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The Learning App For

AAJ KA NEET AIIMS JEE RAS
CHAPTER
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter ll
Chapter 12
Chapter 13
CONTENTS
SKELETONS WITII EXAMS
MATHEMATICAL INTRODUCTION
l.l Planar Vectors. Scientific Notation, and Units / l.2 Three-Dimensional Vectors;
Dot
and Cross Products
EQUILIBRIUM OF CONCURRENT FORCES
2.1 Ropes. Knots, and Frictionless Pulleys / 2.2 Friction and Inclined Planes /
2.3 Graphical and Other Problems
KINEMATICS IN ONE DIMENSION
3.1 Dimensions and Units; Constant-Acceleration Problems
NEWTON‘S LAWS OF MOTION
4.1 Force, Mass. and Acceleration / 4.2 Friction; Inclined Planes; Vector
Notation / 4.3 'I\vo-Object and Other Problems
MOTION IN A PLANE I
5.1 Projectile Motion / 5.2 Relative Motion
MOTION IN A PLANE II
6.1 Circular Motion; Centripetal Force / 6.2 Law of Universal Gravitation;
Satellite
Motion I 6.3 General Motion in a Plane
WORK AND ENERGY
7.1 Work Done by a Force I 7.2 Work, Kinetic Energy, and Potential
Energy / 7.3 Conservation of Mechanical Energy / 7.4 Additional Problems
POWER AND SIMPLE MACHINES
8.1 Power I 8.2 Simple Machines
IMPULSE AND MOMENTUM
9.1 Elementary Problems / 9.2 Elastic Collisions / 9.3 Inelastic Collisions and
Ballistic
Pendulum: / 9.4 Collisions in Two Dimensions / 9.5 Recoil and Reaction I 9.6 Center
of Mass (see also Chap. 10)
STATICS OF RIGID BODIES
10.1 Equilibrium of Rigid Bodies / 10.2 Center of Mass (Center of Gravity)
ROTATIONAL MOTION I: KINEMATICS AND DYNAMICS
11.1 Angular Motion and Torque / 11.2 Rotational Kinematics I 11.3 Torque and
Rotation / 11.4 Moment of Inertia I 11.5 Translational-Rotational Relationships /
11.6 Problems Involving Cords Around Cylinders, Rolling Objects. etc.
ROTATIONAL MOTION II: KINETIC ENERGY. ANGULAR IMPULSE.
ANGULAR MOMENTUM
12.1 Energy and Power I 12.2 Angular Impulse; the Physical Pendulum /
12.3 Angular Momentum
MATTER IN BULK
13.1 Density and Specific Gravity / 13.2 Elastic Properties
A/-\J K/-\ TOPPER Ii'E°$°§|l?I'§.gJ§EpR§§’§

lv U CONTENTS
Chapter 14
Chapter 15
Chapter 16
Chapter 11
Chapter 18
Chapter 19
Chapter 20
Chapter 21
Chapter 22
Chapter L1
Chapter 24
Chapter 25
Chapter Z6
Chapter 27
Chapter 28
AAJ KA TOPPER
SIMPLE HARMONIC MOTION
14.1 Oscillations of a Mass on a Spring / 14.2 SHM of Pendulums and Other Systems
HYDROSTATICS
15.1 Pressure and Density / 15.2 Pascal's and Archimedes‘ Principles; Surface
Tension
HYDRODYNAMICS
16.1 Equation of Continuity. Bernoulli's Equation. Torricelli's Theorem /
16.2 Viscosity. Stokes‘ Law. Poiseuille‘s Law. Turbulenoe. Reynolds Number
TEMPERATURE AND THERMAL EXPANSION
17.1 Temperature Scales; Linear Expansion I 17.2 Area and Volume Expansion
HEAT AND CALORIMETRY
18.1 Heat and Energy; Mechanical Equivalent of Heat / 18.2 Calorimetry, Specific
Heats. Heats of Fusion and Vaporization
HEAT TRANSFER
19.1 Conduction I 19.2 Convection / 19.3 Radiation
GAS LAWS AND KINETIC THEORY
20.1 The Mole Concept; the Ideal Gas Law / 20.2 Kinetic Theory / Z13 Atmospheric
Properties; Specific Heats of Solids
THE FIRST LAW OF THERMODYNAMICS
21.1 Basic Thennodynamic Concepts / 21.2 The First Law of Thermodynamics, lntemal
Energy. p-V Diagrams. Cyclical Systems
THE SECOND LAW OF THERMODYNAMICS
22.1 Heat Engines; Kelvin - Planck and Clausius Statements of the Second
Law / 22.2 Entropy
WAVE MOTION
23.1 Characteristic Properties I 23.2 Standing Waves and Resonance
SOUND
24.1 Sound Velocity; Beats; Doppler Shift / 24.2 Power. Intensity. Reverberation
Time.
Shock Waves
COULOMB'S LAW AND ELECTRIC FIELDS
25.1 Coulomb's Law of Electrostatic Force / 25.2 The Electric Field; Continuous
Charge
Distributions; Motion of Charged Particles in an Electric Field / 25.3 Electric
Flux and
Gauss's Law
ELECTRIC POTENTIAL AND CAPACITANCE
26.1 Potential Due to Point Charges or Charge Distributions / 26.2 The Potential
Function and the Associated Electric Field / 26.3 Energetics; Problems with Moving
Charges I 26.4 Capacitance and Field Energy I 26.5 Capacitors in Combination
SIMPLE ELECTRIC CIRCUITS
27.1 Ol1m's Law,Current. Resistance I 27.2 Resistors in Combination / 27.3 EMF
and Electrochemical Systems / 27.4 Electric Measurement / 27.5 Electric Power /
27.6 More Complex Circuits. I(irchhot't‘s Circuit Rules. Circuits with Capacitance
THE MAGNETIC FIELD
28.1 Force on a Moving Charge I 28.2 Force on an Electric Current / 28.3 Torque and
Magnetic Dipole Moment / 28.4 Sources of the Magnetic Field; Law of Bio! and
Savart / 28.5 More Complex Geometries; Ampere‘s Law
NEET AIIMS JEE RAS
Chapter 29
Chapter 30
Chapter 31
Chapter 32
Chapter 33
Chapter 34
Chapter 35
Chapter 36
Chapter 37
Chapter 38
Chapter 39
CONTENTS U v
MAGNETIC PROPERTIES OF MATTER
29.! The H and M Fields; Susoeptibility; Relative Permeability ," 29.2 Magnets;
Pole
Strength
INDUCED EMF: GENERATORS AND MOTORS
30.] Change in Magnetic Flux. Faraday‘s Law, benz‘s Law / 30.2 Motional
EMF; Induced Currents and Forces ,-‘ 30.3 Time-Varying Magnetic and Induced
Electric Fields ,' 30.4 Electric Generators and Motors
INDUCTANCE
3l.l Self-Inductance /‘ 3|.2 Mutual Inductance: The Ideal Transformer
ELECTRIC CIRCUITS
32.1 R~C, R-L, L-C and R-L-C Circuits; Time Response / 32.2 AC Circuits in the
Steady State / 32.3 Time Behavior of AC Circuits
ELECTROMAGNETIC WAVES
33.1 Displacement Current. MaxwelI‘s Equations, the Speed of Light I 33.2
Mathematical Description of Waves in One and Three Dimensions / 33.3 The
Component Fields of an Electromagnetic Wave; Induced EMF / 33.4 Energy and
Momentum Fluxes
LIGHT AND OPTICAL PIIENOMENA
34.l Reflection and Refraction ,1 34.2 Dispersion and Color / 34.3 Photometry
and Illumination
MIRRORS. LENSES. AND OPTICAL INSTRUMENTS
35.l Minors / 35.2 Thin Lenses / 35.3 LensmaIrer‘s Equation; Composite
Lens Systems / 35.4 Optical Instruments: Projectors, Cameras, the Eye / 35.5
Optical Instruments: Microscopes and Telescopes
INTERFERENCE, DIFFRACTION, AND POLARIZATION
36.! Interference of Light / 36.2 Diffraction and the Diffraction Grating / 36.3
Polarization of Light
SPECIAL RELATIVITY
37.| Lorentz Transformation, Length Contraction, Time Dilation, and Velocity
Transformation ,' 37.2 Mass-Energy Relation; Relativistic Dynamics
PARTICLES OF LIGHT AND WAVES OF MATTER
38.l Photons and the Photoelectric Effect /' 38.2 Compton Scattering; X-rays; Pair
Production and Annihilation J 38.3 de Broglie Waves and the Uncertainty
Principle
MODERN PHYSICS: ATOMS, NUCLEI, SOLID-STATE ELECTRONICS
39.! Atoms and Molecules ,' 39.2 Nuclei and Radioactivity /' 39.3 Solid-State
Electronics
INDEX
AAJ K/-\ TOPPER iiE°Ei°§i'ii'§gJ'E'EpR':i’§
AAJ K/-\ TQPPER it'Eiz'¥e§|i'ii'égJ§EpR'ii’§

TO THE STUDENT
This book is intended for use by students of general physics, either in calculus-
or noncalculus-
based courses. Problems requiring real calculus (not merely calculus notation) are
marked with a
small superscript c.
The only way to master general physics is to gain ability and sophistication in
problem-solving.
This book is meant to make you a master of the art — and should do so if used
properly. As a
rule, a problem can be solved once you have learned the ideas behind it; sometimes
these very ideas
are brought into sharper focus by looking at sample problems and their solutions.
lfyou have difficulty
with a topic. you can select a few problems in that area, examine the solutions
carefully, and then
try to solve related problems before looking at the printed solutions.
There are numerous ways of posing a problem and, frequently. numerous ways of
solving one. You
should try to gain understanding of how to approach various classes of problems.
rather than memorizing
particular solutions. Understanding is better than memory for success in physics.
The problems in this book cover every important topic in u typical two- or three-
semester general
physics sequence. Ranging from the simple to the complex, they will provide you
with plenty of practice
and food for thought.
The Chapter Skeletons with Exams. beginning on the next page. was devised to help
students with
limited time gain maximum benefit from this book. It is hoped that the use of this
feature is self-
evident; still, the following remarks may help:
v The Chapter Skeletons divide the problems in this book into three categories:
SCAN.
HOMEWORK and EXAMS. (Turn to page ix to see an example.)
I To gain a quick overview of the basic ideas in a chapter, review the SCAN
problems and
study their printed solutions.
0 HQMEWORK problems are for practicing your problem-solving skills: cover the
solution with
an tmlex card as you read. and try to solve. the problem. Do both .ru.r if your
course is
calculus based.
¢ No problem from SCAN or HOMEWORK is duplicated in EXAMS. and no two Exams
overlap.
Calculus-based students are urged also to take the Hard Exam. Exams run about 60
minutes,
unless otherwise indicated.
' Still further problems constitute the two groups of Final Exams. Stay in your
category(ies), and
good luck.
Vll
AAJ K/-\ TOPPER it'i§Ei°§m'§gJ'E'épR'ii’§
A/-\J K/-\ TQPPER ii'EE'¥e§i'm§gJ§EpR'i§'§

CHAPTER 1
Mathematical Introduction
PLANAR VECTORS, SCIENTIFIC NOTATION, AND UNITS
What is a scalar quantity‘?
I A scalar quantity has only magnitude: it is a pure number. positive or negative.
Scalars. being simple
numbers. are added. subtracted. etc.. in the usual way. lt may have a unit after
it. e.g. mass= 3 kg.
What is a vector quantity?
I A vector quantity has both magnitude and direction. For example. a car moving
south at 40km/h has a
vector velocity of 40 kml h southward.
A vector quantity can be represented by an arrow drawn to scale. The length of the
arrow is proportional
to the magnitude of the vector quantity (40 km/h in the above example). The
direction of the arrow
represents the direction of the vector quantity.
What is the ‘resu|tant‘ vector?
I The resultant of a number of similar vectors. lorce vectors. for example. is that
single vector which would
have the same eflect as all the original vectors taken together.
Describe the graphical addition of vectors.
I The method for finding the resultant of several vectors consists in beginning at
any convenient point and
drawing (to scale) each vector arrow in turn. They may be taken in any order of
succession. The tail end of
each arrow is attached to the tip end of the preceding one.
The resultant is represented by an arrow with its tail end at the starting point
and its tip end at the tip of
the last vector added.
Describe the parallelogram method oi addition of two vectors.
I The resultant of two vectors acting at any angle may be represented by the
diagonal of a parallelogram.
The two vectors are drawn as the sides of the parallelogram and the resultant is
its diagonal, as shown in Fig‘
l-1. The direction of the resultant is away from the origin of the two vectors.
\\\“\ / /
Q
1-“ /
/
r-1;. r-r
How do you subtract vectors?
I To subtract a vector B from a vector A. reverse the direction of B and add it
vectorially to vector A. that
is.A—B=A+(—B).
Describe the trigonometric functions.
I For the right triangle shown in Fig. l-2. by definition
sin!)-E oos6=£ tan6=g
Q“,
\"\olV°\‘ o I opposite side
a - adjacent side I-1‘, 1-2
AA-I K/-\ TOPPER I.“Ei.#*:.1?i2“.’;';“Ri(’; ‘

2 J CHAPTER 1
Express each oi the following in scientific notation: (a) 627.4, (bl 0.000365, (e)
20 001, (d) 1.0067, (e) 0.0067
I (ll) 6.274 X 10’. (b) 3.65 X 10“. (c) 2.001 X 10‘. (d) l.lI)67 X 10°. (2) 6.7 X
10' ".
Express each of the following as simple numbers Xl0°: (0) 31.65 X 10"’ (la) 0.415 X
10° (e) 1/(2.05 X 10")
(4) 1/(43 >< 10’).
I (a) 0.03165. (ll) 415.000. (c) tss. ta) o.o0otrza3.
The diameter of the earth is about 1.27 X 10’ m. Find its diameter in (0)
millimeters,
lb) megameters. (c) miles.
I (¢) (1.27 x 10’ m)(1(X)0 mm/1 m) = 1.27 X 10'” mm. (b) Multiply meters by 1Mm/10°
m to obtain 12.7 Mm.
(cl Then use (I km/1000 m)(l mi/1.61 km); the diameter is
A 100-m race is run on a 200-m-circumference circular track. The runners ntn
eastward at the start and bend
south. What is the displacement of the endpoint of the raee from the starting
point?
| The runners move as shown in Fig. 1-3. The race is halfway around the traclt so
the displacement is one
diameter = fllljgij = 63.7 m due south.
N
W E
s t-1;. 1-3
What is a component of a vector‘?
I A component of a vector is its "shadow" (perpendicular drop) on an axis in a
given direction. For
example. the p-component ol a displacement is the distance along the p axis
oorresponding to the given
displacement. it is a scalar quantity. being positive or negative as it is
positively or negatively directed along
the axis in question. In Fig. l~4. A, is positive. (One sometimes defines a vector
component as a vector
pointing along the axis and having the size of the scalar component. if the scalar
component is negative
the vector component points in the negative direction along the axis.) lt is
customary, and useful. to resolve a
vector into components along mutually perpendicular directions (rectangular
£‘0mp0!|tI|l‘_t).
A
_ 1-
avix axis p
.-t,, = component along titisp “L 1'4
What is the component method for adding vectors?
I Each vector is resolved into its x. y. and z components. with negatively directed
components taken as
negative. The x component of the resultant, R,. is the algebraic sum of all the x
components. The y and z
components of the resultant are found in a similar way.
Define the multiplication of a vector by a scalar.
| The quantity bl? is a vector having magnitude |b| F (the absolute value of b
times the magnitude of F); the
direction of bl-‘ is that of F or -F. depending on whether b is positive or
negative.
Using the graphical method. find the resultant of the following two displacements: 2
m at 40° and 4 m at 127°.
the angles being taken relative to the +x axis.
I Choose x. y axes as shown in Fig. 1-5 and lay out the displacements to scale tip
to tail from the origin.
Note that all angles are measured from the +x axis. The resultant vector. R. points
from starting point to
endpoint as shown. Measure its length on the scale diagram to find its magnitude.
4.6 m. Using a protractor.
measure its angle 9 to be 101°. The resultant displacement is therefore 4.6m at
101°.
AAJ K/-\ TOPPER Ii'EeEi°§i'ii'§gJ'E'épR':I’§
Th L ‘ g App For
A/-\J K/-\ TOPPER ~E‘ET°Z|1?l'éJEE RAS
MATHEMATICAL INTRODUCTION U
)'
210‘
‘ 25 cos 30'
V =
5 |z1- .
25
)'
25 sin 30
40‘ ’
X
14;. 1-s Hg. 1-a
1.16 Find the x and y components of a B-m displacement at an angle of 210°.
I The vector displacement and its components are shown in Fig. 1-6. The components
are
x component = -25 cos 30° = L y component = -25 sin 30° = ;l_2_.ig
Note in particular that each component points in the negative coordinate direction
and must therefore be
taken as negative.
1.17 Solve Prob. 1.15 by use of rectangular components.
I Resolve each vector into rectangular components as shown in Fig. l-7(a) and (b).
(Place ii cross-hatch
symbol on the original vector to show that it can be replaced by the sum of its
vector components.) The
resultant has the scalar components
R, = l.53—Z.4O= —0.87t'n R, =1.29+-3.20-4.49m
Note that components pointing in the negative direction must be assigned a negative
value.
The resultant is shown in Fig. 1-7(c); we see that
4.49
R = V(0.87)!+(4.49)i=4,57m tan¢ =(T87
Hence, ¢-Z2, from which 8=180°—¢-jg.
7 I Y
n
2 4 sin S3" ‘
- 1.20
2 sin 40- ‘*9
- |.29 111- ‘ ,
40' sr l
2 cos 40- - 1.53 ‘ 4 cos sr - 2.40 ‘ 0-81 ‘
(0) lb) (o
r-1;. 1-1
1.18 Add the following two force vectors by use of the parallelogram method: 30
pounds at 30° and 20 pounds at
I40". (A pound of/om is chosen such that a 1-kg object weighs 2.21 lb on earth. One
pound is equivalent to
a force of 4.45 N.)
I The force vectois are shown in Fig. 1~8. Construct a parallelogram using them as
sides, as shown in Fig.
1-9. The resultant. R. is then shown as the diagonal. Measurement shows that It is
30 lb at 72°.
/’ \\
w 30 // R \
20 I '
10- ,
Flg. 1-a ng. 1-0
A/-\J K/-\ TOPPER Ii'E°$°§i?i'§gJ§EpR§§§

AAJ R NEET AIIMS JEE RAS
4 I CHAPTER 1
1.19 Find the components of the vector F in Fig. l-10 along the x and y axes.
I ln Fig. l-10 the dashed perpendiculars from P to X and Y determine the magnitudes
and directions of the
vector components I-‘_ and I-‘, of vector F. The signed magnitudes of these vector
components. which are the
scalar components of F. are written as F_. K. It is seen that Ii = Fcos 9. E =
I-"sin 8.
Y
- — — -
-- P
7'
2
1
" ,; .\- Hg. 1-to
1.20 (0) Let I-‘ have a magnitude of JOON and make angle 8 = 30" with the positive
x direction. Find E and Ii.
(b) Suppose that F = Jlll N and 6 = 145° (F is here in the second quadrant). Find
F, and F}.
I (at F. = 3UUCOs 30° =§Q._§§_ E = 3tl0sin 30° = 1.‘-ON. lb) F, = 300cos 145° =
(30U)(-0.8192) = ;Z45.75 N
(in the negative direction of X). I-I = 3(1) sin l45"= (3(ll)(+0.5736) = l72.07 N
1.21 A car goes 5.0 km east. 10 km south. 2.0 km west. and 1.11 km north. (a)
Determine how far north and how
tar cast it has been displaced. (b) Find the displacement vector both graphically
and algebraically.
| (n) Recalling that vectors can be added in any order we can immediately add the
3.0-km south and I.0-km
north displacement vectors to get a net 2.0-km south displacement vector. Similarly
the 5.0-km east and
2.0-km west vectors add to a 3~km east displacement vector. Because the east
displacement contributes no
component along the north~south line and the south displacement has no component
along the east-west line.
the car is -2.0 km north and 3.0 km east of its starting point. (b) Using the head-
to-tail method. we easily can
construct the resultant displacement D as shown in Fig. 1-I l. Algebraically we
note that
D=\/I)f+Df=v'2’+3’=3.6km tanw = —§ or tan 0=§ 6=E south oi east.
\
1
0/X \ ll
\\ II I
. 9 tn
In
5 Fig. 1 ll
1.22 Find the .t' and _v components of a 400-N force at an tingle of I25’ to the x
axis.
I Formal method (uses angle above positive .\' axis):
F; = (Jill N) cos 125" = -229 N F, = Hill N) sin 125° = 327 N
Visual method {uses only acute angles" above or below positive or negative x axis):
IE2 = Fcosip =~l00cos 55"= ZZ9N lfil = Fsin ¢ =4(l)sin 55°=3Z7N
By inspection of Fig. l-12. F, = -IF. = N; I-1 = |!»§| =1
1.23 Add the Iollowing two coplanar forces: 3l)N at 37° and S0 N at I80’.
I Split each into components and find the resultant: R, = 24 — Si] = -26 N. R, =18 +
l)= 18 N. Then
R = 3l.fiN and tztn 0 = l8/-26. \n II = L“.
AAJ K/-\ TOPPER m§Ei°§m'§gJ'E'é°R'ii’§
AAJ KA TOPPER I.“;E$Z.1?.‘2*‘.‘;’.;’R'i{’s'

1.7.4
1.25
1.26
L27
1.28
1.29
1.30
[.31
MATHEMATICAL INTRODUCTION I-' 5
iBl=lZm
\
| |
I I e
__ H = t:.< I ‘fin _\f_°"‘
Q) = §\ I ‘ I ,’)0° \>
1
Fig. t-12 ‘
l('l=9m
t-1;. l-13
For the vectors A and B shown in Fig. 1-13. find (a) A + B. lb) A — B. and (c) B -
A.
I The components are A, =6m. A, =0. B, =12cos60°=6m, and B, = l2sin60°=1U.4m.
(a) (A+B), = 12m and (A+B)_ = l(l.~lm. sothat A+B=l_5.9lat4i‘L°. (6) (A—B),=0
and(A—B),=0—10.4soA—B=10.4mati“. (c) (B-A),=Uand(B—A)_= l0.4—0 so
B—A=l0.4mat9l)__°.
For the vectors shown in Fig. 1-13. find ta) A + B + C and tb) A + B — C.
I The x and y components of C are 4.5 m and -7.8 m. (ill The x component is A, + B,
+ C, = 16.5 and for
the y component we find 2.6, so the vector is 16.7 in at 2Q__°. (b) A, + B, — C. =
7.5 and the y component is
0 + l0.4— (-7.8) = I82; changing this to a magnitude and angle. we find 19.7 m at
6Q.
For the vectors shown in Fig. 1-13. find (0) A - 2C. (bl B — (A + C). and (e) —A — B
— C.
I ta) The x component is A, — 2C, = -3 and the _v component is —2(—7.8) = 15.0,
giving in at ltl’.
(b) The x component = 6 — (6 + 4.5) = -4.5; they component = 10.4 — [0 + (—7,8)] =
18.2; therefore
(4.5: + 182*)" =18.7 m at l_04°. (e) This is the negative of the vector of Prob.
1.25 ta). so that it = 16.7 m
at 9.0” +180“ = Q = -171“.
A displacement of 20m is made in the xy plane at an angle of 70" (i.e.. 70°
counterclockwise from the +.r
axis). Find itsx and _v components. Repeat if the angle is 120°; if the angle is
250°.
I In each case s, =.t cos 8 and s, =.t sin H. The results are Q3. 18.8m: -10.0.
17.3 m: -6.8. —18.8 m.
lt is found that an object will hang properly if an x force of ZON and a y force of
—30N are applied to it.
Find the single force (magnitude and direction) which would do the same job.
I Adding components of the forces yields R, = 20 N and R, = -30 N. R = (441) +
9(1))" = 36 N. Calling 8 the
oounterclockwise angle from the +.r axis. tan 8 = -30/20 and so B = 303.7“ =
-56.3‘.
Find the magnitude and direction of the force which has an x component of —40N and
a y component of
—60 N.
I The resultant of these two forces is R = (16(ll+ 3600)": = 72 N. The angle B is
I80“ + tan '(6/4) = 230.3°.
Find the magnitude and direction of the sum of the following two coplanar
displacement vectors: 20m at 0°
and 10 m at 120°.
I Splitting each into components, R, = 20 - 5 =15 m and R, = 0 + &7 = 8.7 m. Then R
= _|7._3_rgi with
tan 6 = 8.7/15 giving 0 IE.
Four coplanar forces act on a body at point 0 as shown in Fig. 1-l4(a). Find their
resultant graphically.
I Starting from O. the four vectors are plotted in turn as shown in Fig. l.l4(b).
Place the tail end of one
vector at the tip end of the preceding one. The arrow from 0 to the tip of the last
vector represents the
resultant of the vectors.
AA-I KA TOPPER L?E#°:.1?l2“.’;'.;°Ri(’;

6 U CHAPTER 1
"ON l(DN \$ to [I0
to"
,,- .4‘ ‘ \
10° 0 so N \\ 0 \@
W0 N u E '( 2 £ g0“ ,‘
o no
W ta) n;. 1-14
Measure R from the scale drawing in fig. l.14(b) and find it to be 119 N. Angle a is
measured by
protractor and is found to be 37'. Hence the resultant makes an angle 8 = 180‘ —
37° = 143° with rhe positive x
axis. The resultant is 119 N at 112.
1.32 Solve Prob. 1.31 by use of the rectangular component method.
I The vectors and their components are as follows.
magnitude, N x component, N y component, N
80 80 0
I00 l(I)¢0s45°= 71 l(Dsin4S°= 71
110 —1l0cos30"=—95 1l0sin3(f= 55
160 —160oos20°--150 -16OsinZ0"=—S5
Note the sign of each component. To find the resultant, we have
R,=80+7l-9S—l50=—94N R,=0+7l+S5—5$=7lN
The resultant is shown in Fig. l-15: we see that R = \/(94)! + (71): = ll§ N.
Further, tan tr = 71/94, from
which a I 37°. Therefore the resultant is 118 N at 180 — 37 = Q‘.
|»'
R
7|
8
,, , Hg. i-is
1.33 Perform graphically the following vector additions and subtractions. where A.
B, and C are the vectors shown
in Fig. 1-16: (a) A4-B. (6) A-0-B4-C. (c) A-8. (ii) A+B—C.
I See Fig. 1-16(0) through (d). In (c), A — B = A + (-3); that is, to subtract B
from A. reverse the
direction of B and add it vectorially to A. Similarly, in (d), A + B - C = A + B +
(—C), where -C is equal in
magnitude but opposite in direction to C.
-c
~_ 4 1
__3__
> _
. K '
C /I
I
0' 7
| , I
0| ql
+| _a s I
t‘ A ‘ "I
<1 _ I ~
1-‘
(0) (b) (1') (4)
Fl‘. I-I6
A/-\J K/-\ TOPPER Ii'E°$°§i?i'§gJ§EpR§§§

MATHEMATICAL INTRODUCTION U 7
Y
30 sin 45° 30lb
5‘l"'_ _ _ _ _ * "
| 50 sin2
I
J.
‘fit:
x
50 cos 20 os 45“
‘FT
won.» M H,
1.34 Find the resultant It of the following forces all acting on the same point in
the given directions: 30 lb to the
northeast; 70 lb to the south; and S0 lb 20‘ north of west.
I Choose east as the positive x direction (Figure 1-17).
x components, lh y components, lb
30cos45°= 21.2 30sin45‘= 21.2
—50oos20°=—47.0 50sin20‘- 17.1
mu = -2.s.a lb -"'1
Total = -31.7 lb
R = V(—25.8)’ + (—31.7)' = \/665.8 + l(X)4.9 -40.9 lb tan 6 =%}= 0.8139 6 = 39°
west of south
1.35 Find the angle between two vector forces of equal magnitude. such that the
resultant is one-third as much as
either of the original forces.
I ln the vector force diagram (Fig. 1-18), the diagonals of the rhombus bisect each
other. Thus.
oc>s6=FT/6=%=0.1667 0=80.4° 28=160.8°
The angle between the two forces is 160.§°.
F/3
//\\
// \\
/ \\
Fl
6
F F
0 2 0 |-1;. 1-I8
1.36 Find the vector sum of the following four displacements on a map: 60mm north;
30 mm west; 40 mm at 60°
west of north; S0 mm at 30° west of south. Solve (a) graphically and (b)
algebraically.
I (a) With ruler and prou-actor, construct the sum of vector displacements by the
tail-to-head method as
shown in Fig. 1-19. The resultant vector from tail of first to head of last is then
also measured with ruler and
protractor. An.r.: 97 mm at 67.7°W g1N. (b) bet D = resultant displacement.
D,= —30—40sin60°—50sin30’= —89.6mm D, =60+40cos60°—50cos30'= -t-36.7mm
D . .
D =VDf+ Df=fiL§_q1£ tan ¢ = E"=>¢ =22.3° above negative x mus
AAJ K/-\ TQPPER I|'E°Ei°§|Tii|'§gJ'E'EpR':I’§

AAJ NEET AIIMS JEE RAS
8 u CHAPTER 1
' N
ur;
Sn mm '
NI mm
L ,i
n;. 1-19
Two forces. 80N and 100 N acting at an angle of 60° with each other. pull on an
object. What single force
would replace the two forces‘! What single force (called the zquilibranl) would
balance the two forces? Solve
algebraically.
I Choose the x axis along the 80-N force and the y axis so that the l00-N force 60°
above the positive x axis
has a positive y component. Then the single foroe R that replaces the two forces is
the vector sum of these
forces:
R,=S0+llX)COS6(]°=l30N R,=lOOsin6O"=37N
R=\/_Rf??§-lS6N 0=tan"E11=I£ above positive: axis.
The force that balances R is —R with a magnitude of l56N but pointing in the
opposite direction to R: 34°
below negative x axis (or 214“ above positive x axis).
Two forces act on a point object as follows: ltll N at 170" and I00 N at 50°. Find
their resultant.
I F. = l00 N at 170‘ above x axis; F; = l00N at 50" above x axis.
R=I-‘,+I-‘, R,=l00cosl70°+100cos50"=—34.2N R,.=ll)0sinl7t)°+ltDsinS0°=94.0N
R=\/Rf+R;=lO0N 9=tan"%
has two solutions: 290° and ll0°. From a look at its components we see that It lies
in the second quadrant, so
the answer is ml’ (or M above negative x axis). In a less formal approach we can
find: ¢ = tan 'lR,./R,|.
which will give an acute angle solution. in this case of 71’. which always
represents the angle with the posit-ive
or negative x axis. and either above or below that axis. Since we already know from
the components which
quadrant R lies in. we know the direction precisely. ln our case the 70‘ is above
the negative x axis.
A force of 1(1) N makes an angle of 0 with the x axis and has a y component of 30N.
Find both the x
component of the force and the angle B.
I The data are sketched in Fig. l~20. We wish to find F, and 9. We know that
. 0 30
sinfl-5--1-6-0.30
from which U = l7.S°. Then. since n = Ii cos 0, we have
F, = I00 cos l7.5“ = 95.4 N
lm
30
i _-..._
fl " Fl]. 1-no
A boat can travel at a speed of 8 km/h in still water on a lake. tn the flowing
water of a stream, it can move
at 8km/h relative to the water in the stream. If the stream speed is 3 km/h. how
fast can the boat move past
a tree on the shore in traveling (a) upstream? (h) downstream?
AAJ K/-\ TOPPER Ii'EeE§°§i'§i'§gJ'E'épR':I’§

1.41
1.42
1.43
| (a) If the water were standing still, the boat‘s speed past the tree would be 8
km/h. But the stream is
carrying it in the opposite direction at 3 km/h. Therefore the boat's speed
relative to the tree is
8 — 3 = Q km(h. (b) ln this case. the stream is carrying the boat in the same
direction the boat is trying to
move. Hence its speed past the tree is 8 + 3 =l1km[h.
A plane is traveling eastward at an airspeed ot' S00 km/h. But a 90 km] h wind is
blowing southward. What are
the direction and speed of the plane relative to the ground?
I The plane's resultant velocity is the sum of two velocities, 500 km/h eastward
and 90 km/h southward.
These component velocities are shown in Fig. l-21. The plane's resultant velocity
is found by use of
R = W500)’ + (90)! = SQ kmlh
The angle or is given by
90
tana-50-0180
from which a = 10.2“. The plane's velocity relative to the ground is S08 km/h at
l0.2° south of east.
N N
500
50° 90
90 E R E
R
Hg. 1-21 I-1;. 1-U
Vfith the same airspeed as in Prob. 1.41. in what direction must the plane head in
order to move due east
relative to the earth?
I The sum of the plane's velocity through the air and the velocity of the wind must
be the resultant eastward
velocity of the plane relative to the earth. This is shown in the vector diagram of
Fig. l-Z2. It is seen that
sin 6 = 90/500, from which 0 = l0.4°. The plane should head 10.4‘ north of east if
it is to move eastward on
the earth.
If we wish to find the plane's eastward speed. Fig. l-22 tells us that R = SCI) cos
9 = 492 kmlh.
A child pulls a rope attached to a sled with a force of 60 N. The rope makes an
angle of 40° to the ground.
(0) Compute the effective value of the pull tending to move the sled along the
ground. (b) Compute the force
tending to lift the sled vertically.
I As shown in Fig. l-23. the components of the 60 N force are 39 N and 46 N. (1)
The pull along tne ground
is the horizontal component, 46N. (D) The lilting force is the vertical component,
33}
r,-ootnar
wh -aw
F,-6000140‘-46N 1 fig-I-23
1.44 Find the resultant of the coplanar force system shown in Fig. k24.
I |t=t-‘.+|-",+t', R,=—40+80oos30°+0=29.3lb R,=0—80sin30"+60=20lb
R
R =\/R,,’+R,’=35.4lb 0=tan"#=34.3° above +x axis
AAJ K/-\ TQPPER Ii'E°Ei°§i'ii'§gJ'E'EpR':I§

10 J CHAPTER 1
\‘ l
/~. = wnv In = tillh F‘ : qmh
f, = -mlh “I F, - Jilh 50*‘
\ 1
IF .11!‘ (J
r, I wlh
1‘, mlh
“s- 1-" i Fig. 1-25
I.-I5 Repeat Prob. 1.44 for Fig. l-5.
I R=F.+F;+F\+F. R,=Z5+0—9(lC0s5(Y°+0=—32.8|b R,=0—4(l—90sinS0°+60=—48.9|b
R=VR§+R§=S§.9lb ¢=tan 'i'%l=&.l° below -x axis
1.46 Repeat Pmh. 1.44 for Fig 1-26.
| R=Il,+R,+R,+Il. R, = —l25t:0sZ5°-I-0+I80cosZ3°+1S0c0s62°=l22.Blb
R, = -125 sin Z5"— 130- 180 sin 23°+ lSOsin62°= —l20.7lb
R = VR:+ Rf = l72.2lh. ¢ = tan 'i% =44.5° below +x axis
I
il} = littlli
61
,‘| fit“, " r.= twin
r: - unit.
(I
5 *1 _‘_“‘ ‘
Fig. l-Z6
1.47 Compute algebraically the resultant (R) and equilibrant (E) of the following
coplanar forces: 1(1) kN at 30°.
141,4 kN ai 45°. and 100 kN at 240°.
I R = F, + F, + I-1. R,=1lX)cos30" +141.4cos-$5° + 100 cos 240° = 136.0 kN.
Note: lillcos 2-l(f'= —1ll)cos60° Ill) sin 240"= —l(l‘l sin 60“.
R, =100sin 30° +141.-4 sin 45° +100 sin 240" = 63.4 kN
R=VR;+R:=l50.6kN ¢=tan 'g*=24.9" above +x axis.
E= —R = l_§_Q.Q_lQl at 2-t.9°+ t8()°= V abovc +x axis
. 204 9°
I-48 Compute algebraically the resultant of the following displacements: 20 m at
30°. 40m at 120". 25 m at I80“.
42 m at 270°. and 12m at 315‘.
I D = d, + 1|; + 1|, + d, + d, = resultant displacement D_ = 20 cos 30° + 40005120“
—- 25 + 0 + l2 cos3l5° -=
— I913 m
Note: 180° is along —x axis; 270° is along —y axis; cos 120° = -00560“; sin 120°
=sin 60°; cm 3lS‘=cos 45°;
sin 315° = —sin 45°.
D, = 20sin3l.l°+40sin I20‘ +0- 42 + i2sin3l5°= —5.8m
AA‘-I NEET AIIMS JEE RAS

1.49
MATHEMATICAL INTRODUCTION I 11
D,=VDf+D;=20.2m ¢=tan" £1 =i below -1 axis
or 6 = tan" £1 = 196.7“ above +x axis.
Refer to Fig. I-27. ln terms of vectors A and B. express the vectors P. R. S. and
Q.
I We have here a parallelogram. 90 R = B and P = A + R = A + B. Clearly S = -A and
Q is the sum of —B
with A or Q=A-B.
I
A .
3 I-‘lg. 1-27
1.50 Refer to Fig. l~28. ln temrs of vectors A and B, express the vectors E. D — C.
and E + I) - C.
1.51
1.52
1.53
I Clearly —II=A+BorE=—(A+B)=—-A-B. D-C=D+(-C)=A.ThenI-l+D-C=E+A=—ll.
D
C
E
D
r-1;. 1-as
A displacement D of 1(1) m from the origin at an angle of 37° above the x axis is
the result of three successive
displacements: d,, which is 100m along the negative x axis: 11;, which is 200m at
an angle of 150° above the x
axis; and a displacement dt. Find 4..
I D=t'|,+d,+d1 D,=d,,+d,,,+d,,orl00cos37°=—100+200c0slS0°+d,, d,,,=353m
(Note:cosl50°= —cos30°) D,=d., +d,, +d_., or l00sin 37°=0+2lDsin l50’+d,,
d_,,=--40m (Note:sinlS0°=sin30°). d_,=Vd§,+d§,=35§m ¢=tan"|5’1’=§._5° below-0-xaxis
J1
The resultant force due to the action of four forces is ll, which is I00 N along
the negative y axis. Three of
the forces are 100 N, 60" above the x axis; 200 N, 140° above the x axis; 250 N,
320° above the x axis. Find the
founh force.
I R=F,+F,+I-‘,+F. R,=r,,+i-:_+r_.,+n, or 0=100cos60°+200cos140°+L50cos320°+I-1,.
F.,=-88.3N R,=r,,+r,,+r_.,+r,,or-io0=ioostneowzoostni40'>+2s0sma20~+1a,.
F¢,=—l54.5N F.=VFi,+Fi,=180N ¢=tan"':—_'Z =@bel0\v —xaxis
4.
A car whose weight is w is on a ramp which makes an angle 0 to the horizontal. How
large a perpendicular
force must the ramp withstand if it is not to break under the car's weight?
I As shown in Fig. 1-29. the car‘s weight is a force I that pulls straight down on
the car. We take
components of W along the incline and perpendicular to it. The ramp must balance
the force component
w cos 8 if the car is not to crash through the ramp.
AAJ K/-\ TOPPER ii'E°E§°§i'ii'§gJ'E'EpR':I’§

12 J CHAPTER 1
1.54 The fivu coplanar forces shown in Figt 1-30(a) act on an object. Find the
resultant force due to them.
l6N
ISN
wnnfi
\
W"
\ W
\
\ //
/
J’ )
Fig. 1-29
. 60*
'5 t 9 5.1 N
30°
I9N " 9 I
IIN
R —3.ZN
ZZN
V
(I) (M
Fig. 1-30
I (l) Find the x~ and y-components of each force.
magnitude. N x component. N y component. N
4.-._.._-._.
N-~O~v\~0
19.0 0
15cos60°= 7.5 lSsin60°=13.0
—l6cos45°=-11.3 l6sin45°= 11.3
-llcos30°= -9.5 -llsin30°=-5.5
0
22.0
Note the signs to indicate + and — directions.
(2) The resultant R has components
R, = E F, = l9.(]+7.5— ll.3—9.S+0= +5.7N R, =Xfi =0+13.04-1l.3—5.5—22.0=—3.2N
(3) Find the magnitude of the resultant from
R=VRf+Rf=6.5N
(4) Sketch the resultant as shown in Fig. 1-29(b) and find its angle. We see that
3 2
tan ¢ — -
‘“° “ * — 29° - 31!“ Th resultant is 6 5 N at 331" (or —29°).
=57 0,56
I ‘t'I17eLearn‘intj'}-\'|‘>'t$'F8r'=36(F " " ‘ ° ‘ ——‘ —
AAJ KA TOPPER NEETAIIMSJEE RAS

MATHEMATICAL INTRODUCTION J 13
1.55 Find algebraically the resultant (R) and equilibrant (I-I) of the following
coplanar forces: 300N at 0°, 4(1) N at
30°. and 400N at 150°.
|R=r,+r-‘,+r. E=—R R.=300+4oo¢<>§30°+4o0¢<>§1s0==a00.v
Note that 4(X)cos l5(7’= -400 cos 30° 400sin l50°=400sin30°
R,=0+400sin30°+400sinl50°=4tI)N R=\/Rf+R_f=500N ¢=tan"%|=5l’above +xaxis.
Then E =SO0N, 0,; =S_3f below —x axis.
1.2 THREE-DIMENSIONAL VECTORS; DOT AND CROSS PRODUCTS
1.56 Find the magnitude oi the vector A in Fig. l-31. whose tail lies at the origin
and whose head lies at the point
(7.0 m. 4.0 m. 5.0 m).
| First. note that the vector A and its vector component A, are the hypotenuse and
one side of a right
triangle whose plane is perpendicular to the xy plane. lience the pythagorean
theorem gives A’ = B’ +
But the vector B is itself the hypotenuse of the right triangle in the xy plane
whose sides are the vector
components A, and A,. Thus you can use the pythagorean theorem again to obtain B’ =
/11+ A;‘.. Combining
the two equations.
A’=Af+Af+/if or A=\.//t;+A;+A;
which is the fonn taken by the pythagorean theorem in three dimensions.
For the data.
A = \/(7.0 m)‘ + (4.0 m)‘ + (5.0 m)’ = 9.5 m
rm mt
t»
rt_. l.. l_.|=t‘tlm.-l.Uru. $.0ml
5 .-<:—_-_'I:T_ti:tl :. :_i— ~
. .
a " '
l \
‘ 1
.r . 3
\ »\.
' 0..
U
' J
4 ; t.r_.»|,.0r=
'¢= 1. .. .0"
A I Out -{Chit it
, .
I!
t Int ml r-1;, 1.31
1.57 Find the scalar components of the three-dimensional vector F in Fig. 1-32.
I In Fig. I-32. |".. F,. F, are the rectangular vector O0mp0ncnls of F; the scalar
components F}. F,. F, are
given by
l'1=Fcos9, F,=Fcos9, F,=Fco§9,
Or for convenience. writing cos 9, =1, cos 0, = m. cos 9‘ = n.
F. = Fl F; = Fm F, = Fn
l, m. and n are referred to as the direction cosine: of F. By the three-dimensional
pythagorean theorem
(Prob. 1.56). la 4-ml + nz =1.
AAJ K/-\ TOPPER :-l|:EeE|‘ie:lll'IiIilggJll?épR|:fSr

14 U CHAPTER 1
Z
\
\
\
-J
/I‘
/ .
I
/
/
._.\
-It
\
\
\
7'
I/,=t~t-c‘ II I
I
I
n, = t:os"' m |
H, _= ens" I
| O
'_.._
|-I!
I.
I
l
I
|
I
I
I
I
|
\.___
\
\
\
\
\
~<
.\' lit‘. I-32
l-59 ln Fig. 1-32. let F represent a force of 2(1) N. Let 9, = 60°. 0, = 40°. Find
F,. I-1, and E.
I I-0.5 m =0.1oo n =(l -1’-m=)"’=0.404
(assuming that F, is positive; otherwise, n = -0.404), and the rectangular
components of I? are
F'.*(109)(9-5)=@ F Ffiiil
As a check. (ND: + l53.2’ + 803*)": =# 2(Il. Note that 0‘ = 66.17”.
1.99 Find the vector sum (R) of three vectors. I-‘,, F1, l‘,. drawn from the origin
of a three-dimensional coordinate
system such as that ot Fig. L32.
I Following the component method,
R.=F}.+F1.+l'§. R.=Fn+F1,+F,\,» R.=Fr=+F1,+Fn
and
R = l(F.. + E. + Fe)‘ + (E, + F1, + 8,)’ + (H. + R. + R.)’1"’
where F, is the X component of l?,. etc. The direction oosines of R are given by
I=F..+2.+Ft. m=F.,+€.+fi, ”=F..+:';.+F-.
The resultant (magnitude and direction) of any number of vectors drawn from 0 can
be obtained in the
same way.
1.60 Define a unit vector.
I Any nonzero vector I7 may be written as I-‘ = Fe. where F is the magnitude of F
and where e is a uni!
vector (a vector Whose magnitude is 1) in the direction of P. That is, the
magnitude of F is indicated by F and
its direction is that of e. If I-‘ carries units (e.g., N, m/s), F carries the same
units; e is a dimensionless vector.
1.61 Express the vector F in Fig. 1.32 in terms of unit vectors along the
coordinate axes.
I ln Fig. I-32. let us introduce unit vectors i, j. It along X , Y. Z,
respectively. Then the vector components
of F can be written as
F,=I-Ll I-‘,=F,] I-‘,=I-1k
according to Prob. 1.60. [If one of the scalar components is negative—say. F. = —|
l‘l| = —|F,| —then we have
F, = —|F,| l= |F,| (-l), which is still as prescribed by Prob. 1.60.] Since F is
the resultant of its vector
components, we obtain the very important expression F - El + F_‘,j + El. ln this
expression, I-1 = F cos 0, =
Fl, etc., as previously shown; and, as before, the magnitude and direction
(direction oosines, that is) are
obtained as
r=(r3+r§+r})"= :=§ m=§ »=§
AA-1 KA TOPPER NEETAIIMSJEE RAS

