in its secondary coil. If the primary coil
generators at a science fair. The generator made by stu- draws 2.50 A at 110 V, what are (a) the current and (b) the
dent A has a loop area of 100 cm2 rotating in a magnetic output voltage of the secondary coil?
field of 20 mT at 60 Hz. The one made by student B has a 32. ● ● The efficiency e of a transformer is defined as the
loop area of 75 cm2 rotating in a magnetic field of 200 mT ratio of the power output to the power input, or
at 120 Hz. Which one generates the largest maximum e = Ps>Pp. (a) Show that for an ideal transformer, this
emf? Justify your answer mathematically. expression gives an efficiency of 100% 1e = 1.002.
23. IE ● ● A motor has a resistance of 2.50 Æ and is con- (b) Suppose a step-up transformer increased the line
nected to a 110-V line. (a) Is the operating current of the voltage from 120 to 240 V, while at the same time the out-
motor (1) higher than 44 A, (2) 44 A, or (3) lower than put current was reduced to 5.0 A from 12 A. What is the
44 A? Why? (b) If the back emf of the motor at operating transformer’s efficiency? Is it ideal?
speed is 100 V, what is its operating current? 33. IE ● ● The specifications of a transformer used with a
24. ● ● ● The starter motor in an automobile has a resistance small appliance read as follows: input, 120 V, 6.0 W; out-
of 0.40 Æ in its armature windings. The motor operates on put, 9.0 V, 600 mA. (a) Is this transformer (1) an ideal or
12 V and has a back emf of 10 V when running at normal (2) a nonideal transformer? Why? (b) What is its effi-
operating speed. How much current does the motor draw ciency? (See Exercise 32.)
34. An ac generator supplies 20 A at 440 V to a 10 000-V

●● 40. ● A meteorologist for a TV station is using radar to
power line through a step-up transformer that has 150 determine the distance to a cloud. He notes that a time of
turns in its primary coil. (a) If the transformer is 95% effi- 0.24 ms elapses between the sending and the return of a
cient (see Exercise 32), how many turns are in the sec- radar pulse. How far away is the cloud?
ondary coil? (b) What is the current in the power line? 41. ● How long does a laser beam take to travel from the
35. ● ● The electricity supplied in Exercise 34 is transmitted Earth to a reflector on the Moon and back? Take the dis-
over a line 80.0 km long with a resistance of 0.80 Æ>km. tance from the Earth to the Moon to be 2.4 * 105 mi.
(a) How many kilowatt-hours are saved in 5.00 h by (This experiment was done when the Apollo flights of the
stepping up the voltage? (b) At $0.15>kWh, how much early 1970s left laser reflectors on the lunar surface.)
of a savings (to the nearest $10) is this to all the con- 42. ● ● Orange light has a wavelength of 600 nm, and green
sumers the line supplies in a 30-day month, assuming light has a wavelength of 510 nm. (a) What is the differ-
that the energy is supplied continuously? ence in frequency between the two types of light? (b) If
36. ● ● A small plant produces electric energy and, through you doubled the wavelength of both, what type of light
a transformer, sends it out over the transmission lines at would they become?
50 A and 20 kV. The line reaches a small town over 43. ● ● A certain type of radio antenna is called a quarter-
25-km-long transmission lines whose resistance is wavelength antenna, because its length is equal to one-
1.2 Æ>km. (a) What is the power loss in the lines if the quarter of the wavelength to be received. If you were
energy is transmitted at 20 kV? (b) What should be the going to make such antennae for the AM and FM radio
output voltage of the transformer to decrease the power bands by using the middle frequencies of each band,
loss by a factor of 15? Assume the transformer is ideal. what lengths of wire would you use?
(c) What would be the current in the lines in part (b)? 44. IE ● ● ● Microwave ovens can have cold spots and hot
37. ● ● Electrical power from a generator is transmitted spots due to standing electromagnetic waves, analogous
through a power line 175 km long with a resistance of to standing wave nodes and antinodes in strings
1.2 Æ>km. The generator’s output is 50 A at its operating (䉲 Fig. 20.32). (a) The longer the distance between the
voltage of 440 V. This output is increased by a single cold spots, (1) the higher the frequency of the waves,
step-up for transmission at 44 kV. (a) How much power (2) the lower the frequency of the waves, (3) the frequency
is lost as joule heat during the transmission? (b) What of the waves is independent of this distance. Why? (b) In
must be the turn ratio of a transformer at the delivery your microwave the cold spots (nodes) occur approxi-
point in order to provide an output voltage of 220 V? mately every 5.0 cm, but your neighbor’s microwave pro-
(Neglect the voltage drop in the line.) duces them at every 6.0 cm. Which microwave operates at
a higher frequency and by how much?
20.4 ELECTROMAGNETIC WAVES 䉳 F I G U R E 2 0 . 3 2 Cold
38. Find the frequencies of electromagnetic waves with
● spots? See Exercise 44.
wavelengths of (a) 3.0 cm, (b) 650 nm, and (c) 1.2 fm.
(d) Classify the type of light in each case.
39. ● In a small town there are only two AM radio stations,
one at 920 kHz and one at 1280 kHz. What are the wave-
lengths of the radio waves transmitted by each station?
45. A basic telephone has both a speaker–transmitter and a closely packed. As a result, the resistance of the button
receiver (䉲 Fig. 20.33). Until the advent of digital phones changed. The receiver converted these electrical
in the 1990s, the transmitter had a diaphragm coupled to impulses to sound. Applying the principles of electricity
a carbon chamber (called the button), which contained and magnetism that you have learned, explain the basic
loosely packed granules of carbon. As the diaphragm operation of this type of telephone.
vibrated because of incident sound waves, the pressure 46. IE In 䉲 Fig. 20.34, a metal bar moves at constant velocity
on the granules varied, causing them to be more or less in a constant magnetic field. That field is directed into the
Magnet page. (a) The direction of the induced current through the
Sound Carbon with Sound
䉳 FIGURE 20.33 bar is (1) up, (2) down, (3) there is no current. Why? (b) If
waves granules coils waves Telephone
in out the magnitude of the magnetic field is 0.55 T, what is the
operation See
Exercise 45. current in the bar? Neglect the resistances of the bar and
wires. (c) What are the magnitude and direction of the
force on the bar? (d) What is the power required by the
Speaker–
force on the bar? (e) Compare your answer to part (d) to
Receiver
transmitter the rate of joule heating in the resistor. They should be the
Thin metal diaphragm
Diaphragm same. Are they? Explain why they should be.
current compare to the ideal case? (c) At what rate is heat lost
䉳 FIGURE 20.34 in the nonideal transformer? (d) If you wanted to keep the
Basics of motional transformer cool and to do this needed to dissipate half of
B
emf See the joule heating of part (c) using water cooling lines (the
Exercise 46.
other half is taken care of by air cooling), what should be the
L = 0.50 m R = 10 Ω rate of flow (in liters per minute) of water in the lines?
v = 2.0 m/s
Assume the input cool water is at 68 °F and the maximum
allowable output water temperature is 98 °F.
51. Suppose you wanted to build an electric generator using the
Earth’s magnetic field. Assume it has a strength of 0.040 mT
at your location. Your generator design calls for a coil of
1000 windings rotated at exactly 60 Hz. The coil is oriented
47. A transformer is used by a European traveler while she so that the normal to the area lines up with the Earth’s field
is visiting the United States. She primarily uses it to run at the end of each cycle.
a 1200-watt hair dryer she brought with her. When the (a) What must the coil diameter be to generate a maxi-
hair dryer is plugged in to her hotel room outlet in Los mum voltage of 170 V (required in order to average 120 V)?
Angeles, she notices that it runs exactly as it does at Does this seem like a practical design? (b) Some generators
home. The input voltage and current are 120 V and operate at 50 Hz. How would this change the coil diameter?
11.0 A, respectively. (a) Prove that this is not an ideal (c) What number of windings would make this coil arrange-
transformer. (b) What is its efficiency? (c) What is the ment a “manageable” size?
rate at which heat is generated in the transformer itself? 52. (a) In May 2008, the United States successfully landed a
48. A solenoid of length 40.0 cm is made of 10 000 circular coils. spacecraft named Phoenix near the northern polar regions of
It carries a steady current of 12.0 A. Near its center is placed Mars. Immediately upon landing, the craft sent a message
a small, flat, circular metallic coil of 200 circular loops, each indicating that all had gone well. Using the astronomical
with a radius of 2.00 mm. This small coil is oriented so that data in the appendix of this book, determine the shortest
it receives half of the maximum magnetic flux. A switch is amount of time it took this signal to reach the Earth.
opened in the solenoid circuit and its current drops to zero (b) If the Phoenix transmitter sent out spherical electro-
in 25.0 ms. (a) What was the initial flux through the small magnetic waves with a power of 100 W, how many watts
coil? (b) Determine the average induced emf in the small per square meter would arrive at the Earth, assuming that
coil during the 25.0 ms. (c) If you look along the long axis of Mars was in its closest location to Earth? (c) A radio signal
the solenoid so that the initial 12.0 A current is clockwise, was sent to a deep space probe traveling in the plane of the
determine the direction of the induced current in the small solar system. Earth received a response 3.5 days later.
inner coil during the time the current drops to zero. Assuming the probe computers took 4.5 hours to process
(d) During the 25.0 ms, what was the average current in the the signal instructions and to send out the return message,
small coil, assuming it has a resistance of 0.15 Æ ? was the probe within the solar system? (Assume a solar sys-
49. IE A flat coil of copper wire consists of 100 loops and has tem radius of about 40 times the Earth–Sun distance.)
a total resistance of 0.500 Æ . The coil diameter is 4.00 cm 53. Assume that a uniform magnetic field exists perpendicular
and it is in a uniform magnetic field pointing toward you to the plane of this page (into it) and has a strength of
(out the page). The coil orientation is in the plane of the 0.150 T. Assume further that this field ends sharply at the
page. It is then pulled to the right (without rotating) until paper’s edges. A single circular loop of wire is also in the
it is completely out of the field. (a) What is the direction of plane of the paper and moves across it from left to right at a
the induced current in the coil: (1) clockwise, (2) counter- speed of 1.00 m>s. The loop has a radius of 1.50 cm. The
clockwise, or (3) there is no induced current? (b) During loop starts with its center 10.0 cm to the left of the left edge,
the time the coil leaves the field, an average induced cur- in zero field, enters the field, then exits at the right edge back
rent of 20.0 mA is measured. What is the average induced into zero field until its center is 10.0 cm to the right of the
emf in the coil? (c) If the field strength is 5.50 mT, how right edge.
much time did it take to pull the coil out? (a) Make a sketch of the induced emf in the coil versus
50. The transformer on a utility pole steps the voltage down time, putting numbers on the time axis and taking positive
from 10 000 V to 220 V for use in a college science building. emf to indicate clockwise direction and negative emf to indi-
During the day, the transformer delivers electric energy at cate counterclockwise (the emf axis will not have any num-
the rate of 10.0 kW. bers on it.) (b) What is the average emf (magnitude) induced
(a) Assuming the transformer to be ideal, during that in the coil when it is (1) to the left of the left edge, (2) enter-
time, what are the primary and secondary currents in the ing the left side of the field, (3) completely in the field
transformer? (b) If the transformer is only 90% efficient (but region, (4) exiting the right field edge, and (5) out in the zero
still delivers electric power at 10.0 kW), how does its input field region to the right of the right edge.
21 AC Circuits
21.1 Resistance in an
ac circuit (730)
■ peak and rms values of voltage,
current and power
21.2 Capacitive reactance

(733)
■ phase relationship between
voltage and current
21.3 Inductive reactance

(735)
■ phase relationship between
voltage and current
21.4 Impedance:
PHYSICS FACTS
D
RLC circuits (737)
✦ Under ac conditions (alternating
irect current (dc) circuits
■ phase diagrams
■ power factors
voltage direction), a capacitor, have many uses, but the
even with the gap between the
plates, allows current in the circuit airport control tower in the chapter-
during the charging and discharg-
ing stages. Under dc conditions
opening photo operates many
21.5 Circuit resonance (742) (steady voltage across the plates), devices that use alternating current
■ resonance frequency there is no current.
✦ Under dc voltages, a solenoid
(ac). The electric power delivered to
offers no impedance to the flow of our homes and offices is also ac, and
charge and thus can readily con-
duct current. However, under ac most everyday devices and appli-
conditions, a solenoid impedes the
change in current by producing a
ances require alternating current.
reverse emf in accordance with There are several reasons for our
Faraday’s law of induction.
✦ A circuit of a capacitor, inductor
reliance on alternating current. For
(such as a solenoid), and resistor one thing, almost all electric energy
connected in series to an ac power
supply is analogous to a mechani- generators produce electric energy
cal damped, driven spring–mass
using electromagnetic induction,
system. When driven at its natural
frequency, the circuit “resonates,” and thus produce ac outputs
that is, exhibits a current maxi-
mum, just as the mechanical sys-
(Chapter 20). Furthermore, electri-
tem has its largest amplitude when cal energy produced in ac fashion
driven at its natural frequency.
can be transmitted economically
730 21 AC CIRCUITS
over long distances through the use of transformers. But perhaps the most
important reason is that ac currents produce electromagnetic effects that can be
exploited in a variety of devices. For example, when a radio is tuned to a station,
it takes advantage of a special resonance property of ac circuits (studied in this
chapter).
To determine currents in dc circuits, resistance values were of main concern
(Chapter 18). There is, of course, resistance present in ac circuits as well, but addi-
tional factors can affect the flow of charge. For instance, a capacitor in a dc circuit is
equivalent to an infinite resistance (an open circuit). However, in an ac circuit the
alternating voltage continually charges and discharges a capacitor. Under such
conditions, current can exist in a circuit even if it contains a capacitor. Moreover,
wrapped coils of wire can oppose an ac current through the principle of electro-
magnetic induction (Lenz’s law; Section 20.1).
In this chapter, the principles of ac circuits will be studied. More generalized
forms of Ohm’s law and expressions for power, applicable to ac circuits, will be
developed. Finally, the phenomenon and uses of circuit resonance is explored.
21.1 Resistance in an AC Circuit

For a resistor driven by an ac voltage source:

➥ How are the peak current and the peak voltage related in phase and value?
➥ How are rms current and maximum current related?
➥ How are the peak power and time-averaged power related?
R An ac circuit contains an ac voltage source (such as a small generator or simply a

household outlet) and one or more elements. An ac circuit with a single resistive
I = Io sin 2 π ft element is shown in 䉳 Fig. 21.1. If the source’s output voltage varies sinusoidally,
as that from a generator (see Section 20.2), the voltage across the resistor varies
with time in accordance with the equation
V = Vo sin vt = Vo sin 2pft (21.1)

V = Vo sin 2π ft
where v is the angular frequency of the voltage (in rad>s) and is related to its fre-
quency f (in Hz) by v = 2pf. The voltage across the resistor oscillates between
ac source +Vo and -Vo as sin 2pft oscillates between + 1 and -1. The voltage Vo , called the
peak (or maximum) voltage, represents the amplitude (maximum value) of the voltage.
䉱 F I G U R E 2 1 . 1 A purely resistive
circuit The ac source delivers a
sinusoidal voltage to a circuit con- AC CURRENT AND POWER
sisting of a single resistor. The volt-
age across, and current in, the
Under ac conditions, the current through the resistor oscillates in direction and
resistor vary sinusoidally at the fre- magnitude. From Ohm’s law, the ac current in the resistor, as a function of time, is
quency of the applied ac voltage. V Vo
I = = ¢ ≤ sin 2pft
R R
Because Vo represents the peak voltage across the resistor, the expression in the
parentheses represents the maximum current in the resistor. Thus, this expression
can be rewritten as
I = Io sin 2pft (21.2)
where the amplitude of the current is Io = Vo>R and is called the peak (or
maximum) current.
21.1 RESISTANCE IN AN AC CIRCUIT 731
䉴 Figure 21.2 shows both current and voltage as functions of time for a resistor. V, I
Note that they are in step, or in phase. That is, both reach their zero, minimum, and
maximum values at the same time. The current oscillates and takes on both posi- Voltage V
tive and negative values, indicating its directional changes during each cycle. Vo
Current I
Because the current spends equal time in both directions, the average current is zero.
Mathematically, this is because the time-averaged value of the sine function over Io 3
cycle
one or more complete 1360°2 cycles is zero. Using overbars to denote a time-aver-
4
(270°)
aged value, then sin u = sin 2pft = 0. Similarly, cos u = 0. t
1
Even though the average current is zero, this does not mean that there is no joule 4
cycle
heating (I2R losses). This is because the dissipation of electrical energy in a resistor (90°)
does not depend on the current’s direction. The instantaneous power as a function
of time is obtained from the instantaneous current (Eq. 21.2). Thus,
P = I 2R = 1I 2oR2 sin2 2pft (21.3) 䉱 F I G U R E 2 1 . 2 Voltage and cur-
rent in phase In a purely resistive ac
Even though the current changes sign, the square of the current, I2, is always posi- circuit, the voltage and current are
tive. Thus the average value of I2R is not zero. The average, or mean, value of I2 is in step, or in phase.
I 2 = I 2o sin2 2pft = I 2o sin2 2pft

Using the trigonometric identity sin2 u = 12 11 - cos 2u2, then sin2 u =
2 11
1
- cos 2u2. Because cos 2u = 0 (just as cos u = 0), it follows that sin2 u = 12 .
Thus the previous expression for I 2 can be rewritten as
I 2 = I 2o sin2 2pft = 12 I 2o (21.4)
The average power is therefore
P = I 2R = 12 I 2o R (21.5)
It should be emphasized that ac power has the same form as dc power 1P = I 2R2,
P
and is valid at all times. It is customary, however, to work with average power and
a special kind of “average” current defined as follows:
Po
1 2 Io
Irms = 3 I 2 = 22 I o = = 0.707Io (21.6)
P = 21 Po
22
Irms is called the rms current, or effective current. (Here, rms stands for root-mean-
t
square, indicating the square root of the mean value of the square of the current.) T T 3T
t=0 T
The rms current represents the value of a steady (dc) current required to produce 4 2 4
the same power as its ac current counterpart, hence the name effective current.
Using I 2rms = 1Io> 1222 = 12 I 2o, the average power (Eq. 21.5) can be rewritten as 䉱 F I G U R E 2 1 . 3 Power variation
with time in a resistor Although
both current and voltage oscillate in
P = 12 I 2o R = I 2rms R (time-averaged power of a resistor) (21.7) direction (sign), their product
(power) is always a positive oscillat-
The average power is just the time-varying (oscillating) power averaged over time ing quantity. The average power is
(䉴 Fig. 21.3). one-half the peak power.
AC VOLTAGE
The peak values of voltage and current for a resistor are related by Vo = Io R.
Using a development similar to that for rms current, the rms voltage, or effective
voltage, is defined as
Vo
Vrms = = 0.707Vo (21.8)
22
For resistors under ac conditions, then, dc-like relationships can be used—as long
as it is kept in mind that the quantities represent rms values. Thus for ac situations
involving only a resistor, the relationship between rms values of current and voltage is
Vrms = Irms R (rms voltage across resistor) (21.9)

732 21 AC CIRCUITS
I Combining Eqs. 21.9 and 21.7 results in several physically equivalent expressions
for ac power:
V2rms
Io P = I 2rms R = Irms Vrms = (ac power of a resistor) (21.10)
R
Io Irms = 0.707Io
(peak) It is customary to measure and specify rms values when dealing with ac quanti-
0 t
ties. For example, the household line voltage of 120 V is really the rms value of the
voltage. Household voltage actually has a peak value of
–Io
Vo = 22Vrms = 1.4141120 V2 = 170 V
(a) Visual interpretations of peak and rms values of current and voltage are shown in
䉳Fig. 21.4.
V
EXAMPLE 21.1 A Bright Lightbulb: Its RMS and Peak Values

Vo
A lamp with a 60-W bulb is plugged into a 120-V outlet. (a) What are the rms and peak
Vo Vrms = 0.707Vo currents through the lamp? (b) What is the resistance of the bulb under these conditions?
(peak)
T H I N K I N G I T T H R O U G H . (a) Because the average power and rms voltage are known,
0 t
the rms current can be found from Eq. 21.10. From the rms current, Eq. 21.6 can be used
to calculate the peak current. (b) The resistance is found from Eq. 21.9.
SOLUTION. The average power and the rms voltage of the source are given.
–Vo
Given: P = 60 W Find: (a) Irms and Io (rms and peak currents)
(b) Vrms = 120 V (b) R (bulb resistance)
䉱 F I G U R E 2 1 . 4 Root-mean- (a) The rms current is

square (rms) current and voltage P 60 W
The rms values of (a) current and Irms = = = 0.50 A
Vrms 120 V
(b) voltage are 0.707, or 1> 12, times
the peak (maximum) values. and the peak current is determined by rearranging Eq. 21.6:
Io = 22Irms = 11.41210.50 A2 = 0.71 A
(b) The resistance of the bulb is
Vrms 120 V
R = = = 240 Æ
Irms 0.50 A
F O L L O W - U P E X E R C I S E . What would be the (a) rms current and (b) peak current in a
60-W lightbulb in Great Britain, where the house rms voltage is 240 V at 50 Hz? (c) How
would the resistance of a 60-W bulb in Great Britain compare with one designed for
operation at 120 V? Why are the two resistances different? (Answers to all Follow-Up
Exercises are in Appendix VI at the back of the book.)
CONCEPTUAL EXAMPLE 21.2 Across the Pond: British versus American Electrical Systems
In many countries, the line voltage is 240 V. If a British answer is (c). In addition, the decreased current could cause
tourist visiting the United States plugged in a hair dryer the motor to run slower than normal.
from home (where the voltage is 240 V), you would expect it When traveling in a foreign country, most people do not
(a) not to operate, (b) to operate normally, (c) to operate make this mistake, because the shape of plugs and sockets varies
poorly, or (d) to burn out. from country to country. If you are traveling with appliances
from home, a converter>adapter kit can be useful (䉴Fig. 21.5).
REASONING AND ANSWER. British appliances operate at This kit contains a selection of plugs for adapting to foreign
240 V. At a decreased voltage (namely, 120 V), there would be sockets, as well as a voltage converter. The converter is a device
decreased current 1I = V>R2 and reduced joule heating that converts 240 V to 120 V for U.S. travelers and vice versa for
(P = IV). If the resistance of the appliance were constant, tourists visiting the United States who have 240-V electrical sup-
then at half the voltage, there would be only one-fourth the plies at home. (There also exist appliances that can be switched
power output. Thus the heating element of the hair dryer between 120 V and 240 V, negating the need for voltage convert-
might get warm, but it would not work as expected, so the ers. However, plug socket adaptors may still be needed.)
21.2 CAPACITIVE REACTANCE 733
䉳 F I G U R E 2 1 . 5 Converter and adapters

In countries that have 240-V line voltages,
U.S. tourists need converters that convert to
120 V to operate normal U.S. appliances
properly. Note the different types of plugs
for different countries. The small plugs go
into the foreign sockets, and the converter
prongs fit into the back of each plug. A U.S.
standard two-prong plug fits into the con-
verter, which has a 120-V output.
F O L L O W - U P E X E R C I S E . What happens if an American tourist inadvertently plugs a 120-V appliance into a British 240-V outlet
without a converter? Explain.
DID YOU LEARN?

For a resistor driven by an ac voltage source:
➥ The peak current and peak voltage are in phase and proportional.
➥ The rms current is 70.7% of the maximum current.
➥ The peak power is twice the time-averaged power.
21.2 Capacitive Reactance

For a capacitor driven by an ac voltage source:

➥ What happens to the rms current in the circuit if the source frequency increases?
➥ What happens to the rms current in the circuit if the capacitance value increases?
➥ What is the current in the circuit when the maximum energy is stored in the
capacitor?
Discussions in Chapter 16 considered situations (such as RC circuits) in which a

capacitor is connected to a dc voltage source. In these situations, current exists
only for the short time required to charge or discharge the capacitor. As charge ac
C
accumulates on the capacitor’s plates, the voltage across them increases, opposing source
the external voltage and reducing the current. When the capacitor is fully charged,
the current drops to zero.
Things are different when a capacitor is driven by an ac voltage source (a)
(䉴 Fig. 21.6a). Under these conditions, the capacitor limits the current, but doesn’t
completely prevent the flow of charge. This is because the capacitor is alternately
charged and discharged as the current and voltage reverse each half-cycle. V, I
Voltage V
Plots of ac current and voltage versus time for a circuit with just a capacitor are
shown in Fig. 21.6b. Let’s look at how the conditions of the capacitor change with + 1
cycle (90°)
4
time (䉲 Fig. 21.7).
■ In Fig. 21.7a, t = 0 is arbitrarily chosen as the time of maximum voltage t
1V = Vo2.* At the start, the capacitor is assumed to be fully charged
1Qo = CVo2 with the polarity shown. Because the plates cannot accommodate
more charge, there is no current in the circuit. Current I
−
■ As the voltage decreases so that 0 6 V 6 Vo , the capacitor begins to discharge,
giving rise to a counterclockwise current (labeled negative; compare Fig. 21.6b (b)
to Fig. 21.7b).
䉱 F I G U R E 2 1 . 6 A purely capaci-
■ The current reaches its maximum value when the voltage drops to zero and the tive circuit (a) In a circuit with only
capacitor plates are completely discharged (Fig. 21.7c). This occurs one-quarter capacitance, (b) the current leads the
of the way through the cycle 1t = T>42. voltage by 90°, or one-quarter cycle.
Half of a cycle of voltage and cur-
*The polarity of initial capacitor voltage is arbitrarily chosen as positive in Fig. 21.6b. rent, shown, corresponds to Fig. 21.7.
734 21 AC CIRCUITS
■ The ac voltage source now reverses polarity and starts to increase in magni-
t=0 tude, so that - Vo 6 V 6 0. The capacitor begins to charge, this time with the
ac + + ++
Qo C Q = Qo
source − − −− opposite polarity (Fig. 21.7d). With the plates uncharged, there is no opposition
I=0
to the current, so the current is at its maximum value. However, as the plates
accumulate charge, they begin to inhibit the current, and the current decreases
(a) in magnitude.
I ■ Halfway through the cycle 1t = T>22, the capacitor is fully charged, but opposite
in polarity to its starting condition (Fig. 21.7e). The current is zero, and the volt-
+ + Q < Qo age is at its maximum magnitude, but opposite the initial polarity 1V = - Vo2.
− − I>0
During the next half-cycle (not shown), the process is reversed and the circuit
returns to its initial condition.
(b) Note that the current and voltage are not in step (that is, not in phase). The current
reaches its maximum a quarter cycle ahead of the voltage. The phase relationship
Io
between the current and the voltage for a capacitor is commonly stated this way:
t = T/4 In a purely capacitive ac circuit, the current leads the voltage by 90°, or a one-quarter
Q=0 A 14 B cycle.
I = Io
Thus in an ac situation, a capacitor provides opposition to the charging process,
(c) but it is not totally limiting as it would be under dc conditions when it behaves as
an open circuit. The quantitative measure of this “capacitive opposition” to cur-
I rent is called the capacitor’s capacitive reactance (X C). Under ac conditions, the
capacitive reactance is given by
− − Q < Qo
+ + I < Io 1 1
XC = = (capacitive reactance) (21.11)
vC 2pfC
(d)
SI unit of capacitive reactance:
ohm 1Æ2, or seconds per farad 1s>F2
t = T/2
− − −− where, as usual, v = 2pf, C is the capacitance (in farads), and f is the frequency
(in Hz). Like resistance, reactance is measured in ohms 1Æ2. (Using unit analysis,
Q = Qo
+ + ++
I=0
you should show that the ohm is equivalent to seconds per farad.)
I=0 Equation 21.11 shows that the reactance is inversely proportional to both the
(e) capacitance (C) and the voltage frequency (f). Both of these dependencies can be
understood physically as follows.
䉱 F I G U R E 2 1 . 7 A capacitor under Recall that capacitance means “charge stored per volt” 1C = Q>V2. Therefore,
ac conditions This sequence shows for a particular voltage, the greater the capacitance, the more charge the capacitor
the voltage, charge, and current in a
circuit containing only a capacitor can accommodate. This requires a larger charge flow rate, or current. Increasing
and an ac voltage source. All five the capacitance offers less opposition to charge flow (that is, a reduced capacitive
circuit diagrams taken together rep- reactance) at a given frequency.
resent physically what is plotted in To understand the frequency dependence, consider the fact that the greater the
the first half of the cycle (from t = 0 frequency of the voltage, the shorter the time for charging each cycle. A shorter
to t = T>2) in the graph shown in
Fig. 21.6b. charging time means less charge will be able to accumulate on the plates and there
will be less opposition to the current. Increasing the frequency results in a
decrease in capacitive reactance. Hence capacitive reactance is inversely propor-
tional to both frequency and capacitance.
It is always good to check a general relationship to see whether it gives a result
known to be true in a special case [or in several special case(s)]. As a special case
for the capacitor, note that if f = 0 (that is, nonoscillating dc conditions), the
capacitive reactance is infinite. As expected under such conditions, there would be
no current.
Capacitive reactance is related to the voltage across the capacitor and the cur-
rent by an equation that has the same form as V = IR for pure resistances:
Vrms = Irms XC (voltage across a capacitor) (21.12)
Consider Example 21.3, in which a capacitor is connected to an ac voltage source.

21.3 INDUCTIVE REACTANCE 735
EXAMPLE 21.3 Current under ac Conditions: Capacitive Reactance

A 15.0-mF capacitor is connected to a 120-V, 60-Hz source. T H I N K I N G I T T H R O U G H . The capacitive reactance can be
What are (a) the capacitive reactance and (b) the current (rms obtained from the capacitance and the frequency by using
and peak) in the circuit? Eq. 21.11. (b) The rms current can then be determined from
the reactance and rms voltage via Eq. 21.12. Finally, Eq. 21.6
gives the peak current.
SOLUTION. Assuming the 60-Hz frequency to be exact, our answers are to three significant figures.
Given: C = 15.0 mF = 15.0 * 10-6 F Find: (a) XC (capacitive reactance)
Vms = 120 V (b) Io (peak current), Irms (rms current)
f = 60 Hz
(a) The capacitive reactance is
1 1
XC = = = 177 Æ
2pfC 2p160 Hz2115.0 * 10-6 F2
(b) Then, the rms current is
Vrms 120 V
Irms = = = 0.678 A
XC 177 Æ
and therefore, the peak current is
Io = 22Irms = 22 (0.678 A) = 0.959 A
The current oscillates at 60 cycles per second with a magnitude of 0.959 A.

F O L L O W - U P E X E R C I S E . In this Example, (a) what is the peak voltage and (b) what frequency would give the same current if the
capacitance were reduced by half?
DID YOU LEARN?

For a capacitor driven by an ac voltage source:
➥ Capacitive reactance is inversely related to the driving frequency.Thus current
increases with driving frequency.
➥ Capacitive reactance is inversely related to the capacitance.Thus current increases if
capacitance increases.
➥ The current is zero when the maximum energy is stored in the capacitor.
21.3 Inductive Reactance

For an inductor driven by an ac voltage source:

➥ What happens to the rms current in the circuit if the source’s frequency increases?
➥ What happens to the rms current in the circuit if the inductance value increases?
➥ What is the voltage across the inductor when the current is at its maximum value?
Inductance is a measure of the opposition a circuit element presents to a time-

varying current (by Lenz’s law). In principle, all circuit elements (even resistors)
have some inductance. However, a coil of wire with negligible resistance has, in
effect, only inductance. When placed in a circuit with a time-varying current, such
a coil, called an inductor, exhibits a reverse voltage, or back emf, in opposition to
the changing current. The changing current through the coil produces a changing
magnetic field and flux. The back emf is the induced emf in opposition to this
changing flux. Because the back emf is induced in the inductor as a result of its
own changing magnetic field, this phenomenon is called self-induction.
736 21 AC CIRCUITS
The self-induced emf (for a coil consisting of N loops) is given by Faraday’s law
(Eq. 20.2): e = - N¢£>¢t. The time rate of change of the total flux through the coil,
N¢ £>¢t, is proportional to the rate of change of the current in the coil, ¢I> ¢t. This
is because the current produces the magnetic field responsible for the changing flux.
Thus the back emf is proportional to, and oppositely directed to, the rate of current
change. This relationship is expressed using a proportionality constant, L:
¢I
e = -L (21.13)
¢t
where L is the inductance of the coil (more properly, its self-inductance). You should
be able to show, using unit analysis, that the units of inductance are volt-seconds
per ampere 1V # s>A2. This combination is called a henry 1H, 1 H = 1 V # s>A2, in
honor of Joseph Henry (1797–1878), an American physicist and early investigator
of electromagnetic induction. Smaller units, such as the millihenry (mH), are com-
monly used 11 mH = 10-3 H2.
The opposition presented to current by an inductor under ac conditions
depends on the inductance and the voltage frequency. This is expressed quantita-
tively by the circuit’s inductive reactance (XL), which is
XL = vL = 2pfL (inductive reactance) (21.14)
SI unit of inductive reactance:

ohm 1Æ2, or henrys per second 1H>s2
where f is the frequency of the driving voltage, v = 2pf, and L is the inductance.
Like capacitive reactance, inductive reactance is measured in ohms 1Æ2, which is
equivalent to henrys per second.
Note that the inductive reactance is directly proportional to both the coil induc-
tance (L) and the voltage frequency (f). The inductance is a property of the coil that
depends on the number of turns, the coil’s diameter and length, and the material in
the coil (if any). The frequency of the voltage plays a role because the more rapidly the
current in the coil changes, the greater the rate of change of its magnetic flux. This
implies a larger self-induced (back or reverse) emf to oppose the changes in current.
In terms of XL , the voltage across an inductor is related to the current and
inductive reactance by the following:
ac
L
source
Vrms = Irms XL (voltage across an inductor) (21.15)
The circuit symbol for an inductor and the graphs of the voltage across the
(a) inductor and the current in the circuit are shown in 䉳Fig. 21.8. When an inductor
is connected to an ac voltage source, maximum voltage corresponds to zero cur-
rent. When the voltage drops to zero, the current is maximum. This happens
V, I
Voltage V because as the voltage changes polarity (causing the magnetic flux through the
+
inductor to drop to zero), the inductor acts to prevent the change in accordance
with Lenz’s law, so the induced emf creates a current. In an inductor, the current
lags one quarter cycle behind the voltage, a relationship commonly expressed as
t follows:
In a purely inductive ac circuit, the voltage leads the current by 90°, or a one-quarter
A 14 B cycle.
1
4
cycle
− (90°)
Current I
Because the phase relationships between current and voltage for purely induc-
(b) tive and purely capacitive circuits are opposite, there is a phrase that may help
you remember the difference: ELI the ICE man. Here E represents voltage (for emf)
䉱 F I G U R E 2 1 . 8 A purely induc-
tive circuit (a) In a circuit with only and I represents current. The three letters ELI indicate that for inductance (L), the
inductance, (b) the voltage leads the voltage leads the current (I)—reading the acronym from left to right. Similarly, ICE
current by 90°, or one-quarter cycle. means that for capacitance (C), the current leads the voltage.
21.4 IMPEDANCE: RLC CIRCUITS 737
EXAMPLE 21.4 Current Opposition without Resistance: Inductive Reactance

A 125-mH inductor is connected to a 120-V, 60-Hz source. T H I N K I N G I T T H R O U G H . Because the inductance and fre-
What are (a) the inductive reactance and (b) the rms current in quency are known, the inductive reactance can be calculated
the circuit? from Eq. 21.14 and the current from Eq. 21.15.
SOLUTION. Listing the given data:

Given: L = 125 mH = 0.125 H Find: (a) XL (inductive reactance)
Vrms = 120 V (b) Irms
f = 60 Hz
(a) The inductive reactance is
XL = 2pfL = 2p160 Hz210.125 H2 = 47.1 Æ
(b) The rms current is then
Vrms 120 V
Irms = = = 2.55 A
XL 47.1 Æ
F O L L O W - U P E X E R C I S E . In this Example, (a) what is the peak current? (b) What voltage frequency would yield the same current
if the inductance were reduced to one-third the value in this Example?
DID YOU LEARN?

For an inductor driven by an ac voltage source:
➥ The inductive reactance is directly proportional to the frequency.Thus the rms
current decreases as frequency increases.
➥ The rms current decreases as the inductive reactance increases.
➥ The inductor’s voltage is zero when the current is at its maximum.
21.4 Impedance: RLC Circuits

➥ What is the sign of the phase angle in an RC circuit driven by an ac voltage source?
➥ In an RL circuit driven by an ac voltage source, how does the circuit’s resistance com-
R
pare to its impedance?
➥ For an RLC circuit with a resistance R, what can you say about its impedance
compared to R? ac
source
C
In the previous sections, purely capacitive or purely inductive circuits were con-
sidered separately and without resistance present. However, in the real world, it is
impossible to have purely reactive circuits, because there is always some resis- (a) RC circuit diagram
tance—at a minimum, that from the connecting wires. Thus resistances, capacitive
reactances, and inductive reactances combine to impede the current in ac circuits.
An analysis of some combination circuits illustrates these effects.
R
␾
SERIES RC CIRCUIT
Z
Suppose an ac circuit consists of a voltage source, a resistor, and a capacitor con- XC
nected in series (䉴 Fig. 21.9a). The phase relationship between the current and the
voltage is different for each circuit element. As a result, a special graphical method
is needed to find the overall impedance to the current in the circuit. This method
employs a phase diagram.
Z = √ R 2 + X 2C
In a phase diagram, such as in Fig. 21.9b for an RC circuit, the resistance and
reactance in the circuit are endowed with vectorlike properties and their magni- (b) Phase diagram
tudes are represented by arrows called phasors. On a set of x–y coordinate axes, the
resistance is plotted on the positive x-axis (that is, at 0°), because the voltage– 䉱 F I G U R E 2 1 . 9 A series RC circuit
(a) In a series RC circuit, (b) the
current phase difference for a resistor is zero. The capacitive reactance is plotted
along the negative y-axis, to reflect a phase difference 1f2 of - 90° because for a
impedance Z is the phasor sum of
the resistance R and the capacitive
capacitor, the voltage lags behind the current by one-quarter of a cycle. reactance XC.
738 21 AC CIRCUITS
The phasor sum is the effective, or net, impedance to the current, and is called the
circuit’s impedance (Z). Phasors must be added in the same way as vectors because
the effects of the resistor and capacitor are not in phase. For the series RC circuit,
2 2
Z = 3R + XC (series RC circuit impedance) (21.16)
The unit of impedance is the ohm.

The generalization of Ohm’s law to circuits containing capacitors and inductors
along with resistors is
Vrms = Irms Z (Ohm’s law for ac circuits) (21.17)
To illustrate how phasors can be used to analyze an RC circuit, consider

Example 21.5. Take particular note of part (b), in which there is an apparent viola-
tion of Kirchhoff’s loop theorem—explained by phase differences between the
voltages across the two elements in the circuit.
EXAMPLE 21.5 RC Impedance and Kirchhoff’s Loop Theorem

A series RC circuit has a resistance of 100 Æ and a capacitance reactance and the resistance can be combined to determine
of 15.0 mF. (a) What is the (rms) current in the circuit when it the overall impedance (Eq. 21.16). From Eq. 21.17, the
is driven by a 120-V, 60-Hz source? (b) Compute the (rms) impedance and voltage can be used to find the current.
voltage across each circuit element and the two elements (b) Because the current is the same everywhere at any given
combined. Compare it with that of the voltage source. Is time in a series circuit, the result of part (a) can be used to
Kirchhoff’s loop theorem satisfied? Explain your reasoning. calculate the voltages. The rms voltage across both elements
together is found by recalling that the individual voltages
T H I N K I N G I T T H R O U G H . (a) Note that the voltage and capac- are out of phase by 90°. What this means physically is that
itor values are the same as those in Example 21.3 and a resis- they reach their peak values not at the same time, but rather
tor has been added in series. This will help in reducing the one-fourth of a period apart. Thus, the voltages cannot sim-
necessary calculations. Then, using phasors, the capacitive ply be added.
SOLUTION.
Given: R = 100 Æ Find: (a) I (rms current)

C = 15.0 mF = 15.0 * 10-6 F (b) VC (rms voltage across capacitor)
Vrms = 120 V VR (rms voltage across resistor)
f = 60 Hz V1R + C2 (combined rms voltage)
(a) In Example 21.3, we found that the reactance for this The algebraic sum of these two rms voltages is 164 V, which
capacitor at this frequency was XC = 177 Æ . Now Eq. 21.16 is not the same as the rms value of the voltage source (120 V).
can be used to calculate the circuit impedance: This does not mean that Kirchhoff’s loop theorem has been vio-
Z = 3R + XC = 41100 Æ22 + 1177 Æ22 = 203 Æ
2 2 lated. In fact, the source voltage does equal the combined volt-
ages across the capacitor and resistor if you account for phase
Because Vrms = Irms Z, the rms current is differences. The combined voltage must be calculated properly to
take into account the 90° phase difference between the two volt-
Vrms 120 V
Irms = = = 0.591 A ages. Using the Pythagorean theorem to get the total voltage
Z 203 Æ
V1R + C2 = 3VR + VC = 4159.1 V22 + 1105 V22 = 120 V
2 2
(b) Using Eq. 21.17 first for the rms voltage across the resistor
alone 1Z = R2, Thus when the individual voltages are combined properly
VR = Irms R = 10.591 A21100 Æ2 = 59.1 V
(taking into account that the voltages do not peak at the same
time), Kirchhoff’s laws are still valid. Here it has been shown
For the capacitor alone 1Z = XC2, the rms voltage across the that the total rms voltage across both elements is equal to the
capacitor is rms voltage of the source. Care must be taken to always add
VC = Irms XC = 10.591 A21177 Æ2 = 105 V
the voltages this way because they are out of phase in general.
Kirchhoff’s laws are valid at any instant of time, not just for rms
values, but care must be taken to account for phase differences.
F O L L O W - U P E X E R C I S E . (a) How would the result in part (a) of this Example change if the circuit were driven by a voltage source
with the same rms voltage, but oscillating at 120 Hz? (b) Is the resistor or the capacitor responsible for the change?
SERIES RL CIRCUIT
The analysis of a series RL circuit (䉴 Fig. 21.10) is similar to that of a series RC cir-
R
cuit. However, the inductive reactance is plotted along the positive y-axis in the
phase diagram, to reflect a phase difference of + 90° with respect to the resistance.
Remember that a positive phase angle means that the voltage leads the current, as ac
source
is true for an inductor.
Thus the impedance in an RL series circuit is L
2 2
Z = 3R + XL (series RL circuit impedance) (21.18)
(a) RL circuit diagram
SERIES RLC CIRCUIT
More generally, an ac circuit may contain all three circuit elements—a resistor, an Z = √ R 2 + XL2
inductor, and a capacitor—as shown in series in 䉴 Fig. 21.11. Again, phasor addi-
tion must be used to determine the overall circuit impedance. Combining the ver-
tical components (that is, inductive and capacitive reactances) gives the total
reactance, XL - XC. Subtraction is used because the phase difference between XL XL
Z
and XC is 180°. The overall circuit impedance is the phasor sum of the resistance
and the total reactance. Employing the Pythagorean theorem once more on the ␾
phasor diagram,
R
Z = 3R2 + 1XL - X C22 (series RLC circuit impedance) (21.19) (b) Phase diagram
The phase angle 1f2 between the source voltage and the current in the circuit is 䉱 F I G U R E 2 1 . 1 0 A series RL cir-
cuit (a) In a series RL circuit, (b) the
the angle between the overall impedance phasor (Z) and the + x-axis (Fig. 21.11b), or impedance Z is the phasor sum of
the resistance R and the inductive
XL - X C reactance XL.
tan f = (phase angle in series RLC circuit) (21.20)
R
Notice that if XL is greater than XC (as in Fig. 21.11b), the phase angle is positive
1 +f2, and the circuit is said to be inductive, because the nonresistive part of the
impedance (that is, the reactance) is dominated by the inductor. If XC is greater
than XL , the phase angle is negative 1- f2, and the circuit is said to be capacitive,
because capacitive reactance dominates over inductive reactance.
R
A summary of impedances and phase angles for the three circuit elements and
various combinations is given in 䉲 Table 21.1. Example 21.6 analyzes an RLC circuit.
ac
source L
TABLE 21.1 Impedances and Phase Angles for Series Circuits
Circuit Element(s) Impedance Z (in W) Phase Angle F
C
R R 0°
C XC -90°
L XL +90° (a) RLC circuit diagram
RC 2 2 Negative (meaning that f is
3R + XC
between 0° and -90°)
2 2 XL
RL 3R + XL Positive (meaning that f is
RLC 2R
2
+ 1XL - X C2 2
between 0° and +90°)
Positive if XL 7 XC
Negative if XC 7 XL
(XL – XC)
XC
{ ␾
Z
Z = √ R 2 + (X L – X C) 2
䉴 F I G U R E 2 1 . 1 1 A series RLC circuit (a) In a
series RLC circuit, (b) the impedance Z is the phasor XL – X C
tan ␾ =
sum of the resistance R and the total (or net) reac- R
tance 1XL - XC2. Note that the phasor diagram is
drawn for the case of XL 7 XC. (b) Phase diagram
740 21 AC CIRCUITS
EXAMPLE 21.6 All Together Now: Impedance in an RLC Circuit

A series RLC circuit has a resistance of 25.0 Æ , a capacitance T H I N K I N G I T T H R O U G H . (a) To calculate the overall imped-
of 50.0 mF, and an inductance of 0.300 H. If the circuit is dri- ance from Eq. 21.19, the individual reactances must first be
ven by a 120-V, 60-Hz source, what are (a) the total impedance determined. (b) The current is computed from the generaliza-
of the circuit, (b) the rms current in the circuit, and (c) the tion of Ohm’s law, Vrms = Irms Z (Eq. 21.17). (c) The phase
phase angle between the current and the voltage? angle is calculated from Eq. 21.20.
SOLUTION. All the necessary data are given:

Given: R = 25.0 Æ Find: (a) Z (overall circuit impedance)
C = 50.0 mF = 5.00 * 10-5 F (b) Irms
L = 0.300 H (c) f (phase angle)
Vrms = 120 V
f = 60 Hz
(a) The individual reactances are
1 1
XC = = = 53.1 Æ
2pfC 2p160 Hz215.00 * 10-5 F2
and
XL = 2pfL = 2p160 Hz210.300 H2 = 113 Æ
Then,
Z = 4R2 + 1XL - XC22 = 4125.0 Æ22 + 1113 Æ - 53.1 Æ22 = 64.9 Æ
(b) Because Vrms = Irms Z, the rms current can be found:
Vrms 120 V
Irms = = = 1.85 A
Z 64.9 Æ
(c) Solving tan f = 1XL - XC2>R for the phase angle gives
≤ = tan-1 a b = + 67.3°
XL - XC 113 Æ - 53.1 Æ
f = tan-1 ¢
R 25.0 Æ
A positive phase angle was to be expected, because the inductive reactance is greater than the capacitive reactance [see part (a)].
Thus this circuit is inductive in nature.
F O L L O W - U P E X E R C I S E . (a) Consider the RLC circuit in this Example, but with the driving frequency doubled. Reasoning con-
ceptually, should the phase angle f be greater or less than the + 67.3° after the increase? (b) Compute the new phase angle to
show that your reasoning is correct.
By now, you should appreciate the usefulness of phasor diagrams in determin-

ing impedances, voltages, and currents in ac circuits. However, you might still be
wondering what are the use and meaning of the phase angle f. To illustrate its
importance, let’s examine the power loss in an RLC circuit. Note that this power
analysis also depends on the use of phasor diagrams.
POWER FACTOR FOR A SERIES RLC CIRCUIT

In considering an RLC circuit, a crucial fact to realize is that any circuit power loss
(joule heating) can take place only in the resistor. There are no power losses associated
with capacitors and inductors.* Capacitors and inductors simply store energy and
give it back, without loss.
The average (rms) power dissipated by a resistor is Prms = I 2rms R. This rms
power can also be expressed in terms of the rms current and voltage, but the voltage
must be that across the resistor (VR), because it is the only dissipative element. Thus the
average power dissipated in a series RLC circuit is equal to that of the resistor PR:
P = PR = Irms VR
The voltage across the resistor can be found from a voltage triangle that corre-
sponds to the phasor triangle (䉴 Fig. 21.12). The rms voltages across the individual
*Ideally, neither has any resistance, and thus any joule heating attributed to them is zero.
components in an RLC circuit are VR = Irms R, VL = Irms XL, and VC = Irms XC.
Combining the last two voltages, we can write 1VL - VC2 = Irms1XL - XC2. If Z
each leg of the phasor triangle (Fig. 21.12a) is multiplied by the rms current, an (X L – X C)
equivalent voltage triangle results (Fig. 21.12b). As this figure shows, the voltage ␾
across the resistor is
R
VR = Vrms cos f (21.21) (a) Phasor triangle
The term cos f is called the power factor. From Fig. 21.11,
R
cos f = (series RLC power factor) (21.22) V
Z (V L – V C)
␾
The average power, rewritten in terms of the power factor, is
VR
P = Irms Vrms cos f (series RLC power) (21.23) (b) Equivalent voltage triangle
Because power is dissipated only in the resistance 1P = I 2rms R2, Eq. 21.22 enables 䉱 F I G U R E 2 1 . 1 2 Phasor and volt-
us to express the average power as age triangles The rms voltages
across the components of a series
RLC circuit are given by VR = Irms R,
P = I 2rms Z cos f (series RLC power) (21.24) VL = Irms XL, and VC = Irms XC.
Because the current is the same
Note that cos f varies from a maximum of + 1 (when f = 0°) to a minimum of through each, (a) the phasor triangle
zero (when f = ⫾90°). When f = 0°, the circuit is said to be completely resistive. can be converted to (b) a voltage tri-
angle. Note that VR = V cos f. Both
That is, there is maximum power dissipation (as though the circuit contained only phasor diagrams are drawn for the
a resistor). The power factor decreases as the phase angle increases in either direc- case of XL 7 XC.
tion [because cos1-f2 = cos f]—in other words, as the circuit becomes inductive
or capacitive. At f = + 90°, the circuit is said to be inductive; at f = - 90°, it is
capacitive. In these latter two cases, the circuit contains only an inductors and/or
capacitors, but no resistance, so no power is dissipated.
In practice, because there is always some resistance, a circuit can never be com-
pletely inductive or capacitive. It is possible, however, for an RLC circuit to appear to be
completely resistive even if it contains a capacitor and an inductor, as will be seen in
Section 21.5. Let’s look at our previous RLC Example but with an emphasis on power.
EXAMPLE 21.7 Power Factor Revisited

What is the average power dissipated in the circuit described Therefore, its power factor is
in Example 21.6?
R 25.0 Æ
THINKING IT THROUGH. The power factor can be determined cos f = = = 0.385
Z 64.9 Æ
because the resistance (R) and impedance (Z) are known. Once
the power factor is known, the actual power can be calculated. Using the other data from Example 21.6 and Eq. 21.23 gives
SOLUTION. P = Irms Vrms cos f = 11.85 A21120 V210.3852 = 85.5 W

Given: See Example 21.6 Find: P (average power) This is less than the power that would be dissipated without a
In Example 21.6, it was determined that the circuit had an capacitor and an inductor. (Can you show this to be true?
impedance of Z = 64.9 Æ , and its resistance was R = 25.0 Æ . Why is it true?)
FOLLOW-UP EXERCISE. If the frequency were doubled and the capacitor removed from this Example, what would be the rms
power?
DID YOU LEARN?

➥ In a driven RC circuit, the phase angle is always negative because the capacitor
voltage lags behind the resistor voltage.
➥ In a driven RL circuit, the resistance is always less than the impedance because the
latter includes both resistance and inductive reactance.
➥ In an RLC circuit, the minimum possible value of its impedance is its resistance
value, R.
742 21 AC CIRCUITS
21.5 Circuit Resonance

For an RLC circuit driven by an ac voltage source:

X ➥ What happens to the resonance frequency if either the capacitance and inductance
are increased?
➥ What happens to the current at resonance if the resistance is reduced?
Reactance
XC XL
➥ What happens to the power at resonance if the resistance is reduced?
From the previous discussion, it can be seen that when the power factor 1cos f2 of
an RLC series circuit is equal to unity, maximum power is transferred to the cir-
cuit. In this situation, the current in the circuit must be at a maximum, because the
f impedance is at its minimum. This occurs because at this unique frequency, the
fo
inductive and capacitive reactances effectively cancel—that is, they are equal in
Frequency
magnitude and 180° out of phase, or opposite. This situation can
(a) happen in any RLC circuit if the appropriate source frequency is
chosen.
I The key to finding this frequency is to realize that because induc-
tive and capacitive reactances are frequency dependent, so is the
overall impedance. From the expression for the RLC series imped-
ance, Z = 2R2 + 1XL - XC22, it can be seen that the impedance is
R1 a minimum when XL - XC = 0. This occurs at a frequency fo ,
Current
found by setting XL = XC. Using the expressions for the reactances,

this means that 2pfo L = 1>12pfo C2. Solving for fo yields
R2 1
R3 fo = (series RLC resonance frequency) (21.25)
R 3 > R2 > R1 2p2LC
This frequency satisfies the condition of minimum impedance—and
f
fo therefore maximizes the current in the circuit, a situation analogous
Frequency to pumping a swing at just the right frequency or having a violin
string vibrate in one of its normal modes. The frequency fo is called
(b)
the circuit’s resonance frequency. A plot of capacitive and induc-
䉱 F I G U R E 2 1 . 1 3 Resonance fre- tive reactances versus frequency is shown in 䉳 Fig. 21.13a. The curves XC and XL
quency for a series RLC circuit (a) At intersect at fo, the frequency at which their values are equal.
the resonance frequency ( fo), the
capacitive and inductive reactances
are equal 1XL = XC2. On a graph of A PHYSICAL EXPLANATION OF RESONANCE
X versus f, this is the frequency at The physical explanation of resonance in a series RLC circuit is worth exploring. The
which the curves of XC and XL capacitor and inductor voltages are always 180° out of phase, or have opposite polar-
intersect. (b) On a graph of I versus ity. In other words, they tend to cancel out but usually don’t completely do so because
f, the current is a maximum at fo.
The curve becomes sharper and nar- their values are not equal. If this is the case, then the voltage across the resistor is less
rower as the resistance in the circuit than that of the source voltage because there is a net voltage across the combination of
decreases. capacitor and inductor. This means that the power dissipated in the resistor is less
than its maximum value (the value at resonance).
However, in the special situation when the capacitive and inductive voltages do
cancel, the full source voltage appears across the resistor, the power factor becomes 1,
and the resistor dissipates the maximum possible power. This is what is meant when
it is said that the circuit is being driven “at resonance.”
APPLICATIONS OF RESONANCE
The previous discussion showed that when a series RLC circuit is driven at its res-
onance frequency, both the current in the circuit and the power transfer to the cir-
cuit are at a maximum. A graph of rms current versus driving frequency is shown
in Fig. 21.13b for several different values of resistance. As expected, the maximum
current occurs at frequency fo. Notice also that the curve becomes sharper and nar-
rower as the resistance decreases.
Resonant circuits have a variety of applications. One common application is in
the tuning mechanism of a radio. Each radio station has an assigned broadcast
21.5 CIRCUIT RESONANCE 743
frequency at which its radio waves are transmitted (see Insight 21.1, Oscillator Cir-
cuits: Broadcasters of Electromagnetic Radiation). When the waves are received at
the antenna, their oscillating electric and magnetic fields set the electrons in the
antenna into regular back-and-forth motion. In other words, they produce an alter-
nating current in the receiver circuit, just as a regular ac voltage source would do.
In a given geographic area, usually several different radio signals reach an
antenna together, but a good receiver circuit selectively picks up only the signal
with a frequency at or near its resonance frequency. Most radios allow you to alter
this resonance frequency to “tune in” different stations. In the early days of radio,
variable air capacitors were used for this purpose (䉴Fig. 21.14). Today, more compact
variable capacitors in smaller radios have a polymer dielectric between thin plates. 䉱 F I G U R E 2 1 . 1 4 Variable air
The polymer sheets help maintain the plate separation and increase the capacitance, capacitor Rotating the movable plates
between the fixed plates changes the
thus allowing manufacturers to use plates of a much smaller area. (Recall from overlap area and thus the capacitance.
Section 16.4 that C = keo A>d.) In most modern radios, solid-state devices have Such capacitors were common in the
replaced variable capacitors. tuning circuits of older radios.
INSIGHT 21.1 Oscillator Circuits: Broadcasters of Electromagnetic Radiation

To broadcast the high-frequency electromagnetic waves used 4. When the capacitor is again fully charged (but at reverse
in radio communications and television, electric current must polarity), it has its initial energy again (Fig. 2c). This
be made to oscillate at high frequencies in antennas (Fig. 1). occurs halfway through the cycle, or half a period from
This can be accomplished with RLC circuits. Such circuits are the start 1T>22. The magnetic field in the coil is zero, as is
called oscillator circuits, because the current in them oscillates the circuit current.
at a frequency determined by their inductive and capacitive 5. The capacitor again begins
elements. to discharge, and these
When the resistance in an RLC circuit is very small, the cir- four steps are repeated
cuit is essentially an LC circuit. The current in such a circuit over and over. Thus there
oscillates at a frequency f, which is the circuit’s “natural” fre- is both a current and
quency and also its resonance frequency (Eq. 21.25). Any energy oscillation in the
small resistance in the circuit would dissipate energy. How- circuit. In an ideal case of a
ever, in an ideal LC circuit (which we are considering here) circuit without resistance,
with no resistance, the oscillation would continue indefinitely. the oscillations would con-
To understand this oscillation, consider the energy oscillations tinue indefinitely.
in an ideal (resistanceless) parallel LC circuit, shown in Fig. 2a.
Let’s assume that the capacitor is initially charged and the switch FIGURE 1
is then closed 1t = 02. The following sequence of events occurs: A broadcast antenna
1. The capacitor would discharge instantaneously (because
RC = 0) if it were not for the current having to pass S FIGURE 2
through the coil. At t = 0, the current in the coil is zero I=0 UC = E
An oscillating LC
(Fig. 2a). As the current builds, so does the magnetic field ++ ++ circuit If the resis-
in the coil. By Lenz’s law, the increasing magnetic field C E Qmax L tance is negligible,
–––– this circuit will
and change of flux in the coil induce a back emf to
UL = 0 oscillate indefi-
oppose this current increase. Because of this back emf,
the capacitor does takes time to discharge. nitely. Half a full
(a) t=0 cycle is shown
2. When the capacitor is fully discharged (Fig. 2b), all of its between t = 0
energy (in its electric field) has been transferred to the and t = T>2.
inductor in the form of its magnetic field. (Because in this S Imax UL = E Energy is trans-
circuit it is assumed that R = 0, no energy is lost to joule ferred back and
heating.) At this time (one-quarter of a period; T>4), the C Q=0 forth between
magnetic field and the current in the coil are at maximum magnetic and
B U =0 electric types of
values, and all the energy is stored in the inductor. (Refer C
energy (as shown
to the energy “histograms” accompanying the circuit dia-
grams in Figs. 2a, 2b, and 2c to visualize the energy trades (b) t=T by the energy his-
4 tograms to the
as the cycle proceeds.) right). The oscil-
S I=0 UC = E
3. As the magnetic field collapses from its maximum value, lating electrons in
an emf that opposes the collapse is induced in the coil. –––– the wire will give
This emf acts in a direction that tends to continue the cur- C E Qmax L off electromag-
rent in the coil even as it is decreasing (Lenz’s law again). ++ ++ netic radiation at
UL = 0 the circuit’s oscil-
The polarity of the emf is now opposite that in step 1.
lating frequency.
Thus current continues to charge to the capacitor, but the
(c) t=T
result is a polarity reversed from its initial polarity. 2
744 21 AC CIRCUITS
INTEGRATED EXAMPLE 21.8 AM versus FM: Resonance in Radio Reception

(a) When you switch from an AM station (on the “AM Given: fo = 920 kHz Find: C¿>C (ratio of FM
band”—the word band refers to a specific range of frequen- f oœ = 99.7 MHz capacitance to
cies) to one on the FM band, you are effectively changing the AM capacitance)
= 99.7 * 103 kHz
capacitance of the receiving circuit, assuming constant induc-
tance. Is the capacitance (1) increased or (2) decreased when From Eq. 21.25, the two resonant frequencies are given by
you make this change? (b) Suppose you were listening to
news on an AM station at 920 kHz and switched to a music 1 1
fo = and f oœ =
station on the FM band at 99.7 MHz. By what factor would 2p2LC 2p 2LC¿
you have changed the capacitance of the receiving circuit in
the radio, assuming constant inductance? Dividing the first of these equations by the second gives
(A) CONCEPTUAL REASONING. Because FM stations broadcast fo 2p2LC¿ C¿
= =
at significantly higher frequencies than do AM stations (see f oœ 2p2LC AC
Table 20.1), the resonance frequency of the receiver must be
increased to receive signals on the FM band. An increase of Solving for the capacitance ratio by squaring and substituting
the resonance frequency requires reducing the capacitance the numbers gives
since the inductance is fixed. Thus the correct answer is (2). fo 2 2
C¿ 920 kHz
(B) QUANTITATIVE REASONING AND SOLUTION. The resonance = ¢ œ≤ = ¢ 3
≤ = 8.51 * 10-5
C fo 99.7 * 10 kHz
frequency (Eq. 21.25) depends on the inductance and the
capacitance. Because the question asks for a “factor,” it is Thus, C¿ = 8.51 * 10-5 C and the capacitance was decreased
clearly asking for a ratio of the new capacitance to the original by a factor of almost one ten-thousandth 18.51 * 10-5 L 10-42.
capacitance. The frequencies have to be expressed in the same
units, so convert MHz into kHz and use unprimed quantities
for AM and primed quantities for FM.
F O L L O W - U P E X E R C I S E . (a) Based on the resonance curves shown in Fig. 21.13b, can you explain how it is possible to pick up
two radio stations simultaneously on your radio? (You may have encountered this phenomenon, particularly between two cities
located far apart from one another. Two stations are sometimes granted licenses for broadcasting at closely spaced frequencies
under the assumption that they won’t both be received by the same radio. However, under certain atmospheric conditions, this
may not be true.) (b) In part (b) of this Example, if you next increased the capacitance by a factor of two (starting with the news at
920 kHz) to listen to a hockey game, to what new frequency on the AM band would you now be tuned?
DID YOU LEARN?