1.62
1.63
1.61
1.65
1.66
1.6‘!
1.68
1.69
1.70
1.71
1.72
1.73
MATHEMATICAL INTRODUCTION J 15
A force P has components E = ll!) N. F; =153.2 N. I-I =80.8 N. Exprew F in tenns of
unit vectors and find
its magnitude and direction.
I The vector F can be written as F= llDi + 153.2] +80.8k with magnitude F = (100:
+1532‘ + 81:)“ =
2(DN and direction:
100
I-2?)-0.5 m-0.766 n—0.404
Strictly, we should have written
I-‘=(100N)i-+-(153.2 N)j+(80.8N)k or F=1(X)i+l53.2j+80.8lt N
A force A is added to a second force which has x and y components 3 N and -5 N. The
resultant of the two
forces is in the -x direction and has a magnitude of 4 N. Find the x and y
components of A.
I let A=A,l+A,j. ThenA, +s= -4 and/i,—5=O. SoA,=—7N andA,=§_§.
Express A, B. and C of I-“lg. 1-13 in terms of the unit vectors l, j. and lr.
I A==A,i+A,j +A,t=@. B =6i + l0.4|'m. and c=4.sr- 7.8[m.
Find the components of a displacement which when added to a displacement of 7i - 4]
m will give a resultant
displacement of Si - 3j m.
I We haveA,+7=§andA,—4=—3, so.-l_=-Zl and A,=@.
Find the magnitude and direction of the vector sum of the following three vectors:
Zi — 3j, —9l — Sj. 4l + 8].
I We have (2 — 9 + 4)l + (-3 - 5 + 8)] = -3i. Thus. it is Q units along the -1
direction.
What must be the components of a vector which when added to the following two
vectors gives rise to a
vector 6]: l0i — 7j and 4i + Zj?
I Call the vector A. Then A,+l0+4=0andA, —7+2=6. SoA,=i and A,.=Q.
A certain room has a floor which is 5 X 6 m and the ceiling height is 3 m. Write an
expression for the vector
distance from one corner of the room to the corner diagonally opposite it. What is
the magnitude of the
distance?
I The x. y, z displacements in going from one corner to the other are 5. 6. and 3
m. respectively. Therefore.
D = 5| + 6| + 3|: m. Also. D‘ = Df + Df + Df, which gives D = 8.4 m.
Find the displacement vector from the point (0. 3. -1) m to the point (-2. 6. 4) m.
Give your answer in I. ]. It
notation. Also give the magnitude of the displacement.
I The component displacements are D, = -2 — 0= -2. D, = 6 - 3 = 3, and D, = 4 - (-
1) = 5; thus
D=—2l+3|+5km and D=6.2m.
An object, originally at the point (2, 5. I) cm. is given a displacement 8i — 2] +
kcm. Find the coordinates of
its new position.
I The newcoordinates are x =2+8= l0:y =5+ (—2)=3; z = l + 1 =2. The object isat
§l0.3.2)cm.
Find the resultant displacement caused by the following three displacements: Zl —
3k. Sj — Zlt. and
—6l+] + 8k. all in millimeters. Give its magnitude as well as its l. j. It
representation.
I |t=(2-6)|+(5+ l)j +(—3—2+8)lt=—4l+6| +3ltmm. and so R =6l"’=7_.8@.
Give the i, j, k representation and magnitude of the force which must be added to
the following two forces to
give a force 7| — 6] — l: 2i — 7lt and 3] + 2k. All forces are in newtons.
I Call the force F. Then I-}+2=7;I-1 +3=—6; F_. -7+2=—l. ThereloreF=5i—9’|+-tkN and
F=@.
Vectors A and B are in the xy plane. If A is 70N at 90‘ and B is l20N at 210°. find
ta) A — B. and
(b) vector C such that A — B + C = 0.
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1.74
L75
1.76
1.77
1.78
1.79
1.30
rt
_i
CHAPTER 1
I A and B written in terms of unit vectors are A= 70] and B = —l20eos30°i — 120 sin
30°] = -1046 — 60].
(a) A—B=(0+104)i+(70+60)j=166Nat5l°.(bl C=-(A—B)=—l04l—l30j=lQNgtL§|°.
If A= 2i— 3j + 5k mm and B= —i— 2j+ 7k mm. find (in component form) (tr) A-B. lb) B—
A.
(c) Vector C such that A + B + C = U.
I (n) A—B=|2—(—l)]l+|—3—(—2)]j+(S—7)lt=3i—|—2lrmm (b) B-A= —(A—B)=
_—_3i+|‘-t-Zkmm (c)C= -(A+B)=-H-5|— l2kmm.
Vector A = 3| + S] - Zli and vector B = -3] + on. Find a vector C such that 2A +
7|! + 4C=0.
I A vector is zero il and only if each of its components is zero. For example. 2A,,
+ 78,, +4C, = 0. so
C. = —l.5. For (.‘,, solve 2(5) + 7(—3) + 4C, = 0 to obtain C, = 2.75. Similarly,
C, = -9.5. Therefore
C = — 1 .5i + 2.75] — 9.5K.
A certain vector is given by 3i +4] + 7Ii. Find the angle it makes with the z axis.
I First find the projection of the vector in the xy plane; this is (3: + 4‘)“‘= 5.
The magnitude of the vector is
(7: + 5’)"'= = 81> and it makes an angle of tan"(S/7) =35.5" to the z axis.
Otherwise. from Prob. 1.61.
7
cm0.= =().8l4 0;=35.$°
What must be the relation between vectors A and B if the following condition is to
be tniec
A—2B=—3(A+B)
ll vector A = 6i — Zli m. what is B?
I One has A — ZB = —3A — 3B. so 4A = —B. Substituting (ti — Zlt for A gives B = -2-
ii + 8|t m.
What must be the relation between two vectors A and B if the magnitude of A + B
equals the magnitude of
A — B. that is.
IA + Bl = IA - Bl
| A and B must be mutually perpendicular. To SCI.‘ that. let P and Q be two vectors
of the same magnitude.
As shown in Fig. l-33. P and Q form a rhombus whose diagonals. P + Q and P— Q, are
necessarily
perpendicular. Now set P = §(A + B). Q = §(A — B).
P—Q= .-\
I/_.
1"’ I
. I
P '»|'~o=a
Q
t
‘ Fig. i-as
The vector displacements of two points A and B from the origin are
.r,=3i—2j+5ltcm and s,=—i—5]+2lrcm
Find the magnitude and i. j. It representation of the vector from point A to point
B.
I The vector in question has components D, = —l — 3= -4: D, = -5 — (-2) = -3; D, =2
— 5= -3.
Thus D = —4l — 3| -— 3|: cm and from D’ = 4: 4- 3* + 3’. we find D L.
The rectangular components of an acceleration vector 1 are a, = 6, n, = 4. a, = 9
m/s‘. Find the vector
expression for I and its direction cosines.
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MATHEMATICAL INTRODUCTION I 17
I ln vector form. I = 6i + -tj + 9|: m/s‘. The magnitude of n is u = t6‘ + 4’ +
9:)‘ ‘ = l1.S3 mg. and the
direction cosines of a are
hi ,_; ,_i
—L-52 "its; "us;
1.81 Find a vector expression for a line segment.
I 'l1le straight line ab. Fig. l-34. is determined by points P, and P, Regarding
the line segment from P, to P
as a vcctnr s. we can write
s= (X: -mi +(y= -yilj + (:1 — :.)|i
with magnitude s = [(x; —X.): 4- ()5 )3): <1-(:3 — z,):]' : and direction
.r_—x. _i,—_\, :,—:,
I-i m-i rt-
; : s
Z
n
O
l’i
,,_ It
it
“.6
\
I
I
I
I
I
| .1
I
I
I
I
% i
\
:\
.\' Fig. I-34
An important special case of this is the so-called radius vector r. the directed
segment from the origin O to
a point Pix. )'. 2):
r=xi+yj+:lt
with r = (x:-t-y’ + z’)' I and
I = '5 m = ‘L ll = E
I‘ I‘ I
1.82 Consider the velocity vector v = lbl + 30] + 24k m/s. with 1- = ( I6’ + 30‘ +
24")‘ " = 4l.b2 m/s. direction given
by != to/41.62. etc.
Now let us multiply v by Ill: ltlv = 160i + Iltlij + 240k = v,. Find the magnitude
and direction of v,.
| 1-, = [(l60)‘+ 1300): +12-II.l)‘]"=(lll]t4l.62)= ltlv
and the direction cosines of v, are
'_ mo _ to _' _ _
'(l(l)(-ll.62)‘4l.62_ ""_"' "'""
which shows that v. has the direction of v.
1.83 Define the scalar or dot product of two vectors.
I The dot product of any two vectors. as F, and I}. Fig. l-35. is written as F, ~
F, and is defined its the
product of their magnitudes and the cosine of the included angle. That is. F, - F;
= Efieos 0. which is a scalar
quantity. ln Fig. l-35. F. = 75, F} = I00. 0 = 60°. Thus. F, ~ F; = (75)(l00)(l).5)
=
1.84 Find the dot products of the unit vectors along X . Y. Z,
I Since I, j. It are mutually perpendicular and of unit magnitude. the definition oi
the dot product gives
;.;=].]=k.k=| |.j=|.k=].k=()
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18 U CHAPTER1
1.55
1.86
1.87
LN
b __.___ r
~ IX ‘,3 /
'\
Qt“ \‘-3‘ /
w ti’ /
/
x‘ /
\ o=ur /
.-.=ss" /
0 :5 so F, 0 Fig.1-35
Find the dot product of any two vectors in tenns of rectangular components.
I W|'ite any two vectors as
F,=F|,i+F,,]+F,,l l7;=F,,l+F,,]+I'},l
Their dot product is given by
F. ~ F,= (F,,l + F,,j + F,,l) - (Fhi + Fa] + I§,l)
The right-hand side may be simplified by noting that the distributive law holds and
employing the values of
l- I, etc., found in Prob. 1.84.
F: ' F== F--Fa + 5,5, + EIFU
To check that P, - F; is just the quantity F,F,cos 0. where 0 is the angle between
F, and B. multiply and
divide the right side through by F,F,. giving
F,,F F F F,F,
l'_i‘|"z="_|"_z(F.‘F7;."’?{:!?-?+?|‘€) =5"-2(l|':+'"\'"z+'||'l2)
Now. the familiar addition formula in two dimensions.
cosfl =oos(9, — 9,) =cos 8,cos 0,+sin 0,sin 0;=l,l,+m,m;
extends to three dimensions as oos 9 = l,!, + rn,m, + n,n,. Hence, the above
becomes F,F,oos 8, and this
method of multiplication is. as expected. in accord with the definition of the dot
product.
bet F, = 10l — 151- 20k. F, = 6i + 8} — 1211. Find their dot product and the angle
between them.
I F‘-I"1=(10)(6)+(—15)(8)+(—ZU)(—lZ)=w
Now note that F] = (101 +15’ + 20*)" = 26.93. I-Q 1- l5.62. Hence. the angle 0
between F, and F, is given by
I-', -F, l80
0=i=-ii=O.4279 9=64.@°
°°‘ r.r, (26~93)(1S~62)
Of course. the same value can be obtained from cos 0 = (,1, + m,m, + n,n,.
Find the projection of any vector along a straight line.
I The projection of vector A = (A,. A,. A,) along the line detennined by the radius
vector r = (x. y. z) is
A, = A oos 0, where 8 is the angle between r and A. From the definition of the dot
product. A - r =
(Ar oos 6) = A,r. Hence, writing r= (1/r)r = (I, m, n) —a unit vector along r—we
have A, = A - i =
A_l + A,m + A,n. This expression for A, remains valid even when the line does not
pass through the on'gin.
Find the projection of A = l0i + 8] — 6|: along r = SI + 61+ 9k.
I Here r = (s= + 6* +9*)"'= =11.92 and Prob. 1.31 gives
A, =A,,l + 4,,» +A,n =1o(fi)+ s(“%) - o(1—1?;2) = fi=3_.@_g
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1.89 Define the cross product (or vector product) of two vectors.
I The cross product of two vectors. as F, and F1. Fig. 1-36, written as F - F, X
I-‘,, is defined as a vector F
having a magnitude F = I-‘J; sin 0 and a direction which is the direction of
advance of a right-hand screw
when turned from F, to F, through angle 0. it being assumed that the axis of the
screw is normal to the plane
detemtined by P, and P, (the right-hand screw rule). Or, it the curled fingers of
the right hand point from I’,
to F, the extended thumb points in the direction of F (right-hand rule).
Note that in aooord with the right-hand screw rule. F, X I’, - —(I?, X l',).
1.90 Find the cross products of the coordinate unit vectors.
I Since I. 1. k are mutually perpendicular and of unit magnitude. it follows from
the definition of the ems
product that
hl=jxj=|rx|i=0
lXj=k 1xt=| kXi=j ]Xi=—k |tx1=-t iXk=—j
r=r.x|r, Y
I>—>\%h§ \\:\;b —>
B
Advlnc 1;, m1. rt;
C=AlI ¥|.WIi_!It
Pi. rm. In
O | X
F, I‘ B
' Vc¢l0rPr0dut:tC*AXI
Q Z
Fig. 1-36 Hg. 1-37
1.91 Given two vectors, as in Fig. I-37.
A=A,i+A,j+A,lt B=B,l+B,j+B,k
find their cross product in rectangular coordinates.
I C=AXB=(A,l+A,]+A,k)X(B,I+B,j+B,k)
Applying the distributive law to the right-hand side and using the values of iX i.
etc.. found in Prob. L90. we
obtain
C = A X B = (/1,3, — A,B,)l + (A,B, — /LBJ] + (A,B, —A,D,)l
Equivalentiy, A X B may be expressed as a determinant:
I j k
C=A><B= A,,A,A,
B, B,’ B, ~ ‘ <
as may be verified by expanding the determinant with respect to the first row. Note
that the X. Y. Z
components of C are
C,=A,B,-A,B, C,=(A,B, —A,D,) C,=A,B,—A,B,
Hence the magnitude of C is C = (Ci + Cf -0- C E)“ and its direction cosines are
_ C. _ Q _ C.
I ~ C m — C n — C
Vector C is, of course. normal to the plane of vectors A and B.
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20 I CHAPTEH1
1.92 Assuming that vectors A and B. Fig. 1-38. are in the XY plane. determine the
magnitude and direction of
C = A x B.
I C=(200)(100) sin (S5°— 15°) = 20000sin40"= 12855
and by the right-hand rule the direction of C is that of +2. Vectorially we can
write C = 12 885k.
Y
mu A
55‘ 2“,
l
1*‘
1 -\'
I 11
c ; z “W "39
1.93 Referring to Fig. I-37. let A = Z0i - 10] + 3(lt and B = -61 + IS] -— 25lt.
(n) Find the magnitudes of A and B.
(bl Find the direction cosines of A. (c) Find the vector product C = A X B. (d)
Find the magnitude and
direction of C. (e) Find angle 6 between A and B (I) Find the values of the
direction oosines I1, m,, n, of B,
and find angles an. om. an between B and the X, Y, and Z axes. respectively.
I (a) A =(20’+10’+30’)"'=37.42 B=29.77
(b) 20 ~10 30
1,=— »-.=— n.=-
37.42 37.42 37.42
(e) Applying the determinant formula.
i j k
C = 20 -10 30
-6 15 -25
C = ll! -10)(-Z5) - (15)(30)l *ll(Z0)(-25) - (39)(-6)] + |Kl(20)(15) - (-19)(-6)]
= —200i4-320] +240|1= 20U(—l+ l.6j + 1.211)
(d) The magnitude 01C is C = 200(1' +1.6’ + 1.2:)" =4-$7.11
The direction oosines are I = fl m = A n = A-
’ 447.21 ’ 447.21 -‘ 441.21
Nole that C = C(l_J + m,j + n,lt).
(2) C = AB sin 8 447.21 = (37.42)(29.77) sin 8 sin 8 = 0.40145 9 = L’
(f) B=—6l+15j—25k=B(l,i+m,j+n,k)
Thus Bl, = -6 Bm, = 15 Bn, = -25 B = (6’ +15‘ + 25*)“ = 29.766
I1 = —0.20l6 m, = 0.5039 n, = —0.8399
Corresponding angles are an = 101.63” 11,, = 59.74“ an = 147.13°
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CHAPTER 2
Equilibrium of Concurrent Forces
2.1 ROPES. KNOW. AND FRICTIONLESS PULLEYS
2.1 The object in Fig. 2-l(a) weighs 50 N and is supported by a cord. Find the
tension in the cord.
I Two forces act upon the object. the upward pull of the cord and the downward pull
of gravity. Represent
the pull of the cord by T. the tension in the cord. The pull of gravity. the weight
of the object, is w = SON.
These two forces are shown in the free-body diagram. Fig. 2-ltb).
The forces are already in component form and so we can write the first condition for
equilibrium at once.
Z F, = 0 becomes t) = 0
§r,=0 becomes T-s0N=n
from which T = 5Q_N_.
\ )'
T
X
..- - so N
(a) (It) I-‘lg. 2-1
2.2 As shown in Fig. 2-2(a). the tension in the horizontal cord is 30 N. Find the
weight of the object.
40° § ’
cord 2 so T: “n 40, T:
P 3“ \ ‘O,
I
T ° N
cord I I ms 40 so I
r, - ..<
weight - it .
(41) (bl
Fig. 2-2
I As seen in Prob. 2.1. the tension in cord I is equal to the weight of the ohject
hanging from it. Therefore
T, = w. and we wish to find T, or w.
Note that the unknown force. T,. and the known force. SUN. both pull on the knot at
point P. It therefore
makes sense to isolate the knot at P as our object. The free-body diagram showing
the forces on the knot is
drawn as Fig. Z-2(b). The force components are also found there.
Next write the first condition for equilibrium for the knot. From the free-body
diagram.
2F,=0 becomes 30N—T,cos40°=l)
EF_=0 becomes T,sin4()‘-w=0
Solving the first equation for T, gives T, = 39.2 N. Substituting this value in the
second equation gives
w = 25.2 N as the weight of the object.
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22 U CHAPTER 2
1.3 For the system oi‘ Fig. 2-3(a). find the values of T} and 1} it the weight is
GIJN.
t
\\ \ \ \-~’$.».~1J>i.\‘.E1‘-A-"?\€*'\ T,
V
R
Q T; @ K
Ti
§ U “N an N
\ (11)
(bl Hg. M
I Consider the knot to be in equilibrium under the action of three ioroes, as shown
in Fig. 2-3(b).
E F,-0 yields 7}cos50°— T, =0 or o.e4sr,= T,
Zr,=0 yields T,sin50°—600N-0 or o.1c>er,-soon
This gives T, flag. Substituting into the E E. ¢q|-Iltion yields T, I 1.
2.4 The following coplanar forces pull on a ting: 2t'llN at 30°, 500N at 80°. 3(XlN
at 240°, and an unknown force.
Find the magnitude and direction oi the unknown force ii the ring is to be in
equilibrium.
I Assume that the 0° line is the x axis and 90’ specifies the y axis. The three
known forces are then as shown
in Fig. 24. If F. is the unknown force. then F, + F, + F, + F, =0. Let R= F, + II‘,
+ I?,. then R + F, =02}
F. I —R. To find F,, we need only find R.
y
F,=$00N
r.-zmu
.
W
F,=3illN n‘_z_‘
R, =r,,+r,_ +5, =ficos30°+F;cos80“-F,cos60°, R, -200(0.s66)+500(0.174)-
300(0.500)=i10N
R, = F,sin30°+ F,sin80°— F,sin60", R, =-20o(o.s0o)+soo(o.9as) -soo(o.sas)= mu, R -
(R: +R;)"*=
R
ssm. w\a,,=§=>o,,=1|.1°, fi=Rand a,._=e,.+1str=z,§;
2.5 in Fig. 2-5(a) the value of W is l80N. Find the tensions in ropes A and B.
I Refer to Fig. 2-5(b). Summing forces in the x and y directions to zero yields A =
B cos 53' and
B sin 53° == W = 180. From the latter, B = 225 N, which when inserted into the
former gives A I Q5 N.
2.6 ll‘ the identical ropes A and B in Fig. 2-5(0) can each support tensions no
larger than 2(1) N. what is the
maximum value that W can have’! What is the tension in the other rope when W has
this maximum value?
I From Prob. 2.5. B will experience the largest tension. 200N in this case. Solve
for forces along the
vertical: W = Zmsin 53' = 1§Q_I1. and in the horizontal direction find A = 2(1)
oos53° = 120 N.
2.7 A rope extends between two poles. A 90-N boy hangs from it, as shown in Fig. 2-
6(a). Find the tensions in
the two pans of the rope.
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w
\
EQUILIBRIUM OF CONCURRENT FORCES U 23
§ B
.
‘~_ 53" A
\\
53° \
\
_ _ __________ ._1
A
C=W
(nl
I0‘ 5‘
|-1;. 2-s
y
10' 5'
T
I T: Tl T2
T, sin I0‘ Tl '5" 5'
T, cos I0’ T; W! 5' X
tv I 90 N
(0) (bl
n,.z-6
I Label the two tensions T, and 1}, and isolate the rope at the‘ boy's hands as the
object. The free-body
diagram tor this object is shown in Fig. 2~6(b).
2I')=0 becomes T,cos5°-'l§cosl0°=0
2r,=o becomes 'I§sin5°+T,sin 10°-90N=O
Evaluating the sines and cosines, these equations become
0.996T,—0.98ST, =0 and 0.08’7T,+0.174T, —90=0
Solving the first for T, gives 1', =0.989'I}. Substituting this in the second
equation gives
0.0861‘, +0.17-£1] -90-=0
from which T, = E. Then. because 7} = 0.9891}, we have T, = Qi.
2.8 The tension in cordA in Fig, 2-7 is 30N. Find the tension in B and the value of
W.
50° 6°“
4 12
II Fig. 2.-1
| Draw a free-body diagram for the point on the rope where the cords meet; the
equilibrium relations in the
x and y directions are T, cos 50° = T, cos 60° and W = T, sin 50° + T, sin 60°,
where T, = 30N. Solving:
T,=§Qand W=$fl.
2.9 ln Fig. 2-7. how large are 11, and T, if W = 80 N?
I The equilibrium equations have already been obtained in Prob. 2.8. The y-
equilibrium equation when
W = 80 N is 80 = T,[sin 50° + (tan 60")(cos 50")], where we have substituted tor T,
from the horizontal
equation. Thus the tensions are T, = 43 N and T, = fiifl.
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24
2.10
2.11
2.12
2.13
2.14
CHAPTER 2
A boy of weight W hangs from the center of ll clothesline and distorts the line so
that it makes 20° angles with
the horizontal at each end. Find the tension in the clothesline in terms of W.
I From Fig. 2~R, ZT sin 20° = W. Therefore T = l.46W.
In I -I 3"’
“ r-1;. 2-a
ln shooting an arrow from a how. an archer holds the bow vertical and pulls back on
the arrow with a force
of 80 N. The two halves of the string make 25‘ angles with the vertical. What is
the tension in the string‘!
I Setting horizontal forces to zero at the point on the string held by the archer
we obtain 80= 2T sin 25°,
whence T = ‘Q.
ln Fig. 2-9(a), the pulleys are frictionless and the system hangs at equilibrium.
It w, is a 200N weight, what
are the values of w, and w,'?
1.
I. "H
_<l I'
., 5‘ ..,
I.
“ | .
I
M, 1),; Fig. Z-9
I The knot above w, is in equilibrium under the action of three forces. as shown in
Fig. 2-9(b). Since the
pulleys are frictionless. 7} = w,; 7} = wt. Also T, = w,. We are given 71 = iv. =
200 N. From X E, = 0.
T, sin 35° — Tzsin S0“ = 0. (Note the x-component equation involves sine functions
because angles are with
respect to y axis.) Then 2fXl((l.57-1) = 1;(0.766):> 1; =150 N = w,. From E F} = 0.
Ttcos 35° + 7;oos50° — T, =
0 or 200(0.819) + l5()(U.643) = Ti 1> T. = 260N = w,.
Suppose w, in Fig. 2-9(0) weighs SOON. Find the values of w, and w, if the system
is to hang in equilibrium.
l Now T, = 500 N in Fig. 2-9(b). Z F} =0=> Ttsin 35° — Tzsin 50°= 0=>0.574T,=
0.7667}. Or solving for Ti.
71 = 1.337‘, (1)
Z F, = 0:> T. cos 35° + L cos 50° — T, = 0$0.8l9T, + 0.6437‘, = SOON. Substituting
l.331}for 1'. we have
0.XI9(1.33T,)+0.643'l}=500N=>l.73E=500N T,=289N
From (I). we get Ti=384N.
Find the tensions in the ropes shown in Fig. 2-10 if the supported object weighs
600 N.
-. \
,9. "0'
(' B
Tl T,
. soon
|=t;. no
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I Let us select as our object the knot at A because we know one force acting on it.
The weight pulls down
on it with a force of 600N and so the free-body diagram for the knot is as shown in
fig. 2-l1(n). Applying
the first condition for equilibrium to that diagram. we have
2fi=0 or 7§cos60°—T,cos60°=0 2!§=0 or T,sin60°+7}sin60’—600=0
The first equation yields T, = T,. Substitution of T, for T, in the second equation
gives T, =3-$6N. and this is
also 7}.
Let us now isolate knot B as our object. Its free-body diagram is shown in Fig 2-
1l(b). We have already
found that 1} = 346 N and so the equilibrium equations are
EF,=o or T,cos20"-7Z—346sin30°=0
EF,=0 or T,sin20°-346cos30°=0
The last equation yields 1; = 877 N. Substituting this in the prior equation gives
7} = 651 N.
Y 7 )
T, T2
T.
0 T4
T
“ _‘ Ts 29 s t
X I X
GDN
20.
30‘ 30° T,
ta) (h) Ir)
Fm. Z-ll
We can now proeeed to the knot at C and the free~body diagram of Fig 2-11(c).
Recalling that T, = 346 N.
2!-}=0 becomes 1}+346sin30“—'I}cos20°=0
2F;=0 becomes 7}sin2U°—346cos30°=0
The latter equation yields 11 = 877 N.
[Note that from the symmetry of the system we could have deduced T, = 7} and 7} =
T_\.]
2.15 if w = 40N in the equilibrium situation shown in Fig. 2-l2(a) find 7} and T}.
‘ v
V
T, A ,
P W T.
70'
W Tl
(,1) lb)
\
T-
flg. 2-12
I The knot is in equilibrium under the action of three forces. and the free-body
diagram is as shown in Fig.
2-12(0). 1, = w = 4011.
2!-}=0=>T,sin70°—1}oos60°=0 or (o.94o)r,=(o.soo)r,, T.=l.88T,
Z!-}=0=>T,sin60°—7}oos70°—7§=0, (0.866)7}—(0.342)1}=T,=40N
A/-\J K/-\ TOPPER Ii'E°Ei°§i?i'§.gJ§EpR§§’§

26 U CHAPTER 2
Substituting for 7},
(0.866)(l.88T1)—(0.342)1§=4ON 1.291";-40N T§—Q_Ll)_b1 and T,=(l.88)(31.0)=ii
2.16 Refer to Fig. 2-12(0). The oords are strong enough to withstand a maximum
tension of SON. What is the
largest value of W that they can support as shown?
I From Prob. 2.15 the equilibrium equations ate, for any w,,
T, = 1.887} (l)
0.8667‘, — 0.3427‘, = w (2)
From Eq. (l) it is clear that 7} > 1} always. Therefore T, will reach the breaking
point first. We thus sel
1} = 80 N to find the corresponding w. From (1),
1.881} — 80 N => 7} - QLQ
From (2). w - (0.866)(80 N) — (0.342)(42.6 N) -= Si
2.11 The weight W, in Fig. 2‘13(¢) is 300N. Find 1,, 1,. 1,, and W,.
\ \
'~' .—
//' T_ "r.
51‘ w= 1' 17‘ .<.\= r.
I ‘ mu W:
(ll) thi
I From I-‘lg. 2-l3(b): T, sin 37° =300 so 1} =SO0N. Also. T,= 1} oos37°==400 N.
From Fig.
2-l3(c). T,oosS3°= 73. so T, =§E_N. But T,sin 53°- W;. so W, =530 N. (Note answers
are to two-place
accuracy).
(rl |-1;, 1.13
2.18 If 6, = 0, in Fig. 2-14, what can be said about 1,, 7;, T,, W,, and W,
provided the pulley i5 frictionless?
-_ I 1
\_‘ 0| 10 I,’
. I 1
T, \\ I.’ T;
.3.
In Fig. L14
I Where the three ropes join, T, sin 9, = 1§sin 9,. so T, -= T,. Also T,= T, and
W,= T,. Further. the
equilibrium condition for the vertical direction is W, = 21', oos 9,. ‘Therefore T,
= T, = 1‘, = W, = W,/(2 cos 0,).
2.19 ln reference to Fig. 2-14. if 0, =53‘ and 6,= 37‘, how large is W, in
comparison to W,'.'
I Since T, = 7} = W2, the equilibrium equations are W, sin 37° = T, sin 53° and T,
oos 53° + W; cos 37°= “Q.
Solving for 7} in the first equation and placing it into the seoond yields W, =
1.25W;.
2.20 Suppose that W, = W, in I-“rg. 2-14 and that 0, = 53°. Find 8,.
I The equilibrium equations are W, sin 9, = T, sin 53° and 7‘, cos 53° + W, cos 0,
= W, Eliminating 1} in the
AAJ K/-\ TOPPER Ii'EeEi°§i'§ii'§gi'E'épR':I’§

second equation by use oi the first yields
sin 0; oos 53‘ + cos 8, sin 53° = sin 53° or sin (9; + 53") = sin $3‘
Thus, either 6; =0 [the two weights are on the same vertical line and T} = 0] or
else (0, + 53°) + 53° I l80°$
0; = 74° [the “nonrtal" answer].
2.21 Refer to Fig. 2-15(0). What is the tension in cord I473”?
\\ \ .\ ._.;
W __
Tm '_
34 I-
o ' ‘
in) (Ir) 2,15
I First find the tension in the cords below the cord in question by balancing
vertical forces at the lower
junction in Fig. 2-l5(a): 2T cos 30° = 70. so T = 40.4 N. Equilibrium conditions
for junction B [Fig. 2-l5(b)]
are T‘ oos 40° = T4, + T sin 30°; T’ sin 40° = T cos 30°. Substituting for 1' and
eliminating T’ from the two
relations yields T4, =21.5 N. From symmetry the same equations are found at
junction A.
2.22 The weight win Fig. Z-l6(a) is 80N and is in equilibrium. Find T,, T, T}. and
1}.
\\\ \ \§
T:
T
‘T ‘ Ev ss» 7'
\ .-
T1 Tl I ‘ B
I A
., O " 25-»:
H11 N '
V//,
F
M U1! us
Fig. 2-l6
I Labeling the lower knot A and the upper knot B, we have the free~body diagrams of
Fig. 2-lo(b) and (c).
We address the equilibrium of knot A first because it involves the only known force.
Z F, =0$T,sinZ5"- T. =0 1, =n.4zan
Z I-',=0$T=cos25°—8l)N=0 0.90er,=suN T,=as,3N
Then from before. T, = Q7.-i Q.
Tuming now to knot B. and remembering that we know 1},
Zr, =0=>T.sin ss"— 'R— T,sin2S°=0
or
0.819TL— T\=0.423T}=37.4N (1)
Z I-1 =o=> T.cos55" - r,wszs'=0=>u.s1sT.=s0Ne~r.=14oi~:
From (1). "r. = 0.8191, - 31.4 = mg.
A/-\J KA TOPPER Ii'E°E#°Zl?l'§”J'|§‘é"R'§i’é

28 L7 CHAPTER2
2.23 The pulleys shown in Fig. 2-l7(a) have negligible weight and friction. What is
the value of w if it remains
supported as shown by the 70-N weight?
§\ \ \ '\ §\\ s -»
. U U
7. T\
I "Kn
. 40' 10 N O ._ _ “_'°_
.
\
nn thi Hg. 2-17
4 . l
I‘ i
or
I Consider the lower pulley system (just above weight W) as the system in
equilibrium. Since it is weightless,
it is in equilibrium under the action of five forces. as shown in Fig. 2-17(b). T,
and 71 and T, are due to the
common oord wrapped about the ceiling pulley directly above, around the system
pulley, and around the
other ceiling pulley. and finally connected to the 70~N weight. Since the pulleys
are frictionless. T, = T, = 71 =
70 N. For the body in equilibrium.
SF. =0=>T,cos40°-73=0=>B=(0-766)(70N)=53.-.6N
2r,=o:1,+ T,+1]sin40°—w=0:>w=70N+70N+(70N)(0.642)I 15513
2.24 How large is the force that stretches the patient‘s leg in Fig. 2-18? How
large an upward force does the device
exert on foot and leg together? Assume frictionless and massless pulleys.
\
1!»
-g \
J
IFIW///-\4l”=* ' yr \
K
U 14,. 2-18
I 3 ltg weighs about 30 N. Since the pulleys are frictionless and with negligible
mass, the tension T in the
cord is the same everywhere. T holds up the weight, so T = 30 N. The forces on the
leg and foot from the
device are caused by the tensions in the cord. The horizontal or stretching force
is T + T cos 30°- fifi. while
the upward force is T + T sin 30° = Q31.
2.25 For the situation shown in Fig. Z-19, with what force must the 6(1)-N man pull
downward on the rope to
support himself free from the floor’? Assume the pulleys have negligible friction
and weight.
I Call T the tension in the rope the man is holding; T is the same throughout the
one piece of rope. The
other vertical force on the man is the tension in the rope attached to the pulley
above the man's head. which
must be 2T for the pulley in equilibrium. The net vertical force is 3T, which is
balanced by his Weight of
600 N. Therefore the man exerts a downward pull of fig.
2.2‘ In the setup of Fig. 2-20. the mobile pulley and the fixed pulley. both
frictionless, are associated with equal
weights w. Find the angle 6.
I Since the tension in the cord is w, the condition for vertical equilibrium of the
mobile pulley is
2wsin6=w, orsin0=§. or 0=30°.
AAJ K/-\ TQPPER ii'E°Ei°§i'ii'§gJ'E'EpR':I’§

111;. 2-19
.__ J________
u
R “ R l1g.Z-N
2.2 nttcrtou AND tncttmzn PLANES
2.27 A 200-N wagon is to be pulled up a 30° incline at constant speed. How large a
force parallel to the incline is
needed if friction efiects are negligible?
guppon Pulling Y
by incline '07“ ‘
Y
\ P
t
\\ 0.87 w
10- S 10' W
,
Weight / / 0 so
. IV
(0) (b)
Fig. 2-21
I The situation is shown in Fig. 2-2l(a). Because the wagon moves at constant speed
along a straight line.
its velocity vector is constant. Therefore the wagon is in translational
equilibrium. and the first condition for
equilibrium applies to it.
We isolate the wagon as the object. Three nonnegligible forces act on it: (l) the
pull of gravity w (its
weight). directed straight down; (2) the force P exerted on the wagon parallel to
the incline to pull it up the
incline; (3) the push Yof the incline that supports the wagon. These three forces
are shown in the free-body
diagram. fig. 2-2l(b).
For situations involving inclines. it is convenient to take the x axis parallel to
the incline and the y axis
perpendicular to it. After taking components along these axes. we can write the
first condition for equilibrium.
2F,=0 becomes P—0.50w=0 ZF,=0 becomes Y—O.87wx0
Solving the first equation and recalling that w = 200 N. we find that P = 0.50w = 100
N. The required pulling
force is 100 N.
AAJ K/-\ TOPPER Ii'E°Ei°§i?i'§gJ'EEpR':§’§
30 C7 CHAPTER 2
2.28 A box weighing l00N is at rest on a horizontal floor. The ooeficient of static
friction between the box and
the floor is 0.4. What is the smallest force F exerted eastward and upward at an
angle of 30° with the
horizontal that can start the box in motion?
F \Il\ 0 ___ F
N ‘ 1 u = fur
Fcmrl
I
w Fig. 2-22
| First draw a force diagram. Fig. 2-22. Next, consider the forces in the x
direction and apply the conditions
for equilibrium. noting f equals its maximum value to start motion.
2F_=O F|:0sl7—/'=0 Foos0=f O.866F=_f=u,N=0.4N
Now apply the conditions for equilibrium to the forces in y direction.
2!-}=0 N+FsinB—W=0 N+0.5F—l00=0 N=l00—0.5F
Substituting this equation for N in 0.866F = 0.4N above.
0.866F=0.4(lC0-0.5F) 0.866F+0.2F=40 Fm
2.29 A block on an inclined plane just begins to slip if the inclination of the
plane is 50°. (0) What is the
ooelficient of static friction? (6) lf the block has a mass oi 2 kg. what is the
actual frictional force just before
it begins to slip?
\" \.
'-.§
..~'
L.
-.\‘ . \“'
so‘ 51 |~
1 \ ’ ,
§ll' M Hg. 2-29
I (ll) From Fig. 2-23:
/1....-r='"8 $5" 59° -’V=mg c0650‘ it, =f,‘,,.,.,/N=tan 5U°= L192
(b) f, = mg sin 50° = 2(9.8)(0.766)=1§.0 N
2.30 A 50-N box is slid straight across the floor at constant speed hy a force of 25
N. as shown in Fig. 2-24(0).
How large a friction force impedes the motion of the box? How large is the normal
force’!
)'
15 N 15 N
25 sin 40‘
Y
40° [ 40’
1
I 25 cos 40' X
\
Y
"' so N
(11) (b)
Hg. 2-24
AAJ KA TOPPER NEET AIIMSJEE RAS
EQUILIBRIUM OF CONCURRENT FORCES ‘.7 31
I Note the forces acting on the box as shown in Fig. 2-24(a). The friction is I and
the normal force. the
supporting force exerted by the floor, is Y. The free-body diagram and components
are shown in Fig. 2-24(b).
Because the box is moving with constant velocity. it is in equilibrium. The first
condition for equilibrium tells
“""“' §£_=o or 25ccs40"-[=0
We can solve forfat once to find that I = l9 N. The friction force is Q.
To find Y we use the fact that
Zr,=0 or Y+25sin40°—50=0
Solving gives the monnal foroc as Y = 34 N.
2.31 Each of the objects in Fig. 2-25 is in equilibrium. Find the normal force. Y,
in each case.
Vi
/1%‘.
M """
L
’ H H ’ '
s \‘\ s s\
Y >vIl5ON Y w_2mN
w-SOON
ta) <1» to
r-1;. 2-zs
I We apply E F; =0 in each case.
(n) Y4-2iIlsin30°—5iX)=U from which Y=400N
(D) Y—2(X7Sin3Q°—l50=0 from Wl'liCh Y=Z50N
(C) Y—200oos8=0 from which Y=§Z00cos6)N
2.32 For the situations of Prob. 2.31. find the coefficient of kinetic friction if
the object is moving with constant
speed.
I We have already found Y for each case in Prob. 2.31. To find f. the sliding
friction force. we use Z E = 0.
(a) 200o0s30°—f=0 andsc f=173N
Then, it, =f/Y= 173/4(1) =Q._fl.
(b) 200oos30°-/=0 andso /=l73N
Then, 44,, =f/Y = I73/250 =@.
(c) —200sin0+_f=0 andso f=(200sin0)N
Then, ii, =)'/Y = (200 sin 0)/(2[X] cos 0) = tan 8.
2.33 Suppose that in Fig. 2~25(c) the block is at rest. The angle of the incline is
slowly increased. At an angle
9 = 42°, the block begins to slide. What is the coefficient of static friction
between the block and the incline?
(The block and surface are not the same as in Probs. 2.31 and 2.32.)
I At the instant the block begins to slide, the friction foroe has its critical
value. Therefore. M. =1’/Y at that
instant. Following the method of Probs. 2.31 and 2.32. we have
Y=wcos9 and f=wsin0
Therefore. when sliding just starts.
Z wsinfi
M‘ Y woosfl “"6
But 8 was found by experiment to be 42"‘. Therefore, u, = tan 42° =@.
32 U CHAPTER 2
2.34 Two weights are hung over two frictionless pulleys. as shown in Fig. 2-
300-lh block to just start moving to the right?
\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\‘
45“ 300 In M1‘
n_ = 0.3
40 lh W
hi)
Mg. z-as
I From the force diagram. Fig. 2-26(b).
26(a). What weight W will cause the
1 N
-llllh
_~1
3
E
1
i
__:€__
T, F_ W_
14!! lb
(I1)
I=40o0§45°=28.3lb 7;=40sin4S°=28.3lb
2!-;=0 T;+W,+N-300lb=0 28.3lb+Wsin30°+N=300lb N=3(X1lb—28.3lb—O.SW
Zr.=0 W,—11—u,N=0 |A,=0.3 Wcos30°—28.3lb—0.3N=0
We substitute N = 271.7 —0.SW into the last equation to get
W cos 30° —- 28.3 lb -— 0.3(271.7 —- O,5W) = 0.
Solving for W gives W = 108 lb.
2.35 Assume that W = 60lb, 0 = 43°, and u, = 0.3 in Fig. 2-27(a). What push
directed up the plane will move the
block with constant speed (a) up the plane and (b) down the plane?
‘Z
r
W‘ W,
H I
43° 9
vi‘ W
(M un 1‘ >
I The weight components are:
N 1-‘A P
' Fig. Z-2'7
W, = (60 lb) sin 43° =40.92 lb W, = (60 lb) oos43° = 43.88 lb
tn) Usins F1s- 2-27(b).
2r,=o 1v-w,=o 1v=4;.§m
F. = u,,N =0.3(-43.88) F, - 13.161»
Efi=0 P—F.—W.=o P=13416+w492 P-gnu
(b) The push is now just enough to keep motion down plane to constant speed. From
Fig. 2-27(c),
25'-=9 P+E—W}=0 P=40.92lb—l3.l6lb P=27.§lb
AAJ K/-\ TQPPER I|'E°Ei°§|'|$|'§gJ'E'EpR':I’§

2.36 What horizontal push P is required to just hold a 200-N block on a 60°
inclined plane if it, = 0.4‘!
F.
N N
P
P
W
W. I pp_ ‘W P p_
F .
w 200 N 2”“ N
my (b) |'-1‘, 1.13
I Figure 2-28(0) is the free-body diagram.
Vl/,=200aos30°=l73i W,=200sin30"=@ I-}=p,N
Zr,-0 1v-P,-w,-o P,=Pcos30° N=Pcos30"+W,-0.866P+l(XlN
Zr,=o P,+;4,N—W,=0 |l,=0.4
Psin30'+0.4(0.866P+llX)N)—l73.2N=0 0.5P+0.346P+40N—l73.2N=0 P=l57N
2.37 ln Prob. 2.36, what horimntal push would just start the block moving up the
plane‘!
| Now Fig. 2-28(b) applies.
2F,=0 N=l(XlN+0.866P §jF.=o Psin30°—W,—;a,N=0
0.5?-173.2N-(0.4)(0.866P+100N)=0 0.SP—l73.2N-0.3-$6P—40lb=0 P=1;gN
2.38 In Fig. 2-29(0). the system is in equilibrium. (a) What is the maximum value
that w can have if the friction
force on the 40-N block cannot exceed 12.0 N? (la) What is the ooeflicient of static
friction between the block
and tabletop?
xr 7
IEIII 1,
T: T. 30‘
0 II
I N
40 N _
(a) (M (1')
Fl}. 2-29
I (0) ‘Die free-body diagrams for the block and the knot are shown in Fig. 2-29(b)
and (c).
_f,(,,,,, = 12.0 N implies T;(,,_., = 12.0 N
For the knot:
1} = T, cos 30“ w I 1} sin 30°
Eliminating 73, we get w = 1} tan 30° = 0.5771}. Thus w(,,,,, - 0.577T,4,,,,, =
6.92 N.
(b) ll. =f.<--i/N = 12/40 'Q~.E~
2.39 The block of Fig. 2-29(0) is just on the verge of slipping. it w - 8.0 N, what
is the coefficient of static friction
between the block and tabletop?
A/-\J K/-\ TOPPER i|'E°$°§i?i'§.gJ§EpR§§§

34 LT CHAPTER 2
I At the verge of slipping. f, =f,,,,_,, and the corresponding hanging weight is w
= 8.0 N. As in Pmh. 2.38.
w =0.S771; and T, =fl. Thus [,,,,__,, = w/0.577 = 8.0/0.577 =13.9 N; u, = 13.9/40
=0.347.
2.40 Find the normal force acting on the block in each of the equilibrium
situations shown in Fig. 2-30.
‘ION é
,0.
20>:
53 '
m‘ ‘O _
\ \\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\
\\\\\\\\\\\\\\\\@
(a) (b) (I) Hg. 2-30
I ln each case friction is necessary for equilibrium. But friction does not enter
in solving for N.
F I
es; I» N ,_ lb N
y
It \
W ' ll
Ml (M to Fig. 2-31
(n) Free-body diagram is Fig. 2-31(a) (F = 20 N, w = S0 N).
2F,=0$N—w+Fsin55°=0 N=SON-(20N)(0.8l9)=33i
(ll) Free~body diagram is Fig. 2-31(b) (w = 60N).
Z F, =0=>~- wcos40°=0 N=(60N)(0.166)=4@
(c) Free-body diagram is Fig. 2-3l(c): (F = 70N. w = 60N).
2!-;=l):>N-wcos40°-—Fsin40°=0 N=60(0.766)+70(0.643)=9li.
2.41 The block shown in Fig. 2-30(0) slides with constant speed under the action of
the force shown. (a) How
large is the retarding friction force? (b) What is the coeflicient of kinetic
friction between the block and the
floor’!
I Use fig. 2~3](a) and Prob. 2.40(a).
(d)2F,=0$FCOS$5°—f=0 (N,=W,=0) 0|‘ ZO(U.573)=[=ll.5N
l1.5N
tb)ut=£=uZN=&
2.42 The block shown in Fig. 2-30(b) slides at constant speed down the incline. (a)
How large is the friction force
that opposes its motion? (b) What is the cueflicient of sliding (kinetic) friction
between the block and plane?
I Use Fig. 2~3l(b) and Prob. 2.400»).
(n) Z F, =o=>; - wsin40°=O f =60N(0.643) =38.§N
3. N
<1n~.=,{,=%;=w
2.43 The block in Fig. 2-30(c) just begins to slide up the incline when the pushing
force shown is increased to
Th L rnin A For
A/-\J K/-\ TOPPER NEeETe:||MSgJEEpRAS