For an RLC circuit driven by an ac voltage source:
➥ The resonance frequency is inversely related to its inductance and capacitance.
➥ The current at resonance is inversely related to its resistance.
➥ The power at resonance is inversely related to the resistance.
PULLING IT TOGETHER Making Effective Use of Phasor Diagrams For ac Circuit Analysis
A series RLC circuit consists of two identical resistors in par- T H I N K I N G I T T H R O U G H : This situation requires the application
allel (each 100 Æ ), in series with two identical capacitors in of circuit sketching, transformer action, capacitors in parallel,
parallel (each 15.0 mF) followed by a 0.125-H inductor. The ac inductive and capacitive reactances, phasor diagrams, circuit
voltage source is the output of an ideal step-down trans- impedance and phase angle.
former with a winding ratio of 10:1. The transformer is (a) The series sketch will enable us to find the equivalent
plugged into a 120 V (North American) household outlet. parallel resistance and parallel capacitance.
(a) Sketch the circuit. (b) What are the frequency and rms (b) Recalling the discussion in Section 20.4, the frequency will
output voltage of the transformer? (c) Determine the circuit’s be unchanged but the voltage stepped down by a factor of 10.
capacitive and inductive reactances. Use these to make a cor- (c) The reactances can be determined from the frequency
rectly scaled phasor diagram for this circuit, including axes and capacitance and reactance values. The phasor diagram
with units. From the scale drawing, estimate the phase angle should give an idea of the phase angle and impedance.
and circuit impedance. (d) Determine the phase angle and cir- (d) The reactance from (c) can be used to determine f and
cuit impedance exactly by calculation. Compare your Z exactly.
answers to part (c). (e) Find the average power (joule heating) (e) The joule heating can be calculated using the power fac-
in each resistor. tor, rms voltage and resistance values.
21.5 CIRCUIT RESONANCE 745
SOLUTION.
Given: R = 100 Æ (each resistor in parallel) Find: (a) sketch the circuit
C = 15.0 mF = 1.50 * 10-5 F (each capacitors in parallel) (b) f (frequency) and Vrms (rms voltage) of the
L = 0.125 H (ideal step-down transformer with 10:1 transformer output
winding ratio) (c) reactances and draw phasor diagram to scale,
V = 120 V use it to estimate f (phase angle) and Z (circuit
impedance)
(d) calculate f (phase angle) and Z (impedance)
(e) 1P2 (average power in each resistor)
(a) The accompanying figure (䉴 Fig. 21.15) shows the circuit 100 Ω
using the techniques from Chapter 18. Recall also how to com-
bine resistors and capacitors in parallel. You should be able to
show that the two resistors are equivalent to a single resistor of
R = 50.0 Æ . Similarly, the equivalent capacitance is
C = 3.00 * 10-5 F. These values will be used in part (c). 120 V 12 V 100 Ω
(b) The output frequency will be unchanged at 60 Hz. 15.0 µ F 15.0 µ F
Because it as a step-down transformer, the output voltage will
be one-tenth of the input voltage, or 12.0 V.
transformer
(c) The capacitive reactance for the equivalent capacitance is
0.125 H
1 1
XC = = = 88.4 Æ
2pfC 2p160 Hz213.00 * 10-5 F2 䉱 F I G U R E 2 1 . 1 5 Pulling It Together: The Circuit Diagram.
The inductive reactance is
XL = 2pfL = 2p160 Hz210.125 H2 = 47.1 Æ XL (Ω)
50 X (Ω)
The equivalent resistance of the two parallel resistors (see Sec- XL = 47.1 Ω
tion 18.1) is 50.0 Ω
0 R (Ω) 0 R (Ω)
R = 50.0 Æ 50 ␾
R = 50.0 Ω Z
The impedance and phasor diagrams are shown in 䉴Fig. 21.16
50 50
with the axes scaled in ohms. A quick look at the second one
shows that the phase angle is negative and approximately - 40°, XC = 88.4 Ω XL– XC = -41.3 Ω
while the impedance is on the order of 60 to 70 Æ .
100
(d) Using the previous diagram as a guide, the circuit’s phase
XC (Ω)
angle can be calculated exactly, since
(1) (2)
XL - XC
tan f =
R 䉱 F I G U R E 2 1 . 1 6 Pulling It Together: (1) The impedances
47.1 Æ - 88.4 Æ - 41.3 Æ and (2) the phase angle.
= = = - 0.826
50.0 Æ 50.0 Æ
Therefore, f = tan-1 1 -0.8262 = - 39.6°, in good agreement alent resistor. One way is to first find the voltage across the
with the visual assessment in part (c). combination from the phase angle and the rms voltage (being
The impedance is found from the Pythagorean theorem: careful to use the stepped-down voltage of 12.0 V, not 120 V):
Z = 41XL - XC22 + R2 VR = Vrms cos f = 112.0 V2 cos1 - 39.6°2 = 9.25 V
= 4147.1 Æ - 88.4 Æ22 + 150.0 Æ22

Then
VR 19.25 V22
= 64.9 Æ PR = = = 1.71 W
R 50.0 Æ
also in good agreement with part (c).
Since the resistors are in parallel and identical, each one dissi-
(e) There are several ways to find the power loss in the equiv- pates half this amount, or 0.86 W.
746 21 AC CIRCUITS
■ An ac voltage is described by ■ Phasors are vectorlike quantities that allow resistances and
reactances to be represented graphically.
V = Vo sin vt = Vo sin 2pft (21.1)
■ For a sinusoidally varying current, called ac current, the R
␾
peak current Io and the rms (root-mean-square or effective) Z
XC
current Irms are related by
Io
Irms = = 0.707 Io (21.6) Z = √ R2 + X2
22
C
I
■ Impedance (Z) is the total, or effective, opposition to cur-
rent that takes into account both resistances and reactances.
Io Impedance is related to current and circuit voltage by a gen-
Io
(peak)
Irms = 0.707Io eralization of Ohm’s law:
0 t
Vrms = Irms Z (21.17)

–Io
■ The impedance for a series RLC circuit is
■ For an ac voltage, the peak voltage Vo is related to its rms Z = 2R2 + 1XL - XC22 (21.19)
(root-mean-square) voltage Vrms by
Vo
Vrms = = 0.707Vo (21.8) XL
22
V
(XL – XC)
XC
{ ␾
Z
Vo
Z = √ R 2 + (X L – X C) 2
Vo XL – XC
Vrms = 0.707Vo tan ␾ =
(peak) R
0 t
■ The phase angle (F) between the rms voltage and the rms
current in a series RLC circuit is
–Vo
XL - XC
■ The current in a resistor is in phase with the voltage across tan f = (21.20)
it. For a capacitor, the current is 90° (one-quarter of a cycle) R
ahead of the voltage. For an inductor, the current lags ■ The power factor (cos F) for a series RLC circuit is a mea-
behind the voltage by 90°. sure of how close to the maximum power dissipation the
■ In ac circuits, joule heating is due entirely to the resistive circuit is. The power factor is
elements, and the time-averaged power dissipation is R
cos f = (21.22)
P = 2
I rms R (21.10) Z
■ In an ac circuit, capacitors and inductors allow current and The average power dissipated (joule heating in the resistor) is
create opposition to current. This opposition is character- P = Irms Vrms cos f (21.23)
ized by capacitive reactance (XC) and inductive reactance or
(XL), respectively. The capacitive reactance is given by
P = I 2rms Z cos f (21.24)
1 1
XC = = (21.11) ■ The resonance frequency ( fo) of an RLC circuit is the fre-
vC 2pfC
quency at which the circuit dissipates maximum power.
The inductive reactance is given by This frequency is
XL = vL = 2pfL (21.14) 1
fo = (21.25)
■ Ohm’s law, as applied to each type of circuit element, is a 2p2LC
generalization of the version from dc circuits. The relation-
I
ship between rms current and the rms voltage for a resistor is:
Vrms = Irms R (21.9) R1
Current
The relationship between rms current and the rms voltage R2
for a capacitor is R3
R 3 > R2 > R1
f
fo
Vrms = Irms XC (21.12) Frequency
The relationship between rms current and the rms voltage

for an inductor is
Vrms = Irms XL (21.15)

21.1 RESISTANCE IN AN AC CIRCUIT energy stored in the capacitor is (a) zero, (b) at a maximum,
(c) neither of the preceding, but somewhere in between.
1. Which of the following voltages is larger for a sinu-
soidally varying ac voltage: (a) Vo, (b) Vrms, or (c) they 9. A single inductor is connected to an ac voltage source.
have the same value? When the voltage across the inductor is at a maximum,
the current in it is not changing. Is this statement (a) true,
2. During the course of one ac voltage cycle (in the United
(b) false, or (c) cannot be determined from the given
States) how long does the direction of the current stay
information?
constant in a resistor: (a) 1>60 s, (b) 1>120 s, or (c) 1>30 s?
10. A single inductor is connected to an ac voltage source.
3. During seven complete ac voltage cycles (in the United
When the current in the inductor is at a maximum, the
States) what is the average voltage: (a) 0 V, (b) 60 V,
voltage across the inductor is not changing. Is this state-
(c) 120 V, or (d) 170 V?
ment (a) true, (b) false, or (c) cannot be determined from
4. During five complete ac voltage cycles (in the United the given information?
States) how many times does the power dissipated in a
resistor reach its maximum value: (a) once, (b) five times,
(c) ten times, or (d) twice? 21.4 IMPEDANCE: RLC CIRCUITS
5. In ac operation in the Unites States, how much time
AND
elapses between successive maximum power values in a 21.5 CIRCUIT RESONANCE
resistor: (a) 1>60 s, (b) 1>120 s, or (c) 1>30 s? 11. The impedance of an RLC circuit depends on (a) frequency,
(b) inductance, (c) capacitance, (d) all of the preceding.
12. If the capacitance of a series RLC circuit is decreased,
21.2 CAPACITIVE REACTANCE
(a) the capacitive reactance increases, (b) the inductive
AND
reactance increases, (c) the current remains constant,
21.3 INDUCTIVE REACTANCE (d) the power factor remains constant.
6. In a purely capacitive ac circuit, (a) the current and voltage 13. When a series RLC circuit is driven at its resonance fre-
are in phase, (b) the current leads the voltage, (c) the cur- quency, (a) energy is dissipated only by the resistive ele-
rent lags behind the voltage, or (d) none of the preceding. ment, (b) the power factor has a value of one, (c) there is
7. A single capacitor is connected to an ac voltage source. maximum power delivered to the circuit, (d) all of the
When the voltage across the capacitor is at a maximum, preceding.
then the charge on it is (a) zero, (b) at a maximum, 14. When a series RLC circuit is not driven at its resonance
(c) neither of the preceding, but somewhere in between. frequency, energy may be dissipated as joule heat in
8. A single capacitor is connected to an ac voltage source. either the capacitor or the inductor. Is this statement
When the current in the circuit is at a maximum, then the (a) true or (b) false?
21.1 RESISTANCE IN AN AC CIRCUIT 21.2 CAPACITIVE REACTANCE

1. The average current in a resistor in an ac circuit is zero. AND
Explain why the average power delivered to a resistor 21.3 INDUCTIVE REACTANCE
isn’t zero. 7. Explain why, under very low frequency ac conditions, a
2. The voltage and current associated with a resistor in an capacitor acts almost as an open circuit while an induc-
ac circuit are in phase. What does that mean? tor acts almost as a short circuit.
3. A 60-W lightbulb designed to work at 240 V in England 8. Can an inductor oppose dc current? What about a capac-
is instead connected to a 120-V source. Discuss the itor? Explain each and why they are different.
changes in the bulb’s rms current and power when it is 9. If the current on a 10-mF capacitor is described by
at 120 V compared with 240 V. Assume the bulb is ohmic. I = 1120 A2 sin1120pt + p>22, explain why the instan-
4. If the ac voltage and current for a particular circuit ele- taneous voltage across it at t = 0 is zero whereas the cur-
ment are given by V = 120 sin1120pt2 and rent at that time is not.
I = 30 sin1120pt + p>22, respectively, could the circuit 10. An inductor is connected by itself to a 60-Hz ac voltage
element be a resistor? Is the frequency 60 Hz? Explain. source. To ensure it has the same inductive reactance
5. A 25-Æ resistor is wired directly across a 120-V ac source. when it is connected to a 240-Hz source, would you
What happens to the time-average power, rms voltage, decrease or increase its inductance value? By what factor
and rms current when the resistor’s value changes to should it change? Explain your reasoning.
50 Æ ? 11. A capacitor is connected by itself to a 60-Hz ac voltage
6. A 25-Æ resistor is wired directly across a 120-V ac source. source. To ensure the circuit has the same capacitive
What happens to the time-average power, rms voltage, reactance when the capacitor is replaced with one of
and rms current when the ac source is changed to 240 V? twice the capacitance, would you decrease or increase
748 21 AC CIRCUITS
the source frequency? By what factor should it change? 13. For each of the following cases for an RLC circuit, does
Explain your reasoning. the resonance frequency increase, decrease, or stay the
same? If it changes tell by what factor. (a) Only the
capacitance is changed—it is quadrupled. (b) Only the
21.4 IMPEDANCE: RLC CIRCUITS inductance is changed—it is increased by nine times.
AND (c) Only the resistance is changed—it is increased by
21.5 CIRCUIT RESONANCE 10%. (d) Both the inductance and capacitance are dou-
12. An RLC circuit consists of a 25-Æ resistor, a 1.00-mF bled; nothing else changes.
capacitor, and a 250-mH inductor. If it is driven by an ac 14. You have a driven RLC circuit that is at resonance. Then
voltage source whose frequency is 60 Hz, (a) what is the the driving frequency changes and the power output
impedance of an RLC circuit at resonance? (b) How does drops. Can you tell if the frequency increased or
its impedance compare to that in part (a) if the source decreased? Explain.
frequency is doubled: is it more, less, or the same value? 15. If a driven RLC circuit has an inductive reactance of
(c) How does its impedance compare to that in part (a) if 250 Æ and a capacitive reactance of 150 Æ , is the driving
the source frequency is halved; is it more, less, or the frequency exactly at, above, or below the circuit’s reso-
same value? Explain your reasoning for all parts. nant frequency? Explain how you can tell.
EXERCISES
21.1 RESISTANCE IN AN AC CIRCUIT 10. ●● What are the resistance, peak current, and power
level of a computer monitor that draws an rms current of
1. ● What are the peak and rms voltages of a 120-V ac line
0.833 A when connected to a 120-V outlet?
and a 240-V ac line?
11. ● ● Find the rms and peak currents in a 40-W, 120-V
2. ● An ac circuit has an rms current of 5.0 A. What is the
lightbulb. What is its resistance?
peak current? What is the average current?
12. ● ● A 50-kW electric heater is designed to run using a
3. ● How much ac rms current must be in a 10-Æ resistor
240-V ac source. Find its (a) peak current and (b) peak
to produce an average power of 15 W?
voltage. (c) How much energy will you be billed for in a
4. ● An ac circuit contains a resistor with a resistance of
30-day month if it operates 2.0 h per day?
5.0 Æ . The resistor has an rms current of 0.75 A. (a) Find its
13. ● ● The current in a resistor is given by
I = 18.0 A2 sin140pt2 when a voltage given by

rms voltage and peak voltage. (b) Find the average power
V = 160 V2 sin140pt2 is applied to it. (a) What is the
delivered to the resistor.
5. ● A hair dryer is rated at 1200 W when plugged into a
resistance value? (b) What are the frequency and period
120-V outlet. Find (a) its rms current, (b) its peak current, of the voltage source? (c) What is the average power
and (c) its resistance. delivered to the resistor?
6. IE ● ● The voltage across a 10-Æ resistor varies as 14. ● ● The current and voltage outputs of an operating ac
V = 1170 V2 sin1100pt2. (a) Is the current in the resistor generator have peak values of 2.5 A and 16 V, respectively.
(1) in phase with the voltage, (2) ahead of the voltage by (a) What is the average power output of the generator?
90°, or (3) lagging behind the voltage by 90°? (b) Write (b) What is the effective resistance of the circuit it is in?
the expression for the current in the resistor as a function
15. ● ● ● The current in a 60-Æ resistor is given by
I = 12.0 A2 sin1380t2. (a) What is the frequency of the

of time and determine the voltage frequency.
7. ● ● An ac voltage is applied to a 25.0-Æ resistor so that it
current? (b) What is the rms current? (c) How much aver-
dissipates 500 W of power. Find the resistor’s (a) rms age power is delivered to the resistor? (d) Write an equa-
and peak currents and (b) rms and peak voltages. tion for the voltage across the resistor as a function of
8. IE ● ● An ac voltage source has a peak voltage of 85 V time. (e) Write an equation for the power delivered to the
and a frequency of 60 Hz. The voltage at t = 0 is zero. resistor as a function of time. (f) Show that the rms power
(a) If a student measures the voltage at t = 1>240 s, how obtained in part (e) is the same as your answer to part (c).
many possible results are there: (1) one, (2) two, or
(3) three? Why? (b) Determine all possible voltages the
21.2 CAPACITIVE REACTANCE
student might measure.
AND
9. ● ● An ac voltage source has an rms voltage of 120 V. Its
21.3 INDUCTIVE REACTANCE
voltage goes from zero to its maximum positive value in
4.20 ms. Write an expression for the voltage as a function 16. ● At what frequency does a 25-mF capacitor have a reac-
of time. tance of 25 Æ ?
EXERCISES 749
17. ● A single 2.0-mF capacitor is connected across the ter- 31. ●● A series RL circuit has a resistance of 100 Æ and an
minals of a 60-Hz voltage source, and a current of inductance of 100 mH and is driven by a 120-V, 60-Hz
2.0 mA is measured on an ac ammeter. What is the source. (a) Find the inductive reactance and the impedance
capacitive reactance of the capacitor? of the circuit. (b) How much current is drawn from the
18. ● What capacitance value would give a reactance of source?
100 Æ in a 60-Hz ac circuit? 32. ● ● A series RC circuit has a resistance of 250 Æ and a
19. ● How much current is in a circuit containing only a capacitance of 6.0 mF. If the circuit is driven by a 60-Hz
50-mF capacitor connected to an ac generator with an source, find (a) the capacitive reactance and (b) the
output of 120 V and 60 Hz? impedance of the circuit.
20. ● ● A single 50-mH inductor forms a complete circuit 33. IE ● ● A series RC circuit has a resistance of 100 Æ and a
when connected to an ac voltage source at 120 V and capacitive reactance of 50 Æ . (a) Will the phase angle be
60 Hz. (a) What is the inductive reactance of the circuit? (1) positive, (2) zero, or (3) negative? Why? (b) What is
(b) How much current is in the circuit? (c) What is the the phase angle of this circuit?
phase angle between the current and the applied volt- 34. ● ● A series RLC circuit has a resistance of 25 Æ , an
age? (Assume negligible resistance.) inductance of 0.30 H, and a capacitance of 8.0 mF.
21. ● ● A variable capacitor in a circuit with a 120-V, 60-Hz (a) At what frequency should the circuit be driven for the
source initially has a capacitance of 0.25 mF. The capaci- maximum power to be transferred from the driving
tance is then increased to 0.40 mF. (a) What is the per- source? (b) What is the impedance at that frequency?
centage change in the capacitive reactance? (b) What is 35. IE ● ● In a series RLC circuit, R = XC = XL = 40 Æ for a
the percentage change in the current in the circuit? particular driving frequency. (a) This circuit is (1) induc-
22. ● ● (a) An inductor has a reactance of 90 Æ in a 60-Hz ac tive, (2) capacitive, (3) in resonance. Explain your reason-
circuit. What is its inductance? (b) What frequency ing. (b) If the driving frequency is doubled, what will be
would be required to double its reactance? the impedance of the circuit?
23. ● ● (a) Find the frequency at which a 250-mH inductor 36. IE ● ● (a) An RLC series circuit is in resonance. Which one
has a reactance of 400 Æ . (b) At what frequency would a of the following can you change without upsetting the
0.40 mF capacitor have the same reactance? resonance: (1) resistance, (2) capacitance, (3) inductance,
24. IE ● ● A capacitor is connected to a variable-frequency ac or (4) frequency? (b) A resistor, an inductor, and a capaci-
voltage source. (a) If the frequency increases by a factor tor have values of 500 Æ , 500 mH, and 3.5 mF, respec-
of 3, the capacitive reactance will be (1) 3, (2) 31 , (3) 9, (4) 19 tively. They are connected in series to a power supply of
times the original reactance. Why? (b) If the capacitive 240 V with a frequency of 60 Hz. What values of resis-
reactance of a capacitor at 120 Hz is 100 Æ , what is its tance and inductance would be required for this circuit to
reactance if the frequency is changed to 60 Hz? be in resonance (without changing the capacitor)?
37. ● ● (a) How much power is dissipated in the circuit
25. ● ● With a single 150-mH inductor in a circuit with a 60-Hz
voltage source, a current of 1.6 A is measured on an ac described in Exercise 36b using the initial values of resis-
ammeter. (a) What is the rms voltage of the source? (b) What tance, inductance, and capacitance? (b) How much
is the phase angle between the current and that voltage? power is dissipated in the same circuit at resonance?
38. ● ● (a) What is the resonance frequency of an RLC circuit
26. ● ● (a) What inductance has the same reactance in a 120-
V, 60-Hz circuit as a capacitance of 10 mF? (b) What with a resistance of 100 Æ , an inductance of 100 mH, and
would be the ratio of inductive reactance to capacitive a capacitance of 5.00 mF? (b) What is the resonance fre-
reactance if the frequency were changed to 120 Hz? quency if all the values in part (a) are doubled?
39. ● ● A tuning circuit in a radio receiver has a fixed induc-
27. ● ● A circuit with a single capacitor is connected to a 120-
V, 60-Hz source. (a) What is its capacitance if there is a tance of 0.50 mH and a variable capacitor. (a) If the cir-
current of 0.20 A in the circuit? (b) What would be the cuit is tuned to a radio station broadcasting at 980 kHz
current if the source frequency were halved? on the AM dial, what is the capacitance of the capacitor?
(b) What value of capacitance is required to tune into a
28. IE ● ● An inductor is connected to a variable-frequency
station broadcasting at 1280 kHz?
ac voltage source. (a) If the frequency decreases by a fac-
tor of 2, the rms current will be (1) 2, (2) 12 , (3) 4, (4) 14 40. IE ● ● A coil with a resistance of 30 Æ and an inductance
times the original rms current. Why? (b) If the rms cur- of 0.15 H is connected to a 120-V, 60-Hz source. (a) Is the
rent in an inductor at 40 Hz is 9.0 A, what is its rms cur- phase angle of this circuit (1) positive, (2) zero, or
rent if the frequency is changed to 120 Hz? (3) negative? Why? (b) What is the phase angle of the cir-
cuit? (c) How much rms current is in the circuit?
(d) What is the average power delivered to the circuit?
21.4 IMPEDANCE: RLC CIRCUITS
41. ● ● A small welding machine uses a voltage source of
AND 120 V at 60 Hz. When the source is operating, it requires
21.5 CIRCUIT RESONANCE 1200 W of power, and the power factor is 0.75. (a) What
29. ●A coil in a 60-Hz circuit has a resistance of 100 Æ and is the machine’s impedance? (b) Find the rms current in
an inductance of 0.45 H. Calculate (a) the coil’s reactance the machine while operating.
and (b) the circuit’s impedance. 42. ●● A series circuit is connected to a 220-V, 60-Hz power sup-
30. ● ● A series RC circuit has a resistance of 200 Æ and a capac- ply. The circuit has the following components: a 10-Æ resistor,
itance of 25 mF and is driven by a 120-V, 60-Hz source. a coil with an inductive reactance of 120 Æ , and a capacitor
(a) Find the capacitive reactance and impedance of the cir- with a reactance of 120 Æ . Compute the rms voltage across
cuit. (b) How much current is drawn from the source? (a) the resistor, (b) the inductor, and (c) the capacitor.
750 21 AC CIRCUITS
43. ●● A series RLC circuit has a resistance of 25 Æ , a capac- (2) equal to 25 Æ , (3) less than 25 Æ . Why? (b) If the dri-
itance of 0.80 mF, and an inductance of 250 mH. The cir- ving frequency is 60 Hz, what is the circuit’s impedance?
cuit is connected to a variable-frequency source with a 46. ● ● ● A series RLC circuit with a resistance of 400 Æ has
fixed rms voltage output of 12 V. If the frequency that is capacitive and inductive reactances of 300 Æ and 500 Æ ,
supplied is set at the circuit’s resonance frequency, what respectively. (a) What is the power factor of the circuit?
is the rms voltage across each of the circuit elements? (b) If the circuit operates at 60 Hz, what additional
44. ● ● (a) In Exercises 42 and 43, determine the numerical capacitance should be connected to the original capaci-
(scalar) sum of the rms voltages across the three circuit tance to give a power factor of unity, and how should the
elements and explain why it is much larger than the capacitors be connected?
source voltage. (b) Determine the sum of these voltages 47. ● ● ● A series RLC circuit has components with R = 50 Æ ,
using the proper phasor techniques and show that your L = 0.15 H, and C = 20 mF. The circuit is driven by a
result is equal to the source voltage. 120-V, 60-Hz source. (a) What is the current in the circuit,
45. IE ● ● ● (a) If the circuit in 䉲 Fig. 21.17 is in resonance, the expressed as a percentage of the maximum possible cur-
impedance of the circuit is (1) greater than 25 Æ , rent? (b) What is the power delivered to the circuit,
expressed as a percentage of the power delivered when
25.0 Ω 䉳 FIGURE 21.17 the circuit is in resonance?
Tune to resonance
See Exercises 45, 54,
Signal and 55.
generator 2.50 µ F 2.50 µ F
0.450 H
48. A series RLC radio receiver circuit with an inductance of R1 R1

1.50 mH is tuned to an FM station at 98.9 MHz by adjust-
ing a variable capacitor. When the circuit is tuned to this L
ac ac C
source C RL source L RL
station, (a) what is its inductive reactance? (b) What is its
capacitive reactance? (c) What is its capacitance?
49. A circuit connected to a 110-V, 60-Hz source contains a (a) Low-pass filter (b) High-pass filter
50-Æ resistor and a coil with an inductance of 100 mH.
Find (a) the reactance of the coil, (b) the impedance of 䉱 F I G U R E 2 1 . 1 8 Low-pass and high-pass filters
the circuit, (c) the current in the circuit, and (d) the See Exercise 52.
power dissipated by the coil, and (e) calculate the phase 53. An ideal transformer is plugged into a 12-V, 60-Hz ac
angle between the current and the applied voltage. outlet in a motor home, thus enabling the owners to use
50. A 1.0-mF capacitor is connected to a 120-V, 60-Hz source. a 1500-W, 120-V hair dryer. (Ignore any inductance or
(a) What is the capacitive reactance of the circuit? (b) How capacitance associated with the hair dryer.) (a) What
much current is in the circuit? (c) What is the phase angle type of transformer should be used and what should its
between the current and the applied voltage? (d) What is turn ratio be? When the hair dryer is in operation,
the maximum energy stored in the capacitor? (f) What is (b) what is its resistance? (c) What are its frequency, rms
the power dissipated by this circuit? voltage, and rms current? (d) What are its peak current
and peak voltage and peak power output? (e) What are
51. IE (a) If an RLC circuit with a resistance of 450 Æ is in the peak power, current, and voltage values on the input
resonance, the phase angle of the circuit is (1) positive, side of the transformer?
(2) zero, (3) negative. (b) A circuit has an inductive reac- 54. (a) Determine the resonance frequency for the circuit in
tance of 280 Æ at 60 Hz. What value of capacitance Fig. 21.17. (b) If the signal generator was capable of a peak
would set this circuit into resonance? (c) What is the voltage output of 24 V, what is the maximum power out-
inductance value? (d) If the inductance doubles from the put of the resistor? (c) Determine the resonance frequency
value in part (c), what is the circuit’s phase angle now? and maximum power output if another identical resistor
52. The circuit in 䉴 Fig. 21.18a is called a low-pass filter is wired in parallel with the existing one.
because a large current and voltage (and thus a lot of 55. (a) For the circuit in Fig. 21.17, what would be the effect
power) are delivered to the load resistor (RL) only by a on the resonance frequency if one of the capacitors had a
low-frequency source. The circuit in Fig. 21.18b is called capacitance value smaller than that shown: (1) it would
a high-pass filter because a large current and voltage (and decrease, (2) it would increase, or (3) it would remain the
thus a lot of power) are delivered to the load only by a same? Explain your reasoning. (b) Determine the new
high-frequency source. Describe conceptually why the resonance frequency if one of the capacitors is replaced
circuits have these characteristics. by one with a capacitance of 1.50 mF.
Reflection and
22 Refraction of Light
22.1 Wave fronts
and rays (752)
■ plane wave fronts
■ rays
22.2 Reflection (753)

■ law of reflection
22.3 Refraction (756)

■ index of refraction
■ law of refraction
22.4 Total internal reflection

and fiber optics (764)
PHYSICS FACTS
W
■ critical angle
✦ Most camera lenses are coated with a
e live in a visual world, sur-
thin film to reduce light loss due to rounded by eye-catching
reflection. For a typical seven-element
22.5 Dispersion (768) camera lens, about 50% of the light
images such as that refractive
■ index of refraction would be lost to reflection if the lens image of the turtle shown in the
and wavelength were not coated with thin films.
chapter-opening photograph. How
✦ Every day, installers lay enough new
fiber-optic cables for computer net- these images are formed is some-
works to circle the Earth three times. thing taken largely for granted—
Optical fibers can be drawn to diam-
eters smaller than copper wire. Fibers
until something is seen that can’t
can be as small as 10 mm in diameter. be easily explained. Optics is the
In comparison, the average human
hair is about 100 mm in diameter.
study of light and vision. Human
✦ The saying “diamonds are forever” is vision requires visible light of wave-
not without a reason. Diamond has lengths from 400 nm to 700 nm
an exceptionally high index of refrac-
tion (n = 2.42) among transparent
(see Fig. 20.23). Optical phenom-
materials, which contributes to its ena, such as reflection and
brilliance due to the small critical
angle for total internal reflection.
refraction, are shared by all electro-
✦ In 1998, scientists at MIT made a magnetic waves. Light acts as a
perfect mirror, a mirror with 100% wave in its propagation (Chapter 24)
reflection. A tube lined with this
type of mirror would transmit light and as a particle (photon) when
over long distances better than it interacts with matter
optical fibers.
(Chapters 27–30).
752 22 REFLECTION AND REFRACTION OF LIGHT
In this chapter, the basic optical phenomena of reflection, refraction, total

internal reflection, and dispersion will be investigated. The principles that govern
reflection explain the behavior of mirrors, while those that govern refraction
explain the properties of lenses. With the aid of these and other optical principles,
we can understand many optical phenomena experienced every day—why a
glass prism spreads light into a spectrum of colors, what causes mirages, how rain-
bows are formed, and why the legs of a person standing in a swimming pool seem
to shorten. Some less familiar but increasingly useful subjects, including the fasci-
Point source
nating field of fiber optics, will also be explored.
Wave fronts
A simple geometrical approach involving straight lines and angles can be used
Ray
to investigate many aspects of the properties of light, especially how light propa-
Wave gates. For these purposes, we need not be concerned with the physical (wave)
Ray nature of electromagnetic waves described in Chapter 20. The principles of geo-
l
metrical optics will be introduced here and applied in greater detail in the study of
mirrors and lenses in Chapter 23.
(a)
Ray
22.1 Wave Fronts and Rays
l Wave front LEARNING PATH QUESTIONS
➥ What is a wave front?

(b)
➥ What is a wave ray?
䉱 F I G U R E 2 2 . 1 Wave fronts and ➥ Can our eyes tell whether rays are actually coming from objects or appear to come
rays A wave front is defined by from objects?
adjacent points on a wave that are in
phase, such as those along wave Waves, electromagnetic or otherwise, are conveniently described in terms of wave
crests or troughs. A line perpendicu-
lar to a wave front in the direction of
fronts. A wave front is the line or surface defined by adjacent portions of a wave
the wave’s propagation is called a that are in phase (Section 13.2). If an arc is drawn along one of the crests of a circu-
ray. (a) Near a point source, the lar water wave moving out from a point source, all the particles on the arc will be
wave fronts are circular in two in phase (䉳 Fig. 22.1a). An arc along a wave trough would work equally well. For a
dimensions and spherical in three three-dimensional spherical wave, such as a sound or light wave emitted from a
dimensions. (b) Very far from a
point source, the wave fronts are
point source, the wave front is a spherical surface rather than a circle.
approximately linear or planar and Very far from the source, the curvature of a short segment of a circular or spher-
the rays nearly parallel. ical wave front is extremely small. Such a segment may be approximated as a
linear wave front (in two dimensions) or a plane wave front (in three dimensions),
just as we take the surface of the Earth to be locally flat (Fig. 22.1b). A plane wave
front can also be produced directly by a large luminous flat surface. In a uniform
Rays
medium, wave fronts propagate outward from the source at a speed characteristic
of the medium. This was seen for sound waves in Section 14.2, and the same
occurs for light, although at a much faster speed. The speed of light is greatest in
a vacuum: c = 3.00 * 108 m>s. The speed of light in air, for all practical purposes,
is the same as that in vacuum.
The geometrical description of a wave in terms of wave fronts tends to neglect
the fact that the wave is actually oscillating, like those studied in Chapter 13. This
simplification is carried a step further with the concept of a ray. As illustrated in
Plane wave fronts Fig. 22.1, a line drawn perpendicular to a series of wave fronts and pointing in the
direction of propagation is called a ray. Note that a ray points in the direction of
䉱 F I G U R E 2 2 . 2 Light rays A the energy flow of a wave. A plane wave is assumed to travel in a straight line in a
plane wave travels in a direction medium in the direction of its rays, perpendicular to its plane wave fronts. A beam
perpendicular to its wave fronts. A
beam of light can be represented by
of light can be represented by a group of rays or simply as a single ray (䉳 Fig. 22.2).
a group of parallel rays (or by a sin- The representation of light as rays is adequate and convenient for describing
gle ray). many optical phenomena.
22.2 REFLECTION 753
How do we see things and objects around us? They are seen because rays from
the objects, or rays that appear to come from the objects, enter our eyes (䉴 Fig. 22.3)
and form images of the objects on the retina (Chapter 23). The rays could be com-
ing directly from the objects, as in the case of light sources, or could be reflected or (a)
refracted by the objects or other optical systems. Our eyes and brain working
together, however, cannot tell whether the rays actually come from the objects or
only appear to come from the objects. This is one way magicians can fool our eyes
with seemingly impossible illusions. Mirror
The use of the geometrical representations of wave fronts and rays to explain
phenomena such as the reflection and refraction of light is called geometrical
optics. However, certain other phenomena, such as the interference of light, can-
not be treated in this manner and must be explained in terms of actual wave char-
acteristics. These phenomena will be considered in Chapter 24.
(b)
DID YOU LEARN? 䉱 F I G U R E 2 2 . 3 How things are

➥ A wave front is defined by adjacent points on a wave that are in phase. For example, seen We see things because (a) rays
all crests of a water wave have the same phase; so do all the trough points. from the objects or (b) rays appear-
➥ A wave ray is a line drawn perpendicular to a series of wave fronts in the direction ing to come from the objects enter
of a wave’s propagation. our eyes.
➥ Our eyes cannot tell whether rays are actually coming from objects or appear to
come from objects.That is why various optical images look real to us.
22.2 Reflection
➥ How are the angles of incidence and reflection measured?

➥ What is the law of reflection?
➥ What is the difference between specular (regular) reflection and diffuse (irregular)
reflection?
The reflection of light is an optical phenomenon of enormous θ i = θr

importance: If light were not reflected by objects around us to our Normal
eyes, we wouldn’t see the objects at all. Reflection involves the
absorption and re-emission of light by means of complex electro-
magnetic oscillations in the atoms of the reflecting medium. How-
ever, the phenomenon is easily described by using rays. Plane of
A light ray incident on a surface is described by an angle of incidence
θi θr
incidence (Ui). This angle is measured relative to a normal—a line
perpendicular to the reflecting surface (䉴 Fig. 22.4). Similarly, the
reflected ray is described by an angle of reflection (Ur), also mea-
sured from the normal. The relationship between these angles is
given by the law of reflection: The angle of incidence is equal to Reflecti
ng surfa
the angle of reflection, or ce
ui = ur (law of reflection) (22.1)

䉱 F I G U R E 2 2 . 4 The law of reflec-
tion According to the law of reflec-
tion, the angle of incidence 1ui2 is
Two other attributes of reflection are that the incident ray, the reflected ray, and the
normal all lie in the same plane, which is sometimes called the plane of incidence, equal to the angle of reflection 1ur2.
and that the incident ray and the reflected ray are on opposite sides of the normal. Note that the angles are measured
When the reflecting surface is smooth and flat, the reflected rays from parallel inci- relative to a normal (a line perpen-
dent rays are also parallel (䉲 Fig. 22.5a). This type of reflection is called specular, or dicular to the reflecting surface).
regular, reflection. The reflection from a smooth water surface is an example of spec- The normal and the incident and
reflected rays always lie in the same
ular (regular) reflection (Fig. 22.5b). If the reflecting surface is rough, however, the plane.
reflected rays are not parallel (䉲 Fig. 22.6), because of the irregular nature of the sur-
face. This type of reflection is termed diffuse, or irregular, reflection. The reflection
of light from this page is an example of diffuse reflection because the paper is micro-
scopically rough. Insight 22.1, A Dark, Rainy Night further discusses the difference
between specular and diffuse reflection in a real-life situation.
ui ur
θi θr θi θr θi θr
(a) Specular (regular) (b) Specular (regular)

reflection (diagram) reflection (photo)
䉱 F I G U R E 2 2 . 5 Specular (regular) reflection (a) When a light 䉱 F I G U R E 2 2 . 6 Diffuse (irregular) reflection

beam is reflected from a smooth surface and the reflected rays are Reflected rays from a relatively rough surface, such
parallel, the reflection is said to be specular or regular. (b) Specular as this page, are not parallel; the reflection is said to
(regular) reflection from a smooth water surface produces an almost be diffuse or irregular. (Note that the law of reflec-
perfect mirror image of salt mounds at this Australian salt mine. tion still applies locally to each individual ray.)
INSIGHT 22.1 A Dark, Rainy Night

When you are driving on a dry night, the road and the street road can be clearly seen (just as you can read this page, because
signs can be clearly seen in the headlights. However, on a the paper is microscopically rough). However, when the road
dark, rainy night, even with headlights, you can hardly see surface is wet, water fills the crevices, turning the road into a
the road ahead. When a car approaches, the situation relatively smooth reflecting surface (Fig. 1a). Light from your
becomes even worse. You see the reflections of the approach- headlights then reflects ahead. The normally diffuse reflection
ing car’s headlights from the surface of the road, and they is replaced by specular reflection. Reflected images of lighted
appear brighter than usual. Often nothing can be seen except buildings and road lights form, blurring the view of the sur-
the reflective glare of the oncoming headlights. face, and the specular reflection of oncoming cars’ headlights
What causes these conditions? When the road surface is dry, may make it difficult for you to see the road (Fig. 1b).
the reflection of light off the road is diffuse (irregular), because Besides wet and slippery surfaces, specular reflection is a
the surface is rough. Light from your headlights hits the road major cause of accidents on rainy nights. Thus, extra caution
and reflects in all directions. Some of it reflects back, and the is advised under such conditions.
θi θr
Water
Road
(a) (b)
F I G U R E 1 Diffuse to specular (a) The diffuse reflection from a dry road is turned into specular reflection by water
on the road’s surface. (b) Instead of seeing the road, a driver sees the reflected images of lights, buildings, etc.
22.2 REFLECTION 755
Note in Fig. 22.5a and Fig. 22.6 that the law of reflection still applies locally to
both specular and diffuse reflection. However, the type of reflection involved
determines whether we see images from a reflecting surface. In specular reflec-
M2
tion, the reflected, parallel rays produce an image when they are viewed by an
optical system such as an eye or a camera. Diffuse reflection does not produce an
image, because the light is reflected in various directions. Ray
Experience with friction and direct investigations show that all surfaces are rough
on a microscopic scale. What, then, determines whether reflection is specular or dif-
30°
fuse? In general, if the dimensions of the surface irregularities are greater than the
wavelength of the light, the reflection is diffuse. Therefore, to make a good mirror, M1
glass (with a metal coating) or metal must be polished at least until the surface
irregularities are about the same size as the wavelength of light. Recall from Section 䉱 F I G U R E 2 2 . 7 Trace the ray See
Example 22.1.
20.4 that the wavelength of visible light is on the order of 10-7 m. (You will learn
more about reflection from a mirror in the Learn by Drawing, Tracing the Reflected
Rays, presented in Example 22.1.)
Diffuse reflection enables us to see illuminated objects, such as the Moon. If the
Moon’s spherical surface were smooth, only the reflected sunlight from a small
region would come to an observer on the Earth, and only that small illuminated LEARN BY DRAWING
area would be seen. Also, you can see the beam of light from a flashlight or spot-
light because of diffuse reflection from dust and particles in the air. tracing the reflected
rays
EXAMPLE 22.1 Tracing the Reflected Ray 1
Two mirrors, M1 and M2 , are perpendicular to each other, with a light ray incident on
one of the mirrors as shown in 䉴 Fig. 22.7. (a) Sketch a diagram to trace the path of the
reflected light ray. (b) Find the direction of the ray after it is reflected by M2. M2
Ray
T H I N K I N G I T T H R O U G H . The law of reflection can be used to determine the direction of θ i1 = 60°
the ray after it leaves the first and then the second mirror.
30°
SOLUTION.
M1
Given: u = 30° (angle relative to M1) Find: (a) Sketch a diagram tracing the light ray
(b) ur2 (angle of reflection from M2) 2
Follow steps 1–4 in Learn by Drawing:
(a) 1. Since the incident and reflected rays are measured from the normal (a line perpen-
M2
dicular to the reflecting surface), we draw the normal to mirror M1 at the point
Ray
where the incident ray hits M1. From geometry, it can be seen that the angle of
θ i1 = 60° θ r1 = 60°
incidence on M1 is ui1 = 60°.
2. According to the law of reflection, the angle of reflection from M1 is also ur1 = 60°. 30°
Next, draw this reflected ray with an angle of reflection of 60°, and extend it until it
M1
hits M2.
3. Draw another normal to M2 at the point where the ray hits M2. Also from geometry
3
(focus on the triangle in the diagram), the angle of incidence on M2 is ui2 = 30°. (Why?)
M2
(b) The angle of reflection off M2 is ur2 = ui2 = 30° (Step 4). This is the final direction of θ i2 = 30°
the ray reflected after both mirrors.
What if the directions of the rays are reversed? In other words, if a ray is first incident Ray
on M2, in the direction opposite that of the one drawn in step 4, will all the rays reverse θ i1= 60° θ r = 60°
1
their directions? Draw another diagram to prove that this is indeed the case. Light rays
30°
are reversible.
M1
FOLLOW-UP EXERCISE. When following an eighteen-wheel semitrailer, there may be a
sign on the back stating, “If you can’t see my mirror, I can’t see you.” What does this mean?
4
DID YOU LEARN?

➥ The angles of incidence and reflection are measured relative to a normal to the M2
θ r2 = 30°
reflecting surface. Ray
➥ The law of reflection states that the angle of incidence is equal to the angle of θ i1= 60° θ r = 60°
1 θ i2 = 30°
reflection.
➥ Specular (regular) reflection refers to reflection from a smooth surface, while diffuse 30°
(irregular) reflection occurs on a rough surface. Both reflections obey the law of M1
reflection, however.
22.3 Refraction
➥ How is the index of refraction of a medium defined?

➥ What is the fundamental physical cause of refraction, and how is the law of refrac-
tion written in terms of indices of refraction?
➥ When light is refracted from one medium to another, what quantity remains con-
stant and what changes?
Refraction refers to the change in direction of a wave at a boundary where the wave
passes from one medium into another. In general, when a wave is incident on a
boundary between media, some of the wave’s energy is reflected and some is trans-
mitted. For example, when light traveling in air is incident on a transparent material
such as glass, it is partially reflected and partially transmitted (䉳 Fig. 22.8). But the
direction of the transmitted light is different from the direction of the incident light,
so the light is said to have been refracted; in other words, it has changed direction.
This change in direction is caused by the fact that light travels with different
speeds in different media. Intuitively, you might expect the passage of light to take
longer through a medium with more atoms per volume, and the speed of light is,
in fact, generally less in denser media. For example, the speed of light in water is
about 75% of that in air or a vacuum. 䉲 Fig. 22.9a shows the refraction of light at an
air–water boundary.
䉱 F I G U R E 2 2 . 8 Reflection and The change in the direction of wave propagation is described by the angle of
refraction A beam of light is inci- refraction. In Fig. 22.9b, u1 is the angle of incidence and u2 is the angle of refrac-
dent on a trapezoidal glass prism tion. The symbols u1 and u2 are used for the angles of incidence and refraction so
from the left. Part of the beam is
reflected, and part is refracted at the as to avoid confusion with ui and ur, the angles of incidence and reflection.
air–glass surface. The refracted Willebrord Snell (1580–1626), a Dutch physicist, discovered a relationship
beam is again partially reflected and between the angles 1u2 and the speeds (v) of light in two media (Fig. 22.9b):
partially refracted at the bottom
glass–air surface. sin u1 v1
= (Snell’s law) (22.2)
sin u2 v2
This expression is known as Snell’s law or the law of refraction. Note that u1 and
u2 are always taken with respect to the normal.
Thus, light is refracted when passing from one medium into another because the
speed of light is different in the two media. The speed of light is greatest in a vacuum,
and it is therefore convenient to compare the speed of light in other media with this
constant value (c). This is done by defining a ratio called the index of refraction (n):
c speed of light in a vacuum

n = a b (index of refraction) (22.3)
v speed of light in a medium
As a ratio of speeds, the index of refraction is a unitless quantity. The indices of

refraction of several substances are given in 䉴 Table 22.1. Note that these values are
for a specific wavelength of light. The wavelength is specified because v, and
consequently n, are slightly different for different wavelengths within a particular
䉴 F I G U R E 2 2 . 9 Refraction Incident ray Normal

(a) Light changes direction on enter- θ1
ing a different medium. (b) The
direction of the refracted ray is Boundary Medium 1
described by the angle of refraction, Medium 2
u2 , measured from the normal.
θ2
Refracted ray
(a) (b)
22.3 REFRACTION 757
TABLE 22.1 Indices of Refraction (at l = 590 nm)*

Substance n
Air 1.000 29
Water 1.33
Ice 1.31
Ethyl alcohol 1.36
Fused quartz 1.46
Human eye 1.336–1.406
Polystyrene 1.49
Oil (typical value) 1.50
Glass (by type)† 1.45–1.70
crown 1.52
flint 1.66
Zircon 1.92
Diamond 2.42
*One nanometer (nm) is 10-9 m.

†
Crown glass is a soda–lime silicate glass; flint glass is a lead–alkali silicate
glass. Flint glass is more dispersive than crown glass (Section 22.5).
medium. (This is the cause of dispersion, to be discussed in Section 22.5.) The

values of n given in the table will be used in Examples and Exercises in this chap-
ter for all wavelengths of light in the visible region, unless otherwise noted.
Observe that n is always greater than 1, because the speed of light in a vacuum is
always greater than the speed of light in any material 1c 7 v2.
The frequency ( f ) of light does not change when the light enters another
medium, but the wavelength of light in a material 1lm2 differs from the wave-
length of that light in a vacuum 1l2, as can be easily shown:
c lf
n = =
v lm f
or
l
n = (22.4)
lm
The wavelength of light in the medium is then lm = l>n. Since n 7 1, it follows

that lm 6 l. Thus, the wavelength is the longest in a vacuum.
EXAMPLE 22.2 The Speed of Light in Water: Index of Refraction

Light from a He–Ne laser with a wavelength of 632.8 nm Since n = c>v,
travels from air into water. What are the speed and wave- c 3.00 * 108 m>s
length of the laser light in water? v = = = 2.26 * 108 m>s
n 1.33
T H I N K I N G I T T H R O U G H . If the index of refraction (n) of a
Note that 1>n = v>c = 1>1.33 = 0.75; therefore, v is 75% of
medium is known, the speed and wavelength of light in the the speed of light in a vacuum. Also, n = l>lm, so
medium can be obtained from Eq. 22.3 and Eq. 22.4.
l 632.8 nm
SOLUTION. lm = = = 476 nm
n 1.33
Given: n = 1.33 (from Table 22.1) Find: v and lm F O L L O W - U P E X E R C I S E . The speed of light of wavelength
l = 632.8 nm (speed and 500 nm (in air) in a particular liquid is 2.40 * 108 m>s. What
c = 3.00 * 108 m>s wavelength of are the index of refraction of the liquid and the wavelength of
(speed of light in air) light in water) light in the liquid?
The index of refraction, n, is a measure of the speed of light in a transparent

material, or technically, a measure of the optical density of the material. For example,
the speed of light in water is less than that in air, so water is said to be optically
denser than air. (Optical density in general correlates with mass density. However, in
some instances, a material with a greater optical density than another can have a
lower mass density.) Thus, the greater the index of refraction of a material, the greater
the material’s optical density and the smaller the speed of light in the material.
For practical purposes, the index of refraction is measured in air rather than in a
vacuum, since the speed of light in air is very close to c, and
c c
nair = L = 1
vair c
(From Table 22.1, nair = 1.000 29, but for calculations nair = 1 may be used.)
A more practical form of the law of refraction can be rewritten as
sin u1 v1 c>n1 n2
= = =
sin u2 v2 c>n2 n1
or
n1 sin u1 = n2 sin u2 (law of refraction) (22.5)
where n1 and n2 are the indices of refraction for the first and second media,
respectively.
Note that Eq. 22.5 can be used to measure the index of refraction. If the first
medium is air, then n1 L 1 and n2 L sin u1>sin u2. Thus, only the angles of inci-
dence and refraction need to be measured to determine the index of refraction of a
material experimentally. On the other hand, if the index of refraction of a material
is known, then the law of refraction can be used to find the angle of refraction for
any angle of incidence.
Note also that the sine of the refraction angle is inversely proportional to the
index of refraction: sin u2 = n1 sin u1>n2. Hence, for a given angle of incidence,
the greater the index of refraction of the second medium, the smaller the sin u2
and the smaller the angle of refraction, u2.
More generally, the following relationships hold:
■ If the second medium has a higher index of refraction than the first medium
1n2 7 n12, the ray is refracted toward the normal 1u2 6 u12, as illustrated in
䉲 Fig. 22.10a.
■ If the second medium has a lower index of refraction than the first medium
1n2 6 n12, the ray is refracted away from the normal 1u2 7 u12, as illustrated in
Fig. 22.10b.
Normal Normal
䉴 F I G U R E 2 2 . 1 0 Index of refrac-
tion and ray deviation (a) When the
second medium has a higher index
of refraction than the first 1n2 7 n12,
the ray is refracted toward the nor-
mal, as in the case of light entering θ1 θ1
water from air. (b) When the second
medium has a lower index of refrac-
tion than the first 1n2 6 n12, the ray Medium 1 n1 Medium 1 n1
is refracted away from the normal. Medium 2 n2 Medium 2 n2
[This is the case if the ray in part (a)
is traced in reverse, going from n2 > n1
θ2 θ2
medium 2 to medium 1.] θ2 < θ1 n2 < n1
(Bent toward θ2 > θ1
normal) (Bent away
from normal)
(a) (b)
22.3 REFRACTION 759
INTEGRATED EXAMPLE 22.3 Angle of Refraction: The Law of Refraction

Light in water is incident on a piece of crown glass at an angle (B) QUANTITATIVE REASONING AND SOLUTION. Again, the law
of 37° (relative to the normal). (a) Will the transmitted ray be of refraction (Eq. 22.5) is most practical in this case. (Why?)
(1) bent toward the normal, (2) bent away from the normal, or Listing the given quantities,
(3) not bent at all? Use a diagram to illustrate. (b) What is the
angle of refraction? Given: u1 = 37° Find: (b) u2 (angle of
(A) CONCEPTUAL REASONING. The indices of refraction of n1 = 1.33 (water) refraction)
water and crown glass can be found from Table 22.1. Accord- n2 = 1.52 (crown glass)
ing to the law of refraction (Eq. 22.5), n1 sin u1 = n2 sin u2, (1)
is the correct answer. Since n2 7 n1, the angle of refraction The angle of refraction is found by using Eq. 22.5,
must be smaller than the angle of incidence 1u2 6 u12.
n1 sin u1 11.332sin 37°
Because both u1 and u2 are measured from the normal, the sin u2 = = = 0.53
refracted ray will bend toward the normal. The ray diagram n2 1.52
in this case is identical to Fig. 22.10a. and
u2 = sin-110.532 = 32°
F O L L O W - U P E X E R C I S E . It is found experimentally that a beam of light entering a liquid from air at an angle of incidence of 37°
exhibits an angle of refraction of 29° in the liquid. What is the speed of light in the liquid?
EXAMPLE 22.4 A Glass Tabletop: More about Refraction

A beam of light traveling in air strikes
the glass top of a coffee table at an angle θ 1 = 45°
of incidence of 45° (䉴 Fig. 22.11). The
glass has an index of refraction of 1.5.
(a) What is the angle of refraction for n1 = 1.0
Air
the light transmitted into the glass? θ 1 – θ2
n2 = 1.5 θ2
(b) Prove that the beam emerging from
the other side of the glass is parallel to y
y = 2.0 cm θ2 r
the incident beam—that is, u4 = u1.
(c) If the glass is 2.0 cm thick, what is r θ3
d
the lateral displacement between the Glass d
ray entering and the ray emerging from Air
the glass (the perpendicular distance
between the two rays—d in the figure)? θ4
Original path
T H I N K I N G I T T H R O U G H . Since two New path
refractions are involved in this Exam- 䉱 F I G U R E 2 2 . 1 1 Two refractions In the glass, the refracted ray is displaced later-
ple, the law of refraction is used in ally (sideways) a distance d from the incident ray, and the emergent ray is parallel to
parts (a) and (b), and then some the original ray.
geometry and trigonometry in part (c).
SOLUTION. Listing the data: (b) If u1 = u4, then the emergent ray is parallel to the incident
Given: u1 = 45° Find: (a) u2 (angle of refraction) ray. Applying the law of refraction to the ray at both surfaces,
n1 = 1.0 (air) (b) Show that u4 = u1 n1 sin u1 = n2 sin u2
n2 = 1.5 (c) d (lateral displacement) and
y = 2.0 cm
n2 sin u3 = n1 sin u4
(a) Using the law of refraction, Eq. 22.5, with n1 = 1.0 for air From the figure, u2 = u3. Therefore,
gives n1 sin u1 = n1 sin u4
n1 sin u1 11.02 sin 45° 0.707 or
sin u2 = = = = 0.47
n2 1.5 1.5 u1 = u4
Thus, Thus, the emergent beam is parallel to the incident beam but
u2 = sin 10.472 = 28°
-1 displaced laterally or perpendicularly to the incident direc-
tion at a distance d.
Note that the beam is refracted toward the normal.
(c) It can be seen from the inset in Fig. 22.11 that, to find d, we In the yellow right triangle, d = r sin1u1 - u22. Substituting r
need to first find r from the known information in the pink from the previous step yields
right triangle. Then, y sin1u1 - u22 12.0 cm2 sin145° - 28°2
y y d = = = 0.66 cm
= cos u2 or r = cos u2 cos 28°
r cos u2
F O L L O W - U P E X E R C I S E . If the glass in this Example had n = 1.6, would the lateral displacement be the same, larger, or smaller?
Explain your answer conceptually, and then calculate the actual value to verify your reasoning.
EXAMPLE 22.5 The Human Eye: Refraction and Wavelength

A simplified representation of the crystalline lens in a human front of the eye and into the crystalline lens, qualitatively
eye shows it to have a cortex (an outer layer) of ncortex = 1.386 compare and list the frequency, speed, and wavelength of
and a nucleus (core) of nnucleus = 1.406. (See Fig. 25.1b.) If a light in air, in the cortex, and in the nucleus. First do the com-
beam of monochromatic (single-frequency or -wavelength) parison without numbers, and then calculate the actual
light of wavelength 590 nm is directed from air through the values to verify your reasoning.
REASONING AND ANSWER. First, the relative magnitudes of the indices of refraction are needed, where nair 6 ncortex 6 nnucleus.
As learned earlier in this section, the frequency ( f ) of light is the same in all three media: air, the cortex, and the nucleus. Thus,
the frequency can be calculated by using the speed and the wavelength of light in any of these materials, but it is easiest in air.
(Why?) From the wave relationship c = lf (Eq. 13.17),
c 3.00 * 108 m>s
f = fair = fcortex = fnucleus = = = 5.08 * 1014 Hz
l 590 * 10-9 m
The speed of light in a medium depends on its index of refraction, since v = c>n. The smaller the index of refraction, the
higher the speed. Therefore, the speed of light is the highest in air 1n = 1.002 and lowest in the nucleus 1n = 1.4062.
The speed of light in the cortex is
c 3.00 * 108 m>s
vcortex = = = 2.16 * 108 m>s
ncortex 1.386
and the speed of light in the nucleus is
3.00 * 108 m>s
vnucleus = = 2.13 * 108 m>s
1.406
We also know that the wavelength of light in a medium depends on the index of refraction of the medium 1lm = l>n2. The
smaller the index of refraction, the longer the wavelength. Therefore, the wavelength of light is the longest in air (n = 1.00 and
l = 590 nm) and the shortest in the nucleus 1n = 1.4062.
The wavelength in the cortex can be calculated from Eq. 22.4:
l 590 nm
lcortex = = = 426 nm
ncortex 1.386
and the wavelength in the nucleus is
590 nm
lnucleus = = 420 nm
1.406
Finally, a table is constructed to compare the index of refraction, frequency, speed, and wavelength of light in the three media:
Index of refraction Frequency (Hz) Speed (m/s) Wavelength (nm)
Air 1.00 5.08 * 1014 3.00 * 108 590

Cortex 1.386 5.08 * 1014 2.16 * 108 426
Nucleus 1.406 5.08 * 1014 2.13 * 108 420
F O L L O W - U P E X E R C I S E . A light source of a single frequency is submerged in water in a fish tank. The beam travels in the water,
through the glass pane at the side of the tank, and into the air. In general, what happens to (a) the frequency and (b) the wave-
length of the light when it emerges into the outside air?
22.3 REFRACTION 761
n1 > n 2
u 2 > u1
u1
larger n
Cooler air
(larger n) smaller n
u2
Warm air
(smaller n)
Road surface
(a) (b)
䉱 F I G U R E 2 2 . 1 2 Refraction in action (a) An inverted car on a “wet” road—a mirage.

(b) The mirage is formed when light from the object is refracted by layers of air at different
temperatures near the surface of the road.
Refraction is common in everyday life and can be used to explain many

phenomena.
Mirage: A common example of this phenomenon sometimes occurs on a high-
way on a hot summer day. The refraction of light is caused by layers of air that
have different temperatures (the layer closer to the road has a higher temperature,
lower density, and therefore lower index of refraction). This variation in indices of
refraction gives rise to the observed “wet” spot and an inverted image of an object
such as a car (䉱 Fig. 22.12a). The term mirage generally brings to mind a thirsty per-
son in the desert “seeing” a pool of water that really isn’t there. This optical illu-
sion plays tricks on the mind, with the image usually seen as a pool of water and
our eye’s past experience unconsciously leading us to conclude that there is water
on the road.
As shown in Fig. 22.12b, there are two ways for light to get to our eyes from the
car. First, the horizontal rays come directly from the car to our eyes, so we see the
car above the ground. Also, the rays from the car that travel toward the road sur-
face will be gradually refracted by the layered air. After hitting the road, these rays
will be refracted again and travel toward our eyes. (See the inset in the figure.)
Cooler air has a higher density and so a higher index of refraction. A ray travel-
ing toward the road surface will be gradually refracted with an increasing angle of
refraction until it hits the surface. It will then be refracted with a decreasing angle
of refraction, going toward our eyes. As a consequence, we also see an inverted
image of the car, appearing below the road surface. In other words, the surface of
the road acts almost as a mirror. The “pool of water” is actually skylight being
refracted—an image of the sky. This layering of air of different temperatures,
creating different indices of refraction, causes us to “see” the rising hot air as a
result of continually changing refraction.
The opposite of this is the mirage at sea (the looming effect). At sea, the air
above is warmer than that below. This causes the light to be refracted opposite to
what is in Fig. 22.12b, causing objects to be seen in the air above the water.
Not where it should be: You may have experienced a refractive effect while try-
ing to reach for something underwater, such as a fish (䉲 Fig. 22.13a). We are used to
light traveling in straight lines from objects to our eyes, but the light reaching our
eyes from a submerged object has had a directional change at the water–air inter-
face. (Note in the figure that the ray is refracted away from the normal.) As a
result, the object appears closer to the surface than it actually is, and therefore we
θ1
Apparent position
θ2
Actual position
(a) (b) (c)
䉱 F I G U R E 2 2 . 1 3 Refractive effects (a) The light is refracted, and because we tend to

think of light as traveling in straight lines, the fish is below where we think it is. (b) The
chopstick appears bent at the air–water boundary. If the cup is transparent, we see a differ-
ent refraction. (See Conceptual Question 7.) (c) Because of refraction, the coin appears to be
closer than it actually is.
tend to miss the object when reaching for it. For the same reason, a chopstick in a
cup appears bent (Fig. 22.13b), a coin in a glass of water appears closer than it
really is (Fig. 22.13c), and the legs of a person standing in water seem shorter than
their actual length. The relationship between the true depth and the apparent
depth can be calculated. (See Exercise 25.)
Atmospheric effects: The Sun on the horizon sometimes appears flattened, with
its horizontal dimension greater than its vertical dimension (䉲 Fig. 22.14a). This
effect is the result of temperature and density variations in the denser air along the
horizon. These variations occur predominantly vertically, so light from the top and
bottom portions of the Sun are refracted differently as the two sets of beams pass
through different atmospheric densities with different indices of refraction.
Atmospheric refraction lengthens the day, so to speak, by allowing us to see the
Sun (or the Moon, for that matter) just before it actually rises above the horizon
and just after it actually sets below the horizon (lengthening the day by as much as
20 min on both ends). The denser air near the Earth refracts the light over the hori-
zon toward us (Fig. 22.14b).
䉴 F I G U R E 2 2 . 1 4 Atmospheric You Your

effects (a) The Sun on the horizon horizon
commonly appears flattened as a
result of atmospheric refraction. Apparent
(b) Before rising and after setting, Sun
the Sun can be seen briefly also
because of atmospheric refraction.
(Exaggerated for illustration).
Actual sun
(a) (b)
22.3 REFRACTION 763
The twinkling of stars is due to atmospheric turbulence, which distorts the light
from the stars. The turbulence refracts light in random directions and causes the
stars to appear to “twinkle.” Stars on the horizon appear to twinkle more than
stars directly overhead, because the light from those stars has to pass through
more of the Earth’s atmosphere.
DID YOU LEARN?

➥ The index of refraction of a medium is defined as a ratio of the speed of light in a
vacuum to the speed of light in the medium; that is, n = c>v, where c is the speed in
a vacuum and v is the speed in the medium.
➥ The fundamental physical cause of refraction is that light has different speeds in
different media.The law of refraction written in terms of indices of refractions is
n1 sin u1 = n2 sin u2.
➥ The frequency of light is a constant.The speed of light (v) and the wavelength of
light (l) change from one medium to another. v = lf is valid for any medium.
INSIGHT 22.2 Negative Index of Refraction and the Superlens

In 1968 physicists predicted the existence of a material with a incident on a positive index material is refracted to the other
negative index of refraction. They expected that, in the pres- side of the normal to the interface. However, if the material
ence of such a material, nearly all known wave propagation has a negative index of refraction, the same incident light is
and optical phenomena would be substantially altered. refracted to the same side of the normal to the interface
Negative index materials were not known to exist at the time, (Fig. 2b). Due to this “abnormal” refraction, negative index
however. materials with flat surfaces can focus light as shown in Fig. 2c,
At the beginning of the twenty-first century, a new class of resulting in a new class of lenses (Chapter 23).
artificially structured materials that were found to have nega- The undesirable characteristics of lenses made of materials
tive indices of refraction was created (Fig. 1). In early 2007, with a positive index of refraction are energy loss due to
scientists fabricated a sample of such a material for the visible reflection, aberrations, and low resolution due to diffraction
region by etching an array of holes roughly 100 nm wide into limit (more on this in Chapter 24). The latest experiments pro-
layers of silver and magnesium fluoride on a glass substrate. vide strong evidence that negative index lenses have an
For some perspective, a human hair is about 100 000 nm in important future in imaging, as they offer a new degree of
diameter. The material had an index of ⫺0.6 at a wavelength flexibility that could lead to more compact lenses with
of 780 nm. reduced lens aberration. The diffraction limit—which is the
Figure 2 illustrates the difference between materials with most fundamental limitation to image resolution—may be
positive and negative indices of refraction. In Fig. 2a, light circumvented by negative index materials. Furthermore, total
negative refraction—that is, absence of a reflection—has been
observed in materials with a negative index of refraction.
Such a lens would truly be a superlens.
(a) (c)
(b)
F I G U R E 2 Reflection in positive index versus negative index

materials (a) Light incident on the interface between air and a
positive index material is bent toward the other side of the nor-
mal, (b) whereas in a negative index material light is bent
toward the same side of the normal. (c) If a light source is
F I G U R E 1 Material with a negative index of refraction This placed on one side of a slab with a refractive index of n = - 1,
artificial material made from grids of rings and wires has a the waves are refracted in such a way as to converge inside the
negative index of refraction. material and then again just outside the material.
Normal
Total
u2 internal
u2 = 90° reflection
Air
Water
uc u1 u2
u1
u1 ⫽ u2 and
u1 ⬎ uc
Light source
(a) (b)
䉱 F I G U R E 2 2 . 1 5 Internal reflection (a) When light enters a medium with a lower index
of refraction, it is refracted away from the normal. At a critical angle (uc), the light is
refracted along the interface (common boundary) of the media. At an angle greater than the
critical angle (u1 7 uc), there is total internal reflection. (b) Can you estimate the critical
angle in the photograph?
22.4 Total Internal Reflection and Fiber Optics

➥ What is total internal reflection, and how is the critical angle calculated?
➥ What conditions must be satisfied in order for total internal reflection to occur?
90° Normal ➥ What is the basis of fiber optics?
An interesting phenomenon occurs when light travels from a medium with a

90° higher index of refraction into a medium with a lower one, such as when light
goes from water into air. As you know, in such a case a ray will be refracted away
from the normal. (The angle of refraction is larger than the angle of incidence.)
45° Normal Furthermore, the law of refraction states that the greater the angle of incidence,
the greater the angle of refraction. That is, as the angle of incidence increases, the
(a) refracted ray diverges farther from the normal.
However, there is a limit. For a certain angle of incidence called the critical
angle (Uc), the angle of refraction is 90°, and the refracted ray is directed along the
boundary between the media. But what happens if the angle of incidence is even
larger? If the angle of incidence is greater than the critical angle 1u1 7 uc2, the light
isn’t refracted at all, but is internally reflected (䉱 Fig. 22.15). This condition is called
total internal reflection. The reflection process is about 100% efficient. (There is
always some absorption of light in the materials.) Because of total internal reflec-
tion, glass prisms can be used as mirrors (䉳 Fig. 22.16). In summary, where
n1 7 n2 , reflection and refraction occur at all angles for u1 … uc, but the refracted
or transmitted ray disappears for u1 7 uc.
An expression for the critical angle can be obtained from the law of refraction. If
u1 = uc in the medium with a higher index of refraction, u2 = 90°, and it follows that
(b)
n1 sin u1 = n2 sin u2 or n1 sin uc = n2 sin 90°
䉱 F I G U R E 2 2 . 1 6 Internal reflec-
Since sin 90° = 1,
tion in a prism (a) Because the criti-
cal angle of glass is less than 45°,
prisms with 45° and 90° angles can n2
sin uc = (critical angle, where n1 7 n2) (22.6)
be used to reflect light through 180°. n1
(b) Internal reflection of light by
prisms in binoculars makes this
If the second medium is air, n2 L 1, then the critical angle at the boundary from a
instrument much shorter than a
telescope because the rays are medium into air is given by sin uc = 1>n, where n is the index of refraction of the first
“folded” by the prisms. medium.
22.4 TOTAL INTERNAL REFLECTION AND FIBER OPTICS 765
EXAMPLE 22.6 A View from the Pool: Critical Angle

(a) What is the critical angle for light traveling in water and
incident on a water–air boundary? (b) If a diver submerged in
a pool looked up at the surface of the water at an angle of
u 6 uc, what would she see?
T H I N K I N G I T T H R O U G H . (a) The critical angle is given by
Eq. 22.6. (b) As shown in Fig. 22.15a, uc forms a cone of vision
for viewing from below the water.
SOLUTION.
Given: n1 = 1.33 (water, from Find: (a) uc (critical angle)

Table 22.1) (b) view for u 6 uc
n2 L 1 (air)
(a) The critical angle is 䉱 F I G U R E 2 2 . 1 7 Panoramic and distorted An
underwater view of the surface of a swimming
≤ = sin-1 a b = 48.8°
n2 1 pool in Hawaii.
uc = sin-1 ¢
n1 1.33
(b) Using Fig. 22.15a, trace the rays in reverse for light com- face would also appear distorted. An underwater
ing from all angles outside the pool. Light coming from the panoramic view is seen in 䉱 Fig. 22.17. Now can you explain
above-water 180° panorama could be viewed only in a cone why wading birds like herons usually keep their bodies low
with a half-angle of 48.8°. As a result, objects above the sur- while trying to catch a fish?
FOLLOW-UP EXERCISE. What would the diver see when looking up at the water surface at an angle of u 7 uc?
Internal reflections enhance the brilliance of cut diamonds. (Brilliance is a mea-

sure of the amount of light returning back to the viewer. Brilliance is reduced if
light leaks out the back of a diamond—that is, if the reflection is not total.) The crit-
ical angle for a diamond–air surface is
uc = sin-1 a b = sin-1 a b = 24.4°

1 1
n 2.42
A so-called brilliant-cut diamond has many facets, or faces (fifty-eight in all—
thirty-three on the upper face and twenty-five on the lower). Light from above hit-
ting the lower facets at angles greater than the critical angle is internally reflected
in the diamond. The light then emerges from the upper facets, giving rise to the
diamond’s brilliance (䉲 Fig. 22.18).
FIBER OPTICS
When a fountain is illuminated from below, the light is transmitted along the
curved streams of water. This phenomenon was first demonstrated in 1870 by the
䉳 F I G U R E 2 2 . 1 8 Diamond bril-
liance (a) Internal reflection gives
rise to a diamond’s brilliance.
(b) The “cut,” or the depth propor-
tions, of the facets is critical. If a
stone is too shallow or too deep,
light will be lost (refracted out)
through the lower facets.
1
3
2
3
(a) (b)
British scientist John Tyndall (1820–1893), who showed that light was “con-
ducted” along the curved path of a stream of water flowing from a hole in the side
of a container. The phenomenon is observed because light undergoes total internal
reflection along the stream.
Total internal reflection forms the basis of fiber optics, a fascinating technology
centered on the use of transparent fibers to transmit light. Multiple total internal
reflections make it possible to “pipe” light along a transparent rod (as in streams of
water), even if the rod is curved (䉳 Fig. 22.19). Note from the figure that the smaller
(a)
the diameter of the light pipe, the more total internal reflections it has. A small fiber
can produce as many as several hundred total internal reflections per centimeter.
Total internal reflection is an exceptionally efficient process. Compared with
copper electrical wires, optical fibers can be used to transmit light over very long
(b) distances with much less signal loss. These losses are due primarily to impurities
in the fiber, which scatter the light. Transparent materials have different degrees of
transmission. Fibers are made of special plastics and glasses for maximum trans-
mission efficiency. The greatest efficiency is achieved with infrared radiation,
because there is less scattering, as will be learned in Section 24.5.
The greater efficiency of multiple total internal reflections compared with mul-
tiple mirror reflections can be illustrated by a good reflecting plane mirror, which
(c)
has a typical reflectivity of about 95%. After each reflection, the beam intensity is
95% of that of the incident beam from the preceding reflection 1I1 = 0.95Io ,
I2 = 0.95I1 = 0.952Io , Á 2. Therefore, the intensity I of the reflected
beam after n reflections is given by
I = 0.95nIo
(d)
where Io is the initial intensity of the beam before the first reflection. Thus, after
fourteen reflections,
䉱 F I G U R E 2 2 . 1 9 Light pipes I = 0.9514Io = 0.49Io
(a) Total internal reflection in an
optical fiber. (b) When light is inci- In other words, after 14 reflections, the intensity is reduced to less than half 149%2.
dent on the end of a cylindrical form For 100 reflections, I = 0.006Io and the intensity is only 0.6% of the initial inten-
of transparent material such that the
internal angle of incidence is greater sity! When you compare this to an intensity of about 75% of the initial intensity in
than the critical angle of the mater- optical fibers over a kilometer in length with thousands of reflections, you can see
ial, the light undergoes total internal the advantage of total internal reflection.
reflection down the length of the Fibers whose diameters are about 10 mm 110-5 m2 are grouped together in flexi-
light pipe. (c) Light is also transmit- ble bundles that are 4 to 10 mm in diameter, depending on the application
ted along curved light pipes by total
internal reflection. (d) As the diame- (䉲 Fig. 22.20). A fiber bundle with a cross-sectional area of 1 cm2 can contain as
ter of the rod or fiber becomes many as 50 000 individual fibers. (A coating on each fiber is needed to keep the
smaller, the number of reflections fibers from touching each other.)
per unit length increases.
䉴 F I G U R E 2 2 . 2 0 Fiber-optic
bundle (a) Hundreds or even thou-
sands of extremely thin fibers are
grouped together (b) to make an
optical fiber, here colored blue by a
laser. (a) (b)
22.4 TOTAL INTERNAL REFLECTION AND FIBER OPTICS 767
There are many important and interesting applications of fiber optics, including
communications, computer networking, and medical instruments. (See Insight 22.3,
Fiber Optics: Medical Applications.) Light signals, converted from electrical signals,
are transmitted through optical telephone lines and computer networks. At the other
end, they are converted back to electrical signals. Optical fibers have lower energy
losses than electric current–carrying wires, particularly at higher frequencies, and
can carry far more data. Also, optical fibers are lighter than metal wires, have greater
flexibility, and are not affected by electromagnetic disturbances (electric and mag-
netic fields), because they are made of materials that are electrical insulators.
DID YOU LEARN?

➥ Total internal reflection refers to a phenomenon of no externally refracted light and
only reflected light in the medium.The critical angle is calculated from the equation
1n 7 n22.
n2
sin uc =
n1 1
➥ For total internal reflection to occur, light has to travel from a medium with a higher
index of refraction to a medium with a lower index of refraction (n1 7 n2), and the
angle of incidence must be greater than the critical angle (u1 7 uc).
➥ Light travels in an optic fiber with many total internal reflections. Since there is no
refraction, no energy is lost to the outside of the fiber and all the energy is confined
to the fiber.
INSIGHT 22.3 Fiber Optics: Medical Applications

Before fiber optics, endoscopes—instruments used to view fibers. Mechanical linkages allow maneuverability. The end of
internal portions of the human body—consisted of lens sys- a fiber endoscope can be equipped with devices to obtain
tems in long, narrow tubes. Some endoscopes contained a specimens of the viewed tissues for biopsy (diagnostic exami-
dozen or more lenses and produced relatively poor images. nation) or even to perform surgical procedures. For example,
Because the lenses had to be aligned in certain ways, the tubes arthroscopic surgery is performed on injured joints (Fig. 1).
had to have rigid sections, which limited the endoscope’s The arthroscope that is now routinely used for inspecting and
maneuverability. Such an endoscope could be inserted down repairing damaged joints is simply a fiber endoscope fitted
a patient’s throat into the stomach to observe the stomach with appropriate surgical implements.
lining. However, there were blind spots due to the curvature A fiber-optic cardioscope (for direct observation of heart
of the stomach and the inflexibility of the instrument. valves) typically is a fiber bundle about 4 mm in diameter and
Fiber-optic bundles have eliminated these problems. 30 cm long. It passes easily to the heart through the jugular
Lenses placed at the end of the fiber bundles focus the light, vein, which is about 15 mm in diameter, in the neck. To dis-
and a prism is used to change the direction for its return. The place the blood and provide a clear field of view for observing
incident light is usually transmitted by an outer layer of fiber and photographing, a transparent balloon at the tip of the car-
bundles, and the image is returned through a central core of dioscope is inflated with saline (saltwater) solution.
F I G U R E 1 Arthroscopy
(a) A fiber-optic arthroscope
used to perform surgery.
(b) An arthroscopic view of a
torn knee meniscus.
(a) (b)
22.5 Dispersion
➥ What is dispersion?
➥ What is the fundamental physical cause of dispersion?
➥ In a glass prism, which color of the visible spectrum experiences the greatest devia-
tion? Which the least deviation?
Light of a single frequency, and consequently a single wavelength, is called

monochromatic light (from the Greek mono, meaning “one,” and chroma, meaning
“color”). Visible light that contains all the component frequencies, or colors, at
about the same intensities (such as sunlight) is called white light. When a beam of
white light passes through a glass prism, as shown in 䉲 Fig. 22.21a, it is spread out,
or dispersed, into a spectrum of colors. This phenomenon led Newton to believe
that sunlight is a mixture of colors. When the beam enters the prism, the compo-
nent colors, corresponding to different wavelengths of light, are refracted at
slightly different angles, so they spread out into a spectrum (Fig. 22.21b).
The emergence of a spectrum indicates that the index of refraction of glass is
slightly different for different wavelengths, which is true for many transparent
media (Fig. 22.21c). The reason has to do with the fact that in a dispersive medium
the speed of light is slightly different for different wavelengths. Since the index of
refraction of a medium is a function of the speed of light in that medium 1n = c>v2,
the index of refraction is different for different wavelengths. It follows from the
law of refraction that light of different wavelengths will be refracted at different
angles of refraction even if the angle of incidence is the same.
The preceding discussion can be summarized by saying that in a transparent
material with different indices of refraction for different wavelengths of light,
refraction causes a separation of light according to wavelength, and the material is
said to be dispersive and exhibit dispersion. Dispersion varies with different media
(Fig. 22.21c). Also, because the differences in the indices of refraction for different
wavelengths are small, a representative value at some specified wavelength can
be used for general purposes. (See Table 22.1.)
A good example of a dispersive material is diamond, which is about five times as
dispersive as glass. In addition to revealing the brilliance resulting from internal
1.7
Flint glass
Refractive index
δ red 1.6
Quartz
Red
θ1 Crown glass
Orange
θ2 Yellow 1.5
Fused quartz
Green
Blue
White Indigo Blue Red
light Violet 1.4
400 500 600 700
Prism Wavelength (nm)
(a) (b) (c)
䉱 F I G U R E 2 2 . 2 1 Dispersion (a) White light is dispersed into a spectrum of colors by

glass prisms. (b) In a dispersive medium, the index of refraction varies slightly with wave-
length. Red light, longest in wavelength, has the smallest index of refraction and is
refracted least. The angle between the incident beam and an emergent ray is the angle of
deviation 1d2 for that ray. (The angles are exaggerated for clarity.) (c) Variation in the index
of refraction with wavelength for some common transparent media.
22.5 DISPERSION 769
reflections off many facets, a cut diamond shows a display of colors, or “fire,” result-
ing from the dispersion of the refracted light.
Dispersion is a cause of chromatic aberration in lenses, which is described more
fully in Section 23.4. Optical systems in cameras often consist of several lenses to
minimize this problem.
Another dramatic example of dispersion is the production of a rainbow, as dis-
cussed in Insight 22.4, The Rainbow.
INSIGHT 22.4 The Rainbow

Everyone has been fascinated by the beautiful array of colors Red appears on the top of the rainbow because light of
of a rainbow. With the optical principles learned in this chap- shorter wavelengths from those water droplets will pass over
ter, we are now in a position to understand the formation of our eyes (Fig. 2b). Similarly, violet is at the bottom of the rain-
this spectacular display. bow because light of longer wavelengths passes under our
A rainbow is formed by refraction, dispersion, and reflec- eyes. Rainbows are generally seen only as arcs, because their
tion of light within water droplets. When sunlight shines on formation is cut off at the ground. The secondary rainbow has
millions of water droplets in the air during and after a rain, a reversed color orders because of the extra reflection.
multicolored arc is seen whose colors run, in order of wave-
length, from violet along the lower part to red along the
upper part. Occasionally, more than one rainbow is seen: The
main, or primary, rainbow is sometimes accompanied by a
fainter and higher secondary rainbow (Fig. 1). The secondary
rainbow is caused by two reflections within the water
droplets.
The light that forms the primary rainbow is first refracted
and dispersed in each water droplet, then reflected once at
the back surface of each droplet. Finally, it is refracted and
dispersed again upon exiting each droplet, resulting in the
light being spread out in different directions into a
spectrum of colors (Fig. 2a). However, because of the
conditions for refraction and reflection in water, the angles
between incoming and outgoing rays for violet to red
light lie within a narrow range of 40° to 42°. This means
that you can see a rainbow only when the Sun is behind
you, so that the dispersed light travels to you only at these F I G U R E 1 Rainbow The colors of the primary rainbow
angles. run vertically from red (top) to violet (bottom).
Sunli
ght
Violet 40°
42°
Red
Red
Sun
ligh
t 40° 42°
Violet
Re
d Red
42° Vi
ol
et
40° Red Primary rainbow Horizo
n
Violet
Violet
Red
(a) (b)
F I G U R E 2 The rainbow Rainbows are created by the refraction, dispersion, and internal reflection of sunlight within water
droplets. (a) Light of different colors emerges from the droplet in different directions. (b) An observer sees red light at the top of
the rainbow and violet at the bottom.
INTEGRATED EXAMPLE 22.7 Refraction and Dispersion

The index of refraction of a particular transparent material is Given: (red) nr = 1.4503 Find: ¢u2 (angular
1.4503 for the red end 1lr = 640 nm2 of the visible spectrum for lr = 700 nm separation)
and 1.4698 for the blue end 1lb = 434 nm2. White light is inci- (blue) nb = 1.4698
dent on a prism of this material, as in Fig. 22.21b, at an angle for lb = 400 nm
of incidence u1 of 45.00. (a) Inside the prism, the angle of
u1 = 45.00
refraction of the red light is (a) larger than, (2) smaller than, or
(3) the same as the angle of refraction of the blue light. Using Eq. 22.5 with n1 = 1.00 (air),
Explain. (b) What is the angular separation of the visible spec-
trum inside the prism? sin u1 sin 45.00
sin u2r = = = 0.48756 and u2r = 29.180°
n2r 1.4503
(A) CONCEPTUAL REASONING. This example illustrates the
essence of the cause of dispersion, that is, different colors Similarly,
have different indices of refraction in a material. Since red sin u1 sin 45.00
light has a smaller index of refraction than blue light, the sin u2b = = = 0.48109 and u2b = 28.757°
n2b 1.4698
angle of refraction of red light is larger than that of blue light
for the same angle of incidence. Sometimes it is also said that The angle of refraction of the red light is indeed larger than that
red light is “refracted less” than blue light because the larger of the blue light, as discussed in (a). The angular separation is
angle of refraction of red light means that it is closer to the
direction of the original incident ray. So the answer is (1). ¢u2 = u2r - u2b = 29.180° - 28.757° = 0.423°
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . The law of This is not much of a deviation, but as the light travels to the
refraction is used to compute the angle of refraction for the other side of the prism, it is refracted and dispersed again by
red and blue ends of the visible spectrum. The angular sepa- the second boundary. Thus the colors spread out even farther.
ration of the visible spectrum inside the prism is the differ- When the light emerges from the prism, the dispersion
ence between the angles of refraction of the two colors. becomes evident (Fig. 22.21a, b).
F O L L O W - U P E X E R C I S E . In the prism in this Example, if the green light exhibits an angular separation of 0.156° from the red
light, what is the index of refraction for green light in the material? Will the green light refract more or less than the red light?
Explain.
DID YOU LEARN?

➥ Dispersion is the phenomenon of white light being separated into the component
colors or wavelengths.
➥ The fundamental physical cause of dispersion is the fact that different colors
(wavelengths) of light have different speeds, and therefore different indices of
refraction, for a particular medium.The different indices of refraction will result in
different angles of refraction, and therefore a separation of the colors.
➥ In a glass prism, violet (shortest wavelength) experiences the greatest deviation
and red (longest wavelength) has the smallest deviation.
PULLING IT TOGETHER Reflection, Refraction, and Dispersion

A beam of light consisting of two different wavelengths in air (b) Since the angle of refraction for light of wavelength A is
is incident on the surface of a transparent and dispersive smaller than that for light of wavelength B, their indices of
material. When the angle of incidence of the light is 35.00°, the refraction can be compared using the law of refraction (Eq.
angle of refraction for light of wavelength A is 0.30° smaller 22.5) and then their wavelengths can be further compared.
than that for light of wavelength B. The index of refraction of (c) Since the index of refraction of the material for light of
the material for wavelength B is 1.5210. (a) What are the wavelength B is known, its angle of refraction can be deter-
angles of reflection for light of both wavelengths? (b) If the mined from the law of refraction. Then the angle of refraction
transparent material has a larger index of refraction for for light of wavelength A can also be calculated because it is
shorter wavelength, which wavelength is shorter? (c) What 0.30° smaller. (d) Once the angle of refraction for light of
are the angles of refraction for light of both wavelengths? wavelength A is known, its index of refraction can be calcu-
(d) What is the index of refraction of the material for light of lated from the law of refraction. (e) The definition of the index
wavelength A? (e) What are the speeds of light for both wave- of refraction (Eq. 22.3) enables the calculations of the speeds
lengths in the material? of light of both wavelengths.
T H I N K I N G I T T H R O U G H . This Example combines reflection, SOLUTION. Listing the data, using the subscripts A and B for
refraction, the definition of the index of refraction, and dis- the two wavelengths, i and r for incidence and reflection, and
persion. (a) Dispersion occurs only for refraction so light of 1 and 2 for incidence and refraction:
both wavelengths should have the same angle of reflection.
Given: uiA = uiB = u1A = u1B = 35.00° (angle of incidence) Find: (a) urA and urB (angles of reflection)
n1A = n1B = 1.0003 (air, from Table 22.1) (b) which wavelength is shorter, A or B
n2B = 1.5210 (c) u2A and u2B
u2B - u2A = 0.30° (d) n2A
c = 3.00 * 108 m>s (air) (e) v2A and v2B
(a) Dispersion occurs only in refraction for light of different wavelengths, so light of both wavelengths has the exact same angle of
reflection. That is, urA = urB = uiA = uiB = 35.00°, according to the law of reflection (Eq. 22.1).
(b) Light of wavelength A has a smaller angle of refraction than light of wavelength B. From the law of refraction,
n1 sin u1 = n2 sin u2 , the index of refraction of the material for light of wavelength A is therefore greater than that for light of
wavelength B. (The index of refraction of air and the angle of incidence u1 are the same for light of both wavelengths.) Further-
more, since the transparent material has a larger index of refraction for light of shorter wavelength, wavelength A is shorter.
(c) Since the index of refraction of the material for light of wavelength B is known, its angle of refraction can be determined
directly from the law of refraction.
n1B sin u1B 11.00032 sin 35.00°
sin u2B = = = 0.3771 so u2B = sin-1 0.3771 = 22.15°
n2B 1.5210
The angle of refraction for light of wavelength A is then
u2A = u2B - 0.30° = 22.15° - 0.30° = 21.85°
(d) Using the law of refraction again, the index of refraction of the material for light of wavelength A is
n1A sin u1A 11.00032 sin 35.00°
n2A = = = 1.5411
sin u2A sin 21.85°
As predicted in (b), the index of refraction for light of wavelength A is indeed greater than that for light of wavelength B.
(e) Using the definition of the index of refraction, Eq. 22.3, the speeds of light of both wavelengths are
c 3.00 * 108 m>s
v2A = = = 1.95 * 108 m>s
n2A 1.5411
and
c 3.00 * 108 m>s
v2B = = = 1.97 * 108 m>s
n2B 1.5210
■ Law of reflection: The angle of incidence equals the angle ■ The refraction of light as it enters a medium from another is
of reflection (as measured from the normal to the reflecting given by Snell’s law or the law of refraction:
surface):
sin u1 v1
ui = ur (22.1) = (22.2)
sin u2 v2
θ i = θr
n1 sin u1 = n2 sin u2 (22.5)
Normal
If the second medium has a higher index of refraction
1n2 7 n12, the ray is refracted toward the normal; if the sec-
Plane of
incidence
ond medium has a lower index of refraction 1n2 6 n12, the
θi θr ray is refracted away from the normal.
Incident ray Normal

θ1
Reflectin
g surfac
e
Boundary Medium 1
■ The index of refraction (n) of any medium is the ratio of the Medium 2
speed of light in a vacuum to its speed in that medium:
θ2
c l
n = = (22.3, 22.4)
v lm Refracted ray
■ Total internal reflection occurs if the second medium has a ■ Dispersion of light occurs in a medium because different
lower index of refraction than the first and the angle of inci- wavelengths have slightly different speeds and hence dif-
dence exceeds the critical angle, which is given by ferent indices of refraction in the medium. This results in
slightly different refraction angles for different wave-
1n1 7 n22
n2
sin uc = (22.6) lengths.
n1
Normal
Total δ red
θ2 internal
Air θ 2 = 90° reflection θ1
Red
Orange
Water θ2 Yellow
Green
θc θ1 θ2 Blue
θ1 White Indigo
θ 1= θ 2 light Violet
Prism
Light source

22.1 WAVE FRONTS AND RAYS given medium has different indices of refraction for dif-
AND ferent angles of incidence.
22.2 REFLECTION 5. Light refracted at the boundary of two different media
(a) is bent toward the normal when n1 7 n2, (b) is bent
1. A ray is (a) perpendicular to the direction of energy flow,
away from the normal when n1 7 n2, (c) is bent away
(b) parallel to the direction of energy flow, (c) always
from the normal when n1 6 n2, (d) has the same angle of
parallel to other rays, (d) parallel to a series of wave
refraction as the angle of incidence.
fronts.
6. The index of refraction (a) is always greater than or equal
2. The angle of reflection is the angle between (a) the
to 1, (b) is inversely proportional to the speed of light in a
reflected ray and the reflecting surface, (b) the incident
medium, (c) is inversely proportional to the wavelength
ray and the normal to the surface, (c) the reflected ray
of light in the medium, (d) all of the preceding.
and the incident ray, (d) the reflected ray and the normal
to the reflecting surface. 7. Which of the following must be satisfied for total inter-
nal reflection to occur: (a) n1 7 n2, (b) n2 7 n1,
3. For both specular (regular) and diffuse (irregular) reflec-
(c) u1 7 uc, or (d) uc 7 u1?
tions, (a) the angle of incidence equals the angle of reflec-
tion, (b) the incident and reflected rays are on opposite
sides of the normal, (c) the incident ray, the reflected ray,
and the local normal lie in the same plane, (d) all of the 22.5 DISPERSION
preceding. 8. Dispersion can occur only if the light is (a) monochro-
matic, (b) polychromatic, (c) white light, (d) both b and c.
22.3 REFRACTION 9. Dispersion can occur only during (a) specular reflection,
(b) diffuse reflection, (c) refraction, (d) total internal
AND
reflection.
22.4 TOTAL INTERNAL REFLECTION AND
10. Dispersion is caused by (a) the difference in the speed of
FIBER OPTICS
light in different media, (b) the difference in the speed of
4. Refraction is caused by the fact that (a) different media light for different wavelengths of light in a given medium,
have different speeds of light, (b) a given medium has (c) the difference in the angle of incidence for different
different speeds of light for different wavelengths, (c) the wavelengths of light in a given medium, (d) the difference
angle of incidence is greater than the critical angle, (d) a in the indices of refraction of light in different media.
EXERCISES 773
22.1 WAVE FRONTS AND RAYS 8. The photos in 䉲 Fig. 22.24 were taken with a camera on a
AND tripod at a fixed angle. There is a penny in the container,
22.2 REFLECTION but only its tip is seen initially. However, when water is
added, more of the coin is seen. Why? Use a diagram to
1. Under what circumstances is the angle of reflection not explain.
equal to the angle of incidence?
2. The book you are reading is not, itself, a light source, so
it must be reflecting light from other sources. What type
of reflection is this?
3. After a rain, what kind of reflection can take place off of
the road surface?
4. When you see the Sun over a lake or the ocean, you often
observe a long swath of light (䉲 Fig. 22.22). What causes
this effect, sometimes called a “glitter path”?
䉳 FIGURE 22.22 䉱 F I G U R E 2 2 . 2 4 You barely see it, but then you
A glitter path See do See Conceptual Question 8 and Exercise 39.
9. When light enters medium 2 from medium 1, the angle

of incidence is always greater than the angle of refrac-
tion. Could total internal reflection take place from
medium 1 to medium 2? Explain.
22.3 REFRACTION
AND 22.5 DISPERSION
22.4 TOTAL INTERNAL REFLECTION AND 10. Both refraction and dispersion are caused by the differ-
FIBER OPTICS ence in the speed of light. What is the difference in the
5. Two hunters, one with a bow and arrow and the other physical cause (reason)?
with a laser gun, see a fish under water. They both aim 11. Why is dispersion more prominent when using a
directly for the location where they see the fish. Does the triangular prism rather than a square block?
arrow or the laser beam have a better chance of hitting 12. If white light is incident on a square block of dispersive
the fish? Explain. material, will there be a spectrum? How about the angles
6. As light travels from one medium to another, does its of the colors when they exit the block?
wavelength change? Its frequency? Its speed? 13. A glass prism disperses white light into a spectrum. Can
7. Explain why the pencil in a second glass prism be used to recombine the spectral
䉴 Fig. 22.23 appears almost components? Explain.
severed. Also, compare this 14. A light beam consisting of two colors, A and B, is sent
figure with Fig. 22.13b and through a prism. Color A is refracted more than color B.
explain the difference. Which color has a longer wavelength? Explain.
15. If glass is dispersive, why don’t we normally see a spec-
trum of colors when sunlight passes through a glass
window? Explain. (Are the speeds of each color of light
the same in the glass?)
䉴 FIGURE 22.23
Refraction effect See
EXERCISES*
*Assume all angles to be exact.
22.1 WAVE FRONTS AND RAYS 11. ● The speed of light in the core of the crystalline lens in
AND a human eye is 2.13 * 108 m>s. What is the index of
22.2 REFLECTION refraction of the core?
12. IE ● The indices of refraction for zircon and fused quartz
1. ● The angle of incidence of a light ray on a mirrored sur-
can be found in Table 22.1. (a) The speed of light in fused
face is 30°. What is the angle between the incident and
quartz is (1) greater than, (2) less than, (3) the same as the
reflected rays?
speed of light in zircon. Explain. (b) Compute the ratio of
2. ● A beam of light is incident on a plane mirror at an
the speed of light in fused quartz to that in zircon.
angle of 25° relative to the normal. What is the angle
13. ● A beam of light traveling in air is incident on a trans-
between the reflected ray and the surface of the mirror?
parent plastic material at an angle of incidence of 50°.
3. IE ● A beam of light is incident on a plane mirror at an The angle of refraction is 35°. What is the index of refrac-
angle a relative to the surface of the mirror. (a) Will the tion of the plastic?
angle between the reflected ray and the normal be (1) a,
14. IE ● A beam of light enters water from air. (a) Will the
(2) 90° - a, or (3) 2a? (b) If a = 33°, what is the angle
angle of refraction be (1) greater than, (2) equal to, or
between the reflected ray and the normal?
(3) less than the angle of incidence? Explain. (b) If the
4. IE ● ● Two upright plane mirrors touch along one edge, angle of incidence is 50°, find the angle of refraction.
where their planes make an angle of a. A beam of light is
15. IE ● Light passes from a crown glass container into
directed onto one of the mirrors at an angle of incidence
of b 1b 6 a2 and is reflected onto the other mirror.
water. (a) Will the angle of refraction be (1) greater than,
(2) equal to, or (3) less than the angle of incidence?
(a) Will the angle of reflection of the beam from the
Explain. (b) If the angle of refraction is 20°, what is the
second mirror be (1) a, (2) b , (3) a + b , or (4) a - b ?
angle of incidence?
(b) If a = 60° and b = 40°, what will be the angle of
16. ● The critical angle for a certain type of glass in air is
reflection of the beam from the second mirror?
41.8°. What is the index of refraction of the glass?
5. IE ● ● Two identical plane mirrors of width w are placed a
distance d apart with their mirrored surfaces parallel and 17. IE ● (a) For total internal reflection to occur, should the
facing each other. (a) A beam of light is incident at one end light be directed from (1) air to water or (2) water to air?
of one mirror so that the light just strikes the far end of the Explain. (b) What is the critical angle of water in air?
other mirror after reflection. Will the angle of incidence be 18. ● What is the critical angle of a diamond in water?
(1) sin-11w>d2, (2) cos -11w>d2, or (3) tan-11w>d2? (b) If 19. ● ● A beam of light in air is incident on the surface of a
d = 50 cm and w = 25 cm, what is the angle of incidence? slab of fused quartz. Part of the beam is transmitted into
6. ● ● Two people stand 5.0 m apart and 3.0 m away from a the quartz at an angle of refraction of 30° relative to a
large plane mirror in a dark room. At what angle of inci- normal to the surface, and part is reflected. What is the
dence should one of them shine a flashlight on the mirror angle of reflection?
so that the reflected beam directly strikes the other person? 20. ● ● A beam of light is incident from air onto a flat piece
7. ● ● A beam of light is incident on a plane mirror at an angle of polystyrene at an angle of 55° relative to a normal to
of incidence of 35°. If the mirror rotates through a small the surface. What angle does the refracted ray make with
angle of u, through what angle will the reflected ray rotate? the plane of the surface?
8. ● ● ● Two plane mirrors, M1 and M2 , are placed together 21. ● ● The laser used in cornea surgery to treat corneal
as illustrated in 䉲 Fig. 22.25. (a) If the angle a between the disease is the excimer laser, which emits ultraviolet light
mirrors is 70° and the angle of incidence, ui1 of a light ray at a wavelength of 193 nm in air. The index of refraction
incident on M1 is 35°, what is the angle of reflection, ur2, of the cornea is 1.376. What are the wavelength and
from M2? (b) If a = 115° and ui1 = 60°, what is ur2? frequency of the light in the cornea?
22. IE ● ● Light passes from material A, which has an index
䉳 FIGURE 22.25 of refraction of 34, into material B, which has an index of
Plane mirrors refraction of 54. (a) The speed of light in material A is
α
M1 M2 together See Exer- (1) greater than, (2) the same as, (3) less than the speed of
cises 8 and 9. light in material B. Explain. (b) Find the ratio of the speed
θ i1 of light in material A to the speed of light in material B.
23. IE ● ● In Exercise 22, (a) the wavelength of light in material
Top view A is (1) greater than, (2) the same as, (3) less than the wave-
length of light in material B. Explain. (b) What is the ratio of
9. ●●● For the plane mirrors in Fig. 22.25, what angles a and the light’s wavelength in material A to that in material B?
ui1 would allow a ray to be reflected back in the direction 24. ● ● The critical angle between two materials is 43°. If the
from which it came (parallel to the incident ray)? angle of incidence is 35°, what is the angle of refraction?
(Consider that light can travel to the interface from
either material.)
22.3 REFRACTION
25. ● ● (a) An object immersed in water appears closer to the
AND
surface than it actually is. What is the cause of this illu-
22.4 TOTAL INTERNAL REFLECTION AND
sion? (b) Using 䉴 Fig. 22.26, show that the apparent
FIBER OPTICS
depth for small angles of refraction is d¿ = d>n, where n
10. ● The index of refraction in a certain precious transpar- is the index of refraction of the water. [Hint: Recall that
ent stone is 2.35. What is the speed of light in that stone? for small angles, tan u L sin u.]
EXERCISES 775
␪1 34. ●● A 45°–90°–45° prism (Fig. 22.27) is made of a

material with an index of refraction of 1.85. Can the
Air a prism be used to deflect a beam of light by 90° (a) in air?
Water
(b) What about in water?
d d′ ␪2 35. ●● A coin lies on the bottom of a pool under 1.5 m of
␪1
␪1 water and 0.90 m from the side wall (䉲 Fig. 22.28). If a
light beam is incident on the water surface at the wall, at
what angle u relative to the wall must the beam be
␪2 directed so that it will illuminate the coin?
θ
䉳 FIGURE 22.28
Find the coin See
䉱 F I G U R E 2 2 . 2 6 Apparent depth? See Exercise 25. Exercise 35 (not
(For small angles only; angles enlarged for clarity.) drawn to scale).
Water
26. ●● A person looks over the edge of a pool and sees a 1.5 m
bottle cap on the bottom directly below, where the depth
is 3.2 m. How far below the water surface does the bottle
cap appear to be? (See Exercise 25b.)
27. ●● What percentage of the actual depth is the apparent
0.90 m
depth of an object submerged in oil if the observer is
looking almost straight downward? (See Exercise 25b.)
36. ●● A flint glass plate 3.5 cm thick is placed over a news-
28. ●● A light ray in air is incident on a glass plate 10.0 cm
paper. How far beneath the top surface of the plate would
thick at an angle of incidence of 40°. The glass has an
the print appear to be if you were looking almost verti-
index of refraction of 1.65. The emerging ray on the other
cally downward through the plate? (See Exercise 25b.)
side of the plate is parallel to the incident ray, but is later-
37. ● ● Yellow-green light of wavelength 550 nm in air is
ally displaced. What is the perpendicular distance
between the original direction of the ray and the direc- incident on the surface of a flat piece of crown glass at an
tion of the emerging ray? [Hint: See Example 22.4.] angle of 40°. (a) What is the angle of refraction of the
light? (b) What is the speed of the light in the glass?
29. IE ● ● To a submerged diver looking upward through the (c) What is the wavelength of the light in the glass?
water, the altitude of the Sun (the angle between the Sun
38. IE ● ● ● A light beam traveling upward in a plastic material
and the horizon) appears to be 45°. (a) The actual altitude
with an index of refraction of 1.60 is incident on an upper
of the Sun is (1) greater than, (2) the same as, (3) less than
horizontal air interface. (a) At certain angles of incidence,
45°. Explain. (b) What is the Sun’s actual altitude?
the light is not transmitted into air. The cause of this is
30. ●● At what angle to the surface must a diver submerged (1) reflection, (2) refraction, (3) total internal reflection.
in a lake look toward the surface to see the setting Sun Explain. (b) If the angle of incidence is 45°, is some of the
just along the horizon? beam transmitted into air? (c) Suppose the upper surface
31. ●● A submerged diver shines a light toward the surface of the plastic material is covered with a layer of liquid with
of a body of water at angles of incidence of 40° and 50°. an index of refraction of 1.20. What happens in this case?
Can a person on the shore see a beam of light emerging 39. ● ● ● A 15-cm-deep opaque container is empty except for a
from the surface in either case? Justify your answer single coin resting on its bottom surface. When looking
mathematically. into the container at a viewing angle of 50° relative to the
32. ●● Light in air is incident on a transparent material. It is vertical side of the container, you see nothing on the bot-
found that the angle of reflection is twice the angle of tom. When the container is filled with water, you see the
refraction. What is the range of the index of refraction of coin (from the same viewing angle) on the bottom of, and
the material? just beyond, the side of the container. (See Fig. 22.24.)
33. IE ● ● A beam of light is to undergo total internal reflection How far is the coin from the side of the container?
through a 45° – 90° – 45° prism (䉲 Fig. 22.27). (a) Will this 40. ● ● ● An outdoor circular fish pond has a diameter of
arrangement depend on (1) the index of refraction of the 4.00 m and a uniform full depth of 1.50 m. A fish halfway
prism, (2) the index of refraction of the surrounding down in the pond and 0.50 m from the near side can just
medium, or (3) the indices of refraction of both? Explain. see the full height of a 1.80-m-tall person. How far away
(b) Calculate the minimum index of refraction of the prism from the edge of the pond is the person?
if the surrounding medium is air. Repeat if it is water. 41. ● ● ● For total internal reflection to occur inside an optic
fiber as shown in 䉲 Fig. 22.29, the angle u must be greater

䉳 F I G U R E 2 2 . 2 7 Total
internal reflection in a prism 䉳 F I G U R E 2 2 . 2 9 Optic
fiber See Exercise 41.
θ u u
θ
ui
than the critical angle for the fiber–air interface. At the amount as the red color. Explain. (b) Find the angle sepa-
end of the fiber, the incident light undergoes a refraction rating rays of the two colors in a piece of crown glass if
to enter the fiber. If total internal reflection is to occur for their angle of incidence is 37°.
any angle of incidence 1ui2 outside the end of the fiber, 45. ● ● A beam of light with red and blue components of
what is the minimum index of refraction of the fiber? wavelengths 670 nm and 425 nm, respectively, strikes a
42. ●●● Two glass prisms are placed together (䉲 Fig. 22.30). slab of fused quartz at an incident angle of 30°. On refrac-
(a) If a beam of light strikes the face of one of the prisms tion, the different components are separated by an angle
at normal incidence as shown, at what angle u does the of 0.001 31 rad. If the index of refraction of the red light is
beam emerge from the other prism? (b) At what angle of 1.4925, what is the index of refraction of the blue light?
incidence would the beam be refracted along the inter- 46. ● ● White light passes through a piece of crown glass
face of the prisms? and strikes an interface with air at an angle of 41.15°.
Assume the indices of refraction of crown glass are the
same as given in Exercise 44. Which color(s) of light will
n = 1.60 45° 䉳 FIGURE 22.30 be refracted out into the air?
θ Joined prisms See
Exercise 42. 47. ● ● ● A beam of red light is incident on an equilateral
prism as shown in 䉲 Fig. 22.31. (a) If the index of refrac-
tion of red light of the prism is 1.400, at what angle u
does the beam emerge from the other face of the prism?
(b) Suppose the incident beam were white light. What
45° would be the angular separation of the red and blue
n = 1.40
components in the emergent beam if the index of refrac-
tion of blue light were 1.403? (c) What would it be if the
22.5 DISPERSION index of refraction of blue light were 1.405?
43. ●White light is incident from air onto a transparent

material at an angle of incidence of 40°. The angles of 䉳 FIGURE 22.31
refraction for the red and blue colors are 28.15° and 60° Prism revisited See
27.95°, respectively. What are the indices of refraction for Exercise 47.
the two colors? θ
44. IE ● ● The index of refraction of crown glass is 1.515 for 80.0°
red light and 1.523 for blue light. (a) If light of both
colors is incident on crown glass from air, the blue color
will be refracted (1) more, (2) less, or (3) the same 60° 60°
48. The angle of incidence and angle of refraction along a partic- blue light travels 1.000 m, the red light will travel
ular interface between two media are 30° and 40°, respec- (1) more than, (2) less than, (3) exactly 1.000 m. Explain.
tively. What is the critical angle for the same interface? (b) Calculate the difference in distance traveled by the
49. Light passes from medium A into medium B at an angle of two colors.
incidence of 30°. The index of refraction of A is 1.5 times 52. In Exercise 47, if the angle of incidence is too small, light
that of B. (a) What is the angle of refraction? (b) What is the will not emerge from the other side of the prism. How
ratio of the speed of light in B to the speed of light in A? (c) could this happen? Calculate the minimum angle of inci-
What is the ratio of the frequency of light in B to the fre- dence for the red light so that it does not emerge from
quency of light in A? (d) At what angle of incidence would the other side of the prism.
the light be internally reflected? 53. An optic fiber of index of refraction of 1.6 is surrounded
50. The critical angle between two materials is 43°. If the by a layer called the cladding layer. The index of refrac-
angle of incidence is twice the angle of refraction, what tion of the cladding is 1.3. If the angle of incidence inside
are the angles of incidence and refraction? the fiber is 40°, will there be light leaking into air? What
51. IE The critical angle for a glass–air interface is 41.11° for if the angle of incidence is 50°?
red light and 41.04° for blue light. (a) During the time the
23 Mirrors and Lenses
23.1 Plane mirrors (778)
■ image location
■ image characteristics
23.2 Spherical mirrors (782)

■ focal length
■ spherical mirror equation
23.3 Lenses (790)

■ thin-lens equation
23.4 The Lens Maker’s

equation (798)
■ thin lens in air
PHYSICS FACTS
W
■ lens power
✦ In July 2005, the world’s largest
hat would life be like if
mirror was cast successfully at Uni- there were no mirrors in
versity of Arizona Steward Obser-
*23.5 Lens aberrations (800) vatory Mirror Lab. The off-axis bathrooms or cars, and if eye-
parabolic mirror has a diameter of
■ spherical, chromatic,
8.4 m (27 ft) and is made for the
glasses did not exist? Imagine a
and astigmatism
Giant Magellan Telescope (GMT) in world without optical images of
Las Campanas, Chile.
✦ The largest refracting optical lens
any kind—no photographs, no
in the world measures 1.827 m movies, no TV. Think about how lit-
(5.99 ft) in diameter. It was con-
structed by a team at the Univer- tle we would know about the uni-
sity of Arizona in Tucson, Arizona,
and completed in January 2000.
verse if there were no telescopes to
✦ Silicon carbide’s hardness, rigidity, observe distant planets and stars—
and lightness make it a desirable
mirror material. The European
or how little we would know about
Space Agency’s Herschel Space biology and medicine if there were
Observatory plans to launch the
largest space telescope in 2009. Its no microscopes to see bacteria and
silicon carbide mirror is 3.5 m
(11.5 ft) in diameter and the tele-
cells. It is often forgotten how
scope has a mass of 320 kg. By dependent we are on mirrors and
comparison, the Hubble Telescope
uses a 2.4-m (7.9 ft) mirror but has lenses.
a mass of 1500 kg.
The first mirror was probably the
reflecting surface of a pool of water.
778 23 MIRRORS AND LENSES
Later, people discovered that polished metals and glass also have reflective prop-
erties. They must also have noticed that when they looked at things through glass,
the objects looked different than when viewed directly, depending on the shape
of the glass. In some cases, the objects appeared to be reduced, as is the case in
the interesting chapter-opening photograph. It shows the reflected image of a
person by the cornea of an eye, which normally forms refracted images that we
see. In time, people learned to shape glass into lenses, paving the way for the
eventual development of the many optical devices we now take for granted.
The optical properties of mirrors and lenses are based on the principles of
reflection and refraction of light, as introduced in Chapter 22. In this chapter, the
principles of mirrors and lenses will be discussed. Among other things, you’ll dis-
cover why the image in the photo is upright and reduced, whereas your image in
an ordinary flat mirror is the same size as you—but the image doesn’t seem to
Plane mirror comb your hair with the same hand you use.
Ray appears to
Eye originate from
θ behind mirror
θ 23.1 Plane Mirrors
➥ How are images formed?

➥ Where is the image formed by a plane mirror and how does the image distance
Object
compare with the object distance?
➥ What are the characteristics of images formed by a plane mirror?
(a)
Mirrors are smooth reflecting surfaces, usually made of polished metal or glass
that has been coated with some metallic substance. As you know, even an
uncoated piece of glass, such as a window pane, can act as a mirror. However,
Eye when one side of a piece of glass is coated with a compound of tin, aluminum, or
silver, the reflectivity of the glass is increased, as light is not transmitted through
Virtual the coating. A mirror may be front coated or back coated, but most mirrors are
image backcoated.
When you look directly into a mirror, you see the images of yourself and objects
around you (apparently on the other side of the surface of the mirror). The geome-
try of a mirror’s surface affects the size, orientation, and type of image. In general,
do di an image is the visual counterpart of an object, produced by reflection (mirrors) or
refraction (lenses).
Object Image
distance distance A mirror with a flat surface is called a plane mirror. How images are formed
by a plane mirror is illustrated by the ray diagram in 䉳 Fig. 23.1. An image
(b) appears to be behind or “inside” the mirror. This is because when the mirror
reflects a ray of light from the object to the eye (Fig. 23.1a), the ray appears to
䉱 F I G U R E 2 3 . 1 Image formed by
a plane mirror (a) A ray from a point originate from behind the mirror. Reflected rays from the top and bottom of an
on the object is reflected in the mir- object are shown in Fig. 23.1b. In actuality, light rays coming from all points on
ror according to the law of reflec- the side of the object facing the mirror are reflected, and an image of the com-
tion. (b) Rays from various points plete object is observed.
on the object produce an image. The image formed in this way appears to be behind the mirror. Such an image is
Because the two shaded triangles
are identical, the image distance di called a virtual image. Light rays appear to diverge from virtual images, but do
(the distance of the image from the not actually do so. No light rays actually come from or pass through the image.
mirror) is equal to the object dis- However, spherical mirrors (discussed in Section 23.2) can project images in front
tance do. That is, the image appears of the mirror where light actually passes through and then diverge from the
to be the same distance behind the image. This type of image is called a real image. An example of a real image is
mirror as the object is in front of the
mirror. The rays appear to diverge the image produced on a screen by an overhead projector in a classroom.
from the image position. In this Notice in Fig. 23.1b the distances of the object and image from the mirror. Quite
case, the image is said to be virtual. logically, the distance of an object from a mirror is called the object distance (do),
23.1 PLANE MIRRORS 779
and the distance its image appears to be behind the mirror is called the image
distance (di). By geometry of identical triangles and the law of reflection, ui = ur
(Section 22.2), it can be shown that do = ƒ di ƒ , which means that the image formed by a
plane mirror appears to be at a distance behind the mirror that is equal to the distance
between the object and the front of the mirror. (See Exercise 9.) The absolute value sign
on di will be discussed in detail later; it is used because di is negative.
We are interested in various characteristics of images. Two of these features are
the height and orientation of an image compared with those of its object. Both are
expressed in terms of the lateral magnification factor (M), which is defined as a
ratio of heights of the image (hi) and object (ho):
image height hi
M = = (23.1)
object height ho
A lighted candle used as an object allows us to address an important image

characteristic: orientation, that is, whether the image is upright or inverted with
respect to the orientation of the object. (In sketching ray diagrams, an arrow is a
convenient object for this purpose.) For a plane mirror, the image is always
upright. This means that the image is oriented in the same direction as the object.
We say that hi and ho have the same sign (both positive or both negative), so M is
positive. Note that M is a dimensionless quantity, as it is a ratio of heights.
In 䉴 Fig. 23.2, you should also be able to see that the image and object have the
hi
same sizes (heights). Therefore, M = = + 1 for a plane mirror, the image is
ho Mirror
upright (M is positive), and there is no magnification (hi = ho). That is, the object
and the image in a plane mirror are the same size.
With other types of mirrors, such as spherical mirrors (which will be considered
shortly), it is possible to have inverted images where M is negative. In summary, θ
the sign of M tells us the orientation of the image relative to the object, and the θ
absolute value of M gives the magnification.
Another characteristic of reflected images of plane mirrors is the so-called ho hi
right–left reversal. When you look at yourself in a mirror and raise your right hand, it
appears that your image raises its left hand. However, this right–left reversal is actu- do di
ally caused by the front–back reversal. For example, if your front faces south, then
your back “faces” north. Your image, on the other hand, has its front to the north 䉱 F I G U R E 2 3 . 2 Magnification
and back to the south—a front–back reversal. You can demonstrate this reversal by The lateral, or height magnification,
asking one of your friends to stand facing you (without a mirror). If your friend factor is given by M = hi>ho . For a
plane mirror, M = + 1, which means
raises his right hand, you can see that that hand is actually on your left side. that hi = ho , that is, the image is the
The main characteristics of an image formed by a plane mirror are summarized same height as the object, and the
in 䉲 Table 23.1. See also Insight 23.1, It’s All Done with Mirrors. image is upright.
TABLE 23.1 Characteristics of Images Formed by Plane Mirrors

do = ƒ di ƒ The object distance is equal to the image distance; that is, the image
appears to be as far behind the mirror as the object is in front.
M = +1 The image is virtual, upright, and unmagnified.
EXAMPLE 23.1 All of Me: Minimum Mirror Length

What is the minimum vertical length of a plane mirror SOLUTION. To determine this length, consider the situation
needed for a woman to be able to see a complete (head-to-toe) shown in Fig. 23.3. With a mirror of minimum length, a ray
image of herself (䉲 Fig. 23.3)? from the top of the woman’s head would be reflected at the
T H I N K I N G I T T H R O U G H . Applying the law of reflection, It
top of the mirror, and a ray from her feet would be reflected at
can be seen in the figure that two triangles are formed by the the bottom of the mirror, to her eyes. The length L of the mir-
rays needed for the image to be complete. These triangles ror is then the distance between the dashed horizontal lines
relate the woman’s height to the minimum mirror length. perpendicular to the mirror at its top and bottom.
However, these lines are also the normals for the ray h2
reflections. By the law of reflection, they bisect the angles
between incident and reflected rays; that is, ui = ur . Then,
because their respective triangles on each side of the dashed
normal are identical, the length of the mirror from its bottom h1/2 L
to a point even with the woman’s eyes is h1>2, where h1 is the
woman’s height from her feet to her eyes. Similarly, the small
upper length of the mirror is h2>2 (the vertical distance θr
between the woman’s eyes and the top of mirror). Then, h1
θi
h1 h2 h1 + h2 h
L = + = =
2 2 2 2
where h is the woman’s total height.
Hence, for the woman to see her complete image (head-
to-toe) in a plane mirror, the minimum height, or vertical
length, of the mirror must be half her height.
Object Image
You can do a simple experiment to prove this conclu- h = h1 + h2
sion. Get some newspaper and tape, and find a full-length
mirror. Gradually cover parts of the mirror with the news- 䉱 F I G U R E 2 3 . 3 Seeing it all The minimum height, or vertical
paper until you cannot see your complete image. You will length, of a plane mirror needed for a person to see his or her
find you need a mirror length that is only half your height complete (head-to-toe) image turns out to be half the person’s
to see a complete image. height.
F O L L O W - U P E X E R C I S E . What effect does a woman’s distance from the mirror have on the minimum mirror length required to
produce her complete image? (Answers to all Follow-Up Exercises are given in Appendix VI at the back of the book.)
INSIGHT 23.1 It’s All Done with Mirrors

Most of us are fascinated by magicians’ sensational tricks that The very first mirror illusion, “the Sphinx,” was invented
appear to make objects and animals suddenly appear or dis- for magicians by Thomas William Tobin in 1876. His inven-
appear. Of course, they do not really appear or disappear. The tion used mirrors to conceal a person or an object, as shown in
magicians have special skills enabling them to make the per- Fig. 1, and it appeared on the front cover of Modern Magic in
formance quick and smooth enough to “fool” the audience. 1876. In this illusion, two plane mirrors were placed between
It’s all done with mirrors, as they say. the legs of a three-legged table to hide the person’s body.
F I G U R E 1 An illustration of Tobin’s sensational illusion, the

Sphinx The person’s body was concealed by two plane mirrors F I G U R E 2 Houdini and Jennie, the disappearing elephant
between the legs of the three-legged table. The elephant vanished from view when Houdini fired a pistol.
23.1 PLANE MIRRORS 781
Harry Houdini, the world-famous master of illusion, felt it was too X

easy to make a pigeon fly out of a hat or a rabbit disappear into thin air.
In 1918, Houdini made a 10 000-lb elephant named Jennie “disappear” on
the stage of the Hippodrome Theater in New York City (Fig. 2). The act
was called “the Vanishing Elephant.” or
M
irr
irr
When the time for the elephant’s disappearance came, two large plane M
or
mirrors at right angles to each other were slid quickly into place (Fig. 3).
A strobe light was used to conceal the brief motion of the mirrors. When X
properly aligned, the mirrors reflected light (red lines) from the side
walls of the stage to form virtual images that matched the pattern of the
stage backdrop. Thus the audience apparently saw the stage with no ele-
phant visible. Unseen by the audience, the elephant was quickly led off
stage.
Audience X
F I G U R E 3 The disappearing elephant Two

large mirrors at right angles to each other were
used to conceal the elephant.
DID YOU LEARN?