EQUILIBRIUM OF CONCURRENT FORCES L7 35
T \ T
2m». ’
Zuni N
tin rm um Fig. 2-32
7ON. (a) What is the critical static friction force on it? (b) What is the value of
the coefficient of static
friction?
I Use Fig. 2-am) and Prob. 2.409).
(0) E F, = 0:> F cos 40" - w sin 40° -1 = 0. F = 10 N and amt just starts to move
=>f =f.r...., = 10(o.7r,e) -
6O(0.643) = l5.0 E
_f|(mul _ 15_0N_
u»w.- N -gm -w
2.40 The system in Fig. 2-32(0) remains at rest when the hanging weight w is 220 N.
What are the magnitude and
direction of the friction force on the 200N block? The pulley is frictionless.
| Since the pulley is frictionless, the tension is the same throughout the cord.
The free-body diagrams for
the two blocks are as in Fig. 2-32(b) and (c). ln principle the frictional force f
could be either down the
incline or up the incline depending on the details of the problem. In this case we
can quickly assert that it is
down the incline with the following reasoning: T = 220 N. Opposing T along the
incline is the component of
the 200-N weight down the incline. This is surely less than the full Z00 N and thus
is insuflicicnt to balance T.
Therefore the help of friction down the incline is necessary. To obtain f we solve
2 I-1 = 0 (along incline) T — 200 sin 35° —f = Oéf = 220 — 200(0.574) = 105 N
(Note that for this problem the normal force. N, does not enter the calculation.)
AAJ K/-\ TOPPER It'E°E§°§i'ii'§gJ'E'EpR':I§
I.“E‘*E#°2.1?l;“J*;’;’R'i{’;
CHAPTER 3
Kinematics in One Dimension
3.1 DIMENSIONS AND UNITS; CONSTANT-ACCELERATION PROBLEMS
A car's odometer reads 22 687 km at the start of a trip and 22 79] km at the end.
The trip took 4 h. What was
the car's average speed in kilometers per hour‘! in meters per second?
distance traveled (22 79l —- 22 687) km
I Average speed "me “ken ‘ h 26 ltm(h
To convert; Average speed = 26 = 7.2 m[s
An auto travels at a rate of 25 km/h for 4 min, then at 50 km/h for 8 min. and
finally at 20 km/h for 2 min.
Find (a) the total distance covered in kilometers and (b) the average speed for the
complete trip in meters
per second.
I (tr) Distance traveled = d, + d; + d,, where
4, = (25 "T':)(4 m;n)(‘%) = 1; km 4, = (s0%)(s mi»-)(6c%n) =6; km
4, = (2o%)(2 mtn)(6T‘;"—5) = 5 km
Thus d, +d,+d,=Q@.
(fr) Average speed = %- <9 Rot >< 1000m/Nn)/(1-hhin X oos/thin) -it1.1_-iig
A nmner travels 1.5 laps around a circular track in a time of 50s. The diameter of
the track is 40 tn and its
circumference is 126 m. Find ta) the average speed of the runner and (b) the
magnitude of the runners
average velocity.
| (tr) Average speed = = ( (126 m/Yap) = ILA.
(0) Average velocity is a vector. It is the displacement vector for the time lapse
of interest divided by the
time lapse—in this case, S05. Since 1% laps have taken place, the displacement
points from the starting point
on the track to a point on the track Q lap away, which is of course directly across
a diameter of the track. The
magnitude of the average velocity is therefore the magnitude of the displacement
divided by the time lapse.
Thus magnitude of average velocity = 40 m/50 s = 0.80 mls.
Use dimensional analysis to detennine which of the following equations is certainly
wrong:
_ - E E 1’: _ W
1-v! F-a F I It 28 v-(Zgh)
where 1 and h are lengths and [F] = [MLT'*]. The other symbols have their usual
meaning.
I |ut]= [LT' '|[T]-[L], but |/1] =[L], so equation /1=ul can be correct. [m/n]=|M]
[T‘L '] =[MT’L"'],
but [F] = [MLT"]; hence F = m/a is incorrect. [mu/!] = [M|[LT"][T"'| = |MLT"| and
F=mv/I is
dimensionally correct. [vz/lg] = [(L'/T‘)/(L/T’)] = [L]; and since [It] = [L], It =
u’/2g is dimensionally
correct. Since [u] = |LT"]. [(2,gh)"‘] = [(L"’7“‘)L"‘] = [LT"], u = (2gh)"’ is also
dimensionally eorreet. We
note that pure numbers are dimensionless.
If s is distance and 1 is time. what must be the dimensions of C,. C1. C,. and C.
in each of the following
' ')
°qum°m' .t = C.t J == §C,t’ r - C, sin C.t
(Hint: The argument of any trigonometric function must be dimensionles.)
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KINEMATICS IN ONE DIMENSION U 37
I The dimension of s is [L], so all expressions on the right-hand sides of the
equations must also have the
dimensions of length. [C,] = [L/T], since [C,t] are then [(L/T)T] = L. [C1] =
[LT“], [C,] = [L] because the
sine is dimensionless, and since the argument of a trigonometric function has no
units, [C.] = [T"].
The speed v of a wave on a suing depends on the tension F in the string and the
mass per unit length ml 6’ of
the string. lf it is known that [F] = [ML][T]", find the constants n and b in the
following equation for the
speed of a wave on a string: v - (constant)F‘(m/I)’.
I It is given that [u] = [F]'[m/(]’. After inserting into this expression [v] and
[F], we have, recalling M"
means no mass units involved, [M°L‘T"| = [MLT"]‘[ML"]' = [M]""[L]""[T"]'. A
fundamental
dimension must appear to the same power on both sides of an equation; hence, a + I:
= 0, a — b = l,
and —7A=—1. From thisa=]andb=—§.
The frequency of vibration [of a mass m at the end of a spring that has a stiffness
constant k is related to m
and k by a relation of the form f = (constant)m‘k’. Use dimensional analysis to find
a and b. lt is known that
lfl = lTl" and [kl = (MllT1"¢
I As in Prob. 3.6,
I 1 mlk§ :> [Mar-1] = [M-HM»-I--2:] = [M..sT-la]
so
a+b=0 and -2b=-l or b=—a=§
A body with initial velocity 8 m/s moves along a straight line with constant
acceleration and travels 640 m in
40$. For the 4-Us interval, find (u) the average velocity. (b) the final velocity,
and (cl the acceleration.
I Assumex=0att=0.
di l 640
tn) Average velocity = € = T? = 16 mls
(b, c) v, = vn + al. We know vo = 8 m/s and t= 40 s, but we don't know a. However,
x = vat + §a|‘ or 640 m = (8 in/s)(40 s) + §u(40 s)’
and solving we have
640-320
a= =0.40m[s'
Substituting into our velocity fonnula tr, = 8 m/s + (0.40 m/s’)(40 s) = 241111;.
A truck starts from rest and moves with a constant acceleration of 5 m/s’. find its
speed and the distance
traveled after 4 s has elapsed.
I 0, =uo+at, v°=0, a=Sm/s’, l=4s
v, = 0 + (5 m/s‘)(4 s) = 20 mls
To get distance, which in this case is the same as the magnitude of the
displacement, we have
it = not + ht‘ = 0 + §(5 m/s’)(4 5)’ =40rn
A box slides down an incline with uniform acceleration. It starts from rest and
attains a speed of 2.7 m/s in
3s. Find (a) the acceleration and (b) the distance moved in the first 6 s.
I bet x, I-I represent the displacement and velocity down the incline,
respectively. Then. (cl u, = 11,, + at and
we are given u,, = 0, v, = 2.7 m/s, I = 3.0 s; thus 2.7 mls = 0 + a(3.0s), or n -
gggmgj. Now we are
interested in time I = 6.0 s. So (b) x = uni + Qat‘ = 0 + ](0.90 m/s‘)(6.0s)’ =
16.2 gt.
A car starts from rest and coasts down a hill with a constant acceleration. If it
goes 90 m in 8s, find (a) the
acceleration and (I1) the velocity after 8s.
|(a)s=%al‘ (sincev,,=0) 90=%a(8)' a=%°-2.amg§
(b)v-at-2.8x8-Q.4m{s
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38 U
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CHAPTER 3
A car is accelerating uniformly as it passes two checkpoints that are 30 m apart.
The time taken between
checkpoints is 4.0s and the cat's speed at the first checkpoint is 5.0 mls. Find the
car's acceleration and its
speed at the second checkpoint.
I Calling the first checkpoint the initial position and the second the final
position. we have, u,,= 5.0 mls.
x = 30 m. t= 4.0 s. To find a we use the displacement equation
x = uot + lat’ or 30 m = (5.0 m/s)(4.0 s) + §a(4.0 s)’ or a = 1.5 mlsz
Then u, = on + at yields v, = 5.0 m/s + (1.25 m/s’)(4.0s) = 10 m[s.
An auto‘s velocity increases unifonnly from 6.0 mls to 20 m/s while covering 70 m.
Find the acceleration and
the time taken.
I We are given u, = 20 mls and 0, = 6.0 tn/s and the associated displacement is 70
m. Since the elapsed time
I is not given, we find a using u} = of, + lax. or (20 m/s)‘ = (6.0 mls)’ + 2a(70
m). Thus a = 2.6 mls‘. Now I
follows immediately from
u, = 0., + at or 20m/s = 6.0 m/s + (2.6 m/s’)! and 1 = 533
A plane starts from rest and accelerates along the ground before takeoff. lt moves
600m in l2 s. Find (a) the
acceleration. (b) speed at the end of 12 s, (c) distance moved during the twelfth
second.
I We are given v,,==0, 1 ==6lI)m. t= 12s.
(4) x = uut + Qat‘. or 6(X)m= 0 + §a(12s)’. So a = §._3igjs*.
(b) u, = ua + at. or u, = 0 + (8.33 rnls‘)(l2 s) and 0, -= gym];
(e) Remembering that the first second is between 1- 0 and t= l s. we realize that
the twelfth second is
between I - 11 and I = l2 s. Since we already know x(! = 12s), we solve for x(t =
11 s):
x =u,,H-int’ or x -0+ §(8.33tn/s’)(l1s)‘=504m
Our answer is then Ax(twelfth s) = x(| = 12 s) — x(t = 11 s) =91m.
A train running at 30 m/s is slowed uniformly to a stop in 44 s. Find the
acceleration and the stopping
distance.
I Here vu=3Om/s and v, =0 at t=44s. o, =u.,+at yields 0- 30m/s+n(44s). or
a = -Mg mils’. x I u.,t + éat’ = (30 m/s)(44 s) + §(—0.68 m/s’)(44 s)‘ = film.
An object moving at 13 m/s slows uniformly at the rate of 2.0 m/s each second for a
time of 6s. Detemiine
(n) its final speed, lb) its average speed during the 6 s, and (c) the distance
moved in the 6s.
I Slowing “at the rate of 2.0 m/s each second" means a = -2.0 m/s‘. We are given
vo= I3 m/s and 1 = 6.0.
(ii) Then u, = u,, + at = 13 mls + (-2.0 m/s’)(6.0 s) = 1.0 m[s. Because
instantaneous speed is the magnitude
of instantaneous velocity. our answer is 1.0 mls.
(b, c) Average speed =distanceltime. The distance will be the same as the magnitude
of the displacement as
long as the object does not backtrack. i.e.. reverse direction. ln our case the
velocity at the end of 6 s is still
positive. so there indeed has been no backtracking. For displacement, x = uot +
§at' = (13 m/s)(6.0 s) +
!(—2.0 m/s")(6.0 s)’ = Ln [which incidentally solves part (c)]. Average speed = 42
m/6.0 s = 7.0 m[s.
A rocket-propelled cat starts from test at x = 0 and moves in the + direction of X
with constant acceleration
n - 5 m/s’ for 8 s until the fuel is exhausted. It then continues with constant
velocity. What distance does the
car cover in l2 s?
I The distance from xu at the moment fuel is exhausted is x, = (0)(8) + i(5)(8)’ =
160 m, and at this point
v = (2a.x,)"‘ = 40 m/s. Hence the distance covered in l2 s is x, =x, + v(12 — 8) =
160 + (40)(4) = 320 m.
The particle shown in fig. 3-1 moves along X with a constant acceleration of -4
mls’. As it passes the origin.
moving in the + direction of X. its velocity is 20m/s. In this problem. time t is
measured from the moment
the particle is first at the origin. (n) At what distance x’ and time t‘ does v = 0?
(b) At what time is the
particle at x = l5 m, and what is its velocity at this point? (c) What is the
velocity of the particle at
x = +25 m? at x = -25 m? Try finding the velocity of the particle at x = 55 m.
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I (a) Applying v = no + at, 0 = 20 + (-4):’, or i. Then x’ = tlol' + int" = (20)(5)
+ §(—4)(5)’ =
Q Or, from tr’ = vf,+ 24x:
0=(20)z+2(—4)x' or x'=$0m.
(D) 15 =ZUt+ §(—4)lz 0|’ Z12-K)! + l5=U
Solving this quadratic,
\/ ! _
,= =%(2°t16_7)
Thus I, -=0.82s, 1, -9.17 s, where I, is the time from the origin to x =15 m and I,
is the time to go from O
out beyond x = l5 rn and retum to that point. At x = 15 m,
ti, =20-4(0.82)= +16.7m/s v,=20—4(9.17)=—16.7 m/s
Observe that the speeds are equal.
(c) At x I 25m, 11' = (20): + 2(—4)(25), or u =;_Lgg;Ls; and at x = —25m, v’ = 20'
+ 2(—4)(—25),
or u = -24.5 mls. (Why has the root u = +2-1.5 mls been discarded?)
Assuming that x = S5 rn. ti‘ = 20’ + 2(-4)(55), from which u = iv-40. The imaginary
value of v indicates
that x never reaches 55 In, as expected from the result oi pan (n).
Y
al -4 IIIIS2
qi
,'\|.--0
1:;
\_¢ X
u. = 20 mls 0 = -4 In/$1 |_-
.i'.i> go ii’
/"\
tr;
“-’ Plrlitk
A body falls freely from rest. Find (a) its acceleration. (It) the distance it
falls in 3s, (e) its speed after falling
70 m. (4) the time required to reach a speed of 25 m/s. (2) the time taken to fall
3(1) m.
I ta) Choose y downward as positive. Then a =3 = 9.8 mls’. (b) For l = 3.0s, y =
v,,t + §at’ =
0 + §(9.8 m/s’)(3.0 s)’ =iw. (c) Letting y = 70 m, we have
of — 11,} + Zay = 0 + 2(9.8 m/s’)(70 m) -= 1372 ml/s‘ or u, - Q15
(d) betting u, now equal 25 m/s, we have
u, = v,, + al yields 25 in/s = 0 + (9.8 m/s‘)! or I = N
(e) Now we let y =3(I]tn and we have
y = U0! + int‘ yields 3(1) m = 0 + §(9.8 m/s')!’ or 1= i
I.
Hg. 3-1
A ball dropped from a bridge strikes the water in 5 s. Calculate (n) the speed with
which it strikes and
(b) the height of the bridge.
I Choose y downward as positive. Then a =g = 9.8 m/s‘. We are given v,, = 0, and l=
S s to strike the
water. Let u I u,.
in U - v., + III - 0 + (9.s in/=*)(s 5) =-gmg (1,) y = u,,! + lat’ = 0 + l(9.s
in/s’>(5 s)’ = 1;3_m
A ball is thrown vertically downward from the edge of a high cliti with an initial
velocity of 25 ftls. (a) How
fast is it moving after 1.5 s? (b) How far has it moved after 1.5 s?
I (I) u = v,,+a! =15 + 32(l.S) or u = 73 ftls (where n =3 = 32ft/s’)
(b) For constant acceleration um = §(u + vn) = §(73 + Z5) = 49 ft/s
1 = v..,i = 49(1.s)=-lgm or 1 = 11,! + gm’ =2s(1.s) + l(3z)(1.s)’=s1.s + ao= 73.5
it
Sir
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CHAPTER 3
A stone is thrown downward with initial speed 8m/s from a height of 25 m. find (a)
the time it takes to
reach the ground and (Ir) the speed with which it strikes.
I Choose downward positive. a =g = 9.8 m/sz. We are given u,, = 8 mls.
(e) One might directly solve y = vol + int’. or Z5 m = (8 m/s)! + §(9.8 m/s‘)t’ for
t; but it is easier first to find
the final velocity:
(b) ti} = vf, + My = (8 m/s)’ + 2(9.8 m/s)(25 m) = 554 m’/s’. or v, = is Returning
to (a). u, = u,, + at
yields 23.5 m/s = 8 m/s + (9.8 m/s’)t. or t= Qb.
A ball thrown vertically upward returns to its starting point in 4s. Find its
initial speed.
I Let us take up as positive. For the trip from beginning to end. y = 0. a = -9.8
m/s’, t I 4s. Note that the
start and the endpoint for the trip are the same, so the displacement is zero. Use
y = vol + §al’ to find
0 = u,.(4 s) + §(-9.8 m/s‘)(4 s)‘. from which u,, = 19.6 mls.
An antiaircratt shell is fired vertically upward with an initial velocity of 500
m/s. Neglecting friction, compute
(n) the maximum height it can reach, (la) the time taken to reach that height. and
(e) the instantaneous
velocity at the end of 60$. (d) When will its height be 10 km’!
I Take up as positive. At the highest point. the velocity of the shell will be
zero.
(tr) u} = ti}, + Zay or 0 = (500 m/s)’ + 2(—9.8 m/s’)y or y = l2.§ km
tb) u, = u,, + at or 0 = S00 m/s + (-9.8 m/s’)! or 1 = §l_s_
(e) ti, = tn, + at or v, = 500 m/s + (—9.8)(60 s) = -88 mfls
Because v, is negative. and because we are taking up as positive, the velodty is
directed downward. The shell
is on its way down at I = 60 s.
(d) y = u,,I + §a!* or 10000 m = ($00 m/s)r + §(—9.8 m/s’)t‘ or 4.9!’ — 5(1)! +
10000 = 0
The quadratic formula.
x=—b:l:Vb=—4ac
20
gives r =2ls and 'L5_s. At I = 27s. the shell is at 10 km and ascending; at t= 75
s, it is at the same height but
descending.
A ballast hag is dropped from a balloon that is 3tX)m above the ground and rising
at 13 m/s. For the bag. find
(tr) the maximum height reached. (b) its position and velocity 5s after being
released. and (e) the time before
it hits the ground.
I The initial velocity of the bag when released is the same as that of the balloon.
13 m/s upward. Let us
choose up as positive and take y =0 at the point of release.
(4) At the highest point. it, = 0. From v} = vfi + Zay. 0 = (l3 m/s)’ + 2(-9.8
m/s‘)y. or y = QLQ. The
maximum height is 300 + 8.6 = 3&6 m. (la) Take the endpoint to be its position at I
= 5 s. Then. from
y = ti“! + Quiz. y = (I3 m/s)(S s) + §(—9.8 m/s’)(5 s)‘= -58 m. So its height is
300- 58 W. Also, from
1', = v., + at, u, = l3 mls + (-9.8 m/s’)(5 s) = It is moving downward at 36m/s.
(c) Just before it hits the ground. the bag's displacement is -300 m. y = 1101+
lat” becomes —3tX)m =
(13 m/s)1+ §(—9.8 m/s’)t’. or 4.91’ - l3! — 300 = 0. The quadratic formula gives r
= $25 (and -6.6s. which is
not physically meaningful).
A stone is thrown vertically upward with velocity 40m/s at the edge of a clilf
having a height of 110 m.
Neglecting air resistance, compute the time required to strike the ground at the
base of the clifl. With what
velocity does it strike?
I Choose upward as positive. a = —g = -9.8 rnls‘. u,,=40m/s. y,,,_,,,,,,,, = — I10
m. First obtain
the final velocity: tr: = ufi + My = 40‘ + 2(—9.8)(—1l0)= 3756 m’/5:. whence u =
-61.1 mls (downward
motion). Then. from u = tn, +111,
—6l.3=40+(—9.8)t or |—l0.3s
The hammer ol a pile driver strikes the pile with a speed of 25 ft/s. From what
height above the top of the
pile did it fall? Neglect friction forces.
I Downward is positive. We assume it falls from rest. so tn, = 0, v = 25 ft/s. a =g
= 32 ft/sz. We need not
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KINEMATICS IN ONE DIMENSION J 41
conoem ourselves with the time of the fall 1, since we can solve using
tr‘ = uf, + Zay or (25 ft/s): = 2(32 ft/s=)_v
Thfll Y = £8}
A baseball is thrown straight upward with a speed ol 30mls. tn) How long will it
rise? (b) How high will it
rise? (e) I-low long after it leaves the hand will it retum to the starting point?
(ll) When will its speed be
16 m/s?
| We choose upward as positive. a = —g = -9.8 m/st. u,,= 30 mls.
ta) At the highest point v = 0; so for time to reach highest point. v = v.» + at
yields
0 -= 30 m/s + (-9.8 m/s’):, or 1 = is (D) y = v.,t + 50:’ = (30 m/s)(3.06 s) + §(-
9.8 m/s’)(3.06s)‘ = ‘Q.
[or ll’ = vf, + 2ay yields 0 = (30 m/s)’ +2(-9.8 rn/s’)y. and y = 46 ml. (c)
Without calculation: Since the time
up equals the time down. we double the time up to get 6. I2 s for the round tnp.
With calculation: For the
final displacement. we have y =0. so
y = Uni + lat‘ yields 0 = (30 m/s)! + §(-9.8 m/s’):‘
This is a quadratic equation but easy to solve. One solution is 1= 0, but this
corresponds to y = 0 as it leaves
the hand. The other solution is not 0. so we can divide out by 1. leaving
0=30m/s-(4.9m/s:)t or t=6.l2s
(d) Recalling that the speed is the magnitude of the velocity. we must consider the
possible velocities:
v = :tl6 mls. u = v.,+ at yields il6m/s- 30m/s + (-9,8 m/s)1: and solving for r: I.
= L43 s. I = -t._‘Is.
The acceleration due to gravity at the surface of Mars is roughly 4 m/s‘. If an
astronaut on Mars were to toss
a wrench upward with a speed of 10 mls. find (a) how long it would rise; (Ir) how
high it would go; (e) its
speed at 1» 3 s; and (d) its displacement at r = 3s.
I Let us choose up as positive. Then at equals -4 m/s’. and 0., equals +10 mls.
(n) By equation of motion a = (ti — Ufl)!I| we have
, — 0
-4 m/s‘ = g ‘me = 1-ii
(b) By equation of motion ti: = vi + 20:. we have
0‘ = (l0)’ + 2(—4)!..... and rm = 12.5 m
(c) By equation of motion a = (v — v,,)/1, we have
-4 = w and U1 = —2 m[s or 2 m/s downward
(d) .r = U9! + lat’ and thus
S,=(l()>(3)+ §(—4)(9)=30- l8=l@
The acceleration due to gravity on the moon is 1.67 mls‘. If a person can thmw a
stone 12.Um straight
upward on the earth, how high should the person be able to throw a stone on the
moon? Assume that the
throwing speeds are the same in the two cases.
I On earth we can write vi I Zg;(l2) while on the moon vfl, - 2g,,h,,. The throwing
velocities are the same.
so the second expresion can be divided by the first to give It, = l2(g£/g_.,) =
12(9.80/1.67) =70 m.
A proton in a unilonn electric field moves along a straight line with constant
acceleration. Starting from rest it
attains a velocity of 1011.1 ltrn/s in a distance of I cm. (al What is its
acceleration? lb) What time is required to
teach the given velocity?
I (it) u,,=0, u = l0“m/s in x =10" m of displacement. Then tr: = 11.3,-+ Zax yields
(l0° m/s)’ =
O + 2a(l0" m), or a = 5.0 x10" mls‘. (6) u = u,, + at yields 10" m/s = 0 + (5.0
X10" m/s=)t. or
r=2.!@x1Q"‘;.
A bottle dropped from a balloon reaches the ground in 20s. Detennine the height of
the balloon if tn) it was
at rest in the air and (b) it was ascending with a speed of 50 m/s when the bottle
was dropped.
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CHAPTER 3
I Choose upward as positive for both (a) and (b). a I —g - -9.8 m/sz.
ta) u0=0. To find the height. let y be the displacement at time r (remember the y =0
point is at the balloon)
and we are given I = 20 s. Then y = uni + §at’ = 0 + §(—9.8m/s')(20 s)’ = -1960 m.
The height is |y| = 1960 m.
(b) Here the bottle initially has the velocity of the balloon so v., = 50m/s. Now.
y = u,,r + iat‘ =
(S0 m/s)(20 s) + §(—9.8 mls=)(20 s)’ == -960 m. Again the height = |y| = 960 m.
A sandbag ballast is dropped from a balloon that is ascending with a velocity of
40ft/s. If the sandbag reaches
the ground in 20 s. how high was the balloon when the bag was dropped? Neglect air
resistance.
I Take downward as positive. Then a =g = 32 ft/s‘ and v.,= -40 ft/s.
.r = vat + §a|’ = —40(20) + l(32)(20)’ = -800 + 6400 = S600 ft
The balloon was S600ft above the ground.
A stone is shot straight upward with a speed of 80 ft/s from a tower 224ft high.
Find the speed with which it
strikes the ground.
I Choose upward as positive. a = g = -32 ft/s‘. 11,, = 80 ft/s. The presumption is
that the stone just misses
the edge of the tower on the way down and strikes the ground 224 ft below. We can
avoid all reference to the
time in the problem by setting y = -22-4ft in the equation
v’ = ufi + Zay = (80 it/s)‘ + 2(—32 ft/s’)(—224 ft) = 20 736 ft’/s‘
And u = 1144 ft/s, where the minus sign gives the physical solution. The speed is
lvl =144h[§.
A nut oomes loose from a bolt on the bottom of an elevator as the elevator is
moving up the shaft at 3.0 mls.
The nut strikes the bottom of the shaft in 2s. (a) How far from the bottom of the
shaft was the elevator
when the nut fell off? (b) How far above the bottom was the nut 0.25 s after it
tell ofl‘?
| Here the nut initially has the velocity of the elevator. so choosing upward as
positive. un = 3.0 mls. Also
n = —-g = -9.8 mlsz.
(a) The time to hit bottom is 2.0 s. so y = 1101+ £0!’ = (3.0 m/s)(2.0 s) + §(—9.8
m/s‘)(2.0 s)’ = -13.6 m. The
bottom of the shaft is l3.6m below where the elevator was when the nut fell off.
(b) The new displacement y is for 1 = 0.25 s. so y = U“! + §al: = (3.0 m/$)(0.25 s)
4-
§(—9.8 m/s’)(0.25 s)’ =0.44 m. Thus the nut is above its starting position. (This
makes sense if we remember
the initial velocity.) The height above the bottom is thus 0.4 + 13.6 =1-4.0 m.
3.2 GRAPHICAL AND OTHER PROBLEMS
3.36
The graph of an object's motion (along a line) is shown in Fig. 3~2. Find the
instantaneous velocity of the
object at points A and B. What is the object's average velocity? lts aooeleration?
I Because the velocity is given by the slope, Ax/At, of the tangent line. we take a
tangent to the curve at
point A. The tangent line is the curve itself in this case. For the triangle shown
at A, we have
Ax 4m
E=fi-0.50m/s
This is also the velocity at point B and at every other point on the straight-line
graph. lt follows that a = 0
and fl, = u, = 0.50 m[s.
1| Ill
I!
’ A
|Ax-4m
.At=u'
0
n to 20 I, 1 I118, 3-2
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3.37
3.38‘
3.39
Refer to Fig. 3-3. Find the instantaneous velocity at point F for the object whose
motion the curve represents.
I The tangent at F is the dashed line GH. Taking triangle GHJ, we have
A!=24—4=2Os Ax=0—15=-15m
Ax - 15
Hence slope at Fis v,.=;=Tm= -0.75 mls
The negative sign tells us that the object is moving in the —.t direction.
x, in
__._.._-743
/
/
B
to \
\
\
S C F
I E
I \
| \
o
o J to :0 H 1. s |.1'_ 34
Refer back to Fig. 3.3 for the motion of an object along the x axis. What is the
instantaneous velocity of the
object (a) at point D? (b) at point C‘! (e) at point E?
| (0) Point d is a maximum of the x-vs.-I curve. Therelore v = dx/dt = 0.
(b) Without the exact equation for x as function of I one cannot get a precise
answer. The best we can do is
to draw the tangent line at point c and get the slope in the same way as in Prob.
3.37. This yields the answer
v¢=%: =-1.3m[s
(e) We proceed as in part (b), but here the tangent line has a negative slope and
the answer should be
tr, =%: = -0.13 mls
E
A girl walks along an east-west street, and a graph of her displacement from home
is shown in Hg. 3-4. Find
her average velocity for the whole time interval shown as well as her instantaneous
velocity at points A, B.
and C.
I The average velocity is zero, since the displacement vector is zero. The
instantaneous velocities are just
the slopes of the curve at each point. At A the velocity is 40/6 = 6.7 m/min east.
At B it is zero. At C it is
-65/5 = -13 m/min east. or +13 m/min west.
Distance
east (m) 5
40
to
o
>
'3
O
S IO Z0 I
(min)
20 ______________ __
l7Ig.34
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Th L ‘ A F
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44 U CHAPTER 3
3.40 Referring to Fig. 3-4. find (n) average velocity for the time interval I = 7
min to I = I4 min: the instantaneous
velocity at (b) 1= 13.5 min and (e) t= 15 min.
I (a) t7 = (-25 - 40)l(i4 — 7) = -9.3 m/min east; (D) the same as at point C. -13
m/min east; (1') the slope
is +25/(l9- 14) = 5.0m[min east. Note that negative velocity east means motion is
west.
saw .|W._
1
. »—4>
.\+»e§.
¢+<k'?‘
. .1>.,_+_
w.\,_._?.:.
»~ \§~ -
K
(cm! __ ,;;,_,
.. m-F.’
4 7 .(.‘Li0 W
_- ..-.\.
.4,..... .1---
_ ....._.
L)
II
_.
is
--t-;_4»
i~—r+
.-¢
44
J4
"4.
y . .
0 »
S I0
Its) Fig.3-5
3.41 The graph of a particle's motion along the x axis is given in Fig. 3-5.
Estimate the (I) average velocity for the
interval from A to C; instantaneous velocity at (b) D and at (c) A.
I ta) D = (4.8 -0)/(8.0 -0)=§L@_g|15. From the slope at each point (bl u = ;QL-Q8
and (e) Lungs.
3.42 Figure 3-6 shows the velocity of a particle as it moves along the x axis. Find
its acceleration at (a) A and at
(D) C.
vim/sl + ' ' ' ' '
" _' |
1 4 ; ‘ -4
. ._
.._ s._
.-.._
..-. t
...e_>.
-.1.1
~'1
oyrf
°*_. ..li
, :_. .-__
It
4 t .§ . ‘ ._..-e-._»4
_i........,.,.
o ...
A .;...,
_ 4+‘;
._,,./2.’
,0"-' ./
._,,_.
I 44
.03..
.*Y‘
IJ \.$-.
.»*$
T.-
. ..,
3+.
l-I w
a-—
1
1-—»
J
4
1(5) Fl]. 36
I The acceleration at any time is the slope of the v-vs.~t curve. (0) At A the
slope is, from where the
tangent line through A cuts the coordinate axes, a = -7.0/0.73 = -2.6 mlsz. (b) The
slope and therefore n is
zero.
3.43 For the motion described by Fig. 3-6. find the acceleration at (0) B and at (b)
D.
| Taking slopes at B and D we find that the acceleration = (0) -2.4 and (b) 3.2
rnls’.
3.44 A ball is thrown vertically upward with a velocity of 20 m/s from the top of a
tower having a height of 50m.
Fig. 3-7. On its retum it misses the tower and finally strikes the ground. (a) What
time t, elapses from the
instant the ball was thrown until it passes the edge of the tower? What velocity v,
does it have at this time?
(b) What total time I, is required for the ball to reach the ground? With what
velocity v, does it strike the
ground?
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Y
a = -9.8 m/5:
I-‘at = 20 ml;
ll
X Tower
Pi
50m
P:
1':
\\\\\\\\\\ \\\\\\\\\ \\\\\\\\\\\‘ in 3.7
I (a) For the coordinate system shown in Fig. 3-7. y = v,,r + flat’. But at the edge
of the roofy = 0, and thus
0 = 201, + §(—9.8)l§. from which t, = 0, indicating the instant at which the hall
is released. and also t, =4.08 s.
which is the time to go up and return to the edge. Then. from v = 11,, + al, u, =
20 + (-9.8)(-4.08) = -20 mls.
which is the negative of the initial velocity.
(b) —50=20t1+§(—-9.8).'§ or t;=gl_s v,=20+(—9.8)($.8)= -37 mls
3.45 Refer to Prob. 3.44 and Fig. 3-7. (at What is the maximum height above ground
reached by the ball?
(b) Points P, and P, are l5 and 30m, respectively, below the top of the tower. What
time interval is required
for the ball to travel from P, to Pf! (0) It is desired that after passing the
edge. the ball will reach the ground
in 3s. With what velocity must it be thrown upward from the roof?
I (a) Maximum height above ground: h =y,,,,, + 50. From vfi + 2a_v,,,_, = 0.
>'.-..- = = IL-43
Thus. h =70.4 m.
(b) If I, and I, are the times to reach P, and P2, respectively. -15 = 201, — 4.9!?
and -30 = 201, —4.9t§.
Solving, r, = 4.723 s. I, = 5.2485, and the time from P, to P, is I, -1, = 0.525 s.
(c) If vn is the desired initial velocity. then —v,, is the velocity upon passing
the edge. Then. applying
y = v,,t + int’ to the trip down the tower. we find -50 = (—u.,)(3) — 49(3)’. or vu=
1.96m[s.
3.46 A man runs at a speed of-1.0m/s to overtake a standing bus. When he is 6.0m
behind the door (at r = 0), the
bus moves forward and continues with a constant acceleration of 1.2 m/s’. (n) How
long does it take for the
rnan to gain the door? (b) If in the beginning he is l0.0m from the door, will he
(running at the same speed)
ever catch up?
I At r- 0 let the man's position be the origin, x,,,,, = 0. The bus door is then at
x,,., = 6.0 m. The equations of
motion for the man and the bus are
1,, = x,,,,, + um"! + ¥;a,,,l‘ x, = x,,,, + 11,0! + §a,,t2
Now v,,,,,=4.0m/s tI,..,=0 a,,=0 n,,= 1.2 m/s’
Thus .r,,, = 4.0: x,, = 6.0 + 0.6:‘
When the man catches the bus, x,,. = x,, or 4.01 = 6.0 + 0.6!’.
This can be reexpressed as 3|‘ — 20: + 30 = 0. Solving by the quadratic formula.
20:\/400-360 l0:l:\/l_0
I=¢=?=@,4.4s
Note that there are two positive time solutions. This can be understood as follows.
The first time. 1, = 2.3 s.
corresponds to his first reaching the door. This is the real answer to the problem.
However. the equations we
have solved “don't know" he will stop mnning and board the bus; the equations have
him continuing mnning
at constant speed. He thus goes on past the bus; but since the bus is accelerating,
it eventually builds up a
larger velocity than the man and will catch up with him, 1, = 4.4 s.
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46
3.47
3.48
3.69
3.50
3.51
3.52
CHAPTER 3
(b) lf the initial position of the bus is lU.U m. then .r,,, =.r,, yields 31’ —
201+ 50= 0. which has only complex
roots. Thus no real time exists at which the man catches up.
A hall is thrown straight upward with a speed ti from a height h above the ground.
Show that the time taken
for the ball to strike the ground is
5(l+V'll+%)
I Assuming the positive direction to be upward. the uniform acceleration equation
is —h = lll-gl:/21 this
can be rewritten in the form 1’ - Zttl/g — 2h/g = U. Solving for I using the
quadratic formula gives the desired
answer. Since we seek positive values for time. the + root must be chosen.
A ball is dropped from the top of a building. The ball takes U.5s to fall past the
3-m length of a window some
distance from the top of the huilding. ta) How fast was the ball going as it passed
the top of the window?
(bl How far is the top of the window from the point at which the ball was dropped?
| The velocities 1', at the top and t-,, at the bottom of the window are related by
the following equations:
13 = (tr, + vi!)/2 = 3/0.5 =6. so v, + tI,, =12. and tt,, = v, +g(0.5). so ti, —
ll, =-4.9. Eliminating v,, between
these two expressions yields 1', = 3.55 m(s. The distance needed to reach this
speed is
_L’,_(3.55)’_
It - Zg - —--W -0.64 m
A truck is moving forward at a constant speed of 2! m/s. The driver sees a
stationary car directly ahead at a
distance of ll0 m. After a "reaction time" of A1. he applies the brakes. which
gives the truck an acceleration
of -3 m/s:. (a) What is the maximum allowable A! to avoid a collision. and what
distance will the truck have
moved before the brakes take hold‘? (b) Assuming a reaction time of l.-*1 s. how
far behind the car will the
truck stop. and in how many seconds from the time the driver first saw the car?
| The total displacement of the truck from the instant the driver sees the car is x
= u,, AI + x4, where x4 is
the displacement from the point of deceleration to the point of rest. We are given
v..= 2| m/s and the
deceleration is a = -3 m/s:. x. can be obtained from the equation vf = vf, +
2a.x,,, with ti, = 0. Thus
x,. = —v.1./Za, or x,. = —(2l m/s)‘/(-6 m/s‘) = 73.5 m.
(a) To find the maximum At. we note that x____ = 110 m and x,,,,, = (21 m/s) Arm, +
x,,. or 110m =
(21 m/s) A1,“, + 73.5 m. and A1,“, = is The distance moved before braking started
is of course just
v., Ar = 30.5 m.
(b) lf AI = L4 s. then x = (2l m/s)(l.-4 s) + 73.5 m = 102.9 m. The distance to the
car is just ll0 m — ltl2.9 m =
L. To find the time. we need to know the time | during which the truck accelerates.
We have tr, = v.,+ at.
with again u, = U and u = —3 m/s‘. Then 0 = 21m/s — (3 m/s’)! and 1 = 7s. The total
time is 1 + A1= 7 + l.4 =
8.4 s.
Just as a car starts to accelerate from rest with acceleration 1.4m/sz. a bus
moving with constant speed of
12 ml§ passe! it in B Parallel lint‘ (I) How long before the car overtakes the bus?
(b) How fast will the car
then be going? (c) How far will the car then have gone?
| The car starts with initial velocity zero and acceleration a_. = 1.4 mls‘. while
the bus has constant velocity
u,, = I2 m/s. (bl Both travel the same distance x in time I. so set 0,!‘/2 = tut to
give r= l_7_§. (b) The final
velocity of the car is u = 0,! = 24 m[s. (e) The average velocity of the car (or
the hus's fixed velocity) times
l7s yields x = 204 m.
A monkey in a perch 20m high in ti tree drops a coconut directly above your head as
you run with speed
l.5 m/s beneath the tree. (0) How far behind you does the coconut hit the ground?
(b) If the monkey had
really wanted to hit your toes. how much earlier should the coconut have been
dropped’!
I The time for the coconut to fall 20m is given by 20=gI’/2, or 2.02 s. Distance x
= (1.5 m/s)(2.02 s) =
@ m. Since you are moving at a fixed speed the monkey should have dropped the
coconut 2.02s earlier.
Two balls are dropped to the ground from different heights. One ball is dropped 2 s
after the other but they
both strike the ground at the same time. 5 s after the first is dropped. (a) What is
the difference in the heights
at which they were dropped? (b) From what height was the first dropped?
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3.53
3.54
3.55
3.56
I The time to fall from the greater height h. is r= 5 s; the time from the lesser
height h, is I — 2 = 3 s. Using
y I at’/2, we find (0) the height difference h, — h, =g(5’/2 — 3‘/2) =@ and (b) h, =
9.8(25)/2 = 123 m.
Two boys start running straight toward each other from rwo points that are 100 m
apan. One runs with a
speed of 5 rn/s. while the other moves at 7 mls. How close are they to the slower
one's starting point when
they reach each other?
I The two boys meet at the same place and time. The time for the slower one to
travel a distance x is 1/5.
while the other boy has I = (IW — X)/7. Equating the times yields x =»fl.7 m.
Q-§°Q
of |
gs O
K *“ Q,
// A
(0) (b) “F 3'3
Body 1. Fig. 3-8, is released from rest at the top of a smooth inclined plane. and
at the same instant body 2 is
projected upward from the foot of the plane with such velocity that they meet
halfway up the plane.
Determine (0) the velocity of projection and (b) the velocity of each body when
they meet.
I (rr) ln the common time r. body l travels the distance
I
5= (0): +%(g sin 6):‘
and body 2 travels the distance
I l .
i= um! + 5 (-—g sin 0):’
Adding these two equations gives I = v I or r= I/Um. Substituting this value of r
in the first equation and
solving for um, we obtain no, = \/gl sin 8 =
. I .
(0) t»i=0*+2(gsma)i or v,==\/gIsin6=m.
v§=ufi,+2(—gsin0)%=glsin6—g1sin0=0 or v,=0
Two trains are headed toward each other on the same track with equal speeds of 20
m/s. When they are 2 km
apart, they see each other and begin to decelerate. (a) If their deoelerations are
uniform and equal, how large
must the acceleration be if the trains are to barely avoid collision? (b) If only
one train slows with this
acceleration, how far will it go before collision occurs?
I (a) Each train stops in I000 m. so using v‘ — v5 = Zax yields. for Ug = 20 m/s.
an acceleration of -0.2 rn[s‘.
(b) The decelerating train travels a distance x = 20! — 01!’, while the other train
travels (2011) -x) = 201.
Because they collide. times are the same. Eliminating I and solving for x by the
quadratic formula yields
828 m.
A ball after having fallen from rest under the influence of gravity for 6 s cruhes
through a horizontal glass
plate, thereby losing two-thirds of its velocityi If it then reaches the ground in
2 s, find the height of the plate
above the ground.
I From u = on + al, the velocity just before striking the glass is v, = 0 — 9.8(6)
= -58.8 rn/5. and so the
velocity after passing through glass is (1/3)v, = —l9.6m/s. Thus —h = (—19.6)(2) —
48(2)’. or h = 5&8m.
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48 .7 CHAPTERS
3.57 An inclined plane. Fig. 3-9. makes an angle 6 with the horizontal. A groove 0A
cut in the plane makes an
angle tr with OX. A short smooth cylinder is free to slide down the groove under
the influence of gravity.
starting from rest at the point (x... y..)t Find: (al its downward acceleration
along the groove. (b) the time to
reach O, and le) its velocity at O. Let 0 = 30‘, x.,= 3 m, _v.,= 4 m.
Y
I A
XI!
Y ’ lg - 9.1; m/s‘
)'n
Short
Cylinder
11
O ‘ X
Fig. 3-9
I (0) The downward component of g parallel to OY is g sin U: hence. the downward
component along the
groove is a = g sin 9 sin an Since
sina= =U.8 and sin 9=O.5 a =(9.8)(0.5)(O.8) =Q,9'Z m[s:
n _ ll
(b) .t = t.-01+ tat’. where s = (xf, +y§)"'* = 5 m and uu= 0. Thus. s = §(3.92)r’
or r =1.597s.
(cl u = 0 + (3.92)(1.597) = @115.
3.58 A bead. Fig. 3-10. is free to slide down a smooth wire tightly stretched
between points P, and R on a vertical
circle of radius R. II the head starts from rest at P,, the highest point on the
circle. find (0) its velocity v on
arriving at P; and lb) the time to arrive at P, and show that this time is the same
for any chord drawn from
P,.
/ s
§
Q
_____ ____1___ F
/ ._
x H
Hg 3-I0
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3.59‘
3.60‘
3.61‘
3.62‘
I (ii) The acceleration of the bead down the wire is g cos 0 and the length of the
wire is 2R cos 6. Hence.
U’ = 0* + zq; cos 9)(ZR cos 0). or v = 2(\/H) cos 0.
_1'_2\/F?==°§*"_2\/F
(bl |- ; - S cos 9 - E
which is thc same regardless of where P; is located on the circle.
An object is forced to move along the X axis in such a way that its displacement is
given by
x = 30 + 201 — 15:’ where x is in rn and 1 is in s. (a) Find expressions for the
velocity ii and acceleration ls
the acceleration constant? (b) What are the initial position and the initial
velocity of the object? (c) At what
time and distance from the origin is the velocity zero‘! (d) At what time and
location is the velocity -50 m/s?
I Remembering that all units are S1 and that
i=2 ,_£2
_ dz df
(a) tt = X = (20 — 30!) m/s; a = X = -30 m/s‘. Acceleration is constant.
(b) At |=0. x=30m=x... .i =20m/s=v,,.
(e) From the velocity equation. for u = 0, 0 = 20 - 301 and r = _% s. Substituting
r = § s into the displacement
equation. x = 30 + 20(§) — lS(§)‘ =36.7 m.
(ll) Setting ll = -50 m/s. -50 = 20 — 301. and r = 2i s. Then. x = 30 + 20(2§)—
15(2§)’ = —5.0m.
(Note that a comparison of the displacement equation with x =x,, + u.,r + int‘
would have made the use of
calculus unnecessary.)
A particle moving along the x axis has a velocity given by u = 4t — 2.50!‘ cm/s for
I in seconds. Find its
acceleration at (tr) I = 0.50s and (b) I = 3.0 s.
I tr = dv/dl = 4 — 5.0!. giving (a) a = l.50cm[s: and (b) -11.0 cm[s1.
A ball is released from rest at the edge of a deep ravine. Assume that air
resistance gives it an acceleration of
-by, where y is measured posilivc downward. (This negative acceleration is
proportional to its speed. y; the
positive constant b can he found by experiment.) The hall has a total acceleration
of —by + g, and so
5‘ = -hi‘ +8 (1)
is the differential equation of motion‘ (n) Show by diflerentiation and substitution
that
)‘=k(¢"‘-1l+(x/bl! (2)
is a solution of (1) for an arbitrary value of the constant k and that (2) gives y
= 0 for I = 0. (bl Since at I = 0.
y =0. prove that k =g/b’ and show that as 1-» ==, _y"—>g/b; that is. the velocity
reaches a limiting value such
that the negative acceleration due to air resistance exactly offsets the positive
acceleration of gravity and thus
Y = 0. (c) Assuming that b = 0.1 s '. find the distance fallen and the speed reached
after 10 s. (J) Show that
after 1 min the ball will have essentially reached its ten-ninal velocity of 98
m/s.
I (tr) Differentiating (2) once. _t" = —bke "' +g/b. Differentiating once more. _t"
= b’ke"". Multiplying our
expression for y by —b and adding g yields b’ke""’. Thus (1) is satisfied.
Substituting l= 0 into (2) and
recalling that e" = l. we gct y = ll.
db) Since the hall is released from rest. y» = 0 at I = 0. Using our expression for
y from (a). we have
0= —-bk +g/b. which yields It =g/bl. As I—>== the first term in becomes
infinitesimal and _t"—>g/b. Thus ji
must approach zero. as can he seen directly from the expression for
(c) lf b = 0.1/s and using g = 9.8 m/s’. we have k = g/b’ = 980 m. Then at r =
ltls. y = (980 m)(e '— 1) +
(98 n1/s)(l0 s) = 360 m. tr = = (-98 m/s)e" + (9.8 m/s‘)/(0.1/s) = 62 m(s.
(d) At r= 60 5. S’ = ('98 m/s)e'°+ 98 mls. e °= 0.0025. so y‘ =98 m/s.
A ball is thrown vertically upward from the origin of axes (Y regarded + upward).
with initial velocity y},.
Assuming as in Prob. 3.61 an acceleration —by due to air resistance. we write
i‘=—l>.\"-s (9
(Note that when y changes sign. so does —by": hence (1) is valid for the trip down
as well as the trip up]. (a)
Show that y = k(c "' — l) — (3/b)t is a solution of (1) for any value of k. (b)
Prove that
*=‘i(.Y*-*5)
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50 :7 CHAPTER3
(c) Assuming that b = 0.1s " and yo-=50m/s, find the height and speed at r = 3s.
(if) How long does it take
for the ball to attain its maximum height and what is the height?
I (d) )" = —bke""—g/b, ji = b'ke"". Multiplying y by —b and subtracting g yields Y,
which proves y is a
solution ot (1). (b) At l= 0, y =y.,= -bk —g/b. Solving for k. k = -(;-0+;/u)/u.
(c) Again b = 0.1 s, and we are told y'., = 50m/s. Then k = —(50+98)/0.1 = -1480 m.
At I = 3s, we have
y = -14ao(¢ "‘-1)-(9s)(s)=§3;§_m.y = +148e'“’—98=1l.§m[s.
(d) For maximum height. y = 0, which yields 0 = 148e"“'- 98, or e°"'= 1.51. Thus I
L. Substituting
r= 4.12 s into the y equation yields y = —l480(e""‘" — 1) — (98)(4. 12) =§.0 m.
3.63‘ A mass at the end of a spring vibrates up and down according to the equation
y = 8 sin 1.51 cm. where r is the
time in seconds and the complete argument (angle) of the sine function. 1.5!. is in
radians. (0) What is the
velocity of the mass at I = 0.75 s? (b) At t= 3.0s’! (c) What is the maximum
velocity of the mass? (Hint: To
express the angle in degrees. multiply it by 180/1:.)
I The velocity v = dy/dl = (l.S)(8) cos l.St cm/s; remember that 1.5! is expressed
in radians. (a) u =
12oos 1.13 =§.2cm[s- (b) v =12cos4.5 = —2.5 cmls. (c) Maximum occurs where the
cosine is 11. so
u = 112.0 cmls.
AAJ K/-\ TOPPER I|'E°E§°§|T§i|'§gJ'E'EpR':I’§

4.1
4.1
4.2
4.3
4.4
4.5
4.6
CHAPTER 4
Newton’s Laws of Motion
FORCE, MASS, AND ACCELERATION
A force acts on a 2-kg mass and gives it an acceleration of 3 m/5*. What
acceleration is produced by the same
foroe when acting on ti mass of (it) lkg? (b) 4 kg? (cl How large is the force?
I We first find the force F using F = ma (one dimension). F = (2 kg)(3 m/s’) =@. Then
noting ll = F/m.
we have for the different masses m = lkg, a = (6 N)/(1 kg) = 6 m[s’; m = 4 kg. -1 =
L5 m[s*. The answers are
thus (at 6m[s= (bl 1.5 mlsl (c) (Q.
(Fill in the blanks.) The mass of a 3(X)g object is i (cl. its weight on earth is i
(bl. An object that
weighs 20N on earth has a mass on the moon equal to ___. (e). The mass of an object
that weighs 5 lb on
earth is __ (d).
I (4) 3ll)g (b) w = (300 g)(980 cm/s2) = 2.94 X 10‘ dyn =2.94 (c) m = wig = (20
N)/(9.8 m/s3)=
2.04 kg; mass is the same anywhere. til) m = w/g = (5 lb)/(32.2 ft/s’) =0.l55 slug.
A resultant external force of 7t0 lb acts on an object that weighs 40lb on earth.
What is the object's
acceleration (n) on earth? (b) on the moon’?
I (a) We use F =ma, where we recall that F stands for the resultant of all forces
acting on the mass m. To
get m we note that m = w/5 = (40lb)/(32.2 it/s‘) = l.Z4 slug. Then a = F/m =
(7.0lb)/(1.24 slug) = 5t64 it[s‘.
lb) The acceleration on the moon is the same since the resultant foroe is still
7.0lb and the mass is the same
anywhere.
A horizontal cable pulls a 200-kg can along a horizontal track. The tension in the
cable is 500 NA Starting
from rest. (cl how long will it take the cart to reach a speed of Sm/s? (b) How tar
will it have gone?
I Assuming no friction, the tension in the cable is the only horizontal force. Then
from F, = ma, we get
a, = (5(X) N)/(200 kg) = 2.50 m/$2. We now use kinematics to solve the problem.
(a) t/, = v.,, + a,|; and since the cart starts from rest. u._ = 0. Then l= v,lu, =
(8 m/s)/(2.S0m/s’) =
(b) Letting the starting point be the origin we have x = u.,r + §ut‘ =0 + §(24S0
m/s‘)(3.2 s)‘ = l2.8m
A 9(1)-kg car is going 20 m/s along a level road. How large a constant retarding
force is required to stop it in
a distance of 30 m?
I Here we start with the kinematical equation that allows us to find the
acceleration u,: vf = uf, + 211,1.
where tr, =0 when 1 = 30m and v,,, = 20 m/s. Solving we obtain a, = -6.67 mls’.
Finally we solve for the
retarding force using E = mu, = (900 kg)(—6.67 m/s’) = —6000 N.
How much force does it take to give a 20000-kg locomotive an acceleration of l.5
m/s‘ on a level track with a
coefficient of rolling friction of 0.03?
I Refer to Fig. 4-l. F — u,,N =ma, N= mg, F = iqmg -4-ma = 0.03(20000)(9.8) +
20000(l.S) =
5880+30000=35.880 kN
8
is
i ii
7m7l ¥
F
'"9 t-1;. 4-t
_ 51
AAJ K/-\ TOPPER Ir'E°E§°§|T§'|'§gJ'E'EpR':I’§