➥ Images are formed when light rays from an object actually converge (forming a real
image) or appear to diverge from an image (forming a virtual image). Where the
rays converge or appear to diverge from are the image positions.
➥ The image formed by a plane mirror appears behind the mirror (virtual image).The
distance from the image to the mirror (image distance) is equal to the distance
from the object to the mirror (object distance).
➥ The images formed by a plane mirror are always virtual, upright, and the same size
as the object.
DEMONSTRATION 5 A Candle Burning Underwater?

It would appear so, but you know this is not possible. It’s a reflection of an image.
(a) The black frame holds a pane of glass, which acts both as (b) The effect can be removed by tilting the glass—the image
a window and as a plane mirror. The burning candle seen in can no longer be seen from this viewing point. (Something to
the water is the image of the candle on the front stand. There do with the law of reflection. What?)
is a container of water on a similar stand behind the glass,
but no burning candle.
23.2 Spherical Mirrors

Vertex ➥ What are the differences between converging and diverging spherical mirrors?
R ➥ What are the three key rays that can be used in a ray diagram for a spherical mirror?
➥ How does the lateral magnification describe various image characteristics?
C
Optic
Concave surface axis As the name implies, a spherical mirror is a reflecting surface with spherical
geometry. 䉳 Figure 23.4 shows that if a portion of a sphere of radius R is sliced off
along a plane, the severed section has the shape of a spherical mirror. Either the
inside or the outside of such a section can be the reflecting surface. For reflections
Convex surface on the inside surface, the section behaves as a concave mirror. (Think of looking
Spherical section into a cave in order to help yourself remember that a concave mirror has a recessed
surface.) For reflections from the outside surface, the section behaves as a convex
䉱 F I G U R E 2 3 . 4 Spherical mirrors mirror.
A spherical mirror is a section of a The radial line through the center of the spherical mirror that intersects the
sphere. Either the outside (convex)
surface of the mirror at the vertex of the spherical section is called the optic axis
surface or the inside (concave) sur-
face of the spherical section may be (Fig. 23.4). The point on the optic axis that corresponds to the center of the sphere
the reflecting surface. of which the mirror forms a section is called the center of curvature (C). The dis-
tance between the vertex and the center of curvature is equal to the radius of the
sphere and is called the radius of curvature (R).
When rays parallel and close to the optic axis are incident on a concave mirror,
θi the reflected rays intersect, or converge, at a common point called the focal point
θr
F
(F). As a result, a concave mirror acts as a converging mirror (䉳 Fig. 23.5a). Note
Axis C
that the law of reflection, ui = ur, is satisfied for each ray.
Similarly, rays parallel and close to the optic axis of a convex mirror diverge on
reflection, as though the reflected rays came from a focal point behind the mirror’s
surface (Fig. 23.5b). Thus, a convex mirror acts as a diverging mirror (䉲 Fig. 23.6).
When you see diverging rays, your brain assumes that there is an image from
f which the rays appear to diverge, even though no such image is actually there.
R The distance from the vertex to the focal point (F) is called the focal length
( f ). (See Fig. 23.5.) The focal length can be shown to be half the radius of
(a) Concave, or converging, mirror
curvature:
θr R
θi f = (focal length of spherical mirror) (23.2)
2
Axis F C
The preceding result is valid only when the rays are close to the optic axis—that
is, for small-angle approximation. Rays far away from the optic axis will focus at
different focal points, resulting in some image distortion. In optics, this type of
䉳 F I G U R E 2 3 . 6 Diverging mirror
f
Note by reverse-ray tracing in Fig 23.5b
R that a diverging (convex) spherical
mirror gives an expanded, although
(b) Convex, or diverging, mirror distorted, field of view, as can be seen
in this store-monitoring mirror.
䉱 F I G U R E 2 3 . 5 Focal point
(a) Rays parallel and close to the
optic axis of a concave spherical
mirror converge at the focal point F.
(b) Rays parallel and close to the
optic axis of a convex spherical mir-
ror are reflected along paths as
though they diverge from a focal
point behind the mirror. Note that
the law of reflection, ui = ur, is satis-
fied for each ray in each diagram.
23.2 SPHERICAL MIRRORS 783
distortion is an example of a spherical aberration. Some telescope mirrors are para-

bolic in shape, rather than spherical, in which case all rays parallel to the optic axis
are focused at the same focal point, thus eliminating spherical aberration.
RAY DIAGRAMS
The characteristics of images formed by spherical mirrors can be determined from
geometrical optics (Chapter 22). The method involves drawing rays emanating
from one or more points on an object. The law of reflection 1ui = ur2 applies, and
three key rays are defined with respect to the mirror’s geometry as follows: LEARN BY DRAWING 23.1
1. A parallel ray is a ray that is incident along a path parallel to the optic axis
and is reflected through (or appears to go through) the focal point F (as do all
a mirror ray diagram
rays near and parallel to the axis). Converging (concave) mirrior
2. A chief ray, or radial ray, is a ray that is incident through the center of curva-
ture (C) of the spherical mirror. Since the chief ray is incident normal to the Object
mirror’s surface, this ray is reflected back along its incident path, through 1
point C. F
3. A focal ray is a ray that passes through (or appears to go through) the focal C
point and is reflected parallel to the optic axis. (It is a reversed parallel ray, so Optic axis
to speak.)
Using any two of these three rays, the image can be located to find the image 1 Parallel ray
distance and determine its type (real or virtual), orientation (upright or inverted),
and size (magnified or reduced). It is customary to use the tip of an asymmetrical Object
object (for example, the head of an arrow or the flame of a candle) as the origin
1
point of the rays. The corresponding point of the image is at the point of intersec-
2 F
tion of the rays. This makes it easy to see whether the image is upright or inverted.
Keep in mind, however, that properly traced rays from any point on the object can be C
used to find the image. Every point on a visible object acts as an emitter of light. For
example, for a candle, the flame emits its own light, and many other points on the
candle surface reflect light.
2 Chief (radial) ray
EXAMPLE 23.2 Learn by Drawing: A Mirror Ray Diagram
Object
Real image
An object is placed 39 cm in front of a converging spherical mirror of radius 24 cm.
1
(a) Use a ray diagram to locate the image formed by this mirror. (b) Discuss the charac-
2 F
teristics of the image.
C
T H I N K I N G I T T H R O U G H . A ray diagram, drawn accurately, can by itself provide “quan-
titative” information about image location and image characteristics that might other-
wise be determined mathematically.
SOLUTION.
Given: R = 24 cm Find: (a) image location 3 Locating image
do = 39 cm (b) image characteristics
(a) Since a ray diagram (drawing) is to be used to locate the image, a scale for the draw- Object
ing is needed. If a scale of 1 cm (on the drawing) represents 10 cm, the object would be 1
drawn 3.9 cm in front of the mirror. 3
2 F
First draw the optic axis, the mirror, the object (a lighted candle), and the center of
curvature (C). From Eq. 23.2, the focal length f = 24 cm>2 = 12 cm, so the focal point C
(F) is halfway from the vertex to the center of curvature. Real image
To locate the image, follow steps 1–4 in the accompanying Learn by Drawing 23.l, A
Mirror Ray Diagram.
R
1. The first ray drawn is the parallel ray (➀ in the drawing). From the tip of the
di
flame, draw a ray parallel to the optic axis. After reflecting, this ray goes through
do
the focal point, F.
2. Then draw the chief ray (➁ in the drawing). From the tip of the flame, draw a ray 10 cm
going through the center of curvature, C. This ray will be reflected back along the
original direction. (Why?) 4 Can also use focal ray to
(continued on next page) confirm image
3. It can be seen that these two rays actually intersect. The point of intersection is
the tip of the image of the candle. From this point, draw the image by extending
the tip of the flame to the optic axis. The image distance di = 17 cm, as measured
from the diagram.
4. Only two rays are needed to locate the image. However, the third ray can be
drawn as a double check. In this case, the focal ray (➂ in the drawing) should go
through the same image point at which the other two rays intersect (if drawn
carefully). The focal ray from the tip of the flame going through the focal point, F,
after reflection, will travel out parallel to the optic axis.
(b) From the ray diagram drawn in part (a), it can clearly be seen that the image is real
(because the reflected rays intersect). As a result, the real image could be seen on a
screen (for example, a piece of white paper) that is positioned at the image point. The
image is also inverted (the image of the candle points downward) and is reduced, that
is, it is smaller in size than the object.
F O L L O W - U P E X E R C I S E . In this example, what would the characteristics of the image be
if the object were 15 cm in front of the mirror? Locate the image and discuss its charac-
teristics.
An example of a ray diagram using the same three rays for a convex (diverging)
mirror will be shown in Integrated Example 23.4.
A converging mirror does not always form a real image. For a converging
spherical mirror, the characteristics of the image change with the object distance.
Dramatic changes take place at two points: C (the center of curvature) and F (the
focal point). These points divide the optic axis into three regions (䉴 Fig. 23.7a):
do 7 R, R 7 do 7 f, and do 6 f.
Let’s start with an object in the region farthest from the mirror 1do 7 R2 and
move toward the mirror:
■ The case of do 7 R was shown in Example 23.2.
■ When do = R = 2f, the image is real, inverted, and the same size as the object.
■ When R 7 do 7 f, a real, inverted, and magnified image is formed (Fig. 23.7b).
The image is magnified when the object is inside the center of curvature, C.
■ When do = f, the object is at the focal point (Fig. 23.7c). The reflected rays are
parallel, and the image is said to be “formed at infinity.” The focal point F is a
special “crossover” point between real and virtual images.
■ When do 6 f, the object is inside the focal point (between the focal point and the
mirror’s surface). A virtual, upright, and magnified image is formed (Fig. 23.7d).
When do 7 f, the image is real; when do 6 f, the image is virtual. For do = f, the
image is said to be formed at infinity (Fig. 23.7c). When an object is at “infinity”—
when it is so far away that rays emanating from it and falling on the mirror are
essentially parallel—its image is formed at the focal plane. This fact provides an
easy method for determining the focal length of a converging mirror.
As we have seen, the position, orientation, and size of the image can be approxi-
mately determined graphically from ray diagrams drawn to scale. However, these
characteristics can be determined more accurately by analytical methods. It can be
shown by means of geometry that the object distance (do), the image distance (di),
and the focal length ( f ) are related through the spherical mirror equation:
1 1 1 2
+ = = (spherical mirror equation) (23.3)
do di f R
Note that this equation can be written in terms of either the radius of curvature, R,
or the focal distance, f, since by Eq. 23.2, f = R>2. Both R and f can be either posi-
tive or negative, as will be discussed shortly.
do = R do = f
Real, Image at
inverted, infinity
same size
Real, Real, Virtual, Real, Object 1
inverted, inverted, upright, inverted,
magnified 2
reduced magnified magnified
C F
(do > R) C F (do < f )
(R > do > f ) f
R
(a) Concave Mirror R

do
di
(b) R > do > f
Rays "converge"
at ∞
Virtual,
Object Object upright,
magnified
1 1 2
C 2 F C F
do di ∞ do di
(c) do = f (d) do < f
䉱 F I G U R E 2 3 . 7 Converging mirrors (a) For a converging (concave) mirror, the object is

located within one of three regions defined by the center of curvature (C) and the focal
point (F), or at one of these two points. For do 7 R, the image is real, inverted, and
reduced, as shown by the ray diagrams in Example 23.2. (b) For R 7 do 7 f, the image
will also be real and inverted but magnified. (c) For an object at the focal point F, or do = f,
the image is said to be formed at infinity. (d) For do 6 f , the image will be virtual, upright,
and magnified.
If di is the quantity to be found for a spherical mirror, it may be convenient to

use an alternative form of the spherical mirror equation:
do f
di = (23.3a)
do - f
The signs of the various quantities are very important in the application of
Eq. 23.3. The sign conventions summarized in 䉲 Table 23.2 will be used. For
example, for a real object, a positive di indicates a real image and a negative di cor-
responds to a virtual image.
The lateral magnification factor (M), also called the magnification factor, or sim-
ply magnification, defined in Eq. 23.1 can also be found analytically for a spherical
mirror. Again, by using geometry, it can be expressed in terms of the image and
object distances:
hi di
M = = - (magnification factor) (23.4)
ho do
(The derivations of Eqs. 23.3 and 23.4 follow as optional content.)

TABLE 23.2 Sign Conventions for Spherical Mirrors

Focal length ( f )
Converging (concave) mirror f (or R) is positive
Diverging (convex) mirror f (or R) is negative
Object distance (do)

Object is in front of the mirror (real object) do is positive
Object is behind the mirror (virtual object)* do is negative
Image distance (di) and image type

Image is formed in front of the mirror (real image) di is positive
Image is formed behind the mirror (virtual image) di is negative
Image orientation (M)

Image is upright with respect to the object M is positive
Image is inverted with respect to the object M is negative
*In a combination of two (or more) mirrors, the image formed by the first mirror is the object of
the second mirror (and so on). If this image–object falls behind the second mirror, it is referred to
as a virtual object, and the object distance is taken to be negative. This concept is more important
for lens combinations, as we will see in Section 23.3, and is mentioned here only for completeness.
The negative sign is added by convention to indicate the orientation of the image:
A positive value for M indicates an upright image, whereas a negative M implies
an inverted image. Also, if ƒ M ƒ 7 1, the image is magnified; if ƒ M ƒ 6 1, the image
is reduced; if ƒ M ƒ = 1, the image is the same size as the object.
*(Optional) Derivation of the Spherical Mirror Equation You might wonder

from where Eqs. 23.3 and 23.4 originate. The spherical mirror equation can be
derived with the aid of a little geometry. Consider the ray diagram in 䉲 Fig. 23.8. The
object and image distances (do and di) and the heights of the object and image (ho and
hi) are shown. Note that these lengths make up the bases and heights of triangles
formed by the ray reflected at the vertex (V). These triangles (O¿VO and I¿VI) are
similar, since, by the law of reflection, their angles at V are equal. Hence,
hi di
= - (1)
ho do
This equation is Eq. 23.4, from the definition of Eq. 23.1. The negative sign inserted
here signifies the fact that the image is inverted, so hi is negative.
The (focal) ray through F also forms similar triangles, O¿FO and AVF in
the approximation that the mirror is small compared with its radius. (Why are the
䉴 F I G U R E 2 3 . 8 Spherical mirror
equation The rays provide the O'
geometry, through similar triangles,
for the derivation of the spherical ho
I C V
mirror equation.
O F
hi
I' A
f
do
di
triangles similar?) The bases of these triangles are VF = f and OF = do - f.

Then, if VA is taken to be hi ,
hi VF f
= - = - (2)
ho OF do - f
Again, the negative sign inserted here signifies the fact that the image is inverted,
so hi is negative.
Equating Eqs. 1 and 2,
di f
= (3)
do do - f
Algebraic manipulation yields
1 1 1
+ =
do di f
which is the spherical mirror equation (Eq. 23.3).
Example 23.3 and Integrated Example 23.4 show how these equations and sign
conventions are used for spherical mirrors. In general, this approach usually
involves finding the image of an object; you will be asked where the image is
formed (di) and what the image characteristics are (M). These characteristics tell
whether the image is real or virtual, upright or inverted, and larger or smaller
than the object (that is, magnified or reduced).
EXAMPLE 23.3 What Kind of Image? Characteristics of Images from a Converging Mirror
A converging mirror has a radius of curvature of 30 cm. If an Thus, the image is real (positive di), inverted (negative M), and
object is placed (a) 45 cm, (b) 20 cm, and (c) 10 cm from the reduced 1 ƒ M ƒ = 1>22.
mirror, where is the image formed, and what are its character- (b) Here, R 7 do 7 f, and the object is between the focal
istics? (Specify whether each image is real or virtual, upright point and the center of curvature:
or inverted, and magnified or reduced.)
1 1 1 1
THINKING IT THROUGH. Here the radius R is given so the = - =
di 15 cm 20 cm 60 cm
focal length f is R>2. Also given are three different object dis-
tances, which can be used in Eqs. 23.3 and 23.4 to determine Thus,
the image location and characteristics. 60 cm
di = + 60 cm and M = - = - 3.0
20 cm
SOLUTION.
In this case, the image is real (positive di), inverted (negative M),
and magnified 1 ƒ M ƒ = 3.02.
Given: R = 30 cm, so Find: di , M, and image char-
f = R>2 = 15 cm acteristics for each
(a) do = 45 cm given object distance (c) For this case, do 6 f, and the object is inside the focal
(b) do = 20 cm point.
(c) do = 10 cm Using the alternate form of Eq. 23.3 for illustration:
Note that the given object distances correspond to the regions do f 110 cm2115 cm2
di = = = - 30 cm
shown in Fig. 23.7a. There is no need to convert the distances to do - f 10 cm - 15 cm
meters as long as all distances are expressed in the same unit
(centimeters in this case). Ray diagrams can be drawn for each Then
of these cases in order to find the characteristics of each image. di 1- 30 cm2
(a) In this case, the object distance is greater than the radius of M = - = - = + 3.0
do 10 cm
curvature 1do 7 R2, and
In this case, the image is virtual (negative di), upright (posi-
tive M), and magnified 1 ƒ M ƒ = 3.02.
1 1 1 1 1 1 1 1 2
+ = or = - = - =
do di f di f do 15 cm 45 cm 45 cm From the denominator of the expression for di of this alter-
Then nate form, it can be seen that di will always be negative when
45 cm di 22.5 cm 1 do is less than f. Therefore, a virtual image is always formed
di = = + 22.5 cm and M = - = - = - for an object inside the focal point of a converging mirror.
2 do 45 cm 2
F O L L O W - U P E X E R C I S E . For the converging mirror in this Example, where is the image formed and what are its characteristics if
the object is at 30 cm, or do = R?
When using the spherical mirror equations to find image characteristics, it is helpful to
first make a quick sketch (approximate, not necessarily to scale) of the ray diagram for
the situation. This sketch shows the image characteristics and helps to avoid making
mistakes when applying the sign conventions. The ray diagram and the mathematical solu-
tion must agree.
INTEGRATED EXAMPLE 23.4 Big Differences: Characteristics of a Diverging Mirror

An object (in this case, a candle) is 20 cm in front of a diverg-
ing mirror that has a focal length of - 15 cm (see the sign con-
ventions in Table 23.2). (a) Use a ray diagram to determine
whether the image formed is (1) real, upright, magnified, Virtual image
(2) virtual, upright, magnified, (3) real, upright, reduced,
(4) virtual, upright, reduced, (5) real, inverted, magnified, or 1
(6) virtual, inverted, reduced. (b) Find the location and char- 2
3
acteristics of the image by using the mirror equations.
(A) CONCEPTUAL REASONING. Since the object distance and
F C
the focal length of the diverging mirror are known, a ray dia-
Object
gram can be drawn and the image characteristics can be
determined. The first thing to decide on is a scale for the ray
diagram. In this example, 1 cm (on the drawing) could be
used to represent 10 cm. That way, the object would be 2.0 cm
in front of the mirror in the drawing. Draw the optic axis, the
mirror, the object (a lighted candle), and the focal point (F).
Since this mirror is convex, the focal point (F) and the center
of curvature (C) are behind the mirror. From Eq. 23.2,
R = 2f = 21 - 15 cm2 = - 30 cm. So C is drawn at twice the R
distance of F from the vertex (䉴 Fig. 23.9).
Only two out of the three key rays are necessary to locate do di
the image. The parallel ray ➀ starts from the tip of the
flame, travels parallel to the optic axis, and then diverges 䉱 F I G U R E 2 3 . 9 Diverging mirror Ray diagram of a diverg-
from the mirror after reflection, appearing to come from F. ing mirror.
The chief ray ➁ originates from the tip of the flame, appears
to go through C, and then reflects straight back, but appears Note that the focal length is negative for a convex mirror. (See
to come from C. It is clearly seen that these two rays, after Table 23.2.) Using Eq. 23.3,
reflection, diverge from each other, and there is no chance 1 1 1
for them to intersect. However, they appear to start from a + =
20 cm di -15 cm
common point behind the mirror: the image point of the
tip of the flame. The focal ray ➂ can also be drawn to verify or
that all three rays appear to emanate from the same image 1 1 1 7
point. di
=
-15 cm
-
20 cm
= -
60 cm
The image is virtual (the reflected rays don’t actually
come from the image point), upright, and reduced (the so
image is smaller than the object). Therefore, the answer is 60 cm
(4) virtual, upright, reduced. Measuring from the diagram di = - = - 8.6 cm
7
(keep in mind the drawing scale used), di L - 9.0 cm, and
hi 0.5 cm Then
M = L = + 0.4. di -8.6 cm
ho 1.2 cm M = - = - = + 0.43
( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . The image do 20 cm
position and characteristics can be calculated with the mirror Thus, the image is virtual (di is negative), upright (M is posi-
equations by using the given object distance and focal length. tive), and reduced 1ƒMƒ = 0.432. These results agree well with
those from the ray diagram. The image of an object is always
Given: do = 20 cm Find: di , M, and image virtual for a diverging (convex) mirror for a real object. (Can
f = - 15 cm characteristics you prove this using either a ray diagram or the mirror equa-
tion?)
F O L L O W - U P E X E R C I S E . As has been pointed out, a diverging mirror always forms a virtual image of a real object. What about
the other characteristics of the image—its orientation and magnification? Can any general statements be made about them?
SPHERICAL MIRROR ABERRATIONS

Technically, our descriptions of image characteristics for spherical mirrors are
true only for objects near the optic axis—that is, only for small angles of inci-
dence and reflection. If these conditions do not hold, the images will be blurred
(out of focus) or distorted, because not all of the parallel rays will converge in
the same plane. As illustrated in 䉴 Fig. 23.10, incident parallel rays far from the F
optic axis do not converge at the focal point. The farther the incident ray from
the axis, the farther the reflected ray from the focal point. This effect is called C
Axis
spherical aberration.
Spherical aberration does not occur with a parabolic mirror. (As the name
parabolic mirror implies, a parabolic mirror has the form of a paraboloid.) All of the
incident rays parallel to the optic axis of such a mirror have a common focal point.
For this reason, parabolic mirrors are used in most astronomical telescopes
(Chapter 24). However, these mirrors are more difficult to make than spherical
mirrors and are therefore more expensive. Focal
plane
䉱 F I G U R E 2 3 . 1 0 Spherical aber-
DID YOU LEARN?
ration for a mirror According to the
➥ A converging spherical mirror uses the inside surface (the concave side), has a small-angle approximation, rays
positive focal length, and can form images with a variety of image characteristics. parallel to and near the mirror’s axis
A diverging spherical mirror uses the outside surface (the convex side), has a negative converge at the focal point. How-
focal length, and can form only virtual, upright, and reduced images (for real objects). ever, when parallel rays not near the
➥ The three key rays used in a spherical mirror ray diagram are the parallel ray, the axis are reflected, they converge in
chief ray or radial ray, and the focal ray. (Only two are actually needed.) front of the focal point. This effect,
➥ The sign of the lateral magnification indicates whether an image is upright (M 7 0) called spherical aberration, gives rise
or inverted (M 6 0).The absolute value indicates whether an image is magnified to blurred images.
1 ƒ M ƒ 7 12 or reduced 1 ƒ M ƒ 6 12. For real objects, the sign of the lateral
magnification indicates whether an image is real (M 6 0) or virtual (M 7 0).
DEMONSTRATION 6 A Candle Burning at Both Ends

Or is it? Notice that one flame is burning downward, which is rather strange. It’s an illu-
sion done with a spherical concave mirror
(a) When an object is at the center of curvature of a spherical (b) A side view showing the burning end of the horizontal
concave mirror, a real image is formed that is inverted and candle in front of the spherical mirror.
the same size as the object, and the image distance is the
same as the object distance. What is seen here is a horizontal
candle burning at one end (flame up) and its overlapping
image (flame down). Viewed at the same level, the inverted
flame image appears to be at the opposite end of the candle.
䉴 F I G U R E 2 3 . 1 1 Spherical lenses (a) Biconvex (converging) lens (b) Biconcave (diverging) lens
Spherical lenses have surfaces
defined by two spheres, and the sur-
faces are either convex or concave. R2
(a) Biconvex and (b) biconcave R1 R1 R2
lenses are shown here. If R1 = R2 , a Principal axis
lens is spherically symmetric.
23.3 Lenses
➥ What are the differences between converging and diverging lenses?

F
➥ What are the three key rays that can be used in a ray diagram for a thin lens?
➥ How does the lateral magnification describe various image characteristics?
Converging lens The word lens is from the Latin lentil, which is a round, flattened, edible seed of a
pea-like plant. Its shape is similar to that of a lens. An optical lens is made from
(a) Biconvex (converging) lens transparent material such as glass or plastic. One or both surfaces usually have a
spherical contour. Biconvex spherical lenses (with both surfaces convex) and
biconcave spherical lenses (with both surfaces concave) are illustrated in 䉱 Fig. 23.11.
Lenses can form images by refracting the light that passes through them.
A biconvex lens is an example of a converging lens. Incident light rays parallel
to the axis of the lens converge at a focal point (F) on the opposite side of the lens
(䉳 Fig. 23.12a). This fact provides a way to experimentally determine the focal
length of a converging lens. You may have focused the Sun’s rays with a magnify-
ing glass (a biconvex, or converging, lens) and thereby witnessed the concentra-
tion of radiation energy that results (Fig. 23.12b).
A biconcave lens is an example of a diverging lens. Incident parallel rays
emerge from the lens as though they emanated from a focal point on the incident
side of the lens (䉲 Fig. 23.13).
There are several types of converging and diverging lenses (䉴 Fig. 23.14). Con-
vex and concave meniscus lenses are the type most commonly used for corrective
eyeglasses. In general, a converging lens is thicker at its center than at its periph-
ery, and a diverging lens is thinner at its center than at its periphery. Most of our
discussions will be limited to spherically symmetric biconvex and biconcave
lenses, for which both surfaces have the same radius of curvature.
When light passes through a lens, it is refracted and displaced laterally (see
(b)
Example 22.4 and Fig. 22.11). If a lens is thick, this displacement may be fairly
䉱 F I G U R E 2 3 . 1 2 Converging large and can complicate analysis of the lens’s characteristics. This problem does
lens (a) For a thin converging (con- not arise as much with thin lenses, for which the refractive displacement of trans-
vex) lens, rays parallel to the axis mitted light is negligible. Our discussion will be limited to thin lenses. A thin lens
converge at the focal point F. (b) A
magnifying glass (converging lens) is a lens whose thickness is assumed to be negligible compared with the lens’s
can be used to focus the Sun’s rays focal length.
to a spot—with incendiary results.
Do not try this at home! 䉳 FIGURE 23.13
Diverging lens Rays
parallel to the axis of a
F diverging (concave) lens
appear to diverge from
a focal point on the inci-
dent side of the lens.
Diverging lens
Biconcave (diverging) lens

23.3 LENSES 791
A lens with spherical geometry has, for each lens surface, a center of curvature
(C), a radius of curvature (R), a focal point (F), and a focal length ( f ). The focal
points are at equal distances on either side of a thin lens. However, for a spherical
lens, the focal length is not simply related to R by f = R>2, as it is for spherical
mirrors. Because the focal length also depends on the lens’s index of refraction, the
focal length of a lens is usually specified, rather than its radius of curvature. This Double Plano- Convex
will be discussed in Section 23.4. convex convex meniscus
The general rules for drawing ray diagrams for lenses are similar to those for Converging lenses
spherical mirrors. But some modifications are necessary, since light passes
through a lens. Opposite sides of a lens are generally distinguished as the object
side and the image side. The object side is the side on which an object is positioned,
and the image side is the opposite side of the lens (where a real image would be
formed). The three key rays from a point on an object are drawn as follows:
1. A parallel ray is a ray that is parallel to the lens’s optic axis on incidence and,
after refraction, passes through the focal point on the image side of a con-
Double Plano- Concave
verging lens (or appears to diverge from the focal point on the object side of a
concave concave meniscus
diverging lens).
2. A chief ray, or central ray, is a ray that passes through the center of the lens Diverging lenses
and is undeviated because the lens is “thin.” 䉱 F I G U R E 2 3 . 1 4 Lens shapes
3. A focal ray is a ray that passes through the focal point on the object side of a Lens shapes vary widely and are
converging lens (or appears to pass through the focal point on the image side normally categorized as converging
or diverging. In general, a converg-
of a diverging lens) and, after refraction, is parallel to the lens’s optic axis.
ing lens is thicker at its center than
As with spherical mirrors, only two rays are needed to determine the image. at the periphery, and a diverging
lens is thinner at its center than at
(However, it is also generally a good idea to include the third ray as a check.)
the periphery.
EXAMPLE 23.5 Learn by Drawing: A Lens Diagram

An object is placed 30 cm in front of a thin converging lens of 2. Then draw the chief ray (➁ in the drawing). From the
focal length 20 cm. (a) Use a ray diagram to locate the image. tip of the flame, draw a ray passing through the center
(b) Discuss the characteristics of the image. of the lens. This ray will go undeviated through the thin
lens to the image side.
T H I N K I N G I T T H R O U G H . Follow the steps for lens ray dia-
3. It can be clearly seen that these two rays intersect on
grams, as given previously.
the image side. The point of intersection is the image
SOLUTION. point of the tip of the candle. From this point, draw
Given: do = 30 cm Find: (a) di (location of the image, the image by extending the tip of the flame to the
f = 20 cm using a ray diagram) optic axis.
(b) the image’s characteristics 4. Only two rays are needed to locate the image. However,
if the third ray, in this case the focal ray (➂ in the draw-
(a) Since a ray diagram is to be used to locate the image (see
ing), is drawn, it must go through the same point on the
Learn by Drawing 23.2), the first thing to decide on is a scale
image at which the other two rays intersect (if drawn
for the drawing. In this example, a scale of 1 cm to represent
carefully). The ray from the tip of the flame passing
10 cm is used. That way, the object would be 3.0 cm in front of
through the focal point F on the object side will travel
the mirror in the drawing.
parallel to the optic axis on the image side.
First the optic axis, the lens, the object (a lighted candle), and
the focal points (F) are drawn. A vertical dashed line through (b) From the ray diagram in part (a), the image is real
the center of the lens is drawn because, for simplicity, the refrac- (because the rays intersect). As a result, this real image could
tion is depicted as if it occurs at the center of the lens. In reality, be seen on a screen (for example, a piece of white paper) that
it would occur at the air–glass and glass–air surfaces of the lens. is positioned at the image point. The image is also inverted
Follow steps 1–4 in the accompanying Learn by Drawing (the image of the candle points downward) and is magnified
23.2, A Lens Ray Diagram. (the image is larger than the object).
1. The first ray drawn is the parallel ray (➀ in the draw- In this case, do = 30 cm and f = 20 cm, so 2f 7 do 7 f.
ing). From the tip of the flame, draw a horizontal ray Using similar ray diagrams, it can be proven that for any do in
(parallel to the optic axis). After passing through the this range, the image is always real, inverted, and magnified.
lens, this ray goes through the focal point F on the Actually, the overhead projector in your classroom uses this
image side. type of arrangement.
F O L L O W - U P E X E R C I S E . In this Example, what does the image look like if the object is 10 cm in front of the lens? Locate the image
graphically and discuss the characteristics of the image.
a lens ray diagram

1 Object 1
Parallel ray 1
F F
do
Object 1
2
Chief (central) ray 2 2
F F
do
Object 1
3 Real image
Locating image 2
F F
do di
Object 1
4 Real image
Can also use 2
focal ray 3 to
confirm image F F
3
do di
23.3 LENSES 793
Object Image (real, inverted,

1 and reduced)
2 F
F 3
do di
(a) Convex lens, do > 2f
F F
Image Object
(virtual,
upright, and
magnified) 2
do
di
(b) Convex lens, do < f
䉱 F I G U R E 2 3 . 1 5 Ray diagrams for lenses (a) A converging biconvex lens forms a real
image when do 7 2f. The image is real, inverted, and reduced. (b) Ray diagram for a con-
verging lens with do 6 f. The image is virtual, upright, and magnified. Practical examples
are shown for both cases.
䉱 Figure 23.15 shows other ray diagrams with different object distances for a
converging lens, along with real-life applications. The image of an object is real
when it is formed on the side of the lens opposite the object’s side (see Fig. 23.15a).
A virtual image is said to be formed on the same side of the lens as the object (see
Fig. 23.15b).
Regions could be similarly divided for the object distance for a converging lens
as was done for a converging mirror in Fig. 23.7a. Here, an object distance of
do = 2f for a converging lens has significance similar to that of do = R = 2f for a
converging mirror (䉲 Fig. 23.16).
The ray diagram for a diverging lens will be discussed shortly. Like diverging
mirrors, diverging lenses can form only virtual images of real objects.
䉳 F I G U R E 2 3 . 1 6 Converging
do = 2f do = f
lens For a converging lens, the
Real, Image at
inverted, infinity
object is located within one of three
same size regions defined by the focal length
( f ) and twice the focal length (2f ) or
Real, Real, Virtual, at one of these two points. For
inverted, inverted, upright, Lens do 7 2f, the image is real, inverted,
reduced magnified magnified and reduced (Fig. 23.15a). For
2f 7 do 7 f, the image will also be
2F F real and inverted, but magnified, as
(do > 2f ) (2f > do > f ) (do < f ) shown by the ray diagrams in
Example 23.5. For do 6 f, the image
will be virtual, upright, and magni-
Convex lens fied (Fig. 23.15b).
TABLE 23.3 Sign Conventions for Thin Lenses

Focal length ( f )
Converging (convex) lens f is positive
Diverging (concave) lens f is negative
Object distance (do)

Object is in front of the lens (real object) do is positive
Object is behind the lens (virtual object)* do is negative
Image distance (di) and image type

Image is formed on the image side of the lens—opposite di is positive
to the object (real image)
Image is formed on the object side of the lens—same side di is negative
as the object (virtual image)
Image orientation (M)

Image is upright with respect to the object M is positive
Image is inverted with respect to the object M is negative
*In a combination of two (or more) lenses, the image formed by the first lens is taken as the object
of the second lens (and so on). If this image–object falls behind the second lens, it is referred to as
a virtual object, and the object distance is taken to be negative 1 -2.
The image distances and characteristics for a thin lens can also be found analyti-
cally. The equations for thin lenses are identical to those for spherical mirrors. The
thin lens equation is
1 1 1
+ = (thin lens equation) (23.5)
do di f
As in the case for spherical mirrors, an alternative form of the thin lens
equation is,
do f
di = (23.5a)
do - f
gives a quick and easy way to find di.
The lateral magnification factor, like that for spherical mirrors, is given by
hi di
M = = - (magnification factor) (23.6)
ho do
The sign conventions for these thin lens equations are given in 䉱 Table 23.3.
Just as when you are working with mirrors, it is helpful to sketch a ray diagram
before working a lens problem analytically.
EXAMPLE 23.6 Three Images: Behavior of a Converging Lens

A converging lens has a focal length of 12 cm. For an object of the image characteristics. The diagrams should be in good
(a) 60 cm, (b) 15 cm, and (c) 8.0 cm from the lens, where is the agreement with the calculations.
image formed, and what are its characteristics? SOLUTION.
THINKING IT THROUGH. With the focal length ( f ) and the Given: f = 12 cm Find: di , M, and the image
object distances (do) given, Eq. 23.5 can be applied to find the (a) do = 60 cm characteristics for all three
image distances (di) and Eq. 23.6 can be used to determine the (b) do = 15 cm cases
image characteristics. Sketch ray diagrams first to get an idea (c) do = 8.0 cm
23.3 LENSES 795
(a) The object distance is greater than twice the focal length Then
1do 7 2f2. Using Eq. 23.5,
di 60 cm
di = 60 cm and M = - = - = - 4.0
1 1 1 do 15 cm
+ =
do di f
The image is real (positive di), inverted (negative M), and
or magnified 1ƒMƒ = 4.02. A similar situation applies to overhead
projectors and slide projectors 12f 7 do 7 f2.
1 1 1
= - (c) For this case, do 6 f. Using the alternative form
di f do
(Eq. 23.5a),
1 1 5 1 4 1
= - = - = = do f 18.0 cm2112 cm2
12 cm 60 cm 60 cm 60 cm 60 cm 15 cm di = = = - 24 cm
do - f 8.0 cm - 12 cm
Then
Then
di 15 cm
di = 15 cm and M = - = - = - 0.25 di 1- 24 cm2
do 60 cm M = - = - = + 3.0
do 8.0 cm
The image is real (positive di), inverted (negative M), and
reduced 1ƒMƒ = 0.252. A camera uses a similar arrangement The image is virtual (negative di), upright (positive M), and
when the object distance is usually much greater than magnified 1ƒMƒ = 3.02. This situation is an example of a sim-
2f1do W 2f2. ple microscope or magnifying glass 1do 6 f2.
As you can see, a converging lens is very versatile.
(b) Here, 2f 7 do 7 f. Using Eq. 23.5,
Depending on the object distance (relative to the focal length),
1 1 1 5 4 1 the lens can be used as a camera, projector, or magnifying
= - = - =
di 12 cm 15 cm 60 cm 60 cm 60 cm glass.
F O L L O W - U P E X E R C I S E . If the object distance of a converging lens is allowed to vary, at what object distance does the real image
change from being reduced to being magnified?
CONCEPTUAL EXAMPLE 23.7 Half an Image?

A converging lens forms an image on a screen, as shown in the image. However, rays from every point on the object pass
䉲 Fig. 23.17a. Then the lower half of the lens is blocked, as through all parts of the lens. Thus, the upper half of the lens
shown in Fig. 23.17b. As a result, (a) only the top half of the can form a total image (as could the lower half), so the answer
original image will be visible on the screen; (b) only the bot- is (c).
tom half of the original image will be visible on the screen; You might confirm this conclusion by drawing a chief ray
(c) the entire image will still be visible. in Fig. 23.17b. Or you might use the scientific method and
experiment—particularly if you wear eyeglasses. Block off the
REASONING AND ANSWER. At first thought, you might imag- bottom part of your glasses, and you will find that you can
ine that blocking off half of the lens would eliminate half of still read through the top part (unless you wear bifocals).
Screen Screen
(a) (b)
䉱 F I G U R E 2 3 . 1 7 Half a lens, half an image? (a) A converging lens forms an image on a screen. (b) The
lower half of the lens is blocked. What happens to the image?
FOLLOW-UP EXERCISE. Can you think of any property of the image that would be affected by blocking off half of the lens?
Explain.
INTEGRATED EXAMPLE 23.8 Time for a Change: Behavior of a Diverging Lens

An object is 24 cm in front of a diverging lens that has a focal that these rays appear to come from the same image point.
length of - 15 cm. (a) Use a ray diagram to determine whether The focal ray appears to go through the focal point on the
the image is (1) real and magnified, (2) virtual and reduced, image side and travels parallel to the optic axis after refrac-
(3) real and upright, or (4) upright and magnified. (b) Find the tion from the lens.
location and characteristics of the image with the thin lens The image is virtual (why?), upright, and reduced, so the
equations. answer is (2) virtual and reduced. Measuring from the dia-
(A) CONCEPTUAL REASONING. (See the sign conventions in gram (keeping in mind the drawing scale used), di L - 9 cm
Table 23.3.) Use a scale of 1 cm (in the drawing of 䉲 Fig. 23.18) hi 0.5 cm
(virtual image) and M = L = + 0.4.
to represent 10 cm. The object will be 2.4 cm in front of the ho 1.4 cm
lens in the drawing. Draw the optic axis, the lens, the object (B) QUANTITATIVE REASONING AND SOLUTION.
(in this case, a lighted candle), the focal point (F), and a verti-
Given: do = 24 cm Find: di , M, and image
cal dashed line through the center of the lens.
f = - 15 cm (diverging characteristics
The parallel ray ➀ starts from the tip of the flame, travels
parallel to the optic axis, diverges from the lens after refrac- lens)
tion, and appears to diverge from F on the object side. The Note that the focal length is negative for a diverging lens. (See
chief ray ➁ originates from the tip of the flame and goes Table 23.3.) From Eq. 23.5,
through the center of the lens, with no direction change.
These two rays, after refraction, diverge and do not intersect. 1 1 1
+ =
However, they appear to come from in front of the lens (object 24 cm di -15 cm
side), and that apparent intersection is the image point of the 1 1 1 13
tip of the flame. The focal ray ➂ can also be drawn to verify or = - = -
di -15 cm 24 cm 120 cm
so
䉳 FIGURE 23.18
Object Diverging lens Ray 120 cm
1 di = - = - 9.2 cm
diagram of a 13
diverging lens.
3 Here, the image is Then
F F virtual and in front di -9.2 cm
of the lens, upright, M = = - = + 0.38
2 and reduced. do 24 cm
Image (virtual, Thus, the image is virtual (di is negative) and upright (M is
upright, and
reduced) positive), and reduced (0.38 times the height of the object).
Due to the fact that f is negative for a diverging lens, di is
di
always negative for any positive value of do, so the image of a
do
real object is always virtual.
F O L L O W - U P E X E R C I S E . A diverging lens always forms a virtual image of a real object. What general statements can be made
about the image’s orientation and magnification?
A special type of lens that you may have encountered is discussed in Insight
23.2, Fresnel Lenses.
COMBINATIONS OF LENSES
Many optical instruments, such as microscopes and telescopes (Chapter 25), use a
combination of lenses, or a compound lens system. When two or more lenses are
used in combination, the overall image produced can be determined by consider-
ing the lenses individually in sequence. That is, the image formed by the first lens
becomes the object for the second lens, and so on.
If the first lens produces an image in front of the second lens, that image is
treated as a real object (do is positive) for the second lens (䉴 Fig. 23.19a). If, how-
ever, the lenses are close enough, the image from the first lens is not formed before
the rays pass through the second lens (Fig. 23.19b). In this case, the image from the
first lens is treated as a virtual object for the second lens. The virtual object distance
is taken to be negative in the lens equation (Table 23.3).
23.3 LENSES 797
L1 L2 䉳 F I G U R E 2 3 . 1 9 Lens combina-
Image from L1: tions The final image produced by
1 Real object for L2 a compound-lens system can be
found by treating the image of one
3 2 F1 F2 3⬘ lens as the object for the adjacent
Object F1 2⬘ F2 Final image lens. (a) If the image of the first lens
1⬘ from L2 (L1) is formed in front of the second
lens (L2), the object for the second
lens is said to be real. (Note that
(a) rays 1¿ , 2¿ , and 3¿ are the parallel,
chief, and focal rays, respectively,
for L2. They are not continuations of
L1 L2 rays 1, 2, and 3—the parallel, chief,
and focal rays, respectively, for L1.)
Final
(b) If the rays pass through the
image second lens before the image is
1 2⬘ from
1⬘ formed, the object for the second
Object 3 2 F1 F2 L2
lens is said to be virtual, and the
F1 3⬘ F2 Image from L1: object distance for the second lens is
Virtual object taken to be negative.
for L2 (do2< 0)
(b)
It can be shown that the total magnification (Mtotal) of a compound lens system
is the product of the individual magnification factors of the component lenses. For
example, for a two-lens system, as in Fig. 23.19,
Mtotal = M1 M2 (23.7)
The usual signs for M1 and M2 carry through to the product to indicate, from the
sign of Mtotal, whether the final image is upright or inverted. (See Exercise 65.)
INSIGHT 23.2 Fresnel Lenses

To focus parallel light or to produce a large beam of parallel
light rays, a sizable converging lens is sometimes neces-
sary. The large mass of glass necessary to form such a lens
is bulky and heavy. Moreover, a thick lens absorbs more
light and is likely to show distortions. A French physicist
named Augustin Fresnel (Fray-nel’, 1788–1827) devel-
oped a solution to this problem for the lenses used in
lighthouses. Fresnel recognized that the refraction of light
takes place at the surfaces of a lens. Hence, a lens could be
made thinner—and almost flat—by removing glass from
the interior, as long as the refracting properties of the sur-
(a)
faces were not changed.
This can be accomplished by cutting a series of concen-
tric grooves in the surface of the lens (Fig. 1a). Note that
the surface of each remaining curved segment has the
same radius as the surface of the original lens. Together,
the concentric segments refract light as does the original
converging lens. In effect, the lens has simply been
slimmed down by the removal of unnecessary glass
between the refracting surfaces.
This type of lens is called a Fresnel lens and it is widely (b)
used in overhead projectors and in beacons (Fig. 1b). A
Fresnel lens is very thin and therefore much lighter than a F I G U R E 1 Fresnel lens (a) The focusing action of a lens comes
conventional lens with the same optical properties. Also, from refraction at its surfaces. It is therefore possible to reduce the
Fresnel lenses are easily molded from plastic—often with thickness of a lens by cutting away glass in concentric grooves,
leaving a set of curved surfaces with the same refractive properties
one flat side (plano-convex) so that the lens can be
as the lens from which they were derived. (b) An array of Fresnel
attached to a flat surface. lenses produces focused beams in a Boston Harbor light.
EXAMPLE 23.9 A Special Offer: A Lens Combo and a Virtual Object

Consider two lenses similar to those illustrated in Fig. 23.19b. and
Suppose the object is 20 cm in front of lens L1, which has a di1 60 cm
focal length of 15 cm. Lens L2 , with a focal length of 12 cm, is M1 = = - = - 3.0 (inverted and magnified)
do1 20 cm
26 cm from L1. What is the location of the final image, and
what are its characteristics? The image from lens L1 becomes the object for lens L2. This
image is then 60 cm - 26 cm = 34 cm on the right, or image,
T H I N K I N G I T T H R O U G H . This is a double application of the
side of L2. Therefore, it is a virtual object (see Table 23.3), and
thin lens equation. The lenses are treated successively: The
do2 = - 34 cm. (Remember that do for virtual objects is taken
image of lens L1 becomes the object of lens L2. The quantities
to be negative.)
must be distinctly labeled and the distances appropriately ref-
Then applying the equations to the second lens, L2:
erenced (with signs!).
1 1 1 1 1 23
1- 34 cm2
SOLUTION. Listing the known quantities and what is to be = - = - =
found: di2 f2 do2 12 cm 204 cm
Given: do1 = + 20 cm Find: di2 , Mtotal , or

f1 = + 15 cm and image 204 cm
di2 = = 8.9 cm (real image)
f2 = + 12 cm characteristics 23
D = 26 cm (distance and
between lenses) di2
= 0.26 1upright and reduced2
8.9 cm
1- 34 cm2
M2 = - = -
The first step is to apply the thin lens equation (Eq. 23.5) and do2
the magnification factor for thin lenses (Eq. 23.6) to L1: (Note: The virtual object for L2 was inverted, and thus the
1 1 1 term upright means that the final image is also inverted.) The
di1
=
f1
-
do1 total magnification Mtotal is then
1 1 4 3 1 Mtotal = M1 M2 = 1-3.0210.262 = - 0.78
= - = - =
15 cm 20 cm 60 cm 60 cm 60 cm The sign is carried through with the magnifications. We
determine that the final real image is located at 8.9 cm on the
or
right (image) side of L2 and that it is inverted relative to the
di1 = 60 cm (real image from L1) initial object 1Mtotal 6 02 and reduced 1ƒMtotalƒ = 0.782.
F O L L O W - U P E X E R C I S E . Suppose the object in Fig. 23.19b were located 30 cm in front of L1. Where would the final image be
formed in this case, and what would be its characteristics?
DID YOU LEARN?

➥ A converging lens is generally thicker at its center than at its periphery, has a
positive focal length, and can form images with a variety of image characteristics.
A diverging lens is generally thinner at its center than at its periphery, has a
negative focal length, and can only form virtual, upright, and reduced images (for
real objects).
➥ The three key rays that can be used in a ray diagram for a thin lens are the parallel
ray, the chief ray or central ray, and the focal ray. (Only two are actually needed.)
➥ The sign of the lateral magnification describes if an image is upright 1M 7 02 or
inverted 1M 6 02.The absolute value describes if an image is magnified 1 ƒ M ƒ 7 12
or reduced 1 ƒ M ƒ 6 12. For real objects, the sign of the lateral magnification also
describes if an image is real 1M 6 02 or virtual 1M 7 02.
23.4 The Lens Maker’s Equation

➥ On what does the focal length of a thin lens in air depend?

➥ Do different media surrounding a thin lens affect the focal length?
➥ How is the lens power in diopters defined?
There are a variety of other shapes of lenses, as illustrated in Fig. 23.14. Lens
refraction depends on the shapes of the lens’s surfaces and on the index of refrac-
23.4 THE LENS MAKER’S EQUATION 799
TABLE 23.4 Sign Conventions for the Lens Maker’s Equation

Convex surface R is positive
Double (bi)
Concave surface R is negative convex lens
Plane (flat) surface R = q Incident light
Converging (convex) lens f is positive R1
Diverging (concave) lens f is negative R2 C1
C2
R1: positive (convex)

tion of the lens. These properties together determine the focal length of a thin lens.
(a)
The thin lens focal length is given by the lens maker’s equation, which enables us
to calculate the focal length of a thin lens in air 1nair = 12 as
Double (bi)
concave lens
= 1n- 12 ¢
1 1 1
+ ≤ (for thin lens in air) (23.8)
f R1 R2 R1
C1 C2
R2
where n is the index of refraction of the lens material and R1 and R2 are the
radii of curvature of the first (front side) and second (back side) lens surfaces, R1: negative (concave)
respectively. R2: negative (concave)
A sign convention is required for the lens maker’s equation, and a common one (b)
is summarized in 䉱 Table 23.4. The signs depend only on the shape of the surface,
that is, convex or concave (䉴 Fig. 23.20). For the biconvex lens in Fig. 23.20a, both 䉱 F I G U R E 2 3 . 2 0 Centers of
R1 and R2 are positive (both surfaces are convex), and for the biconcave lens in Fig. curvature Lenses, such as (a) a
23.20b, both R1 and R2 are negative (both surfaces are concave). biconvex lens and (b) a biconcave
lens, have two centers of curvature,
If the lens is surrounded by a medium other than air, then the first term in
parentheses in Eq. 23.8 becomes 1n>nm2 - 1, where n and nm are the indices of
which define the signs of the radii of
curvature. See Table 23.4.
refraction of the lens material and the surrounding medium, respectively. Now we
can see why some converging lenses in air become diverging when submerged in
water: If nm 7 n, then f is negative, and the lens is diverging.
LENS POWER: DIOPTERS

Notice that the lens maker’s equation (Eq. 23.8) gives the inverse focal length 1>f.
Optometrists use this inverse relationship to express the lens power (P) of a lens in
units called diopters (abbreviated as D). The lens power is the reciprocal or
inverse of the focal length of the lens expressed in meters:
P 1expressed in diopters2 =
1
f 1expressed in meters2
(23.9)
So, 1 D = 1 m-1. The lens maker’s equation gives a lens’s power 11>f2 in diopters
if the radii of curvature are expressed in meters.
If you wear glasses, you may have noticed that the prescription the optometrist
gave you for your eyeglass lenses was written in terms of diopters. Converging
and diverging lenses are referred to as positive 1+ 2 and negative 1-2 lenses,
respectively. Thus, if an optometrist prescribes a corrective lens with a power of
+ 2.0 diopters, it is a converging lens with a focal length of
1 1 1
f = = = = + 0.50 m = + 50 cm
P +2.0 D +2.0 m-1
The greater the power of the lens in diopters, the shorter its focal length and the
more strongly converging or diverging it is. Thus, a “stronger” prescription lens
(greater lens power) has a shorter f than does a “weaker” prescription lens (lesser
lens power).
INTEGRATED EXAMPLE 23.10 A Convex Meniscus Lens: Converging

or Diverging
The convex meniscus lens shown in Fig. 23.14 has a 15-cm radius for the convex surface
and a 20-cm radius for the concave surface. The lens is made of crown glass and is sur-
rounded by air. (a) Is this lens a (1) converging or (2) diverging lens? Explain. (b) What
are the focal length and the power of the lens?
( A ) C O N C E P T U A L R E A S O N I N G . The index of refraction of crown glass can be obtained
from Table 22.1: n = 1.52. For a convex meniscus lens, the first surface is convex, so R1
is positive; the second surface is concave, so R2 is negative. Since
R1 = 15 cm 6 ƒ R2 ƒ = 20 cm, 1>R1 + 1>R2 will be positive. Therefore, the lens is a con-
verging (positive) lens, according to Eq. 23.8. Thus the answer is (1) converging.
(B) QUANTITATIVE REASONING AND SOLUTION.
Given: R1 = 15 cm = 0.15 m Find: f (focal length) and

R2 = - 20 cm = - 0.20 m P (lens power)
n = 1.52 (from Table 22.1 for crown glass)
(b) From Eq. 23.8,
= 1n - 12 ¢ ≤ = 11.52 - 12a b = 0.867 m-1

1 1 1 1 1
+ +
f R1 R2 0.15 m -0.20 m
Hence,
1
f = = + 1.15 m
0.867 m-1
and the power of the lens therefore is
1
P = = + 0.867 D.
f
FOLLOW-UP EXERCISE. In this Example, if this lens were is immersed in water, what
would your answers be?
DID YOU LEARN?

➥ The focal length of a thin lens in air depends on the lens material’s index of
refraction and the radii of curvatures of the two surfaces.
➥ Different media surrounding a thin lens affects the focal length of the lens. In
extreme cases, a different surrounding medium can make a converging lens
diverging or vice versa.
➥ The power of a lens in diopters is defined as the inverse focal length when the focal
length is expressed in meters.
*23.5 Lens Aberrations

➥ What are three common lens aberrations?

➥ What causes chromatic aberration?
➥ Can lens aberrations be minimized?
Lenses, like mirrors, can also have aberrations. Some common aberrations are as
follows.
SPHERICAL ABERRATION
The discussion of lenses thus far has concentrated on rays that are near the optic
axis. Like spherical mirrors, however, lenses may show spherical aberration, an
effect that occurs when parallel rays passing through different regions of a lens do
not come together on a common focal plane. In general, rays close to the axis of a
*23.5 LENS ABERRATIONS 801
White light
Optic axis
F2 Fred
F1 Fblue
(a) Spherical aberration (b) Chromatic aberration
䉱 F I G U R E 2 3 . 2 1 Lens aberra-
converging lens are refracted less and come together at a point farther from the tions (a) Spherical aberration. In
lens than do rays passing through the periphery of the lens (䉱 Fig. 23.21a). general, rays closer to the axis of a
lens are refracted less and come
Spherical aberration can be minimized by using an aperture to reduce the effec- together at a point farther from the
tive area of the lens, so that only light rays near the axis are transmitted. Also, lens than do rays passing through
combinations of converging and diverging lenses can be used, as the aberration of the periphery of the lens. (b) Chro-
one lens can be compensated for by the optical properties of another lens. matic aberration. Because of disper-
sion, different wavelengths (colors)
of light are focused in different
CHROMATIC ABERRATION planes, which results in distortion of
the overall image.
Chromatic aberration is an effect that occurs because the index of refraction of the
lens material is not the same for all wavelengths of light (that is, the material is dis-
persive). When white light is incident on a lens, the transmitted rays of different
wavelengths (colors) do not have a common focal point, and images of different
colors are produced at different locations (Fig. 23.21b).
This dispersive aberration can be minimized, but not eliminated, by using a
compound lens system consisting of lenses of different materials, such as crown
glass and flint glass. The lenses are chosen so that the dispersion produced by one
is approximately compensated for by the opposite dispersion produced by the
other. With a properly constructed two-component lens system, called an
achromatic doublet (achromatic means “without color”), the images of any two
selected colors can be made to coincide.
ASTIGMATISM
A circular beam of light along the lens axis forms a circular illuminated area on the
lens. When incident on a converging lens, the parallel beam converges at the focal
point. However, when a circular beam of light from an off-axis source falls on the
lens, the light forms an elliptical illuminated area on the lens. The rays entering
along the major and minor axes of the ellipse then focus at different points after
passing through the lens. This condition is called astigmatism.
With different focal points in different planes, the images in both planes are
blurred. For example, the image of a point is no longer a point, but rather two
separated short-line images (blurred points). Astigmatism can be reduced by
decreasing the effective area of the lens with an aperture or by adding a cylindrical
lens to compensate.
DID YOU LEARN?

➥ Three common lens aberrations are spherical aberration, chromatic aberration, and
astigmatism.
➥ Chromatic aberration arises because the index of refraction of the lens material is
not the same for all wavelengths (colors) of light, that is, the material is dispersive.
➥ Lens aberration can be minimized. Some common practices are to use smaller
aperture, multiple lenses, and lenses using different materials.
PULLING IT TOGETHER Making Lenses to Correct Nearsightedness

A certain corrective lens is made of polycarbonate that has an T H I N K I N G I T T H R O U G H . This Example incorporates the
index of refraction of 1.59. The radii of the two surfaces of the shapes of various lenses, the lens maker’s equation, the rela-
lens are + 0.200 m and - 0.100 m. The lens is thinner at its center tion between focal length and power of a lens, the thin-lens
than at its edge. (a) Referring to Fig. 23.14, identify the shape of equation, and lateral magnification. The focal length and
the lens. Is the lens converging or diverging? (b) What are the power of a lens can be calculated using the lens maker’s equa-
focal length and the power of the lens? (c) If there is an object tion. Then the thin lens and magnification equations can be
that is 10.0 m away from the lens, where is the image located used to determine the image distance and lateral magnifica-
and what is its lateral magnification? (d) Repeat the calculations tion. When the lens is in water, the modified version of the
for parts (b) and (c) if the lens is under water and the object is lens maker’s equation must be used.
also under water and still 10.0 m away.
SOLUTION.
Given: R1 = 0.200 m Find: (a) lens shape and type

R2 = - 0.100 m (b) f (focal length) and P (power)
n = 1.59 (index of refraction of lens) (c) di (image distance) and M (magnification)
nair = 1.00 and nwater = 1.33 (from Table 22.1) (d) repeat (b) and (c) with lens in water
do = 10.0 m
(a) Since one surface of the lens has a negative radius and the other one is positive, the lens is either convex meniscus or concave
meniscus, according to the sign convention and Fig. 23.14. However, the lens is known to be thinner at the center than at the
edge, therefore it must be concave meniscus. (A convex meniscus lens is thicker at the center than at the edge.) A concave
meniscus lens is a diverging lens.
(b) From the lens maker’s equation (Eq. 23.8),
= 1n - 12 ¢ ≤ = 11.59 - 12a b = - 2.95 m-1

1 1 1 1 1
+ -
f R1 R2 0.200 m -0.100 m
Hence
1
f = = - 0.339 m (diverging lens)
-2.95 m-1
The negative focal length means it is a diverging lens as expected from the reasoning in part (a).
Therefore, the power of the lens is (Eq. 23.9)
1
P = = - 2.95 D.
f
(c) The thin-lens equation (Eq. 23.5) can be solved for the image distance, and using the given data yields
do f 110.0 m21 -0.339 m2

10.0 m - 1 -0.339 m2
di = = = - 0.328 m
do - f
(It is a virtual image; how do you know this from the answer?)
The lateral magnification follows directly (Eq. 23.6) since the image distance is now known.
di -0.328 m
M = - = - = + 0.0328
do 10.0 m
(The image is upright and reduced; how do you know this from the answer?)
(d) When the lens is surrounded by water rather than air, the lens maker’s equation (Eq. 23.8) needs to be modified. The index of
refraction n must be replaced by n>nm , where nm is the index of refraction of the surrounding medium (water).
≤ = a - 1b a b = - 0.977 m-1
1 n 1 1 1.59 1 1
= ¢ - 1≤ ¢ + +
f nm R1 R2 1.33 0.200 m -0.100 m
Hence
1
= - 1.02 m
f =
-0.977 m-1
Using the new focal length, the calculations in part (c) can be repeated.
The power of the lens is
1
P = = - 0.977 D
f
The image distance and lateral magnification are
110.0 m21 - 1.02 m2
= - 0.926 m 1virtual image2
10.0 m - 1 -1.02 m2
di =
= + 0.0926 1upright and reduced2

-0.926 m
M = -
10.0 m
■ Plane mirrors form virtual, upright, and unmagnified ■ The thin lens equation relates focal length, object distance,
images. The object distance is equal to the absolute value of and image distance:
the image distance 1do = ƒ di ƒ 2. 1 1 1
+ = (23.5)
■ The lateral magnification factor for all mirrors and lenses is do di f
di Alternative form:
M = - (23.4, 23.6)
do do f
di = (23.5a)
■ Spherical mirrors are either converging (concave) or do - f
diverging (convex). Diverging spherical mirrors always do = 2 f do = f
form virtual, upright, and reduced images. Real,
inverted,
Image at
infinity
same size
Focal length of a spherical mirror: Real, Real, Virtual,
Lens
inverted, inverted, upright,
reduced magnified magnified
R
f = (23.2) 2F F
2 (do > 2f ) (2f > do > f ) (do < f )
Spherical mirror equation: Convex lens
1 1 1 2 ■ The lens maker’s equation determines the focal length of a

+ = = (23.3)
do di f R lens based on the radii and index of refraction of the lens.
= 1n - 12 ¢ 1thin lens in air only2

Alternative form: 1 1 1
+ ≤ (23.8)
do f f R1 R2
di = (23.3a)
do - f Double (bi)
convex lens
O' Incident light
R1
ho
I C V R2 C1
C2
O F
hi
I' A
f
di
do ■ Lens power in diopters (where f is in meters) is given by
■ Lenses are either convex (converging) or concave (diverg- 1
P = (23.9)
ing). Diverging lenses always form virtual, upright, and f
reduced images.
Converging lens

23.1 PLANE MIRRORS 23.3 LENSES

1. A plane mirror has (a) a greater image distance than 9. If an object is placed at the focal point of a converging lens,
object distance, (b) a greater object distance than image the image is (a) at zero, (b) also at the focal point, (c) at a
distance, (c) the same object and image distance. distance equal to twice the focal length, (d) at infinity.
2. The image of a person formed by a plane mirror is 10. A converging thin lens can form (a) both magnified and
(a) real, upright, and unmagnified, (b) virtual, upright, reduced images, (b) only magnified images, (c) only
and magnified, (c) real, upright, and magnified, (d) vir- reduced images.
tual, upright, and unmagnified. 11. The image produced by a diverging lens is always
3. A plane mirror (a) produces both real and virtual (a) virtual and magnified, (b) real and magnified, (c) vir-
images, (b) always produces a virtual image, (c) always tual and reduced, (d) real and reduced.
produces a real image, (d) forms images by diffuse 12. A converging lens (a) must have at least one convex sur-
reflection. face, (b) cannot produce a virtual and reduced image,
4. The lateral magnification of a plane mirror is (a) greater (c) is thicker at its center than at the periphery, (d) all of
than 1, (b) less than 1, (c) equal to + 1, (d) equal to -1. the preceding.
23.2 SPHERICAL MIRRORS 23.4 THE LENS MAKER’S EQUATION

AND
5. A concave spherical mirror can form (a) both real and
*23.5 LENS ABERRATIONS
virtual images, (b) only virtual images, (c) only real
images. 13. The focal length of a rectangular glass block is (a) zero,
6. Which of the following statements concerning spherical (b) infinity, (c) not defined.
mirrors is correct? (a) A converging mirror can produce 14. The focal length of a thin lens depends on (a) the radii of
an inverted virtual image. (b) A diverging mirror can both surfaces, (b) the index of refraction of the lens
produce an inverted virtual image. (c) A diverging mir- material, (c) the index of refraction of the surrounding
ror can produce an inverted real image. (d) A converging material, (d) all of the preceding.
mirror can produce an inverted real image. 15. The power of a lens is expressed in units of (a) watts,
7. The image produced by a convex mirror is always (b) joules, (c) diopters, (d) meters.
(a) virtual and upright, (b) real and upright, (c) virtual 16. If the focal length of a lens increases, the lens power will
and inverted, (d) real and inverted. (a) also increase, (b) decrease, (c) remain the same.
8. A shaving>makeup mirror is used to form an image that 17. A lens aberration that is caused by dispersion is called
is larger than the object, so it is a (a) concave mirror (a) spherical aberration, (b) chromatic aberration,
(b) convex mirror (c) plane mirror. (c) astigmatism, (d) none of the preceding.
23.1 PLANE MIRRORS Incident light

Silvered 䉳 FIGURE 23.22
side Automobile
1. Can a virtual image be projected onto a screen? Why or
day–night mirror
why not? See Conceptual
2. What is the focal length of a plane mirror? Explain. Question 3.
3. Day–night rearview mirrors are common in cars. At
night, when you tilt the mirror backward, the intensity Dim
and glare of headlights behind you are reduced (a) Day
(䉴 Fig. 23.22). The mirror is wedge shaped and is silvered
on the back. The unsilvered front surface reflects about
5% of incident light; the silvered back surface reflects Tilted
about 90% of the incident light. Explain how the
day–night mirror works. Headlight
4. When you stand in front of a plane mirror, there is a
right–left reversal. (a) Why is there not a top–bottom
reversal of your body? (b) Could you create an apparent Dim
top–bottom reversal by positioning your body
differently? (b) Night
5. Why do some emergency vehicles have “Ambulance” 9. (a) A 10-cm-tall mirror bears the following advertise-
printed backward and reversed on the front of the ment: “Full-view mini mirror. See your full body in
vehicle (䉲 Fig. 23.23)? 10 cm.” How can this be? (b) A popular novelty item
䉳 FIGURE 23.23 consists of a concave mirror with a ball suspended at or
Backward and slightly inside the center of curvature (䉲 Fig. 23.26).
reversed See Con- When the ball swings toward the mirror, its image grows
ceptual Question 5. larger and suddenly fills the whole mirror. The image
appears to be jumping out of the mirror. Explain what is
happening.
䉳 FIGURE 23.26
Spherical mirror toy
See Conceptual Ques-
tion 9b.
23.2 SPHERICAL MIRRORS

6. How can the focal length be quickly determined experi-
mentally for a concave mirror? Can you do the same
thing for a convex mirror?
7. (a) If you look into a shiny spoon, you see an upright
image on one side and an inverted image on the other
(䉲 Fig. 23.24). (Try it.) Why? (b) Could you see upright 10. Can a convex mirror produce an image that is taller than
images on both sides? Explain. the object? Why or why not?
11. When a girl looks at herself in a cosmetic mirror, the lateral
magnification is + 2. Discuss the image characteristics.
23.3 LENSES
12. How can you quickly determine the focal length of a
converging lens? Will the same method work for a
diverging lens?
13. If you want to use a converging lens to design a simple
overhead projector to project the magnified image of
an object containing small writing onto a screen on a
wall, how far should you place the object in front of
the lens?
14. Explain why a fish in a spherical fish bowl, viewed from
the side, appears larger than it really is.
15. How would you use a converging lens as a magnifying
䉱 F I G U R E 2 3 . 2 4 Images from convex and concave surfaces glass? Can you do the same with a diverging lens?
16. The lateral magnification of an image formed by a lens
of a chair is - 0.50. Discuss the image characteristics.
8. (a) What is the purpose of using two mirrors on a car or
truck, such as the one shown in 䉲 Fig. 23.25? (b) Some
side rearview mirrors have the warning “OBJECTS IN
MIRROR ARE CLOSER THAN THEY APPEAR.” 23.4 THE LENS MAKER’S EQUATION
Explain why. (c) Could a TV satellite dish be considered AND
a converging mirror? Explain. *23.5 LENS ABERRATIONS
17. Determine the signs of R1 and R2 for each lens shown in
Fig. 23.14.
䉳 FIGURE 23.25
Mirror applications 18. When you open your eyes underwater, everything is
See Conceptual blurry. However, when you wear goggles, you can see
Question 8. clearly. Explain.
19. A lens that is converging in air is submerged in a fluid
whose index of refraction is greater than that of the lens.
Is the lens still converging?
20. If a farsighted person is prescribed with a “stronger” or
more “powerful” lens, does the lens have a longer or
shorter focal length? Explain.
21. What is the cause of spherical aberration?
EXERCISES
23.1 PLANE MIRRORS I1 I3

1. ● Standing 2.5 m in front of a plane mirror with your
camera, you decide to take a picture of yourself. To what
distance should the camera be focused to get a sharp Object I2
image?
2. ● A man stands 2.0 m away from a plane mirror.
(a) What is the distance between the mirror and the

man’s image? (b) What are the image characteristics? (a) (b)
3. ● An object 5.0 cm tall is placed 40 cm from a plane mirror.

䉱 F I G U R E 2 3 . 2 7 Two mirrors—multiple images
Find (a) the distance from the object to the image, (b) the See Exercise 10.
height of the image, and (c) the image’s magnification.
2
4. ● ● If you hold a 900-cm square plane mirror 45 cm 23.2 SPHERICAL MIRRORS
from your eyes and can just see the full length of an
8.5-m flagpole behind you, how far are you from the 11. IE ● An object is 100 cm in front of a concave mirror that
pole? [Hint: A diagram is helpful.] has a radius of 80 cm. (a) Use a ray diagram to determine
whether the image is (1) real or virtual, (2) upright or
5. ● ● A small dog sits 3.0 m in front of a plane mirror.
inverted, and (3) magnified or reduced. (b) Calculate the
(a) Where is the dog’s image in relation to the mirror?
image distance and lateral magnification.
(b) If the dog jumps at the mirror at a speed of 1.0 m>s,
how fast does the dog approach its image? 12. ● A candle with a flame 1.5 cm tall is placed 5.0 cm from
the front of a concave mirror. A virtual image is formed
6. IE ● ● A woman fixing the hair on the back of her head
10 cm behind the mirror. (a) Find the focal length and
holds a plane mirror 30 cm in front of her face so as to
radius of curvature of the mirror. (b) How tall is the
look into a plane mirror on the bathroom wall behind
image of the flame?
her. She is 90 cm from the wall mirror. (a) The image of
the back of her head will be from (1) only the front mir- 13. ● An object is placed 50 cm in front of a convex mirror
ror, (2) only the wall mirror, or (3) both mirrors. and its image is found to be 20 cm behind the mirror.
(b) Approximately how far does the image of the back of What is the focal length of the mirror? What is the lateral
her head appear in front of her? magnification?
7. IE ● ● (a) When you stand between two plane mirrors on 14. ● An object 3.0 cm tall is placed 20 cm from the front of a
opposite walls in a dance studio, you observe (1) one, concave mirror with a radius of curvature of 30 cm.
(2) two, or (3) multiple images. Explain. (b) If you stand Where is the image formed, and how tall is it?
3.0 m from the mirror on the north wall and 5.0 m from 15. ● If the object in Exercise 14 is moved to a position 10 cm
the mirror on the south wall, what are the image dis- from the front of the mirror, what will be the characteris-
tances for the first two images in both mirrors? tics of the image?
8. ● ● A woman 1.7 m tall stands 3.0 m in front of a plane 16. ● ● An object 3.0 cm tall is placed at different locations in
mirror. (a) What is the minimum height the mirror must front of a concave mirror whose radius of curvature is
be to allow the woman to view her complete image from 30 cm. Determine the location of the image and its char-
head to foot? Assume that her eyes are 10 cm below the acteristics when the object distance is 40 cm, 30 cm,
top of her head. (b) What would be the required mini- 15 cm, and 5.0 cm, using (a) a ray diagram and (b) the
mum height of the mirror if she were to stand 5.0 m mirror equation.
away? 17. ● ● Use the mirror equation and the magnification factor
9. ● ● Prove that do = ƒ di ƒ (equal magnitude) for a plane to show that when do = R = 2f for a concave mirror, the
mirror. [Hint: Refer to Fig. 23.2 and use similar and iden- image is real, inverted, and the same size as the object.
tical triangles.] 18. IE ● ● An object is 120 cm in front of a convex mirror that
10. ● ● ● Draw ray diagrams that show how three images of has a focal length of 50 cm. (a) Use a ray diagram to
an object are formed in two plane mirrors at right angles, determine whether the image is (1) real or virtual,
as shown in 䉴 Fig. 23.27a. [Hint: Consider rays from both (2) upright or inverted, and (3) magnified or reduced.
ends of the object in the drawing for each image.] Figure (b) Calculate the image distance and image height.
23.27b shows a similar situation from a different point of 19. ● ● A bottle 6.0 cm tall is located 75 cm from the concave
view that gives four images. Explain the extra image in surface of a mirror with a radius of curvature of 50 cm.
this case. Where is the image located, and what are its characteristics?
EXERCISES 807
20. IE ● ● A virtual image of magnification + 2.0 is produced tion and height of the image of a car that is 2.0 m high and
when an object is placed 7.0 cm in front of a spherical (a) 100 m and (b) 10.0 m behind the truck mirror?
mirror. (a) The mirror is (1) convex, (2) concave, (3) flat. 35. ● ● ● For values of do from 0 to q , (a) sketch graphs of
Explain. (b) Find the radius of curvature of the mirror. (1) di versus do and (2) M versus do for a converging mir-
21. IE ● ● A virtual image of magnification + 0.50 is pro- ror, (b) Sketch similar graphs for a diverging mirror.
duced when an object is placed 5.0 cm in front of a 36. ● ● ● The front surface of a glass cube 5.00 cm on each
spherical mirror. (a) The mirror is (1) convex, (2) con- side is placed a distance of 30.0 cm in front of a converg-
cave, (3) flat. Explain. (b) Find the radius of curvature of ing mirror that has a focal length of 20.0 cm. (a) Where is
the mirror. the image of the front and back surface of the cube
22. ● ● Using the spherical mirror equation and the magnifi- located, and what are the image characteristics? (b) Is the
cation factor, show that for a convex mirror, the image of image of the cube still a cube?
an object is always virtual, upright, and reduced. 37. ● ● ● A section of a sphere is mirrored on both sides. If
23. IE ● ● When a man’s face is in front of a concave mirror the magnification of the image of an object is + 1.8 when
of radius 100 cm, the lateral magnification of the image the section is used as a concave mirror, what is the mag-
is + 1.5. What is the image distance? nification of the same object at the same object distance
24. ● ● A convex mirror in a department store produces an in front of the convex side?
upright image 0.25 times the size of a person who is 38. IE ● ● ● A concave mirror of radius of curvature of 20 cm
standing 200 cm from the mirror. What is the focal length forms an image of an object that is twice the height of the
of the mirror? object. (a) There could be (1) one, (2) two, (3) three object
25. IE ● ● The image of an object located 30 cm from a mirror distance(s) that satisfy the image characteristics. Explain.
is formed on a screen located 20 cm from the mirror. (b) What are the object distances?
(a) The mirror is (1) convex, (2) concave, (3) flat. Explain. 39. ● ● ● Two students in a physics laboratory each have a
(b) What is the mirror’s radius of curvature? concave mirror with the same radius of curvature, 40 cm.
26. IE ● ● The upright image of an object 18 cm in front of a Each student places an object in front of their mirror. The
mirror is half the size of the object. (a) The mirror is image in both mirrors is three times the size of the object.
(1) convex, (2) concave, (3) flat. Explain. (b) What is the However, when the students compare notes, they find
focal length of the mirror? that the object distances are not the same. Is this possi-
27. IE ● ● A concave mirror has a magnification of +3.0 for ble? If so, what are the object distances?
an object placed 50 cm in front of it. (a) The type of 40. ● ● ● When an object is moved closer to a convex mirror,
image produced is (1) virtual and upright, (2) real and its image size (1) increases, (2) remains the same,
upright, (3) virtual and inverted, (4) real and inverted. (3) decreases. Prove your answer mathematically.
Explain. (b) Find the radius of curvature of the mirror.
28. ● ● A concave mirror is constructed so that a man at a
distance of 20 cm from the mirror sees his image magni- 23.3 LENSES
fied 2.5 times. What is the radius of curvature of the
mirror? 41. ● An object is placed 50.0 cm in front of a converging
lens of focal length 10.0 cm. What are the image distance
29. ● ● A child looks at a reflective Christmas tree ball orna-
and the lateral magnification?
ment that has a diameter of 9.0 cm and sees an image of
her face that is half the real size. How far is the child’s 42. ● An object placed 30 cm in front of a converging lens
face from the ball? forms an image 15 cm behind the lens. What are the focal
30. IE ● ● A dentist uses a spherical mirror that produces an length of the lens and the lateral magnification of the
upright image of a tooth that is magnified four times. image?
(a) The mirror is (1) converging, (2) diverging, (3) flat. 43. ● A converging lens with a focal length of 20 cm is used
Explain. (b) What is the mirror’s focal length in terms of to produce an image on a screen that is 2.0 m from the
the object distance? lens. What are the object distance and the lateral magni-
31. ● ● A 15-cm-long pencil is placed with its eraser on the fication of the image?
optic axis of a concave mirror and its point directed 44. ●● When an object is placed at 2.0 m in front of a diverg-
upward at a distance of 20 cm in front of the mirror. The ing lens, a virtual image is formed at 30 cm in front of the
radius of curvature of the mirror is 30 cm. Use (a) a ray lens. What are the focal length of the lens and the lateral
diagram and (b) the mirror equation to locate the image magnification of the image?
and determine the image characteristics. 45. IE ● ● An object 4.0 cm tall is in front of a converging
32. ● ● A spherical mirror at an amusement park has a radius lens of focal length 22 cm. The object is 15 cm away from
of 10 m. If it forms an image that has a lateral magnifica- the lens. (a) Use a ray diagram to determine whether the
tion of + 2.0, what are the object and image distances? image is (1) real or virtual, (2) upright or inverted, and
33. IE ● ● A pill bottle 3.0 cm tall is placed 12 cm in front of a (3) magnified or reduced. (b) Calculate the image dis-
mirror. A 9.0-cm-tall upright image is formed. (a) The tance and lateral magnification.
mirror is (1) convex, (2) concave, (3) flat. Explain. 46. ●● (a) Design the lens in a single-lens slide projector that
(b) What is its radius of curvature? will form a sharp image on a screen 4.0 m away with the
34. ● ● A convex mirror is on the exterior of the passenger side transparent slides 6.0 cm from the lens. (b) If the object
of many trucks (see Conceptional Question 8a). If the focal on a slide is 1.0 cm tall, how tall will the image on the
length of such a mirror is - 40.0 cm, what will be the loca- screen be?
47. ●● An object is placed in front of a concave lens whose 59. ●● To correct myopia (nearsightedness), concave lenses
focal length is - 18 cm. Where is the image located and are prescribed. If a student can read her physics book
what are its characteristics, if the object distance is only when she holds it no farther than 18 cm away, what
(a) 10 cm and (b) 25 cm? Sketch ray diagrams for each case. focal length of lens should be prescribed so she can read
48. ● ● A convex lens produces a real, inverted image of an when she holds the book 35 cm away?
object that is magnified 2.5 times when the object is 60. ● ● To correct hyperopia (farsightedness), convex lenses
20 cm from the lens. What are the image distance and the are prescribed. If a senior citizen can read a newspaper
focal length of the lens? only when he holds it no closer than 50 cm away, what
49. ● ● A convex lens has a focal length of 0.12 m. Where on focal length of lens should be prescribed so he can read
the lens axis should an object be placed in order to get when he holds the newspaper 25 cm away?
(a) a real, magnified image with a magnification of 2.0 and 61. IE ● ● A biology student wants to examine a bug at a
(b) a virtual, magnified image with a magnification of 2.0? magnification of + 5.00 (a) The lens should be (1) convex,
50. ● ● Using the thin lens equation and the magnification (2) concave, (3) flat. Explain. (b) If the bug is 5.00 cm
factor, show that for a spherical diverging lens, the from the lens, what is the focal length of the lens?
image of a real object is always virtual, upright, and 62. ● ● The human eye is a complex multiple-lens system.
reduced. However, it can be approximated to an equivalent single

51. ● ● (a) For values of do from 0 to q , sketch graphs of converging lens with an average focal length about 1.7 cm
(1) di versus do and (2) M versus do for a converging lens. when the eye is relaxed. If an eye is viewing a 2.0-m-tall
(b) Sketch similar graphs for a diverging lens. (Compare tree located 15 m in front of the eye, what are the height
to Exercise 35.) and orientation of the image of the tree on the retina?
52. ● ● A simple single-lens camera (convex lens) is used to 63. ● ● ● The geometry of a compound microscope, which
photograph a man 1.7 m tall who is standing 4.0 m from consists of two converging lenses, is shown in
the camera. If the man’s image fills the height of a frame 䉲 Fig. 23.29. (More detail on microscopes is given in
of film (35 mm), what is the focal length of the lens? Chapter 25.) The objective lens and the eyepiece lens
53. ● ● To photograph a full moon, a photographer uses a
have focal lengths of 2.8 mm and 3.3 cm, respectively. If
single-lens camera with a focal length of 60 mm. What an object is located 3.0 mm from the objective lens,
will be the diameter of the Moon’s image on the film? where is the final image located, and what are the image
(Note: Data about the Moon are given inside the back characteristics?
cover of this text.) Eyepiece
Objective
54. ● ● An object 5.0 cm tall is 10 cm from a concave lens.
7.0 cm
The resulting virtual image is one-fifth as large as the
object. What is the focal length of the lens and the image Fo Fo Fe Fe
distance?
55. ● ● An object is placed 80 cm from a screen. (a) At what
point from the object should a converging lens with a focal

length of 20 cm be placed so that it will produce a sharp
image on the screen? (b) What is the image’s magnification? Final
image
56. ● ● (a) For a convex lens, what is the minimum distance
between an object and its image if the image is real? 䉱 F I G U R E 2 3 . 2 9 Compound microscope See Exercise 63.
(b) What is the minimum distance if the image is virtual?
57. ● ● Using 䉲 Fig. 23.28, derive (a) the thin lens equation 64. ●●● Two converging lenses L1 and L2 have focal lengths of
and (b) the magnification equation for a thin lens. [Hint: 30 cm and 20 cm, respectively. The lenses are placed 60 cm
Use similar triangles.] apart along the same axis, and an object is placed 50 cm
from L1 (110 cm from L2). Where is the image formed rela-
Object (di − f )
tive to L2, and what are its characteristics?
65. ● ● ● For a lens combination, show that the total magnifi-
ho ho
F cation Mtotal = M1 M2. [Hint: Think about the definition
hi
of magnification.]
66. ● ● ● Show that for thin lenses that have focal lengths f1
Image and f2 and are in contact, the effective focal length ( f) is
given by
f
1 1 1
do di = +
f f1 f2
䉱 F I G U R E 2 3 . 2 8 The thin lens equation The geometry for
deriving the thin lens equation and magnification factor. Note 23.4 THE LENS MAKER’S EQUATION
the two sets of similar triangles. See Exercise 57. AND
*23.5 LENS ABERRATIONS
58. ●● (a) If a book is held 30 cm from an eyeglass lens with
a focal length of -45 cm, where is the image of the print 67. ● An optometrist prescribes glasses with a power of
formed? (b) If an eyeglass lens with a focal length of -4.0 D for a nearsighted student. What is the focal
+ 57 cm is used, where is the image formed? length of the glass lenses?
68. ●A farsighted senior citizen needs glasses with a focal 72. IE ● ● A converging glass lens with an index of refraction
length of 45 cm. What is the power of the lens? of 1.62 has a focal length of 30 cm in air. (a) When the
69. IE ● ● A plastic convex meniscus (Fig. 23.14) contact lens lens is immersed in water, the focal length of the lens
is made of plastic with an index of refraction of 1.55. The will (1) increase, (2) remain the same, (3) decrease.
lens has a front radius of 2.50 cm and a back radius of Explain. (b) What is the focal length when the lens is
3.00 cm. (a) The signs of R1 and R2 are (1) + , + , (2) + , - , submerged in water?
(3) - , + , (4) - , - . Explain. (b) What is the focal length of 73. IE ● ● A biconvex lens is made of glass whose index of
the lens? refraction is 1.6. The lens has a radius of curvature of
70. ● ● A plastic plano-concave lens has a radius of curvature 30 cm for one surface and 40 cm for the other. Calculate
of 50 cm for its concave surface. If the index of refraction the focal length of this lens as used in air and under
of the plastic is 1.35, what is the power of the lens? water.
71. ● ● An optometrist prescribes a corrective lens with a 74. IE ● ● ● A lens made of fused quartz 1n = 1.462 has a
power of +1.5 D. The lens maker starts with a glass focal length of + 45 cm when it is in air. (a) If the lens is
blank that has an index of refraction of 1.6 and a convex immersed in oil (n = 1.50), the lens will (1) remain con-
front surface whose radius of curvature is 20 cm. To verging, (2) become diverging, (3) have an infinite focal
what radius of curvature should the other surface be length. Explain. (b) What is the focal length when it is
ground? Is the surface convex or concave? in oil?
75. (a) Use ray diagrams to show that a ray parallel to the reflected by the mirror back through the compound-lens
optic axis of a biconvex lens is refracted toward the axis system, and an image will be formed on a screen near
at the incident surface and again at the exit surface. the light source. This image is sharpened by adjusting
(b) Show that this also holds for a biconcave lens, but the distance between the diverging lens and the mirror.
with both refractions pointing away from the axis. The distance at which the image is clearest is equal to the
76. An object is 40 cm from a converging lens whose focal focal length of the lens. Explain why this method works.
length is 20 cm. On the opposite side of that lens, at a 78. For the arrangement shown in 䉲 Fig. 23.31, an object is
distance of 60 cm, is a plane mirror. Where is the final placed 0.40 m in front of the converging lens, which has
image measured from the lens, and what are its a focal length of 0.15 m. If the concave mirror has a focal
characteristics? length of 0.13 m, where is the final image formed, and
77. A method of determining the focal length of a diverging what are its characteristics?
lens is called autocollimation. As 䉲 Fig. 23.30 shows, first a
Object
sharp image of a light source is projected onto a screen
by a converging lens. Second, the screen is replaced with
a plane mirror. Third, a diverging lens is placed between
the converging lens and the mirror. Light will then be
Light
source 0.50 m
Screen
䉱 F I G U R E 2 3 . 3 1 Lens–mirror combination See Exercise 78.
(a) 79. A student is given a piece of glass 1n = 1.602 to make a

symmetrical biconvex lens so it can form an image with
a magnification of -0.0101 when the object is 5.00 m
Mirror from the lens. What should be the radius of curvature of
the lens surfaces?
80. Two lenses, each having a power of + 10 D, are placed
(b) 20 cm apart along the same axis. If an object is 60 cm
from the first lens (not in between the lenses), where is
Screen the final image relative to the first lens, and what are its
Mirror characteristics?
81. Certain people wear bifocal glasses to improve their near
and far vision. If a person can only see things clearly from
(c) a distance of 40.0 cm to 80.0 cm from his eyes without
glasses, what powers of bifocal glasses should she wear so
䉱 F I G U R E 2 3 . 3 0 Autocollimation See Exercise 77. she can see things clearly from 25.0 cm to infinity?
Physical Optics: The Wave
24 Nature of Light
24.1 Young’s double-slit
experiment (811)
■ condition for constructive
interference
■ condition for destructive
interference
24.2 Thin-film
interference (815)
■ 180° phase change
■ nonreflecting coating
24.3 Diffraction (819)

■ single slit diffraction
PHYSICS FACTS
■
24.4 Polarization (827)
■ Malus’ law
Brewster (polarization) angle
✦ The track-to-track distance on a DVD
(Digital Video Disc) is 0.74 mm , and it
is 1.6 mm on a CD (Compact Disc). In
comparison, the diameter of human
I t’s always intriguing to see bril-
liant colors produced by objects
that don’t have any colors of their
hair is about 100 mm . DVD and CD
tracks really split hairs. own. A glass prism, for example,
*24.5 Atmospheric Scattering ✦ AM radio can be heard better in
which is clear and transparent by
of Light (833) some areas than FM radio. This is
■ Rayleigh scattering
because the longer AM waves are itself, nevertheless gives rise to a
more easily diffracted (bent) around
buildings and other obstacles. whole array of colors when white
✦ Skylight is partially polarized. It is
believed that some insects, such as
light passes through it. Prisms, like
bees, use polarized skylight to deter- the water droplets that produce
mine navigational directions relative
to the Sun. rainbows, don’t create color. They
✦ To an observer on the Earth, the “red
merely separate the different colors
planet” Mars appears reddish
because surface material contains that make up white light.
iron oxide (iron rust).
✦ The explanation for the beautiful
The phenomena of reflection
color patterns of an open peacock tail and refraction are conveniently
is complicated. However, it is interfer-
ence that causes the beautiful colors, analyzed by using geometrical
provided that the feathers have black
pigment, which absorbs most of the optics (Chapter 22). Ray diagrams
incident light, allowing only the inter-
(Chapter 23) show what happens
ference colors to be seen.
24.1 YO U N G ’ S D O U B L E - S L I T E X P E R I M E N T 811
when light is reflected from a mirror or refracted through a lens. However, other
phenomena involving light, such as the colorful patterns of the peacock feathers
in the chapter-opening photograph, cannot be adequately explained or described
using the ray concept. They can only be explained with the wave theory of light.
The prominent wave phenomena are interference, diffraction, and polarization.
Physical optics, or wave optics, takes into account wave properties that geo-
metrical optics ignores. The wave theory of light leads to satisfactory explanations
of those phenomena that cannot be analyzed with rays. Thus, in this chapter, the
wave nature of light must be used to analyze phenomena such as interference and
diffraction.
Wave optics must be used to explain how light propagates around small
objects or through small openings. This is seen in everyday life with the narrow
track-to-track distances in CDs, DVDs, and other items. An object or opening is
considered small if it is on the order of magnitude of the wavelength of light.
24.1 Young’s Double-Slit Experiment

➥ What did Young’s double-slit experiment prove in terms of the nature of light?
➥ How can the wavelength of light be determined from Young’s double-slit
experiment?
➥ If the distance between the two slits increases in Young’s experiment, what happens
to the distance between the maxima an interference pattern?
It has been stated that light behaves like a wave, but no proof of this assertion has
been discussed. How would you go about demonstrating the wave nature of
light? One method that involves the use of interference was first devised in 1801
by the English scientist Thomas Young (1773–1829). Young’s double-slit
experiment not only demonstrates the wave nature of light, but also allows the
measurement of its wavelengths. Essentially, light can be shown to be a wave if it
exhibits wave properties such as interference and diffraction.
Recall from the discussion of wave interference in Sections 13.4 and 14.4 that
superimposed waves may interfere constructively or destructively. Constructive
interference occurs when two crests are superimposed. If a crest and a trough are
superimposed, then destructive interference occurs. Interference can be easily
observed with water waves, for which constructive and destructive interference
produce obvious interference patterns (䉴 Fig. 24.1).
The interference of (visible) light waves is not as easily observed, because of
their relatively short wavelengths 1 L10-7 m2 and the fact that they usually are not
monochromatic (single frequency). Also, stationary interference patterns are pro-
duced only with coherent sources—sources that produce light waves that have a
constant phase relationship to one another. For example, for constructive interfer-
ence to occur at some point, the waves meeting at that point must be in phase. As 䉱 F I G U R E 2 4 . 1 Water wave inter-
the waves meet, a crest must always overlap a crest, and a trough must always ference The interference of water
overlap a trough. If a phase difference develops between the waves over time, the waves from two coherent sources in
a ripple tank produces patterns of
interference pattern changes, and a stable or stationary pattern will not be constructive and destructive
established. interferences (bright and dark
In an ordinary light source, the atoms are excited randomly, and the emitted regions).
light waves fluctuate in amplitude and frequency. Thus, light from two such
sources is incoherent and cannot produce a stationary interference pattern. Interfer-
ence does occur, but the phase difference between the interfering waves changes
so fast that the interference effects are not discernible.
812 24 PHYSICAL OPTICS: THE WAVE NATURE OF LIGHT
䉴 F I G U R E 2 4 . 2 Double-slit Intensity
interference (a) The coherent waves
from two slits are shown in blue (top
slit) and red (bottom slit). The waves Max n=2
spread out as a result of diffraction (n = 2)
from narrow slits. The waves inter- Min
fere, producing alternating maxima
and minima, on the screen. (b) An Max n=1
Light (n = 1)
interference pattern. Note the sym- S1
source Min
metry of the pattern about the cen-
tral maximum 1n = 02. Max n=0
(n = 0)
S2 Min
Max n=1
Single (n = 1)
slit Double Min
slit
Max n=2
(n = 2)
(b)
L Screen
(a)
To obtain the equivalency of two coherent sources, a barrier with one narrow
slit is placed in front of a single light source, and a barrier with two very narrow
slits is positioned symmetrically in front of the first barrier (䉱 Fig. 24.2a).
Waves propagating out from the single slit are in phase, and the double slits
then act as two coherent sources by separating one single wave into two parts.
Any random changes in the light from the original source will thus occur for both
sources passing through the two slits, and the phase difference will be constant.
The modern laser beam, a coherent light source, makes the observation of a stable
interference pattern much easier. A series of maxima or bright positions can be
observed on a screen placed relatively far from the slits (Fig. 24.2b).
To help analyze Young’s experiment, imagine that light with a single wave-
length (monochromatic light) is used. Because of diffraction (see Sections 13.4 and
14.4 and, in this chapter, Section 24.3), or the spreading of light as it passes
through a slit, the waves spread out and interfere as illustrated in Fig. 24.2a. Com-
ing from two coherent “sources,” the interfering waves produce a stable interfer-
ence pattern on the screen. The pattern consists of a bright central maximum
(䉲 Fig. 24.3a) and a series of symmetrical side minima (Fig. 24.3b) and maxima
Maximum
Central maximum Minimum (constructive
(constructive (destructive interference)
interference) interference)
S1
θ θ
d d θ d θ
S2 ␭
␭
Screen 2 Screen Screen
L L L
(a) (b) (c)
䉱 F I G U R E 2 4 . 3 Interference The interference that produces a maximum or minimum

depends on the difference in the path lengths of the light from the two slits. (a) The path
length difference at the position of the central maximum is zero, so the waves arrive in
phase and interfere constructively. (b) At the position of the first side minimum, the path
length difference is l>2, and the waves interfere destructively. (c) At the position of the first
side maximum, the path length difference is l, and the interference is constructive.
24.1 YO U N G ’ S D O U B L E - S L I T E X P E R I M E N T 813
P 䉳 F I G U R E 2 4 . 4 Geometry of
Young’s double-slit experiment The
r1 difference in the path lengths for
light traveling from the two slits to a
point P is r2 - r1 = ¢L, which
y forms a side of the small shaded tri-
S1
r2 angle. Because the barrier with the
θ slits is parallel to the screen, the
θ angle between r2 and the barrier (at
d S2 , in the small shaded triangle) is
Central max equal to the angle between r2 and the
screen. When L is much greater than
ΔL = d sin θ y, that angle is almost identical to the
S2 L angle between the screen and the
dashed line, which is an angle in the
Screen
large shaded triangle. The two
shaded triangles are then almost
similar, and the angle at S1 in the
(Fig. 24.3c), which mark the positions at which destructive and constructive inter- small triangle is almost exactly equal
to u. Thus, ¢L = d sin u. (Not drawn
ference occur. The existence of this interference pattern clearly demonstrates the to scale. Assume that d V L.)
wave nature of light. The intensities of each side maximum decrease with distance
from the central maximum.
Measuring the wavelength of light requires the use of geometry in Young’s
experiment, as shown in 䉱 Fig. 24.4. Let the screen be a distance L from the slits and
P be an arbitrary point on the screen. P is located a distance y from the center of the
central maximum and at an angle u relative to a normal line between the slits. The
slits S1 and S2 are separated by a distance d. Note that the light path from slit S2 to P
is longer than the path from slit S1 to P. As the figure shows, the path length differ-
ence 1¢L2 is approximately
¢L = d sin u
The fact that the angle in the small shaded triangle is almost equal to u can be
shown by a simple geometrical argument involving similar triangles when d V L,
as described in the caption of Fig. 24.4.
The relationship of the phase difference of two waves to their path length differ-
ence was discussed in Section 14.4 for sound waves. These conditions hold for any
wave, including light. Constructive interference occurs at any point where the path
length difference between the two waves is an integral number of wavelengths:
(condition for
¢L = nl for n = 0, 1, 2, 3, Á (24.1)
constructive interference)
Similarly, for destructive interference, the path length difference is an odd number
of half-wavelengths:
ml (condition for
¢L = for m = 1, 3, 5, Á (24.2)
2 destructive interference)
Thus, in Fig. 24.4, the maxima (constructive interference) satisfy
(condition for
d sin u = nl for n = 0, 1, 2, 3, Á (24.3)
interference maxima)
where n is called the order number. The zeroth order 1n = 02 corresponds to the
central maximum; the first order 1n = 12 is the first maximum on either side of the
central maximum; and so on. As the path length difference varies from point to
point, so does the phase difference and the resulting type of interference (con-
structive or destructive).
The wavelength can therefore be determined by measuring d and u for a maxi-
mum of a particular order (other than the central maximum), because Eq. 24.3 can
be solved as l = 1d sin u2>n.
The angle u locates a side maximum relative to the central maximum. This can
be measured from a photograph of the interference pattern, such as shown in
Fig. 24.2b. If u is small 1y V L2, sin u L tan u = y>L.* Substituting this y>L for
sin u into Eq. 24.3 and solving for y gives a good approximation of the distance of
the nth maximum (yn) from the central maximum on either side:
nLl (lateral distance to

yn L for n = 0, 1, 2, 3, Á (24.4)
d maxima for small u only)
A similar analysis gives the locations of the minima. (See Exercise 6a.)
From Eq. 24.3, it can be seen that, except for the zeroth order, n = 0 (the central
maximum), the positions of the maxima depend on wavelength—that is, different
wavelengths 1l2 give different values of u and y. Hence, when white light is used,
the central maximum is white because all wavelengths are at the same location,
but the other orders become a “spread out” spectrum of colors. Because y is pro-
portional to l 1y r l2, in a given order, red is farther out than blue (red has a
longer wavelength than blue).
By measuring the positions of the color maxima within a particular order,
Young was able to determine the wavelengths of the colors of visible light. Note
also that the size or “spread” of the interference pattern, yn, depends inversely on
the slit separation d. The smaller the slit separation d, the more spread out the pat-
tern. For large d, the interference pattern is so compressed that it appears to us as a
single white spot (all maxima together at the center).
In this analysis, the word destructive does not imply that energy is destroyed.
Destructive interference means that light energy is not present at a particular loca-
tion (minima). By energy conservation, the light energy must be somewhere else
(maxima). This is observed with sound waves as well.
INTEGRATED EXAMPLE 24.1 Measuring the Wavelength of Light:Young’s Double-Slit Experiment

In a lab experiment similar to the one shown in Fig. 24.4, ( A ) C O N C E P T U A L R E A S O N I N G . According to the condition for
monochromatic light (having only one wavelength or fre- constructive interference, d sin u = nl, the product of d and
quency) passes through two narrow slits that are 0.050 mm sin u is a constant, for a given wavelength l and order num-
apart. The interference pattern is observed on a white wall ber n. Therefore, if d decreases, sin u increases, as does u. Thus
1.0 m from the slits, and the second-order maximum is at an the answer is (1).
angle of u2 = 1.5°. (a) If the slit separation decreases, the
(B) AND (C) QUANTITATIVE REASONING AND SOLUTION.
second-order maximum will be seen at an angle of (1) greater
Eq. 24.3 can be used to find the wavelength. Since L W d, that
than 1.5°, (2) 1.5°, (3) less than 1.5°. Explain. (b) What is the
is, 1.0 m W 0.050 mm, u is small. We could compute y2 and y3
wavelength of the light and what is the distance between the
from Eq. 24.4 and determine the distance between the second-
order and third-order maxima 1y3 - y22. However, the maxima
second-order and third-order maxima? (c) If d = 0.040 mm,
what is u2?
for a given wavelength of light are evenly spaced (for a small u).
That is, the distance between adjacent maxima is a constant.
Given: L = 1.0 m Find: (b) l (wavelength) and y3 - y2 (distance between n = 2 and n = 3)

n = 3 (c) u2 (for d = 0.040 mm)
d = 0.050 mm = 5.0 * 10-5 m
(b) u2 = 1.5°
d = 5.0 * 10-5 m
(c) d = 4.0 * 10-5 m
(b) Using Eq. 24.3,
d sin u 15.0 * 10-5 m2 sin 1.5°
l = = = 6.5 * 10-7 m = 650 nm
n 2
This value is 650 nm, which is the wavelength of orange-red light (see Fig. 20.23). From Eq. 24.4, and a general approach for n
and n + 1,
1n + 12Ll nLl Ll
yn + 1 - yn = - =
d d d
*For u = 10° , the percentage difference between sin u and tan u is only 1.5%.
24.2 THIN FILM INTERFERENCE 815
In this case, the distance between successive maxima is

Ll 11.0 m216.5 * 10-7 m2
y3 - y2 = = = 1.3 * 10-2 m = 1.3 cm
d 5.0 * 10-5 m
(c)
1221650 * 10-9 m2
= 0.0325 so u2 = sin-110.03252 = 1.9° 7 1.5°
nl
sin u2 =
14.0 * 10-5 m2
=
d
As is reasoned in (a), u2 is indeed greater than 1.5°.
F O L L O W - U P E X E R C I S E . Suppose white light were used instead of monochromatic light in this Example. What would be the
separation distance of the red 1l = 700 nm2 and blue 1l = 400 nm2 components in the second-order maximum? (Answers to all
Follow-Up Exercises are given in Appendix VI at the back of the book.)
DID YOU LEARN?

➥ Young’s double-slit experiment proved that light behaves like a wave.Young used
wave interference to successfully explain the observed interference pattern.
➥ The wavelength of light can be determined by measuring the angles or positions of
the maxima on an interference pattern.
➥ The distance between the maxima on an interference pattern will decrease if the
distance between the two slits increases.
24.2 Thin Film Inter ference

➥ Under what condition is there a 180° phase change (half wave shift)?
➥ How can soap bubbles produce colorful displays?
➥ What is the purpose of a nonreflective or antireflecting coating on lenses?
Have you ever wondered what causes the rainbowlike colors that occur when
䉲 F I G U R E 2 4 . 5 Reflection and
white light is reflected from a thin film of oil or a soap bubble? This effect—known phase shifts The phase changes that
as thin film interference—is a result of the interference of light reflected from oppo- light waves undergo on reflection are
site surfaces of the film and may be understood in terms of wave interference. analogous to those for pulses in
First, however, you need to know how the phase of a light wave is affected by strings. (a) The phase of a pulse in a
reflection. Recall from Section 13.4 that a wave pulse on a rope undergoes a 180° string is shifted by 180° on reflection
phase change [or a half wave shift 1l>22] when reflected from a rigid support and
from a fixed end, and so is the phase
of a light wave when it is reflected
no phase shift when reflected from a free support (䉲 Fig. 24.5). Similarly, as the from a medium of higher index of
figure shows, the phase change for the reflection of light waves at a boundary refraction. (b) A pulse in a string has
depends on the indices of refraction (n) of the two materials:* a phase shift of zero (it is not shifted)
when reflected from a free end. Anal-
■ A light wave undergoes a 180° phase change on reflection if n1 6 n2. ogously, a light wave is not phase
shifted when reflected from a
■ There is no phase change on reflection if n1 7 n2. medium of lower index of refraction.
Incident Incident
pulse pulse
Reflected
pulse
Reflected n1 n2 n1 n2
pulse
n1 < n2 n1 > n2
(a) Fixed end: 180° phase shift (b) Free end: zero phase shift
*The refracted wave does not shift in phase.

Incoming Sees no Incoming Sees light

light light light
1
1
2
2
180°
Air no shift
180°
shift
λ
t = λ'/2
Oil n1 No shift λ
t = λ'/4
No shift
Water n2 < n1
(a) (b) (c)
䉱 F I G U R E 2 4 . 6 Thin film interference For an oil film on water, there is a 180° phase shift
for light reflected from the air–oil interface and a zero phase shift at the oil–water interface.
l¿ is the wavelength in the oil. (a) Destructive interference occurs if the oil film has a mini-
mum thickness of l¿>2 for normal incidence. (Waves are displaced and angled for clarity.)
(b) Constructive interference occurs with a minimum film thickness of l¿>4. (c) Thin film
interference in an oil slick. Different film thicknesses give rise to the reflections of different
colors.
To understand why you see colors from a soap bubble or an oil film (for example,
floating on water or on a wet road), consider the reflection of monochromatic light
from a thin film in 䉱 Fig. 24.6. The path length of the wave in the film depends on the
angle of incidence (why?), but for simplicity, normal (perpendicular) incidence is
assumed, even though the rays are drawn at an angle in the figure for clarity.
The oil film has a greater index of refraction than that of air, and the light
reflected from the air–oil interface (wave 1 in the figure) undergoes a 180° phase
shift. The transmitted waves pass through the oil film and are reflected at the
oil–water interface. In general, the index of refraction of oil is greater than that of
(a) water (see Table 22.1)—that is, n1 7 n2—so a reflected wave in this instance
(wave 2) does not undergo a phase shift.
You might think that if the path length difference of the waves in the oil film
(2t, twice the thickness—down and back up) were an integral number of wave-
lengths—for example, if 2t = 21l¿>22 = l¿ in Fig. 24.6a, where l¿ = l>n is the
wavelength in the oil—then the waves reflected from the two surfaces would
interfere constructively. But keep in mind that the wave reflected from the top
surface (wave 1) undergoes a 180° phase shift. The reflected waves from the two
surfaces are therefore actually out of phase and would interfere destructively for
this condition. This means that no reflected light for this wavelength would be
observed. (The light would be transmitted.)
Similarly, if the path length difference of the waves in the film were an odd
number of half-wavelengths 32t = 21l¿>42 = l¿>24 in Fig. 24.6b, again where l¿ is
the wavelength in the oil, then the reflected waves would actually be in phase (as a
result of the 180° phase shift of wave 1) and would interfere constructively.
Reflected light for this wavelength would be observed from above the oil film.
(b)
Because oil films generally have different thicknesses in different locations, par-
䉱 F I G U R E 2 4 . 7 Thin film interfer- ticular wavelengths (colors) of white light interfere constructively in different
ence (a) A thin air film between locations after reflection. As a result, a vivid display of various colors appears (Fig.
microscope slides gives colorful pat- 24.6c). Thin film interference may be seen when two glass slides are stuck together
terns. (b) Multilayer interference in with an air film between them (䉳 Fig. 24.7a). The bright colors of a peacock’s tail,
a peacock’s feathers gives rise to
an example of colorful interference in nature, are a result of layers of fibers in its
bright colors. The brilliant throat
colors of hummingbirds are pro- feathers, which mostly lack pigments (chapter–opening photograph). Light
duced in the same way. reflected from successive layers interferes constructively, giving bright colors.
24.2 THIN FILM INTERFERENCE 817
Since the condition for constructive interference depends on the angle of inci-
dence, the color pattern changes somewhat with the viewing angle and motion of 1
2
the bird (Fig. 24.7b).
A soap bubble in the air is slightly thicker near the bottom than on the top, due Air no
to the Earth’s gravitational force. The gradual increase in thickness from the top to
the bottom of the bubble causes the constructive interferences of different colors. Thin film n1 > no t
A practical application of thin film interference is nonreflective coatings for
lenses. (See Insight 24.1, Nonreflecting Lenses.) In this situation, a film coating is
used to create destructive interference between the reflected waves so as to Glass n2 > n1
increase the light transmission into the glass lens (䉴 Fig. 24.8). The index of refraction lens
of the film has a value between that of air and glass 1no 6 n1 6 n22. Conse-
quently, phase shifts of incident light take place at both surfaces of the film.
䉱 F I G U R E 2 4 . 8 Thin film interfer-
In such a case, the condition for constructive interference of the reflected light is ence For a thin film on a glass lens,
there is a 180° phase shift at each
ml¿ ml (condition for
interface when the index of refrac-
¢L = 2t = ml¿ or t = = m = 1, 2, Á constructive interference (24.5)
2 2n1 tion of the film is less than that of
when no 6 n1 6 n2) the glass. The waves reflected off
and the condition for destructive interference is the top and bottom surfaces of the
film interfere. For clarity, the angle
(condition for of incidence is drawn to be large,
ml¿ ml¿ ml but, in reality, it is almost zero.
¢L = 2t = or t = = m = 1, 3, 5, Á destructive interference (24.6)
2 4 4n1
when no 6 n1 6 n2)
The minimum film thickness for destructive interference occurs when m = 1, so
l (minimum film thickness

tmin = (24.7)
4n1 for no 6 n1 6 n2)
INSIGHT 24.1 Nonreflecting Lenses

You may have noticed the blue-purple tint of the coated opti-
cal lenses used in cameras and binoculars (Fig. 1). The coating
makes the lenses almost “nonreflecting” for all the colors you
do not see in the reflected light. The incident light is mostly
transmitted through the lens. Maximum transmission of light
is desirable for all optical instruments, especially under low
light conditions.
For a typical air–glass interface, about 4% of the light is
reflected and 96% is transmitted. A camera lens is actually
made up of a group of lenses (elements) in order to minimize
aberrations and improve image quality. For instance, a
35–70-mm zoom lens might consist of up to thirteen elements,
thus having twenty-six reflective surfaces. 䉱 F I G U R E 1 Coated lenses The nonreflective coating on
After one reflection, 0.96 = 96% of the light is transmitted. binocular and camera lenses generally produces a characteris-
After two reflections, or one lens, the transmitted light is only tic bluish-purple hue (Why?).
0.96 * 0.96 = 0.962 = 0.92, or 92%, of the incident light. Thus
after twenty-six reflections, the transmitted light is only
0.9626 = 0.35, or 35%, of the incident light, if the lenses are not light. The thickness is usually chosen to be a quarter-wave-
coated. Therefore, almost all camera and binocular lenses are length of yellow-green light 1l L 550 nm2, to which the
coated with nonreflecting film. human eye is most sensitive. The wavelengths at the red and
A lens is made nonreflecting by coating it with a thin film blue ends of the visible region are still partially reflected, giv-
material with an index of refraction between the indices of ing the coated lens its bluish-purple tint. Sometimes other
refraction of air and glass (Fig. 24.8). If the coating is a quarter- quarter-wavelength thicknesses are chosen, giving rise to
wavelength l¿>4 thick, the difference in path length between other hues, such as amber or reddish-purple, depending on
the reflected rays is l¿>2, where l¿ is the wavelength of light in the application of the lens.
the coating. In this case, both reflected waves undergo a phase Nonreflective coatings are also applied to the surfaces of
shift, and therefore they are out of phase for a path length dif- solar cells, which convert light into electrical energy
ference of l¿>2 and interfere destructively. That is, the incident (Section 27.2). Because the thickness of such a coating is
light is transmitted, and the coated lens is nonreflecting. wavelength dependent, the overall losses due to reflection
Note that the actual thickness of a quarter-wavelength can be decreased from around 30% to only 10%. Even so, the
thickness of film is specific to the particular wavelength of process improves the cell’s efficiency.
If the index of refraction of the film is greater than that of air and glass, then
only the reflection at the air–film interface has the 180° phase shift. Therefore,
2t = ml¿ will actually create destructive interference, and 2t = ml¿>2 will create
constructive interference. (Why?)
EXAMPLE 24.2 Nonreflective Coatings: Thin Film Interference

A glass lens 1n = 1.602 is coated with a thin, transparent film of magnesium fluoride
1n = 1.382 to make the lens nonreflecting. (a) What is the minimum film thickness so
that the lens will not reflect normally incident light of wavelength 550 nm? (b) Will a
film thickness of 996 nm make the lens nonreflecting?
T H I N K I N G I T T H R O U G H . (a) Equation 24.7 can be used directly to get an idea of the
minimum film thickness needed for a nonreflective coating. (b) We need to determine
whether 996 nm satisfies the condition in Eq. 24.6.
SOLUTION.
Given: no = 1.00 1air2 Find: (a) tmin (minimum film thickness)
n1 = 1.38 1film2 (b) whether t = 996 nm gives a
n2 = 1.60 1lens2 nonreflecting lens
l = 550 nm
(a) Because n2 7 n1 7 no ,
l 550 nm
tmin = = = 99.6 nm
4n1 411.382
which is quite thin 1L10-5 cm2. In terms of atoms, which have diameters on the order
1 2
of 10-10 m, or 10-1 nm, the film is about 1000 atoms thick.
l l
(b) t = 996 nm = 10199.6 nm2 = 10tmin = 10 ¢ ≤ = 5¢ ≤
4n1 2n1
t Air wedge This means that this film thickness does not satisfy the nonreflective condition (destruc-
O tive interference). Actually, it satisfies the requirement for constructive interference
(Eq. 24.5) with m = 5. Such a coating specific for infrared radiation could be useful in
(a) hot climates on car and house windows, because it maximizes reflection and minimizes
transmission.
Bright F O L L O W - U P E X E R C I S E . What would be the minimum film thickness for the glass lens
band
in this Example to reflect, rather than transmit, the incident light through the lens?
Dark OPTICAL FLATS AND NEWTON’S RINGS

band
The phenomenon of thin film interference can be used to check the smoothness
and uniformity of optical components such as mirrors and lenses. Optical flats are
made by grinding and polishing glass plates until they are as flat and smooth as
O
possible. (The surface roughness is usually on the order of l>20.) The degree of
(b) flatness can be checked by putting two such plates together at a slight angle so
that a very thin air wedge is between them (䉳 Fig. 24.9a).
䉱 F I G U R E 2 4 . 9 Optical flatness The reflected waves off the bottom of the top plate (wave 1) and top of the bot-
(a) An optical flat is used to check tom plate (wave 2) interfere. Note that wave 2 has a 180° phase shift as it is
the smoothness of a reflecting sur- reflected from an air–plate interface, whereas wave 1 does not. Therefore, at cer-
face. The flat is placed so that there tain points from where the plates touch (point O), the condition for constructive
is a small air wedge between it and
interference is 2t = ml>2 1m = 1, 3, 5, Á 2, and the condition for destructive inter-
ference is 2t = ml 1m = 0, 1, 2, Á 2. The thickness t determines the type of inter-
the surface. The waves reflected
from the two plates interfere, and
the thickness of the air wedge at cer- ference (constructive or destructive). If the plates are smooth and flat, a regular
tain points determines whether interference pattern of bright and dark bands appears (Fig. 24.9b). This pattern is a
bright or dark bands are seen. (b) If result of the uniformly varying differences in path lengths between the plates.
the surfaces are smooth, a regular or
Any irregularity in the pattern indicates an irregularity in at least one plate. Once
symmetrical interference pattern is
seen. Note that a dark band is at a good optical flat is verified, it can be used to check the flatness of a reflecting sur-
point O where t = 0. face, such as that of a precision mirror.
24.3 DIFFRACTION 819
Light source 䉳 F I G U R E 2 4 . 1 0 Newton’s rings

(a) A lens placed on an optical flat
forms a ring-shaped air wedge,
Eye
which gives rise to interference of
the waves reflected from the top
(wave 1) and the bottom (wave 2) of
the air wedge. (b) The resulting
1 2 interference pattern is a set of con-
centric rings called Newton’s rings.
Note that at the center of the pattern
is a dark spot. Lens irregularities
Lens produce a distorted pattern.
t
Optical flat
(a) (b)
Direct evidence of the 180° phase shift can be clearly seen in Fig. 24.9b. At the
point where the two plates touch 1t = 02, we see a dark band. If there were no
phase shift, t = 0 would correspond to ¢L = 0, and a bright band would appear.
The fact that it is a dark band proves that there is a phase shift in reflection from a
material of higher index of refraction.
A similar technique is used to check the smoothness and symmetry of lenses.
When a curved lens is placed on an optical flat, a radially symmetric air wedge is
formed between the lens and the optical flat (䉱 Fig. 24.10a). Since the thickness of
the air wedge again determines the condition for constructive and destructive
interference, the regular interference pattern in this case is a set of concentric circu-
lar bright and dark rings (Fig. 24.10b). They are called Newton’s rings, after Isaac
Newton, who first described this interference effect. Note that at the point where
the lens and the optical flat touch 1t = 02, there is, once again, a dark spot. (Why?)
Lens irregularities give rise to a distorted fringe pattern, and the radii of these
rings can be used to calculate the radius of curvature of the lens.
DID YOU LEARN?