52 U
4.1
4.3
4.9
4.10
4.11
4.12
4.13
4.14
CHAPTER 4
A 12.0-g bullet is accelerated from rest to a speed of 7(1) m/s as it travels 20cm
in a gun banel. Assuming the
acceleration to be constant. how large was the accelerating force?
I From kinematics we have uf — vi, + 2n,x. For our case U(h = 0. and v, = 700 m/s
when x = 0.20 m. Solving
we get a, = 1.23 - 10" mls‘. We then obtain I-I by noting m =0.0l2 kg and using E,
= ma, = l4.§ kN.
A 20-kg crate hangs at the end of a long rope. Find its acceleration when the
tension in the rope is (0) 250 N,
(6) 150 N. (c) Lew. (J) 196 N.
I The crate is acted on by two vertical forces—the tension in the rope, T. upward
and the weight of the
crate. w = mg. downward. Noting that w = (20 ltg)(9.8 m/s‘) = 196 N and rising T —
w = may, we get:
(n) a, = Z-Li; (bi II, = ;Z_.gji‘; (c) a, = imli‘; (d) a, = 0. Note that negative
acceleration is
downward for our case.
A 40~kg mink sliding across a floor slows down from 5.0 to 2.0 m/s in 6.0 s.
Assuming that the force acting on
the trunk is constant. find its magnitude and its direction relative to the velocity
vector of the tnink.
I Letting the x axis be along the direction of motion. we have for the magnitude of
the resultant force
E = ma, To find a, we use the kinematical relationship v, = um + a,r, with tr“, =5.0
m/s. u, = 2.0 m/s.
t = 6.0 s. Solving we get tr, = -0.50 mlsz. Then E, = (40 ltg)(—0.50 m/s’) = -20 N.
Noting that the resultant
force in the y direction is zero since a, = 0. we have our answer. F is 20 N in the
direction opposite to the
velocity.
A resultant force of 20 N gives a body of mass m an acceleration of 8.0 mls‘. and a
body of mass m’ art
acceleration of 24 m/s’. What acceleration will this force cause the two masses to
acquire if fastened together?
I From I-‘= ma. with F=20N and a —8.0m/sz. we get m -2.50kg. From F=m‘a' and
a’ = 24.0 m/s‘ we get m’ = 0.83 kg. Combining the two masses yields M =m + m’ =
3.33 kg and
F = MA yields A = 6.0 m(s’.
An 1l0tHtg car travels on a straight highway with a speed of 30 m/s. The driver
sees a red light ahead and
applies her brakes. which exert a constant braking force of 4 ltN. ta) What is the
deceleration ct‘ the car?
(D) ln how many seconds will the car stop?
I (0) Use Newton‘s second law of motion. Take a retarding force to be negative.
F = ma -4 X l0‘ - ll00a a = -3.636 mls‘
Thus. the deceleration is SQ 5315*.
(D) a = lL:l—--U" where v equals zero after t seconds. and vi. equals 30m/s.
-3.636=();!%-0 t=3O/3.636=8.25s
The car will come to a stop in 8.25 s.
A force of 70 N gives an object of unknown mass an acceleration of 20ft/s’. What is
the object's mass?
F _ 70N 70 kg - m/s=_
' F " ""' .1 ‘ "' '" ' (201:/s=)(0.3os m/ft) atom/s‘ 'i5“'s k
A boy having a mass of 75 kg holds in his hands a bag of flour weighing 40 N (Fig.
4-2). With what force does
the floor push up on his feet?
I For the boy to be in equilibrium. the floor must push up on the boy‘s feet with a
force F equal and
opposite to the combined weight of the flour and the boy. Let m equal the mass of
the boy and w the weight
of the flour:
F=mg+w=75(9.8)+40=73S+4O=775N
Apply Newton's third law in the following situation: Two drivers. one owning a
large Cadillac and the other
owning a small Volkswagen. make a bet. The VW owner bets that his car can pull as
hard as the Cadillac.
They chain the two rear bumpers together in a large empty parking lot. Each driver
gets into his car and
applies full power. The Cadillac pulls the VW backward all over the lot. The driver
of the VW later claims
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NEWTON'$ LAWS OF MOTION J 53
m-75 kg 1
w-40 N
Fig. 4-2
that his car was pulling on the chain as hard as the Cadillac all the time. What
does Newton's third law say in
this case? Assume that the chain has negligible mass.
I Newton's third law says that the VW owner is right. Each car must pull on the
chain with the same
magnitude of toroe, one being the action and the other the reaction. The motion ot
the VW is a consequence
of all the foroes acting on it. not just the force of the chain. These other forces
include. in particular. the
frictional force between tires and road, which is quite difierent for the VW and the
Cadillac.
A 2-slug mass pulls horizontally on a 3-slug mass by means of a lightly stretched
spring (Fig. 4-3). If at one
instant the 3~slug mass has an acceleration toward the 2'slug mass of 1.8 ft/5'.
find the net force on the 2»slug
mass and its acceleration at that instant.
I The spring pulls on each mass with the same force F but in opposite directions.
Using Newton's second
law,
F=m,a, =m,n, F=m;a;=3slugX1.8ft/s’=5.4slug-ft/s’ F=5.4lb
Using Newton's second law on the mas m,.
F=m,a,=5.4lb 2a,=5.4 a,=2.7ft[s=
m, m,
Fig. 4-3
A 96-lb boy is standing in an elevator. Find the force on the boy's feet when the
elevator (a) stands still (b)
moves downward at a constant velocity of 3 ftls (c) accelerates downward with an
acceleration of 4.0 it/st.
and (d) accelerates upward with an acceleration of 4.0 ft/s’.
I (4) Find the mass of the boy in the proper unit. the slug:
= weight(w) _ 96lb _ _ , _
m acceleration of gravity (g) 32 ft/s’ Mb S/n_3slu8
When an elevator stands still, the force on the boy's feet. according to Newton's
first law, is equal and
opposite to his weight. The answer is therefore 9Q: upward.
(b) A constant velocity means zero acceleration. For zero acceleration the force of
the elevator halanoes the
earth‘: downward gravitational force. that is, the weight of the boy. Again by
Newton‘s first law, the answer
is 9_6_Il3 upward.
(c) When the acceleration of the elevator is downward, the boy's weight exceeds the
upward toroe E of the
elevator, yielding a net downward toroe F. From Fig. 4-4
F-ma-3slugX4.l)ft/s’=l2lbs'/ftxtt/s’ F=l2lb E=w_F=96_|2=gfl_,
By Newton's second law, the floor of the elevator pushes upward on the boy‘s feel
with a tome of 84 lb.
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54 J CHAPTER 4
4.17
4.l8
4.19
4.20
1.21
4.22
m-3 slug a
E rt;.u
(J) When the elevator accelerates at 4.0 ft/s‘ upward. the force E acting on the
boy is greater than his weight
w by an amount F = I2 lb:
E=w+F=96+l2=l08Ibupward
An elevator starts from rest with a constant upward acceleration. lt moves 2.0m in
the first 0.60s. A
pasenger in the elevator is holding a 3-kg package by a vertical string. What is
the tension in the string
during the accelerating process?
I To obtain the tension T in the string we apply the second law to the m = 3.0 kg
package: T — w = ma,,
with w = mg = 29.4 N. The acceleration a of the package is the same as that of the
elevator; it is obtained
trom the displacement fonnula y = um! + lag’. where u.,,. = 0 and y = 2.0 m, when I
=0.60 s. Solving we get
av = 11.1 m(s=. Substituting into our equation for T, we get T =62.7 N.
Just as her parachute opens, a I50-lb parachutist is falling at a speed of I60
ft/s. After 0.805 has passed. the
chute is hilly open and her speed has dropped to 35 ft/s. Find the average
retarding force exerted upon the
chutist during this time.
| We choose downward as the positive y direction. The average acceleration of the
chutist over the 0.80-s
interval is
(35 — l60)ft/s
ti, = = - lS6 ft/s‘
Next we use: =m&,. or w — 1‘ = mi,, with n:= 150 lb, m = w/g = 4.69 slug. and T‘ =
the average
retarding force due to the chute. Solving we get T = 881 N.
A boy who normally weighs 300 N on a bathroom scale crouches on the scale and
suddenly jumps upward.
His companion notices that the scale reading momentarily jumps up to MD N as the
boy springs upward.
Estimate the boy's maximum acceleration in this process.
I The maximum upward force exerted by the scale on the boy is 40t]N. The net force
on the boy is
(MI) — 300) N and this equals ma. Using m = 300/9.8 yields a = 3.3 mlsi.
Shortly after leaping from an airplane a 91.8-kg man has an upward force of 225 N
exerted on him by the air.
Find the resultant force on the man.
I The resultant force on the man is the vector sum of two l'orees—the weight w = mg
= (91.8 kg)(9.8 m/s’) =
900 N downward. and the 225-N force upward. Then the resultant foroe is 900 — 225 =
675 N downward.
To measure the mass of a box, we push it along a smooth surface, exerting a net
horizontal force of l50lh_
The acceleration is observed to be 3.0 m/s’. What is the mass of the box?
I Using E = ma, for the horizontal direction we get ISO lb X 4.45 N/lb = m(3.0
m/$2) and m = 223 kg.
A book sits on a horizontal top of a car as the car accelerates horizontally from
rest. lf the static coeflicient of
friction between car top and book is 0.45. what is the maximum acceleration the car
can have if the book is
not to slip‘!
| When the book of mass m is about to slide, the friction I = itmg. Friction is the
only horizontal force
acting, thus] = ma. Inserting it = 0.45 yields a = ug = 1.41 mfsz.
AAJ K/-\ TOPPER li'EeEi°§i'ii'§gJ'E'é°R':I§
AAJ KA TOPPER I.“;E$Z.1?i2“.‘;’.;’R'i{’;

NEWTON'S LAWS OF MOTION 5' 55
Prove the following for a car moving on a horizontal road: The magnitude of the
car's acceleration cannot
exceed ug, where it is the coefficient of friction between tires and road. What is
the similar expression for the
acceleration of a car going up an incline whose angle is 6?
I Friction between tires and road supplies the force moving the car. so f = ma. On
a horizontal road
F,,, = mg. so f,,,,. = l4F~ = umg. therefore a,,,,. = pg. On the incline. equations
of motion parallel and
perpendicular to the surface are 141'}, — mg sin 8 = mam... and F~ - mg ¢°$ 9 = 0-
3°"‘° 7°" "1 11...... =
(p cos 9 — sin 9)g.
A 5-kg mass hang at the end of a cord. Find the tension in the cord it the
acceleration of the mass is
(ti) L5 In/S2 "P. (b) 1.5 m/s: down. and (c) 9.8 m/s’ down.
I Choosing upward as positive we have T — w = mu,, where Tis the tension in the
cord, m = S kg. and
w = mg = 49 N is the weight of the mass.
(a)T=56.5fl(b) T=41.5N(c) T=Q. _
A 700-N man stands on a scale on the floor of an elevator. The scale records the
foroe it exerts on whatever is
on it. What is the scale reading if the elevator has an acceleration of (a) 1.8
m/s’ up? (b) 1.8 mls’ down?
(c) 9.8 m/s’ down?
I Again choosing upward as positive and letting N represent the force of the scale
on the man. we have
N — w = ma,. Noting that w = 700 N, and that m = w/g = 71.4 kg, we solve for N
using the values of a, given:
(a) N=§2_9.I!- (b) 1v=s71 N. up N=Q.
Using the scale described in Prob. 4.25. a 65-kg atronaut weighs himself on the
moon, where g = 1.60 m/s’.
What does the scale read?
I Since g,,,._,,, is the acceleration of free fall on the moon. and w,,_,,,,, is
the force of gravity on the moon's
surface. we have w,,,,_,,, = mg,,,,,,,,, = (65 1tg)(1.60 mls‘) = 104 N. By Newton‘s
first law scale reads 104 N.
A rough rule of thumb states that the frictional force between dry concrete and a
skidding car's tires is about
equal to nine-tenths of the car‘s weight. If the skid marks left by a car in coming
to rest are 20m long, about
how fast was the car going just before the brakes were applied?
I Since [fl = it |F,,-|, and since F,., = W on level ground, we have for it =0.9, [
= -0.9W. Find the car's
acceleration from F = ma. which isf = (W/g)a with)‘ = -0.9W. Then a = —0.9g. Since
x is 20m. use
ti‘ — u§,= Zax with v =0. to give 11., =63“ =18.8m[s.
It the coeflicient of friction between a car's wheels and a roadway is 0.70. what is
the least distance in which
the car can accelerate from rest to a speed of 15 m/s?
I Using n =;tg (see Probs. 4.23 and 4.27) in the kinematical formula ta’ = vf, +
Zax.
u -0.. I5 —-0’ _
X _ T _ 2(o.10)(9.s) ' ——1°"m
A constant force accelerates an electron (m = 9.1 X 10 '-" kg) from rest to a speed
of 5 X 10’ m/s in a distance
of 0.80em. Determine this force. How many times larger than mg is it?
I First find the constant acceleration from ti‘ — t'.l= 2a.x; then a = 25 X 10"/1.6
X 10 ‘ = 1.56 X 10" m/s’. so
F = ml! =(9.1X 10'”) (1.56 X 10”) = L43 X 10 '3 N. Then F/mg =0/g = 1.56 X 10"/9.8
= 1.6 X10“.
The 4.0-kg head of a sledge hammer is moving at 6.0 m/s when it strikes a spike,
driving it into a log; the
duration of the impact (or the time for the sledge hammer to stop after contact) is
0.0020s. Find (tr) the time
average of the impact force, (bl the distance the spike penetrates the log.
I la) The average force is a constant force that would effect the same result as
the actual time-varying force
over the time interval involved. For a constant force we can use tr, = u,,,, + a,t
to find the corresponding
constant acceleration a,. Here u,._ = 6.0m/s. ti, =0. and r =0.(X)20 s. Solving we
get a, = -3011) mls’. '11ien
E, =ma, yields E = -12 ltN. Here F, represents the force of the spike on the
hammer-
head. The reaction force on the spike is the impact force, which is 12 ltN.
(b) The spike moves the same distance as the hammerhead in the time interval in
question. For the
hammerhead we have x = t-.._! + §a,t‘ = (6.0 m/s)(U.0020 s) + §(-3000 m/s')(0.0020
s)’ = 0.006 m or x =
6.0 mm.
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56 .7
4.si'
4.32‘
4.33‘
4.34‘
CHAPTER 4
A body of mass m moves along Y such that at time tits position is y(1) = at“ — bt +
c. where a. b. c are
constants. (a) Calculate the acceleration of the body. (la) What is the force
acting on it’?
I ta) n, = d’y /df. We first obtain v, =dy/dl = in!“ - b and then dzy/dig = ia|"".
To obtain the force we
have (b) F = may = §ma!' "'2
Measurements on a 300-g object moving along the x axis show its position (in
centimeters) to be given by
x = 0.20! — 5.0:‘ + 7.5!’, where t is the time in seconds. find the net force that
acted on the object during the
time for which this expression applies.
I Because E... = ma. we must find n; a = du/dl and v = dx/dr. Doing the
ditferentiations. v -= 0.20 - 10.01 +
22.51’. a = —l0.0 + 45/; so Fm = 300(- 10.0 + 45!) dyn = 0.lI)3(X)(— 10.0 + 45!) N
A 50-g mass vibrating up and down at the end of a spring has its position given by
y = 0. l50sin 3| m for I in
seconds. Find the net force that acts on the mass to give it this motion.
| The acceleration is d*y/Jr’ = -1.35 sin 3: m/s’; so Fm, = 0.050(-1.35 sin 31) =
-Q.@7§ sin 3| N. The
minus sign indicates a restoring force.
A body of mass m moves along X such that at time t its position is x(|) = m‘—- fit’
+ yr, where tr, fl, y are
constants. (0) Calculate the acceleration of the body. (la) What is the force
acting on it?
I ta) it = Am’ -35:’ + y and ii = 12.11’ - 65:
(b) E. =nu't =12mm‘-omflt
4.2 FRICTION; INCLINED PLANES; VECTOR NOTATION
4.35
4.36
4.37
4.38
The breaking strength of a steel cable is 20 ltN. If one pulls horizontally with
this cable. what is the maximum
horizontal acceleration which can be given to an 8-ton (metric) body resting on a
rough horizontal surface if
the coetficient of kinetic friction is 0.15?
I Let T be the cable force. Then Z F = ma becomes 7' — |4,,mg =ma. For maximum
acceleration T=
2.0 x l0‘ N, so that 2.0 X 10‘ N — 0.15(8000 kg)(9.8 m/s=) = (8000 ltg)a. Solving
we get a = 1.03 mls’.
A 20-kg wagon is pulled along the level ground by a rope inclined at 30° above the
horizontal. A friction
force of 30 N opposes the motion. How large is the pulling force if the wagon is
moving with (a) constant
speed. and (la) an acceleration of 0.40 m/s’?
I Let the pulling force of the rope be T. Using E F, =ma, we have for our case
Toos30°— 30N = ma,,,
where m = 20 kg. la) For a, = 0, T =3-4.6 N. (b) For a, =0.40 mls’. T =43.9N
Suppose. as shown in Fig. 4-5, that a 70-kg box is pulled by a 400-N force at an
angle of 30° to the horizontal.
The coeflicient of sliding friction is 0.50. Find the acceleration of the box.
I Because the box does not move vertically, Z F, = ma, = 0. From Fig. 4-5. we see
that this equation is
Y + Zll) N - mg = 0. But mg = (70 kg)(9.8 m/s’) = 686 N. lt follows that Y = 486 N.
We can find the friction force acting on the box by writing f = |4Y I (0.50)(486 N)
= 243 N. Now let us write
Z F, =ma, for the box. It is (346 - 243) N = (70 kg)"-. from which a, = 1.47 m(s’.
“DN
ZWN
Y
mg “Q. I-S
As shown in Fig. 46, a toroe of 400 N pushes on a 25-kg box. Starting from rest,
the box achieves a velocity
of 2.0 m/s in a time of 4s. Find the coeflicient of sliding friction between box and
floor.
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4.39
4.40
4.41
4.42
NEWTON'S LAWS OF MOTION Q’ 57
306N
2S7N
]<i
Y
<2s)(9.a) N Fig. 4-6
I We must find f by use of F = ma. But first we must find a from a motion problem. We
know that 0., = 0,
u, = 2 mls. t= 4 s. Using v, = Uri + at gives
a=9%"“=2_‘ls/‘=0.50m/§=
Now we can Write X F, = ma,, where a, = a =().$0 m/s’. From Fig. 4-6. this equation
is 257 N -/ =
(25 kg)(0.50m/s’), orf = 245 N. We now wish to use it =/'/Y. To find Y, we write Z
F, = ma, = 0. since no
vertical motion occurs. From Fig. 4-6. Y —306N — (25)(9.8) N =0. or Y= S51 N.
Then
f 245
”=Y'=3§=fli
A 12-kg box is released from the top of an incline that is 5.0 m long and makes an
angle of 40° to the
horizontal. A 60-N friction force impedes the motion of the box. (a) What will be
the acceleration of the box
and (b) how long will it take to reach the bottom of the incline?
I In Fig. 4~7 we show the three forces acting on the block: the frictional force f
= 60N; the normal force N.
which is perpendicular to the incline; and the weight of the block. w = mg = (12
kg)(9.8 m/s’) = I18 N. We
choose the x axis along the incline with downward as positive. Using Z E = ma,, we
have w sin 40° —f = man
or (118 N)(0.642) — (60 N) = (12 kg)a,. Solving we have a, =1.31 mls’. To find the
time to reach the bottom
of the incline. starting from rest, we use z = u,,,r + gm, with t/,,, = 0 and x =
5.0 m. Solving we get
I = (1.63 5*)" = 1.
For the situation outlined in Prob. 4.39. what is the coelficient of friction
between box and incline?
I Again referring to Fig. 4-7 we have from Z E = 0, N — wcos 40° = 0. or N = 90N.
Then. recalling that the
coefficient of kinetic friction is given by it, =]/N. we have it, =()._67.
I
in " Hg. 4-1
I
An inclined plane makes an angle of 30° with the horizontal. Find the constant
toroe. applied parallel to the
plane, required to cause a 15-kg box to slide (a) up the plane with acceleration
l.2 m/s’ and (bl down the
incline with acceleration 1.2 m/s’. Neglect friction forces.
I Here we assume that the x axis is along the incline and positive upward. lf P is
the constant force referred
to, then from Z I-1 = ma, we have P — w sin 30°= ma,, with m =15 kg and w = mg =
I47 N.
(a) For n, = 1.2 m/s‘. P = 91.5 N. (b) a, = -1.2 m/s‘, P =55.$ N
A 400-g block originally moving at 120 cm/s coasts 70 cm along a tabletop before
coming to rest, What is the
ooelficient of friction between block and table?
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58 U
4.43
4.“
4.45
4.46
4.47
4.45
CHAPTER 4
I For the block E 1-} = 0 yields I-1, -= W == mg and E F,‘ =ma yields —uF,., = ma;
hence, it = —a/g. The
uniform acceleration, from u’ — 115- 241, is a = -1.2’/2(0.7) = -1.03 mls’; then it
= 1.03/9.8 =(L19_§.
How large a force parallel to a 30‘ incline is needed to give a 5.0-kg box an
acceleration of 0.20 m/s’ up the
incline (0) if friction is negligible? (b) If the coefficient of friction is 0.30?
I (n) The component of the weight down the incline = mg sin 30° = 5(9.8)(0.5) =
24.5 N, while the external
force up the plane is F. So Fm = ma becomes F — 24.5 -= S(0.20) from which F =Z§j_E
(It) A friction
force = uF,. must be added to 24.5 N down the incline. Ft, = mg cos30° = 42.4N and
uF,, = 12.7 N, so F is
larger by this amount; thus F = 3§.2 N.
An 8.0-kg box is released on a 30° incline and accelerates down the incline at 0.30
m/s’. Find the frictional
force impeding its motion. I-low large is the coefficient of friction in this
situation?
I The component of the weight down the incline = 8(9.8)(0.S) - 39.2 N. Now F = ma
leads to 39.2 —f =
8(0.3), so that I = 36.8 N. The nonnal force equals the component of W
perpendicular to the incline.
8(9.8)(0.867) = 67.9 N. Therefore tr = 36.8/67.9 =lQ4;.
A horizontal force P is exerted on a 20-kg box in order to slide it up a 30°
incline. The friction force retarding
the motion is 80 N. How large must P be if the acceleration of the moving box is to
be (a) zero and
(D) 0.75 In/s’?
I Here we choose the x axis along the incline with positive upward. All the forces
on the block are shown in
Fig. 4-8. From E E =ma,, we have Pcos30°— wsin30°—f=ma,, where m =20kg. w=mg =l96N,
and
f =80N. (ct) For n, -0. P= 206N. (b) For a, =0.75m/s’, P=Z§_l§
I’ l
N
I.
.@ t=t;.4-s
A horizontal force of 2(1) N is required to cause a I5-kg block to slide up a 20°
incline with an acceleration of
25 cm/s’. Find (a) the friction force on the block and (b) the coefitcient of
friction.
I The situation is as shown in Fig. 4-9. We again choose the x axis along the
incline, and the y axis
perpendicular to the incline. Then from E I-1 = ma, we get P cos 20" —f — w sin 20"
= (15 lrg)(0.25 m/s’) and
from equilibrium in the y direction we get N — P sin 20° — w cos 20" = 0. Noting
that P = ZOON and
w = mg = 147 N. we solve the equations. yielding f = i and N = ZQ_7_lj.
respectively. Finally it, =f/N =
(L25 is the eoeflicienl of friction. Note how P contributes to the nonnal foroe.
"8
What is the smallest force parallel to a 37° incline needed to keep a ill)-N weight
from sliding down the
incline if the coeflicients of static and kinetic friction are both 0.30?
I As usual, choose the x and y axes parallel and perpendicular to the incline. Let
F be the unknown force, f
the friction force. and N the normal force. If the weight is about to slide down
the incline. then the frictional
force is maximum and in the upward direction. 'lhen
F — w sin 37" +f,,,, = O, Where f,,,, = ;4,N and N — w cos 37° = 0.
Noting that w = ll!) N and it, = 0.3. we get N = 80N and [,,,, - 24 N; and finally F
= }@ is the minimum
force needed along the incline.
Referring to Problem 4.47, indicate what parallel force is required to keep the
weight moving up the incline at
constant speed.
AAJ K/-\ TOPPER Ii'E°Ei°§i'ii'§gJ'E'EpR':I§

AAJ NEETAIIMS JEE RAS
NEWTON'S LAWS OF MOTION 5' 59
I Here we have kinetic friction down the incline. and f = u,N. N is still 80N
(why?) and sinoe ii, =0.30, We
havef = 24 N. Then Z E, = 0 yields F — wsin 37° —f = 0; and solving, we get F = Q
4.49 For the same conditions as in Prob. 4.48 assume that the force F. up the
incline. is 94 N. What is the
acceleration of the object? If the object starts from rest. how far will it move in
10s?
I As before. f = ii,,N = 24 N and is down the incline opposing the motion. Then
using E F, = ma“ we have
F — w sin 37°—f = ma,. Noting m = w/g = 10.2 kg we solve. getting a, =0.98 mlsz up
the incline. Next we use
the kinematic formula x = v,,,r + 541,1’. with v,,, = 0. Setting I =10 s and using
our a,, we get x = 49 m.
4.50 A S-kg block rests on a 30" incline. The coelficient of static friction between
the block and incline is 0.20.
How large a horizontal force must push on the block it the block is to be on the
verge of sliding up the
incline?
I Lct P be the horizontal force. We choose x and y axes [| and J. to the incline.
Since the block is on the
verge of moving. it is in equilibrium: Z F, = 0 and E F} = 0. Here we have maximum
static frictional force
down the incline. P cos 30° - w sin 30° —[.,,,, = 0; N — w cos 30° — P sin 30° = 0.
where N and f are the normal
and friction forces. respectively. w = mg = 49 N. Substituting f,,,,, = ;t,N in the
first equation, and putting in all
known quantities. we get
0.866P——0.20N=24.5N N—0.50I’=42.4N
These two equations in two unknowns can be solved in a variety of ways. We multiply
the first equation by 5
and add to the second equation to eliminate N. This yields 3.83? = 165 N. or P =
43.1 N.
4.51 Rework Prob. 4.50 for incipient motion down the incline.
I The only diflercnce from Prob. 4.50 is that the maximum frictional force is up the
incline. We therefore
change the sign in front of f,,_,_ in the E F, =0 equation. We then get for our two
equations:
0.866? + ().20N = 24.5 N N — 0.50P = 42.4 N
We again multiply the first equation by 5. but now subtract the second equation to
eliminate N. This yields
4.83? =80. l N, or P = 16.6 N. Note that in Probs. 4.50 and 4.51 the numerical
valves of N are different,
reflecting their dependence on P.
4.52 ln Fig. 4-10. the 8-kg object is subject to the forces F, = 30N and F, =40N.
Find the acceleration of the
object.
Ft
,__
‘~—i at
' ct
‘ .
l
l
|
l
v
I
t
t
1
,4
/‘ 70°
F’ ng. 4-to
| Newton's second law in component form is F,, + F1, = ma, and F,, + F2, = rnn,, or
30 cos 40°+ 4000s 70°=
Ba, and 30 sin 40° — 40 sin 70° = 8a,.. Solving. n = 4.6! — 2.3] mls‘.
4.53 A 7-kg object is subjected to two forces. F, = 20i + 30] N and F, = 8i — 50j
N. Find the acceleration of the
object.
I I-'=I-',+F==28l—20]N; a=iF=4l—(20/7)jm/s’.
4.54 The forces F, and F, shown in Fig. 4-10 give the 8-kg object the acceleration
tt = 3.01 m/5’. Find F, and F}.
AA-I K/-\ TOPPER I,%°E#°:.1?i2“.’;'.;"Ri(’;

60 5' CHAPTER 4
I Our motion equations take the form I-1,, + F1, = 8(3). F,, 4- I-'2, =0. From the
latter, F, sin 40" =
fisin 70°; from the former. F. cos 40“ = 24 — F, cos 70°. Dividing these relations
we have tan 40° =
F, sin 70°/(24 — F, cos 70°). Using these. Ii = 16.4 N and Ii = (sin 70°/sin 40°)F}
= Z4 N.
4.55 Find the force needed to give a proton (m = 1.67 X l0'” kg) an acceleration 2
X ll)°i — 3 X 10“j m/s‘.
I r~‘=m-= 3.34>< l0’"‘l - 5.01 ><10*“‘1 N
4.56 A 200-g object is subjected to a force 0.30! — 0.40j N. lf the object starts
from test. what will be the velocity
vector of the object after 6 s‘?
e
I v=v,,+ru=0+iI"=‘-fl(0.30l-0.40j)=9l-l2|m/s
4.57 If the object of Prob. 4.56 started at the origin. what was its location at
the end of the 6-s period?
I $=l|,+!V,,+1l‘:l=0+U+i|"=i(0.30i-0.40])=Z7i—$6j|'n
' 2m 2(O.200)
Thus the object was found at the point (27 m. -36 m).
4.3 TWO-OBJECT AND OTHER PROBLEMS
4.58 ln Fig. 4-ll. find the acceleration of the cart that is required to prevent
block B from falling. The coeflicient
of static friction between the block and the cart is ;4,.
I It the block is not to tall. the friction force. I. must balance the block‘s
weight: / = mg. But the horizontal
motion of the block is given by N = ma. Therelore.
/ >1 E _ _§_
/v ' .1 °' “ ' f/N
Since the maximum value off/N is "'] we must have a Zg/it, if the block is not to
fall.
a I '
--zp
B
N
"III
fig. 4-ll
4.59 A passenger on a large ship sailing in a quiet sea hang a ball from the
ceiling of her cabin by means of a long
thread. Whenever the ship accelerates. she notes that the pendulum ball lags behind
the point of suspension
and so the pendulum no longer hangs vertically. How large is the ship's
acceleration when the pendulum
stands at an angle of 5° to the vertical?
I See Fig. 4-I2. The ball is accelerated by the force Tsin 5°. Therefore Tsin 5°=
mu. Vertically E F =0, so
T cos 5° = mg. Solving for n =g tan 5° gives a = 0.U8’7Sg = 0.86m[s‘.
I
\
my fig‘ ‘J2
4.60 A rectangular block of mass m sits on top of another similar block. which in
turn sits on a flat table. The
maximum possible frictional force of one block on the other is 2.0m N. What is the
largest possible
acceleration which can be given the lower block without the upper block sliding ofi?
What is the coeflicient of
friction between the two blocks?
I rm, = 2.0», = mam, so um.‘ = 2.0 m[s". Also ,4 =f/mg = 2.0m/9.8m =i@.
AAJ KA TOPPE R NEET AIIMS JEE RAS

NEWTON’S LAWS OF MOTION 5' 61
I S I 1:‘
a = 3 m}.\*
37» mg cm .17 mg "'" 37'
uh my Fll- 4-13
4.61 A block sits on an incline as shown in Fig. 4-l3(n). (a) What must be the
frictional force between block and
incline if the block is not to slide along the incline when the incline is
accelerating to the right at 3 m/$2?
(Ia) What is the least value ti, can have for this to happen?
I Resolve the forces and the 3-rn/s’ acceleration into components perpendicular and
parallel to the plane
[Fig. 4—l3(b)]. Write F = ma for each direction: 0.6mg —f = 3(0.8)m and Ft, -
0.80:3 = 3(0.6)m. which yield
(ii)/'=§3.48m!N. F,.,= (9.64m)N
3.48”!
(bl u.=,_L=9—(_m=9i6
N ‘
412 In the absenoe of friction. would the hlock of Prob. 4.6l accelerate up or down
the incline?
I Down [at 3.48 m/s’ relative to incline, since total acceleration down incline is
then 0.6 g by Prob. 4.ol(a)].
4.8 The inclined plane shown in Fig. 4-14 has an acceleration 1 to the right. Show
that the block will slide on the
plane if a >g tan (0 — 11), where ig I tan 9 is the coefficient of static friction
for the contacting surfaces.
Y
N
i—-
4 , X
\\
I lf the block is not to slide. it must have the same acceleration as the planer
Hence
/cosa—Nsina=ma /sin<r+Ncosa—mg=0
From these,
/=m(acosa+gsina) N=m(gcosa—nsinar)
and
I tlCOS0'+gSil‘lfi' a+-gtannr
__=?_____,___i
N gcosa—asina g—ntana
Now the maximum value of f /N in the absence of slipping is pi, = tan 0. Thus the
acceleration a must satisfy
<tang or ,,< _ “mm Q)
g—atana_ _gl+tan8tanm 8
If a >g tan (0 — a), the block will slide.
4.61 Objects A and B, each of mass m, are connected by a light inextensible cord.
They are constrained to move
on a frictionless ring in a vertical plane, as shown in Fig. 4-15. The objects are
released from rest at the
positions shown. Find the tension in the cord just after release.
I At the moment of release, A is constrained to move horizontally and B vertically.
so that the two initial
AAJ The Learning App For
NEET AIIMS JEE RAS

62 U CHAPTER 4
Ni
IA
Ifil
'1
B
I?‘
M:
Hg. 4-is
accelerations are tangential as shown. Furthermore, the two accelerations have the
same magnitude. a, since
otherwise the cord would have to stretch. Thus, the horizontal force equation for A
and the vertical force
equation for B, at the indicated positions. are
Tsin4S°=ma mg—Tsin4$°=ma
» - - -__'"¢ -2
Eliminating 0. T 2 sin 45° v.2.
4.65 If the system in Fig. 4-l6(n) is given an acceleration. find the forces on the
sphere, assuming no friction.
I From Fig. 4-tow). Z F.,, = R, CO530° - w = m¢_,,=0 and E R... - R; - R, sin30°=
ma. Thus. the acting
forces are
R, = $5-‘YT,=1.1sw R, =- R, $11130“ + go = (1.15w)(0.5) + Lin = w(0.58-+2-)
and the weight. w.
30'
.L “ e_"._.
4 "'
\
R:
w
(n) (b)
r-1;. no
4.66 ln Fig. 4-17. mass A is l5 kg and mass B is ll kg. If they are given an upward
acceleration of 3m/s’ by
pulling up on A. find the tensions T, and 1}.
| First apply Newton's second law to the system as a whole to find the force H
accelerating both masses
upward.
F,=(m,+m,)a=(l5+ll)3 F,=78N
Sinoe F, is the resultant force. F, = T. — m,g —m,g, and the tension T, is the sum
of the weights ofA and B
plus F..
T, = m,,g + m,g + F, =15(9.8)+ll(9.8)+ 78=147 + 107.8 + 78 1} -311$ N

Similarly for mass B only,
8
I,
NEWTON'$ LAWS OF MOTION ‘.7 63
T1
Hg. 4-17
r,=m,¢= 11(3)=33N T,=m,.g+r,= u(9.s)+ss= l07.8+33=140.§N
To check. for block A only.
T,=m,g+mAa+T}= l47+45+ 14-0.8=33Z.8N
4.67 Referring to Fig. 4-18, find the acceleration of the blocks and the tension in
the connecting string if the
applied force is F and the frictional forces on the blocks are negligible.
I Apply F = ma to each block in tum to obtain F — T = m,n and T = m,a. Solve for
Tand a to obtain
a = F/(m, + m,) and T =m,F/(m, +m,).
Mil In Fig. 4—l8, if F = EN, m, =m; = 3 kg. and the acceleration is 0.50 m/sz. what
will be the tension in the
connecting cord if the frictional forces on the two blocks are equal? How large is
the frictional force on either
‘ml F
I Write F = ma for each block using I as the friction force on each block. Then we
obtain F -— f — T = m,a
and T —f -m,a. Use the given values and solve to find T =10N and [ =8.5 N.
block ?
ng. us
4.69 The device diagramed in Fig. 4-l9 is called an Atwood‘: machine. In terms of
m, and m, with m, > m,,
(1) how far will m, fall in time 1 after the system is released? (b) What is the
tension in the light cord
that connects the two masses’! Assume the pulley to be frictionless and massless.
I
T
IIl|R
.,.
ll]
...,., fig. no
I (4) Isolate the forces on each mass and write Newton's second law, choosing up as
positive:
T — m,g = m,n and T — mzg = —m,a. Eliminating Tgives n - (m, — m,)g/(m, +m;). Now
use y =11!‘/2 to
find the distance fallen in time t. (b) From the above equations‘ T = 2m,m,g/(m, +
m,).
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64 U
4.10
4.71
4.72
CHAPTER 4
A cord passing over a frictionless, massless pulley (Atwood's machine) has a 4-kg
block tied to one end and a
l2-kg block tied to the other. Compute the aooeleration and the tension in the
cord.
I Using the fonnulas derived in Prob. 4.69,
_12-4 _ , _2(4)(12) _
a-il2+4(9.s) 4.9m/s T-——4H2(9.s)-ss.sN
For an Atwood‘s machine (Prob. 4.69) with masses 10 and 12 kg, find (n) the
velocities at the end of 3 s and
(b) the distances moved in 3 s. (c) If at the end of 3 s the string is cut, find the
distances moved by the mases
in the next 6s.
‘(m,—m,)g_12—l0 _
' (Q) a m,+m, _l2+l0(9'8)_hliz
Since the acceleration is constant, the common speed at the end of 3 s is v = v.,+
at = 0 + (0.89)(3) = 2.67 mls.
Mass 2 moves down and mass l moves up. (b) The distance moved by each mass in 3 s
is
s = vat + ia!’ = (0)(3) + §(O.89)(3)' = Iii.
(e) If the string is cut. the masses fall freely with initial velocities um = -2.67
m/s and u“, = +2.67 m/s, with
up taken as positive. For mass 2, the displacement in 6 s is then
y, = out — lg!‘ = (—2.67)(6) — §(9.B)(6)’ = —l92.4 m
i.e., a downward distance of 12.4 m. Mass 1 travels upward a distance
1 1
4' =@= —(2'°7) - 0.4m
Z8 2(9-3)
before coming to a stop and then falling downward. The time of travel upward before
coming to a stop for
mass 1 is
1.,=%’=%=0.27§
It then travels downward 5.73 s for a distance
d" = ll(—g)r’| = {(9.8)(5.73)’ = 160.9 m.
The total distance traveled by mass 1 is then d = d’ + d" = 0.4 + 160.9 = i.
In Fig. 4-20, the weights of the objects are 200 and 3(1) N. The pulleys are
essentially frictionless and
massless. Pulley P, has a stationary axle but pulley P, is free to move up and
down. Find the tensions T, and
7}. and the acceleration of each body.
§
P.
1, '1
A
1.
zoo N B
3w N Hg. 4-20
A/-\J KA TOPPER Ii'E°$°Zl?l'§”J§E”R'§I’é
4.73
4.74
4.75
NEWTON‘S LAWS OF MOTION U 65
I Mass B will rise and mass A will fall. You can see this by noting that the forces
acting on pulley P, are 21;
up and T. down. Therefore 7} I 27; (the inertialess object transmits the tension).
Twice as large a force is
pulling upward on B as on A.
Let a = downward acceleration of A. Then §u = upward acceleration of B. [As the
cord between P, and A
lengthens by l unit. the segments on either side of P, each shorten by Q unit.
Hence. §= s,/:4 =
(§a,t')l(§a,t‘) = a,/n,.] Write E F} = ma, for each mass in tum, taking the
direction of motion as positive in
each case. We have
T}—300N=m,(§a) and 200N—T}=m,,a
But m = wig and so m, = (20)/9.8) kg and m, = (300/9.8) kg. Further. T, = 2T,.
Substitution of these values
in the two equations allows us to compute 7} and then 7} and a. The results are
T,=327N T;=l64N a=1.78m(sz
An inclined plane making an angle of 25° with the horizontal has a pulley at its
top. A 30-kg block on the
plane is connected to a freely hanging 20-kg block by means of a cord passing over
the pulley. Compute the
distance the Z0-kg block will (all in 2s starting from rest. Neglect friction.
/H
A
I The situation is as shown in Fig. 4-21. We apply Newton's second law to each
block separately. For block
B we choose downward as positive. while for block A we choose our x axis along the
incline with the positive
sense upward. This choice allows us to use the same symbol, a, for the acceleration
of each block. Then for
bloclt B, W, — T =m,a. where m, = 20kg and w, = I96N and T is the tension in the
cord. Since the pulley is
frictionless, the same tension T will exist on both sides of the pulley. Then for
block A, T — w, sin 25° = m,a,
where m, = 30 kg and w, = 294 N. We can eliminate the tension T by adding the two
equations. which yields
w, — W, sin 25° = (m, + m,)a. Substituting in the known values we solve. getting a
= 1.44 mlsz. The equation
for fall from rest is y = t/,,,t + 511,1‘, with u,,, = 0. Substituting in a, = 1.44
mls’ and t = 2s, we get y =2.§§ m.
n;. 4-21
Repeat Prob. 4.73 if the coeflicient of friction between block and plane is 0.20.
I The equation for block A is now T — w, sin 25° — f = m,a. The block B equation is
the same as before:
w, - T = m,a. Adding the two equations we get this time w, — w, sin 25‘ —f = (m, +
m,)n. As soon as we
obtain f we can solve for a. To obtain f we note that f = ;i,N, where ti, = 0.20 is
the ooeflicienl of kinetic
friction and N is the normal force exerted on the block by the incline. Noting Z F,
= 0 for the direction
perpendicular to the incline, we have w, oos 15° — N = 0, or N = 266 N. Then f = 53
N and solving for the
acceleration a = 0.38 rnlsz. Again using y = v,,,| + §n,r" with v,,, = 0, a, = 0.38
m/s’. and I = 2 s. we get
y = 0.76 m.
In Fig. 4-22, the two boxes have identical masses, 40 kg. Both experience a sliding
friction force with
it = 0.15. Find the acceleration of the boxes and the tension in the tie cord.
T
/. ‘ ,
0.87 mg 30-
“: "22 0.5 mg "'5
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4.76
4.77
CHAPTER 4
I Using f = id’. where Y = normal force, we find that the friction forces on the two
boxes are
In = (0-15)("1s) fa = (9-15)(9-87mg)
But m=-40kg and so/,,=59N andf,,=5l N.
Let us apply Z F, = ma, to each block in turn, taking the direction of motion as
positive.
T-59N=(40kg)a and O.5mg-T-5lN=(40kg)a
Solving these two equations for a and T gives a = LQB mlsz and T = L.
'l\No bodies. of masses m, and ml, are released from the position shown in Fig.
4.23(a). If the mass of the
smooth-topped table is m,, find the reaction of the floor on the table while the two
bodies are in motion.
Assume that the table does not move.
I From Fig. 4-23(b). the force equations for the bodies are
Body 1: EF,,,,=w,—T=m|a
Body 2: X FM, = T = mln
Table: X F.¢,=N-T-w,-w,=(l. XfiK.,=T—f=0
1 4L
2
T
1'
1. 1.
W: Ri=w:
Smooth
surface T W}
W} T
wt T m
“I W L //2 //2
\ \\\\\ \\\\\\\\\\\\\\\\\\\ \\‘ ' Na ~n
(H) (b)
Fig. 4-23
where N and f are the vertical and horizontal (frictional) components of the force
exerted by the floor on the
table. [We assume in Fig. 4.?3(b) that left and right legs share load equally. This
does not affect our analysis.)
From the first two equations.
0: Wt = mtg
m, +m; m,+m;
"I M
Then. ;=r=m,a=-1-“?
m,+m,
Mimi
and. finally. N = T + mzg + m,g = (W + m, + m,)g
I 2
Three identical blocks, each of mass 0.6 kg. are connected by light strings as
shown in Fig. 4-24. Assume that
they lie on a smooth. horizontal surface and are observed to have an acceleration
of 4.0 rn/s‘ under the action
of a force F. Calculate F and the two tensions.
T». T.»
* ng. 4-24
I Let the tensions in the cord be T,, and T,,. respectively. and let us write
Newton's second law for each
block separately. choosing positive to the right for each. Sinoe the cords are
inextensible. we know that they
AAJ K/-\ TOPPER ii'i§Ei°§i?i'§gJ'E'épR':I’§

4.78
4.19
NEWTON'S LAWS OF MOTION J 67
have the same acceleration. which we denote by a. Then for blocks A. B. C.
respectively (letting
m =m.-ms=m.)-
F—T,,=ma 'I§,,—T,,=ma T,,=ma
(Note how the tensions appear with opposite signs in adjacent equations.) To solve
these equations for ll we
add the equations. and the tensions cancel in pairs, leaving F = 3ma = 3(0.6 kg)
(4.0 mls’) =7.2 N. The
tensions can now be obtained by substituting back into the individual equations:
T,,, = 7.2 N — (0.6 ltg)(-4.0 m/s’)=~1;ul T,‘ = (0.6 kg)(4.0 m/$2) = Qfi
Three blocks with masses 6 kg, 9 kg. and 10 kg are connected as shown in Fig. 4-25.
The coefficient of friction
between the table and the 10-kg block is 0.2. Find (a) the acceleration of the
system and lb) the tensions in
the cord on the left and in the cord on the right.
H
._-—O
Ill kg
= 1
A - _l| ll _ L_
ft Lg Fl}. 4-L5
I (tr) Let the tension in the cord on the left be T, and on the right be T1. The
pulleys are assumed to be
frictionless. We apply Newton‘s second law to each of the three blocks. choosing
the positive sense of the axis
for each block consistently. Thus we choose downward as positive for block C, to
the right positive for bloclt
B, and upward as positive for block A. The frictional force on block B is to the
left and can be obtained from
[ = urN, where at = 0.20 and the nonnal force N equals the weight w,, = 98 N from
vertical equilibrium. ‘Huts
f = 19.6 N. For our three equations we have
w, - 7} = m,a
where a is the acceleration. m, = 9 kg. and w‘ = 88.2 N.
E — Tl -I = mm
T, — w, = m_a
where m, = 6 kg and w, = 58.8 N.
As with earlier problems involving cords connecting blocks, the tensions in
adjacent equations appear with
opposite signs. Adding the three equations eliminates the tensions completely: w, —
[ — w, = (m, + m, + m,)a.
Note that this is equivalent to a one-dimensional problem involving a single block
of mass m, + m, + m, acted
on by a force w, to the right and forces f and w, to the left. Substituting the
known masses, weights. and f
gives a = 0. 39 mlsl. (lb) Substitute a hack into the equations of motion for each
block. rnost conveniently the
first and third. to obtain T, = L1. 7} = 85 N. The remaining equation can be used to
check the results.
ln Fig. 4-26. the ooefficient of sliding friction between block A and the table is
0.20. Also. m, I 25 kg.
m, = 15 kg. How far will block B drop in the first 3 s after the system is released?
1 ‘ T
<— -9
in r
Y o
"'4! "'08
Hg. 4-26
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68 5'
4.00
4.8K
4.02
CHAPTER 4
I Since, for block A, there is no motion vertically. the normal force is Y = m,g =
(25 kg)(9.8 m/s’) = 245 N.
Then f I uY = (0.20)(2-15 N) = 49 N.
We must first find the acceleration of the system and then we can describe its
motion. Let us apply F = ma
to each block in tum. Taking the motion direction as positive. we have
T—/'=m,,a or T—49N=(25ltg)a
and m,g—T=m,n or —T+(l5)(9.8)N=(l5kg)a
We can eliminate T by adding the two equations. Then. 50]‘/"ll! f°l' 4, WC find 0 =
2-45 ml!‘-
Now we can work a motion problem with a = 2.45 rn/s‘. vi, - 0. r= 3 s.
y = uot + lat’ gives y = 0 + §(2.45 m/s’)(3 s)’ = 11.0 m
as the distance B falls in the first 3s.
How large a horizontal force in addition to T must pull on block A in Fig. 4-26 to
give it an acceleration of
0.75 m/s‘ toward the lefl? Assume. as in Prob. 4.79, that it = 0.20, m, == 25 kg.
and m, = 15 kg.
| It we were to redraw Fig. 4-26 in this case. we should show a force P pulling
toward the left on A. In
addition, the retarding friction foroe f should be reversed in direction in the
figure. As in Prob. 4.79, f = 49N
We write F == ma for each block in tum. taking the motion direction to be positive.
We have
P — T — 49 N = (7.5 kg)(0.75 m/s’) and T — (15)(9.8) N = (15 kg)(0.75 tn/5:)
Solve the last equation for T and substitute in the previous equation. We can then
solve for the single
unknown, P, and find it to be 226 N.
The two blocks shown in Fig. 4-27 have equal masses. The coefficients of static and
dynamic friction are
equal, 0.30 for both blocks. ll the system is given an initial speed of 0.90 m/s to
the left. how far will it move
before coming to rest if the inclines are quite long?
! /R
§ [qt 447
I Again we assume that the pulley is frictionless and the tension in the cord is
the same everywhere. We
apply Nev/t0n‘s second law to each block to get the acceleration of the blocks. We
choose our x axis for each
block || to the inclines and choose the positive sense to the left. Then we have w
sin 53° —f, - T = ma and
T — w sin 30° -f, = ma, Where m is the common mass and w = mg, the oommon weights
of blocks A and B.
The frictional forces on the two blocks are determined from the equilibrium
conditions perpendicular to the
inclines. Thus for block A, I, = ;4,N_, with normal force N, = w cos 53°; and for
block B, ft = u,N,, with
N, I w cos 30". We eliminate the tension by adding our two equations to yield w sin
53° -1} -1‘, — w sin 30° I
baa; or substituting for w. [,, and 11,, mg sin 53° - tqmg oos 53° — uimg cos 30° —
mg sin 30° = Zma. Dividing
out by m and solving, we get a =g(sin 53° — ti, obs 53" — ii, oos 30° — sin 30°)/2
= -0.694 m[s’. We now apply
the ltinematical equation uf = ufi, + 20,1 to either block with 11., = 0.90 m/s. a,
= -0.694 m/s‘. and u, = 0 to
get x = QSQ m.
lf the blocks in Fig. 4-27 are momentarily at rest. what is the smallest ooeflicient
oi friction for which the
blocks will remain at rest?
>9
I The tendency to motion will be to the left sinoe that slope is steeper. For
minimum coelficient of friction
the frictional foroes will be their maximum value (the verge of slipping). Thus. [1
= ;4,mg cos 53° and
f, = u,mg cos 30°. both acting to the right. Then the equations of equilibrium are
Block A: mgsin53°— tt,mgcos53°— T=0. Block B: T-mgsin30°—tt,mgcos30°=0
Adding the two equations yields mg(sin 53° — ti, cos 53° — sin 30° — ii, cos 30°) =
0. Dividing out mg
and rearranging terms. we get u,(cos 53' + cos 30°) = sin 53° - sin 3lT, and ti, =
2.
The Learnin A For
AAJ K/-\ TOPPER NEETAIIN-ISQJEEPRAS