➥ There is a 180° phase shift in the reflected light when light travels from a medium
with a lower index of refraction to a medium with a higher index of refraction
(n1 6 n2).
➥ Due to the Earth’s gravitational force, a soap bubble in the air is slightly thicker on
the bottom than on the top. Different thicknesses satisfy the condition of
constructive interference for different wavelengths (color).
➥ The nonreflective coating on lenses causes destructive interference in the reflected
light so as to increase the transmitted light into the lenses.
24.3 Diffraction
➥ How does the size of an opening or obstacle affect optical diffraction?

➥ If the width of a single slit decreases, what happens to the width of the central maxi-
mum on a diffraction pattern?
➥ A beam of light has the colors red and blue in it.When it passes through a diffraction
grating, which color of the first order on a diffraction pattern is closer to the central
maximum?
In geometrical optics, light is represented by rays and pictured as traveling in

straight lines. If this model were to represent the real nature of light, however,
there would be no interference effects in Young’s double-slit experiment. Instead,
there would be only two bright images of slits on the screen, with a well-defined
shadow area where no light enters. But we do see interference patterns, which
means that the light must deviate from a straight-line path and enter the regions
that would otherwise be in shadow. The waves actually “spread out” as they pass
through the slits. This spreading is called diffraction. Diffraction generally occurs
when waves pass through small openings or around sharp edges or corners. The
diffraction of water waves is shown in 䉳 Fig. 24.11. (See also Fig. 13.18.)
As Fig. 13.18 shows, the amount of diffraction depends on the wavelength in
relation to the size of the opening or object. In general, the longer the wavelength
compared to the width of the opening or object, the greater the diffraction. This effect is
also shown in 䉲 Fig. 24.12. For example, in Fig. 24.12a, the width of the opening w is
much greater than the wavelength 1w W l2, and there is little diffraction—the wave
keeps traveling without much spreading. (There is some degree of diffraction around
the edges of the opening.) In Fig. 24.12b, with the wavelength and opening width on
䉱 F I G U R E 2 4 . 1 1 Water wave the same order of magnitude 1w L l2, there is noticeable diffraction—the wave
diffraction This photograph of a spreads out and deviates from its original direction. Part of the wave keeps traveling
beach dramatically shows single-slit in its original direction, but the rest bends around the opening and clearly spreads out.
diffraction of ocean waves through
The diffraction of sound is quite evident (Section 14.4). Someone can talk to you
the barrier opening. Note that the
beach has been shaped by the circu- from another room or around the corner of a building, and even in the absence of
lar wave front. reflections, you can easily hear the person. Recall that audible sound wavelengths
are on the order of centimeters to meters. Thus, the widths of ordinary objects and
openings are about the same as or narrower than the wavelengths of sound, and
diffraction will readily occur under these conditions.
Visible light waves, however, have wavelengths on the order of 10-7 m. Therefore,
diffraction phenomena for these waves often go unnoticed, especially through large
openings such as doors where sound readily diffracts. However, close inspection of
the area around a sharp razor blade will show a pattern of bright and dark bands
(䉴 Fig. 24.13). Diffraction can lead to interference, and thus these interference patterns
are evidence of the diffraction of the light around the edge of the blade.
As an illustration of “single-slit” diffraction, consider a slit in a barrier
(䉴 Fig. 24.14). Suppose that the slit (width w) is illuminated with monochromatic
light. A diffraction pattern consisting of a bright central maximum and a symmet-
rical array of side maxima (regions of constructive interference) on both sides is
observed on a screen at a distance L from the slit (we will assume L W w).
Thus a diffraction pattern results from the fact that various points on the wave
front passing through the slit can be considered to be small point sources of light. The
interference of those waves gives rise to the diffraction maxima and minima.
(a) (b)
䉱 F I G U R E 2 4 . 1 2 Wavelength and opening dimensions In general, the narrower the

opening compared to the wavelength, the greater the diffraction. (a) Without much diffrac-
tion 1w W l2, the wave would keep traveling in its original direction. (b) With noticeable
diffraction 1w L l2, the wave bends around the opening and spreads out.
Intensity
m=3
m=2
(a)
m=1
Monochromatic
light θ
w
m=1
m=2
Single
slit m=3
L
L >> w
(b)
Physical
boundary Screen
䉱 F I G U R E 2 4 . 1 3 Diffraction in action 䉱 F I G U R E 2 4 . 1 4 Single-slit diffraction The diffraction of light by a single slit

(a) Diffraction patterns produced by a gives rise to a diffraction pattern consisting of a wide and bright central maximum
razor blade. (b) A close-up view of the dif- and a symmetric array of side maxima. The order number m corresponds to the
fraction pattern formed at the edge of the minima or dark positions. (See text for description.)
blade.
The fairly complex analysis is not done here; however, from geometry, it can be
proven that the minima (regions of destructive interference) satisfy the relationship
w sin u = ml for m = 1, 2, 3, Á (condition for diffraction minima) (24.8)
where u is the angle of a particular minimum, designated by m = 1, 2, 3, Á , on

either side of the central maximum and m is called the order number. (There is no
m = 0. Why?)
Although this result is similar in form to that for Young’s double-slit experiment
(Eq. 24.3), it is extremely important to realize that for the single-slit experiment, dif-
fraction minima, rather than interference maxima, are analyzed. Also, note that the
width of the slit (w) is used in diffraction. Physically, this is diffraction from a single
slit, not interference from two slits.
The small-angle approximation, sin u L tan u = y>L, can be used when y V L.
In this case, the distances of the minima relative to the center of the central maxi-
mum are given by
b
Ll
ym = ma for m = 1, 2, 3, Á (location for diffraction minima) (24.9)
w
The qualitative predictions from Eq. 24.9 are interesting and instructive:
■ For a given slit width (w), the longer the wavelength 1l2, the wider (more
“spread out”) the diffraction pattern.
■ For a given wavelength 1l2, the narrower the slit width (w), the wider the dif-
fraction pattern.
■ The width of the central maximum is twice the width of any side maximum.
Let’s look in detail at these results. As the slit is made narrower, the central
maximum and the side maxima spread out and become wider. Equation 24.9 is
not applicable to very small slit widths (because of the small-angle approxima-
tion). If the slit width is decreased until it is of the same order of magnitude as the
wavelength of the light, then the central maximum spreads out over the whole
screen. That is, diffraction becomes dramatically evident when the width of the
slit is about the same as the wavelength of the light used. Diffraction effects are
most easily observed when l>w L 1, or w L l.
Conversely, if the slit is made wider for a given wavelength, then the diffraction
pattern becomes less spread out. The maxima move closer together and eventu-
ally become difficult to distinguish when w is much wider than l 1w W l2. The
pattern then appears as a fuzzy shadow around the central maximum, which is
the illuminated image of the slit. This type of pattern is observed for the image
produced by sunlight entering a dark room through a hole in a curtain. Such an
observation led early experimenters to investigate the wave nature of light. The
acceptance of this concept was due, in large part, to the explanation of diffraction
offered by physical optics.
The central maximum is twice as wide as any of the side maxima. The width of
the central maximum is simply the distance between the first minima on each side
1m = 12, or a value of 2y1. From Eq. 24.9, y1 = Ll>w, so
2Ll
2y1 = (width of central maximum) (24.10)
w
Similarly, the width of the side maxima is given by
ym + 1 - ym = 1m + 12a b - ma b =
Ll Ll Ll
= y1 (24.11)
w w w
Thus, the width of the central maximum is twice that of the side maxima.
CONCEPTUAL EXAMPLE 24.3 Diffraction and Radio Reception

When you drive through a city or mountainous areas, the However, because of diffraction, radio waves can also
quality of your radio reception varies sharply from place to “wrap around” obstacles or “fan out” as they pass by obsta-
place, with stations seeming to fade out and reappear. Could cles and through openings, provided that their wavelength is
diffraction be a cause of this phenomenon? Which of the fol- at least roughly the size of the obstacle or opening. The longer
lowing frequency bands would you expect to be least affected: the wavelength, the greater the amount of diffraction, and so
(a) weather (162 MHz); (b) FM (88–108 MHz); or (c) AM the less likely the radio waves are to be obstructed.
(525–1610 kHz)? To determine which band benefits most by such diffrac-
tion, we need the wavelengths that correspond to the given
REASONING AND ANSWER. Radio waves, like visible light,
frequencies, as given by c = lf. AM radio waves, with
are electromagnetic waves and so tend to travel in straight
l = 186 - 571 m, are the longest of the three bands (by a fac-
lines when they are long distances from their sources. They
tor of about 100). Thus AM broadcasts are more likely to be
can be blocked by objects in their path—especially if the
diffracted around such objects as buildings or mountains or
objects are massive (such as hills and buildings).
through the openings between them, and the answer is (c).
F O L L O W - U P E X E R C I S E . Woodwind instruments, such as the clarinet and the flute, usually have smaller openings than brass
instruments, such as the trumpet and trombone. During halftime at a football game, when a marching band faces you, you can
easily hear both the woodwind instruments and the brass instruments. Yet when the band marches away from you, the brass
instruments sound muted, but you can hear the woodwinds quite well. Why?
INTEGRATED EXAMPLE 24.4 Single-Slit Diffraction: Wavelength and Central Maximum

Monochromatic light passes through a slit whose width is (A) CONCEPTUAL REASONING. The general size of the diffrac-
0.050 mm. (a) The resulting diffraction pattern is generally tion pattern can be characterized by the position and width of
(1) wider for longer wavelengths, (2) wider for shorter wave- a particular maximum or minimum. From Eq. 24.8, it can be
lengths, (3) the same width for all wavelengths. Explain. seen that for a given width w and order number m, the posi-
(b) At what angle will the third order minimum be seen and tion of a minimum 1sin u2 is directly proportional to the
what is the width of the central maximum on a screen located wavelength l. Therefore, a longer wavelength will correspond
1.0 m from the slit, for l = 400 nm and 550 nm, respectively? to a greater sin u or a greater u, and the answer is (1).
(B) QUANTITATIVE REASONING AND SOLUTION. This part is a direct application of Eq. 24.8 and Eq. 24.10.
Given: l1 = 400 nm = 4.00 * 10-7 m Find: u3 and 2y1 (width of central maximum)
l2 = 550 nm = 5.50 * 10-7 m
w = 0.050 mm = 5.0 * 10-5 m
m = 3
L = 1.0 m
For l = 400 nm:

From Eq. 24.8,
ml 314.00 * 10-7 m2
sin u3 = = = 0.024 so u3 = sin-1 0.024 = 1.4°
w 5.0 * 10-5 m
Equation 24.10 gives
2Ll 211.0 m214.00 * 10-7 m2
2y1 = = = 1.6 * 10-2 m = 1.6 cm
w 5.0 * 10-5 m
For l = 550 nm:
ml 315.50 * 10-7 m2
sin u3 = = = 0.033 so u3 = sin-1 0.033 = 1.9°
w 5.0 * 10-5 m
2Ll 211.0 m215.50 * 10-7 m2
2y1 = = = 2.2 * 10-2 m = 2.2 cm
w 5.0 * 10-5 m
As is reasoned in (a), the diffraction pattern for 550 nm is wider (larger u3 and 2y1).
FOLLOW-UP EXERCISE. By what factor would the width of the central maximum change if red light 1l = 700 nm2 were used
instead of light with l = 550 nm?
DIFFRACTION GRATINGS
We have seen that maxima and minima result from diffraction followed by interfer-
ence when monochromatic light passes through a set of double slits. As the number
of slits is increased, the maxima become sharper (narrower) and the minima become
wider. The sharp maxima are very useful in optical analysis of light sources and
other applications. 䉲 Fig. 24.15 shows a typical experiment with monochromatic
light incident on a diffraction grating, which consists of large numbers of parallel,
closely spaced slits. Two parameters define a diffraction grating: the slit separa-
tion between successive slits, the grating constant, d, and the individual slit
width, w. The resulting pattern of interference and diffraction is shown in
䉲 Fig. 24.16.
Intensity 䉳 F I G U R E 2 4 . 1 5 Diffraction
grating A diffraction grating pro-
duces a sharply defined interference>
n=3 diffraction pattern. Two parameters
define a grating: the slit separation d
n=2 and the slit width w. The combina-
tion of multiple-slit interference and
n=1
Monochromatic single-slit diffraction determines the
light θ n=0 intensity distribution of the various
orders of maxima.
n=1
Grating
n=2
w
d n=3
Screen
䉴 F I G U R E 2 4 . 1 6 Intensity distri- Double-slit interference

n
bution of interference and diffrac- 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7
tion (a) Interference determines the
positions of the interference max-
ima: d sin u = nl, n = 0, 1, 2, 3, Á I
(b) Diffraction locates the positions
of the diffraction minima: (a)
u
w sin u = ml, m = 1, 2, 3, Á , and
the relative intensity of the maxima.
Single-slit diffraction
(c) The combination of interference
and diffraction determine the over-
m=2 m=1 m=1 m=2
all intensity distribution.
I
(b)
u
Combined
I
(c)
0 u
If light is transmitted through a grating, it is called a transmission grating. However,

reflection gratings are also common. The closely spaced tracks of a compact disc or a
DVD act as a reflection grating, giving rise to their familiar iridescent sheen (䉳 Fig.
24.17). Commercial master gratings are made by depositing a thin film of aluminum
on an optically flat surface and then removing some of the reflecting metal by cut-
ting regularly spaced, parallel lines. Precision diffraction gratings are made using
laser beams that expose a layer of photosensitive material, which is then etched. Pre-
cision gratings may have 30 000 or more lines per centimeter and are therefore
expensive and difficult to fabricate. Most gratings used in laboratory instruments
are replica gratings, which are plastic castings of high-precision master gratings.
It can be shown that the condition for interference maxima for a grating illumi-
nated with monochromatic light is identical to that for a double slit:
䉱 F I G U R E 2 4 . 1 7 Diffraction d sin u = nl for n = 0, 1, 2, 3, Á (grating interference maxima) (24.12)

effects The narrow tracks of com-
pact discs (CDs) act as reflection dif-
where n is called the order number and u is the angle at which that maximum occurs
fraction gratings, producing
colorful displays. for a particular wavelength. The zeroth-order maximum is coincident with the cen-
tral maximum. The grating constant d is obtained from the number of lines or slits
per unit length of the grating: d = 1>N. For example, if N = 5000 lines>cm, then
1 1
5000 >cm
d L = = 2.0 * 10-4 cm
N
If the light incident on a grating is white light (polychromatic), then the maxima
are multicolored (䉴 Fig. 24.18a). For the zeroth order, all color components are at
the same location (sin u = 0 for all wavelengths), so the central maximum is
white. However, the colors separate for higher orders, since the position of the
maximum depends on wavelength (Eq. 24.12). Because longer wavelengths have
a larger u, this produces a spectrum. Note that it is possible for higher orders pro-
duced by a diffraction grating to overlap. That is, the angles for different orders
may be the same for two different wavelengths. For example, the red color in the
second order might be at the same location as the blue color in the third order.
Only a limited number of spectral orders can be obtained using a diffraction
grating. The number depends on the wavelength of the light and on the grating
constant (d). From Eq. 24.12, u cannot exceed 90° (that is, sin u … 1), so
nl d
sin u = … 1 or nmax …
d l
Telescope
Slit
n=2 Distant
n=1 n=0 n=1 screen
R V R V 0˚ n=2 θ
V R V
R
h Source
Collimator Grating
(a) (b)
䉱 F I G U R E 2 4 . 1 8 Spectroscopy (a) In each side maximum, components of different

wavelengths (R = red and V = violet) are separated, because the angle depends on wave-
length: u = sin-11nl>d2. (b) As a result, gratings are used in spectrometers to determine
the wavelengths present in a beam of light by measuring their angles and to separate the
various wavelengths for further analysis.
Diffraction gratings have almost completely replaced prisms in spectroscopy.

The creation of a spectrum and the measurement of wavelengths by a grating
depend only on geometrical measurements such as lengths and>or angles. Wave-
length determination using a prism, in contrast, depends on the dispersive charac-
teristics of the material of which the prism is made. Thus, it is crucial to know
precisely how the index of refraction depends on the wavelength of light. In con-
trast to a prism, which bends red light the least and violet light the most, a diffrac-
tion grating produces the smallest angle for violet light (short l) and the greatest
angle for red light (long l). Notice that a prism disperses white light into a single
spectrum. A diffraction grating, however, produces a number of spectra, one for
each order other than n = 0, and the higher the order, the more spread out the
spectrum.
The sharp spectra produced by gratings are used in instruments called
spectrometers (Fig. 24.18b). With a spectrometer, materials can be illuminated with
light of various wavelengths to find which wavelengths are strongly transmitted
or reflected. Their absorption can then be measured and material characteristics
determined.
EXAMPLE 24.5 A Diffraction Grating: Grating Constant and Spectral Orders

A particular diffraction grating produces an n = 2 spectral T H I N K I N G I T T H R O U G H . Equation 24.12 can be used for all
order at an angle of 32° for light with a wavelength of 500 nm. three questions. (a) To find the number of lines per centimeter
(a) How many lines per centimeter does the grating have? (N) the grating has, the grating constant (d) needs to be found,
(b) At what angle can the n = 3 spectral order be seen? since N = 1>d. (b) The angle u can be computed for n = 3.
(c) What is the highest order maximum that can be observed? (c) The maximum angle is 90°, which corresponds to the
highest spectral order.
SOLUTION.
Given: l = 500 nm = 5.00 * 10-7 m Find: (a) N 1lines>cm2

u = 32° for n = 2 (b) u for n = 3
(b) n = 3 (c) nmax
(c) umax = 90°
(a) Using Eq. 24.12, the grating constant is
nl 215.00 * 10-7 m2
d = = = 1.887 * 10-6 m = 1.89 * 10-4 cm
sin u sin 32°
Then so
1 1 u = sin-1 0.794 = 52.6°
N = = = 5300 lines>cm
d 1.89 * 10-4 cm (c)
(b) d sin umax 11.89 * 10-6m2 sin 90°
nmax = = = 3.8
nl 315.00 * 10-7 m2 l 5.00 * 10-7 m
sin u = = = 0.794
d 1.89 * 10-6 m This means the n = 3 order is seen, but the n = 4 order is
not seen.
F O L L O W - U P E X E R C I S E . If white light of wavelengths ranging from 400 to 700 nm were used, what would be the angular width
of the spectrum for the second order?
X-RAY DIFFRACTION
In principle, the wavelength of any electromagnetic wave can be determined by using
a diffraction grating with the appropriate grating constant. Diffraction was used to
determine the wavelengths of X-rays early in the twentieth century. Experimental evi-
dence indicated that the wavelengths of X-rays were probably around 10-10 m or 0.1
nm (much shorter than visible light wavelengths), but it is impossible to construct a
diffraction grating with line spacing this close. Around 1913, Max von Laue
(1879–1960), a German physicist, suggested that the regular spacing of the atoms in a
crystalline solid might make the crystal act as a diffraction grating for X-rays, since the
atomic spacing is on the order of 0.1 nm (䉲 Fig. 24.19). When X-rays were directed at
crystals, diffraction patterns were indeed observed. (See Fig. 24.19b.)
Figure 24.19a illustrates diffraction by the planes of atoms in a crystal such as
sodium chloride. The path length difference is 2d sin u, where d is the distance
between the crystal’s atomic planes. Thus, the condition for constructive interference is
(constructive interference,
2d sin u = nl for n = 1, 2, 3, Á (24.13)
X-ray diffraction)
d
䉴 F I G U R E 2 4 . 1 9 Crystal diffrac-
tion (a) The array of atoms in a d
crystal lattice structure acts as a
diffraction grating, and X-rays are
diffracted from the planes of atoms. θ
With a lattice spacing of d, the path
length difference for the X-rays d θ θ
diffracted from adjacent planes is d
2d sin u. (b) X-ray diffraction pat-
tern of a crystal of potassium sul-
fate. By analyzing the geometry of d sin θ d sin θ
such patterns, investigators can Atomic cubic lattice
deduce the structure of the crystal
and the position of its various (a)
atoms. (c) X-ray diffraction pattern
of the protein hemoglobin, which
carries oxygen in blood.
(b) (c)
24.4 POLARIZATION 827
This relationship is known as Bragg’s law, after W. L. Bragg (1890–1971), the

British physicist who first derived it. Note that u is not measured from the normal.
X-ray diffraction is now routinely used to investigate the internal structure not
only of simple crystals, but also of large, complex biological molecules such as
proteins and DNA (Fig. 24.19c). Because of their short wavelengths, which are
comparable with interatomic distances within a molecule, X-rays provide a
method for investigating atomic structures within molecules.
DID YOU LEARN?

➥ Diffraction is most prominent when the wavelength of light is comparable to the
size of the opening the light is passing through or the size of the obstacle the light
is passing around.
➥ The width of the central maximum on a diffraction pattern will increase if the width
of the slit decreases.
➥ Blue light is closer to the central maximum than red light on a diffraction pattern
because it has a shorter wavelength.
24.4 Polarization
➥ What is meant by the polarization of light?

➥ What are the three common processes that produce polarized light?
➥ When two polarizing sheets have transmission axes at an angle of 90° to each other,
what percentage of unpolarized light will go through the two sheets?
When you think of polarized light, you may visualize polarizing (or Polaroid)
sunglasses, since this is one of the more common applications of polarization.
When something is polarized, it has a preferential direction, or orientation. In
terms of the transverse light waves, polarization refers to the orientation of elec-
tric field oscillations.
Recall from Section 20.4 that light is an electromagnetic wave with oscillating
B B
electric and magnetic field vectors (E and B, respectively) perpendicular (trans-
verse) to the direction of propagation. Light from most sources consists of a very
large number of electromagnetic waves emitted by the atoms of the source. Each
B
atom produces a wave with a particular E orientation, corresponding to the
direction of the atomic vibration. However, since electromagnetic waves from a
typical source are produced by many atoms, many ran-
B
dom orientations of the E fields are in the emitted com- Light is coming at you Light is traveling to the right
B
posite light. When the E vectors are randomly oriented,
the light is said to be unpolarized. This situation is com- E E
monly represented schematically in terms of the electric
field vector as shown in 䉴 Fig. 24.20a.
B
As viewed along the direction of propagation, the E is
equally distributed in all directions. However, as viewed (a) Unpolarized
parallel to the direction of propagation, this random or
E E
䉴 F I G U R E 2 4 . 2 0 Polarization Polarization is represented by

the orientationB
of the plane of vibration of the electric field vectors.
(a) When the E vectors are randomly oriented, the light is unpolar- (b) Partially polarized
ized. The dots represent an electric field direction perpendicular to
the paper, and the vertical arrows denote the up-and-down direc-
E E
tion of the electric field. Equal numbers of dots and arrows are
used to represent
B
unpolarized light. (b) With preferential orienta-
tion of the E vectors, the light is partially
B
polarized. Here, there are
fewer dots than arrows. (c) When the E vectors are in one direc-
tion, the light is linearly polarized, or plane polarized. No dots are
seen here. (c) Linearly (plane) polarized
equal distribution can be represented by two directions (such as the x- and y-direc-
tions in a two-dimensional coordinate system). Here, the vertical arrows denote
B B
the E components in that direction, and the dots represent the E components
going in and out of the paper. This notation will be used throughout this section.
B
If there is some preferential orientation of the E vectors, the light is said to be
B
partially polarized. Both representations in Fig. 24.20b show that there are more E
B
vectors in the vertical direction than in the horizontal direction. If the E vectors
oscillate in only one plane, the light is linearly polarized or plane polarized. In Fig.
B
24.20c, the E is entirely in the vertical direction and there is no horizontal compo-
nent. Note that polarization is evidence that light is a transverse wave. True longi-
tudinal waves, such as sound waves, cannot be polarized, because the molecules
of the media do not vibrate perpendicular to the direction of propagation.
Light can be polarized in many ways. Polarization by selective absorption,
reflection, and double refraction will be discussed here. Polarization by scattering
will be considered in Section 24.5.
POLARIZATION BY SELECTIVE ABSORPTION (DICHROISM)

Incident Vertical component Some crystals, such as those of the mineral tourmaline, exhibit the interesting
B
light is absorbed in property of absorbing one of the E components more than the other. This property
unpolarized crystal is called dichroism. If a dichroic crystal is sufficiently thick, the more strongly
absorbed component may be completely absorbed. In that case, the emerging
beam is linearly polarized (䉳 Fig. 24.21).
Another dichroic crystal is quinine sulfide periodide (commonly called
herapathite, after W. Herapath, an English physician who discovered its polarizing
Transmitted
light is properties in 1852). This crystal was of great practical importance in the develop-
linearly ment of modern polarizers. Around 1930, Edwin H. Land (1909–1991), an Ameri-
polarized can scientist, found a way to align tiny, needle-shaped dichroic crystals in sheets
Partially Crystal of transparent celluloid. The result was a thin sheet of polarizing material that was
polarized given the commercial name Polaroid.
Better polarizing films have been developed that use synthetic polymer materi-
䉱 F I G U R E 2 4 . 2 1 Selective als instead of celluloid. During the manufacturing process, this kind of film is
absorption (dichroism) Dichroic
crystals selectively absorb one stretched to align the long molecular chains of the polymer. With proper treat-
polarized component (the vertical ment, the outer (valence) electrons of the molecules can move along the oriented
B
component) more than the other. If chains. As a result, light with E vectors parallel to the oriented chains is readily
B
the crystal is thick enough, the absorbed, but light with E vectors perpendicular to the chains is transmitted. The
emerging beam is linearly direction perpendicular to the orientation of the molecular chains is called the
polarized.
transmission axis, or the polarization direction. Thus, when unpolarized light
falls on a polarizing sheet, the sheet acts as a polarizer and transmits polarized
light (䉴 Fig. 24.22).
B
Since one of the two E components is absorbed, the light intensity after the polar-
izer is half of the intensity incident on it 1Io>22. The human eye cannot distinguish
between polarized and unpolarized light. To tell whether light is polarized, we must
use a second polarizer, or an analyzer. As shown in Fig. 24.22a, if the transmission
axis of an analyzer is parallel to the direction of polarization of polarized light, there
is maximum transmission. If the transmission axis of the analyzer is perpendicular
to the direction of polarization, little light (ideally, none) will be transmitted.
In general, the light intensity through the analyzer is given by
I = Io cos2 u (Malus’ law) (24.14)
where Io is the light intensity after the first polarizer and u is the angle between the
transmission axes of the polarizer and analyzer. This expression is known as
Malus’ law, after its discoverer, French physicist E. L. Malus (1775–1812).
Polarizing glasses whose lenses have different transmission axes are used to
view some 3D movies. The pictures are projected on the screen by two projectors
that transmit slightly different images, photographed by two cameras a short dis-
tance apart. The projected light from each projector is linearly polarized, but in
Polarizer Analyzer
Io Io /2 Io /2
Transmitted
polarized light
Light
source
(a)
Polarizer Analyzer
Io Io /2 0
No light
Light
source
(b) (c)
䉱 F I G U R E 2 4 . 2 2 Polarizing sheets (a) When polarizing sheets are oriented so that their
transmission axes are in the same direction, the emerging light is polarized. The first sheet
acts as a polarizer, and the second acts as an analyzer. (b) When one of the sheets is rotated
90° and the transmission axes are perpendicular (crossed polarizers), little light (ideally,
none) is transmitted. (c) Crossed polarizers made using polarizing sunglasses.
mutually perpendicular directions. The lenses of the 3D glasses also have trans-
mission axes that are perpendicular. Thus, one eye sees the image from one projec-
tor, and the other eye sees the image from the other projector. The brain receives a
slight difference in perspective (or “viewing angle”) from the two images, and
interprets the image as having depth, or a third dimension, just as in normal
vision.
INTEGRATED EXAMPLE 24.6 Make Something Out of Nothing: Three Polarizers

In Fig. 24.22b, no light is transmitted through the analyzer, because the angle between the transmission axes is 90°. How-
because the transmission axes of the polarizer and analyzer ever, when an additional polarizer is inserted in between the
are perpendicular. Assume that the unpolarized light incident first polarizer and the analyzer, some light can actually pass
on the first polarizer has an intensity of Io. Another polarizer through the system. For example, if the transmission axis of
is then inserted between the first polarizer and analyzer, and the additional polarizer makes an angle of u with that of the
the transmission axis of this additional polarizer makes an first polarizer, then the angle between the transmission axes
angle of u with the first polarizer. (a) Is it possible for some of the additional polarizer and the analyzer will be 90° - u.
light to go through this arrangement? If yes, does it occur at (Why?)
(1) u = 0°, (2) u = 30°, (3) u = 45°, or (4) u = 90°? Explain. When unpolarized light of intensity Io is incident on the
(b) When u = 30°, what is the light intensity through the ana- first polarizer, the light intensity after the first polarizer is Io>2,
B
lyzer in terms of the incident light intensity? because only one of the two E components is transmitted.
(A) CONCEPTUAL REASONING. Yes, it is possible for some light After the additional polarizer, the intensity is decreased by a
to go through this arrangement at any angle other than 0° or factor of cos2 u. After the analyzer, the intensity is decreased
90°. The accompanying Learn by Drawing 24.1 can help in further by a factor of cos2190° - u2 = sin2 u. So the overall
understand this situation. transmitted intensity is I = 1Io>221cos 2 u21sin2 u2. Therefore,
With just the first polarizer and the analyzer, no light is as long as u is not 0° or 90°, some light will be transmitted
transmitted, according to Malus’ law (Equation 24.14), through the system.
(B) QUANTITATIVE REASONING AND SOLUTION. Once this situation is understood, part (b) is a straightforward calculation.
Given: u = 30° Find: (b) I (after analyzer in terms of Io)
Io 2 3 2 1 2
1cos2 30°21sin2 30°2 = ¢ ≤ a b =
Io 3Io
When u = 30°, I = .
2 2 2 2 32
FOLLOW-UP EXERCISE. For what value of u will the transmitted intensity be a maximum in this Example?
three polarizers (see integrated example 24.6)

θ
Polarizer
θ Analyzer
Additional 90°– θ
Io polarizer (Io /2) cos2θ
Io /2
θ
(Io /2) cos2θ cos2 (90°– θ )
POLARIZATION BY REFLECTION
When a beam of unpolarized light strikes a smooth, transparent medium such as
glass, the beam is partially reflected and partially transmitted. The reflected light
may be linearly polarized, partially polarized, or unpolarized, depending on the
angle of incidence. The unpolarized case occurs for 0°, or normal incidence. As
the angle of incidence is changed from 0°, both the reflected and refracted light
B
become partially polarized. For example, the E components perpendicular to the
plane of incidence (the plane containing the incident, reflected, and refracted
rays) are reflected more strongly, producing partial polarization (䉲 Fig. 24.23a).
Unpolarized Partially polarized Unpolarized Linearly polarized

light light light light
θ1
θp θp
θ1
n1 n1 90°
n2 n2
θ2
Partially
θ2 polarized light
Partially
polarized light
(a) (b)
䉱 F I G U R E 2 4 . 2 3 Polarization by reflection (a) When light is incident on a boundary, the

reflected and refracted lights are normally partially polarized. (b) When the reflected and
refracted rays are 90° apart, the reflected light is linearly polarized, and the refracted light
n2
is partially polarized. This situation occurs when u1 = up = tan-1 ¢ ≤ .
n1
However, at one particular angle of incidence, the reflected light is linearly polar-
ized (Fig. 24.23b). (At this angle, though, the refracted light is still only partially
polarized.)
David Brewster (1781–1868), a Scottish physicist, found that the linear polariza-
tion of the reflected light occurs when the reflected and refracted rays are perpen-
dicular. The angle of incidence at which linear polarization occurs is the
polarizing angle (Up), or the Brewster angle, and it depends on the indices of
refraction of the two media. In Fig. 24.23b, the reflected and refracted rays are at
90° and the angle of incidence u1 is thus the polarizing angle up : u1 = up.
By the law of refraction (Section 22.3),
n1 sin u1 = n2 sin u2
Since u1 + 90° + u2 = 180° or u2 = 90° - u1, sin u2 = sin190° - u12 = cos u1.
Therefore,
sin u1 sin u1 n2
= = tan u1 =
sin u2 cos u1 n1
With u1 = up ,
n2 n2
tan up = or up = tan-1 ¢ ≤ (24.15)
n1 n1
If the first medium is air 1n1 = 12, then tan up =

n2
= n2 = n, where n is the
1
index of refraction of the second medium.
Now you can understand the reason for inventing polarizing glasses. Light
reflected from a smooth surface is partially polarized. The direction of polariza-
tion is mostly perpendicular to the plane of incidence (horizontal direction in
Fig. 24.23b). Reflected light can be so intense that it gives rise to visual glare
(䉲 Fig. 24.24a). To reduce this effect, polarizing glasses are oriented with their
transmission axes vertical so that some of the partially polarized light from reflec-
tion is absorbed. Polarizing filters also enable cameras to take “clean” pictures
without interference from glare (Fig. 24.24b).
䉳 F I G U R E 2 4 . 2 4 Glare reduction (a) Light

reflected from a horizontal surface is partially
polarized in the horizontal direction. When
glasses are oriented so that their transmission
axis is vertical, the horizontally polarized com-
ponent of such light is not transmitted, so glare
is reduced. (b) Polarizing filters for cameras
use the same principle. The photo at right was
taken with such a filter. Note the reduction in
reflections from the store window.
(a)
(b)
EXAMPLE 24.7 Calling Brewster: Polarization by Reflection

ZnSe (zinc selenide) is a material used for the windows of high- up is the Brewster (polarization) angle. Using Eq. 24.15, we
power CO2 lasers. The index of refraction of ZnSe is 2.40 at a find that
wavelength of 10.6 mm, and the index of refaction of CO2 is
≤ = tan-1 a b = 67.4°
n2 2.40
1.00. What should be the angle of incidence when the polariza- up = tan-1 ¢
tion of the reflected laser light is greatest? n1 1.00
T H I N K I N G I T T H R O U G H . Incident light at the Brewster (polar- The ZnSe window in CO2 lasers is therefore installed at this
ization) angle has the greatest polarization upon reflection, so angle of incidence so as to maximize the polarization of the
it is the simple application of Eq. 24.15. laser beam.
SOLUTION.
Given: n1 = 1.00 Find: u1 = up (angle of incidence for
n2 = 2.40 greatest polarization)
F O L L O W - U P E X E R C I S E . Light is incident on a flat, transparent material with an index of refraction of 1.52. At what angle of
refraction would the transmitted light have the greatest polarization if the transparent material is in water?
POLARIZATION BY DOUBLE REFRACTION (BIREFRINGENCE)

When monochromatic light travels through glass, its speed is the same in all direc-
tions and is characterized by a single index of refraction. Any material that has
such a property is said to be isotropic, meaning that it has the same optical charac-
teristics in all directions. Some crystalline materials, such as quartz, calcite, and
ice, are anisotropic; that is, the speed of light, and therefore the index of refraction,
are different for different directions within the material. Anisotropy gives rise to
some interesting optical properties. Anisotropic materials are said to be doubly
refracting, or to exhibit birefringence, and polarization is involved.
For example, unpolarized light incident on a birefringent crystal of calcite
(CaCO3, calcium carbonate) is illustrated in 䉲 Fig. 24.25. When the light propagates
at an angle to a particular crystal axis, the beam is doubly refracted and separated
into two components, or rays, upon refraction. These two rays are linearly polar-
ized in mutually perpendicular directions. One ray, called the ordinary (o) ray,
passes straight through the crystal and is characterized by an index of refraction
no . The second ray, called the extraordinary (e) ray, is refracted and is characterized
by an index of refraction ne . The particular axis direction indicated by dashed
lines in Fig. 24.25a is called the optic axis. Along this direction, no = ne , and
nothing extraordinary is noted about the transmitted light.
Some transparent materials have the ability to rotate the plane of polarization of
linearly polarized light. This property, called optical activity, is due to the molecu-
lar structure of the material (䉴 Fig. 24.26a). Optically active molecules include
those of certain proteins, amino acids, and sugars.
Optic axis
䉴 F I G U R E 2 4 . 2 5 Double refrac-
tion or birefringence (a) Unpolar-
ized light incident normal to the
surface of a birefringent crystal and
at an angle to a particular direction Unpolarized
in the crystal (dashed lines) is sepa- light
rated into two components. The
ordinary (o) ray and the extraordi- o ray
nary (e) ray are linearly polarized in
mutually perpendicular directions.
e ray
(b) Double refraction seen through a
calcite crystal.
(a) (b)
*24.5 ATMOSPHERIC SCATTERING OF LIGHT 833
Polarized
light
(a) (b)
䉱 F I G U R E 2 4 . 2 6 Optical activity and stress detection (a) Some substances have the
property of rotating the polarization direction of linearly polarized light. This ability, which
depends on the molecular structure of the substance, is called optical activity. (b) Glasses and
plastics become optically active under stress, and the points of greatest stress are apparent
when the material is viewed through crossed polarizers. Engineers can thus test plastic
models of structural elements to see where the greatest stresses will occur when the models
are “loaded.” Here, a model of a suspension bridge strut is being analyzed.
Glasses and plastics become optically active under stress. The greatest rotation
of the direction of polarization occurs in the regions where the stress is the great-
est. Viewing the stressed piece of material through crossed polarizers allows the
points of greatest stress to be identified. This determination is called optical stress
analysis (Fig. 24.26b). Another use of polarizing films, the liquid crystal display
(LCD), is described in accompanying Insight 24.2, LCDs and Polarized Light.
DID YOU LEARN?

➥ The polarization of light refers to the direction of oscillation of the electric field of
light (an electromagnetic wave). As light is a transverse wave, the direction of the
electric field is perpendicular to the direction of light propagation.
➥ The three common processes that produce polarized light are selective absorption,
reflection, and double refraction.
➥ Ideally no light will pass through the two polarizing sheets when their transmission
axes are at an angle of 90° to each other.
*24.5 Atmospheric Scattering of Light

➥ What occurs in a scattering process?

➥ How does the intensity of Rayleigh scattering depend on the wavelength of light?
➥ Why is the sky blue?
When light is incident on a suspension of particles, such as the molecules in air,

some of the light may be absorbed and reradiated in all directions. This process is
called scattering. The scattering of sunlight in the atmosphere produces some
interesting effects, including the polarization of skylight (that is, sunlight that has
been scattered by the atmosphere), the blueness of the sky, and the redness of sun-
sets and sunrises.
Atmospheric scattering causes skylight to be polarized. When unpolarized sun-
light is incident on air molecules, the electric field of the light wave sets electrons of
the molecules into vibration. The vibrations are complex, but these accelerated
charges emit radiation, like the vibrating electrons in the antenna of a radio broad-
cast station (see Section 20.4). As illustrated in 䉲 Fig. 24.27, an observer viewing from
an angle of 90° with respect to the direction of the sunlight will receive linearly
polarized light, because the electric field component in the direction of travel cannot
INSIGHT 24.2 LCDs and Polarized Light

Today, liquid crystal displays (LCDs) are commonplace in items pattern are applied to the liquid crystal. Each block, or dis-
such as watches, calculators, televisions, and computer screens. play segment, has a separate electrical connection. The dark
The name “liquid crystal” may seem self-contradictory. Nor- numbers or letters on an LCD are formed by applying an elec-
mally, when a crystalline solid melts, the resulting liquid no tric voltage to certain blocks of the liquid crystal. Note that
longer has an orderly atomic or molecular arrangement. Some the numerals 0 through 9 can be formed out of pieces of the
organic compounds, however, pass through an intermediate segmented display.
state in which the molecules may rearrange somewhat but still By using another polarizer (an analyzer), you can readily
maintain the overall order that is characteristic of a crystal. show that the light from the LCD is polarized (Fig. 2). You can
A common type of LCD, called a twisted nematic display, either see or not see the display by rotating the analyzer over
makes use of polarized light (Fig. 1). These special liquid the watch. You have noticed this effect if you have ever not
crystals are optically active and will rotate the direction of been able to see the time on the LCD of a wristwatch while
polarization of light by 90° if no voltage is applied across wearing polarizing sunglasses.
them. However, if voltage is applied, the crystals will lose this One of the major advantages of LCDs is their low power
optical activity. consumption. Similar displays, such as those using light-
The liquid crystals are then placed between crossed emitting diodes (LEDs), produce light themselves, using rela-
polarizing sheets and backed with a mirrored surface. With tively large amounts of electric energy. LCDs produce no light
voltage off, light entering and passing through the first but instead use reflected light.
polarizer is polarized, rotated 90°, reflected, and again rotated Color flat-panel computer and TV screens, which rely on
90°. After the return trip through the liquid crystal, the direc- LCD technology, are popular today. They are about one-
tion of polarization of the light is the same as that of the initial quarter the size of, consume less than half the energy of, and
polarizer. Thus, the light coming out of the display unit and are easier on the eyes than CRT monitors and traditional TV
the display appears to be a light color (usually light gray) screens of the same size. Computer displays and TVs are
when illuminated with unpolarized light. usually specified in pixels, much like the smallest square on
With voltage on, the polarized light passing through the graph paper. To produce color, three LCD segments (red,
liquid crystal is absorbed by the second polarizer. Thus, the green, and blue) are grouped on each pixel. By controlling
liquid crystal is opaque and appears dark. Transparent, elec- the intensities of the three colors, each pixel can generate
trically conductive film coatings arranged in a seven-block every color in the visible spectrum.
Mirror Polarizer Liquid crystal Polarizer

Incident
light
Reflected
light
Voltage off: optical activity allowed
Incident
light
No reflected
Voltage on: no optical activity light
F I G U R E 2 Polarized light The light from an LCD is polar-
ized, as can be shown by using polarizing sunglasses as an
analyzer. When the analyzer is rotated through 90°, the
Voltage off: Voltage on: numbers on the watch are no longer visible.
no display affected segments
appears darken, number
appears
F I G U R E 1 Liquid crystal display (LCD) A twisted nematic

display is an application involving the optical activity of a
liquid crystal and crossed polarizing sheets. When the crys-
talline order is disoriented by an electric field from an
applied voltage, the liquid crystal loses its optical activity in
that region, and light is reflected. Numerals and letters are
formed by applying voltages to segments of a block display.
*24.5 ATMOSPHERIC SCATTERING OF LIGHT 835
exist in a transverse wave. At other viewing angles, both components are Unpolarized Air Unpolarized
present, and skylight seen through a polarizing filter appears partially light molecule
polarized.
Since the scattering of light with the greatest degree of polarization
occurs at a right angle to the direction of the Sun, at sunrise and sunset
the scattered light from directly overhead has the greatest degree of
polarization. The polarization of skylight can be observed by viewing
the sky through a polarizing filter (or a polarizing sunglass lens) and Partially
polarized
rotating the filter. Light from different regions of the sky will be trans-
mitted in different degrees, depending on its degree of polarization. It
is believed that some insects, such as bees, use polarized skylight to
Linearly
determine navigational directions relative to the Sun.
polarized
WHY THE SKY IS BLUE

The scattering of sunlight by air molecules is the reason why the sky
looks blue. This effect is not due to polarization, but to the selective
absorption of light. As oscillators, air molecules have resonant frequen-
cies (at which they scatter most efficiently) in the blue-violet region.
Consequently, when sunlight is scattered, the blue end of the visible
䉱 F I G U R E 2 4 . 2 7 Polarization by scattering
spectrum is scattered more than the red end.
When incident unpolarized sunlight is scat-
For particles such as air molecules, which are much smaller than the tered by a gas molecule in the air, the light
wavelength of light, the intensity of the scattered light is inversely pro- perpendicular to the direction of the incident
portional to the wavelength to the fourth power 11>l42. This relation- ray is linearly polarized. Light scattered at
ship between wavelength and scattering intensity is called Rayleigh some arbitrary angle is partially polarized. An
observer at a right angle 190°2 to the direction
scattering, after Lord Rayleigh (1842–1919), a British physicist who
of the incident sunlight receives linearly
derived it. This inverse relationship predicts that light of the shorter- polarized light.
wavelength, or blue, end of the spectrum will be scattered much more
than light of the longer-wavelength, or red, end. The scattered blue
light is rescattered in the atmosphere and eventually is directed toward
the ground. This is why the sky appears blue.
EXAMPLE 24.8 Red and Blue: Rayleigh Scattering

How much more is light at the blue end (400 nm) of the visible spectrum scattered by
air molecules than light at the red end (700 nm)?
4
T H I N K I N G I T T H R O U G H . We know that Rayleigh scattering is proportional to 1>l and
that light from the blue end of the spectrum (shorter wavelength) is scattered more than
light from the red end. The wording “how much more” implies a factor or ratio.
SOLUTION. The Rayleigh scattering relationship is I r 1>l4, where I is the intensity of
scattering for a particular wavelength. Thus, forming a ratio:
Iblue lred 4
= ¢ ≤
Ired lblue
Inserting the wavelengths:

lred 4 700 nm 4
= a b = a b = 9.4 or Iblue = 9.4Ired
Iblue
Ired lblue 400 nm
And so, blue light is scattered almost ten times as much as red light.
F O L L O W - U P E X E R C I S E . What wavelength of light is scattered twice as much as red
light? What color light is this?
WHY SUNSETS AND SUNRISES ARE RED

Beautiful red sunsets and sunrises are sometimes observed. When the Sun is near
the horizon, sunlight travels a greater distance through the denser air near the
Earth’s surface. Since the light therefore undergoes a great deal of scattering, you
might think that only the least scattered light, the red light, would reach observers
on the Earth’s surface. This would explain red sunsets. However, it has been
shown that the dominant color of white light after only molecular scattering is
orange. Thus, other types of scattering must shift the light from the setting (or
rising) Sun toward the red end of the spectrum (䉳 Fig. 24.28).
Red sunsets have been found to result from the scattering of sunlight by atmos-
pheric gases and by small dust particles. These particles are not necessary for the
blueness of the sky, but are compulsory for deep-red sunsets and sunrises. (This is
why spectacular red sunsets are observed in the months after large volcanic erup-
tions that put tons of particulate matter into the atmosphere.) Red sunsets occur
most often when there is a high-pressure air mass to the west, since the concentra-
䉱 F I G U R E 2 4 . 2 8 Red sky at night tion of dust particles is generally greater in high-pressure air masses than in low-
A spectacular red sunset over a pressure air masses. Similarly, red sunrises occur most often when there is a
mountaintop observatory in Chile. high-pressure air mass to the east.
The red sky results from the scatter- Now you can understand the old saying “Red sky at night, sailors’ delight; red
ing of sunlight by atmospheric
gases and small solid particles. A sky in the morning, sailors take warning.” Fair weather generally accompanies
directly observed reddening Sun is high-pressure air masses, because they are associated with reduced cloud forma-
due to the scattering of light in the tion. Most of the United States lies in the westerlies wind zone, in which air
blue end of the spectrum out of the masses generally move from west to east. A red sky at night is thus likely to indi-
direct line of sight. cate a fair weather, high-pressure air mass to the west that will be coming your
way. A red sky in the morning means that the high-pressure air mass has passed
and poor weather may set in.
As a final note, how would you like a sky that is normally red? On Mars, the
“red planet,” the thin atmosphere is about 95% carbon dioxide (CO2). The CO2
molecule is more massive than an oxygen (O2) or a nitrogen (N2) molecule. As a
result, CO2 molecules have a lower resonant frequency (longer wavelength) and
preferentially scatter the red end of the visible spectrum. Hence, the Martian sky is
red during the day. And what of the color of sunrises and sunsets on Mars? Think
about it . . . .
And finally, the use of light in biomedical application is discussed in Insight
24.3, Optical Biopsy.
DID YOU LEARN?

➥ Scattering is the process of molecules absorbing light and then reradiating it in
different directions.The reradiated light can have a different polarization from the
absorbed light.
➥ The intensity of Rayleigh scattering is inversely proportional to the wavelength to
the fourth power, that is, 1>l4.
➥ The sky is blue because of light scattered by air molecules. Shorter-wavelength
(blue) light has a stronger scattering intensity than longer-wavelength (red) light.
INSIGHT 24.3 Optical Biopsy

One of the most reliable ways to detect disease is to perform a tures, such as the presence of collagen fibers and the level of
surgical biopsy—the removal of tissue samples—and then hydration in the tissue. (A major component of skin and bone,
look for abnormal changes in the samples. “Optical biopsy,” or collagen is a fibrous protein found in animal cells.) The mea-
biomedical scattering, appears to be a promising tool for diag- surement of the scattered light as a function of wavelength,
nosing and monitoring diseases such as cancer without such polarization, or angle can be an important diagnostic tool.
surgery. An example of an optical biopsy is the diagnosis and mea-
Optical biopsies are based on the following physical princi- surement of collagen fibers. The fibers in the dormant form of
ple. The particles in the tissue absorb and re-emit light; thus, collagen (about 2–3 mm in diameter) are composed of bundles
the scattered light contains information about the makeup of of smaller collagen fibrils, about 0.3 mm in diameter, as shown
the tissue. Scattering from a tissue depends on internal struc- in Fig. 1. The fibrils are made up of (entwined tropocollagen)
24.5 ATMOSPHERIC SCATTERING OF LIGHT 837
molecules and present a banded pattern of striations with A recent advance in mouth cancer detection using optical
70-nm periodicity due to the staggered alignment of the scattering is the invention of VELscope (Visually Enhanced
tropocollagen molecules. Each of these molecules has an Lesion scope) in 2007. The VELscope emits a cone of blue
electron-dense “head group” that appears dark in the electron light into the mouth, and the scattered light (as visible fluo-
micrograph. This periodic variation in refractive index at this rescence) is then analyzed. Abnormal cells can then be identi-
level scatters light strongly in the visible and ultraviolet fied because they appear black (Fig. 2), while healthy tissue
regions. The information contained in the scattered light can shows up with a green glow. The VELscope examination
reveal abnormal conditions in the collagen fibers. takes less than 3 minutes.
F I G U R E 1 An electron micrograph of collagen fibers The F I G U R E 2 VELscope Irregular, dark area visible under fluo-
details of collagen fibers show the presence of collagen fibrils rescence visualization. Biopsy-confirmed Carcinoma.
and tropocollagen molecules.
PULLING IT TOGETHER Sunlight on a Pond: Polarization, Reflection, and Refraction

Sunlight is reflected from the smooth surface of a pond. from the normal and the altitude angle is measured from the
(a) What is the Sun’s altitude (the angle between the Sun and horizon, the angle of incidence is the angle complementary to
the horizon) when the polarization of the reflected light is the the altitude angle (draw a sketch to help visualize this situa-
greatest? (b) What is the angle of reflection? (c) What is the tion). Incident light at the polarizing (Brewster) angle has the
angle of refraction in water? greatest polarization for reflected light, so the Sun’s altitude is
at 90° - up from the horizon. (b) and (c) Once the angle of
T H I N K I N G I T T H R O U G H . The concepts involved in this Exam- incidence is known, the laws of reflection and refraction can
ple are polarization by reflection and the laws of reflection be used to calculate the angle of reflection in air and refraction
and refraction. (a) Since the angle of incidence is measured in water.
SOLUTION. The index of refraction of water is listed in Table 22.1.

Given: n1 = 1 Find: (a) u (altitude angle for greatest polarization)
n2 = 1.33 (Table 22.1) (b) u1œ (angle of reflection)
(c) u2 (angle of refraction)
(a) The Sun needs to be at an altitude angle of u = 90° - up, where up is the polarizing (Brewster) angle. From Eq. 24.15,
≤ = tan-1 a b = 53.1°
n2 1.33
up = tan-1 ¢
n1 1
So
u = 90° - up = 90° - 53.1° = 36.9°
(b) The angle of incidence is the polarizing angle, u1 = up. The angle of reflection is then u1œ = u1 = up = 53.1°.
(c) Using the law of refraction.
n1 sin u1 112 sin 53.1°
sin u2 = = = 0.601 so u2 = sin-1 0.601 = 36.9°
n2 1.33
The angle of refraction is equal to the Sun’s altitude angle. (Why?)
■ Young’s double-slit experiment provides evidence of the ■ For a diffraction grating, the maxima satisfy
wave nature of light and a way to measure the wavelength
of light 1 L 10-7 m2.
d sin u = nl for n = 0, 1, 2, Á (24.12)
The angular position 1u2 of the maxima satisfies the where d = 1>N and N is the number of lines per unit length.
condition
Intensity
d sin u = nl for n = 0, 1, 2, 3, Á (24.3)

n=3
where d is the slit separation. Mono-

n=2
chromatic n=1
For small u, the distance between the nth maximum and light θ n=0
the central maximum is
Grating n=1
n=2
nLl w
yn L for n = 0, 1, 2, 3, Á (24.4) d n=3
d
Intensity Screen
Max ■ Polarization is the preferential orientation of the electric field

(n = 2)
Min vectors that make up a light wave and is evidence that light is
Light
Max
(n = 1)
a transverse wave. Light can be polarized by selective
S1
source Min
Max
absorption, reflection, double refraction (birefringence), and
Min
(n = 0) scattering.
S2
Max When the transmission axes of a polarizer and an
Single (n = 1)
slit Double Min analyzer make an angle of u, the light intensity through the
slit
Max
(n = 2)
analyzer is given by Malus’ law:
I = Io cos 2 u (24.14)
Screen
L
In reflection, if the angle of incidence is equal to the
■ Light reflected at a media boundary for which n2 7 n1 polarizing (Brewster) angle up, then the reflected light is
undergoes a 180° phase change. If n2 6 n1, there is no linearly polarized:
phase change on reflection. The phase changes affect thin- n2 n2
film interference, which also depends on film thickness and tan up = or up = tan-1 ¢ ≤ (24.15)
n1 n1
index of refraction.
The minimum thickness for a nonreflecting film is Unpolarized Linearly polarized
light light
1for n2 7 n1 7 no2
l
tmin = (24.7) θp θp
4n1
n1 90°
1 n2
2
Air no
θ2
Thin film n 1 > no t Partially

polarized light
Glass n 2 > n1
lens
■ The intensity of Rayleigh scattering is inversely propor-
■ In a single-slit diffraction experiment, the minima at angle u tional to the fourth power of the wavelength of the light.
satisfy The blueness of the Earth’s sky results from the preferential
scattering of sunlight by air molecules.
w sin u = ml for m = 1, 2, 3, Á (24.8)
where w is the slit width. In general, the longer the wave-
length as compared with the width of an opening or object,
the greater the diffraction.
Intensity
m=3
m=2
Mono-
chromatic m=1
light θ
w
m=1
Single m=2
slit m=3
L
L >> w
Screen

24.1 YOUNG’S DOUBLE-SLIT the side maxima, (c) the side maxima are twice as wide as
EXPERIMENT the central maximum, (d) none of the preceding.
8. In a single-slit diffraction pattern, if the wavelength of
1. If the path length difference between two identical and
light decreases, the width of the central maximum will
coherent beams is 2.5l when they arrive at a point on a
(a) increase, (b) decrease, (c) remain the same.
screen, the point will be (a) bright, (b) dark, (c) multi-
colored, (d) gray. 9. As the number of lines per unit length of a diffraction
2. In a Young’s double-slit experiment using monochro- grating increases, the spacing between the maxima
matic light, if the slit spacing d increases, the interference (a) increases, (b) decreases, (c) remains unchanged.
maxima spacing will (a) decrease, (b) increase, (c) remain
unchanged, (d) disappear. 24.4 POLARIZATION
3. When white light is used in Young’s double-slit experi-
ment, many maxima with a spectrum of colors are seen. 10. A sound wave cannot be polarized. This is because
In a given maximum, the color farthest from the central sound is (a) a transverse wave, (b) a longitudinal wave,
maximum is (a) red, (b) blue, (c) other colors. (c) none of the preceding.
11. Light can be polarized by (a) reflection, (b) refraction,
(c) selective absorption, (d) all of the preceding.
24.2 THIN-FILM INTERFERENCE
12. The polarizing (Brewster) angle depends on (a) the
4. When a thin film of kerosene spreads out on water, the
indices of refraction of materials, (b) Bragg’s law,
thinnest part looks bright. The index of refraction of
(c) internal reflection, (d) interference.
kerosene is (a) greater than, (b) less than, (c) the same as
that of water. 13. The percentage of unpolarized light that will pass through
5. For a thin film with no 6 n1 6 n2, where n1 is the index two polarizing sheets with their transmission axes parallel
of refraction of the film, the minimum film thickness for to each other ideally is (a) 100%, (b) 50%, (c) 25%, (d) 0%.
destructive interference of the reflected light is (a) l¿>4,
(b) l¿>2, (c) l¿ . *24.5 ATMOSPHERIC SCATTERING OF
6. For a thin film with no 6 n1 7 n2, where n1 is the index LIGHT
of refraction of the film, a film thickness for destructive
interference of the reflected light is (a) l¿>4, (b) l¿>2, 14. Scattering involves (a) the reflection of light off particles,
(c) l¿ , (d) both (b) and (c). (b) the refraction of light off particles, (c) the absorption
and reradiation of light by particles, (d) the interference
of light with particles.
24.3 DIFFRACTION 15. Which of the following colors is scattered the least in the
7. In a single-slit diffraction pattern, (a) all maxima have the atmosphere: (a) blue, (b) yellow, (c) red, or (d) color
same width, (b) the central maximum is twice as wide as makes no difference?
24.1 YOUNG’S DOUBLE-SLIT situation a violation of the law of conservation of

EXPERIMENT energy? Explain.
5. At the center of a Newton’s rings arrangement
1. Discuss how the interference pattern in Young’s double-
(Fig. 24.10a), the air wedge has a thickness of zero. Why
slit experiment would change if the distance between the
is this area always dark (Fig. 24.10b)?
double slits decreases.
6. Most lenses used in cameras are coated with thin films
2. Describe what would happen to the interference pattern
and appear bluish-purple when viewed in the reflected
in Young’s double-slit experiment if the wavelength of
light. What wavelengths are refracting through the lens?
the monochromatic light were to decrease.
3. The intensity of the central maximum in the interference
pattern of a Young’s double-slit experiment is about four 24.3 DIFFRACTION
times that of either light wave. Is this a violation of the
7. In our discussion of single-slit diffraction, the length of the
law of conservation of energy?
slit was assumed to be much greater than the width. What
changes would be observed in the diffraction pattern if the
24.2 THIN-FILM INTERFERENCE length were comparable with the width of the slit?
4. When destructive interference of two waves occurs at a 8. From Eq. 24.8, can the m = 2 minimum be seen if w = l?
certain location, there is no energy at that location. Is this How about the m = 1 minimum?
9. In a diffraction grating, the slits are very closely spaced. 12. How does selective absorption produce polarized light?
What is the advantage of this design? 13. If you place a pair of polarizing sunglasses in front of
your calculator’s LCD display and rotate them, what
24.4 POLARIZATION would you observe?
10. Given two pairs of sunglasses, could you tell whether

one or both were polarizing?
*24.5 ATMOSPHERIC SCATTERING OF
11. Suppose that you held two polarizing sheets in front of
you and looked through both of them. How many times LIGHT
would you see the sheets lighten and darken (a) if one 14. Explain why the sky may be red in the morning and
were rotated through one complete rotation, (b) if both evening and blue during the day.
were rotated through one complete rotation at the same 15. What color would an astronaut on the Moon see when
speed in opposite directions, (c) if both were rotated looking at the sky or into space?
through one complete rotation at the same speed in the
same direction, and (d) if one were rotated twice as fast
as the other and the slower one were rotated through
one complete rotation?
EXERCISES*
24.1 YOUNG’S DOUBLE-SLIT the screen is 1.5 m away from the slits, and light of wave-
EXPERIMENT length 550 nm is used, what is the distance from the center
of the central maximum to the center of the third-order
1. ● To study wave interference, a student uses two speak- maximum? (c) What if the wavelength is 680 nm?
ers driven by the same sound wave of wavelength 0.50 m.
7. ● ● (a) If the wavelength used in a double-slit experi-
If the distances from a point to the speakers differ by
ment is decreased, the distance between adjacent
0.75 m, will the waves interfere constructively or destruc-
maxima will (1) increase, (2) decrease, (3) remain the
tively at that point? What if the distances differ by 1.0 m?
same. Explain. (b) If the separation between the two slits
2. ● In the development of Young’s double-slit experiment, a
small-angle approximation 1tan u L sin u2 was used to
is 0.20 mm and the adjacent maxima of the interference
pattern on a screen 1.5 m away from the slits are 0.45 cm
find the lateral displacements of the maxima (bright) and apart, what is the wavelength and color of the light?
minima (dark) positions. How good is this approximation? (c) If the wavelength is 550 nm, what is the distance
For example, what is the percentage error for u = 10°? between adjacent maxima?
3. ● Two parallel slits 0.075 mm apart are illuminated with 8. ● ● In a double-slit experiment using monochromatic
monochromatic light of wavelength 480 nm. Find the light, a screen is placed 1.25 m away from the slits, which
angle between the center of the central maximum and have a separation distance of 0.0250 mm. The position of
the center of the first side maximum. the third-order maximum is 6.60 cm from the center of
4. ● When two parallel slits are illuminated with mono- the central maximum. Find (a) the wavelength of the
chromatic light of wavelength 632.8 nm, the angle light and (b) the position of the second-order maximum.
between the center of the central maximum and the cen- 9. ● ● In a double-slit experiment with monochromatic light
ter of the second side maximum is 0.45°. What is the dis- and a screen at a distance of 1.50 m from the slits, the angle
tance between the parallel slits? between the second-order maximum and the central
5. ● In a double-slit experiment that uses monochromatic maximum is 0.0230 rad. If the separation distance of the
light, the angular separation between the central maxi- slits is 0.0350 mm, what are (a) the wavelength and color of
mum and the second-order maximum is 0.160°. What is the light and (b) the lateral displacement of this maximum?
the wavelength of the light if the distance between the 10. IE ● ● Two parallel slits are illuminated with monochro-
slits is 0.350 mm? matic light, and an interference pattern is observed on a
6. IE ● ● Monochromatic light passes through two narrow screen. (a) If the distance between the slits were decreased,
slits and forms an interference pattern on a screen. (a) If would the distance between the maxima (1) increase,
the wavelength of light used increases, will the distance (2) remain the same, or (3) decrease? Explain. (b) If the
between the maxima (1) increase, (2) remain the same, or slit separation is 1.0 mm, the wavelength is 640 nm, and
(3) decrease? Explain. (b) If the slit separation is 0.25 mm, the distance from the slits to the screen is 3.00 m, what is
*Assume all angles to be exact.
EXERCISES 841
the separation between adjacent interference maxima? 20. ● A film on a lens with an index of refraction of 1.5 is
(c) What if the slit separation is 0.80 mm? 1.0 * 10-7 m thick and is illuminated with white light.
11. IE ● ● (a) In a double-slit experiment, if the distance from The index of refraction of the film is 1.4. (a) The number
the double slits to the screen is increased, the separation of waves that experience the 180° phase shift is (1) zero,
between the adiacent maxima will (1) increase, (2) one, (3) two. Explain. (b) For what wavelength of visi-
(2) decrease, (3) remain the same. Explain. (b) Yellow-green ble light will the lens be nonreflecting?
light 1l = 550 nm2 illuminates a double-slit separated by 21. ●● A solar cell is designed to have a nonreflective film of
1.75 * 10-4 m. If the screen is located 2.00 m from the slits, a transparent material for a wavelength of 550 nm.
determine the separation between the adjacent maxima. (a) Will the thickness of the film depend on the index of
(c) What if the screen is located 3.00 m from the slits? refraction of the underlying material in the solar cell?
12. ● ● (a) Derive a relationship that gives the locations of Discuss the possible scenarios. (b) If nsolar 7 nfilm and
the minima in a Young’s double-slit experiment. What is nfilm = 1.22, what is the minimum thickness of the film?
the distance between adjacent minima? (b) For a third- (c) Repeat the calculation in (b) if nsolar 6 nfilm and
order minimum (the third side dark position from the nfilm = 1.40.
central maximum), what is the path length difference 22. IE ● ● A thin layer of oil 1n = 1.502 floats on water.
between that location and the two slits? Destructive interference is observed for reflected light of
13. ● ● When a double-slit setup is illuminated with light of wavelengths 480 nm and 600 nm, each at a different loca-
wavelength 632.8 nm, the distance between the center of tion. (a) If the order number is the same for both wave-
the central bright position and the second side dark posi- lengths, which wavelength is at a greater thickness:
tion is 4.5 cm on a screen that is 2.0 m from the slits. (1) 480 nm, or (2) 600 nm? Explain. (b) Write the general
What is the distance between the slits? condition of destructive intereference for reflected light.
14. IE ● ● ● (a) If the apparatus for a Young’s double-slit (c) Find the two minimum thicknesses of the oil film,
experiment were completely immersed in water, would assuming normal incidence.
the spacing of the interference maxima (1) increase, 23. ●● Two parallel plates are separated by a small distance
(2) remain the same, or (3) decrease? Explain. (b) What as illustrated in 䉲 Fig. 24.29. If the top plate is illuminated
would the lateral displacements in Exercise 6 be if the with light from a He–Ne laser 1l = 632.8 nm2, for what
entire system were immersed in still water? minimum separation distances will the light be (a) con-
15. ● ● ● Light of two different wavelengths is used in a dou- structively reflected and (b) destructively reflected?
ble-slit experiment. The location of the third-order maxi- [Note: t = 0 is not an answer for part (b).]
mum for the first light, yellow-orange light 1l = 600 nm2,
coincides with the location of the fourth-order maximum
for the other color’s light. What is the wavelength of the
other light?
Light
24.2 THIN-FILM INTERFERENCE

16. ●Light of wavelength 550 nm in air is normally incident
on a glass plate 1n = 1.52 whose thickness is
Glass
t
1.1 * 10-5 m. (a) What is the thickness of the glass Glass
expressed in terms of the wavelength of light in glass?
(b) How many reflected waves will experience the 180° 䉱 F I G U R E 2 4 . 2 9 Reflection or transmission? See Exercise 23.
phase shift? (c) Will the reflected waves interfere con-
structively or destructively? 24. IE ● ● An air wedge such as that shown in 䉲 Fig. 24.30
17. ● A film of index of refraction of 1.4 and thickness of can be used to measure small dimensions, such as the
1.2 * 10-5 m is on a lens with an index of refraction of diameter of a thin wire. (a) If the top glass plate is illumi-
1.6. Light of wavelength 600 nm is incident normally nated with monochromatic light, the interference pattern
from air to the film. Consider only reflections from the observed will be (1) bright, (2) dark, (3) bright and dark
top and bottom surfaces of the film. (a) How many lines based on the air film thickness. Explain. (b) Express
reflected waves will experience the 180° phase shift? the locations of the bright interference maxima in terms
(b) What is the path length difference between the two of wedge thickness measured from the apex of the
reflected waves? (c) Will the reflected waves interfere wedge.
constructively or destructively?
18. ● A lens with an index of refraction of 1.60 is to be
coated with a material 1n = 1.402 that will make the lens
nonreflecting for yellow-orange light 1l = 515 nm2 nor-
mally incident on the lens. What is the minimum Light
required thickness of the coating?
19. ● ● Magnesium fluoride 1n = 1.382 is frequently used as
a lens coating to make nonreflecting lenses. What is the t Glass
difference in the minimum film thickness required for
maximum transmission of blue light 1l = 400 nm2 and
Glass
of red light 1l = 700 nm2?