AAJ KA NEETAIIMS JEE RAS
NEWTON'S LAWS OF MOTION U 69
4.83 A blimp is descending with an acceleration a. How much ballast must be
jettisioned for the blimp to rise with
the same acceleration a? There is a buoyant force acting upward on the blimp which
is equal to the weight of
the air displaced by the blimp; assume that the buoyant force is the same in both
cases.
I From Fig. 4-28, the equations of motion are
Descending: m,g — F, = m,a. Ascending: F, — mzg = m,n
Adding gives (m, — m,)g = (m, + m,)a. But m, — m, = m. the mass of the discarded
ballast. Therefore,
mg—[m,+(m,—m)]n or m-=<%)m,
1.
Fa F:
"'22
"HR
l.
(a) Descending (b) Ascending |7i‘,4.23
4.8!‘ Show that the acceleration of the center of mass in Prob. 4.83 does not
change when the ballast is ejected.
Use this fact to confirm the value of m found in Prob. 4.83.
I Choose up u positive. Since Z F on the system of blimp and ballast is the same
before and after the
ballast was thrown out. E l7= m,a,,,,, we must have am = -a, before and after. Now,
measured from some
reference level. ya = ['":.vt-..... + (mi - m1)ys.,....]/mi. Then 7... = lmJn....,
+ (m. — m,)y'.......]/m-. B"! 9'“... =
a, y,,_,,_,, = —g, and }Q_,=n,,,, = -a; so —a = [man — (m, — m,)g]/m, or (m, + m,)a
I (m, -- m,)g as before.
yielding the same value for m I (m, — m1).
4.85 Three blocks, of masses 2.0, 4.0, and 6.0 kg, arranged in the order lower,
middle, and upper. respectively. are
connected by strings on a frictionless inclined plane of 60°. A force of 120 N is
applied upward along the
incline to the uppermost block, causing an upward movement of the blocks. The
connecting oords are light.
What is the acceleration oi’ the blocks‘!
r
UH
T:
m,
T.
nt,
Hg. 4-29
I The situation is depicted in Fig. 4-29 with F = 120 N.
m, =2.0 kg m;=4.0 kg and m,=6.0 kg
Applying Newton's second law to each block. we have
F—1}—m,gsin60°=m,a T}—1}-m,gsin60°=m;a T,—m,gsin60°=m,a
AAJ K/-\ TOPPER Ii'E°E§°§i?i'§gJ'EEpR':§’§

70 .7 CHAPTER 4
Adding these equations, F — (m, + m, + m,)g sin 60" = (m, + nt, + m,)a; I20 N —
(12.0 |tg)(9.B m/sz)(0.866) =
(I2-0 REM: I =
Refer to Prob. 4.85. What are the tensions between the upper and middle blocks. and
the lower and middle
blocks?
I Continuing from Prob. 4.85. substitute the value of a into the individual block
equations. and solve for T,
and 7}.
For block l 1 T, = (2.0 kg)(9.s m/s*)(0.aso) + (2.0 ks)( 1.51 in/S‘) =- Qg
For hloclt 3: T, = I20 N — (6.0 ltg)(9.8 m/s’)(0.866) — (6.0 kg)(l.5l mls’) = 1
This can be checked by substituting into the equation for block 2.
A skier goes down a hillside. which makes an angle 0 with respect to the
horizontal. ll’ it, is the ooeflicient of
sliding friction between skis and slope. show that the acceleration of the skier is
a = g(sin 6 — u, cos 8).
I Reverse the direction of motion in Prob. 4.23. and apply to kinetic rather than
static friction.
Refer to Fig. 4-30. Find T, and T; if the blocks are to aocelerate (0) upward at
6.0 m/s’ and (b) downward at
0.60 m/sz.
1",
it
#5
ii:
. I00 v
T1
I-‘lg. no
| Treating the two masses and the connecting cord as an isolated object. T, — 1.03
= l.0a; then isolating the
800~g mass. obtain T, — 0.8g = 0.8a: a is the same in both expressions. (n) a = 6.0
mls‘. so T, = 6.0 + 9.8 =
l5.§N and 7} = 12.6 N. (b) n = —0.6Um/s’. so T, = 9.8 -0.6 = W and T, =7i4. As a
check on the
answers. isolate the Zill-g mass and observe that T, — 0.2g — T} = 0.2a.
The cords holding the two masses shown in Fig. 4-30 will break if the tension
exceeds 15.0 N. What is the
maximum upward acceleration one can give the masses without the cord breaking?
Repeat if the strength is
only 7.0N.
I Note. 7', > T, from Prob. 4.88. Consider the free body made up of both masses and
the massless cord
between them: T, -1.0g = 1.00; for T, = l5.0 N. a = 5.2 m[s:: for T, = 7.0 N. a = —
Z.§ mjs‘. (The system
must he accelerating downward. since T, could not support the 9.8-N weight.)
A 6.0-kg block rests on a smooth frictionless table. A string attached to the block
passes over a frictionless
pulley, and a 3.0-ltg mass hangs from the string as shown in Fig. 4-31. (4) What is
the acceleration a?
(b) What is the tension T in the string?
I (a) This type of problem. as seen in Prob. 4.78. can be treated as if it were in
one dimension. Thus.
Newton's second law takes the fonn
F=ma m,g =(m,+m;)a 3(9.8)=(6+3)a Z9.4=9a a =3.27m[s‘
(b) Applying Newton's second law to mass m, alone.
T =m,a = a(3.21)= l9.6N
AAJ K/-\ TOPPER ii'EeEi°§i'ii'§gJ'E'é°R'§I’§

NEWTON'S LAWS OF MOTION L7 71
H T
3kg
"'1 Hg. 4-at
To check this. apply Newton's second law to m, alone.
m;g— T=m;0 3(9.8)— T=3(3.27) 29.4— T=9.8 T=l9.6N
As expected. the tension is the same.
4.91 A 6.0-kg block rests on a horizontal surface. lts coelficient of kinetic
friction is 0.22. The block is connected
by a string passing over a pulley to a 3.0-kg mass. as in Fig. 4-32. (a) What is
the acceleration 2? (b) What is
the tension T in the string?
N
--—>
file is
“KN
mz
W9 I-‘lg. 4-sz
I We treat the problem as if it were in one dimension.
(a) For the system as a Whole.
netl-‘=ma m=m,+m,=6.0+3.0=9.0kg netF=m;g-p,N=ma N=m,g
ma = "|;8 — i|km,g 9.0a = 3.l)(9.8) — 0. 22(6.0)(9.8) = l.68(9.8) a = 1.83 m[s=
(6) The force the string exerts on m, is
T = p,N + m,a = 0.22(6.0)(9.8) + 6.0(l.83) = 12.94 + I098 T = Zigfl
4.92 Suppose that blocks A and B have masses of 2 and 6kg. respectively. and are in
contact on a smooth
horizontal surface. If a horizontal force of 6 N pushes them. calculate (a) the
acceleration of the system and
(b) the force that the 2-kg block exerts on the other block.
I (u) See Fig. 4-3310). Considering the blocks to move as a unit. M = m, + m, = 8
kg, F = Ma = 6 N
a = 0.75 Q15’.
/» = 1» .\‘
F =0 N R i> B
fi A A
wt 1"! I13. 4-33
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72 5'
4.93
4.94
4.95
4.96
4.97
CHAPTER 4
(b) lf we now consider block B to be our system, the only force acting on it is the
force due to block A. F...
Then since the acceleration is the same as in part (a). we have F, = M,a =§iN.
However. we must consider
also the case in which we reverse blocks A and B as in Fig. 4-33(b). As before,
considering the blocks as a
unit we have a = 0.75 rn/s‘. Now, however. if we consider block B as our system we
have two forces acting,
the force F to the right and the force F, to the left. Then F - F, = M,,a and
solving we get F, = Q in
magnitude and points to the left.
We could verify these results by considering block A to be the system for the two
eases.
ln Fig. 4-34, the pulley is assumed massless and frictionless. Find the
acceleration of the mass m in terms of F
if there is no friction between the surface and m. Repeat if the frictional force
on m is f.
F
t-1;. 4-34
I Note T = F /2. Newton's law for the block gives T = ma. hence a = F l2m. When
friction is involved we
have FIZ —[=ma. so a = (F/2m) — (f/m).
In Fig. 4-35, assume that there is negligible friction between the blocks and
table. Compute the tension in the
cord and the acceleration of m; if m, = 300g, m,= 200g. and F =0.40 N.
I Write F =ma for each block. Let a be the aomleration of m,. The acceleration of
m, is then a/2 (compare
Prob. 4.72). Then T =m,a and F — 2T =m.(a/2). One finds n =0.73 rn[s= and T =0. 145
N.
r-1;. 4-as
How large must F be in Fig. 4-36 to give the 700-g block an acceleration of
Jflern/s’? The coeflicient of
friction between the two blocks and also between block and table is 0.150.
tail! .
'i7‘?7‘"'~
I Converting to Sl units, isolate each mass and note the forces that act on each.
Vertically only the weights
and normal forces are involved, F~ = 0.2g on the upper block and 0.9g on the lower
one. Friction forces are:
u(0.2g) between blocks and p(0.9g) at the table. F= ma for the blocks: T — i4(0.2g)
= 0.741 and F- T —
;4(0.2,g) — ;4(0.9g) = 0.7a; after eliminating tension T between these. F = 0.9a +
u(l.3g) = 0.9(0.30) +
0.1S0(l.3)(9.8) = 2.18 N.
Assume in Fig. +36 that the coefficient of friction is the same at the top and
bottom of the 700-g block. lf
n =70cmIs’ when F I 1.30 N. how large is the coeflicient of friction?
I Following Prob. 4.95, F =0.9a + ;4(I.3g); then we use F = 1.30 N and a = 0.700
m/s‘ to find it =0.0§3.
rig. we
ln Fig. 4-37, when m is 3.0 kg, the acceleration of the block m is 0.6 m/s‘, while
a = 1.6 m/s’ if m = 4.0 kg.
Find the frictional force on block M as well as its mass. Neglect the mass and
friction of the pulleys.
I Apply Newton's second law to m for each case: 3(9.8) — 21} = 3(0.6). so 1‘, =
13.8 N; and 4(9.8) — 27} =
4(1.6). so T, = 16.4 N. Applying Newton's second law to M for each case, 13.8 —f =
M(1.2) and 16.4-f =
M(3.2). Solve to find M =1.3 kg and I = 12.2 N.
AAJ K/-\ TOPPER iiEeEi°§i?i'§gJ'E'épR':i§

uswrotrs utws OF MOTION :1 1a
u‘;> 1'
‘ .
as
Hg. 4-31
4.98 In Fig. 4-38-(a), block 1 is one-fourth the length of block 2 and weighs one-
fourth as much. Assume that there
is no friction between block 2 and the surface on which it moves and that the
coeflicient of sliding friction
between blocks 1 and 2 is pt, = 0.2. After the system is released, find the distance
block 2 has moved when
only one-fourth of block 1 is still on block 2. Block l and block 3 have the same
mass.
I From Fig. 4-38(b), the equations of motion are
2Ft=T—utwt=IMt EFt=mwt=4»wt 2Ft=w.—T=nwt
Solve the first and third equations simultaneously to get a, I (g/2)(l - 44,); from
the second equation.
a,= (3/4);“. Then the displacements of blocks 1 and 2 are given by x = Q01’. i.e.,
<\'|=%(1“Ih)/2 -‘z=%#|J2
At the instant that one-fourth of block 1 remains on block 2. 1,4-I =x, + (I/I6).
where I is the length of
block 2. Therefore.
I 15!
%)l;Iz+l=%(1_P|t)l2+fi> Oi’ I
_ 5 15! _ 15¢, _L
‘M “'(8"‘)2g<1-am"wtz-1u.>"Lel
‘ I
.
\
(n) Configuration at I = U
0:
Q, Z> T
i’ 'i> /a = Inw-
av, Z
It |Wt 1‘
wt
(b) R: ' 5":
’ ll
1-? d;= at
\v| R| 1 W;
4w,
n|. 4-as
4.99 A dinner plate rests on a tablecloth. with its center 0.3 m from the edge of
the table. The tablecloth is
suddenly yanked horizontally with a constant acceleration of 9.2 m/s‘ [Fig. 4-
39(a)]. The coefficient of sliding
friction between the tablecloth and the plate is ti, = 0.75. Find (I) the
acceleration. tb) the velocity. and
(e) the distance of the plate from the edge of the table. when the edge of the
tablecloth passes under the
center of the plate. Assume that the tablecloth just fits the tabletop.
AAJ K/-\ TOPPER it'E°Ei°§i'§i'§gJ'E'EpR':I§

74 U CHAPTER 4
/
=91 m/5’
Q "—~
/
\
?__ ____ ___ _ / \ \
j ; \
Plate
i—>
ll N I ' BAN ‘ Bunl
(ti) (5)
Hg. 4-39
I (4) from Fig. 4-39(b), the force equation for the plate is lung = ma,, or a, = pg
= (0.75)(9.8) = 7.35 m[s‘.
'l'he plate slips, since n, is less than 9.2 mls’. (b) At the time the edge at the
tablecloth is at the center of the
plate. the cloth and the plate are at the same distance from the edge of the table:
x, = 1, 0.3 + §(7.4)l’ = 0 + i(9.2)I’
Solving. I = 0'53 s and u, = 0 + (7.35)(0.S8) = ;4._26m[s.
(e) 1,, = 0.3 + 0(0.58) + §(7.35)(0.58)’ = LQ4 m
4.1“ In the pulley system shown in Fig. 4-40. the movable pulleys A. B. C are of
mass 1 kg each. D and E are
fixed pulleys. The strings are vertical and inextensible. Find the tension in the
string and the accelerations of
the lrietionless pulleys.
I Write y,, y,, yr for the positions of the centers of the pulleys A. B. C at time
I; 11,, a,. a.» are the
accelerations at time I.
\ \
Positive I I
direction 1 u u
downwlrd
Y4 \
0 " T T J
Vt
T . 4 .
Position at time I
Q w = -llllg
)
~t ~t
N
I--> --
:4
F
= tllltlg
J
r _ _
| itiul posttinrt
ms
Fig.4-40 Fig.4-41
Th L ' A F
AAJ K/-\ TQPPER ~E°ETe§|T§'|'§gJE'EpRX§

NEWTON'S LAWS OF MOTION £7 75
Following the string from the end at the center of A to the end at the center of B.
we get
(Yr: ' Y4) ‘l’ Y! "' ZYA '*Yc ‘* (Yc -75) = ¢°"5l3'“ °|' Y4 +7: 4' 2)’r = Wllillfll
Take the second time-derivative of this equation to get a, + a, + 2a; = 0.
There is just one string and. thus, one tension T. The force equations are
T+mg—2T=ma,, T+mg—2T=ma, mg—2T=ma¢
Substituting m - l kg and solving the four equations for the four unknowns 11,, a,,
n<;. T, we obtain
a,=a,,=—a¢=§=3.3m[s 7'-6.5N
4.101‘ A body of mass 400 kg is suspended at the lower end of a light vertical
chain and is being pulled up vertically
(see Fig. 4-41). Initially the body is at rest and the pull on the chain is 60(1);
N. The pull gets smaller
uniformly at the rate of 360g N per each meter through which the body is raised.
What is the velocity of the
body when it has been raised 10m?
I At time I, let y be the height (in meters) of the body above its initial
position. The pull in the chain is then
T = (6000 — 360y)g and Newton's second law gives
T—400g=-400)? or (56(I)—360y)g=-4(1))?
This equation may be changed into one for y = u (the velocity of the body) by use
of the identity
-~_ Q’- ELY- .di'_ 1
2y—2dl—2d.vd’-2udy—d(u )/dy
Z
Thus Z00%;)=(S6iX)—360y)g or d(t'*)=g(28—1.8y)dy
Let V be the velocity at height 10 m. Then, on integrating
v1 in
L d(v‘) =gL (28 — l.8y) dy V’ = g[2/By — 0.9y’]},° = g[28(10) — 0.9(100)] = 190g
V = +\/1905 = +412 m[s.
The choice of the + sign for V (upward motion) should be checked. For Osy s l0, the
net force.
(5600 — 360y)g, is positive, and so the acceleration is positive. Then, since the
body started from rest, V must
be positive.
AAJ K/-\ TQPPER It'E°Ei°§i?i'§gJ§'EpR':I’§
L“E°E#°§.1?l2°f;‘é"R'i{’;
CHAPTER 5
Motion in a Plane I
PROJECITLE MOTION
A marble with speed 20cm/s rolls off the edge of a table 80 cm high. How long does
it take to drop to the
floor? How far, horizontally. from the table edge does the marble strike the floor?
| Choose downward as positive with origin at edge ol table top.
ii“, = t.-.,= 20 cm/s t',,, = 0 a, = +g = +980cm/s‘ a, = 0
To find time of fall. y = u,,,l + §gI:, or 80cm = 0 + (490 cm/s’)r*; I = 0.40 s. The
horizontal distance is gotten
from x = tt...I = (20 cm/s)(0.40 s) =8.0 cm.
How fast must a ball be rolled along a 70-cm-high table so that when it rolls ofl
the edge it will strike the
floor at this same distance (70 cm) from the point directly below the table edge?
I In the horizontal problem. x = v,l gives v, = 0. 70/ I. ln the vertical problem.
choosing down as positive.
t/(,= 0. y = 0.7 m. and a = 9.8 m/s:. Use these values in y = v,,r + all/2 to give
I = 0.378 s. Then u, = Q5_mg.
A marble traveling at llltlcm/s rolls off the edge of a level table. ll it hits the
floor 30cm away from the spot
directly below the edge of the table. how high is the table?
I This is at projectile problem with v,,, = l()llcm/s. For the horizontal motion:
s, = um! 3()=100t t= 0.30s
For the vertical motion:
s, = 11...! + Qtu‘ = 0 + §(98(l)(0.30)= = -ti (height of the table)
ln an ordinary television set. the electron beam consists of electrons shot
horizontally at the television screen
with a specd of about S X I0’ mls. How tar does a typical electron fall as it moves
the approximately 40cm
from the electron gun to the screen? For comparison, how far would a droplet of
water shot horizontally at
2 mls from a hose drop as it moves a horizontal distance of 40cm?
I For the electron. the horizontal problem yields the time to hit the screen as t
=x/tt, =0.40/(5 x 10’) =
8 x10 “s. Then in the vertical problem. y = v.,,t + at:/2, so y = 0 + 4.9(64 >< l0
"‘) =3.l x l0"° m. For a
droplet. I = 0.40/2 = 0.20 s. and so y =0.196 m.
A body projected upward from the level ground at an angle of 50° with the
horizontal has an initial speed of
40m/s. How long will it be before it hits the ground?
H
_ _ . _ _
___ \
4 ~
@ \<m r-1;. s-1
| Choose upward as positive. and place the origin at the launch point (Fig. 5-1).
ti... = tn. cos 50° = (40 m/s)((l.642) = 25.7 m/s tv,,,. = U0 sin 50° = (40 m/s)
(0.766) = 30.6 m/s
rt, = -g = -9.8m/5’ a, =(l
To find the time in air. we have y = t'.,,I — 13!’ and since _v = ll at the end of
flight. 0 = (30.6m/s)r —
(4.9 m/s’)|’. or 4.9:’ = 306:. The first solution I = 0 corresponds to the starting
point. y = 0. The second
solution is not zero and is obtained by dividing out by t. 4.9! = 30.6 and I =6.24
s.
ln Prob. 5.5. how far from the starting point will the body hit the ground. and at
what angle with the
horizontal?
AAJ K/-\ TOPPER It'EeEi°§|i?i'§gJ'E'épR':I’§
AAJ K/-\ TOPPER ii'EE'ie§|i'ii'§gJ'§EpR'ii’§

MOTION IN A PLANE I J 77
I The horizontal distance traveled. or range R. is obtained from R = v.,,| = (25.7
m/s)(6.24s)==160 m.
By symmetry rt stnltes at 50° (with negative x axis). This can also be seen by
noting that u, = —v.,,,, v, = um,
so tan 9 = —tan 0,, and therefore 180° — 0 = 0,,.
A body is projected downward at an angle of 30° with the horizontal from the top of
a building I70 m high.
"S il1i!i=I| $P¢=d is 40m/s. How long will it take before striking the ground?
'Tu?T"_
\
\
\
\
| t
[70 m \
\
\
\
\
\
n \
\\/0
/ Fig. s-2
I Choose downward as positive and origin at the top edge of building (Fig. S-2).
v.,, = u,, cos 30" = (40 m/s)(0.866) = 34.6 m/s um. = v,, sin 30° = (40 m/s)(0.500)
= 20.0 m/s
a,=g=9.8m/s: a,=0
We can solve for the time in different ways. Method l:
y = t-..,! + £31’ or I70 m = (20.0 m/s): + (4.9 m/s=lI:
We can solve the quadratic to yield
(We keep only the positive solution; the negative time corresponds to a time before
1= 0 when it would have
been at ground level if it were a projectile launched so as to reach the starting
position and velocity at 1= 0.)
Method 2: We avoid the quadratic. First find u, just before impact:
uf = ufi, + Zgy or of = (20.0 m/s)’ + 2(9.8 m/s’)(l70 m) v, = 161 m/s
For our case v,. = bl m/s. Next we find II
1,-‘ =13,‘ + gt or bl m/s = 20.0 m/s + (9.8 m/s=)! or r = Al
ln Prob. 5.7. find out how far from the foot of the building the body will strike
and at what angle with the
horizontal.
I x = um! = (34.6 m/s)(4.2 s) = 14$ rn. We need the angle that the vector velocity
makes with the x axis just
before hitting the ground. We avoid having to directly deduce the quadrant this
angle is in by solving for the
acute angle 8 made with the x axis (positive or negative; above or below).
tan8=iS1=1.76 and e=r>0.4=
Since tr, is negative and ll, is positive, this is clearly the angle below the
positive x axis (see Fig. 5-Z) and
equals the angle we are looking for.
A body is projected from the ground at an angle of 30° with the horizontal at an
initial speed of I28 ft/s.
Ignoring air friction. determine ta) in how many seconds it will strike the ground.
(b) how high it will go. and
tr) what its range will be.
I (a) Find ti... and then determine the time for a freely falling body projected
upward at this velocity to
return to the ground; that is. s, = 0. From Fig. S-3.
v,,_. =u sin 9= l28sin 30°=6-Ht/s r, =v...I+ Qgr‘ U=64|+ §(—32)r’=(64— 16!): t=4_s
lb) Since time of ascent is equal to time of descent. the projectile reaches
maximum height H at I = 2s. Thus.
H =tr..1 + ggfi = 64(2) + it-s2>(2>‘ =6Lft
The Learntn A For
KA TOPPER NEETAl|NlSgJEEpRAS

78 J CHAPTER 5
Y
_u_= L28 ft/s
vo, H
0 ' 36 x
Um I = 2 S
-Y1
I = 4 S r-1;. as
(c) The projectile travels with a constant velocity u, = v cos 8 in the x
direction. It reaches the ground in 4 s.
The range is
s, = v,! = (v cos 0)! = l28(oos 30°)4 = 51Z(0.866) =443 ft
5.10 A hose lying on the ground shoots a stream of water upward at an angle of 40“
to the horizontal. The speed
of the water is 20 m/s as it leaves the hose. How high up will it strike a wall
which is Sm away?
\
Wall
\||
‘____it ll mi» \ Fig. 5-4
I Setting coordinates as shoum in Fig. 5-4, with uu = 20 m/s and 0,, = 40‘, we get
u.,, = 21., cos 9., = (20 m/s) cos 40° = 15.3 n1/s 110, = u,, sin 9,, = (20 m/s)
sin40° = 12.8 m/s
x = v.,,1. and setting x = 8 m we find the time to hit the wall: 8 m = (15.3 m/s)r,
yielding r = 0.52 s. To find the
height at which it hits the wall. we use y = vi,,1 — §g!', with |= 0.52 s. This
yields y = (12.8 m/5)(0.52 s) —
(4.9 m/s:)(0.5Z 5)! = 5.33 ITI-
5.ll A baseball batter hits a home run ball with a velocity of 132 ft/s at an angle
of 26° above the horizontal. A
fielder who has a reach of 711 above the ground is backed up against the bleacher
wall, which is 386ft from
home plate. The ball was 3 ft above the ground when hit. How high above the
fielder's glove does the ball
pass‘!
I
i/<
ti _.’E . . _ _ _ _ _ _ _ _ . . .
--i
I
l
— 3 ll
Ground H‘, S-5
| The situation is depicted in Fig. S-5, with the origin and x axis 3ft above the
ground. We must find the
value of y on the ball‘s trajectory corresponding to x = 3861't. Then we can
subtract the height of the ficlder's
glove above the x axis, i.e., 7ft — 3 ft = 4ft. To find y we note that
u,,, = 11,, cos 9., = (132 ft/s) cos 26° = 119 ft/s U9, = vo sin 8,, = (132 ft/s)
sin 26° = 57.9 ft/s
The time to reach x = 386ft is given by 1 = u,,,r, or 38611 = (119 ft/s)|, and 1 =
3.24 s.
AA-I K/-\ TOPPER L%°E#°:.1?l2“.’;'.;"Ri(’;

5.12
5.13
5.14
5.15
5.16
MOTION IN A PLANE I J’ 79
Then y = u.,,I — lg!’ = (57.9 tt/s)(3.24 s) — §(32 ft/s’)(3.24 s)’ = 19.6ft. Height
above glove = l9.6 ft — 4 ft =
Q [Note: The trajectory equation y = (tan 80):: — gx 1/(20% cos’ 8,) can also be
used to find y.]
A ball is thrown upward at an angle of 30° to the horizontal and lands on the top
edge of a building that is
20 rn away. The top edge is S m above the throwing point. How fast was the ball
thrown?
u.
Sm
<i2lin\—-—> Fig.5‘
| The situation is depicted in Fig. 5-6 with 00 = 30". We can use the trajectory
equation y = tan 60x — gx'/
(luf, cos’ 80), setting x = 20 m and y = 5 m. Then 5 m = (0.S8)(20 m) -— (9.8 m/s’)
(20 m)‘/(2v§ x0.75),
or u.,=20m[s. (Note: It you didn't remember the trajectory equation. you could
still solve the problem by
using the x-vs.-I and y-vs.-I equations.)
A projectile is fired with initial velocity 11,, = 95 m/s at an angle 0 = 50°. After
5 s it strikes the top of a hill.
What is the elevation of the hill above the point of firing? At what horizontal
distance from the gun does the
projectile land?
lo ,’
»
I
1
A fig. 5-7
I The situation is as shown in Fig. 5-7. un = 95 mls; 9., = 50°. At any time 1, y =
v,,,.t - jgl’, where
v,,, = t/0 sin 0., = 72.8 m/s. For I = 5 s. we get y = (72.8 mls)(5.0 s) — {(9.8
m/s‘)(S.0s)’ = 241 m. The horizontal
distance. x, is given by x = u,,,r = vacos 0,,r =(61.1m/s)t. At 1 = 5 s. we have x
= 305 m.
A ball is throum with a speed of 20m/s at an angle of 37° above the horizontal. lt
lands on the root of a
building at a point displaced 24 m horizontally from the throwing point. How high
above the throwing point is
the root?
I The velocity components are um = 16 m/s and u.,, = l2 m/s. ln the horizontal
problem. r=x/v, = 24ll6=
1.5 s. Then in the vertical problem we have y = u‘,,r 4-at‘/2, so y = l2(l.5) -
4.9(2.2S) ==7.0m.
A hit baseball leaves the bat with a velocity of 110 ft/s at 45° above the
horizontal. The ball hits the top of a
screen at the 320-ft mark and bounces into the crowd for a home run. How high above
the ground is the top
of the screen? (Neglect air resistance.)
I s, = v,1 v, = no cos4S° = a constant 320 = ll0(0.707)t I= 4.11;
s, = v.,,r+iur1 = (110 sin 4s")(4.11)+ it-a2)(4.it)* = (l10)(0.707)(4.ll) + it-s2)
(4.i1)=
= 319.6 — 270.3 s, =i% (height of screen)
A ball is thrown upward from a point on the side of a hill which slopes upward
unitonnly at an angle of 28°.
Initial velocity of ball: 11.. = 33 mls. at an angle 6° = 65° (with respect to the
horizontal). At what distance up
the slope does the ball strike and in what time?
I The situation is depicted in Fig. 5-8. u‘. = 33 m/s and 9,, = 65°. The trajectory
equation of the ball is
y,, = (tan 0,,)x — gx’/Zuf, cos’ 9») = 114‘ - 0.02512
AAJ K/-\ TOPPER il'EeEi°§i'ii'§gJ'E'épR':I’§

80 Q’ CHAPTER 5
5.l'!
5.18
5.19
t
\-
,4
/
I
mo _ ,W
The equation for the incline is y, = (tan 28“)x = 0.S3x. At the value of x for
which the ball hits the incline
y,, = y,. or 0.53; = 2.14:: — l).(l25x’. and 0025:’ = Loh, which yields .r = 64.4
m. The distance along the
incline. S. obeys x = S cos 28° or S = 72.9 m. ‘Hie time to reach any x value is
given by x = v.,,! =
(v..cos 8")! = 13.91. So for Jr =6-1.4 m. I = 593;.
A projectile is to be shot at 50 m/s over level ground in such at way that it will
land N0 m from the shooting
point. At what angle should the projectile be shot‘?
I In the horizontal problem. x = v...r gives 200 = (50 cos 0)t. where 0 is the
angle we seek. ln the vertical
problem. y = t'.>,I + at‘/2 gives 0 = 5Osin 6 — 4.91. But I = 200/(S0 cos 8). and
so 50sin 6 = 4.9(4/cos 6). This
simplifies to 2 sin 9 cos 9 = 0.784, and so sin 29 = 0.784, from which 6 = 25.8°.
As shown in Fig. 5-9. a ball is thrown from the top of one building toward a tall
building 50 ft away. The
initial velocity of the ball is 20l‘t/s at 40° above the horizontal. How far above
or below its original level will
the ball strike the opposite wall?
20 \“
J _ _ _ _ \ _
\ 1’
so lt fig’ 5_,
I tr“, =(20ft/s)cos40"= 15.3 ft/s u,,, =(20ftls) sin40°= l2.9ft/s
ln the horizontal motion. tr“, = u,, = 12', = 15.3 ft/s. Then x = 0,1 gives 50ft =
(15.3 ft/s)!, or 1 = 3.27 s. ln the
vertical motion. taking down as positive.
y = Um! + §tI,.I: = (-12.9 ft/s)(3.27 s) + {(32.2 ft/s=)(3.27 s)‘ = 130 ft below
(a) Find the range x of a gun which fires a shell Mth muzzle velocity v at an angle
of elevation 0. (b) Find
the angle oi elevation B of a gun which fires a shell with a muzzle velocity ol 1.2
kmls at it target on the same
level but 15 km distant. See Fig. 5-10.
U
"0,
7
' L
x x _
eel ng 5 to
I ta) Let I be the time it takes the shell to hit the target. Then. x = v.,,r or I
= x/u.,,. Consider the vertical
motion alone. and take up as positive. When the shell strikes the target. vertical
displacement =0 =
v,,,.1+ §(—g)|’. Solving this equation gives I = 2v.,,/g. But r=x/t'.\.. so
;=% or ,=%, ,Mi°
vth 8 S B 8
AAJ K/-\ TOPPER 1-l'IIEeE|Te:ll:Ii|ggJll?épR|2)Sr

MOTION IN A PLANE I U 81
(b) From the range equation found in (4),
, _£:_(9.8x 10 ‘km/s’)(15km)=
Hfl26—u€— l_2km/S); 0.l02
whence 20 = 5.9‘ and 6 =E.
A rifle bullet has a muzzle velocity of 680 it/s. (a) At what angle (ignoring air
resistance) should the rifle be
pointed to give the maximum range? (b) Evaluate the maximum range.
I (a) By Prob. 5.19(a), the range is a maximum when sin 20 = 1, or 6 = 45°.
I
Z
(la) rm, - %'- %‘/L51 - 14 450 rt.
A golf ball leaves the golf club at an angle of 60° above the horizontal with a
velocity of 30 mls. (a) How high
does it go? (b) Assuming a level fairway, determine how far away it hits the
ground.
I (it) With up positive, U0, - 30sin60°= 1S\/3 mls and. at maximum height. u,
equals zero. 'l1ius.
u§=v?,,-t-Zah 0=67S—l9.6h h=%=34.4m
(b) By Prob. 5.l9(a).
U’ sin 20 9000/5/2)
0
=i=i=79.s
X g 9.8 "'_m
Prove that a gun will shoot three times as high when its angle oi elevation is 60"
as when it is 30°, but will
carry the same horizontal distance.
I We assume that tn, is the same at both angles. Maximum height is given by the
condition v, = 0. Then.
vi = u§, — Zgy yields y,,,, = v§,,l(2g). Noting that v,,, = uo sin 80. we have
finally y,,,,, = (vf, sin: 8)/2g.
Then y,,,,(6()°)/y,,,,,(30°) =sin‘ 60°/sin‘ 30°= 3. To show that the horizontal
range is the same, we note that in
the range formula [Prob. 5.l9(a)] cos 60" sin 60° = cos 30°sin 30° since 30° and
60° are complementary angles.
As shown in Fig. S-1 l. a projectile is fired with a horizontal velocity of 330 mls
from the top of a clifl 80 m
high. (n) How long will it take for the projectile to strike the level ground at
the base of the cliif! lb) How
far from the foot of the cliff will it strike? (c) With what velocity will it
strike’!
330 in/s
—i>
I-?—-1
Li>=—-———>l _ n;.s-11
I (a) The horizontal and vertical motions are independent of each other. Consider
first the vertical motion.
Taking down as positive we have y — 00,! + §a,t', or 80 m = 0 + §(9.8 tn/s’)|’,
from which l=-W. Note that
the initial velocity had zero vertical component. and so uo = 0 in the vertical
motion.
(It) Now consider the horizontal motion. For it, a = 0 and so D, - u,,, = 1-,, =
330 m/s. Then, using the value
of I found in (I), x = 0,1 - (330 tn/s)(4.04s) = l330m.
(c) The final velocity has a horizontal component of 330 mls. But its vertical
velocity at 1 = 4.04: is given by
u,, = 0°, + 11,! as u,, -1 0 + (9.8 mIs')(4.04 s) = 40 mls. The resultant oi these
two components is labeled v in
Fig. 5-ll; we have
u = \/(40 mls)! + (330 m/s)’ = 332 mls
Angle 8 shown is given by tan 9 = 40/330 to be QL
AA-I KA TOPPER I.?E#°:.1?l2?.’;'.;"Ri(’;
AAJ K/-\ TQPPER ii'E‘E'ie§|i'ii§gJ'§EpR'j§’§

CHAPTER 5
A stunt flier is moving at l5 m/s parallel to the flat ground 100m below. How large
must the horizontal
distance x from plane to target be if a sack of flour released from the plane is to
strike the target?
I Following the same procedure as in Prob. 5.23. use y = u,,_| + §a_,!* to get
1(Xlm= 0+ {(9.8 m/s’)l’, or
I = 4.5s. Now use x = tI,I = (15 m/s)(4.5 s) =Q m, since the sack's initial
velocity is that oi the plane.
A baseball is thrown with an initial velocity of 100 m/s at an angle of 30° above
the horizontal. How far from
the throwing point will the baseball attain its original level?
I By the range formula,
:15 sin 20 10‘(\/3/2)
*="T=T=*M
A cart is moving horizontally along a straight line with constant speed 30 m/s. A
projectile is to be fired from
the moving cart in such a way that it will retum to the cart after the cart has
moved 80 m. At what speed
(relative to the cart) and at what angle (to the horizontal) must the projectile be
fired?
I To move horizontally with the cart. the projectile must be fired vertically with a
flight time = x/u, =
(80 m)/(30 m/s) = 2.67 s. The initial velocity tn, must satisfy y = vol + at’/Z,
with y = 0 and r = 2.67 s; thus
4.9! = v.,, and so tn, = l3.l n1j_§ at 8 =9l.
ln Fig. 5-12. a particles from a hit of radioactive material enter through slit S
into the space between two
large parallel metal plates. A and B. connected to a source of voltage. As a result
of the unifomt electric field
between the plates. each particle has a constant acceleration a = 4 X l0"m/s’
normal to and toward B. If
u‘, = 6 x 10“ m/s and H = 45°. determine h and R.
I Here the electric force takes the place of gravity. but otherwise the analysis is
the same. Choosing upward
as positive. i.e.. the direction from B to A, we have a regular trajectory problem
with u,_.= 6 X l0” m/s;
0., = 45°; a, = -4 X 10” m/s’; a, = 0. Then x = um! = v.,oos 0,,l = (4.24 X 10"
m/s)t; and similarly
y = v,,,t + ‘,a,r’ = (4.24 x 10“ m/s): — (2.0 X I0" rn/6):’
t', = u.,, + 11,! = (4.24 X l0"m/s) — (4.0 X 10" m/s’):
At the highest point.y = Ir. u_, = 0, and I = 1.06 X l0 ’ s. Thuslt = (4.24X l0")
(1.06X l0 7) “ (2.(lX 10") x
(L06 X IO ')‘=0.225 rn. The horizontal range R corresponds to x at the time the m
particle returns to plate
B. By symmetry. this is 21 = Z. l2 X l0 ’s. Then R =1-$.24 X l0" m/s)(2.l2 X I0 ’s)
=0.90 m.
0 0 4 K
v,.
~
* napartrcle
- It
0
Slit S \ \ I I B
Source
£7? R r Tl
Fig.5-12
A ball is thrown upward from the top ol a 35-m tower, Fig. 5-13, with initial
velocity u‘,= gum/5 3| an angk
0 = 7.57 (I) find the time to reach the ground and the distance R from P to the
point of impact. (6) Find the
magnitude and direction of the velocity at the moment of impact.
I (I) At the point of impact. y = -35 m and x = R. From y = -35 = (80sin L5“)! —
§(9.8)r’, I = 7.§]4 §. Then
1 = R = (so oos 25°)(7.s14) = 506.55 m.
(bl At impart. v, = 80sin 25° — (9.8)(7.8l4) = -42.77 m/s and v, = v,,, = 8000s 25°
= 72.5 m/s. Thus tr =
(42.77: + 72.52)" = 84.18 rn/s and tan B = -42.77/72.5, or [3 = -30.54‘.
AAJ K/-\ TOPPER It'EeEi°§|T§i'§gJ'E'é°R':I’§
I.";E$2.1?l2°.‘é‘;’R'i{’;
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X
\
\
35 m 1 \
I
\\\\\\\Pl '
\ \1\:\\\\\ \\\>}>:>\\\\\\\\\\ :,\\\\\\\\
t-1;. 5-ta
5.29 The arrangement in Fig. 5-l4 is the same as that in Fig. 5-l2 except that a
particles enter slit S from two
sources. A, and AZ, at angles 0, and 6,, respectively. u.. and a are the same for
both groups. Given that
uo = 6 X ll?" mls. a =4 X I0" mls’, 6, = 45° + 1’, 8; = 45° — 1°. show that all
particles are "focused" at a single
point P and find the value of R.
I Here we can use the horizontal range formula R = Zuf, cos 6,, sin 9.,/n. Noting
that for the first a particle
0., = 0, = 46° and for the second a particle 9., = 0; = 44°. we see that the two
are complementary angles. Then
cos 9, sin 0, =sin 0, cos 0; and the ranges are the same. Indeed R = 2(6 X
l0"m/s)’(0.7l9)(0.695)/
(4 X 10" mls‘) =0.90 m.
5.30 Again referring to Fig. 5-14, find h, — h,.
I For the vertical heights we have of = vfi, + 2.a,y with v, = 0: a, = —|a| = 4 x
10"‘ mls’. Then
h,= =0.233m h;= =0.217m h,—/|;=0.l)l6m= lfimrn
Zlal Zia]
o 0 < 4 A
0
_; n ’
11 45* ti,
sms\ 9: a.‘ K P B
4: 1: 3 :I
fig. s-14
5.31 A ball is thrown upward with initial velocity vn= 15.0 mls at an angle of 30“
with the horizontal. The thrower
stands near the top of a long hill which slopes downward at an angle of 20°. When
does the ball strike the
slope?
I We choose the launch point at the origin (sec Fig. 5-I5). The equations of motion
of the ball are
x = vt. cos 30°! = (13.0 mls)! y = 11,. sin 30° — igt’ = (7.5 mls)! — (4.9 mls‘)?
The equation of the straight line incline is y = -—x tan 20° = —0.364x. We want the
time at which the (x. y)
values for the ball satisfy this equation. We thus substitute the time expressions
for y and x:
AAJ K/-\ TOPPER 1-l'IIEeE|‘Te:|llrl:Ii|ggJll?épR|:)Sr

84 5' CHAPTER 5
nt
-- ------<----t
t|‘\1"° x
L
Q
Hg. 5-15
(7.5 m/s)1—(4.9 m/s’)t‘ = -0.36-t[(13.0 m/s)1], or 12.21 = 4.91’. The solutions are
I = 0 (corresponding to
x=y=0) and l=Z.49s.
5.32 Referring to Prob. 5.31 determine how far down the slope the ball strikes.
I Referring to Fig. 5-l5 we need the x and y components of the dis lacement to
position A: x =
(13.0 mls)(2.49 s) = 32.4 m; y = -0.364: = -ll.8 m. Then L = Vxz + ya -34.5 m (or
directly from x, L =
x/cos 20° == 34.5 n1).
5.33 Referring to Prob. 5.31 indicate with what velocity the ball hits.
I For the velocity of the ball just before impact, v,:
IIA, = uo cos 30° = l3.0m/s u,,, = tr, sin 30°- gt = 7.5 m/s - (9.8m/s')(2.49 s) =
-16.9 m/s
Thus
v, = Vvf,,+ vi, = 21.3 mls 9, = tan" 21 ==52.4° below positive: axis
V41
5.34 A bomber. Fig. 5-16, is flying level at a speed u, = 72 m/s (about 161 mi/h).
at an elevation of h = 103 m.
When directly over the origin bomb B is released and strikes the tnick T, which is
moving along a level mad
(the X axis) with oonstant speed. At the instant the bomb is released the tnick is
at a distance xu = 125 m
from 0. Find the value of u, and the time of flight of B. (Assume that the truck is
3 m high.)
I The equations for x and y motion of the bomb are
x = Um! = vi! = (72 m/s)! y =y., + v.,,| — 53:‘ = Ii - 5;!‘ = 103m —(-1.9 m/6):‘
The time for hitting the truck corresponds to y = 3 tn. so
00
3m=l03m—(4.9m/s‘)1’ bu‘ |=4.52s
Then x = (72 m/s)(4.52 s) = 352 m.
Y
L'|
B .
h
T .
. ... >
O X
n;.s-is
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MOTION IN A PLANE I .7 85
5.35 A projectile. Fig. 5-17, is fired upward with velocity v., at an angle 9. (a)
At what point P(x, y) does it strilte
the roof of the building. and in what time? (bl Find the magnitude and direction of
v at P. Let 0 = 35°,
vn=40m/S, a=30°, and h = 15m.
I First note that (using notation 3 I v,,. y I 11,)
in = U“ cos 35° = 32.7661 m/S ji,, = Lu, sin 35° = 22.943 m/s
and. from the equation of the roof.
y=h—xtantr=15-(0.S7735)x (1)
(0) Eliminating I from y = )1”: — 4.9:’ by x I i‘,|, we have
y=({§): <2)
for the path of the projectile. Equating y in (1) to y in (2) and inserting
numerical values. 0.(l)4564x‘ —
1.2775582: +15 = 0, from which x =12.28 m. Then y = h — (12.28) tan a =7.90 m. The
time to strike is given
by 12.28 = 32.76611, or I =0.375 s.
(II) At P, i = it, = 32.766 m/S )1 = y‘., — 9.8| = 22.943 — (9.8)(0.375) = 19.268
m/s
Thus u = (iz + y’)"’ = 38.0 m[s and tan 5 = )"/X = 0.588. or ti = $.46“. where 5 is
the angle that v makes with
X at P.
Y
lino!‘ )'=h—1tana
P
h
Y
6 .
j: 5'
5.36‘ In Prob. 5.35, angle 6 can be adjusted. Find the value of 6 for which the
projectile strikes the roof in a
minimum time.
‘ x Fig. s-i1
I Again y =)"4tl — lg!’ = (vt, sin 8)! — égt’ and y = h — x tan a
Equating these two expressions for y and eliminating x by x = X01 = (ut, cos 9)|,
we obtain the following
equation for the time of striking: jg!‘ — u..(ms 0 tan tr + sin 6): + h == 0. Or.
using the addition formula
sin(0+nr)=sin8oosa+cos0sinor.
1 : _ "H ~ =
:31 [cTasin(8+a)]!+h 0 (1)
For a minimum t, we must have dlldfl = U. Diiferentiating (l) with respect to 9 and
setting dt/d8 = 0, we
obtain
-[lees (0 + a)]t,,,,, = 0
cos a
which implies that (since I,,,,, #0) oos (0 + or) = 0, or 0 =29: — a.
This result means that the projectile should be aimed in the direction of minimum
distance. just as though
the acceleration of gravity did not exist. However, gravity cannot be ignored in
this problem. lt we seek to
determine the value of1,.,.,, by substituting 6 + a = 90‘ into (1) and solving, we
obtain
I =u,—Vu,,1—2ghoos!a
ml" gcosa
AAJ K/-\ TOPPER li'l§Ei°§ill'§gJ'E'épR'li’§
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86 5‘
5.37
5.38
5.39
5.40‘
CHAPTER 5
which is complex if 11., < \/Zgh cos a. In other words. if u,, < \/Zgh cos a. the
projectile never reaches the roof.
whatever the value of 6, and the concept of a minimum time becomes meaningless.
With reference to Fig. 5-18. the projectile is fired with an initial velocity v,, =
35 m/s at an angle 6 = 23°. ‘Hie
truck is moving along X with a constant speed of l5 m/s. At the instant the
projectile is fired. the back of the
tntclt is at x = 45 nt. Find the time for the projectile to strike the hack of the
tntclt. if the tmclt is very tall.
I ln this case. the projectile hits the back of the truck at the moment of
overtaking it, which is the moment
at which the distance of the back of the tntclt, x, = 45 + 151, equals the
horizontal distance of the projectile.
x = (vn oos 0)! = 32.221.
45
“"5 '-Tarn-%
Y I5 m/\
____>
W Truck
. T
\. ll .
Hg. 5-18
What will happen if the truck of Prob. 5.37 is only 2.0m tall?
/
I At t = 2.614 s, when the projectile overtakes the back of the truck, its height
is, noting v,, sin 9 =
13.67 m/s: y = (l3.67)(2.6l4) - i(9.8)(2.6l4)’ = 2.25m. i.e.. 25cm above the top of
the truck. Since the
projectile travels faster horizontally than does the truck, it is clear that
thereafter the projectile remains ahead
of the back of the truck, and so never hits the back.
The projectile will reach (for the second time) a height of 2 m in a total time I,
given by 2 =
(1167):, — l(9.8)r§. or I; = 2.635 s. that is. 2.635 — 2.614 = 0.021 s after
overtaking the back oi the truck. Thus
the projectile hits the top of the truck of a distance of (32.22 — l5)(0.02l) =
0.36m = g in front of the rear
edge.
Relerring to Pr0bt 5-37. find a value of v<.. all other conditions remaining the
same. for which the projectile
hits the truck at y = 3 m.
I The time taken to overtake the back of the truck is given by
5
-1S+l5t=(vncos0)t or |=?v°o:9_|5
45 45 ‘
- » . S = - _ 1 : = - 2 _ 1 ____i
at which time , 3 (u..stn 0)| ,(9.s)1 (t1.,stn e)(mos 8 _ 5) ,(9_s)(Mos 6 _ ls)
Inserting the numerical values of sin 8 and cos 8. we obtain the following
quadratic equation for u..:
11514-55) ' v..(60.3) — 3532 = 0, Solving, v,. = L5.
The motion of a particle in the XY plane is given by x = 25 + 6!’: y = -50 — 20! +
8!‘. Find the following
initial values: x.,, y.,, .i.,, j?.,, u.,.
I At !=0, x =x,,=25m; y =y.,=—50m.
v, ii = 121 and v.,, =i.,= llmls u_, =9 = -20 + l6l and u.,, =y,,= -20 m/s
v,, = (vfi, + u§,,)"" = 20 mls
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MOTION IN A PLANE I E7 87
5.41‘ Find magnitude and direction of a. the acceleration of the particle in Prob.
5.40.
I a, = 1), = 12 mls’; a, = ii, = 16 m/s’.
= it = 19 .= 5
tan 0,, al 12 3
implies that 9, = §_3f above the positive x axis. n = (12: + 162)“ = 20 mlsl.
5.42 Write an equation for the particle's path (find y as a function of x) in Prob.
5.40.
I We eliminate time between the two equations as follows: I’ = (x — 25)/6, or I =
[(x — 25)/6]“. Then
substitltéting into the y equation: y = -50 — 20[(x — 25)/6]"' + 8(x — 25)l6, or 6y
= -500 + 8x — l20[(x —
25)/6] .
5.43‘ A particle moving in the XY plane has X and Yoomponents of velocity given by
i=b.+c,i y=b,+c,r (1)
where x and y are measured in meters and t in seconds. (a) What are the units and
dimensions of the
constants b, and b,? of c, and C2? (b) Integrate the above relations to obtain x
and y as functions of time.
(c) Denoting total acceleration as a and total velocity as v, find expressions for
the magnitude and direction of
a and of v. (ii) Write v in terms of the unit vectors.
I (a) Inspection of (1) shows that bl and b; must represent velocities in meters
per second; unit-
dimensionally c, and c, must be lm/s‘|, thus accelerations.
(bi x =x,,+ b,1+ §c,:‘ y =y,,+ b,r + it-,:’
where XIII Yo are the values of x and y at != O.
(c) Differentiating (1) with respect to r, i =v,, y =c,. Then
a=(i"+ji')'”=(cf+c§)'° tana=‘¥=Q
x c,
where a is the angle 1 makes with X. Note that a is constant in magnitude and
direction. For the velocity,
u = (i1 + y"')"* = [(0, + ¢,r)= + (b, + ¢,|)*]'” tan 5 = g
where Ii is the angle v makes with X. (d) v = (b, + c,t)i+ (b, + c,1)j.
$.44‘ A particle moves in the XY plane along the path given by y = 10 + 31 + Sx‘.
The X component of velocity,
X =4 mls, is constant, and at r = 0, x =x,,= 6m. (a) Write y and x as functions of
I. (I2) Find ji and it‘, the
components of acceleration of the particle.
I (0) y = 3: + tau; i =4 m/S at all times and X, - ts tn. Then
x =x,.+il=6m +(4m/s)l.
y =10 + (18 + 12!) + 5(6 + 4!)’ = 208 m + (252 m/s)t + (80 m/s‘)|‘
(0) i=0 y=160m[s’
5.45 A ball, 8,, is fired upward from the origin of X, Y with initial velocity v, =
ll!) m/s at an angle 0, =40‘.
After I, = 10 s, as can easily be shown, the ball is at point P(x,, y,), where x, =
766.0 m, y, = 152.8 m. Some
time later. another ball, B,, is fired upward, also from the origin, with velodty u,
at angle 0, = 35“. (n) Find a
value of v, such that B, will pass through the point P(x,, y,). (b) Find the time
when B, must be fired in
order that the two balls will collide at P(1,, y,).
I (a) Let (xl, y|, 1,) refer to the coordinates and time of B, and (X2, y,, 1,) to
those of 8,. Since B, is to pass
through P(x., yl).
:1 = (u,cos35°)r, = 766.0 y; = (v, sin 35°)!, — 4.91% = 152.8
Eliminating 1,, 766 O 2
152.8 - (166.0) l6l1 as‘ - (4.9)(i')
v, cos 35°
from which u, = 105.69 mls.
Th L ' A F
AAJ KA TOPPER NEeETe:lli:I|lggJEEpR:Sr
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88 I
s.4o'
5.47‘
5.48
5.49‘
CHAPTER 5
(b) lnmrting the value of u, in x; = (tr; cos 3S°)l1 = 766.0 m, we find I; = 8.848
s. Hence. with ti, = 105.69 m/s
and 9, = 35°. B, passes through P(x.. y,) 8.8-48$ after it is fired. But B, arrives
at this point 10s after starting
llence. if the two are to collide. the firing of B; must be delayed l0 — 8.848 =
l.l52 s.
The motion of at particle in the XY plane is given by
x=l04-121-201’ y=25+15/+301’
Find values of 1... .i.,; yin y., and the magnitude and direction of Vq.
| At I = U. x =x,, = ll); y =y.t = Z5. Differentiating we get :2 =12 — 401; y = l5
+ 60:; in = 12; yy, =15.
1'.» = (m + fill": = l9.2 9., = Ian '?'= 5l.3° above positive x axis
ll
Referring to Prob. 5.46. find and a. the vector acceleration.
I We differentiate the expressions for i and y to get X = -40, = 60. Acceleration
is thus constant and
a = V40’ + ht)’ = Q. The direction of I is given by
60
B = tan ' =56.3° above the negative x axis
Referring to Pmbs. 5.46 and 5.47. state whether or not the motion is along a
straight line.
I No: Since the lines of v,, and a do not coincide the path must be p3l'3b0liC, as
in projectile motion. See
Prob. 5.57.
Can the motion oi a particle be given by
x=5+i()r+l7r‘+-tr‘ y=8+91+20t°—6r‘
if the particle is acted on by a constant ftirce‘?
I i = 10+ 341 + 121:; _i" =9+ 40: - 181-’; ii = 34 + 24r; ji =40— 36:. But the
acceleration is not constant. as it
must be for a constant force. Hence. the motion is impossible.
5.2 RELATIVE MOTION
5.50
5.5l
An elevator is moving upward at a constant speed of 4 m/s. A light bulb falls out
of a socket in the ceiling of
the elevator. A man in the building watching the cage sees the bulb rise for
(4/9.8)s and then fall for
(4/9.8) s; at t= (4/9.8) s the bulb appears to be at rest to the man. Compute the
velocity of the bulb at
r = (4/9.8) s from the view of an observer in the elevator.
I Taking up as positive v,,,,,,,,,,v = i',,_,,,,_m — 1-,,“ W‘ = 0 — 4 = -4 mls.
Alternatively. in the inertial frame of
the elevator.
U = i-,, + ll! = 0+ (-9.s)(9‘—8) = -4 mls
A ship is traveling due east at 10 km/h. What must be the speed of a second ship
heading 30° east of north if
it is always due north from the first ship?
"' \
‘i
\\ l
‘i
.< fig. s-19
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5.52
5.53
$.54
MOTION IN A PLANE I 1' 89
I v, = velocity of first ship relative to the earth; v; = velocity of second ship
relative to the earth. Let
v,, = relative velocity of second ship to first ship. Then V3 = v1, + v,. (Fig. 5-
l9). where v1, is due north. Thus
tr, sin 30° = u, =10 km/h. and 0, = 20 kmlh.
During a rainstonn. raindrops are observed to be striking the ground at an angle of
35° with the vertical. The
wind speed is 4.5 m/s. Assuming that the horizontal velocity component of the
raindrops is the same as the
speed of the air. what is the vertical velocity component of the raindrops‘? What
is their speed?
| Let v = velocity of raindrops relative to earth. v,_ = wind velocity. t'_. = 4.5
m/s. tt, = u_.. From Fig. 5-20.
u, =%=6.43m(s v=Vuf+uf=7.84m[s
‘T’ l
.-_ =-,_ Fig. s-20
A rowboat is pointing perpendicular to the hank of a river. The rower can propel
the boat with a speed of
3.0 m/s with respect to the water. The river has a current of 4.0m/s. (a) Construct
a diagram in which the two
velocities are represented as vectors. (bl Find the vector which represents the
boat‘s velocity with respect to
the shore. (c) At what angle is this vector inclined to the direction in which the
boat is pointing? What is the
boat's speed with respect to the launch point? (ti) ll the river is 100m widc,
detemiine how far downstream
of the launch point the rowboat is when it reaches the opposite bank.
-.-. ¢\~\.»\!~ ~..~ Dimfinn v.,,,
.»~»- ,_ ,5 i v,,
til Iltm
“MW MM
I
'-
ti-» it-> Fig. s-21
I (a) See Fig. 5-21(1).
lb, cl The velocit with res ct to the shore is given by v,,_, = v, + v,,. Since v,
and v, are perpendicular‘ we
have um = Vv§,+ vi = V3‘ + 4" = Sig. The angle Q: shown in Figr S-2l(b) is
determined by tan tp = ti,/t',,.
For the speeds given. we find tp =5_3.L’. The boat moves along a line directed 53.l°
downstream from
“straight across.“
id) Letting D = distance downstream. we have D/100m = v,/v,, = 4/3, so that D = in
A swimmer can swim at a speed ol‘ 0.70 m/s with respect to the water. She wants to
cross il river which is
50m wide and has a current of 0.50 m/s. ta) If she wishes to land on the other bank
at a point directly across
the river from her starting point. in what direction must shc swim? How rapidly
will she increase her distance
from the near bank’! How long will it take her to cross‘! (b) ll‘. she instead
decides to cross in the shortest
possible time. in what direction must she swim? How rapidly will she increase her
distance lmm the near
bank? How long will it take her to cross? How tar downstream will she be when shc
lands?
I la) Let v, be the velocity of the current. v,,. be the velocity of the swimmer
with respect to the water. and
v, be the velocity of the swimmer with respect to the shore. Then v, = v_ + v,. For
a direct crossing. v, must
be perpendicular to v,. Therefore sin 0 = u,/tv_. as shown in Fig. 5-Z2: We are
given the values iv, = 0.50 m/s.
u_. =0r70 mls, and d = 50m; we find 0 = 45.6’ upstream from the direction "straight
across." The swimmer
will increase her distance from the near shore at the rate v, = tr, cos 6 = 0.49
m/s. She will cross the river in a
time t =4/v, = l02s.
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90 J CHAPTER 5
v_
rl v, \ .
4 .
Hg 5-22
(bl To maximize the component of her velocity perpendicular to the river bank. the
swimmer should head
straight across the stream. She will cross in a time r = d/v,,. = 7l.4 s. and she
will land a distance
v,1 = (v,/u._)d = 35.7 m downstream from her starting point.
5.55 An armored car 2 m long and 3 m wide is moving at 13 m/s when a bullet hits it
in a direction making an
angle arctan (3/4) with the car as seen from the street (Fig. 5-230). The bullet
enters one edge of the car at
the comer and passes out at the diagonally opposite comer, Neglecting any
interaction between bullet and
car. find the time for the bullet to cross the car.
I Call the speed of the bullet V. Because of the motion of the car. the velocity of
the bullet relative to the
car in the direction of the length of the car is V cos 9 - 13. and the velocity in
the direction oi the width of
the car is V sin 6 (Fig. S-23b). Then, from s = vi.
2=(Vc0s0—l3)I 3=(Vsin6)|
Eliminate V to find
1 3 ' Z
“E (m"2l-5-°—*~‘5
l'—3 "\—'1
it-= I3 m/s /
3 m i.—> /
/ / \
/ /
,4- -mu-(3/4) "°°‘°' *3
I
/
v /
ta) <1») I71‘. 5.1;
5.56 Rain. pouring down at an angle a with the vertical, has a constant speed of 10
m/s. A woman runs against the
rain with a speed of 8m/s and sees that the rain makes an angle B with the
vertical. Find the relation between
ix and 5.
I From the vector diagram, Fig. S-2,4.
“mg = u_,,,,,, + v,,,,, sin at: 8 + l0sin or
u,_,, cos a 10 cos a
B a
D.____
vnlltlimtlln
W-
t--..___"' Fig.5-IA
AAJ K/-\ TQPPER Ii'E°E§°§|i?i'§gJ'E'EpR':I’§

5.57 Verity that the trajectory of Prob, 5.46 is a parabola by choosing coordinate
axes parallel and perpendicular
to the constant acceleration vector.
V
.\"‘ '
\
\
,,‘ -- so
§"'
.
I
-tn"
n’ \
e’ \
\ I-13.5-2.5
Q:
\
' \
‘X
§\ <
\=
\
I Figure 5-25 shows the new coordinate system; from analytic geometry we have the
relations
3
x'=xoos0+ysin9=\/Tx+\%y y'=—xsin0+ycos8=—\/—€_:;x+;/—:i_—3-y
Hence the equations of motion in the primed coordinates are
\/T3}'=3(to+12t-2ni=)+z(2s+1si+a0t*)=a0+6et (1)
\/Ely’ = —2(l0 +12! — 20¢’) + 3(25 + 151+ 301’) = 55 + 21! +130? (2)
We can now solve (1) for I in tenns of x’ and substitute the result in (2).
obtaining an equation
y'=ax“+bx‘+c (3)
for definite constants n. b. c. Finally. completing the square on the right of (3).
we transform it to
y' — B = Kw — tr)’ <4)
which is recognized as the equation of a parabola. with vertex at x’ = a. y’ = B
and with axis parallel to the y’
axis.
5.58 Observer 0 drops a stone from the thirtieth floor of a skyscraper. Observer 0',
descending in an elevator at
constant speed V = 5.0 m/s. passes the thirtieth floor just as the stone is
released. At the time r= 3.0s after
the stone is dropped. find the position. the velocity. and the acceleration of the
stone relative to 0. Then find
the position. the velocity. and the acceleration of the stone relative to 0'.
I For 0, the position of the stone is given by
— + 1 + i:
x - x., U, 2
where x = 0 at the thirtieth floor with the downward direction as the positive x
direction. Thus. at r = 3.0s,
,=0+0+ =+Mm
Also. u =v., + fll gives u'= 0 + 9.8 m/s’ X 3.0; = +29 m/S. '
The acceleration of a freely falling body. as seen by the observer 0 who is
stationary with respect to the
earth. is known to be the constant gravitational acceleration. (Indeed, this
underlies the validity of the two
calculations immediately above.) You thus have a = +3 = +9.8 m/s’.
0' measures position x‘, related to x via x’ =x — V1. Hence. after 3.0 s. x’ = 44 m
— 5.0 m/s X 3.0 s =
+29 m. That is. the stone is located 29 m below observer 0' at the end of 3.0 s.
The stone's velodty relative
to O‘ is v‘ =v — V; hence. at r =3.0s.
u’ == 29 m/s - 5.0m/s = +24 m/s
Since V is constant, a’ == a. and a’ I +3 = +9.8 m/s’. Observer 0’ sees the stone
to have the same downward
acceleration as that seen by 0. (In general. accelerations are the same in all
inertial frames.)
AAJ K/-\ TOPPER ii'E°Ei°§i'ii'§gJ'E'EpR':I§
92:7
5.5!
5.60‘
5.61‘
CHAPTER 5
A truck is traveling due north and descending a 10 percent grade (angle of slope =
tan" 0. l0 = 5.7‘) at a
constant speed of 90 km/h. At the base of the hill there is a gentle curve, and
beyond that the road is level
and heads 30" east of the north. A southbound police car with a radar unit is
traveling at 80km/ h along the
level road at the base of the hill, approaching the tniclt. What is the velocity
vector of the tntck with respect
to the police car’!
I We use coordinate axes with 2 east, y north. and 2 vertically upward. We let v1
be the velocity of the
truck with respect to the ground and V, be the velocity of the police car with
respect to the ground.
According to the information given. we have
v, = 02 + (90 km/h)(cos 5.7°)9 — (90 km/h)(sin 5.7")2 and
1, = (—80km/h)(sin 30°)i - (80 km/h)(cos 30°)? + (1!
Reducing these, we find v, = (89.6? — 8.942) km/h and v,. = (—4(l.0i‘t — 693$) kmlh.
The velocity I1 of the
truck with respect to the police car is given by
I1 = v, — v, = (4t).0l + 158.8? - 8.9i) km/h
A bird, in level flight at constant acceleration an relative to the ground frame X,
Y (Fig. S-26), lets fall a
womt from its beak. What is the path of the worm, as seen by the bird?
I ln the hird‘s noninertial coordinate system X ', Y’ (Fig. 5-26). the equation of
motion of the wonn is
M E “ III "I Of L’ 1 In ' COIISIQIII
dt ‘ '° di g
Thus the acceleration of the worm is constant, and its path is a straight line
(supposing that it was dropped
from rest). The slope of the line with respect to the horizontal is tan 0 = g/a,,.
Y
Y.
/
—mm/ an
IO
/ 0 x'
° X Flg. s-24
Refer to Prob. 5.60 and Fig. 5-26. (I) Determine the path of the worm as seen from
the ground. (b) Verify
that the two desaiptions of the path are equivalent.
I (4) In the ground frame X, Y, the wonn has constant acceleration Y = —g and an
initial velocity io= vu.
where v,, is the speed of the bird at the instant the worm is released (call this
time t= 0). Hence
X=»\'o*‘Un| )'=Y0—i87z (1)
and the path is a parabola.
(b) Let us suppose that at 1 = 0 the two coordinate frames coincide. At time I. O’
will have advanced a
distance vol + loot‘ along the X axis. so that the coordinates (x. y) and (x', y’)
of the worm in the two systems
are related by
x=x'+(v,,r+§a,,l') y=y' (2)
The path in the X ‘, Y’ system is obtained by substituting the expressions (2) into
(1):
1' +M+ l¢<>r’=x»+vrf y‘ =n— ix"
or
.v' “Yo 8
x’ — m an
which is a straight line of slope g/an, as found in Prob. 5.60.
AAJ K/-\ TQPPER Ii'E°Ei°§i?i'§gJ'E'EpR':I§

Z
MOTION IN A PLANE I U 93
N I Y s
I
x Hg. 5-27
5.62 A helicopter is trying to land on a submarine deck which is moving south at I7
m/s. A 12 m/s wind is blowing
into the west. If to the submarine crew the helicopter is descending vertically at
5 m/s. what is its speed
(1) relative to the water and (la) relative to the air. See Fig. 5-27.
I (I) '»..»~..m = VIUBIWIIH 4‘ Vt.»-» = 17.i 4' (-5)|l =(17i - 5*) m/S
lb) '»ev-4| = vlkl/Wilt! + V-..m.r = "M---,. - Y-..~.m = (171 - 55) - |3| = (-12! +
17] - 5|!) m/S
A/-\J K/-\ TOPPER I|'E°$°§|'|'J|§gJ§EpR§§’§
L‘A°E#°§.1?l2“.é‘é"R'i{’;
CHAPTER 6
Motion in a Plane II
6.1 CIRCULAR MOTION; CENTRIPETAL FORCE
6.1 A 0.3-kg mass attached to a 1.5 m-long string is whirled around in a horizontal
circle at ti speed of 6 m/s. (0)
What is the centripetal acceleration of the mass? (la) What is the tension in the
string? (Neglect gravity.)
2 Z
(fiml)
I (u) '1=%=-l—'%=@./_Sf
(b) The tension in the string exerts the centripetal foroe required to keep the
mass in circular motion. This
loroe is T = ma = (0.3 kg)(24 m/s‘) =7.2 N.
6.2 A small ball is fastened to a string 24 cm long and suspended from a fixed point
P to make a conical
pendulum, as shown in Fig. 6-1. The ball describes a horizontal circle about a
center vertically under point P,
and the string makes an angle of 15° with the vertical. Find the speed of the ball.
P
| T
|15°
/’_+~\
\ f m
\___v
I
I
ms nan
2 I
I Tcos lS°=mg Tsin 15%? and hence tan 15°-:2
Since r = 24 sin 15° = 24(0.259) = 6.22 cm,
B U1
tan l5 -622680) v - 40.4gg[§
6.3 ln the Bohr model of the hydrogen atom an electron is pictured rotating in a
circle (with a radius of
0.5 >4 10"" m) about the positive nucleus of the atom. The centripetal force is
fumished by the electric
attraction of the positive nucleus for the negative electron. How large is this
force it the electron is moving
with a speed of 2.3 X 10° mls? (The mass of an electron is 9 X 10"‘ kg.)
I Force = (9 X 10*” kg)(2.3 X l0°m/s)’/(5.0 X 10-" m) =9.§ X 1Q"N
6.4 Find the maximum speed with which an automobile can round a curve of 80-m
radius without slipping ii the
road is unbanked and the coelficient oi friction between the road and the tires is
0.81.
I First draw a diagram showing the forces (Fig. 6-2). If mg is the weight of the
automobile, then the nonnal
force is N = mg. The frictional force supplies the centripetal force E.
E = ;i,N =0.8l mg
2 Z
Mm. n=§§ um»@=%% w=o3|xwxas=u2mp
AA-I K/-\ i0PPER I.?.i.#*:.1?li:.“.’.;'.;“Ri(’.§

MOTION IN A PLANE ll 5 95
N
i E'|“sN
Li,
l
"'9 Fig. (>2
6.5 What is the maximum velocity. in miles per hour, for an automobile rounding a
level curve of 200-ft radius if
u, between tires and roadbed is 1.0.
I [4,mg==E ,1, =5 v = \/)|,gr= \/t.o><32.2n/§’><2o0n=s0n/5-son/3><a=.&;muh
r lg 44 fl/S
6.6 A car is traveling 25 mls (56 mi/h) around a level curve of radius l20 m. What
is the minimum value of the
coeflicient of static friction between the tires and the road required to prevent
the car from skidding?
iv‘ iv‘ (25 m/s 1
_ __ ___i)__,
I F mr_ES“'mg "’2gr_(9.8m/s')(l20m) (i
6.7 Tralfic is expected to move around a curve of radius 2(1) m at 90 km/h. What
should be the value of the
banking angle if no dependence is to be placed on friction?
tr
V
Fig. 6-3
I See!-‘ig.6-34 w=mg=Ncos0 and fi="~l;U-=Nsin0
Dividing the second by the first equation. tan B = iv’/rg. Substitute the data.
changing kilometers per hour
into meters per second:
lift 0= =0.3t9 so e= t7.7"
(90 km/h X 1000 m/km)’
6.8 As indicated in Fig. 6»-l, a plane flying at constant speed is banked at angle 9
in order to fly in a horizontal
circle of radius r. The aerodynamic lift force acts generally upward at right
angles to the plane's wings and
fuselage. This lift force corresponds to the tension provided by the string in a
conical pendulum. or the
normal force of a banked road. ta) Obtain the equation for the required banking
angle 8 in terms of u, r.
and g. (b) What is the required angle for u = 60m/s (216 km/h) and r = l.0km?
i ii,
_ __ _ _ _ _ I,‘ _ _
Ir I-‘lg. 64
I (a) As in Probs. 0.2 and 0.7, tan 0 = 1-=/,3.
m= 0
"" " " ""' -Ii
A/-\J KA TOPPER NEET AIIMSJEE RAS
AAJ K/-\ TQPPER ii'EeE'ie§|i'ii§gJ'§EpR'i§’§

96 U
6.9
6. 10
6.11
6.12
6.13
6.14
CHAPTER 6
A car goes around a curve of radius 48 m. lf the road is banked at an angle of I5’
with the horizontal. at what
maximum speed in kilometers per hour may the car travel if there is to be no
tendency to skid even on very
slippery pavement?
I Use the equation for banking of highways:
u: Q 0‘
tan8-E tanl5 =4m
v = 0.Z68(48)(9.8) = 11.2 m/s = (ll.2 m/s)(0.wl km/m)(3600 slh) = 40.3 ltm(h
A certain car of mass m has a maximum frictional foroe of 0.7 mg between it and
pavement as it rounds a
curve on a flat road (14 = 0.7). How fast can the car be moving if it is to
successfully negotiate a curve of lS-m
radius?
I The centripetal force (mu:/r) must be supplied by the frictional force. ln the
limiting case. mu‘/r =f with
f = 0.7mg. Thus. tr‘ = 0.7rg and v = 10 mls.
A crate sits on the lloor of a boxcar. The coeffieient of friction between the crate
and the floor is 0.6. What is
the maximum speed that the boxcar can go around a curve of radius 200 m without
causing the crate to slide?
I As in other “unbanked-curve" problems (e.g., Prob. 6.5).
ufm = ;4,gr = (0.6)(9.8 mls1)(2(X) m) = 1l76 mils‘ v,,,,, = V1176 ml/s‘ = 15.3 mls
A boy on a bicycle pedals around a circle of Z2-m radius at a speed ol I0 m/s. The
combined mass of the boy
and the bicycle is 80 kg. (a) What is the centripetal force exerted by the pavement
on the bicycle? (b) What is
the upward force exened by the pavement on the bicycle? See Fig. 6-5.
I <¢>r..="'—,":=§’%1=s¢><~ tb>~=»rg=wm>=zm1.
.C
4
..%'
m=80kg
r=22m
mg I-‘lg. 6-S
Refer to Prob. 6.12. What is the angle that the bicycle makes with the venical?
I For the bicycle not to fall. the torque about the center 0! gravity must be zero-
—see Chaps. 9 and
1()—which means that the vector force exerted by the ground must have a line of
action passing through the
center of gravity. Thus
ta»o=§=%=o.ms a=E
A fly of mass 0.2g sits 12cm from the center of a phonograph record revolving at 331
rpm. la) What is the
AAJ K/-\ TOPPER ii'EeEi°§|i'ii'§gJ'E'é°R'ii§

MOTION IN A PLANE ll L7 97
magnitude of the centripetal force on the fly? (b) What is the minimum value of the
coefficient of static
friction between the fly and the record required to prevent the fly from sliding
off"!
2 as-as ' s‘
I (0) v=-fl’-_5=zufr=2ii(-5s7'£i1'n-)(1zx1o ‘rn)=0.419m/s
: s 2
f=,,m=m1= =2_92X 1(]*4N
r 0.12m “T
2.92 x 10"N
r = .9 0-‘ _ =
(bi 1 2 1 *1 N < runs u. 2(0_2 X10-_ik8)(9‘8m/5:) 4%
6.15 Find (a) the speed and (b) the period of a spaceship orbiting around the moon.
The moon's radius is
l.74 X 10° m, and the acceleration due to gravity on the moon is 1.63 m/s’. (Assume
that the spaceship is
orbiting just above the moon's surface.)
I (a) = I g,., v = Vg,,,R,, = V(l.63 m/s‘)(1.74 X 10° m) = 1.68 X ll)’ mls = Lfi
km[s
(Ii) The circumference of the orbit is
d = 2nR,,, = (6.28)(l.74 X 10" m) = 1.09 X 10‘ km
so the period is
d l.09XlO‘ltm
=-=a.=_ 0-‘-= '
I u l.68km/s 65x1 s T
6.16 At the equator. the efiective value of g is smaller than at the poles. One
reason for this is the centripetal
acceleration due to the earth's rotation. The magnitude of the centripetal
acceleration must be subtracted
from the magnitude of the acceleration due purely to gravity in order to obtain the
eflective value of g.
(n) Calculate the fractional diminution of g at the equator as a result of the
earth’s rotation. Express your
result as a percentage. (b) How short would the earth's period of rotation have to
be in order for objects at
the equator to be "weightless" (that is, in order for the effective value of g to
be zero)? (c) How would the
period found in part (b) compare with that of a satellite skimming the surface of
an airless earth?
| (0) Using R, = 637 X l0"m and T = 24 h = 86400 s. we find a = U:/R, = ‘MIR,/T’ =
3.37 >< l0": m/$2.
Therefore a/g = 3.44 x ll) ‘. Since g, =g — u, the fractional dimunition is (3 —
g,,)/g = alg =0}-$4 rcent.
(b) In order that g,,,=0. we need u :3 = 4:r‘R,/Ti. Solving for T,, we find T, =
2:rVR,/g = 5.06 X lg s =
84.4 min. (e) Since an orbiting satellite has ma = mg and g‘,,=0. its period equals
T,.
6.17 A particle is to slide along the horizontal circular path on the inside of the
funnel shown in Fig. 6-6. The
surface of the funnel is frictionless. How fast must the particle be moving (in
terms of r and 0) if it is to
execute this motion?
w q-I: "’~\
\/
‘St '~ - - - -
\‘\
\
\
\\ //
F
‘.
if-'
i Fig.6-6
AAJ KA TOPPER li'i§Ei°§i?i'§”J'|§E”R'§i’§

98 U CHAPTER 6
I The funnel is e uivalent to a road banked at an angle 90° — 9. Hence ti‘/rg = tan
(90°— 6) =
cot 8. or u = Vrg cot 0.
6.18 An automobile moves around a curve of radius 300 m at a constant speed of 60
m/s [Fig. 67(0)]. (I)
Calculate the resultant change in velocity (magnitude and direction) when the car
goes around the arc of 60°.
(b) Compare the magnitude of the instantaneous acceleration of the car to the
magnitude of the average
acceleration over the 60° arc.
I (a) From Fig. 6-7(b), Av =QQ_l't__l§ and Av makes a 120" ange with V4. (D) The
instantaneous acceleration
has magnitude
2
,=%=%,.,;m(,1
The time average acceleration is i -= Av/Ar. Sinoe
lug): 300(:r/3) _5)r
A’ u u so ' 3 ‘
A 60
wehave ii=X7=%=i1iz
v, 160m/s
v. =60m/s F "‘
\’$f
r= Jlllm
>1
<0 (b)
(G) rt; a-1
6.19 While driving around a curve of 200 m radius. an engineer notes that a
pendulum in the car hangs at an angle
of 15° to the vertical. What should the speedometer read (in kilometers per hour)?
1 Z
I Tsin6=%:Tcos0=mg\vhere T=tension.Thustan0=% or u=-Vrgtan0-23m/s-82-$ltm[h
6.3) The bug shown in Fig. 6-8(a) has just lost its footing near the top of the
stationary bowling ball. lt slides
down the ball without appredable friction. Show that it will leave the surface of
the ball at the angle
0==arcoos§=~48°.
\
_
6/s
\\ _~ \
I w
\
\
K] / mg
lu) lh) 6-8
AAJ K/-\ TQPPER Ii'E°Ei°§|i?i'§gJ'E'EpR':I’§
m)“">
\
=2: \

MOTION IN A PLANE ll S’ 99
I The centripetal force is given by
mu:/r=mgcos0-F,‘ (1)
At angle 8. the decrease in potential energy. mgh = mgr(l - cos 0). must equal the
increase in kinetic energy,
mu:/2; hence.
mu’
-T=2mg(l—oos6) (Z)
Together. (I) and (2) give
3mg oos 0 — Ee = Zmg (3)
At the instant the bug loses contact with the ball. EN = 0 and (3) yields cos 6 =
i.
6.21 A I80-lb pilot is executing a vertical loop of radius 20(1) ft at 350 mi/h.
With what force does the seat press
upward against him at the bottom of the loop?
3 2
I F—mg=fl or F=fl+mg
I I
First we change miles per hour to feet per second:
350 mi/h = 513 ft/s
Substitute values:
_1a0it>><(s13n/5;‘ _
F ‘ 32.2 ftls‘ >< 2000 n + '8“ lb ' L lb
6.22 How many g‘s must the pilot of the preceding problem withstand at the bottom
of the loop?
I From the centripetal acceleration formula a, = v‘/r. Substituting.
S13 * 1
Dividing this result by g (32.2 ft/5‘). we obtain
,, =i/*’_,,
' 32.20/§"—5
ls
6.23 The designer of a roller coaster wishes the riders to experience
"weightlessness" as they round the top of one
hill‘ How fast must the car be going if the radius of curvature at the hilltop is
20m?
| To experience weightlessness. the gravitational force mg must exactly equal the
required centripetal force
mu‘/r. Equating the two and solving for v gives l4 m[s.
6.14 A huge pendulum consists of a 2(1)-kg ball at the end of a cable 15 m long. It
the pendulum is drawn back to
an angle of 37“ and released. what maximum force must the cable withstand as the
pendulum swings back and
forth?
l The maximum tension will occur at the bottom when the cable must fumish a force
mg + mu’/r. To reach
the bottom. the mass falls a distance h = (l5 — l5cos 37") = 3.0 m. Its speed there
will be u = (2gh)"’ = (6g)"'=.
Therefore the tension will be T = 200g + 200(6g)/15 = 2740 N.
6.2 LAW OF UNIVERSAL GRAVITATION; SATELLITE MOTION
6.25 Two wlb shot spheres (as used in track meets) are held 2 ft apart. What is the
lnrce of attraction between
them’!
I ln American engineering units. l6lb=>0.497 slug and Newton's law of gravitation
has the form
F= G[(m|m;)/dz], with G = 3.44 X 10 ‘lb - ft:/slug‘. Thus.
r = (3.44 X10 ' “’+i‘;‘,:)(°ii"97 5'": X $497 "“g) = 2.12 >< to ° lb
slu 2ft Z

1 O0
6.26
6.27
6-Z8
6.3
6.30
6.31
6.32
T7
L;
CHAPTER 6
Calculate the force of attraction between two 90-kg spheres of metal spaced so that
their centers are 40cm
apart.
I In SI units the gravitational constant has the value G = 6.67 X 10'" N - m‘/kg’.
_ 9090 _,
F=G%=(6.67x10")w_%)1=3.3X10 N
Compute the mass of the earth, assuming it to be a sphere of radius 6370 km.
I Let M be the mass of the earth. and m the mass of a certain object on the earth's
surface. The weight of
the object is equal to mg. It is also equal to the gravitational force G(Mm)/r’.
where r is the earth's radius.
Hence. mg = G[Mm/r*]. from which
* (9.8 /’)(o.37><10° )1 ,
”=%= *°*=-°*'“*
The average density of solids near the surface of the earth is p = 4 X 10’ kg/m‘.
On the (crude) assumption of
a spherical planet of uniform density p. calculate the gravitational constant G.
I The mass of the spherical earth is given by m, = pV I §npRf. lnsert this value
into g = Gm,/Rf and solve
for G. obtaining:
_ 3g 3X9.8m/s‘ _ ,, , =
G—4!tpR,. 4:rx4><l0‘kg/m’X6.4x10‘m_9Xm N m/k '
This calculation most certainly overestimates G, since p (and hence m,) are
underestimates.
A mass m, = 1 kg weighs one-sixth as much on the surface of the moon as on the
earth. Calculate the mass
m, of the moon. The radius of the moon is 1.738 X 10° m.
I On the moon. m, weighs I-,(9.8 N).
_ "hm: 1 _ -ll hmz
w,—G ,2 6(9.8)-6.67xl0 (L738Xm,),
.,- ...4X.et.
The earth's radius is about 6370 km. An object that has a mass of Z0 kg is taken to
a height of 160 km above
the earth's surface. (1) What is the object's mass at this height? (b) How much
does the object weigh (i.e.,
how large a gravitational force does it experience) at this height?
I (a) The mass is the same as that on the earth's surface. (lt) As long as we are
outside the earth's surface,
the weight (force of gravity) varies inversely as the square of the distance from
the center of the earth. Indeed
w = GmM/1‘. where m, M are the masses of object and earth, respectively, and r is
the distance to the center
of the earth. Thus W;/W. =r§/r§, since G, m, M are constant in this problem. For
our case we set r, -6370 km
and r, = 6530 km and w, = (20 lrg)(9.8 mls’) = l96N. This gives w, = 1&5 N. Note
that we could use the fact
that w = mg and that m is constant to find the acceleration of gravity at the two
heights. Tltat is. g,/g, = r§/r§.
The radius of the earth is about 6370 km. while that of Mars is about 3440 km. lf
an object weighs Z00 N on
earth, what would it weigh, and what would be the acceleration due to gravity on
Mars? Mars has a mass 0.11
that of earth.
I Newton's law of gravitation, w = GmM/I’, gives w,/w, = (M,/M,)(rf/Ii). Letting 1
refer to earth and 2
refer to Mars, we have av, = 0.1l(6370/3440)'(2IJO N) =Z§l4. The acceleration is
gotten from w,/w, = g,/3,, or
g, = (75/Z00)(9.8 N) = 3.7 U115‘.
The moon orbits the earth in an approximately circular path of radius 3.8 X 10" m.
It takes about 27 days to
complete one orbit. What is the mass of the earth as obtained from these data?
I The gravitational attraction between the earth and moon provides the centripetal
force; therefore.
mu’/r = GMm/r'. where M is the earth's mass. Then M = uzr/G = w‘r’/G. Now m = 1
rev/27 days I
2.7 x l0‘° rad/s. I = 3.8 X 10' m, and G =6.7 ><10"' in SI. Solving for M. it is
6.0 X 10“ kg. (Compare Prob.
6.27.)
R NEET AIIMS JEE RAS
AAJ KA TO
MOTION IN A PLANE ll L7 101
The sun's mass is about 3.2 X l0‘ times the earth's mass. The sun is about 4(1)
times as far from the earth u
the moon is. What is the ratio of the magnitude of the pull of the sun on the moon
to that of the pull of the
earth on the moon? lt may be assumed that the sun—moon distance is constant and
equal to the sun-earth
distance.)
I Let m denote the moon's mass, M, the sun's mass. M, the earth's mass. 1,, the
center-to-center distance
from the sun to the moon. and r,,, the center-to-center distance from the earth to
the moon. We let F,
denote the magnitude of the gravitational toroe exerted on the moon by the sun, and
F, denote the
magnitude of the gravitational force exerted on the moon by the earth. Then F,“ =
GM,m/3,, and
F, = GM,m/r’,_,,, so that
5 _ M»;
F... Mai.
Using the given numerical values. we find F.../F,,,, = Z.
Estimate the size of a rocky sphere with a density of 31) g/cm’ from the surfaoe of
which you could just barely
throw away a golt ball and have it never retum. (Assume your best throw is 40m/s.)
| The escape speed 0,, from a sphere of radius R and mass M is given by the energy-
oonservation equation
lm I _ GmM
2 ”“ ' R
Substitution of M = p§:rR" and solution for R gives
__ I 3
R - tluvancp
lt p = 3 X 10’ kg/m’, then R = 0.77 X 10’ v.,, where R is in meters and 1.1., is in
meters per second. Estimating
the highest speed at which a human can throw a golf ball as about 40 m/s. we find R
~ 3 x 10‘ m = 30 lttn.
Newton. without knowledge of the numerical value of the gravitational constant G.
was nevertheless able to
calculate the ratio of the mass of the sun to the mass of any planet. provided the
planet has a moon.
ta) Show that for a circular orbits
L <fi>‘<@>*
M» R- T»
where M. is the mass of the sun. M, the mass ot the planet, R, the distance of the
planet from the sun. R,,,
the distance of the moon from the planet. T,_ the period of the moon around the
planet. and L the period of
the planet around the sun. (bl lf the planet is the earth, R, = 1.50 x 10" km. R,,,
= 3.85 x I0’ km.
T... = 27.3 days. and T, = 365.2 days. Calculate M, /M,,.
I la) Applying Ne\vton‘s second law and the law of gravitation to each orbit. we
find (expressing centripetal
lorce in terms of period. T, using tr = 2:rR/T).
4.~t‘M,,R, = GM,M,, and 42r‘mR,., = GM,m
Tj R} TL Rf,
where m is the mass of the satellite. Solving the above equations for M,/M,, we
obtain
@= (4:t‘R-Q)/GT; = (Lt£)~‘(L,)‘
M, (4:r’R1,)/GT§, R, 1;
as desired. (b) inserting the given numerical values, we find
(u) Find the orbital period of a satellite in a circular orbit ot radius r about a
spherical planet of mass M. (la)
For a low-altitude orbit (r==r,), show that for a given average planetary density
(p) the orbital period is
independent of the size of the planet.

236 U CHAPTER 12
12.“
12.41
___-[U A
il
‘.1
Hg. 12-ll
Pd = 1,00. where 1,, = fml‘)/3 is the moment of inertia of the bar about the pivot.
Since the distance from A
to C is l/2. we also have u, = (wl)/2. By combining the equations. we find
m[(wl)/2]d = [(ml‘)/3]w, which
implies that d = QIQ. [Note the reciprocity with Prob. 12.36: a blow to the end
(c.p.) causes rotation about
the c.p. (end).|
Refer to Prob. 12.39. la) What is the period of oscillation of the rod when it is
suspended from A? (b) What
is the length of the simple pendulum having the same period? The length you obtain
here should be the same
as the distance you obtained in Prob. 12.39. The center of percussion relative to A
is also called the cenler of
oscillation relative to A.
I (0) The oscillation frequency is given by (see Prob. 12.38)
V gi /(I/Z)£
" 2:: G3,
where G} = I,/m = I:/3. Therefore we obtain
T, =2:r\/gz
(bl The period of a simple pendulum of length L is given by 2.vr\/L/g. Therefore
the length of a simple
pendulum with the oscillation period T, is L = LL/__3, which is equal to the
distance found in Prob. 12.39.
A ring of mass M and radius R is hung from a knife edge, so that the ring can swing
in its own plane as a
physical pendulum. Find the period T, of small oscillations.
I The ring is shown in Fig. 12-12, with the knife edge at point A. We must find the
period T, of small
oscillations in the plane of the paper. Taking the origin of a coordinate system at
O, the equilibrium position
of the ring's center. with the positive z axis emerging toward the viewer, the
moment of inertia I," = MR’. By
the parallel-axis theorem, the moment of inertia about the knife edge is given by
1,, = l,,, + MR’ = 2MR';
hence. G}, = ZR’.
._.\
\ _~
\
\\
R \\ _
t
2
fig. I1-l2
Substituting this and D = R in Eq. (1) of Prob. l2.38, we find
1 1- 1 /2R
v,=5\/2% and T.=-;;=2.-r E-
IZAZ Refer to Prob. 12.41. (u) Suppose that an identical ring is pivoted from an
axis PP’ lying in the ring plane
Th L ' A F
A/-\J K/-\ TQPPER ~E°ET°§|i'i'|'§gJE'épRX’§
AAJ K/-\ TOPPER ii'EE'ie§|i'm'§gJ'§’épR'ii’§

11.43
12.“
B-6
ROTATIONAL MOTION ll: J 237
P .'\ I"
\ ~‘ i
k\\\
‘ A
1 \
_ ‘La.
P’:
t\t
n;. mu Fig. 11-14
and tangent to the circumference (Fig. I2-I3). This ring can execute oscillations
in and out of the plane. I-“ind
the period 7} of those small oscillations. (b) Which oscillation has the longer
period? How much longer?
I (a) Now the ring is pivoted about the axis PP‘ as shown at right. By the
parallel-axis theorem.
l,,- = I,,o+ MR’. As shown in Prob. ll.41. I = (MR2)/2. so l,-,- = (3MR’)/2. With
Gf, = 3R‘l2 and D = R.
(l) of Prob. l2.38 gives T, = l/v; = 2:r\/3R/Zg. (II) T./T; = \/{= 1.1547. The
period for oscillations in the
plane is lS.5 percent longer than the period for oscillations about the axis PP’.
A unifonn right-angle iron is hung over a thin nail so that the iron pivots freely
at the bend (Fig. l2-14). Each
arm ol the iron has mass m and length I. Find the period Tof small oscillations (in
the plane of the iron).
I The center of mass of the system is located on the angle hisector. at a distance
D = (l\/5/4) from the pivot
A. The moment of inertia I about an axis through A and perpendicular to the xy
plane is given by
(ml*)/3 + (ml’)/3 = (2mI‘)/3. ‘Therefore the square of the gyration radius is given
by G‘ E I/M =
[(2 ml‘)/3](Zm) 1-l’l3. We can now apply (1) of Prob. l2.38:
2
1 I0‘; { 1/3 /2\/ii
=_= _=3,¢ _i= _
T v 2” Be k1\/E)/41s Z” 32
Figure 12-15 represents a three-dimensional object (not necessarily of uniform
density) whose center of mass
is at point C. The axis ZCZ' passing through point C has been chosen at random
orientation. The body had
gyration radius G¢(ZZ') about the axis ZCZ': the notation G¢(ZZ') makes explicit
the fact that the gyration
radius for an axis through C depemt on the particular choice (ZZ') of axis.
Suppose that a physical pendulum is constnicted by pivoting the body about the axis
PP’. which is parallel
to ZCZ' at a distance D. Prove that the frequency v of small oscillations about
equilibrium is given by
v=i Ii
2.1 c;(zz')+0=
fa
,. ,
1 W;
§'\:_“\
J’?/~~__ '4‘ // F
I V
I
“T5;
T’!
Fig. 12.15
I The desired result follows at once from (1) of Prob. 12.38. It is seen that the
frequency of pendulum
oscillations is the same for any choice of axis PP’ on a cylinder of radius D
centered on ZZ’.
Figure l2-16 depicts the Kaur pendulum, used to measure the acceleration of gravity
with high accuracy. lt
consists of a rigid rod on which a bob is mounted. The mass of the bob is
suhiciently large so that the center
of mass of the pendulum is fairly far from the middle of the rod. The bob is
mounted on a slide; by moving
the boh and then clamping it in position. the location of the center of mass of the
pendulum can be adjusted.
There are two very precisely honed knife-edges mounted on the rod. The pendulum can
be swung from
AAJ K/-\ TOPPER ii'EeEi°§i'ii'§gJ'E'é°R':I’§

238 U CHAPTER 12
Kmle-edge I
J
i
i
DI
0, ~ 0,
,_- Center at mm
_.-
i D T
1;-i_._A Knifoedge 2
--' Slide
T;-1;
_ \_ _
/Clamp
/
-,aas
D ____’ Rod
n;. 12,14
knife-edge 1. and then reversed and swung from knife-edge 2. If G. is the gyration
radius of the pendulum
about an axis through its center of mass. what are the expressions for the periods
of small oscillations about
knife-edge 1 and knife-edge 2?
I From (1) of Prob. 12.38,
1 105+ 0% 1 /05+ 0;
1- = _ = 1,; i 7' = _ = 1,, __
' v. Dix ’ '2 Du;
12.46 Refer to Prob‘ 12.45; can T, I 1} for D, ¢D,?
I Yes. Consider Tfor an arbitrary D and fixed Ga, T = 2Jr\/(G05 + Di)/(Dg). lf we
plot T against D for D
between 0 and w. we note that T~== as D-v0 and T—>w as D->==. Thus T has a minimum
value. Ta... for
some D I D". Thus for any T > T,,,,,, there are two D values, one greater and one
less than D‘.
12.3 ANGULAR MOMENTUM
11.47 Find the rotational energy and angular momentum due to the daily rotation of
the earth about its axis. Data:
M, = 6 X l0“ kg. R, =6.-1 X l0°IIi. to =1/86-100rev/5. Assume the earth to be a
uniform sphere.
I K, = Iw’/2 =1 [(2Mr=)/S][(22r/86 4(X))’]/2 = LLB]. Angular momentum = la» =
[(2Mrz)/5](Z¢t/86 400) =
7.l X I0” kg - m‘/s. (The equivalent unit .I ' s is also used for angular momentum,
especially in the
atomic domain.)
12.4! Each of the wheels on a certain four-wheel vehicle has a mas of 30 kg and a
radius of gyration of 30cm.
When the car is going forward and the wheels are turning at 5.0 rev/s, what is the
rotational kinetic energy
stored in the four wheels? What is the angular momentum of the vehicle about an
axis parallel to the wheel
axis and through the center of mass? ls the angular momentum vector directed toward
the driver's right or
left?
I K, = 4(!w'/2) = 2(30)(O.30)'(l0:r)’ = §3(X] 1. The angular momentum vector for
each wheel is along the line
described. so total L = 4(!w) = 4(30)(0.09)(l0:r) - 340 kg - m‘/s. Note that ui
must be in radians per second.
The total L will be in the same direction as the angular velocity. that is, to the
driver's left.
12.49 ln a common physics lecture demonstrations. a lecturer sits on a stool that
can rotate freely about a vertical
axis on lmv-friction bearings. The lecturer holds with extended arms two dumbbells,
each of mass m, and
kicks the floor so as to achieve an initial angular speed 01,. The lecturer then
pulls in the dumbbells. so that
their distances from the rotation axis decrease from the initial value R, to the
final value R,. Determine the