25. ● ● ● The glass plates in Fig. 24.30 are separated by a thin, boy is at an angle of 0° and another one at 19.6° from a
round filament. When the top plate is illuminated nor- line normal to the doorway. Taking the speed of sound in
mally with light of wavelength 550 nm, the filament lies air to be 335 m>s, which boy may not hear the whisle?
directly below the sixth maximum (bright) position. Prove your answer.
What is the diameter of the filament? 35. ● ● A diffraction grating is designed to have the second-
order maxima at 10° from the central maximum for red

24.3 DIFFRACTION light 1l = 700 nm2. How many lines per centimeter
26. ● A slit of width 0.15 mm is illuminated with monochro- does the grating have?
● ● Find the angles of the blue 1l = 420 nm2 and red
matic light of wavelength 632.8 nm. At what angle will 36.
the first minimum occur? 1l = 680 nm2 components of the first- and second-order
27. ● In a single-slit diffraction pattern using light of wave- maxima in a pattern produced by a diffraction grating
length 550 nm, the second-order minimum is measured with 7500 lines>cm.
to be at 0.32°. What is the slit width? 37. ● ● A certain crystal gives a deflection angle of 25° for
28. ● A slit of width 0.20 mm is illuminated with monochro- the first-order maximum of monochromatic X-rays with
matic light of wavelength 480 nm, and a diffraction a frequency of 5.0 * 1017 Hz. What is the lattice spacing
pattern is formed on a screen 1.0 m from the slit. of the crystal?
(a) What is the width of the central maximum? (b) What 38. IE ● ● (a) Only a limited number of maxima can be
are the widths of the second- and third-order maxima? observed with a diffraction grating. The factor(s) that
29. ● A slit 0.025 mm wide is illuminated with red light limit(s) the number of maxima seen is (are) (a) (1) the wave-
1l = 680 nm2. How wide are (a) the central maximum length, (2) the grating spacing, (3) both. Explain. (b) How
and (b) the side maxima of the diffraction pattern many maxima appear when monochromatic light of wave-
formed on a screen 1.0 m from the slit? length 560 nm illuminates a diffraction grating that has
30. ● At what angle will the second-order maximum be 10 000 lines>cm, and what are their order numbers?
seen from a diffraction grating of spacing 1.25 mm when 39. ● ● A diffraction grating with 6000 lines>cm is illuminated
illuminated by light of wavelength 550 nm? with a red light from a He–Ne laser 1l = 632.8 nm2. How
31. ● A venetian blind is essentially a diffraction grating—not many side maxima are formed in the diffraction pattern,
for visible light, but for waves with much longer wave- and at what angles are they observed?
lengths. If the spacing between the slats of a blind is 2.5 40. ● ● In a particular diffraction grating pattern, the red
cm, (a) for what wavelength would there be a first-order component (700 nm) in the second-order maximum is
maximum at an angle of 10°, and (b) what type of radia- deviated at an angle of 20°. (a) How many lines per cen-
tion is this? timeter does the grating have? (b) If the grating is illumi-
32. IE ● ● A single slit is illuminated with monochromatic light, nated with white light, how many maxima of the
and a screen is placed behind the slit to observe the diffrac- complete visible spectrum would be produced?
tion pattern. (a) If the width of the slit is increased, will the 41. ● ● The commonly used CD (Compact Disc) consists of
width of the central maximum (1) increase, (2) remain the many closely spaced tracks that can be used as reflecting
same, or (3) decrease? Why? (b) If the width of the slit is gratings. The industry standard for the track-to-track
0.50 mm, the wavelength is 680 nm, and the screen is 1.80 m distance is 1.6 mm. If a He–Ne laser with a wavelength of
from the slit, what would be the width of the central maxi- 632.8 nm is incident normally onto a CD, calculate the
mum? (c) What if the width of the slit is 0.60 mm? angles for all the visible maxima.
33. IE ● ● (a) If the wavelength used in a single-slit diffrac- 42. IE ● ● White light of wavelength ranging from 400 nm to
tion experiment increases, will the width of the central 700 nm is used for a diffraction grating with 6500 lines
maximum (1) increase, (2) remain the same, or per centimeter. (a) In a particular order of maximum, red
(3) decrease? Why? (b) If the width of the slit is 0.45 mm, color will have (1) a larger, (2) the same, or (3) a smaller
the wavelength is 400 nm, and the screen is 2.0 m from angle than blue color. Explain. (b) Calculate the angles
the slit, what would be the width of the central maxi- for 400 nm and 700 nm in the second-order maximum.
mum? (c) What if the wavelength is 700 nm? (c) What is the angular width of the whole spectrum in
34. ● ● A teacher standing in a doorway 1.0 m wide blows a the second order?
whistle with a frequency of 1000 Hz to summon children 43. IE ● ● White light ranging from blue (400 nm) to red
from the playground (䉲 Fig. 24.31). Two boys are playing (700 nm) illuminates a diffraction grating with
on the swings 20 m away from the school building. One 8000 lines>cm. (a) For the first maxima measured from
the central maximum, the blue color is (1) closer to,
1.0 m
䉳 F I G U R E 2 4 . 3 1 Moment (2) farther from, or (3) at the same location as the red
of truth See Exercise 34. (Not color. Explain. (b) What are the angles of the first-order
drawn to scale.) maximum for blue and red?
44. ● ● ● White light whose components have wavelengths
from 400 nm to 700 nm illuminates a diffraction grating

19.6°
20 m with 4000 lines>cm. Do the first- and second-order spec-
tra overlap? Justify your answer.
45. ● ● ● Show that for a diffraction grating, the violet
1l = 400 nm2 portion of the third-order maximum over-

laps the yellow-orange 1l = 600 nm2 portion of the
second-order maximum, regardless of the grating’s spacing.
24.4 POLARIZATION 54. IE ● ● The angle of incidence is adjusted so there is

maximum linear polarization for the reflected light
46. IE ● Unpolarized light is incident on a polarizer–
from a transparent piece of plastic in air. (a) There is
analyzer pair that can have their transmission axes at an
(1) no, (2) maximum, or (3) some light transmitted
angle of either 30° or 45°. (a) The 30° angle will allow
through the plastic. Explain. (b) If the index of refrac-
(1) more, (2) the same, or (3) less light to go through.
tion of the plastic is 1.40, what would be the angle of
(b) Calculate the percentage of light that goes through
refraction in the plastic?
the polarizer–analyzer pair in terms of the incident light
intensity. 55. IE ● ● (a) The polarizing (Brewster) angle of a piece of
flint glass 1n = 1.662 in water is (1) greater than, (2) less
47. IE ● When unpolarized light is incident on a
than, (3) the same as that of the glass in air. Explain.
polarizer–analyzer pair, 30% of the original light inten-
(b) What are the polarizing angles when it is in air and
sity passes the analyzer. What is the angle between the
submerged in water, respectively?
transmission axes of the polarizer and analyzer?
56. ● ● Sunlight is reflected off a vertical plate-glass window
1n = 1.552. What would the Sun’s altitude (angle above
48. ● Some types of glass have a range of indices of refrac-
tion of about 1.4 to 1.7. What is the range of the
the horizon) have to be for the reflected light to be com-
polarizing (Brewster) angle for these glasses when light
pletely polarized?
is incident on them from air?
49. IE ● Light is incident on a certain material in air. (a) If the 57. ● ● ● A plate of crown glass 1n = 1.522 is covered with a
index of refraction of the material increases, the polarizing layer of water. A beam of light traveling in air is incident on
(Brewster) angle will (1) also increase, (2) decrease, the water and partially transmitted. Is there any angle of
(3) remain the same. Explain. (b) What are the polarizing incidence for which the light reflected from the water–glass
angles if the index of refraction is 1.6 and 1.8? interface will have maximum linear polarization? Justify
your answer mathematically.
50. IE ● ● Unpolarized light of intensity Io is incident on a
polarizer–analyzer pair. (a) If the angle between the
polarizer and analyzer increases in the range of 0° to 90°,
the transmitted light intensity will (1) also increase, *24.5 ATMOSPHERIC SCATTERING OF
(2) decrease, (3) remain the same. Explain. (b) If the angle LIGHT
between the polarizer and analyzer is 30°, what light 58. IE ● ● Sunlight is scattered by air molecules. (a) The
intensity would be transmitted through the polarizer and intensity of the scattered blue light is (1) greater, (2) the
the analyzer, respectively? (c) What if the angle is 60°? same as, or (3) less than that of the scattered red light.
51. ● ● A beam of light is incident on a glass plate 1n = 1.622 Explain. (b) Calculate the ratio of the scattered light
in air and the reflected ray is completely polarized. What intensity of blue color (400 nm) to that of red color
is the angle of refraction for the beam? (700 nm).
52. ● ● The critical angle for total internal reflection in a certain 59. IE ● ● When sunlight is scattered by air molecules, the
media boundary is 45°. What is the polarizing (Brewster) intensity of scattered light for a wavelength of 550 nm is
angle for light externally incident on the same boundary? greater than another color by a factor of 5.0. (a) The
53. ● ● The polarizing (Brewster) angle for a certain media wavelength of the other color is (1) longer, (2) the same
boundary is 33°. What is the critical angle for total inter- as, or (3) shorter than 550 nm. Explain. (b) What is the
nal reflection for the same boundary? wavelength of the other color?
60. A thin air wedge between two flat glass plates forms ence maxima? Explain. (b) If not, which interference
bright and dark interference bands when illuminated maxima will be missing? [Hint: See Fig. 24.16.]
with normally incident monochromatic light. (See 63. Show that when the reflected light is completely
Fig. 24.9.) (a) Show that the thickness of the air wedge polarized, the sum of the angle of incidence and the
changes by l>2 from one bright band to the next, where angle of refraction is equal to 90°.
l is the wavelength of the light. (b) What would be the
change in the thickness of the wedge between bright 64. The critical angle for a certain plastic and air interface is
bands if the space were filled with a liquid with an index 39°. If the angle of incidence is adjusted so the reflected
of refraction n? light has maximum polarization, what would be the
61. A salesman tries to sell you an optic fiber that claims to angle of refraction?
give linearly polarized light when light is totally internally 65. A diver under water is looking at the overhead Sun
reflected off the fiber–air interface. (a) Would you buy it? through a diffraction grating that has 5000 lines>cm.
Explain. (b) If total internal reflection occurs at an angle of What is the highest complete spectrum order that can be
42°, what would be the polarizing (Brewster) angle? seen by the diver?
62. Three parallel slits of width w have a slit separation of d,
where d = 3w. (a) Will you be able to see all the interfer-
Vision and Optical
25 Instruments
25.1 The human eye (845)
■ nearsightedness (myopia)
■ farsightedness (hyperopia)
25.2 Microscopes (852)

■ magnifying glass
■ microscope
25.3 Telescopes (856)

■ angular magnification
■ refracting and reflecting
telescopes
25.4 Diffraction and

resolution (862) PHYSICS FACTS
■ minimum angle of resolution

■ Rayleigh criterion
✦ About 80% of the refracting power
of a human eye comes from the
cornea, while the other 20%
comes from the crystalline lens.
V ision is one of our chief means
of acquiring information about
the world around us. However, the
✦ If the human eye were compared images seen by many eyes are not
*25.5 Color (865) to a digital camera, the resolution
■ additive primary colors of the human eye would be equiv- always clear or in focus, so glasses
alent to 500 megapixels. A com-
■ subtractive primary pigments
mon digital camera has a
or some other remedy are needed.
resolution of 5 to 10 megapixels. Great progress has been made in
✦ A red blood cell has a diameter of
about 7 mm 17 * 10-6 m2. When
the last decade in contact lens
viewed with a compound micro- therapy and surgical correction of
scope at 1000* , it appears to be
7 mm 17 * 10-3 m2. vision defects. A popular procedure
✦ Some cameras on satellites have is laser surgery, as shown in the
excellent resolution. From space
they can read the license plates chapter-opening photograph.
on cars.
Laser surgery can be used for such
✦ The Giant Magellan Telescope
(GMT) is planned to be completed procedures as repairing torn
in 2016. Seven 8.4-m diameter
retinas, destroying eye tumors, and
mirrors form a single primary
mirror with an effective diameter stopping abnormal growth of
of 24.5 m (80 ft). It will have ten
times the resolving power of the
blood vessels that can endanger
Hubble Space Telescope (HST). vision.
25.1 THE HUMAN EYE 845
Optical instruments, whose basic function is to improve and extend the power of
observation beyond that of the human eye, augment our vision. Mirrors and lenses
are used in a variety of optical instruments, including microscopes and telescopes.
The earliest magnifying lenses were drops of water captured in small holes. By
the seventeenth century, artisans were able to grind fair-quality lenses for simple
microscopes or magnifying glasses, which were used primarily for botanical
studies. (These early lenses also found a use in spectacles.) Soon thereafter, the
basic compound microscope, which uses two lenses, was developed. Modern
compound microscopes, which can magnify an object up to 2000 times, extended
our vision into the microscopic world.
Around 1609, Galileo used lenses to construct an astronomical telescope that
allowed him to observe valleys and mountains on the Moon, sunspots, and the
four largest moons of Jupiter. Today, huge telescopes that use multiple lenses and
mirrors (to form very large equivalent lenses and mirrors) have extended our vision
far into the past as we look at more distant, and therefore younger, galaxies (the
light of which takes years to reach us).
How much knowledge would we lack if these instruments had never been
invented? Bacteria would still be unknown, and planets, stars, and galaxies would
have remained nothing but mysterious points of light.
Mirrors and lenses were discussed in terms of geometrical optics in
Chapter 23, and the wave nature of light was investigated in Chapter 24. These
principles can be applied to the study of vision and optical instruments. In this
chapter, you will learn about our fundamental optical instrument—the human
eye, without which all others would be of little use. Also microscopes and tele-
scopes will be discussed, along with the factors that limit their resolutions.
25.1 The Human Eye

➥ What are the components of an eye that serve analogously the same purpose as the
aperture, shutter, lens, and film of a camera.
➥ What are the three common vision defects?
➥ What type of lenses are used to correct nearsightedness and farsightedness,
respectively?
The human eye is the most important of all optical instruments. Without it we
would know little about our world and the study of optics would not exist. The
human eye is analogous to a camera in several respects (䉲 Fig. 25.1). A camera con-
sists of a converging lens, which is used to focus images on light-sensitive film
(traditional camera) or a charge-coupled device (CCD) (digital cameras) at the back
of the camera’s interior chamber. (Recall from Chapter 23 that for relatively distant
objects, a converging lens produces a real, inverted, and reduced image.) There is
an adjustable diaphragm opening, or aperture, and a shutter to control the
amount of light entering the camera.
The eye, too, focuses images onto a light-sensitive lining (the retina) on the rear
surface of the eyeball. The eyelid might be thought of as a shutter; however, the
shutter of a camera, which controls the exposure time, is generally opened only
for a fraction of a second, while the eyelid normally remains open for continuous
exposure. The human nervous system actually performs a function analogous to a
846 25 VISION AND OPTICAL INSTRUMENTS
䉴 F I G U R E 2 5 . 1 Camera and eye Film

analogy In some respects, (a) a (and focal plane)
camera is similar to (b) the human Lens
eye. An image is formed on the film Shutter
in a camera and on the retina of the
eye. (The complex refractive proper- Image
ties of the eye are not shown here,
because multiple refractive media
are involved.) See text for a compar-
ative description. Object Aperture
Rods and
cones
(a)
Crystalline
lens
Iris
Pupil Image
Cornea Optic nerve

Object Retina
Aqueous Vitreous
humor humor
(b)
Nucleus
Cortex
shutter by analyzing image signals from the eye at a rate of 20 to 30 times per
second. The eye might therefore be better likened to a movie or video camera,
which exposes a similar number of frames (images) per second.
Although the optical functions of the eye are relatively simple, its physiological
functions are quite complex. As Fig. 25.1b shows, the eyeball is a nearly spherical
chamber. It has an internal diameter of about 1.5 cm and is filled with a transpar-
ent jellylike substance called the vitreous humor. The eyeball has a white outer
covering called the sclera, part of which is visible as the “white” of the eye. Light
enters the eye through a curved, transparent tissue called the cornea and passes
into a clear fluid known as the aqueous humor. Behind the cornea is a circular
diaphragm, the iris, whose central opening is the pupil. The iris contains the pig-
ment that determines eye color. Through muscle action, the area of the pupil can
change (from 2 to 8 mm in diameter), thereby controlling the amount of light
entering the eye.
Behind the iris is a crystalline lens, a converging lens composed of microscopic
glassy fibers. (See Conceptual Example 22.5 about the internal elements, the nucleus
and cortex, inside the crystalline lens.) When tension is exerted on the lens by
attached ciliary muscles, the glassy fibers slide over each other, causing the shape,
and the focal length, of the lens to change, to help focus the image on the retina
properly. Notice that this is an inverted image (Fig. 25.1b). We do not “see” an
inverted image, however, because the brain reinterprets this image as being upright.
On the back interior wall of the eyeball is a light-sensitive surface called the
retina. From the retina, the optic nerve relays retinal signals to the brain. The
retina is composed of nerves and two types of light receptors, or photosensitive
cells, called rods and cones, because of their shapes. The rods are more sensitive to
light than the cones and distinguish light from dark in low light intensities
(twilight vision). The cones can distinguish frequency ranges but require brighter
light. The brain interprets these different frequencies as colors (color vision). Most
of the cones are clustered in a central region of the retina called the macula. The
rods, which are more numerous than the cones, are outside this region and are dis-
tributed nonuniformly over the retina.
The focusing mechanism of the eye differs from that of a simple camera. A non-
zoom camera lens has a fixed focal length, and the image distance is varied by
moving the lens relative to the film to produce sharp images for different object
distances. In the eye, the image distance is constant, and the focal length of the
lens is varied (as the attached ciliary muscles change the lens’s shape) to produce
sharp images on the retina, regardless of object distance. When the eye is focused
on distant objects, the muscles are relaxed, and the crystalline lens is thinnest with
a power of about 20 D (diopters). Recall from Chapter 23 that the power (P) of a
lens in diopters (D) is the reciprocal of its focal length in meters. So 20 D corre-
sponds to a focal length of f = 1>120 D2 = 0.050 m = 5.0 cm. When the eye is
focused on closer objects, the lens becomes thicker. Then its radius of curvature
and hence its focal length are decreased. For close-up vision, the lens power may
increase to 30 D 1f = 3.3 cm2, or even more in young children. The adjustment of
TABLE 25.1 Approximate
the focal length of the crystalline lens is called accommodation. (Look at a nearby
Near Points of the Normal
object and then at an object in the distance, and notice how fast accommodation
Eye at Different Ages
takes place. It’s practically instantaneous.)
The distance extremes over which sharp focus is possible are known as the far Age (years) Near Point (cm)
point and the near point. The far point is the greatest distance at which the eye can
see objects clearly and is infinity for a normal eye. The near point is the position 10 10
closest to the eye at which objects can be seen clearly. This position depends on the 20 12
extent to which the crystalline lens can be deformed (thickened) by accommoda- 30 15
tion. The range of accommodation gradually diminishes with age as the crys- 40 25
talline lens loses its elasticity. Generally, in the normal eye the near point gradually
50 40
recedes (increases) with age. The approximate positions of the near point at vari-
60 100
ous ages are listed in 䉴 Table 25.1.
Children can see sharp images of objects that are within 10 cm of their eyes, and
the crystalline lens of the eye of a normal young adult can do the same for objects
as close as 12 to 15 cm. However, adults at age 40 normally experience a shift in
the near point to about 25 cm. You may have noticed middle-aged people holding
reading material fairly far from their eyes so as to keep it within their range of
accommodation. When the print becomes too small (or the arms too short), correc-
tive reading glasses are one solution. The recession of the near point with age is
not considered an abnormal defect. Since it proceeds at about the same rate in
most normal eyes, it is considered a part of the natural aging process.
VISION DEFECTS
The existence of a “normal” eye (䉲 Fig. 25.2a) implies that some eyes must have
defects. This is indeed the case, as is quite apparent from the number of people
who wear corrective glasses or contact lenses. Many people have eyes that cannot
Uncorrected Corrected Uncorrected Corrected

(a) Normal (b) Nearsightedness (myopia) (c) Farsightedness (hyperopia)
䉱 F I G U R E 2 5 . 2 Nearsightedness and farsightedness (a) The normal eye produces sharp

images on the retina for objects located between its near point and its far point. The image
is real, inverted, and always smaller than the object. (Why?) Here, the object is a distant,
upward-pointing arrow (not shown) and the light rays come from its tip. (b) In a near-
sighted eye, the image of a distant object is focused in front of the retina. This defect is cor-
rected with a diverging lens. (c) In a farsighted eye, the image of a nearby object is focused
behind the retina. This defect is corrected with a converging lens. (Not drawn to scale.)
accommodate within the normal range (25 cm to infinity).* These people usually
have one or both of the two most common visual defects: nearsightedness
(myopia) or farsightedness (hyperopia). Both of these conditions can usually be
corrected with glasses, contact lenses, or surgery.
Nearsightedness (or myopia) is the ability to see nearby objects clearly, but not
distant objects. That is, the far point is less than infinity. When an object is beyond
the far point, the rays focus in front of the retina (Fig. 25.2b). As a result, the image
on the retina is blurred, or out of focus. As the object is moved closer, its image
moves back toward the retina. When the object reaches the far point for that eye, a
sharp image is formed on the retina.
Nearsightedness usually arises because the eyeball is too long (elongated in the
horizontal direction) or the curvature of the cornea is too great (bulging cornea).
In the former, the retina is placed at a farther distance from the cornea–lens sys-
tem, and in the latter, the eyeball converges the light from distant objects to a spot
in front of the retina. Whatever the reason, the image is focused in front of the
retina. Appropriate diverging lenses correct this condition. Such lenses cause the
rays to diverge before reaching the cornea. The image is thus focused farther back
on the retina. The optical purpose of the corrective lens is to form an image of a
distant object (at infinity) at the patient’s far point.
Farsightedness (or hyperopia) is the ability to see distant objects clearly, but not
nearby ones. That is, the near point is farther from the eye than normal. The image
of an object that is closer than the near point is formed behind the retina (Fig.
25.2c). Farsightedness arises because the eyeball is too short, because of insuffi-
cient curvature of the cornea, or because of the weakening of the ciliary muscles
and insufficient elasticity of the crystalline lens. If this occurs as part of the aging
process as previously discussed, it is called presbyopia.
Farsightedness is usually corrected with appropriate converging lenses. Such
lenses cause the rays to converge, and the eye is then able to focus the image on
the retina. Converging lenses are also used in middle-aged people to correct
presbyopia, a vision condition in which the crystalline lens of the eye loses its flex-
ibility, which makes it difficult to focus on close objects. The optical purpose of the
corrective lens is to form an image of an object at 25 cm (the normal near point) at
the patient’s near point.
INTEGRATED EXAMPLE 25.1 Correcting Nearsightedness: Use of Diverging Lenses

An optometrist can give a patient either regular glasses or (b) A certain nearsighted person cannot see objects clearly
contact lenses to correct nearsightedness (䉲 Fig. 25.3). Usually, when they are more than 78.0 cm from either eye. What
regular glasses sit a few centimeters in front of the eye and power must corrective lenses have, for both regular glasses
contact lenses sit right on the eye. (a) Should the power of and contact lenses, if this person is to see distant objects
the contact lenses prescribed be (1) the same as, (2) greater clearly? Assume that the glasses are 2.00 cm in front of
than, or (3) less than that of the regular glasses? Explain. the eye.
do = ∞ 䉳 F I G U R E 2 5 . 3 Correcting near-
di sightedness A diverging lens is used.
Only regular glasses are shown. For
contact lenses, the lens is immediately
d in front of the eye 1d = 02.
Image
Far point Distant

Eyeglass
object
(at ∞ )
df
*A standard near point distance of 25 cm is typically assumed in the design of optical instruments.
( A ) C O N C E P T U A L R E A S O N I N G . For nearsightedness, the cor- Note that di is negative. Recall that the power of a lens is
rective lens is a diverging one (Fig. 25.3). The lens must effec- P = 1>f (Eq. 23.9). We can use the thin-lens equation
tively put the virtual image of a distant object 1do = q 2 at the (Eq. 23.5) to find P if we can determine the object and image
patient’s far point of the eye, that is, df from the eye. The distances, do and di:
image, which acts as an object for the eye, is then within the
1 1 1 1 1 1 1
range of accommodation. Because the image distance is P = = + = + = = -
measured from the lens, a contact lens will have a longer image f do di q di di d
ƒ iƒ
distance. For a contact lens, di = - 1df2. For regular glasses,
That is, a longer ƒ di ƒ will yield a smaller P, so the contact
di = - ƒ df - d ƒ , where d is the distance between the regular
lenses should have a lower power than the regular glasses.
glasses and the eye. A negative sign and absolute values are
Thus, the answer is (3).
used for the image distance because the image is virtual,
being on the object side of the lens. (You may recall from ( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . Once it is
Section 23.3 that diverging lenses can form only virtual understood how corrective lenses work, the calculation for
images.) part (b) is straightforward.
Given: df = 78 cm = 0.780 m 1far point2 Find: P (in diopters for regular glasses)
d = 2.0 cm = 0.0200 m (glass-eye distance) P (in diopters for contact lenses)
For regular glasses,
ƒ di ƒ = ƒ df - d ƒ = 0.780 m - 0.0200 m = 0.760 m

(See Fig 25.3, which is not drawn to scale.) So, di = - 0.760 m.
Then, using the thin-lens equation,
1 1 1 1 1 1
P = = + = + = - = - 1.32 D
f do di q -0.760 m 0.760 m
A negative, or diverging, lens with a power of 1.32 D is needed.

For contact lenses, d = 0,
ƒ di ƒ = ƒ df ƒ = 0.780 m
Since the image is virtual, di = - 0.780 m.
And using the thin-lens equation,
1 1 1
P = + = - = - 1.28 D
q -0.780 m 0.780 m
The contact lenses have lower power, which is in agreement with the result of (a).
FOLLOW-UP EXERCISE. Suppose a mistake was made for the regular glasses in this Example such that a “corrective” lens of
+1.32 D were used. What would happen to the image of objects at infinity? (Answers to all Follow-Up Exercises are given in
Appendix VI at the back of the book.)
If the far point for a nearsighted person is changed using diverging lenses (see
Integrated Example 25.1), the near point will be affected as well. This causes the
close-up vision to worsen, but bifocal lenses can be used in this situation to address
the problem. Bifocals were invented by Benjamin Franklin, who glued two lenses
together. They are now made by grinding or molding lenses with different curva-
tures in two different regions. Both nearsightedness and farsightedness can be
treated at the same time with bifocals. Trifocals, or lenses having three different
curvatures, are also available. The top lens is for far vision and the bottom lens for
near vision. The middle lens is for intermediate vision and is sometimes referred
to as a lens for “computer” vision.
More modern techniques involve contact lens therapy or the use of a laser to cor-
rect nearsightedness. These are discussed in detail in Insight 25.1, Cornea
“Orthodontics” and Surgery. The purpose of either technique is to change the shape
of the exposed surface of the cornea, which changes its refractive characteristics.
INSIGHT 25.1 Cornea “Orthodontics”and Surgery

The imperfect shape of the cornea of the human eye often causes In corneal laser surgery, first, a very precise instrument
refractive errors that result in vision defects. For example, a called a microkeratome is used to create a thin corneal flap with
cornea that is curved too much can cause nearsightedness; a flat- a hinge on one side of the cornea (Fig. 2a). Once the flap is
ter-than-normal cornea can cause farsightedness. A cornea that is folded back, a tightly focused ultraviolet pulsed laser is used
not spherical can cause astigmatism (Section 23.4). to reshape the cornea. Each laser pulse accurately removes a
Recently, a nonsurgical contact lens treatment to improve microscopic layer of the cornea in the targeted area, thus
vision in a matter of hours was developed. This procedure, reshaping the cornea to correct vision defects (Fig. 2b).
called orthokeratology, or Ortho-K, is achieved in a unique way, The flap is then placed back in its original position without
with the wearing of custom-designed contact lenses. These the need for stitches (Fig. 2c). The procedure is usually pain-
contact lenses slowly change the shape of the cornea by means less, and patients typically have only minimal discomfort.
of gentle pressure to improve vision safely and quickly. The Some patients achieve corrected vision within a day after this
best analogy for Ortho-K is “orthodontics for the eye.” procedure.
Laser surgery is also used to reshape the cornea. The surgical Even more exciting advances in vision correction are on the
procedure corrects the defective shape or irregular surface of horizon. For example, researchers have developed techniques
the cornea so that it can better focus light on the retina, thereby for replacing a damaged cornea with freshly bioengineered
reducing or even eliminating vision defects (Fig. 1). tissues. If the patient has one healthy eye, stem cells are har-
vested from it. The cells grow into a sturdy layer of tissue that
surgeons can use to replace the bad corneal tissues of the
other eye by stitching the new tissue onto that damaged eye.
If both of the patient’s eyes are damaged, donor tissues may
be collected from a close relative.
(a) (b) (c)

F I G U R E 1 Eye surgery Laser surgery is performed to F I G U R E 2 Cornea reshaping (a) A flap is made on the corneal
reshape the cornea. Notice that the surgeon is wearing no surface. (b) A laser beam is used to reshape the cornea. (c) The
latex gloves. The fine chalk dust used as a lubricant on flap is placed back.
the gloves could contaminate the eye.
INTEGRATED EXAMPLE 25.2 Correcting Farsightedness: Use of Converging Lens

A farsighted person has a near point of 75 cm for the left eye verging and form an image of an object at 25 cm (the normal
and a near point of 100 cm for the right one. (a) If the person is near point) at the patient’s near point. Since the near point of
prescribed contact lenses, the power of the left lens should be the left eye (75 cm) is closer to the 25-cm normal position than
(1) greater than, (2) the same as, (3) less than the power of the the right eye, the left lens should have lower power so the
right lens. Explain. (b) What powers should contact lenses have answer is (3).
to allow the person to see an object clearly at a distance of 25 cm? ( B ) Q U A N T I T A T I V E R E A S O N I N G A N D S O L U T I O N . The two dif-
(A) CONCEPTUAL REASONING. The normal eye’s near point is ferent eyes are labeled as L (left) and R (right). The image dis-
25 cm. For farsightedness, the corrective lens must be con- tances are negative. (Why?)
Given: diL = - 75 cm = - 0.75 m (left image distance) Find: PL and PR (lens power for each eye)
diR = - 100 cm = - 1.0 m (right image distance)
do = 25 cm = 0.25 m (object distance)
A different lens prescription is usually required for each eye. dation. This situation is similar to a person wearing reading
In this case, each lens is to form an image at its eye’s near glasses (䉴 Fig. 25.4). (For the sake of clarity, the lens in
point of an object that is at a distance (do) of 0.25 m. The image Fig. 25.4a is not in contact with the eye.)
will then act as an object within the eye’s range of accommo-
Image
Object
F Near point
do = 25 cm
di
Eyeglass (a) (b)
䉱 F I G U R E 2 5 . 4 Reading glasses and correcting farsightedness (a) When an object at the

normal near point (25 cm) is viewed through reading glasses with converging lenses, the
image is formed farther away, but within the eye’s range of accommodation (beyond the
receded near point). (b) Small print as viewed through the lens of reading glasses. The
camera used to take this picture is focused past this page onto where the virtual image is.
The image distances are negative, because the images are and
virtual (that is, the image is on the same side as the object).
1 1 1 1 1 3
With contact lenses, the distance from the eye to the object PR = = + = + = = + 3.0 D
and the distance from the lens to the object are the same. Then fR do diR 0.25 m - 1.0 m 1.0 m
1 1 1 1 1 2 Note that the left lens has lower power than the right lens, as
PL = = + = + = = + 2.7 D expected.
fL do diL 0.25 m - 0.75 m 0.75 m
F O L L O W - U P E X E R C I S E . A mistake is made in grinding or molding the corrective lenses in this Example such that the left lens is made
to the prescription intended for the right eye, and vice versa. Discuss what happens to the images of an object at a distance of 25 cm.
Another common defect of vision is astigmatism, which is usually due to a

refractive surface, usually the cornea or crystalline lens, being “out of round”
(nonspherical). As a result, the eye has different focal lengths in different planes
(䉲 Fig. 25.5a). Points may appear as lines, and the image of a line may be distinct in
one direction and blurred in another or blurred in both directions. A test for astig-
matism is given in Fig. 25.5b.
Astigmatism can be corrected with lenses that have greater curvature in the
plane in which the cornea or crystalline lens has deficient curvature (Fig. 25.5c).
Astigmatism is lessened in bright light, because the pupil of the eye becomes
smaller, so only rays near the axis are entering the eye, thus avoiding the outer
edges of the cornea.
v v
Fv
h h
Fh
(a) Uncorrected astigmatism (b) Test for astigmatism (c) Corrected by lens
䉱 F I G U R E 2 5 . 5 Astigmatism When one of the eye’s refracting components is not spher-

ical, the eye has different focal lengths in different planes. (a) The effect occurs because rays
in the vertical plane (red) and horizontal plane (blue) are focused at different points: Fv and
Fh, respectively. (b) To someone with eyes that are astigmatic, some or all of the lines in this
diagram will appear blurred. (c) Nonspherical lenses, such as plano-convex cylindrical
lenses, are used to correct astigmatism.
You have probably heard of 20> 20 vision. But what is it? Visual acuity is a mea-
sure of how vision is affected by object distance. This quantity is commonly deter-
mined by using a chart of letters placed at a given distance (usually 20 ft) from the
eyes. The result is usually expressed as a fraction: The numerator is the distance at
which the test eye sees a standard symbol, such as the letter E, clearly; the
denominator is the distance at which the letter is seen clearly by a normal eye. A
20>20 1test>normal2 rating, which is sometimes called “normal” vision, means
that at a distance of 20 ft, the eye being tested can see standard-sized letters as
clearly as can a normal eye. It is possible (and, in fact, very common) to see better
than that 20>20. For example, a person with 20>15 acuity can see objects as clearly
at 20 feet away as a person with normal vision at 15 feet away from the object.
However, a person with 20>30 acuity indicates that at 20 feet, the person can see
what the normal eye can see at 30 feet.
DID YOU LEARN?

➥ The pupil, eyelid, crystalline lens, and retina of an eye serve analogously the same
purpose as the aperture, shuttle, lens, and film of a camera, respectively.
➥ The three common vision defects are nearsightedness (myopia), farsightedness
(hyperopia), and astigmatism.
➥ Converging lenses are used to correct nearsightedness, and diverging lenses are
used to correct farsightedness.
25.2 Microscopes
➥ How is angular magnification of a magnifying glass defined?

➥ At what image location is the angular magnification of a magnifying glass greater,
the near point or infinity?
➥ In a compound microscope, would a longer or shorter focal length of the objective
have a greater angular magnification?
Microscopes are used to magnify objects so that we can see more detail or see fea-
tures that are normally indiscernible. Two basic types of microscopes will be con-
sidered here.
THE MAGNIFYING GLASS (A SIMPLE MICROSCOPE)

When we look at a distant object, it appears very small. As it moves closer to our
eyes, it appears larger. How large an object appears depends on the size of the
image on the retina. This size is related to the angle subtended by the object, the
greater the angle, the larger the image (䉲 Fig. 25.6).
When wanting to look at something closely, we bring it close to our eyes so that
it subtends a greater angle. For example, you may examine the detail of a figure in
this book by bringing it closer to your eyes. You’ll see the greatest amount of detail
when the book is at your near point. If your eyes were able to accommodate to
shorter distances, an object brought closer would appear even larger. However, as
you can easily prove by bringing this book very close to your eyes, images are
blurred when objects are inside the near point.
䉴 F I G U R E 2 5 . 6 Magnification Virtual image of fly

and angle (a) How large an object
appears is related to the angle sub-
θo θ
tended by the object. (b) The angle
and the size of the virtual image of
an object are increased with a con-
Actual fly
verging lens.
(a) Narrow angle (b) Wider angle
25.2 MICROSCOPES 853
θo yo θ yo
Near point of eye F Image

25 cm do
di = 25 cm
θ
m =
θo
Without lens With lens
䉱 F I G U R E 2 5 . 7 Angular magnification The angular magnification (m) of a lens is

defined as the ratio of the angular size of an object viewed through the lens to the angular
size of the object viewed without the lens: m = u>uo. Both the object (without lens) and the
image (with lens) are at the near point (25 cm) for maximum magnification.
A magnifying glass (sometimes called a simple microscope), which is just a sin-

gle converging lens, forms a magnified image of an object when it is closer than
the focal point (Fig. 23.15b). In such a position, the image of an object subtends a
greater angle than the object itself and therefore appears larger, or magnified
(䉱 Fig. 25.7). The lens produces a virtual image beyond the near point on which the
eye focuses. If a handheld magnifying glass is used, its position is usually
adjusted until this image is seen clearly.
As illustrated in Fig. 25.7, the angle subtended by the virtual image of an object
is much greater when a magnifying glass is used. The angular magnification, or
magnifying power, of an object viewed through a magnifying glass is expressed in
terms of this angle. The angular magnification, m, is defined as the ratio of the
angular size of the object as viewed through the magnifying glass 1u2 to the angu-
lar size of the object as viewed without the magnifying glass 1uo2:
u
m = (angular magnification) (25.1)
uo
This m is not defined the same as M, the lateral magnification, which is a ratio of
heights: M = hi>ho (Section 23.1).
The maximum angular magnification occurs when the image is at the eye’s
near point, di = - 25 cm, since this position is as close as it can be seen clearly. (A
value of 25 cm will be assumed to be typical for the near point of the normal eye.
The negative sign is used because the image is virtual; see Chapter 23.) The corre-
sponding object distance can be calculated from the thin-lens equation, Eq. 23.5, as
dif 1 -25 cm2f
do = =
di - f -25 cm - f
or
125 cm2f
do = (25.2)
25 cm + f
where f must be in centimeters as well.
The angular size of the object is related to its height by
yo yo
tan uo = and tan u =
25 do
(See Fig. 25.7.) Assuming that a small-angle approximation 1tan u L u2 is valid,
yo yo
uo L and u L
25 do
Then the maximum angular magnification can be expressed as

u yo>do 25
yo>25
m = = =
uo do
Substituting for do from Eq. 25.2 gives

25
m =
25f>125 + f2
which simplifies to
25 cm (angular magnification for

m = 1 + (25.3)
f image at near point (25 cm))
where f is in centimeters. Lenses with shorter focal lengths give greater angular
magnifications.
In the derivation of Eq. 25.3, the object being viewed by the unaided eye was
taken to be at the near point, as was the image viewed through the lens. Actually,
the normal eye can focus on an image located anywhere between the near point
and infinity. When the image is at infinity, the eye is more relaxed—the muscles
attached to the crystalline lens are relaxed, and the lens is thin. For the image to be
at infinity, the object must be at the focal point of the lens. In this case,
yo
u L
f
and the angular magnification is
25 cm (angular magnification
m = (25.4)
f for image at infinity)
Mathematically, it seems that the magnifying power can be increased to any

desired value by using lenses that have sufficiently short focal lengths. Physically,
however, lens aberrations limit the practical range of a single magnifying glass to
about 3* or 4* , or a sharp image magnification of three or four times the size of
the object when used normally.
EXAMPLE 25.3 Elementary Holmes: Angular Magnification of a Magnifying Glass

Sherlock Holmes uses a converging lens with a focal length of (a) For Equation 25.3, the near point was taken to be 25 cm:
12 cm to examine the fine detail of some cloth fibers found at
25 cm 25 cm
the scene of a crime. (a) What is the maximum magnification m = 1 + = 1 + = 3.1*
given by the lens? (b) What is the magnification for relaxed- f 12 cm
eye viewing? (b) Equation 25.4 gives the magnification for the image
formed by the lens at infinity:
T H I N K I N G I T T H R O U G H . Equations 25.3 and 25.4 apply here.
Part (a) asks for the maximum magnification, which is dis- 25 cm 25 cm
m = = = 2.1 *
cussed in the derivation of Eq. 25.3 and occurs when the image f 12 cm
formed by the lens is at the near point of the eye. For part (b),
note that the eye is most relaxed when viewing distant objects. F O L L O W - U P E X E R C I S E . Taking the maximum practical mag-
nification of a magnifying glass to be 4* , which would have
SOLUTION. the longer focal length, a lens for near-point viewing or one
Given: f = 12 cm Find: (a) m 1di = near point2 for distant viewing, and how much longer would its focal
(focal length) (b) m 1di = q 2 length be?
THE COMPOUND MICROSCOPE

A compound microscope provides greater magnification than that attained with a
single lens, or a simple microscope. A basic compound microscope consists of a
25.2 MICROSCOPES 855
Eyepiece
Objective Eyepiece
Object Intermediate
image
Ie Fe
Objective
Fo
Io
Stage
do di Diaphragm
Light source
L
Final image
(a) (b)
䉱 F I G U R E 2 5 . 8 The compound
microscope (a) In the optical system
pair of converging lenses, each of which contributes to the overall magnification of a compound microscope, the real
(䉱 Fig. 25.8a). The converging lens with a relatively short focal length 1fo 6 1 cm2 is image formed by the objective falls
just within the focal point of the
known as the objective. It produces a real, inverted, and magnified image of an eyepiece (Fe) and acts as an object
object positioned slightly beyond its focal point. The other lens, called the eyepiece, for this lens. An observer looking
or ocular, has a longer focal length (fe is a few centimeters) and is positioned so that through the eyepiece sees an
the image formed by the objective falls just inside its focal point. The eyepiece forms a enlarged image. (b) A compound
virtual, upright, and magnified image of the image of the objective. Therefore, the microscope.
final image observed is virtual, inverted, and magnified. In essence, the objective
gives a magnified real image, and the eyepiece is a simple magnifying glass.
The total magnification (mtotal) of a lens combination is the product of the mag-
nifications produced by the two lenses. The image formed by the objective is
larger than its object by a factor Mo that is equal to the lateral magnification
1Mo = - di>do2. In Fig. 25.8a, note that the image distance for the objective lens is
approximately equal to the distance between the lenses, L—that is, di L L. This is
because the focal length of the eyepiece is usually much shorter than the distance
between the two lenses and the image Io is formed by the objective just inside the
focal point of the eyepiece. Also, the object is very close to the focal point of the
objective, do L fo. With these approximations,
L
Mo L -
fo
Equation 25.4 gives the angular magnification of the eyepiece for an image at infinity.
25 cm
me =
fe
The total magnification is then equal to
≤a b
L 25 cm
mtotal = Mo me = - ¢
fo fe
or
125 cm2L (angular magnification

mtotal = - (25.5)
fofe of compound microscope)
where fo , fe , and L are in centimeters.

The angular magnification of a compound microscope is negative, indicating
that the final image is inverted compared to the initial orientation of the object.
However, we often state only the magnification (100* , not - 100* ).
EXAMPLE 25.4 A Compound Microscope: Finding the Magnification

A compound microscope has an objective with a focal length Using Eq. 25.5, we get
of 10 mm and an eyepiece with a focal length of 4.0 cm. The 125 cm2L 125 cm2120 cm2
lenses are positioned 20 cm apart in the barrel. Determine the
11.0 cm214.0 cm2
mtotal = - = - = - 125 *
approximate total magnification of the microscope. fofe
THINKING IT THROUGH. This is a direct application of Eq. 25.5. Note the relatively short focal length of the objective. The
SOLUTION. negative sign indicates that the final image is inverted.
Given: fo = 10 mm = 1.0 cm Find: mtotal (total F O L L O W - U P E X E R C I S E . If the focal length of the eyepiece in
(objective focal length) magnification) this Example were doubled, how would the length of the
fe = 4.0 cm microscope change if the same magnification were still
(eyepiece focal length) desired?
L = 20 cm
(objective–eyepiece distance)
A modern compound microscope is shown in Fig. 25.8b. Interchangeable eye-

pieces with magnifications from about 5* to more than 100* are available. For
standard microscopic work in biology or medical laboratories, 5 * and 10* eye-
pieces are normally used. Microscopes are often equipped with rotating turrets,
which usually contain three objectives for different magnifications, such as 10* ,
43* , and 97* . These objectives and the 5* and 10* eyepieces can be used in
various combinations to provide magnifying powers from 50 * to 970* . The maxi-
mum magnification obtained from a compound microscope is about 2000* .
Opaque objects are usually illuminated with a light source placed above them.
Specimens that are transparent, such as cells or thin sections of tissues on glass
slides, are illuminated with a light source beneath the microscope stage so that
light passes through the specimen. A modern microscope is usually equipped
with a light condenser (converging lens) and diaphragm below the stage, which
are used to concentrate the light and control its intensity. A microscope may have
an internal light source. The light is reflected into the condenser from a mirror.
Older microscopes have two mirrors with reflecting surfaces; one is a plane mirror
for reflecting light from a high-intensity external source, and the other is a concave
mirror for converging low-intensity light such as skylight.
DID YOU LEARN?

➥ Angular magnification of a magnifying glass is the ratio of the angular size of the
image of an object viewed through the lens to the angular size of the object
viewed without the lens.
➥ The angular magnification of a magnifying glass is greater when the image is
viewed at the near point (25 cm) than infinity.
➥ In a compound microscope, a shorter focal length of the objective has a greater
angular magnification.
25.3 Telescopes
➥ What is the difference between an astronomical and a terrestrial telescope?

➥ What advantages does a reflecting telescope have over a refracting telescope?
➥ Should the objective of a telescope have a longer or shorter focal length than the
eyepiece?
Telescopes apply the optical principles of mirrors and lenses to allow some objects
to be viewed in greater detail and other fainter or more distant objects simply to be
seen. Basically, there are two types of telescopes—refracting and reflecting—
which are characterized by the gathering and converging of light by lenses or
mirrors, respectively.
25.3 TELESCOPES 857
REFRACTING TELESCOPE
The principle underlying a refracting telescope is similar to fo
that behind a compound microscope. The major components fe fe
of a refracting telescope are the objective and eyepiece lenses,
as illustrated in 䉴 Fig. 25.9. The objective is a large converging
lens with a long focal length, and the movable eyepiece has a
θo Fe Fo Fe
relatively short focal length. Rays from a distant object are
θo θ
essentially parallel and form an image (Io) at the focal point
(Fo) of the objective. This image acts as an object for the eye- Io
piece, which is moved until the image lies just inside its focal
point (Fe). The final image seen by the observer is virtual, Eyepiece
Objective
inverted, and magnified.
For relaxed viewing, the eyepiece is adjusted so that its
image (Ie) is at infinity, which means that the image of the objec-
tive (Io) is at the focal point of the eyepiece ( fe). As Fig. 25.9
shows, the distance between the lenses is then the sum of the
focal lengths 1fo + fe2, which is the length of the telescope tube. Ie
The magnification of a refracting telescope focused for the
final image at infinity can be shown to be 䉱 F I G U R E 2 5 . 9 The refracting
astronomical telescope In an astro-
nomical telescope, rays from a dis-
fo (angular magnification tant object form an intermediate
m = - (25.6) image (Io) at the focal point of the
fe of refracting telescope)
objective (Fo). The eyepiece is
moved so that the image is at or
where the negative sign indicates that the image is inverted, as in our lens sign slightly inside its focal point (Fe). An
observer sees an enlarged image at
convention described in Section 23.3. Thus, to achieve the greatest magnification,
infinity (Ie, shown at a finite dis-
the focal length of the objective should be made as long as possible and the focal tance here for illustration).
length of the eyepiece as short as possible.
The telescope illustrated in Fig. 25.9 is called an astronomical telescope. The
final image produced by an astronomical telescope is inverted, but this condi-
tion poses little problem to astronomers. (Why?) However, someone viewing an
object on Earth through a telescope finds it more convenient to have an upright
image. A telescope in which the final image is upright is called a terrestrial
telescope. An upright final image can be obtained in several ways; two are illus-
trated in 䉲 Fig. 25.10.
In the telescope diagrammed in Fig. 25.10a, a diverging lens is used as an eye-
piece. This type of terrestrial telescope is referred to as a Galilean telescope, because
Galileo built one in 1609. A real and inverted image is formed by the objective to
the left of the eyepiece, and this image acts as a “virtual” object for the eyepiece.
(See Section 23.3.) The diverging lens then forms a virtual, inverted, and magni-
fied image of the image of the objective. Therefore, the final image observed is vir-
tual, upright, and magnified. (The image is inverted twice so the final image is
upright.) Also note that with a diverging lens and negative focal length, Eq. 25.6
gives a positive m, indicating an upright image.
Galilean telescopes have several disadvantages, most notably very narrow
fields of view and limited magnification. Another type of terrestrial telescope,
illustrated in Fig. 25.10b, uses a third lens, called the erecting lens, or inverting lens,
between the converging objective and eyepiece lenses. If the image is formed by
the objective at a distance that is twice the focal length of the intermediate erecting
lens (2fi), then the lens merely inverts the image without magnification, and the
telescope magnification is still given by Eq. 25.6.
However, achieving the upright image in this way requires a longer telescope
length. Using the intermediate erecting lens to invert the image increases the
length of the telescope by four times the focal length of the erecting lens (2fi on
each side). The inconvenient extra length can be decreased by using internally
reflecting prisms. This is the principle behind prism binoculars, which are really
double telescopes—one for each eye (䉲 Fig. 25.11).
䉴 F I G U R E 2 5 . 1 0 Terrestrial tele- Final image

scopes (a) A Galilean telescope uses
a diverging lens as an eyepiece, pro- Rays from
Eyepiece
distant object
ducing upright, virtual images.
(b) Another way to produce upright Fe Intermediate
images is to use a converging “erect- image
ing” lens (focal length fi) between
the objective and eyepiece in an Objective
astronomical telescope. This addi-
tion elongates the telescope, but the
length can be shortened by using
internally reflecting prisms. (a) Galilean telescope
Rays from Final image

distant object
Erecting Fi
lens
Second
First image
image
Objective 2fi 2fi
(b) Terrestrial telescope
䉴 F I G U R E 2 5 . 1 1 Prism binocu- Objective

lars A schematic cutaway view of
one ocular (one half of a pair of
prism binoculars), showing the
internal reflections in the prisms,
which reduce the overall physical
length. The prisms also erect the
image.
Eyepiece
EXAMPLE 25.5 An Astronomical Telescope—and a Longer Terrestrial Telescope

An astronomical telescope has an objective lens with a focal T H I N K I N G I T T H R O U G H . Equation 25.6 applies directly to
length of 30 cm and an eyepiece with a focal length of 9.0 cm. part (a). In part (b), the erecting lens elongates the telescope
(a) What is the magnification of the telescope? (b) If an erect- by four times the focal length of the erecting lens (4fi)
ing lens with a focal length of 7.5 cm is used to convert the (Fig. 25.10b).
telescope to a terrestrial type, what is the overall length of the
telescope tube?