244 U CHAPTER 12
12.67
12.68
12.60
12.70
A student volunteer is sitting stationary on a piano stool with her feet olf the
floor. The stool can tum freely
on its axle.
(0) The volunteer is handed a nonrotating bicycle wheel which has handles on the
axle. Holding the axle
venically with one hand, she grasps the rim of the wheel with the other and spins
the wheel clockwise (as
seen from above). What happens to the volunteer as she does this? (Ii) She now
grasps the ends ot‘ the
vertical axle and turns the wheel until the axle is horizontal. What happens? (e)
Next she gives the rotating
wheel to the instructor. who turns the axle until it is vertical with the wheel
rotating clockwise, as seen from
above. The instnictor now hands the wheel back to the volunteer. What happens? (d)
The volunteer grasps
the ends of the axle and turns the axle until it is horizontal. What happens now?
(e) She continues turning the
axle until it is vertical but with the wheel rotating counterclockwise as viewed
from above. What is the result?
I Since the axle of the piano stool is frictionless, there are no vertical torques
exerted on the stool-volunteer
system. so the vertical component of angular momentum is conserved. (There are
horizontal torques; these
result from forces that the floor exerts on the base of the stool.)
(a) The initial angular momentum is zero, so the final angular momentum mtist also
be zero. Therefore, the
volunteer spins counterclockwise. (ll) The angular momentum of the wheel is now
horizontal. The volunteer‘s
vertical component of angular momentum must now be zero. so she stops spinning. (c)
When the wheel is
handed back to the volunteer. the system of wheel and volunteer has a downward
vertical angular
momentum. all contributed by the wheel. The volunteer remains stationary. (J) Since
the vertical component
of the total angular momentum must not change, the volunteer must rotate clockwise.
(e) The wheel's angular
momentum is now upward. The volunteer must therefore have a downward vertical
angular momentum to
keep the total angular momentum pointing down. She must therefore spin clockwise.
(Her spin rate is twice
as fast as in part (d).)
A top consists of a uniform disk of mass mu and radius r, rigidly attached to an
axial rod of negligible mass.
The top is placed on a smooth table and set spinning about its axis of symmetry
with angular speed w,. How
much work must be done in setting the top spinning? Evaluate your result for m.,-
0.050 kg. r.,- 2.0cm, and
w, = 200:! rad/s (or 6(1)!) rotations per minute).
' The moment of inertia In of the top is given by L, = §m°r§. The work required to
set the top spinning with
angular speed cu, is equal to the spin kinetic energy §!,,wf. For the given
numerical values, we find
In = (0.50)(0.050)(2.0 X 10”)’ = 10” kg t mi. The work required is (0.5)(l0")
(200:r)’ = 1.97 J.
Refer to Prob. 12.68. The center of the disk is ii distance d from the top‘s point
of contact with the table. The
top is observed to precess steadily about the vertical axis with angular speed w,.
Assuming that w, << m,,
write aa, in terms of 1°. d, to,, and g. Evaluate cu, for d = 3.0crn and g = 9.80
m/s‘, with the other quantities
as given in Prob. l2.68. Is your result consistent with the assumption w, << w,'!
I For w, << tn, the angular speed of steady precession is approximated by
_ El
”'ma
where G.§ = L,/mu = lrfi. With the numerical values given, we find
(9.80)(3.(X) X 10-2) l
"*»=m@,0<i>W>=m=Z-’—"E“>@
The ratio of this angular speed to w, is (to,/w,) = 3.72 X 10". The assumption that
(cu,/w,)<< 1 is fulfilled.
As shown in Fig. 12-24, a solid conical top of mass M. height h, and radius R is
spinning about its symmetry
axis 00' with spin angular speed w,. The axis 00' makes an angle or with the
vertical. ‘
We note that for such a system the center of mass of the top_is located along 00:
at a distance 3h/4 from
the vertex O. and the moment of inertia I about the axis 00' is given by = foMR a
_ _
(a) Find the angular speed w, at which the top precesses about the vertical. (bl
Consider a topzfor which
It = 10.0 em and R = 3.0crn. The top is spinning at 580) rotations per minute.
Using g = 9.80 mls . evaluate
the precession angular speed 10,.
I (nl Under the assumption that m, << 0),, we can assume that
(3h/4)
=_flLJ___L=lL
""’ (I/M)w,_(3R2/l0)iu, 2R’w.
AAJ K/-\ TOPPER Il'E°Ei°§i?i'§gJ'EEpR':§’§

13.24
13.3
13.26
13.27
13.28
13.29
13.30
MATTER IN BULK 5 251
I For a force F and displacement s.
k _ f_ 13 _ (1.8 kg)(9.8 m/s’)
J .t 0 02
k - E2 Nlm
F=ks=(882 N/m)(0.05 m) F=-44.1 N m=4.5Oltg
A l0~kg mass is supported by a spring whose constant is I2 N/cm. Compute the
elongation of the spring.
I F=mg r=—=—= .t=§.l7gg1
Note that the mixed units for k need not be convened since mg in newtons cancels
the newtons in k.
*-1
*5
A load of ill) lb is applied to the lower end of a steel rod 3 ft long and 0.20in
in diameter. How much will
the rod stretch? Y = 3.3 X 10" lb/in’ for steel.
= F/A _ Q": (an)(1o01|>) _ _, _ .,.
I Y T/L or AL Ay 'e_lin,)(3’3xwqb/in,) 2.9><10 ti or AL s.s><10 in
Note that the units of L and A need not agree as long as the units of A and Y
agree, since (distanoez) cancels
between these two last quantities.
A wire whose cross section is 4 mm’ is stretched by 0.1 mm by a certain weight. How
far will a wire oi the
same material and length stretch it its cross-sectional area is 8 mm’ and the same
weight is attached?
I If the material and length are fixed as well as the stretching weight, then F, I,
Yare fixed in the relation
Al I (F!)/(A Y). Then
1 A 4mm’
A17 AAl=oonst A.A1.=n,./ti, AI,=’T;Al,=,Wn;0.1mm Al,=0.0smm
A steel wire is 4.0 m long and 2 mm in diameter. How much is it elongated by a
suspended body of mass
20 kg? Young's modulus for steel is l96(X)0 MPa.
I bet AL be the elongation. Then, by Hoolte's law,
F AL
_ = y_
A L
where Y is Young‘s modulus. The elongation is
_1f _fl_ (20)(9.s)(4.0) _ , _
AL—YAL— YA -(i—l96x10,)”(0'm1),-|.27a><1o m-1.;-/gmm
A copper wire 2.0 m long and 2 mm in diameter is stretched 1 mm. What tension is
needed? Young‘s modulus
tor copper is 117 600 MPa.
I §= YATL F=Y%,4=(117.ex10°)9$:r(0.001)’=1§5.1§
A wire is stretched 1 mm by a force of 1 kN. (a) How far would a wire of the same
material and length but
of four times that diameter be stretched? lb) How much work is done in stretching
each wire?
l>~
5
5
I (a) The elongation is inversely proportional to the cross-sectional area, and so
AL = (l)(})’ = M .
(b) The work done in stretching the wire in the two cases is W I Fx. Since the
force varies linearly with
distance,
F=%q so lvY.=@,(0.00l)=@ W,=%W,=0.0313J
Given a 2.0-n1 length oi steel wire with 1.0-mm diameter. about how much will the
wire stretch under a
5.0-kg load? Y = l9$(XD MPa.
I Since tensile modulus is stress/strain. we have AL/L= (F/A)/Y = [5(9.8)]/[rr(S X
10"‘)’(l9S X l0")] =
3.2 X ll)" and AL =6.4 X 10" m.
AAJ K/-\ TOPPER ii'E°Ei°§i'ii'§gJ'E'EpR'ii§

252 U
13.31
13.32
13.33
13.34
13.35
13.36‘
13.37
CHAPTER 13
Approximately how large a force is required to stretch a 2.0-cm-diameter steel rod
by 0.01 percent?
Y = 195 0(1) MPa.
I Solving H00ltc's law for F, it is F =AY(AL/L) = :r(0.01)'(l95 X lO")(10")
=61(X)N.
A platform is suspended by four wires at its comers. The wires are 3 m long and
have a diameter of 2.0 mm.
Young‘s modulus for the material of the wires is 180 000 MPa. How far will the
platform drop (due to
elongation of the wires) if a S0-kg load is placed at the center of the platform?
I AL = (LF)/(AY), where L = 3 m, A = :t(l.0 X10" m)‘ = 3.14 X 10" m’, a.nd since
each wire supports
one-quarter of the load. F = [(50 kg)(9.8 m/s‘)]/4 = 173 N.
(3 m)(lZ3 N)
I
= '5
AL (3.u><1o-"m*)(1_a><10'N/in‘) 65*“) '“ °' °i'65""“
What is the tninimum diameter of a brass rod if it is to support a 400-N load
without exceeding the elastic
limit? Assume that the stress for the elastic limit is 379 MPa.
I T0 find the minimum diameter, and hence minimum cross-sectional area, we assume
that the force
F = 400N brings us to the elastic limit. Then from the stress, F/A = 379 X 10‘ Pa.
we get A =
(400 N)/(379 X l0" Pa) = L0554 X l0'° m’. Then
A=LD2 D1=fl= =]_u4X1Q-¢
4 zt :r
and D=Vl.344Xl0"‘mi=t.l6xl0 ‘m=1.16mm
A No. 18 copper wire has a diameter of 0.04 in and is originally 10 ft long. (a)
What is the greatest load that
can be supported by this wire without exceeding its elastic limit? (b) Compute the
change in length of the
wire under this load. (e) What is the maximum load that can be supported without
breaking the wire? (J)
What is the maximum elongation? (Assume that the elastic limit stress is 73 000
lb/in’ and that the ultimate
strength stress is 49lIl0 lb/in’.)
I (0) F/A = 23 000 lb/in’, so Fm = (23000lb/in’)A.
A = %= it 4” X l0"in' F.._,. = (23 oo0)(4n >< to") = §.9lb
(b) Al = (F/A)(I/Y) = (23(X)0 lb/in)10 ft/17 ><10°lb/in = 0.0l35 it
(c) F‘/A = (49 000 lb/in‘); F’ = (49 000 lb/in‘)(4:r >< 10“ in‘) =§t.Q lb
(dl A!’ = (49000 lb/in’)( 10 ftll7 X 10° lblin’) = 0.0288 It fli
A steel piano wire has an ultimate strength of about 35 000 lb/in‘. How large a
load can a 0.5-in-diameter
steel wire hold before breaking?
a t
I F = (35tXI)lb/in’)A A = lllgsl-)= 1.964 X10 tin‘ F = (3$000lb/in‘)(l.964 ><
l0"in’) =os.7 lb
A wire of original length L and cross-sectional area A is stretched. within the
elastic limit, by a stress 1. Show
that the density of stored elastic energy in the stretched wire is 1':/ZY.
| Let the total deformation be AL. so that r = Y(AL/L). by Hooke‘s law. Again by
Hooke’s law, the
stretching force at deformation x is given by F (1) = [(A Y)/Lh. Hence the stored
elastic energy is
W = L“ F(x)dx =%’J:my1dx =%,AL
But, neglecting a tiny change, the constant volume of the wire is V = AL; hence
E = L‘
V ZY
Note that the energy density is independent of the wire dimensions.
Determine the fractional change in volume as the pressure of the atmosphere (0.1
MPa) around a metal block
is reduced to zero by placing the block in vacuum. The bulk modulus tor the metal
is 12$ 000 MPa.
AAJ K/-\ TOPPER ii'i§Ei°§m'§gJ'E'épR'ii’§

13.38
13.39
U.“
13.41
I342
I343
13.44
13.45
MATTER IN BULK U 253
—Ap gt/=_g_-r-0.0: _
' “Av/v °‘ v a‘12s000 “mi
Compute the volume change of a solid copper cube. 40 mm on each edge. when
subjected to a pressure of
20MPa. The bull: modulus lor copper is 125000 MPa.
_ — VAp _ — (40 mm)’(20 MPa) ,
| AV’ B _ l7_5000MPa Lem
Note that the units of V and Ap need not agree because Ap /B is dimensionless.
The pressure in an explosion chamber is 345 MPa. What would be the percent change
in volume of a piece of
copper subjected to this pressure? The bulk modulus for copper is 138 GPa (=138 X
10° Pa).
I The bulk modulus is defined as B = —Ap/(AV/V). where the minus sign is inserted
because AV is
negative when Ap is positive.
AV Ap 345 x 10°
i00| V I -100 B -|00138X10., 0.25%
The compressibility of water is 5 x 10 “‘ m’lN. Find the decrease in volume of
ll1)niL of water when
subjected to a pressure of l5 MPa.
I We note that the compressihity. k. is simply the reciprocal of the bulk modulus.
Then AV = —Vk Ap =
—(l00 mLl(5 X l0 '”tn’/N)(l5 X ll)’ N/m’) = -0.75 mL. The decrease is thus Q.75 mL.
How large a pressure must be applied to water if it is to be compressed by 0.l
percent? What is the ratio of
this pressure to atmospheric pressure. l0l ltPa? The bulk modulus of water is 2100
MPa.
I The volume strain AV/V.,= l.0 X 10"; set B =p/(AV/V0) to find p == (2.1 x 10°)(1.0
X 10") = 21(1) kPa.
Dividing by the atmospheric pressure. the ratio is Q.
By what fraction will the volume of a steel bar increase as the air is evacuated
from a chamber in which it
rests? Standard atmospheric pressure = 0.101 MPa and B for steel is l60lX10 MPa.
I Av/v..= -AP/B=0.10l/l60000=6.3><10"
What increase in pressure is required to decrease the volume of 200 L of water by
0.(I)4 percent? Find
AV. (B = 2100 MPa)
I AV = 0.0000-t(200 L) = 0.00s L Ap = a( - iv") = (2100 MPa)( ) - 0.004 MPO - 55
tr;
Compute the compressibility of glycerin if a pressure of 290 lb/in: causes a volume
of 64 in’ to decrease by
3 x I0" in’.
I The compressibility is the reciprocal of the hulk modulus. Thus
1 AV 3x10 -‘in’
/t= ———=i.i.= L62 10"‘ 1 lb
Ap v., (29010/m‘)(s4 In") ——x— "‘ /
Two parallel and opposite forces, each 40(1) N. are applied tangentially to the
upper and lower faces of a
cubical metal block 25 cm on a side. Find the angle of shear and the displacement
of the upper surface
relative to the lower surface. The shear modulus for the metal is 80GPa.
I We use the approximate form S = F/(A¢) (Prob. l3.2ll), with S = 8 X l0'° N/m‘, F
=4-(XDN, and
A = (0.25 m)" = 6.25 X 10 = m‘. Solving for ¢, we get
_ (400014) _ _,
4’ " (0.25 X 10 =m=)(a >< 10'" N/in’) "lg-‘Q
The displacement of the upper surface is given by d = Lo. where L is an edge of the
cube; d =
(8.0 X l0")(Z5 cm) = 2.0 X10" cm.

254 U CHAPTER 13
13.46
13.47
13.48
13.49
13.50
The shear modulus for a metal is 500(Il MPa. Suppose that a shear force of 2(1) N
is applied to the upper
surface of a cube of this metal that is 3.0cm on each edge. How far will the top
surface be displaced?
| Shearing strain - AL/L = (F/A)/S = 2(1)/[L'(5 X l0"')] = (4 X l0"')/L’; solve for
AL with L I 0.030m,
AL -1.33 ><10"’m -0.133 Em.
A block of gelatin is 60 mm by 60 mm by 20 mm when unstressed. A force of 0.245 N
is applied tangentially
to the upper surface. causing a S-mm displacement relative to the lower surface
(Fig. 13-2). Find (0) the
shearing stress, (b) the shearing strain, and (e) the shear lI‘l0dlllllS.
l: (=o0mm ;l;:[-d=Srnm
c C‘ 0 I
f" D
la
Ir =_Zi) mm
\\\\\\\\\\\\\\\\\\\§\\\@\\\\\\\\\ \\\@ r-1;. 1&2
F 0.245
I (0) stress=z=W=QJ_Ba
. d 5
(b) strain-tan0-i=2?)-llfi
F/A ss.4_ ,
(C) shear modultls S ‘an 9 0_2S—Z7Z4N m
Two sheets of aluminum on an aircraft wing are to be held together by aluminum
rivets of cross-sectional
area 0.25 in’. The shearing stress on each rivet must not exceed one-tenth of the
elastic limit for aluminum.
How many rivets are needed if each rivet supports the same fraction of a total
shearuig force of 25 (XX) lb?
Assume that the elastic limit stress is 190001»/in’.
I F IA = ,*6(l9000 lb/in’) = 19(1) lb/in‘ maximum stress allowed for each rivet.
This means a shearing force of
F = (1900 lb/in*)(0.7_$ 111*) = 415 lb/rivet. Number of rivets = Z5 000 lb/(475
lb/rivet) = $2.7. or 53 rivets.
A 60-kg motor sits on four cylindrical rubber blocks. Each cylinder has a height of
3cm and a cross-sectional
area of 15 cm’. The shear modulus for this rubber is 2 MPa. (a) If a sideways force
of 3(1) N is applied to the
motor, how far will it move sideways? (b) With what frequency will the motor
vibrate back and forth
sideways if disturbed?
I (I) We assume that the shear force is distributed evenly among the four
cylinders, Then for a given
cylinder F = 75 N.
L _ vsu _ _,
4’ as ' (15 X10" m*)(2 X10" N/tn’) ‘ 2'5 X ‘° ""
The displacement is then d = Lo = (3.0cm)(2.5 X 10" rad) =0.075 cm. (b) Since the
shear force on
each cylinder is proportional to mp, it is also proportional to the horizontal
displacement d.
F = .434» = [(AS)/Lld. Since there are four cylinders, the total extemal horizontal
force, H, is given by
F, = [(4AS)/Lld. This force just balances the elastic restoring force, or effective
“spring” force, of the
system E = —[(4AS)/L]d. lf the shear force is removed, the system osdllates with an
effective force
constant k = (-MS)/L = 4.0 >< 10’ N/m. Assuming that the masses of the cylinders
are negligible, we have
1 \[I?_i !4.0x16‘N/m
f 2» m ' 21 so kg mg‘
The twisting of a cylindrical shaft (Fig. 13-3) through an angle 8 is an example of
a shearing strain. An
AAJ K/-\ TOPPER iiE°Ei°§i'ii'§gJ'EEpR':§’§

-l
analysis of the situation shows that the angle of twist (in radians) is given by
21!
"um
where 1 = applied torque
I= length of cylinder
R = radius of cylinder
S = shear modulus for material
MATTER IN BULK U 255
Hg. 1&3
lf a torque of 1(1) lb - It is applied to the end of a cylindrical steel shalt l0
ft long and 2 in in diameter.
through how many degrees will the shaft twist? Assume that S = 12 X 10° lb/in’.
I We are given S =12 X1(]°lb/in’. R = l in. r = ll!) lb - ft = l200lb - in. I =
l0ft =120ini Using the given
formula we find that
_ 2:1 _ 2(l200lb-in)(l2(lin) _ _, _ ,
0_:rSR‘_:r(l2 >< l0‘lb/in=)(l an)" M“ 1° "‘d'°—L'4 7
13.51 An engine delivers 140 hp at 800 rpm to an 8~ft solid-iron drive shaft 2 in
in diameter. Find the angle of twist
in the drive shaft. Assume that S = l0 X 10° lb/in’.
I First calculate the torque, then apply Prob. l3.50.
P =140 hp(550 = 7.7 X 10‘ ft - lb/s; w = Znf = 2:t(80(] rad/s)(l min/60 s) = 83.78
rad/s.
P 7.7 X l0‘fl - lb/s
P=ftu, SO 1-;= -919th-tt=11mlb|n.
Finally
2r! 2(ll 000 lb - in)(8 ft)(l2 in/ft)
e=—,=—i.——.i=o.os14raa=L3§j
at-SR :t(l0 x 10“ lbItn‘)(l tn)‘
AA-1 KA TOPPER NEET AIIMSJEE RAS
I.“;E$§.'.?i‘2*".‘;‘;".-<'i{’;
CHAPTER 14
Simple Harmonic Motion
14.1 OSCILLATIONS OF A MASS ON A SPRING
14.1
14.2
14-J
14.4
145
14.6
14.7
A spring makes l2 vibrations in 40s. Find the period and frequency of the
vibration.
elapsed time 40s vibrations made 12 _
I T _ vibrations made _ l2 — F I _ elapsed time 40s _o*39i
A 50-g mass hangs at the end of a Hooltean spring. When 20 g more are added to the
end of the spring. it
stretches 7.0cm more. (a) Find the spring constant. (la) If the 20g are now
removed, what will be the period
of the motion?
I (4) Since the spring is linear.
_ g_ (0.020 t;)(9.s in/5') _
k ' Ax 0.07 I'll ' ils N/'“
_ g_ (0.050 kg_
(6) T-2n\f;-2:r 2-i—_8N/m-0.845
A spring is stretched 4 cm when a mass of S0 g is hung on it. If a total of 150 g
is hung on the spring and the
mass is started in a vertical oscillation, what will the period of the oscillation
be?
I First find the spring constant k:
1
k = = 12 250dyn/cm
T=21“/%=2.-it/%=z»i(o.iim>=o.s9ss
A body of weight 27 N hangs on a long spring of such stiffness that an extra force
of 9 N stretches the spring
0.05 m. If the body is pulled downward and released, what is its period?
I The spring constant is k =9/0.05 = 180 N/m. and so
m 27/9.8
T=2:r\/;=2n\iTw—=!@.7§s
A 3-lb weight hangs at the end oi a spring which has k = 25 lb/ft. lf the weight is
displaced slightly and
released, with what frequency will it vibrate?
To find the period use
I Using m = 31b/(32.2 ft/s’) = 0.093 slug gives
_ L ‘/7 _ i _ /Elbi _
/'21 III -2» 0.093slug_L6Ji
A mass m suspended from a spring of constant k has a period T. If a mass M is
added, the period becomes
3T. Find M in tenns of m.
I The period varies as the square root of the mass. Thus the mass must increase
ninefold. making M =@.
A 0.5-kg body performs simple harmonic motion with a frequency of 2 Hz and an
amplitude of 8 mm. Find
the maximum velocity of the body. its maximum acceleration, and the maximum
restoring force to which the
body is subjected.
| We are given frequency v and amplitude R: v = 2 Hz; at = Zzrv = 4:: rad/s; R =
0.008 m. Then we have
256
A/-\J K/-\ TOPPER ii'E°Ei°§i'ii'§gJ'EEpR§§’§

15.15
15.16
15.17
15.18
15.19
15.3)
15.21
HYDROSTATICS U 273
1-‘ind the ratio of a systolic blood pressure of 120 (in mmllg) to atmospheric
pressure. Standard atmospheric
pressure is 1.01 X 10‘ Pa.
I Pressure due to 0.120 m of mercury = pgh = (13 o00)(9.8)(0. 120) = 0.16 X 10’ Pa.
Ratio = 0.16/1.01 = (L115.
It the blood vessels in 1 human being acted as simple pipes (which they do not).
what would be the difierenoe
in blood pressure between the blood in a 1.80-m-tall man's feet and in his head
when he is standing? Assume
the specific gravity of blood to be 1.06.
I Ap=pgh=1060(9.8)(1.8)=1 .7kPa.
What is the pressure due to the Water 1 mi beneath the ocean's surtaoe if we assume
the mean density of
seawater to be 108 kg/m’? lf its compressibility is the same as that of pure water
(bulk modulus =
Ml) MPa), by what peroent has the density changed in going from the surface to this
depth (1 mi = 1609 rn)?
I Ap - pgh = l02S(9.8)(1609) = 16.2 MPa. By definition of the bulk modulus, B = -Ap/
(AV/V). But. for a
fixed mass m,
V
0=Am=pAV+VAp whence A7=—9f
Therefore B = 1311/ (Ap/p), or
Ap Ap 16.2 O
——=—=?= .0081 =0. 1%
P B 2000 _L_
A cylindrical tank 3 ft in diameter and 4 ft high is filled with water whose weight
density is 62.5 lb/ft’. Find
the gauge pressure at the bottom of the tank.
I Use the equation for pressure at a depth y in a liquid. lgnore atmospheric
pressure. p,_ = d.,v =
62.$lb/ft"><4ft=?.i0_ll2lfif.
A tank contains a pool of mercury 0.30 m deep, covered with a layer of water that
is 1.2 m deep. The density
of water is 1.0 x 10’ kg/in’ and that of mercury is 13.6 X 10‘ kg/m’. Find the
pressure exerted by the double
layer of liquids at the bottom of the tank. Ignore the pressure of the atmosphere.
I First find the pressure at the top of the mercury pool. For a point below the
surface of the mercury this
may be regarded as a source of extemal pressure P.-.. Thus
pm =1 p,,,,gh,,,,, = (1.0 X I0’ kg/m")(9.8 m/s’)(l.2 m) = 12 kPa
The pressure P... exerted by the mercury oolumn itself is found in the same manner:
pm, = p,,,,,gh,,,, I (13.6 X 10’ kg/m’)(9.8 m/s‘)(0.30 tn) I 40 kl’:
The total pressure at the bottom is thus S2 ltPa.
How high does a mercury barometer stand on a day when atmospheric pressure is 98.6
kPa?
98.6 x 10’ N/m‘
I Ii = Pl = m? = 740
pmg (13.6 >< 10’ kg/m’)(9.8 m/s’) '““‘
Two liquids which do not react chemically are plaoed in a bent tube, as shown in
Fig. 15-3. Show that the
heights of the liquids above their surfaoe of separation are inversely proportional
to their densities.
i /mi
///
r-1;. 15-3
I The pressure at the interface must be the same, calculated via either tube. Since
both tubes are open to
the atmosphere. we must have Pv8h| = p,gI|;. or h,/h, = p,/p,.
\§\\\1\\\=
-4
AAJ K/-\ TOPPER It'E°Ei°§i?l'§gJ'E'EpR':I’§
AAJ The Learning App Fo

1'
NEET AIIMS JEE RAS
274 U CHAPTER 15
15.22 Assume that the two liquids in the U'shaped tube of Fig. 15-3 are water and
oil. Compute the density of the
oil if the water stands l9crn above the interface and the oil stands 24 cm above
the interface.
I We apply the result of Prob. 15.21:
tr. 19
p_,_. = (';)p_ =% (1000 kg/in’) =792 kggm-‘
15.23 A uniform glass tube is bent into a U shape such as that shown in Fig. 154.
Water is poured into the tube
until it stands 10 cm high in each tube. Benzene (sp gr = 0.879) is then added
slowly to the tube on the left
until the water rises 4cm higher on the right. What length is the column of benzene
when the situation is
reached? (Water and benzene do not mix.)
I An 8-cm column of water balances the benzene column: so by Prob. 15.21.
It 8cm
;, =__;=i=g,|@
” p./p. 0.819
P P
‘._
<‘ " "1 r in
. ‘9___ _ _ ‘ ‘T
Benzene I hp la
\\.|tt~r- "~ ' in i ‘cm.
,, _____ __1__t Al —'—11l
.4
Fit. 15-4 Fig 15-S
The manometer shown in fig 15-5 uses mercu '
. ry as its fluid. lf atmospheric pressure is ll!) ltPa. what rs the
pressure of the gas in the container shown on the left?
I At points A and B pressures must be equal since the fluid is not moving. Therefore
P = P, + pgh =
100 ltPa +[(13.6)(9.8)(0.12)]kPa = 116 kPa.
15.24
15.25 A mercury barometer stands at 762m
m. A gas bubble. whose volume is 33 cm‘ when it is at the bottom of a
lake 45.7 m deep. rises to the surface. What is its volume at the surface of the
lake?
I In tenns of the weight density. pg. of water.
p....... = pry +1»..- = nay + p(L",")g/-... = oxl45~7 +<1a.o<0.1s2>1= 45-7px +
10.4»: = so. Ins
For the bubble. Boyle’: law states that pV = constant. assuming that the
temperature stays fixed. Then.
_ phantom _ 56.198 3
v...... -Mm vs... - ——|0_4p8 >< 11 um
15.26 A small uniform tube is bent into a circle of radius r whose plane is
vertical. Equal volumes of two fluids
whose densities are p and a (p > a) fill half the circl (
e see Fig. 15-6). Find the angle that the radius passing
through the interface makes with the vertical.
Neat) / "V
E
E
G
Q" \
ti-3 ‘~
M5
\fr. I
.\C’
\
~\\\\\\\\
§ L"
5 Z
mt 1-‘1g.l5-6
AAJ K/-\ TOPPER ii'i§$°§i'ii'§gJ'E'é°R'iI’§

15.27
15.28‘
15.29‘
HYDHOSTATICS S‘ 275
I Of the extemal forces acting on the two fluid segments only the two weights. pgV
and ugV, have torques
about the center 0; the forces exerted by the container are purely radial. Thus.
for equilibrium.
0 = pgVr sin (45° — 6) — ogVr sin (45° + 0)
0 = p(sin 45° cos 0 — cos 45° sin 0) — a(sin 45° cos 8 + cos 45° sin 6)
= _ _ J4”
0 p(1 tank?) 0(1+lan9) tan8 p+a
Rework Prob. l5.2b by requiring that the fluid pressures be equal at the interface.
I The pressure at the interface must be the same for the heavy and light liquids to
have equilibrium.
Refen-ing to Fig. 15-6. we calculate the pressure at the interface due to the heavy
liquid and set it equal to
the pressure due to the light liquid. (The added contribution of the gas in the
rest of the tube is the same for
both liquids and can be ignored.) Then pgh. = agh,, where Ir. and Ir, are the
heights of the respective liquids
above the interface. We have, from Fig. l5-6,
h,=roos6—roos(90‘— 0)=r(cos6—sin 6) h,=rcos6+rsin6=r(oos6+sin6)
Substituting into the pressure equation and canceling g and r for both sides. we
have p(cos 0 — sin 0) =
o(cos 6 + sin 0), or dividing by cos B, p(l - tan 6) = a(l + tan 0). Solving for
tan 0, we get tan 0 =
(P - v)/(P + H)-
Water stands at a depth It behind the vertical face of a dam. Fig. 15-7(0). lt
exerts a resultant horizontal force
on the dam. tending to slide it along its foundation. and a torque, tending to
overtum the dam about the
point 0. Find ta) the horizonal force, (b) the torque about O. and ie) the height
at which the resultant force
would have to act to produce the same torque
I .
dy
—>
h T h I'u\l4)' Pa-/dy
h — y
'1
-i-— 1
\ \\\\\\\\\\\\\Q \ \\\\\\\\\\\\\\\\ \
(II) (P1) ls‘)
Fig. 15-7
I (a) Figure 15-7(b) IS a view of the face of the dam from upstream. The pressure
at depth y is p = pgy. We
may neglect the atmospheric pressure. since it acts on the other side of the dam
also. [The construction shown
in Fig. 15-7(c) may be used to justify the neglect of atmospheric pressure] The
force against the shaded strip
is dF = p dA = pgyldy. The total force is
~ M.-
F=nsl’[yd.v=%
(bl The torque of the force dF about an axis through 0 I5. in magnitude. dr = (I: —
y) dF =pgIy(h —y) dy.
The total torque about 0 is
A J
r=flsIJ;y(h-.v)dy=%
(c) lf H is the height above 0 at which the total force F would have to act to
produce this torque.
_ -:_M'fi_fi
HF-r or H-F-pglhz/2-3
A conical cup. r = (b — z) tan ll’. rests open-end-down on a smooth flat surface. as
shown in Fig.l5-8. The cup
is to be filled to a height It with liquid of density p. What will be the lifting
force on the cup’!
AAJ K/-\ TOPPER Il'E°E§°§i?i'§gJ'E'EpR':I§

276 U CHAPTER 15
15.30‘
15.31
15.32
Z
L
fl
T‘
m .......
I imagine that the inside surface of the cup oonsists of an infinite number of
infinitesimal ring-shaped steps
(Fig. 15-8). The pressure p(z) acting on the vertical face of a step does not
contribute to the lifting force,
since it acts horizontally. Thus. the infinitesimal lifting force is only the
pressure foroe on the horizontal face
of the step:
dF, = p(z)dA = pg(h — z)(2m dr) = pg(h — z)[2:r(b — 2) tan a](—dz tan tr)
Integrating to obtain the total lifting foroe, we get
D
F, = -Zztpg tan‘ a'f7(h - z)(b - z)dz = :rpg(b!|' —%) tan’ a
h
Redo Prob. 15.29 using the weight of the fluid and the force it exerts on the fiat
surface.
I The total force exerted by the pressure of a static fluid on its container is
equal to the fluid‘s weight.
Hence, for this problem, B — I-1 = w, where F, is the downward force on the plane
surface; F, is the lifting
force on the cup (by symmetry. the horizontal pressure forces on the cup cancel);
and w is the weight of the
liquid. Now. 2
F, -p(0)A =- pgh(1tb‘ tan or)
and
R I
w = pgV = pg]; arr’ dz I pglt tan’ aJ:(b — 2)’ dz = (pgzr tan’ o')(b’h — bit’ +
Consequently, ha
F, = F, — w = (pgrr tan’ ar)(bI|’ —
as before.
As shown in Fig. 15-9. a weighted piston holds compressed gas in a tank. The piston
and its weights have a
mass of 20 kg. The cross-sectional area of the piston is 8cm’. What is the total
pressure of the gas in the
tank? What would an ordinary pressure gauge on the tank read?
Gas
Hg. 15-9
I The total pressure in the tank will be the pressure of the atmosphere (about 1.0
X 10’ Pa) plus the pressure
due to the piston and weights.
p =1.0 X10’ N/m‘ + = 1.0 x 10’ N/mz + 2.45 X10‘ N/m’ = 3.45 X10’ N/m’ =345 kPa
A pressure gauge on the tank would read the dilferenoe between the prmure inside
and outside the tank:
gauge reading = 2.45 X 10’ N/tn’ = filgjlg. It reads the pressure due to the piston
and weights.
Refer to Prob. 15.31; assume that the area of the tank bottom is 20 cm’. (a) Find
the total force on the
bottom of the tank and oornpare it with the weight of the piston plus the foroe of
atmosphere on the piston.
(The weight of the compressed air is negligible.) (b) How do you explain the
result?

HYDFIOSTATICS U 279
15.2 PASCAIJS AND ARCHIMEDES‘ PRINCIPLES; SURFACE TENSION
[$.39 A hydraulic lift in a service station has a large piston 30cm in diameter and
a small piston 2cm in diameter.
(cl What force is required on the small piston to lilt a load of 1500 kg? (b) What
is the pressure increase due
to the force in the confined liquid?
I (I) Pascal's principle says that the pressure change is uniformly transmitted
throughout the oil. so
Ap = F,/A, = Ii/A1. where F] and B are the forces on the small and on the large
pistons. respectively, and
A, and A, are the respective areas. Thus.
F, l500X 9.8
M2’/4) 100’/4)
Multiplying both sides by zr/4 and solving for F,, we obtain
F es
ts) ap-—‘=i=21N/=m*=2|otP»
Al II(2‘l4)
15.40 A hydraulic lift is to be used to lift a tniclt weighing 5000 lb. lf the
diameter of the large piston of the lift is
lft, what gauge pressure in lb/in’ must be applied to the oil?
I The gauge pressure in the oil acts. by Pascal's principle. on the bottom of the
large piston to produce the
force that lifts the load.
Ap=f=£?=fi=6370lb/ft'=t:%0=44.2lb[in‘
A M’ :r(0.5‘)
15.41 In a hydraulic press the large piston ha cross-sectional area A, = Zmcm’ and
the small piston has
cross-sectional area A, = 5 cm’. If a foroe of ZSON is applied to the small piston.
what is the foroe F} on the
large piston?
I By Pascal’: principle.
F
pressure under large piston = pressure under small piston IT’ — I?
I Z
A 200
F,=;;I-}=?(25ON)=l0l:N
15.42 For the system shown in Fig. 15-14. the cylinder on the left. at L. has a
mass of 600ltg and a cross-sectional
area of 8(X)cm‘. The piston on the right. at S. has cross-sectional area 2.5 cm’
and negligible weight. lf the
apparatus is filled with oil (p = 0.78 g/cm‘). what is the force F required to hold
the system in equilibrium as
shown‘?
\ 8m
L r
~ _ . . . _ .
__i
HI HZ
.r_
F
'///”
vi
I-lg. is-14
I The pressures at points H, and H, are equal since they are at the same level in a
single connected fluid.
Therefore,
pressure at H, = pressure at H,
pressure due to pressure due to F
left piston and right piston
(60°)(9-3)
0.08 m’ ZS X I0 m
Solving for F gives it to be ii.
AA-1 KA TOPPER NEET AIIMS JEE RAS
) = ( )+ (pressure due to 8 m of oil)
N = + (8 m)(780 kg/m"‘)(9.8 m/s’)

280
15.43
15.44
15-45
15.46
15.47
15.48
15.49
I5-50
,"7
_|
CHAPTER 15
A block of wood weighing 71.2 N and of specific gravity 0.75 is tied by a string to
the bottom of a tank of
water in order to have the block totally immersed. What is the tension in the
string?
| The block is in equilibrium under the action of three force&—the weight w = 71.2
N, the tension T. and the
buoyant force B, with B = w + T. We can determine B since B = p, gV,,, where p,_ is
the density of water and
V, is the volume of the totally immersed block. w = p,gV,, where p, is the density
of the block. Then
W/B = p,/p,_ = (sp gr), = 0.75. so B = w/0.75 = 94.9N. Finally. from our
equilibrium equation. 94.9N =
7l.2N+ T. or T=23.7N.
A metal ball weighs 0.096 N. When suspended in water it has an apparent weight of
0.07] N. Find the density
of the metal.
I The desired density is given by p =m/V. But since the volume V of the ball is
also the volume of
displaced water, the buoyant force is g'ven by B = p,,,,,,gV. Thus.
("'8)fl--.. (U¢996N)(1 >< 1°) K5/ml)
P=T=<i1»»6m>~=’°‘°“@/""
A block of material has a density p, and floats three-fourths submerged in a liquid
oi unknown density. Show
that the density p, of the unknown liquid is given by p, = §p..
I By Archimedes‘ principle. p,Vg = p,(3v/4);. which gives p, = ;p,.
The density of ice is 917 kg/m’, and the approximate density of the seawater in
which an iceberg floats is
1025 kg/m‘. What fraction of the iceberg is beneath the water surface?
I By Prob. 15.45. traction submerged = Pi/D; = figs = Oi‘).
A block of wood has a mass of 25 g. When a 5-g metal piece with a volume of 2 cm’
is attached to the bottom
of the block, the wood barely floats in water. What is the volume V of the wood?
I The effective density ol the system is 30/(V + 2) glcm‘; hence. by Probs. 15.45
and 15.46.
30 V 2
l = fraction submerged = —/Lil)
Solving, V = E gm‘.
A piece of wood weighs in mass units 10.0 g in air. When a heavy piece of metal is
suspended below it, the
metal being submerged in water. the "weight" of wood in air plus metal in water is
l4.(Il g. The “weight”
when both wood and metal are submerged in water is 21!] g. Find the volume and the
density of the wood.
I Since in both cases the metal is submerged. the diflerence in weight is only the
buoyant force on the
wood. or (12.(X] X I0‘ ")(9.8) N. Therefore. the volume of the wood can be found by
equating the weight of
displaced water. ltXX)(9.8)V, to this. Then V = l2 X 10”’ m". lts mass is l0 X 10"
ltg and so its density is
(10 X107’)/(12 x l0 ") = 830kg[m’.
What is the minimum volume of a block of wood (density = 850 kg/m’) if it is to
hold a S0-kg woman entirely
above the water when she stands on it‘!
I The woman's weight plus the bloclt‘s weight must be equal to the buoyant tome on
the just barely
submerged block. 50g + 8SOVg = tt1XlVg, which leads to V =0.33 m’.
A man whose weight is 667 N and whose density is 980 kg/m" can just float in water
with his head above the
surface with the help of a lite jacket which is wholly immersed. Assuming that the
volume of his head is {Q of
his whole volume and that the specific gravity of the life jacket is 0.25. find the
volume of the life jacket.
| The man's volume is
_1_ _;’ = -
v‘»>g‘<9s0>t9-8) °‘°7"‘
Equating the buoyant force to the weight of the man plus the weight of the life
jacket,
P.....sll-.iV + V..l = 667 + (0-Z5p.....-)xV..
AAJ K/-\ TOPPER it'E°Ei°§i'§i'§gJ'E'EpR':I’§