Given: fo = 30 cm (objective focal length) Find: (a) m (magnification)
fe = 9.0 cm (eyepiece focal length) (b) L (length of telescope tube)
fi = 7.5 cm (intermediate erecting lens focal length)
25.3 TELESCOPES 859
(a) The magnification is given by Eq. 25.6 as sum of the lenses’ focal lengths (without the erecting lens):
fo 30 cm L1 = fo + fe = 30 cm + 9.0 cm = 39 cm
m = - = - = - 3.3*
fe 9.0 cm
With the erecting lens, the overall length is then
where the negative sign indicates that the final image is
inverted. L = L1 + L2 = 39 cm + 4fi = 39 cm + 417.5 cm2 = 69 cm
(b) Taking the length of the astronomical tube to be the dis- Hence, the telescope length is 77% 130>39 = 0.772 longer, with
tance between the lenses, we find that this length is just the an upright image, but the same magnification, 3.3 * (why?).
F O L L O W - U P E X E R C I S E . A terrestrial telescope 66 cm in length has an intermediate erecting lens with a focal length of 12 cm.
What is the focal length of an erecting lens that would reduce the telescope length to a more manageable 50 cm?
CONCEPTUAL EXAMPLE 25.6 Constructing a Telescope

A student is given two converging lenses, one with a focal If the object is at a great distance, a real image is formed by
length of 5.0 cm and the other with a focal length of 20 cm. To the objective lens in the focal plane of the lens (Fig. 25.9). This
construct a telescope to best view distant objects with these image acts as the object for the eyepiece, which is positioned so
lenses, the student should hold the lenses (a) more than 25 cm that the image by the objective lies just inside its focal point so
apart, (b) less than 25 cm but more than 20 cm apart, (c) less as to produce a second inverted and magnified image.
than 20 cm but more than 5.0 cm apart, (d) less than 5.0 cm The two lenses must be slightly less than 25 cm apart, so
apart. Specify which lens should be used as the eyepiece. answer (a) is not correct. Answers (c) and (d) are also not
correct, because the eyepiece would be too close to the
REASONING AND ANSWER. First, let’s see which lens should objective to produce the second magnified image needed
be used as the eyepiece. The only type of telescope that can be for optimal viewing of a distant object. In these cases, the
constructed with two converging lenses is an astronomical tele- rays would pass through the second lens before the image
scope. In this type of telescope, the lens with the longer focal was formed, and a reduced image might be produced. (See
length is used as an objective lens to produce a real image of a Section 23.3.) Thus, answer (b), with the image by the objec-
distant object. That image is then viewed with the lens with the tive just inside the eyepiece’s focal point, is the correct
shorter focal length, the eyepiece, used as a simple magnifier. answer.
F O L L O W - U P E X E R C I S E . A third converging lens with a focal length of 4.0 cm is used with the aforementioned two lenses to pro-
duce a terrestrial telescope in which the third lens does nothing more than invert the image. How should the lenses be positioned
and how far apart should they be for the final image to be of maximum size and upright?
REFLECTING TELESCOPE
For viewing the Sun, Moon, and nearby planets, large magnifications are important
to see details. However, even with the highest feasible magnification, stars appear
only as faint points of light. For distant stars and galaxies, it is more important to
gather more light than to increase the magnification, so that the object can be seen
and its spectrum analyzed. The intensity of light from a distant source is sometimes
very low. In many instances, such a source can be detected only when the light is
gathered and focused on a photographic plate over a long period of time.
Recall from Section 14.3 that intensity is energy per unit time per unit area. Thus,
more light energy can be gathered if the size of the objective is increased. This
increases the distance at which the telescope can detect faint objects, such as distant
galaxies. (Recall that the light intensity of a point source is inversely proportional to
the square of the distance between the source and the observer.) However, producing
a large lens involves difficulties associated with glass quality, grinding, and polish-
ing. Compound lens systems are required to reduce aberrations, and a very large
lens may sag under its own weight, producing further aberrations. The largest
objective lens in use has a diameter of 40 in. (102 cm) and is part of the refracting
telescope of the Yerkes Observatory at Williams Bay, Wisconsin.
Eyepiece
(a) (b)
䉱 F I G U R E 2 5 . 1 2 Reflecting telescopes A concave parabolic mirror can be used in a tele-

scope to converge light to form an image of a distant object. (a) The image may be at the
prime focus, or (b) a small mirror and lens can be used to focus the image outside the tele-
scope, a configuration called a Newtonian focus.
These problems can be reduced by using a reflecting telescope, which uses a

large, concave, parabolic mirror (䉱 Fig. 25.12). A parabolic mirror does not exhibit
spherical aberration, and a mirror has no inherent chromatic aberration. (Section
23.5, why?) High-quality glass is not needed, because the light is reflected by a
mirrored surface. Only one surface has to be ground, polished, and silvered.
The largest optical reflecting telescope, with a mirror 8.4 m (331 in.) in diameter,
is the Large Binocular Telescope (LBT) in Arizona, USA. The large mirror (8.2 m or
323 in.) is used at the European Southern Observatory in Chile (䉲 Fig 25.13a).
Even though reflecting telescopes have advantages over refracting telescopes,
they do have problems. Like a large lens, a large mirror may sag under its own
weight. The weight factor also increases the costs of construction, because the sup-
porting elements for a heavier mirror must be more massive.
(a) (b)
䉱 F I G U R E 2 5 . 1 3 Large reflecting telescopes (a) An 8.2-m-diameter mirror for the Euro-

pean Southern Observatory, near Paranal, Chile, is undergoing the final phase of polishing.
(b) Seven 8.4-m diameter mirrors forming a single primary mirror with an effective mirror
diameter of 24.5 m (80 ft) is planned for the Giant Magellan Telescope (GMT).
25.3 TELESCOPES 861
These problems are being addressed by new technologies. One approach is to

use an array of mirrors, coordinated to function as a single large mirror. Examples
include the the European Southern Observatory’s four 8.2-m-diameter mirrors
linked to form a VLT (Very Large Telescope) with an equivalent diameter of 16 m.
The Giant Magellan Telescope (GMT) is planned to be completed in 2016. Seven
8.4-m diameter mirrors will form a single primary mirror with an effective mirror
diameter of 24.5 m or 80 ft (Fig. 25.13b).
Another way of extending our view into space is to put telescopes into orbit
around the Earth. Above the atmosphere, the view is unaffected by the twinkling
effect of atmospheric turbulence and refraction, and there is no background prob-
lem from city lights. In 1990, the optical Hubble Space Telescope (HST) was
launched into orbit (䉴 Fig. 25.14). Even with a mirror diameter of only 2.4 m, its
privileged position has allowed the HST to produce images seven times as clear as
those formed by similarly sized Earth-bound telescopes. A flawed optics design of
the HST was repaired in 1993, and it is scheduled for more repairs in 2009. The
repairs, along with the addition of new instruments, will make the HST 90 times
as powerful as it was in 1993.
Lastly, you may know that not all telescopes use the visible region of light. For
more information and examples, read Insight 25.2, Telescopes Using Nonvisible 䉱 F I G U R E 2 5 . 1 4 Hubble Space
Radiation. Telescope (HST) Late in 1993, astro-
nauts from the space shuttle
Endeavor visited the HST in orbit.
They installed corrective equipment
DID YOU LEARN?
that compensated for many of the
➥ The final image of an astronomical telescope is inverted, while the final image of telescope’s optical flaws and
repaired or replaced other malfunc-
a terrestrial telescope is upright.
tioning systems. The HST is cur-
➥ A reflecting telescope can use a larger parabolic objective and does not have rently scheduled for repairs in 2009.
chromatic and spherical aberration. The repairs, along with the addition
➥ The objective of a telescope should have a longer focal length than the of new instruments, will make the
eyepiece. HST ninety times as powerful as it
was in 1993.
INSIGHT 25.2 Telescopes Using Nonvisible Radiation

The word telescope usually brings to mind visual observations. Infrared light is also affected by the Earth’s atmosphere. For
However, the visible region is only a very small part of the example, water vapor is a strong absorber of infrared radiation.
electromagnetic spectrum, and celestial objects emit many
other types of radiation, including radio waves. This fact was
discovered accidentally in 1931 by an American electrical
engineer named Carl Jansky while he was working on static
interference with intercontinental radio communications.
Jansky found an annoying static hiss that came from a fixed
direction in space, apparently from a celestial source. It was
soon clear that radio waves could be another valuable source
of astronomical information, and radio telescopes were built
to investigate this source.
A radio telescope operates similarly to a reflecting visible
light telescope. A reflector with a large area collects and
focuses the radio waves at a point where a detector collects
the signal (Fig. 1). The parabolic collector, called a dish, is
covered with metal wire mesh or metal plates. Since the
wavelengths of radio waves range from a few millimeters to
several meters, wire mesh is “smooth” enough and a good
reflecting surface for such waves. F I G U R E 1 Radio telescopes Several of the dish antennae
that make up the Very Large Array (VLA) radio telescope
Radio telescopes supplement optical telescopes and pro-
near Socorro, New Mexico. There are twenty-seven movable
vide some definite advantages over them. For instance, dishes, each 25 m in diameter, forming the array along a Y-
radio waves pass freely through the huge clouds of dust that shaped railway network. The data from all the antennae are
hide a large part of our galaxy from visual observation. Also, combined to produce a single radio image. In this way, it is
radio waves easily penetrate the Earth’s atmosphere, which possible to attain a resolution equivalent to that of one giant
reflects and scatters a large percentage of the incoming visi- radio dish (a couple hundred feet in diameter).
ble light. (continued on next page)
Thus, observations with infrared telescopes are sometimes The atmosphere is virtually opaque to ultraviolet radia-
made from high-flying aircraft or from orbiting spacecraft, tion, X-rays, and gamma rays, so telescopes that detect these
beyond the influence of atmospheric water vapor. The first types of radiation cannot be Earth based. Orbiting satellites
orbiting infrared observatory was launched in 1983. Not only with telescopes sensitive to these types of radiation have
is atmospheric interference eliminated in space, but the tele- mapped out portions of the sky, and other surveys are
scope may be cooled to a very low temperature without planned. Observations by orbiting satellites in the visible
becoming coated with condensed water vapor. Cooling the region are not affected by air turbulence or refraction. Per-
telescope helps eliminate the interference of infrared radia- haps in the not-too-distant future, a permanently staffed
tion generated by the telescope itself. The orbiting infrared orbiting observatory carrying a variety of telescopes will
telescope launched in 1983 was cooled with liquid helium to replace the uncrewed Hubble Telescope and help expand our
about 10 K; it carried out an infrared survey of the entire sky. knowledge of the universe.
25.4 Diffraction and Resolution

➥ What is the fundamental cause of images of two separate objects not being
resolved?
➥ What is the Rayleigh criterion?
➥ To resolve small details, should an optical system use large lenses or small lenses?
The diffraction of light places a limitation on our ability to distinguish objects that are
close together when microscopes or telescopes are used. This effect can be under-
stood by considering two point sources located far from a narrow slit of width w
(䉲 Fig. 25.15). The sources could represent distant stars, for example. In the absence of
diffraction, two bright spots, or images, would be observed on a screen. As you know
from Section 24.3, however, the slit diffracts the light, and each image is a diffraction
pattern that consists of a central maximum with a pattern of weaker bright and dark
positions on either side. If the sources are close together, the two central maxima may
overlap. In this case, the images cannot be distinguished, or are said to be unresolved.
For the images to be resolved, the central maxima must not overlap appreciably.
In general, images of two sources can be resolved if the center of the central
maximum of one falls at or beyond the first minimum of the other. This limiting
condition for the resolution of two images—that is, the ability to distinguish them
as separate—was first proposed by Lord Rayleigh (1842–1919), a British physicist.
The condition is known as the Rayleigh criterion:
Two images are said to be just resolved when the center of the central maximum of
one image falls on the first minimum of the diffraction pattern of the other image.
The Rayleigh criterion can be expressed in terms of the angular separation 1u2
of the sources. (See Fig. 25.15.) The first minimum 1m = 12 for a single-slit diffrac-
tion pattern satisfies this relationship:
l
w sin u = ml = l or sin u =
w
䉴 F I G U R E 2 5 . 1 5 Resolution
Two light sources in front of a slit
produce diffraction patterns.
(a) When the angle subtended by S1
the sources at the slit is large S1
enough for the images to be distin-
guishable, the images are said to be θ
resolved. (b) At smaller angles, the w θ min
central maxima are closer together.
At umin, the center of the central S2 S2
maximum of one image falls on the
first minimum of the other image,
and the images are said to be just
resolved. For smaller angles, the
patterns are unresolved. Slit Screen Slit Screen
(a) Resolved (b) Just resolved

25.4 DIFFRACTION AND RESOLUTION 863
According to Fig. 25.15, this is the minimum angular separation for two images
to be just resolved according to the Rayleigh criterion. In general, for visible light,
the wavelength is much smaller than the slit width 1l 6 w2, so u is small and
sin u L u. In this case, the limiting, or minimum angle of resolution (Umin) for a
slit of width w is
l (minimum angle of resolution

umin = (25.7)
w for a slit)
Note that umin is dimensionless (a pure number) and is therefore in radians. Thus,
the images of two sources will be distinctly resolved if the angular separation of
the sources is greater than l>w.
The apertures (openings) of cameras, microscopes, and telescopes are generally
circular. Thus, there is a circular diffraction pattern around the central maximum, (a)
in the form of a bright circular disk (䉴 Fig. 25.16). Detailed analysis shows that the
minimum angle of resolution for a circular aperture for the images of two objects
to be just resolved is similar to, but slightly different from, Eq. 25.7. It is
1.22l (minimum angle of resolution

umin = (25.8)
D for a circular aperture)
where D is the diameter of the aperture and umin is again in radians.

Equation 25.8 applies to the objective lens of a microscope or telescope, or the
iris of the eye, all of which may be considered to be circular apertures for light.
According to Eqs. 25.7 and 25.8, the smaller umin , the better the resolution. The
minimum angle of resolution, umin , should be small so that objects close together
can be resolved; therefore, the aperture should be as large as possible. This is yet
another reason for using large lenses (and mirrors) in telescopes.
(b)
For a microscope, it is more convenient to specify the actual separation (s)
between two point sources. Since the objects are usually near the focal point of the 䉱 F I G U R E 2 5 . 1 6 Circular aper-
objective, to a good approximation, ture resolution (a) When the angular
separation of two objects is large
s enough, the images are well resolved.
umin = or s = fumin
f (Compare with Fig. 25.15a.)
(b) Rayleigh criterion: The center of
where f is the focal length of the lens and umin is expressed in radians. (Here, s is the central maximum of one image
taken as the arc length subtended by umin, and s = rumin = fumin.) Then, using falls on the first minimum of the other
Eq. 25.8, we get image. (Compare with Fig. 25.15b.)
The images of objects with smaller
angular separations cannot be clearly
1.22lf distinguished as individual images.
s = fumin = (resolving power of a microscope) (25.9)
D
This minimum distance between two points whose images can be just resolved
is called the resolving power of the microscope. Note that s is directly propor-
tional to l, so shorter wavelengths give better resolution. In practice, the resolv-
ing power of a microscope indicates the ability of the objective to distinguish
fine detail in specimens’ structures. For another real-life example of resolution,
see 䉲 Fig. 25.17.
䉳 F I G U R E 2 5 . 1 7 Real-life resolu-
tion (a), (b), (c) A sequence of an
approaching automobile’s head-
lights. In (a), the headlights are just
resolved through the circular aper-
ture of the camera (or your eye). As
the automobile moves closer, the
headlights are more resolved.
(a) (b) (c)

EXAMPLE 25.7 Viewing from Space: The Great Wall of China

The Great Wall of China was originally about 2400 km
(1500 mi) long, with a base width of about 6.0 m and a top
width of about 3.7 m. Several hundred kilometers of the wall
remain intact (䉴 Fig. 25.18). It is sometimes said that the wall
is the only human construction that can be seen with the
unaided eye by an astronaut in an orbit that is 200 km (125
mi) above the Earth. If the pupil of the eye is 4.0 mm for day-
time visible light with a wavelength of 550 nm, see whether
the wall is visible according to the Rayleigh criterion. (Neglect
any atmospheric effects.)
T H I N K I N G I T T H R O U G H . Despite the great length of the wall,
it would not be visible from space unless its width subtends an
angle that is greater than the minimum angle of resolution for
the eye of an observing astronaut. The angle subtended by the
width of the wall is calculated by s = ru (Eq. 7.3), where s is
the maximum observable width of the wall and r is the dis- 䉱 F I G U R E 2 5 . 1 8 The Great Wall The walkway of the
tance from the wall to the astronaut. This is then compared Great Wall of China, which was built as a fortification along
with umin for the eye—a direct application of Eq. 25.8. China’s northern border.
SOLUTION.
Given: D = 4.0 mm = 4.0 * 10-3 m (diameter) Find: u and umin (minimum angles of resolution)
l = 550 nm = 5.50 * 10-7 m (wavelength)
s = 6.0 m (width)
r = 200 km = 2.0 * 105 m (distance)
The angle subtended by the width of the wall to the astrau- Since u V umin , the wall would not be able to be seen with
naut is the unaided eye.
s 6.0 m Now, here is the living proof. In 2003, Chinese astronaut
u = = = 3.0 * 10-5 rad Yang Liwei went to space in a historic mission. (He was the
r 2.0 * 105 m very first Chinese to go into space.) After he returned to the
The minimum angle of resolution for the eye is Earth, he was asked if he was able to see the Great Wall in
1.2215.50 * 10-7 m2 space, “I did not see our Great Wall from space,” Yang said in
1.22l
umin = = = 1.7 * 10-4 rad an interview with China Central Television.
D 4.0 * 10-3 m
F O L L O W - U P E X E R C I S E . What would be the minimum diameter of the objective of a telescope that would allow an astronaut
orbiting the Earth at an altitude of 300 km to actually see the Great Wall?
Note from Eq. 25.8 that higher resolution can be gained by using radiation of a
shorter wavelength. Thus, a telescope with an objective of a given size will have
greater resolution with violet light than with red light. For microscopes, it is possi-
ble to increase resolving power by shortening the wavelengths of the light used to
create the image. This can be done with a specialized objective called an oil immer-
sion lens. When such a lens is used, a drop of transparent oil fills the space between
the objective and the specimen. Recall that the wavelength of light in oil is
l¿ = l>n, where n is the index of refraction of the oil and l is the wavelength of
light in air. For values of n about 1.50 or higher, the wavelength is significantly
reduced, and the resolution is increased proportionally.
EXAMPLE 25.8 Optical and Radio Telescopes: Resolution and the Rayleigh Criterion
Determine the minimum angle of resolution by the Rayleigh T H I N K I N G I T T H R O U G H . This Example involves the mini-
criterion for (a) the Giant Magellan Telescope (GMT) with an mum angle of resolution umin (Eq. 25.8). Equation 13.17 needs
effective mirror diameter of 24.5 m for visible light of 550 nm to be used to determine the wavelength of radio waves
and (b) the Very Large Array (VLA) radio telescope with an because only the frequency is given. The speed of radio
effective diameter of 36 km at the highest frequency of 43 GHz. waves is 3.00 * 108 m>s.
*25.5 COLOR 865
SOLUTION.
Given: (a) D = 24.5 m (diameter) Find: (a) and (b) umin (minimum angles of resolution)
l = 550 nm = 5.50 * 10-7 m (wavelength)
(b) D = 36 km = 3.6 * 104 m (diameter)
f = 43 GHz = 4.3 * 1010 Hz (frequency)
v = 3.00 * 108 m>s
(a) For the light telescope, (b) The wavelength of the radio telescope is calculated with
1.2215.50 * 10 -7
m2 Eq. 13.17,
umin =
24.5 m
= 2.7 * 10-8 rad v 3.00 * 108 m>s
l = = = 6.98 * 10-3 m
f 4.3 * 1010 Hz
This is significantly smaller (better) than the minimum
angle of resolution for the unaided eye from Example 25.7. The minimum angle of resolution is then
(The smaller the minimum angular of resolution, the better 1.2216.98 * 10-3 m2
the resolution.) umin = = 2.4 * 10-7 rad
3.6 * 104 m
(Note: The resolution of Earth-bound telescopes with large-
diameter objectives is also affected by other effects, such as These are very small angles of resolutions 110-7 rad2 that are
atmospheric turbulence. Thus, in actuality, the GMT will have sufficient to see a golf ball (diameter about 4.0 cm or 1.6 in.)
a umin on the order of 10-7 rad, or a resolution one-tenth as held by a friend 150 km (100 miles) away. (Can you calculate
good as that without the effect of the atmosphere.) the angle subtended by the golf ball at that distance?)
F O L L O W - U P E X E R C I S E . As noted in Section 25.3, the Hubble Space Telescope has a mirror diameter of 2.4 m. How does its reso-
lution compare with that of the GMT in part (a) of this Example?
DID YOU LEARN?

➥ The image of each object is a diffraction pattern of the object through an optical
system. If the central maxima of the two images overlap appreciably, then the two
images are not resolved.
➥ The Rayleigh criterion states that when the center of the central maximum of one
image falls on the first minimum of another image, the two images are just
resolved.
➥ A large lens can resolve small details better because the minimum angle of
resolution is inversely proportional to the diameter of a lens.
*25.5 Color
➥ What are the three additive primary colors?
➥ What are complementary colors?
➥ What are the three subtractive primary pigments?
In general, physical properties are fixed or absolute. For example, a particular

type of electromagnetic radiation has a certain frequency or wavelength. How-
ever, visual perception of this radiation may vary from person to person. How we
“see” (or our brain “interprets”) radiation gives rise to what we call color vision.
COLOR VISION
Color is perceived because of a physiological response to excitation by light of the
cone receptors in the retina of the human eye. (Many animals have no cone cells
and thus live in a black-and-white world.) The cones are sensitive to light with fre-
quencies approximately between 7.5 * 1014 Hz and 4.3 * 1014 Hz (wavelengths
between 400 and 700 nm). The signals representing different frequencies of light
are perceived by the brain as different colors. The association of a color with a par-
ticular frequency is subjective and may vary from person to person.
The details of color vision are not well understood. It is known that there are
three types of cones responding to different parts of the visible spectrum: the red,
䉴 F I G U R E 2 5 . 1 9 Sensitivity of
GE
T
OW
UE
LE
EE
IG
AN
D
cones Different types of cones in
VIO
BL
IND
RE
LL
GR
OR
YE
the retina of the human eye may
Relative cone sensitivity

respond to different frequencies of
light to give three general color
responses: red, green, and blue. Green cones
Red cones
Blue cones
700 600 500 400 ␭ (nm)

4.3 7.5 f (1014 Hz)
green, and blue regions (䉱 Fig. 25.19). Presumably, each cone absorbs light in a spe-
cific range of frequencies and all three functionally overlap to form combinations that
are interpreted by the brain as the various colors of the spectrum. For example, when
red and green cones are stimulated equally, the brain interprets the two super-
imposed signals as yellow. But when the red cones are stimulated more strongly than
the green cones, the brain senses orange (that is, “yellow” but dominated by red).
Color blindness results when one or more type of cone is missing or nonfunctional.
As Fig. 25.19 shows, the human eye is not equally sensitive to all colors. Some
colors evoke a greater response than others and therefore appear brighter at the
same intensity. The wavelength of maximum visual sensitivity is about 550 nm, in
the yellow-green region.
The foregoing theory of color vision (mixing or combining) is based on the
experimental fact that beams of varying intensities of red, green, and blue light
can be arranged to produce most other colors. The red, blue, and green from
which we interpret a full spectrum of colors are called the additive primary
colors. When light beams of the additive primaries are projected and overlapped
on a white screen, other colors are produced, as illustrated in 䉳 Fig. 25.20. This
technique is called the additive method of color production. Triad dots consisting
of three phosphors that emit the additive primary colors are used in television pic-
ture tubes to produce colored images.
Note in Fig. 25.20 that a certain combination of the primary colors appears white
(a) to the eye. Also, many pairs of colors appear white to the eye when combined. The
colors of such pairs are said to be complementary colors. The complement of blue is
yellow, that of red is cyan, and that of green is magenta. As the figure also shows, the
complementary color of a particular primary is the combination, or sum, of the other
two primaries. Hence, the primary and its complement together appear white.
Objects exhibit a color when they are illuminated with white light because they
Red Yellow Green
reflect (scatter) or transmit light predominantly in the frequency range of that
White color. The other frequencies of the white light are mostly absorbed. For example,
when white light strikes a red apple, most of the energy in the red portion of the
spectrum is reflected—most of all the others (and thus all other colors) are
absorbed. Similarly, when white light passes through a piece of transparent red
Blue
glass, or a filter, mostly the light associated with red is transmitted. This occurs
because the color pigments in the glass are selective absorbers.
Magenta Cyan
(purplish–red) (turquoise) Pigments are mixed to form various colors, such as in the production of paints
and dyes. You are probably aware that mixing yellow and cyan (“true” blue) paints
(b)
produces green. This is because the yellow pigment absorbs most of the wave-
䉱 F I G U R E 2 5 . 2 0 Additive lengths except those in the yellow and nearby regions (green plus orange) of the
method of color production When visible spectrum, and the cyan pigment absorbs most of the wavelengths except
light beams of the primary colors those in the blue and nearby regions (violet plus green). The wavelengths in the
(red, blue, and green) are projected intermediate (overlap) green region, between the yellow and cyan range, are not
onto a white screen, mixtures of
them produce other colors. Varying strongly absorbed by either pigment, and therefore the mixture appears green. The
the intensities of the beams allows same effect can be accomplished by passing white light through stacked yellow and
most colors to be produced. cyan filters. The light coming through both filters appears green.
*25.5 COLOR 867
Cyan Magenta
(turquoise) (purplish–red)
Red
RGB R G R
light
Blue Light
Yellow filter Magenta filter
Black
(absorbs B) (absorbs G)
Green Red
Yellow Green
RGB G B G
light
Light
(a) Cyan filter Yellow filter
(absorbs R) (absorbs B)
Blue
RGB R B B light
Light
Magenta filter Cyan filter
(absorbs G) (absorbs R)
(b)
䉱 F I G U R E 2 5 . 2 1 Subtractive method of color production (a) When the primary pig-

ments (cyan, magenta, and yellow) are mixed, different colors are produced by subtractive
absorption; for example, the mixing of yellow and magenta produces red. When all three
pigments are mixed and all the wavelengths of visible light are absorbed, the mixture
appears black. (b) Subtractive color mixing, using filters. The principle is the same as in
part (a). Each pigment selectively absorbs certain colors, removing them from the white
light. The colors that remain are what we see.
Mixing pigments results in the subtraction of colors. The resultant color is cre-
ated by whatever is not absorbed by the pigment—that is, not subtracted from the
original beam. This is the principle of the subtractive method of color production.
Three particular pigments—cyan, magenta, and yellow—are the subtractive
primary pigments. Various combinations of two of the three subtractive primaries
produce the three additive primary colors (red, blue, and green), as illustrated in
䉱 Fig. 25.21. When the subtractive primaries are mixed in the proper proportions,
the mixture appears black (because all wavelengths are absorbed). Painters often
refer to the subtractive primaries as red, yellow, and blue. They are loosely refer-
ring to magenta (purplish-red), yellow, and cyan. Mixing these paints in the
proper proportions produces a broad spectrum of colors.
Note in Fig. 25.21 that the magenta pigment essentially subtracts the color green
where it overlaps with cyan and yellow. As a result, magenta is sometimes referred
to as “minus green.” If a magenta filter were placed in front of a green light, no light
would be transmitted. Similarly, cyan is called “minus red,” and yellow is called
“minus blue.” An example of subtractive color mixing is a photographer’s use of a
yellow filter to bring out white clouds on black and white film. This filter absorbs
blue from the sky, darkening it relative to the clouds, which reflect white light.
Hence, the contrast between the two is enhanced. What type of filter would you use
to darken green vegetation on black and white film? To lighten it?
DID YOU LEARN?

➥ The three additive primary colors are red, green, and blue.
➥ When a pair of combined colors appear white, the colors are said to be
complementary. For example, yellow and blue are complementary because they
appear white when combined.
➥ The three subtractive primary pigments are cyan, magenta, and yellow.
PULLING IT TOGETHER Microscope and Resolution: Dry or Oil Immersion

A compound microscope is designed to have a total magnifi- T H I N K I N G I T T H R O U G H . This Example combines the magnifi-
cation of - 400 * and a minimum resolving power of 0.400 mm cation of a compound microscope, the resolving power and
when it is operated in air (dry). The focal length of the eye- minimum angle of resolution by the Rayleigh criterion, and
piece is 2.50 cm and the distance between the objective and the wavelength dependence on index of refraction. (a) This is
the eyepiece is 20.0 cm. The microscope can also operate as an a direct application of total magnification of a compound
oil immersion microscope when a drop of oil with an index of microscope (Eq. 25.5). (b) The minimum resolving power is
refraction of 1.56 fills the space between the specimen and the the product of the focal length of the objective and the mini-
objective. (a) What should be the focal length of the objective? mum angle of resolution (Eq. 25.9). (c) The diameter of the
(b) Determine the minimum angle of resolution in air. objective can be calculated by the minimum angle of resolu-
(c) What should be the minimum diameter of the objective if tion equation for circular apertures (Eq. 25.8). (d) The ratio of
light with a wavelength of 550 nm in air is used? (d) Calculate the wavelength in air to the wavelength in oil is equal to the
the wavelength of the 550-nm light in air in the oil. (e) What is index of refraction of the oil (Eq. 22.4). (e) The minimum angle
the minimum angle of resolution of the microscope when of resolution with oil can be calculated with Eq. 25.8 again
operated in oil immersion mode? using the wavelength in oil.
SOLUTION. Listing the data, using the subscripts air and oil for the dry and oil immersion operations:
Given: mtotal = - 400 * (total magnification) Find: (a) fo (objective focal length)
fe = 2.50 cm (eyepiece focal length) (b) 1umin2air (minimum angle of resolution in air)
L = 20.0 cm (objective–eyepiece distance) (c) Dmin (minimum diameter of objective)
1smin2air = 0.400 mm = 4.00 * 10-7 m (d) loil (wavelength of light in oil)
noil = 1.56 (oil index of refraction) (e) 1umin2oil (minimum angle of resolution in oil)
lair = 550 nm = 5.50 * 10-7 m (wavelength of light in air)
(a) The focal length of the objective is calculated directly from Eq. 25.5:
125 cm2L 125 cm2120.0 cm2
12.50 cm21- 4002
fo = - = - = 0.500 cm = 5.00 * 10-3 m
fe mtotal
(b) The minimum resolving power is the product of the focal length of the objective and the minimum angle of resolution.
Using Eq. 25.9,
1smin2air
1umin2air =
4.00 * 10-7 m
= = 8.00 * 10-5 rad
fo 5.00 * 10-3 m
This minimum angle is smaller (better resolution) than that of the human eye in Example 25.7.
(c) From the Rayleigh criterion for circular aperture (Eq. 25.8), the minimum diameter of the objective is then
1.22lair 11.2225.50 * 10-7 m
1umin2air
Dmin = = = 8.39 * 10-3 m = 8.39 mm
8.00 * 10-5 m
(d) The wavelength of light in oil is shorter than that in air because oil has a greater index of refraction than air. From Eq. 22.4,
lair 5.50 * 10-7 m
loil = = = 3.53 * 10-7 m
nair 1.56
(e) With this shortened wavelength, the minimum angle of resolution in oil can be determined again from Eq. 25.8.
11.22213.53 * 10-7 m2
1umin2oil =
1.22loil
= = 5.13 * 10-5 rad
Dmin 8.39 * 10-3 m
Notice this minimum angle of resolution is smaller (better resolution) than when the microscope is operated dry. Oil immersion
is a technique to increase the resolution of optical compound microscopes.
■ Nearsighted people cannot see distant objects clearly. Far-

sighted people cannot see nearby objects clearly. These con-
ditions may be corrected by diverging and converging Corrected Uncorrected Corrected
Uncorrected
lenses, respectively. Nearsightedness (myopia) Farsightedness (hyperopia)
■ The magnification of a magnifying glass (or simple micro- fo

fe fe
scope) is expressed in terms of angular magnification (m),
as distinguished from the lateral magnification (M; see
θo Fe Fo Fe
Chapter 23): θo θ
Io
u
m = (25.1) Eyepiece
uo Objective
The magnification of a magnifying glass with the image at

the near point (25 cm) is expressed as Ie
25 cm ■ Diffraction places a limit on resolution—the ability to

m = 1 + (25.3)
f resolve, or distinguish, objects that are close together. Two
images are said to be just resolved when the center of the
The magnification of a magnifying glass with the image at central maximum of one image falls on the first minimum of
infinity (relaxed viewing) is expressed as the other image (the Rayleigh criterion).
25 cm
m = (25.4)
f
Virtual image of fly S1
θ
θ min
Actual fly
S2
■ The objective of a compound microscope has a relatively
short focal length, and the eyepiece, or ocular, has a longer
focal length. Both contribute to the total magnification Slit Screen
mtotal, given by
125 cm2L
■ For a rectangular slit, the minimum angle of resolution is
mtotal = Mo me = - (25.5) l
fo fe umin = (25.7)
w
where L, fo, and fe are in centimeters.
Objective Eyepiece For a circular aperture of diameter D, the minimum angle
of resolution is
Ie Fe
Fo
Io 1.22l
umin = (25.8)
D
do di
L
The resolving power of a microscope is
■ A refracting telescope uses a converging lens to gather light, 1.22lf
and a reflecting telescope uses a converging mirror. The s = fumin = (25.9)
image created by either one is magnified by the eyepiece. D
The magnification of a refracting telescope is
fo
m = - (25.6)
fe

25.1 THE HUMAN EYE 4. The focal length of the crystalline lens of the human eye
varies with muscle action. For close-up vision, the radius
1. The cones of the retina are responsible for (a) 20>20
of the lens is (a) large, (b) small, (c) flat, (d) none of the
vision, (b) black-and-white twilight vision, (c) color
preceding.
vision, (d) close-up vision.
2. An imperfect cornea can cause (a) astigmatism, (b) near-
sightedness, (c) farsightedness, (d) all of the preceding. 25.2 MICROSCOPES
3. The image of an object formed on the retina is 5. A magnifying glass (a) is a converging lens, (b) forms
(a) inverted, (b) upright, (c) the same size as the object, virtual images, (c) magnifies by effectively increasing the
(d) all of the preceding. angle the object subtends, (d) all of the preceding.
6. When using a magnifying glass, the magnification is 12. For a particular wavelength, the minimum angle of reso-
greater when the magnified image is at (a) the near lution is (a) smaller for a lens of a larger radius,
point, (b) the far point, (c) infinity. (b) smaller for a lens of a smaller lens, (c) the same for
7. Compared with the focal length of the eyepiece in a com- lenses of all radii.
pound microscope, the objective has (a) a longer focal 13. The purpose of using oil immersion lenses on microscopes
length, (b) a shorter focal length, (c) the same focal is to (a) reduce the size of the microscope, (b) increase the
length. magnification, (c) increase the wavelength of light so as to
increase resolution, (d) reduce the wavelength of light so
as to increase resolution.
25.3 TELESCOPES
8. An astronomical telescope has (a) unlimited magnifica- *25.5 COLOR
tion, (b) two lenses of the same focal length, (c) an objec- 14. An additive primary color is (a) blue, (b) green, (c) red,
tive of relatively long focal length, (d) an objective of (d) all of the preceding.
relatively short focal length. 15. A subtractive primary color is (a) cyan, (b) yellow,
9. An inverted image is produced by (a) a terrestrial tele- (c) magenta, (d) all of the preceding.
scope, (b) an astronomical telescope, (c) a Galilean tele- 16. White light is incident on two filters as shown in
scope, (d) all of the preceding. 䉲 Fig. 25.22. The color of light that emerges from the yel-
10. Compared with large refracting telescopes, large reflecting low filter is (a) blue, (b) yellow, (c) red, (d) green.
telescopes have the advantage of (a) greater light-
gathering capability, (b) being free from chromatic and Red
spherical aberration, (c) lower cost, (d) all of the preceding. Orange
Yellow
White light Green
25.4 DIFFRACTION AND RESOLUTION Blue
Indigo
11. The images of two sources are said to be just resolved
Violet
when (a) the central maxima of the diffraction patterns
Cyan filter + Yellow filter
fall on each other, (b) the first maxima of the diffraction (pigment) (pigment)
patterns fall on each other, (c) the central maximum of
one diffraction pattern falls on the first minimum of the 䉱 F I G U R E 2 5 . 2 2 Color absorption See Multiple Choice
other, (d) none of the preceding. Question 16.
25.1 THE HUMAN EYE 7. With an object at the focal point of a magnifying glass,
the magnification is given by m = 125 cm2>f (Eq. 25.4).
1. Which parts of a camera correspond to the iris,
According to this equation, the magnification could be
crystalline lens, and retina of the eye?
increased indefinitely by using lenses with shorter focal
2. (a) If an eye has a far point of 15 m and a near point of lengths. Why, then, are compound microscopes needed?
25 cm, is that eye nearsighted or farsighted? (b) How about
8. In a compound microscope, which lens, the objective or
an eye with a far point at infinity and a near point at 50 cm?
the eyepiece, plays the same role as a simple magnifying
(c) What type of corrective lenses (converging or diverging)
glass?
would you use to correct the vision defects in parts (a)
and (b)?
3. Will wearing glasses to correct nearsightedness and far- 25.3 TELESCOPES
sightedness, respectively, affect the size of the image on 9. If you are given two lenses with different focal lengths,
the retina? Explain. how would you decide which should be used as the
4. A fifty-year-old person has a far point of 20 m and near objective and which should be used as the eyepiece for a
point of 45 cm. What type of corrective glasses would be telescope? Explain.
necessary to correct this person’s vision?
10. What are the main differences among the following
5. A person with nearsightedness wishes to switch from refracting telescopes: an astronomical telescope, a
regular glasses to contact lenses. Should the contact Galilean telescope, and a terrestrial telescope?
lenses have a stronger or a weaker prescription than the
11. Why are chromatic and spherical aberrations important
glasses? Explain.
factors in refracting telescopes, but not in reflecting
telescopes?
25.2 MICROSCOPES
12. In Fig. 25.12b, part of the light entering the concave
6. When you use a simple convex lens as a magnifying mirror is obstructed by a small plane mirror that is used
glass to view an object, where should you put the object, to redirect the rays to a viewer. Does this mean that only
farther away than the focal length or closer than the focal a portion of an object can be seen? How does the size of
length? Explain. the obstruction affect the image?
EXERCISES 871
25.4 DIFFRACTION AND RESOLUTION 16. In order to observe fine details of small objects in a
microscope, should you use blue light or red light?
13. When an optical instrument is designed, a high resolu-
tion is often desired so that the instrument may be used
to observe fine details. Does a higher resolution mean a *25.5 COLOR
smaller or larger minimum angle of resolution? Explain.
17. Describe how the American flag would appear if it were
14. A reflecting telescope with a large objective mirror can illuminated with light of each of the primary colors.
collect more light from stars than a reflecting telescope 18. Can white be obtained by the subtractive method of
with a smaller objective mirror. What other advantage is color production? Explain.
gained with a large mirror? Explain.
19. Several beverages, such as root beer, develop a “head” of
15. Modern digital cameras are getting smaller and smaller. foam when poured into a glass. Why is the foam gener-
Discuss the image resolution of these small cameras. ally white or light colored, whereas the liquid is dark?
EXERCISES
25.1 THE HUMAN EYE* Explain. (b) Which type of lens will allow her to see dis-
tant objects clearly, and of what power should the lens be?
1. ● What are the powers of (a) a converging lens of focal
length 20 cm and (b) a diverging lens of focal length 9. IE ● ● A man is unable to focus on objects nearer than
- 50 cm? 1.5 m. (a) Does he have (1) nearsightedness, (2) farsighted-
ness, or (3) astigmatism? Explain. (b) The type of contact
2. ● A person is prescribed with contact lenses that have
lenses that allows him to focus on the print of a book held
powers of - 3.0 D. What type of lenses are these? What is
25 cm from his eyes should be (1) converging, (2) diverging,
the lenses’ focal length?
(3) flat. Explain. (c) What should be the power of the lenses?
3. IE ● The far point of a certain nearsighted person is 90 cm.
10. ● ● A nearsighted student wears contact lenses to correct
(a) Which type of contact lenses, (1) converging, (2) diverg-
for a far point that is 4.00 m from her eyes. When she is
ing, or (3) bifocal, should an optometrist prescribe to enable
not wearing her contact lenses, her near point is 20 cm.
the person to see more distant objects clearly? Explain.
What is her near point when she is wearing her contacts?
(b) What would the power of the lenses be, in diopters?
11. ● ● A nearsighted woman has a far point located 2.00 m
4. IE ● A certain farsighted person has a near point of 50 cm.
from one eye. (a) If a corrective lens is worn 2.00 cm from
(a) Which type of contact lenses, (1) converging, (2) diverg-
the eye, what would be the necessary power of the lens
ing, or (3) bifocal, should an optometrist prescribe to enable
for her to see distant objects? (b) What would be the
the person to see clearly objects as close as 25 cm? Explain.
necessary power if a contact lens were used?
(b) What is the power of the lenses, in diopters?
12. ● ● A nearsighted man wears eyeglasses whose lenses have
5. ● ● A nearsighted person has an uncorrected far point of
a focal length of - 0.25 m. How far away is his far point?
200 cm. Which type of contact lenses would correct this
13. ● ● An eyeglass lens with a power of +2.8 D allows a far-
condition, and of what focal length should it be?
sighted person to read a book held at a distance of 25 cm
6. ● ● A person can just see the print in a book clearly when
from her eyes. At what distance must she hold the book
she holds the book no closer than at arm’s length (0.45 m
to read it without glasses?
from the eyes). (a) Does she have (1) nearsightedness,
14. ● ● A college professor can see objects clearly only if they
(2) farsightedness, or (3) astigmatism? Explain. (b) Which
type of lens will allow her to read the text at the normal are between 70 and 500 cm from her eyes. Her
near point (0.25 m), and what is that lens’s power? optometrist prescribes bifocals (䉲 Fig. 25.23) that enable
7. ● ● To correct a case of farsightedness, an optometrist
prescribes converging contact lenses that effectively

move the patient’s near point from 85 cm to 25 cm.
(a) What is the power of the lenses? (b) To see distant
objects clearly, should the patient wear the contact lenses
Nearsightedness
or take them out? Explain. correction
Farsightedness
8. IE ● ● A woman cannot see objects clearly when they are correction
farther than 12.5 m away. (a) Does she have (1) near-
sightedness, (2) farsightedness, or (3) astigmatism? 䉱 F I G U R E 2 5 . 2 3 Bifocals See Exercises 14, 15, and 18.
*Assume that corrective lenses are in contact with the eye (contact lenses) unless otherwise stated.
her to see distant objects through the top half of the 26. ●● What is the maximum magnification of a magnifying
lenses and read students’ papers at a distance of 25 cm glass with a power of + 3.0 D for (a) a person with a near
through the lower half. What are the respective powers point of 25 cm and (b) a person with a near point of 10 cm?
of the top and bottom lenses? 27. ● ● If a magnifying glass gives an angular magnification
15. ● ● A senior citizen wears bifocals (Fig. 25.23) in which the of 1.5 * when viewed with relaxed eyes, what is the
top half of the lens has a focal length of -0.850 m and power of the lens?
the bottom half of the lens has a focal length of + 0.500 m. 28. ● ● A compound microscope has an objective with a
What are this person’s near point and far point? focal length of 4.00 mm and an eyepiece with a magnifi-
16. ● ● ● A certain man has a far point of 150 cm. (a) What cation of 10.0 * . If the objective and eyepiece are 15.0 cm
power must contact lenses have to allow him to see dis- apart, what is the total magnification of the microscope?
tant objects clearly? (b) If he is able to read a newspaper 29. ● ● A compound microscope has a distance of 15 cm
at 25 cm while wearing his contacts, is his near point less between lenses and an objective with a focal length of
than 25 cm? If so, what is it? (c) Give an approximation 8.0 mm. What power should the eyepiece have to give a
of the man’s age, based on the normal rate of recession of total magnification of -360* ?
the near point. 30. ● ● The focal length of the objective lens of a compound
17. ● ● ● A middle-aged man starts to wear eyeglasses with microscope is 4.5 mm. The eyepiece has a focal length of
lenses of + 2.0 D that allow him to read a book held as 3.0 cm. If the distance between the lenses is 18 cm, what
close as 25 cm. Several years later, he finds that he must is the magnification of a viewed image?
hold a book no closer than 33 cm to read it clearly with 31. ● ● A compound microscope has an objective lens with a
the same glasses, so he gets new glasses. What is the focal length of 0.50 cm and an eyepiece with a focal
power of the new lenses? length of 3.25 cm. The separation distance between the
18. ● ● ● Bifocal glasses are used to correct both nearsighted- lenses is 22 cm. (a) What is the total magnification?
ness and farsightedness at the same time (Fig. 25.23). If (b) Compare (as a percentage) the total magnification
the near points in the right and left eyes are 35.0 cm and with the magnification of the eyepiece alone as a simple
45.0 cm, respectively, and the far point is 220 cm for both magnifying glass.
eyes, what are the powers of the lenses prescribed for the 32. ●● The lenses used in a compound microscope have
glasses? (Assume that the glasses are worn 3.00 cm from
powers of +100 D and +50 D. If a total magnification of
the eyes.) -200 * is desired, what should be the distance between
the two lenses?
25.2 MICROSCOPES* 33. IE ● ● Two lenses of focal length 0.45 cm and 0.35 cm are
available for a compound microscope using an eyepiece of
19. ● Using the small-angle approximation, compare the focal length of 3.0 cm, and the distance between the lenses
angular sizes of a car 1.5 m in height when viewed from has to be 15 cm. (a) Which lens should be used as the objec-
distances of (a) 500 m and (b) 1050 m. tive: (1) the one with the longer focal length, (2) the one
20. ● An object is placed 10 cm in front of a converging lens with the shorter focal length, or (3) either? (b) What are the
with a focal length of 18 cm. What are (a) the lateral mag- two possible total magnifications of the microscope?
nification and (b) the angular magnification? 34. ● ● A - 150* microscope has an eyepiece whose focal
21. ● A biology student uses a converging lens to examine length is 4.4 cm. If the distance between the lenses is
the details of a small insect. If the focal length of the lens 20 cm, find the focal length of the objective.
is 12 cm, what is the maximum angular magnification? 35. ● ● A specimen is 5.0 mm from the objective of a com-
22. ● A converging lens can give a maximum angular mag- pound microscope that has a lens power of + 250 D.
nification of 4.0* . What is the focal length of the lens? What must be the magnifying power of the eyepiece if
23. ● When viewing an object with a magnifying glass the total magnification of the specimen is -100* ?
whose focal length is 10 cm, a student positions the lens 36. ● ● ● A lens with a power of +10 D is used as a simple
so that there is minimum eyestrain. What is the observed microscope. (a) For the image of an object to be seen
magnification? clearly, can the object be placed infinitely close to the
24. IE ● A physics student uses a converging lens with a lens, or is there a limit on how close it can be? Explain.
focal length of 14 cm to read a small measurement scale. (b) Calculate how close an object can be brought to the
(a) Maximum magnification is achieved if the image is at lens. (c) What is the angular magnification at this point?
(1) the near point, (2) infinity, (3) the far point. Explain. 37. IE ● ● ● A modern microscope is equipped with a turret
(b) What are the magnifications when the image is at the that has three objectives with focal lengths of 16 mm,
near point and infinity, respectively? 4.0 mm, and 1.6 mm and interchangeable eyepieces of
25. IE ● A detective wants to achieve maximum magnifica- 5.0 * and 10* . A specimen is positioned such that each
tion when looking at a fingerprint with a magnifying objective produces an image 150 mm from the objective.
glass. (a) He should use a lens with (1) a long focal (a) Which objective and eyepiece combination would
length, (2) a short focal length, (3) a larger size. Explain. you use if you want to have the greatest magnification?
(b) If he uses lenses of focal length + 28 cm and + 40 cm, How about the least magnification? Explain. (b) What
what are the maximum magnifications of the print? are the greatest and least magnifications possible?
*The normal near point should be taken as 25 cm unless otherwise specified.

EXERCISES 873
25.3 TELESCOPES (which has an 8.20-m, or 323-in., diameter) for light with
a wavelength of 550 nm?
38. ● Find the magnification and length of a telescope
49. ● What is the resolution limit due to diffraction for the
whose objective has a focal length of 50 cm and whose
eyepiece has a focal length of 2.0 cm. Hale telescope at Mount Palomar, with its 200-in.-
diameter mirror, for light with a wavelength of 550 nm?
39. ● An astronomical telescope has an objective and an
Compare this value with the resolution limit for the Euro-
eyepiece whose focal lengths are 60 cm and 15 cm,
pean Southern Observatory telescope found in Exercise 48.
respectively. What are the telescope’s (a) magnifying
50. ● ● From a spacecraft in orbit 150 km above the Earth’s
power and (b) length?
surface, an astronaut wishes to observe her hometown as
40. ● ● An astronomical telescope has an eyepiece with a
she passes over it. What size features will she be able to
focal length of 10.0 mm. If the length of the tube is 1.50
identify with the unaided eye, neglecting atmospheric
m, (a) what is the focal length of the objective? (b) What
effects? [Hint: Estimate the diameter of the human iris.]
is the angular magnification of the telescope when it is
focused for an object at infinity? 51. IE ● ● A human eye views small objects of different col-
ors, and the eye’s resolution is measured. (a) The eye sees
41. ● ● A telescope has an angular magnification of - 50 *
the finest details for objects of which color: (1) red, (2)
and a barrel 1.02 m long. What are the focal lengths of
yellow, (3) blue, or (4) any color? Explain. (b) The maxi-
the objective and the eyepiece?
mum diameter of the eye’s pupil at night is about 7.0 mm.
42. IE ● ● A terrestrial telescope has three lenses: an objec- What are the minimum angles of resolution for sources
tive, an erecting lens, and an eyepiece. (a) Does the erect- with wavelengths of 400 nm and 700 nm, respectively?
ing lens (1) increase the magnification, (2) increase the
52. ● ● Some African tribespeople claim to be able to see the
physical length of the telescope, (3) decrease the magnifi-
moons of Jupiter with the unaided eye. If two moons of
cation, or (4) decrease the physical length of the tele-
Jupiter are at a minimum distance of 3.1 * 108 km away
scope? Explain. (b) This terrestrial telescope has focal
from Earth and at a maximum separation distance of
lengths of 40 cm, 10 cm, and 5.0 cm for the objective,
3.0 * 106 km, is this possible in theory? Explain. Assume
erecting lens, and eyepiece, respectively. What is the
that the moons reflect sufficient light and that their obser-
magnification of the telescope for an object at infinity?
vation is not restricted by Jupiter. [Hint: See Exercise 51b.]
(c) What is the length of the telescope barrel?
53. ● ● Assuming that the headlights of a car are point
43. ● ● A terrestrial telescope uses an objective and eyepiece
sources 1.7 m apart, what is the maximum distance from
with focal lengths of 42 cm and 6.0 cm, respectively.
an observer to the car at which the headlights are distin-
(a) What should the focal length of the erecting lens be if
guishable from each other? [Hint: See Exercise 51b.]
the overall length of the telescope is to be 1.0 m? (b) What is
54. ● ● If a camera with a 50-mm lens is to resolve two
the magnification of the telescope for an object at infinity?
objects that are 4.0 mm from each other and both objects
44. ● ● An astronomical telescope uses an objective of power
are 3.5 m from the camera lens, (a) what is the minimum
+ 2.00 D. If the length of the telescope is 52 cm, (a) what
diameter of the camera lens? (b) What is the resolving
is the focal length of the eyepiece? (b) What is the angu-
power? (Assume the wavelength of light is 550 nm.)
lar magnification of the telescope?
55. ● ● The objective of a microscope is 2.50 cm in diameter
45. IE ● ● You are given two objectives and two eyepieces
and has a focal length of 0.80 mm. (a) If blue light with a
and are instructed to make a telescope with them. The
wavelength of 450 nm is used to illuminate a specimen,
focal lengths of the objectives are 60.0 cm and 40.0 cm,
what is the minimum angular separation of two fine
and the focal lengths of the eyepieces are 0.90 cm and
details of the specimen for them to be just resolved?
0.80 cm. (a) Which lens combination would you pick if
(b) What is the resolving power of the lens?
you want to have maximum magnification? How about
56. ● ● A refracting telescope with a lens whose diameter is
minimum magnification? Explain. (b) Calculate the
maximum and minimum magnifications. 30.0 cm is used to view a binary star system that emits
light in the visible region. (a) What is the minimum
angular separation of the two stars for them to be barely
25.4 DIFFRACTION AND RESOLUTION* resolved? (b) If the binary star is a distance of
6.00 * 1020 km from the Earth, what is the distance
46. IE ● (a) For a given wavelength, a wider single slit will between the two stars? (Assume that a line joining the
give (1) a greater, (2) a smaller, (3) the same minimum stars is perpendicular to our line of sight.)
angle of resolution as a narrower slit, according to the
57. ● ● A radio telescope with a diameter of 300 m uses a
Rayleigh criterion. (b) What are the minimum angles of
resolution for two point sources of red light 1l = 680 nm2
wavelength of 4.0 m to observe a binary star system that
is about 2.5 * 1018 km from the Earth. What is the mini-
in the diffraction pattern produced by single slits with
mum distance of two stars that can be distinguished by
widths of 0.55 mm and 0.45 mm, respectively?
the telescope?
47. ● The minimum angle of resolution of the diffraction pat-
58. ● ● A microscope with an objective 1.20 cm in diameter is
terns of two identical monochromatic point sources in a
used to view a specimen via light from a mercury source
single-slit diffraction pattern is 0.0065 rad. If a slit width of
with a wavelength of 546.1 nm. (a) What is the limiting
0.10 mm is used, what is the wavelength of the sources?
angle of resolution? (b) If details finer than those observable
48. ● What is the resolution limit due to diffraction for the in part (a) are to be observed, what color of light in the visi-
European Southern Observatory reflecting telescope ble spectrum would have to be used? (c) If an oil immer-
sion lens were used 1noil = 1.502, what would be the
*Ignore atmospheric blurring unless otherwise stated. change (expressed as a percentage) in the resolving power?
59. A student uses a magnifying glass to examine the details 62. A person with nearsightedness was prescribed with con-
of a microcircuit in the lab. If the lens has a power of tact lenses of power -2.0 D. By mistake, he was given
10 D and a virtual image is formed at the student’s near lenses of power +2.0 D. What is the range of object dis-
point (25 cm), (a) how far from the circuit is the lens tances that this person can see clearly with the wrong
held, and (b) what is the angular magnification? lenses?
60. Referring to 䉲 Fig. 25.24, show that the magnifying 63. Two astronomical telescopes have the characteristics
power of a magnifying glass held at a distance d from the shown in the following table:
eye is given by
Objective Eyepiece Focal Objective
Telescope Focal Length (cm) Length (cm) Diameter (cm)
m = a b a1 - b +
25 d 25
f D D A 90.0 0.840 75.0
when the actual object is located at the near point B 85.0 0.770 60.0
(25 cm). [Hint: Use a small-angle approximation, and
note that yi>yo = - di>do, by similar triangles.]
(a) Which telescope would you choose (1) for best mag-
nification? (2) for best resolution? Explain. (b) Calculate
the maximum magnification and the minimum resolv-
ing angle for a wavelength of 550 nm.
yi 64. A refracting telescope has an objective with a focal
yo length of 50 cm and an eyepiece with a focal length of
θi
F do F 2.0 cm. The telescope is used to view an object that is
10 cm high and located 50 m away. What is the apparent
–d i
angular height, in degrees, of the object as viewed
d through the telescope?
D 65. The amount of light that reaches the film in a camera
depends on the lens aperture (the effective area) as con-
䉱 F I G U R E 2 5 . 2 4 Magnifying power of a magnifying glass trolled by the diaphragm. The f-number is the ratio of
See Exercise 60.
the focal length of the lens to its effective diameter.
For example, an f>8 setting means that the diameter of
61. Referring to 䉲 Fig. 25.25, show that the angular magnifi-
the aperture is one-eighth of the focal length of the lens.
cation of a refracting telescope focused for the final
image at infinity is m = - fo>fe . (Because telescopes are
The lens setting is commonly referred to as the f-stop.
(a) Determine how much light each of the following lens
designed for viewing distant objects, the angular size of
settings admits to the camera as compared with f>8:
an object viewed with the unaided eye is the angular size
(1) f>3.2 and (2) f>16. (b) The exposure time of a camera is
of the object at its actual location rather than at the near
controlled by the shutter speed. If a photographer
point, as is true for a microscope.)
correctly uses a lens setting of f>8 with a film exposure
Objective time of 1>60 s, what exposure time should she use to get
Eyepiece the same amount of light exposure if she sets the f-stop
at f>5.6?
Fo, Fe θi
θo
Fo yi Fe
Object and Intermediate

final image image
at infinity
䉱 F I G U R E 2 5 . 2 5 Angular magnification of a refracting

telescope See Exercise 61.
26 Relativity
26.1 Classical relativity and
the Michelson-Morley
experiment (876)
■ ether reference frame
26.2 The special relativity

postulate and the relativity
of simultaneity (878)
■ constancy of speed of light
The relativity
26.3
of timeand length:
time dilation and length
contraction (882)
Relativistic kinetic energy, PHYSICS FACTS
Y
26.4
momentum, total energy, ✦ In one nanosecond 110-9 s2, light
ou might not think so, but the
and mass–energy travels about 1 ft or about 30 cm. chapter-opening photograph
equivalence (890) ✦ The atomic clocks in orbit in the
Global Positioning System (GPS)
tells

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Library of Congress Cataloging-in-Publication Data

ISBN: 978-0-321-60183-4 (student copy)

J E R R Y D . W I L S O N , a native of A N T H O N Y J . B U F F A received B O L O U is currently Professor of

20 Electromagnetic Induction and Waves 696

Applications (Insights appear in boldface, and “(bio)” indicates a biomedical application)

CHAPTER 1 Leg traction (bio) 120 CHAPTER 7

Computer monitor operation (Ex. 16.19) CHAPTER 20 The Rainbow 769

in recognition of the high percentage of 䉴 F I G U R E 2 Bone density loss with aging

Demonstrations. Six new demonstrations with photos

nRT1 10.100 mol238.31 J>1mol # K241293 K2

Since work was done and the internal energy increased,

End-of-Chapter Exercises. Each section of the end-of- FOR THE INSTRUCTOR

REVIEWERS OF PREVIOUS EDITIONS

David Aaron Raymond D. Benge Michael Browne

Lattie F. Collins Allen Grommet Mark Lindsay

Donald S. Presel Terry Scott Gabriel Umerah

Dan R. Quisenberry Rahim Setoodeh Lorin Vant-Hull

W. Steve Quon Bartlett Sheinberg Pieter B. Visscher

David Rafaelle Jerry Shi Karl Vogler

INSIGHT: 1.1 Why Study Physics? 2

1.6 Significant Figures 17

LEARNING PATH REVIEW 130 EXERCISES 134

3 Motion in Two Dimensions 67 L E A R N B Y D R A W I N G : Energy Exchanges:

3.1 Components of Motion 68

6.2 Impulse 186 8.2 Torque, Equilibrium, and Stability 270

9 Solids and Fluids 311

LEARNING PATH REVIEW 344 EXERCISES 349

7.1 Angular Measure 223 10.1 Temperature and Heat 356

11 Heat 386 13 Vibrations and Waves 455

LEARNING PATH REVIEW 410 EXERCISES 412

LEARNING PATH REVIEW 447 EXERCISES 450

15 Electric Charge, Forces, and Fields 529

LEARNING PATH REVIEW 552 EXERCISES 556

18 Basic Electric Circuits 623

I N S I G H T : 16.1 Electric Potential LEARNING PATH REVIEW 649 EXERCISES 652

LEARNING PATH REVIEW 588 EXERCISES 591

17 Electric Current and Resistance 596

17.1 Batteries and Direct Current 597 in Magnetic Fields 664

INSIGHT: 19.1 The Magnetic Force

23.4 The Lens Maker’s Equation 798

LEARNING PATH REVIEW 722 EXERCISES 725

22.2 Reflection 753

I N S I G H T : 26.1 Relativity in Everyday Living 896 29.2 Radioactivity 969

27 Quantum Physics 910 LEARNING PATH REVIEW 994 EXERCISES 997

27.1 Quantization: Planck’s Hypothesis 911

LEARNING PATH REVIEW 1027 EXERCISES 1030

28.1 Matter Waves: The de Broglie Hypothesis 939

Photo Credits P-1

1.2 SI units of length, mass,

1.3 More about

1.4 Unit analysis (12)

1.5 Unit conversions (14)

Measurements enable us to compute quantities and solve problems. Units of

1.1 Why and How We Measure

➥ What does physics attempt to do?

INSIGHT 1.1 Why Study Physics?

DID YOU LEARN?

1.2 SI Units of Length, Mass, and Time