15.51
15.52
15.53
15.54
15.55
l5.56
15.57
HYDHOSTATICS I] 281
Solving,
_(ee7/p_,,,,)-(14/1s)gv_ 0.601 - (14/l5)(9.8)(0.07) S ,
V" ' 0. 75; ' (0.vs)(9.a) 0'00‘ "‘ ‘" Q
An irregular piece of metal "weighs" l0.(I) g in air and 8.00 g when submerged in
water. (a) Find the volume
of the metal and its density. (bl ll‘ the same piece of metal weighs 8.50 g when
immersed in a particular oil.
what is the density of the oil’!
I Buoyant force = (2 X 10 "‘)g N, with g = 9.8 m/s’. This equals the weight of the
displaced water, p,Vg.
Equating and using p, =1(H)kg/m" gives V - 2 X IQ '“ m". The density I mass/V = (10
>< 10 ‘)/(2 X 10“) I
Silllltglm’. The buoyant tome in oil is (1.50 X l0 ")3 N. 'l'he weight of the
displaced oil is p‘,(2 x l0 “)g,
from which p_, = 750 kg[m".
A beaker partly filled with water has a total mass of 20.00 g. lf a piece of wood
having a density of
0.8lI)g/cm’ and volume 2.0 cm’ is floated on the water in the beaker, how much will
the beaker "weigh" (in
grams)?
I The scale must support both the original 20g and the (0.80 glcrn")(2 cm’) = l.60-
g block. so the
“weight” - 21.6 g.
A beaker when partly filled with water has total mass 20.0) g. lf a piece of metal
with density 3.00 g/cm’ and
volume LID cm’ is suspended by a thin string so that it is submerged in the water
but does not rest on the
bottom of the beaker. how much does the beaker then appear to weigh il it is
resting on a scale?
I The tension in the thread is equal to the weight of the metal less the buoyant
force. The buoyant force will
be (1 X 10" rn’)(l(XD kg/m’)g =10 ’g N. where g = 9.8m/s1. The weiyit of the metal
is 3 X l0“‘g N.
Therefore, the thread exerts an upward force of Z x l0"g N. Hence the scale
supports the total weight less
the tension in the thread. Therefore, the apparent weight read by a scale will be
(23 — 2) g, or 0.206 N.
Equivalently we can get the result by noting that if the water exerts an upward
buoyant force oi
l0"g = 9.8 X 10" N on the metal, by Newton's third law the metal exerts a like
force downward on the
water. Thus the scale balances the weight. 0.196 N, plus the downward force of the
metal, 0.0098 N. for a
total of 0.3% N.
A solid cube of material is 0.75 cm on each edge. lt floats in oil ol density 8%
kg/m’ with one-third of the
block out of the oil. (a) What is the buoyant force on the cube? (b) What is the
density of the material of the
cube?
I (n) The block is in equilibrium so p,V,g = BF = p,,(2V,/3)g. Since V, = 4.22 x
10" m’ and p., =
Hllkg/m’, BF=2.Zl X l0 “ N. (b) Since BF = p,V,g, we find p, =2p°/3 = S33 Kg/m’.
A cubical copper block is 1.50 cm on each edge. (a) What is the buoyant force on it
when it is submerged in
oil for which p = 820 kg/m’? (b) What is the tension in the string that is
supporting the block when
submerged? p¢,, = 8920 kg/m".
I (0) The buoyant force equals the weight of the displaced liquid. BF = p_,,,Vn,g,
with the volume
V = 3.38 X l0"m’ and p = 820 ltg/m". Thus BF - 0.027 N. (b) The forces acting on
the block are the tension
T up. BF up. and the block weight p¢,V¢,,g = 8920(3.38 x l0“)(9.8) = 0.295 N acting
downward. Then
T =1 0.295 — 0.027 =0.Z§_§ N.
A balloon having a mass of 500 kg remains suspended motionless in the air. ll‘ the
air density is L29 kg/m’.
what is the volume of the balloon in cubic meters?
I By equilibrium and Archimedes’ principle. 500g = l.29gV, or V m
A cylindrical wooden buoy. of height 3 m and mass 80 kg, floats vertically in water.
lf its specific gravity is
0.00. how much will it be depressed when a body of mass 10 kg is placed on its
upper surface?
I By Archimedes‘ principle, the submerged height, Ii, of the unloaded buoy is given
by
p.,,,,,gAh = p.__,,,,gA(3) or h = % 3 = (o.s0)(s) = 2.40 In
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282 J CHAPTER 15
Under loading. the submerged height is directly proportional to the total weight or
mass.
It-l>AI| 80+10 ll] l0
T—T 0|‘ Alt-§5h—§Z.4U=0.30m
15.58 A tank eontains water on top of mercury. A cube oi iron, 60 mm along each
edge, is sitting upright in
equilibrium in the liquids. Find how much oi it is in each liquid. The densities of
iron and mercury are
7.7 X 10‘ kg/m‘ and 13.6 X 10‘ltg/m’. respectively.
Water
I’:
V
§
I’;
Mercury H‘. 1545
I Let x equal the distance submerged in mercury. Then (0.06 — x) equals the
distance protmding into the
water. The net vertical force due to the liquids is p,A — p,A, where p=. p, are the
pressures at the lower and
upper lace of the block (see Fig. 15-15) and A is the face area of the block. For
equilibrium we have the
weight of the iron iv, =p,A —p,A = (p, — p,)A. The pressure difierence is just p, —
p, = p,,g(0.06 — x) +
pmgt. Then (p, — p,)A = p_g(0.06 — x)A + p,,,gxA. [Note that the two terms on the
right represent the
weight of displaced water and mercury. respectively: and the expression on the left
is the buoyant force. Thus
Archimedes‘ principle holds even when two (or more) liquids are displaced. This is
true for any shape object.]
We can now solve for 1 from the equilibrium equation. noting that w, = p,gV, =
p.,g(0.06 —x)A + p,,‘gxA;
or canceling the g, we have
(7.7 X 10‘ kg/m‘)(l).06 m)" = (1.0 X I0’ kg/m’)(0.06 m — .r)(0.06 m)’ + (13.6 X10’
kg/m"‘)x(0.06 m)‘
and 7.7(0.06)=(0.l)6-x)+ l3.6x 0.40= 12.6:
x = 0.032 m = 3; mm = depth submerged in mercury. and 28 mm protrudes into water.
15.59 A slender homogeneous rod of length 21 floats partly immersed in water. being
supported by a string fastened
to one of its ends. as pictured in Fig. l5-16. If the specific gravity of the rod is
0.75. what is the fraction ot the
length of the rod that extends out of the water?
T
Q..__ _
/‘/\ 9
___. __..- {Fl -
\/I W Water
_ _ Fig. 15-16
| Since the buoyant force acts through the center of gravity of the displaced
water. the condition for
rotational equilibrium is. taking moments about a point 0 along the line of action
of T.
0= r,,=wIeos9-F, 21-5‘ cos6=p,_,_,gA(21)(lcos8)—p..,.,,gAx 21-5 cost!
2 2
= fl.....g-4 cos 9)(x2 - Ah + 4 ‘fl I2) where A = cross sectional area
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15.60
15.61
l5.62
15.63
15.61
HYDHOSTATICS .7 283
From this. x: — -lh + 3.00!‘ = 0. or x =1. 3!. Discarding the nonphysical root. we
see that one-half the rod
extends out of the water.
Strictly speaking. the above solution is only approximate since the water surface
does not cut the rod
perpendicularly. However. the error will be negligible if A is small.
The weight of a balloon and the gas it contains is 11.12 kN. If the balloon
displaces 1132 m‘ of air and the
weight of l m" of air is 12.3 N. what is the acceleration with which the balloon
begins to rise’!
| The equation of motion of the balloon is
F — 1132 12.3 — 1.112 10' ,
zF=F,,—w=mn or n=’T‘v= =i1£
Note that we have ignored viscous forces. This is justified if the velocity is
negligible.
A small block of wood, of density 0.4 X 10" kg/m’. is submerged in water at a depth
of 2.9 tn. Find (n) the
acceleration of the block toward the surface when the block is released. and (b)
the time for the block to
reach the surface. Ignore visoosity.
I (0) By Archimedes’ principle. the net upward force on the block is F = p,,,,,,gV
— p_,,_,gV, where V is
the volume of the block. Then
0 =5= =(fi- l)g=(0'%‘—1)(9.8)=l4.7m[s°
7-‘
(b) $=%flI: or t=\/1%‘=\2:+79)=o
A body of density p‘ is dropped from rest at a height It into a lake of density p.
where p > p’. Neglect all
dissipative effects and calculate (a) the speed of the body just before entering
the lake. (b) the acceleration of
the body while it is in the lake. and (e) the maximum depth to which the body sinks
before retuming to float
on the surface.
I lo) The speed just before entering can be determined from conservation of
mechanical energy in free fall.
or directly from the kinematic equations of free fall. yielding v = (b) The
bouyant force of the lake. B,
is greater than the weight of the body. w, since p > p’. Choosing upward as
positive, we have
B — w = ma where B = pgV w = p'gV m = p'V and V = volume of body
Then. canceling Von both sides.
I P P P
_ = ' = i,= 1-1 _ d
(P P )8 P '1 °' 4 B p 8(p ) "PW"
(cl To find the maximum depth we have v’ = t1§+ Zay, with v.. the velocity at y =0
(from part a), v =0 is the
velocity at maximum depth, and y is the negative displacement from the surface to
the maximum depth. Then
2 I
' Zsh hp
.§=—2n a - =d m=‘—"=i=—.
v y an (Y) en 20 28(£_1)p_p.
P.
\
A soap bubble has a radius of 5cm. If the soap solution has a surface tension T -
30 X 10" N/m. what is the
gauge pressure within the bubble?
I Consider a hemisphere of the bubble (Fig. 15-'17)‘. The downward force
oflsurtaloe tension on each of the
two bubble surfaces. inside and outside. is Z:rrT. For both surfaces the total
force F is F = 2(Z:rrT) = ApA.
where A is the area of the flat circular face of the hemisphere. (See. e.g.. Prob.
15.36.) Since A = av’.
4r 4T 4(30 x to ‘)
42rrT = Ap(zu*) 7 = Ap Ap = T = = 2.4 Pa
Find an expression for the height h that a liquid of density /1 will rise in a
capillary tube of radius I if the
surface tension of the liquid is y and the meniscus makes an angle 0 with the tube.
as shown in Fig. I5-18.
I y is a force pet‘ unit length and p0inLs in dilierent directions around the tube.
as shown. At equilibrium.
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16.27
16.23
HYDRODYNAMICS J 291
‘ll Area u=I.'¢1
1 ‘Y /[Ig.l_n\:
l>:< -. ./~._"i1l7‘l,
.\ Fig.l(r8
X4934
The opening near the bottom of the vessel in Fig. 16-9 has an area a. A disk is
held against the opening to
keep the liquid. of denisty p, from running out.
(a) With what net force does the liquid press on the disk? tb) The disk is moved
away from the opening a
short distance. The liquid squirts out. striking the disk inelastically. After
striking the disk. the water drops
vertically downward. Show that the force exerted by the water on the disk is twice
the force in pan a.
if
\ dig‘ v
I (ii) The hydrostatic pressure on the inside surface of the disk is given by p,
=p,,,,, + pgh. The air pressure
of the outside of the disk is p.. = p..,... Since the disk has area a, the net
outward force is (p, — p,,)a = egg;
(b) Once the disk is removed. the fluid quickly attains a (relatively) steady flow.
Torrioelli's theorem implies
that the exit speed is v = Since friction is being ignored. this is the exit speed
across the entire opening
Therefore the mass flux 0 = pat’ = pafi and the flux of rightward momentum is (pav)t~
= 2pgha. If the
water loses its entire rightward momentum as it strikes the disk. the disk must
ahsorh momentum at the rate
Zpgha. That is, it will experience a rightward force Zggha, which is twice the
hydrostatic force found in pan
a. Note that the force due to the atmosphere cancels on the left and right of the
disk.
I-‘lg. too
A flat plate moves normally toward a discharging jet of water at the rate of 3 m/s.
11te jet discharges water
at the rate of U.l m‘/s and at it speed of 18 m/s. (4) Find the force on the plate
due to the jet and
(b) compare it with that if the plate were stationary.
I We do part b first. With no other information we assume the plate stops the
forward motion. but there is
no bounce back. i.e.. the water splashes along the plate at right angles to the
original motion. Then the force
normal to the plate equals the time rate of change of momentum along the direction
of the water jet. or
F = (pau)v. Where the term in parentheses is the mass/time hitting the plate and n
is the cross-sectional area
of the jet. We are given it = 18 mls and av = ().l m‘/s. p = llltltl kg/m‘. F
=(l000 kg/m‘)(0.| m“/s)(l8m/s) =
1800 N.
In part a the plate is moving toward the stream at 3 m/s. 'l'\vo things effect a
change in the momentum
change/time. First if the liquid again splashes at right angles lo the plate it has
picked up a velocity of 3 m/s
opposite to the jet‘s direction. The total change in forward velocity is therefore
not u =- 18 m/s hut =
18 + 3 = 21 mls. Second. the mass of water hitting the plate per second increases
from nu to a(u + 3). Noting
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292 fl CHAPTER 16
that a = (0.1 m’/s)/v we have for the force
F = 4%)“, + 3)“, ,, 3) = =24§ON
16.29 A pump draws water from a reservoir and sends it through a horizontal hose.
Since the water starts at rest
and is set into motion by the pump. the pump must deliver power P to the water when
the flow rate is O,
even if fluid friction is negligible. A new pump is to be ordered which will pump
water through the same
system at a rate Q’ = 24>. What must be the power P‘ of the new pump? Assume that
friction is still
negligible.
I By the work-energy theorem,
KE imparted to the water KE vol. water 1 ,
P = _ = X _ I u X u = tr
time vol. water time
The mass flux d> is proportional to v so that P = 0‘. Thus if 0‘ = 21>, the new
power P‘ = SP.
16.30 The U tube of Fig. l6-10 contains a length L of a zero-viscosity fluid. Show
that the fluid column oscillates
like a simple pendulum of length L/Z.
I-‘ig. 16-10
I Let p denote the density of the fluid, M the mass of the fluid. and A the cross-
sectional area of the tube.
lf one end of the fluid column is depressed a distance x below the equilibrium
level. the other end must
(assuming incompressibility) rise an equal distance x. The weight due to the height
diflerence, Zx. is a
restoring force. F = —pgA(Zr) E —kx. Thus we have SHM. of frequency
,_L\/Li l1P.8_/Li /2P.x_A_L is
Zn M 2» M ZR pAL-2:! 1./2
16.2 VISCOSITY, STOKES’ LAW, l'OlSEUILLE’S LAW, TURBULENCE, REYNOLDS
NUMBER
16.31 A number of tiny spheres made of steel with density p,, and having various
radii r,, are released from rest
just under the surface of a tank of water, whose density is p.
(0) Show that the “net gravitational force“ acting on a sphere (the combined effect
of weight and
buoyancy) has magnitude (4:t/3)rf(p, - p)g. (b) Assuming that the fluid flow around
each descending sphere
is laminar, find the terminal speed u of a sphere in terms of r,, p,, p, and the
viscosity n of the water.
. 4 ' 4 4
I on (welsh!) - (bwyanl form) = P.s(3 mi) — ps(5 1"?) = frftn. — ms-
(b) When the sphere is descending at terminal speed. the net force must vanish. so
we can equate the
downward, “net gravitational force“ to the upward viscous drag, which by Stokes‘
law = 6m7r,v. Thus
4 2
% (P. — n)x = 6110'.»
Solving for the terminal speed, we obtain
"=%f(P.-P)! (1)
16.32 Describe an idealized experiment for defining the coeificient of viscosity, 1|.
I Figure 16-11 shows two very large parallel plates A and B, separated by a
distance d. The space between
them is filled with fluid. A constant force F must be applied to plate B to keep it
moving at a constant speed
tn, with respect to plate A. if a thin lamina of fluid at a uniform distance y from
plate A moves with a speed
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16.35
16.31
16.35
l6.36
HYDFIODYNAMICS J 293
n.
Plate ll F
i
I
El} —~"~
l H I—> I‘,
ll l
\
F
Qi
Plttk‘ A fig. 16-ll
u(y) = u.,y/d. the flow is laminar. Then it is found that 0,, the shear stress on
the liquid. is proportional to
ul./d and the proportionality constant is the viscosity, 1). Thus
a, = nu»/d (1)
From (1) it is seen that the S] unit of viscosity is the pascal - second or
poiseuille: l Pl = l Pa -s = 1 N - s/mi.
A rotating-cylinder viscometer is employed to measure the coefficient of viscosity
of castor oil at a
temperature of 20°C. The radius of the inner cylinder is r, = Mllcm. and the radius
of the outer cylinder is
r, = 4.28cm. The inner cylinder is submerged in the oil to a depth h = 10.2 cm.
When the outer cylinder is
rotating at 20.0 rev/min. the torsion balance reads a torque T = 3.24 X 10* N - n1.
find the viscosity of the
castor oil.
I The cylindrical space between the outer and inner cylinders of the viscometer is
not a bad approximation
to the ideal flat-plate system of Fig. l6-ll. Hence. by (I) of Prob. 16.32. 1| = o,
(d/u.,).
‘fire wetted area of the inner cylinder is n == 2:rr,h. Thus you have 0, = F
/2:rr,I|. where F is the drag force
applied to the inner cylinder by the fluid. which is driven by the outer cylinder.
This force results in a torque
T = r,F. which can be read on the scale of the torsion balance. In terms of this
torque, the shear stress is
0, = T/(2:rrfh).
Since the inner cylinder is at rest. the relative speed 11.. of the two cylinders
is given by v.,= 10,}, where in is
the angular speed of the outer cylinder. And the distance between the cylinders is
d = r, - r,.
Using the above values of 0,. 11... and d. you obtain
_ 7-(7: ’ 7t)
W Zirwrirzh
Inserting the numerical values gives
_ (3.24 X 10*’ N - m)(0.28 X104: m) _
7' _ 20.0 rev/min X 2:r rad/rev _ . . _. _, _l”fi
2.-r( )(4.m>< to ‘m)'(4.28 x l0 ‘m)(l0.2 x 10 m)
How last will an aluminum sphere of radius l mm fall through water at 20°C once its
terminal speed has been
reached? Assume laminar flow. [sp gr (Al) = 247; q..,.‘, = 8 X 10" PL]
I Substituting the data in (1) of Prob. 16.31. we lind v = 4.6m/s. (Actually. the
assumption of laminar flow
tor this problem is not realistic; see Proh. l6.5l.)
A typical riverhorne silt particle has a radius oi Z0 um and a density of 2 X I0’
ltglm’. The viscosity of water is
approximated by 1.0 mPl. Find the terminal speed with which such a particle will
settle to the bottom of a
motionless volume of water. (Unless the speed of intemal fluid motions is smaller
than this settling speed. the
silt particles will not settle to the bottom.)
I Using <1) of Prob. |<».3|.
2(2u X 10 *)=[(2.o - 1.0) >< 1u~‘](9.so)
U, = 8.7 X10" m/5 = 0.87 mmls
Refer to Prob. 16.35. Suppose that you filled a l-L bottle oi 50-cm: cross section
with water from a muddy
river. such as the lower Mississippi. After all internal motions of the water
itself had stopped. about how long
would it take for all the silt to settle to the bottom?
I Since the height of the bottle will be about 200 mm. and since all silt particles
(in particular. those at the
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294 .7 CHAPTER 16
16.37
16.38
16.39
16.40
16.41
16.42
16.43
top of the bottle, which govern the overall settling time) take negligible time to
achieve the tenninal speed,
overall settlin time - lgnl ~ 4 min
g 0.87 mm/s i
A tiny glass sphere (density 2600 kg/m") is let fall through a vat of oil (p = 950
kg/m’. vi = 0.21 Pl). In 100s it
is observed to drop 43 em. How large is the sphere?
I Substitute the data and ll = 4.3 X l0"‘m/s in (1) of Prob. 16.31 to find r, =
Qgiqg.
ln a certain centrifuge the liquid is rotated at 20 rev/s at a radius of 10cm from
the axis of rotation. Tiny
spherical panicles of radius b and density 1020 kg/m’ in a dilute water solution
(density 10(1) kg/m’) are
placed in the centrifuge. Find the terminal speed with which they settle out of the
solution. ignore the effect
of gravity» '1-..<. = 8.0 x 10" Pl.
I The particle needs a centripetal force mufr = [(4:rb"‘p)/3](20 >< 2:r)’((l. 10) =
6.75 >< l0“b"‘ N to keep it
moving in a circle. A liquid (water-composed) particle also needs a centripetal
force to keep it moving in a
circle. Since only the surrounding liquid can supply the centripetal force.
particles more dense than water will
pass through to the outer edge of the centifuge. This is then very similar to
particles falling through water in a
gravitational field. The actual static force supplied by the water in the centrifuge
is just the effective “buoyant
force" and must equal the centripetal force on an equivalent volume of water (using
the same reasoning as
that for Archimedes‘ principle in a gravitational field). Hence. BF =
(1000/1(I20)6.75 x 10°11’ =6.6l X ltfb’.
The remaining force is supplied by the viscous friction force (ozrybv). Hence.
omybu = 0.14 X 10°b‘. with
r) = 8.0 X 10 ‘ Pl. yields v = 9.3 X10‘b: mls.
The viscous force on a liquid passing through a length L of pipe in laminar flow is
given hy E, = 4m1Lu,_,
where ry is the liquid viscosity and u,,, is the maximum velocity of the liquid
(i.e., along the central axis of the
pipe). Find an expression for u,,, in a horizontal segment of pipe in tenns ofp,
and p,. the pressure at the
back and forward ends of the pipe. and in ten-ns of :1, L, and r, the radius of the
pipe.
I ln steady flow the viscous force is balanced by the force due to the pressure
difference at the back and
front ends. Thus p,:rr= — p;J!!2 = 4:n1Lv,,,. Solving for v,,, we have u,,, = [(p,
— p,)r‘]/(4r]L).
What is the presure drop (in mmHg) in the blood as it passes through a capillary 1
mm long and 2 pm in
radius if the speed of the blood through the center of the capillary is 0.66 mm/s’!
(The viscosity of whole
blood is 4 X 10" Pl).
I By the result of Prob. 16.39.
pt —p. = = 26O0Pa =(2flDPa)( ) =19.s rnmHg
Using the result of Prob. 16.39, find an expression for the volume flow rate of fluid
through a segment of
pipe. ln laminar how the average flow velocity over a cross section is u,,/2.
I The volume flow rate. H. is the product of the cross-sectional area of the pipe
and the average
flow velocity. Thus H = (:rr’v,,,)l2. Using the expression for v,., from Prob.
16.39. we have H =
[xr‘(p, — p;)]l(8r1L). This is known as Poi.reui!le‘.t law.
How much power is delivered at the back end of the capillary of Prob. 16.40 in
pushing the blood through?
Assume p| = 10.0 kPa.
I P E power = force X velocity = (p,:rr*)(u,.,/2) =p|H, by Prob. 16.41. Thus
(1.0 >< 10‘ N/m’)(3.14)(2 X l0"'m)‘(0.26 X 10‘ N/m’)
P= =4.1 "w
8(4xl0"‘Pl)(l0 ’m] L
Assuming all else remains the same. how would the flow rate. H. change if the
radius of a pipe is doubled?
How would the power necessary to push the fluid through change? How would the fluid
velocity change’!
I Referring to Poiseuille‘s law (Prob. 16.41). and assuming that p.. p,. n. and L
remain constant. we see
that doubling r will increase H sixteen-fold. For the power P we have from Prob.
16.42 P = p,H and the
power increases sixteen-fold as well. The velocity varies as I2 (see Prob. 16.39)
and therefore only quadruples.
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M8
M3
20.10
10.11
NJ2
NJJ
GAS LAWS AND KINETIC THEORY If 327
_g_(i.sz><1o“i(1o~=)_
| “l "_RT_ (8.3l)(298.15) "h
(b) The atomic mass of hydrogen is 1.1118. so that 1 mol of hydrogen (1-1,)
contains 2.016 g. or 2.016 X 10"‘ kg.
The density of the hydrogen is then
M 6.l3 2.016 10"’
p=%= =1.24kg[m’
(cl The atomic mass of oxygen is 16, so that 1 mol of O, contains 32 g, or 32 X 10
‘ kg. The density of the
oxygen is then
M 6.13 32 ><1o" ,
p - '-'7 = ix?)-19.6 kg/m"
A compressor pumps 70 L of air into a 6L tank with the temperature remaining
unchanged. If all the air is
originally at 1 atm, what is the final absolute pressure of the air in the tank‘?
I Use Boyle's law, since the temperature is constant.
p,V.,=pV p¢,=1atm V,=70+6=76L V=6L
1(76) =p(6) p = 12.7 arm absolute pressure
l
A 0.025-m tank contains 0.084 kg of nitrogen gas (N,) at a gauge pressure of 3.17
atm. Find the temperature
of the gas in degrees Celsius. (pm, -= 1.013 x 10’ Pa).
I We note that the mass of lkmol of N, is 28kg. From the ideal gas law, pV =nRT,
with n =
0.084/28 ltmol, p = 3.17 +1= 4.17 atm = 4.17 X 1.013 x 10’ N/m’. R = 83141/kmol -
K. Substituting.
4.110.013 >< io’)(0.o2s) = (0.os4/2s)(s314)1' and
_ 28(4. 17)(l.013 X 10’)(0.02S) _ _ °
T-—————i——o_0a‘(8314) -423K-150 C
A partially inflated balloon contains sou m’ of helium at 27 °C and l-atm pressure.
What is the volume of the
helium at an altitude of 18 000 ft, where the pressure is 0.5 arm and the
temperature is —3 ‘C?
I For a confined gas.
;»._i/,_;;,_t', 1(soo)_o.sv, _soo(21o)_ ,
T, ' T, 300 ‘zvo v"a0o(o.s)'9@
An air bubble released at the bottom of a pond expands to four times its original
volume by the time it
reaches the surface. If atmospheric pressure is 100 kPa. what is the absolute
pressure at the bottom of the
pond? Assume constant T.
I po=p€= (100kPa)(4) =400kPa
A pressure gauge indicates the diflerences between atmospheric pressure and pressure
inside the tank. The
gauge on a 1.00-m’ oxygen tank reads Sflatm. After some use of the oxygen, the gauge
reads 25 atm. I-low
many cubic meters of oxygen at normal atmospheric pressure were used? There is no
temperature change
during the time of consumption.
I Since the temperature is fixed we use Boy|e‘s law to solve the problem. The total
pressure in the tank has
been reduced from 31 to 26 atm. At the latter pressure. the gas originally in the
tank would occupy §-1 m’.
Since 1.00 m’ remains in the tank, the amount of gas used was i m’ at 26 atm. At
the same temperature and
at atmospheric pressure, this would occupy a volume 26 times as large: 5.00 m’.
An air bubble of volume V, is released by a fish at a depth It in a lake. The bubble
rises to the surface.
Assume constant temperature and standard atmospheric pressure above the lake; what
is the volume of the
bubble just before touching the surface? The density of the water is p.
I Equating p°V,, at depth h to pV at the surface leads to the equation (p + pgh)Va
= pV. from which
V = (1 + pgh/p)V,,.
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328:3
20.14
20.1.5
NJ6
11.17
20.18
NJ9
H21
20.22
CHAPTER 20
li 2.1212 g of a monatomic gas occupies 1.49 L when the temperature is 0°C and the
pressure is 810.6 kPa,
what is the gas?
I pV = (m/M)RT. where m = mass of gas and M = molecular weight.
., _
(810.6 X I0, Pa)“-49 X I0 , m_,) = (2.1212 X 10 kg)(8.3l:l4.1/mol K)(273.l5 K)
M = 3.99 X 10 ‘ kg/mol = 3.99 g/mol =>gas is helium
Use the general gas law to compute the density of methane. CH1. at 20°C and 5-atm
pressure. A ltilomole of
methane is 16.0 ltg.
I pV = nRT, with T = 20 + 273 = 293 K. R = 8314.1/kmol - K. n = 1 kmol, andp = S
atm =
S(l.013 X 10‘) N/m‘. Then
s( 1.013 >< i0‘)v = a3i4(29s) and v = -as m‘ 4 = = :—6'89= 3.33 kg/m‘
Solve Prob. 20.15 without finding the volume.
I Use pV = (m/M)RT and p = m/V to give
_&1_ (5 X 1.013 X m‘ N/m‘)(16kg/kmol) _ _,
" _ RT_ (83141/kmol - K)(293 K) ‘ 3:5—'33 k /'“
A llll-it’ volume of nitrogen at 27°C and 15 lb/in‘ is compressed to fill a tank
that is initially empty and has a
volume of Sit". If the final temperature of the nitrogen is 17°C, what is the
absolute pressure in the tank?
I Use the Kelvin scale and the general gas law in the form
PiVi_P=V= 15(1°0) _ P:(5] _1500(290)_ . 1
T, ‘ r. 273+27-2734-17 P" 5000) 290"’ "'
In the preparation of a sealed-ofl' Zl)-mL tube at low temperatures. one drop (50
mg) of liquid nitrogen is
accidentally sealed ofl in the tube. Find the nitrogen (N;) pressure within the tube
when the tube wanns to
27 “C. Assume ideality. Express your answer in atmospheres (1 atm = 101.3 kPa).
I UsepV=(mRT)/M. Mis28for N,. Use T=300K.m=5><1U ‘kg, V=2x1tI ‘in’. and R=
8.31411]/kmol ' K. giving p = 223 kPa = 2.20 atm.
A car tire is filled to a gauge pressure of 24 psi (lb/in’) when the temperature is
20 °C. After the car has been
running at high speed. the tire temperature rises to 60°C. Find the new gauge
pressure within the tire if the
tire's volume does not change.
T _ 333 K . .
I p, = p.,T' = [(14.7 + 24) psi] fi = 44.0 psi or 29.3 psi gauge pressure
0
In a diesel engine. the cylinder compresses air from approximately standard
pressure and temperature to
about one-sixteenth the original volume and a pressure of about 50atm. What is the
temperature of the
compressed air?
I UR (PiVi)/Ti =(R>V»)/77.10 find Ti = (Pi/P.»)(Vi/V.>)7I-= (5")(fi)(273) =853 K.
One way to cool a gas is to let it expand. Typically. a gas at 27°C and a pressure
of 40atm might be
expanded to atmospheric pressure and a volume 13 times larger. Find the new
temperature of the gas.
I As in mu. 20.20. r. = (P./R.)(V./v..)T.. = (.:,)(1s)(300) = 91.5 K g- 116 "Q.
A vertical right cylinder of height h = 30.11) cm and base area A = 12.0cm* is
sitting open under standard
temperature and pressure. A 5.0-kg piston that fits tightly into the cylinder is now
placed into the cylinder and
allowed to fall to an equilibrium height within it. What then is the height of the
piston and what is the
pressure within the cylinder? Assume the final temperature to be 0°C.
I Pressure in the cylinder is increases by Ap = (mg)/A = (5.0 ltg)(9.8 m/s1)/(12.0
X 10" m‘) =41 ltPa. giving
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N."
20.75
20.76
20.77
20.78
20.79
20./U
N31
GAS LAWS AND KINETIC THEORY J 341
A volume of 30 m’ of air with 85% humidity at 20 “C is passed through drying
equipment that removes all the
moisture. How many kilograms of water are removed?
I Use the ideal gas law and Table 20-1 to calculate the density of the water vapor
at 20°C:
_ g _ (18 kg/kmul)(U.8S >< 2.31 its/in’) _
" ' RT ' (8.214 tr»: - m/kmol » r<)(29a to ' Om“ kw",
Hence, (30 m")(0.0l4S kg/m‘) = 0.435 kg water removed.
If the air in a room has a dew point of ll "C. what is its relative humidity at
21°C?
I By the arguments of Probs. 20.71 and 20.73.
_svpatll°C_i4_ p
R'"' —5vp at 21 =c_z.so_5W'
If it sample of air at 68°F and 55% relative humidity is slowly cooled.
condensation will occur at what
temperature’?
I We again use the reasoning of Prob. 20.73. From Table 20-l, the saturated vapor
pressure at 68°F = 20°C
is 2.31 ltPa. (0.$5)t2.3l) = |.27 kPa. which is the saturated vapor pressure at
(flPP°X-) 10°C = 50°F-
1‘he relative humidity of a room is 75% at 23°C. lf the temperature falls to 19°C.
what will the relative
humidity be?
I We again use the reasoning of Prob. 20.73. From Table 20-l the saturated vapor
pressure of water vapor
at 23°C is 2.88 kPa, giving an actual pressure of (0.7$)(2.88) = 2.l6 ltPa. From
the table, the saturated vapor
pressure at l9°C is 2.20 kPa. The new value is therefore relative humidity = 2.16}
2.20 = 98%.
What is the dew point if the air has a relative humidity of 60% at 25 “C?
I From Table 20-l the saturated vapor pressure of water vapor at 25°C is 3.26 ltPa;
hence P = (0.60)(3.26) =
L96 ltPa. The dew point is the temperature at which the corresponding vapor
pressure as the saturated vapor
pressure. From the table. then. dew point *== l7°C.
It the normal lapse rate (decrease of temperature with altitude) prevails, and the
temperature at ground level
is 21°C. What will the air temperature be at height of 1100 m?
I The normal decrease of temperature with altitude is about 1 ‘C per 110 m.
(1100/ll0)(l °C) = 10°C
decrease. 21°C — 10 “C = ll ‘C. (Note that 294 K—-284 K is a small percentage
decrease on the Kelvin scale.)
(I) Pilots of light planes must be careful to calculate the loads on warm days. Why
must pilots leaving from
or landing at high elevation (for example, Denver, Colorado, or Mexico City) be
particularly careful?
(0) Compare the density of the air at 0°C to the density at 30 °C. Assume identical
pressures.
I (n) ln relatively sparse air. there will be less lift on the wings of the
airplane. Other (actors being equal,
take~ofl runs will be longer, climbing rates will he smaller, and the desoent during
landing approaches will be
more rapid in the less dense air at high elevations and/or on warm days. (Ir) At
constant pressure, density
and temperature are inversely proportional p;/p, = T,/1}. With T, = 273.15 K (I;
=0‘C) and
7} = 303.15 K (1, = 30°C). we find p,/p, = 303.15/273.15 = 1.11. That is. at 0°C the
density is
ll Eroent higher than it is at 30°C.
Compare the density of the air at Logan Airport in Boston (elevation 0 m) at 0 ‘C
to the density of air at
Stapleton Field in Denver (elevation 1600 m) at 30 ‘C. At constant temperature. the
atmospheric pressure p
obeys approximately the equation p =p,,z"""" if the elevation z is expressed in
meters. and p, is the
atmospheric pressure at 0 m.
I We begin by determining the pressure at Stapleton Field with the help of the
hydrostatic isothermal
profile. Using the subscripts l and r to refer to Logan and Stapleton, we have
P‘ =p‘t-(1, r,>nt.\u
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342:
N32
20.83
20.84
29.85‘
CHAPTER 20
We now determine the density which corresponds to p, and the given temperature T,.
If the temperatures
were equal (T, = 7)). the density ratio p./p, would equal the pressure ratio p,/p,
(by Boyle's law). Since
7} 1* T,, we must use the more general relationship (p/pT) = constant (since the
gas composition is assumed to
be the same at the two locations). That is.
5 = 51¢ = L t .,.......
Pi KP: 7;
With T, =0“C =273.l5K. T, = 30°C = 303.15 K. and 2. - :,= l600 m. we obtain
‘Z73. '
%:= ttmntnw-i =(0-9O10)e n ms =E
Under the given conditions and assumptions. the density is 26 percent lower at
Stapleton Field than at Logan
Airport. Equivalently. the density at Lo_ga_n_PLirpon is 35 percent higher than
that at $(tlplrl)_n_
Derive the law of atmospheres from the Maxwell-Boltzmann statistical law which
states that the number of
particles (in an equilibrium ensemble) with energy E is proportional to e ' ”.
Assume constant temperature
over the heights being considered. (See Prob. 20.79.)
I Since the only tenn in the energy of air molecules that depends on the vertical
height. :. is the
gravitational potential energy. mg: (measured from ground zero). where m is the
average mass of a molecule
of air. the particle density n_(z) obeys (Maxwell-Boltzmann): n_(zl1 2' """"" = e
""'° "'. and
[n,,(z)]/|n,(o)] =e """“'/l. Since the density of air. p. is proportional to n,,.
we have p(z)/ptn) = e """".
Finally. at constant temperature. p(z) “p(z): so p(z) =p(0)e"""" "r. which is the
law of atmospheres.
Find the unilonn temperature at which the ratio of the densities of mercury vapor
at the top and bottom of a
2.0~m-high tank would he 1/e. (Assume an ideal gas could be obtained.) (Molecular
weight of mercury = 20l.)
| From the law of atmtispheres. p,/p; = expI(—mgh)/(kT)]. Hence l/2.718 = exp {[—
m(9.8)(2)|/(kT))
where m = 201/N, = 3.34 X I0 " kg. lnvert and take natural logarithms of each side
to find l = 19.6":/kT.
Solving for T yields T =0.47 K.
A gas of dust particles fills a 2.0-m-high tank. At equilibrium (27 °C). the density
of particles at the top nf the
tank is I/e the density at the bottom. Find the mast of a typical particle and find
how many times more
massive it is than a nitrogen molecule.
I Proceeding as in Prob. 20.83. (mgh)/(kT) = l. This gives m = (RT)/(gh) = 2.1 X I0
3 kg. The mass of a
nitrogen molecule is 28/NA =4.65 X ll) "‘ kg and so the muss of the dust particle
is gill) times greater.
lnfer the law of atmospheres from the ideal gas law.
Z rlr ~ -\:)
A:
pm Fig. at-1
I Figure 20-7 shows a thin slab or air at altitude z. For equilibrium of the slab.
dr
|p(z)—p(z+Az)]A=pgA AZ or 5= —pg
From the ideal gas law.
P _fl
p mkT or p krp
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20.86
NJ’!
N438‘
20-89'
M30
GAS LAWS AND KINETIC THEORY J 343
where m is the average molecular mass for air. Thus
dP_ ms
.11" /<1”
Ignoring the variation of T and g with altitude,
p£p._.__w : _ (M17):
l["p- kTLdz and p-p.,e ' .
Using the Dulong—Petit law. estimate the high-temperature specific heat capacity (in
J/kg - K) for uranium
metal (M = 238.)
| From the Dulong-Petit law the molar heat capacity at constant volume CV = 3R.
Thus. cv = C,-/M =
[3(8314)]/Z38 = 105 J/kg - IQ.
The Dul0ng—Petit law was used early in this century to determine the molecular
weights of crystalline solids.
A certain pure metal has a specific heat of 2301/ kg - K at high temperatures. What
is the molecular weight of
the metal?
I Again use Cy = 3R and M = CV/cv = 3(83l4.l/kmol - K)/(230J/kg- K) = 108 kg/lrmol.
Show how the equipartition theorem (Prob. 20.38) leads to the law of Dulong and
Pctit.
| At high temperatures. it may be supposed that essentially all the intemal energy
of a metal is due to the
vibrations of the atoms about their equilibrium positions in the crystalline
lattice. lf we picture each atom as
connected to its neighbors by springs. then the atom will have kinetic and
potential energies along three
mutually perpendicular directions-six modes in all. By the equipartition theorem.
its total energy will be
6(§kT) = 3kT, giving a molar energy of E = N,,(3kT) = 3RT and a molar heat capacity
of Cy = dE/dT = 3R.
The Dulong—l’etit law holds tnte surprisingly well for solids if the temperature is
high enough. To predict the
behavior of the heat capacity at lower temperatures, a quantum-mechanical model
must be used. One such
model, originally used by Einstein. assumes that all atoms in a solid vibrate at
the same frequency v. The
total energy of a solid of N atoms is then the same as the energy of 3N one-
dimensional oscillators The
correct quantum-mechanical expression for the average energy of this collection of
oscillators is (E) =
3Nhv[§ + l/(e"" — 1)]. where /3 = 1/kT, and h = 6.63 X l()'"“J ~ s. Show that this
model gives the molar heat
capacity as
8 I (GIT
0- 3R(T)(e&,_1):
where 9 = hv/It.
I Writing A for Avogadro's number, we have. for one mole of material.
l l
(E)=3/""'[§+;mj] (1)
so the molar heat capacity is given by
_d(E)_ (_1) h"T_ B-: ehvtlrf
C ' ar ' 3A"'(¢"""- 1)’ (./<T=)" " 3"“‘)(tr) (¢"'"- I)’ (2)
Since A/r = R. the gas constant. Eq. (2) becomes
v 2 ei')'1
‘ - “(ri <3)
where we have put 9 Ehv/k.
Refer to Prnh. 20.89. Determine the behavior of C in the limit T>> 6. and in the
limit T<< 6.
| For T>> 9, we have 8/T << 1. so Eq. (3) yields
_ . ll+<em+---1 _, (9/T)‘ll+9/Tl__
C“3R(e/T) ([1+(e/r)+---1-1): 3R (9/T)‘ M
which is the Dulong-Petit result.
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350 L7 CHAPTER 21
P
Adiahatit
lsnlhtfm
F" ' ' ' '
' ’ lvnlhctm
Adizrhiitiv
v» v Fig. 21-2
21.26‘ ln a p-V diagram (Fig. Zl-2) an adiabatic and an isothermal curve for an
ideal gas intersect. Show that the
absolute value of the slope of the adiabatic is 7 times that of the isotherm. Hence
the adiabatic curve is
steeper because the specific heat ratio y is greater than 1.
I Denote the intersection point by (p,,, V9). Then the isothermal curve is given by
pV = p,,V,_., or
P-so =PoVnV“' ll)
while the adiabatic curve is described by pV' = p,, V5, or
P-¢*P0viiV"' (Z)
111$, sloae of the isothermal curve at (pm V") is found by differentiating Eq. (1)
and evaluating the derivative
at = H:
£11 ._ -= ,'_P=>
(dv)m p..v..v V0 ‘,0 <1)
Using Eq- (2). the slope of the adiabatic curve at (pg, V0) is
Q) _ _ - —-I __ _YPu
(W _‘- wovsv Y vb-Tn <4)
Equations (3) and (4) show that
.‘.‘—'&).. ='l(.‘I—’;)_..|
for the curves intersecting at (p,,, V").
21.2 THE FIRST LAW OF THERMODYNAMICS, INTERNAL ENERGY, p-V DIAGRAMS,
CYCLICAL SYSTEMS
21.27 What is the change in intemal energy of 0.1(X) moi of nitrogen gas as it is
heated from I0 to 30°C at
(a) constant volume and (b) constant pressure?
I internal energy for an ideal gas is linearly related to the temperature of the
gas with AU -= nC,, ATso the
answer to parts (n) and (b) is the same: AU = (0. 1(1) mol)(4.96cal/moi - K)(20 K)
= 9.92 cal = 41.51.
21.28 When 50 L of air at ST!’ is isothermally compressed to I0 L, how much heat
must flow from the gas?
(pm, = ill) kPa.)
I Since the process is isothermal. and we assume an ideal gas. AU = 0. Then AQ =
AW. But AW =
nRT in (V,/V.) = P,V, ln (V,/V,). The substitution makes use of the ideal gas law.
P,V, = nR1}, and recognizes
that T is constant. in this case, AQ = (1 X 10’)(S X 10") ln I0/50= -8050 J; i.e.,
8.05 kl flows out.
21.29 An ideal gas in a cylinder is compressed adiabatically to one-third its
original volume. During the process.
45 I of work is done on the gas by the compressing agent. (n) By how much did the
internal energy ol the gas
change in the process? (la) How much heat flowed into the gas’!
| in this case. AQ = 0, so AU = -AW = -(-451) =~@; (b) the heat flow in the
adiabatic process is zero.
21.30 ln each of the following situations, find the change in internal energy of the
system. (n) A system absorbs
Sllflcal of heat and at the same time does 4(1)] of work. (b) A system absorbs
300cal and at the same time
4201 of work is done on it. tc) Twelve hundred calories is removed from a gas held
at constant volume.
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21 .31
21.32
21.33
21.34
THE FIRST LAW OF THERMODYNAMICS .» 351
I (a) AU=AQ—AW=(500eal)(4.l84J/cal)—400J=l7(I).I
(II) AU=AQ—AW=(3()0calj(-H84]/eal)—(—420J)= 16801
(cl AU=AQ—AW=(—l20Uc;tl)(4.l8~4.l/caI)—0-=—5000J
Note that AQ is positive when heat is added to the system and AW is positive when
the system does work.
In the reverse cases. AQ and AW must be taken negative.
Rederive the result of Prob. 21.25 using the first law of thennndynamics and the
facts that C, — Ct = R and
C,/Cy = y for an ideal gas. where C denotes molar heat capacity.
I Let the heat flow be denoted by AH. and the internal energy change by AE. Since
the process is
adiabatic, AH = 0 and therefore AW = —AE. But AE = nC.»(T,, — TA). and furthermore
R = C, — Cy =
(y - l)Cv. Hence we can express the energy change as
nR
A5 — pm(Tn - T4)
But the ideal gas law yields nRT =pV. so we obtain
Aw: _AE= 'Pnvn+P4V,i=PAv4‘I7nVn
y-l y—1
in agreement with the result of Prob. 21.25.
Find AW and AU for a 6-cm cube of iron as it is heated from 20 to 300 “C. For iron,
c = 0.11 cal/g - “C and
the volume coetficient of thermal expansion is B = 3.6 >< IO ' °C' ‘. The mass of
the cube is 17tX)g.
I AQ = cm AT = (0.11 cal/g - °C)(l7(l08ll2ll0 “Cl = S2 000 cal
The volume of the cube is V = (ticm)‘ = 216cm‘. Using (AV)/V = )3 AT. we have
AV = Vfl AT = (Zl6X I0 “m")(3.6 X10 “C ‘)(280°C)= 2.18 X ll) "m"‘
Then. assuming atmospheric pressure to be I X 10‘ Pa.
AW =p AV = (l >< l(l' N/m‘)(2.l8 X10 "m"‘) = 0.22]
But the first law tells us that
AU = AQ — AW = (S2000cal)(4.l84 J/cal) — 0.22] = 218000.! -0.22] == 218000]
Note how very small the work of expansion against the atmosphere is in comparison
to AU and AQ. Otten
AW can be neglected when dealing with liquids and solids.
A cubic meter of helium originally at 0°C and 1-atm pressure is cooled at constant
pressure until the volume
is 0.75 rn’. How much heat was removed‘!
I Use the general gas law to find the final temperature. Then use the first law of
thermodynamics.
P»_"~_r'=_"= =
T‘ Pt P:
Then
1 0.75
273 T,
AQ = AU + AW; and for an ideal gas. AU = me, AT. Thus. AQ = me. AT + p AV = n(‘, AT
+p AV. Noting
that at STP. l kmol occupies 22.4 tn“.
_L _ (1.o|3><10‘)(o.75—t)__ _ __
AQ -2“ (3)(205 273) + -i-i—“84 - 9.11 0.05 - _1_5Lg_);;al
T,=2USK
The minus sign means that heat is removed.
The volume of I kg of water at l00°C is about l X t0 ‘ rn‘. The volume of the vapor
formed when it boils at
this temperature and at standard atmospheric pressure is l.67l m". (0) How much
work is done in pushing
back the atmosphere? (bl How rnuch is the increase in the internal energy when the
liquid changes to vapor?
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352 3 CHAPTER 21
21.35
21.36
21.37
21.38
21.39‘
I (n) The work done by the water is given by
AW =p,m AV = (1.013 x 10‘Nlm*)[(l.67l -01001) m’] = 169 kJ.
(b) According to Table 17-2. the latent heat of vaporization of water is L = 540
ltcal/kg. This is the heat per
kilogram that must be added to vaporize 100-°C water at a constant pressure of l
atm. The change in the
intemal energy of 1 kg of water when it is boiled at 1 atm is therefore given by
AE = AH — AW = Lm — AW = (540 ltcal/kg)(l.00 kg)(4.l54 kl/kcal) -— (169 kl) = 2090
kl
A tank contains a fluid that is stirred by a paddle wheel. The power input to the
paddle wheel is 2.24 kW.
Heat is transferred from the tank at the rate of 0.586 kW. Considering the tank and
the fluid as the system.
determine the change in the intemal energy of the system per hour.
I AE=Q—W=—0.586—(—2.24)=l.654lt\\'=1.654kJ/s=59S4kJlh
or about6M.I[h.
A spring having a spring constant 5 N/m is compressed 0.04 m. clamped in this
configuration, and dropped
into a container of acid in which the spring dissolves. How much potential energy
is stored in the spring. and
what happens to it when the spring dissolves?
I The spring in being compressed acquired elastic potential energy of amount
U,,,,,_ = §k.x‘ = §(5)(0.0-1): =
01004 J. When the spring is dissolved. this ordered potential energy is convened
into disordered potential and
kinetic energy of the system. Overall. energy is oonserved.
One pound of fuel, having a heat of combustion of 10000 Btu/lb. was bumed in an
engine that raised 600011:
of water 110 ft. What percentage of the heat was transformed into useful work?
work done by engine (6000 lb)(110 ft)
I °m°i°"°y : work equivalent of heat supplied T (10 000 Btu)(778 11 - lb/Btu) =
0'0“ = L-L
A sample containing 1.00kmol of the nearly ideal gas helium is put through the
cycle of operations shown in
Fig. 21~3. BC is an isothennal. and p, = 1.00 atm. V, = 22.4 m’. p, =2.00 atm. What
are T,,, T,. and V‘-?
p
B
,1 "
Y 11;. 21-s
I Applying the perfect gas law. we have
T =;>,_v, = (1.013 X 10’ 1'a)(22.4 m’) = 273 K
A nR (LU) ltl!'l0l)(8.314 X 101]/ltmol - K) i
Because the process AB shown in Fig. 21-3 is isometric, T, = (p,/p,)T,,. With p, =
21]) atm = 2p,, we find
that T, = S46 K. The process CA is isobaric, so Charles’ law applies. Thus V‘-/ 1}
= VJ TA. so that
V( = (D/T,,)V,,r But BC is an isothermal process, so that TL- = T, = 2T,,.
Therefore V( = 2V, = 44.§m".
Refer to Prob. 21.38 and calculate the work output during the cycle.
I The work done by the gas along AB is zero, since dV = 0. The work done on the gas
along the isotherm
BC is
"" "‘ RT, Vc
Awpt’ =1 P av =J' 7 av = RT, ln (7) = RT, ln 2 = (am kl/ltmol-1()(S46 t<)(o.e9s1) =
3150 u
Va n I
v
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22.14
22.15
22.16
2.17
THE SECOND LAW OF THERMODYNAMICS J 361
- _Q:~-Q¢_T~-71 5411-Q<_590"399
I efl’rc|cncy—-—i-Q‘ -T ism -isw
500 — Q, = 169 keal and Q, = 331 kcal delivered to the sink
W I Q, - Q, I I69 keal = 169(4.18-1) =71!) kJ.
A steam engine operating between a boiler temperature of 220 “C and a condenser
temperature of 35°C
delivers 8 hp. If its efficiency is 30 percent of that for a Camot engine operating
between these temperature
limits, how many calories are absorbed each second by the boiler? How many calories
are exhausted to the
condenser each seoond?
I actual efliciency = (0.30)(Carnot etficiency) =(0.30)(1 — = 0.113
But from the relation
1 calls
efliciency = § input heat/s = - = 12.7 kcalls
To find the energy rejected to the condenser. we use the law of conservation of
energy:
input energy = output work + rejected energy
Thus,
rejected energy/s = (input energy/s) — (output work/s) = (input EIWYBY/$)[l —
(efliciencyfl
= (12.7 kcal/s)(l -— 0.113) = 11.3 kcalis
How many kilograms of water at 0°C can a freezer with a coeflicient of performance S
make into ice cubes at
0°C with a work input of 3.6 MJ (one kilowatt-hour)? Use Table 17-1 for data.
I By Prob. 22.3. the ooeflieient of performance is Q,/ W.
5 _ mL = »-(so kcal/kg)(4184J/kcal)
_3.6><l0“J 3.6><10“J
Solving, m = $4 kg.
A refrigerator removes heat from a freezing chamber at -5 ‘F and discharges it at
95 “F. What is its maximum
coelficient of perfonnance?
I For a Camot refrigerator, Q, I W = 1}/(1; - 7}). Use absolute temperatures
(Rankine). Then
Q -5 + 460 _ Qé = .
W F-95 + 460) _ (_5 + 460)-100 4.55 coeffictent of rformarig
A freezer has a coefficient of performance of 5. If the temperature inside the
freezer is -20 “C. What is I116
temperature at which it rejects heat’? Assume an ideal system.
Q,_ 1; _ 253
' w'1;,-T, 5‘1,-zss
Solving, we get 11 =304K=3l ‘C.
22.2 ENTROPY
2.18‘
Give a mathematical definition of entropy and discuss its relation to the second law
of thermodynamics.
I Any thermodynamic system has I state function S, called the entropy. By this is
meant that S—like p, V.
and U—is always the same for the system when it is in a given equilibrium state.
Entropy may be defined as
follows. Let a system at absolute temperature T undergo an infinitesimal reversible
process in which it absorbs
heat AQ. Then the change in entropy of the system is given by
AS = A12 or d5 = £7? for infinitesimals
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362
22.19
22.20
22.21
22.22‘
CHAPTER 22
Note that dQ is not the differential of a true function. Entropy will have the
units J/K.
The Clau.riu.r equation, db‘ = JQ/T. holds only for reversible processes. However.
since S is a state
function. the entropy change accompanying an irreversible process can be calculated
by integrating dQ/ T
along the path oi an arbitrary reversible process connecting the initial and final
states.
The importance of the entropy function is exhibited in the following form of the
second law of
thermodynamics.‘ In any process, Ike mm! enlmpy of the system and it: Surroundings
increases or (in n
reversible process) does not change. The second law applies to the system alone if
the system is isolated; that
is. it it in no way interacts with its surroundings.
Define entropy in terms of order/disorder, and discuss briefly.
| The second law of thermodynamics indicates that entropy is a measure of
irreversibility. Irreversibility is
associated. on the molecular level, with the increase of disorder. Molecular
systems tend. as time passes, to
become chaotic, and it is extremely unlikely that a more organized state, once
lett. will ever be regained.
Another. fully equivalent. definition of entropy can be given from a detailed
molecular analysis of the system.
It a system can achieve the same state (i.e.. the same values of p, V. T, and U) in
Q dilferent ways (ditferent
arrangements of the molecules, for example), then the entropy of the state is S = k
ln Q, where ln is the
logarithm to base e and k is Boltzmann's constant, 1.38 X 10""!/K.
A state that can occur in only one way (one arrangement of its molecules. for
example) is a state of high
order. But a state that can occur in many ways is a more disordered state. To
associate a number with
disorder. the disorder of a state is taken proportional to S2, the number of ways
the state can occur. Because
S = k ln Q, the entropy is a measure of disorder.
Spontaneous processes in systems that contain many molecules always occur in a
direction from
‘state that can exist state that can exist
->
(in only a few ways) ( in many ways J
Hence systems when left to themselves retain their original state of order or else
increase their disorder.
For a heat engine. over one cycle. AS = AE = 0, since the engine returns to its
original state. The first law of
thermodynamics then gives for the work done by the engine per cycle.
T
w = Q...(1 — A") — 1.... /.\s..... = w' — 1.... As... (I)
Tm:
where W’ is the work that would be done by a Camot engine operating between the
same two temperatures.
and ASW, is tht: entropy change of the universe (in this case. the entropy change
of the hot and cold
reservoirs) during one cycle. Derive (1) and show its significance for the second
law.
I For any engine operating between the temperature reservoirs, we have W = Q,“ —
Q.,,,_,. since AU = 0
over a cycle. (Here QM, is defined as positive when heat leaves the engine. as is
usual in engine and
refrigerator problems.) ln addition the changes in entropy of the hot and cold
reservoirs are related to the
heat transfers at constant temperature by Q,,_,. = —T,_,. AS,..,,: QM, = T‘°]d
ASMII Then. noting AS.“ =
A5’... + A$...a- W have T...» A3»... = Q...» - (T...aQ>....)/Tr..-.- Th\l§ W =
Q»... - Qt.“ = Qmll — (Tm/Teal] -
TM, AS...,,,. Since the etliciency of a Camot cycle is W‘/QM = l — (TM,/TM), we
have the remainder of our
result: W = W‘ — T“... AS,_,._,. We thus see that for a given Q,....- W s W‘cAS 20.
Or. in tertns ot
efliciency, n 5 1;’ ea AS 2 0.
When ll!) coins are tossed, there is one way that all can come up heads. There are
100 ways that only one tail
is up. There are about l >< 10” ways that 50 heads can come up. One hundred coins
are placed in a box with
only one head up. They are shaken and then there are S0 heads up. What was the
change in entropy of the
coins caused by the shaking?
I From Prob. 22.19.
AS = k(ln Q, — In $2,) = (1.38 X 10"’J/K)|ln (1 X 102“) — ln 100]
=(l.38x10'”J/K)(27lnl0)= 8.6 x l0'=.l(|§
using ln l0=2.303.
The number Q of states accessible to N atoms of a monatomic ideal gas with a volume
V, when the energy of
the gas is between E and E + dE, can be shown to be Q =A(N)V"'E"""’. where the
factor A(N) depends only
on N. la) Find the entropy S as a function of V and E. (b) Using this entropy
function and the definition of
the Kelvin temperature. l/T = (SS/BE)“ show that E = §NkT.
AAJ K/-\ TOPPER ii'EeEi°§i'ii'§.gi'E'é°R':i§

AAJ KA TOPPE R NEET AIIMS JEE RAS
372 J CHAPTER 23
while the instantaneous transverse velocity is
g = kv/t sin [k(x — U!) + 6] (3)
Comparing Eqs. (2) and (3), we see that
9y _ 1 8y
8: _ u 5: (4)
as desired.
8.34‘ Give the general mathematical fonn of a wave traveling at constant speed,
without dissipation. along the x
axis, and show that it satisfies the standard wave equation.
I A wave y(x, I) traveling along X with an unchanging form and speed v is given by
y =f(x 1 vi), where the
minus and plus signs refer to wave propagation in the positive and negative X
directions. respectively.
These functions satisfy the one-dimensional wave equation
fir _ 1 iy
ax’ ' U’ afl
as can immediately he seen by setting u = x 1 vr and using
6y dy 8u 8y dy Su . .
— = — — — = — — f h h f
ax du ax at du 8‘ or eit er c once 0 u
23.35 Sketch the profile of the wave/(x. !)= A: """“’: at !=0s and I = l s. using A
= 1.0m. B = 1.0 m‘ :. and
u = +2.0 m/s.
I See Fig. 23-S. Note that peak-to-peak distance divided by time interval equals u.
_\'.m
2
'l'=l\
l=l|s
I ' . .1 x.m
-2 -—l 0 | 2 3 fig, 13.5
23.36‘ Verify by partial diflerentiation that the wave function of Prob. 23.35
satisfies the one-dimensional wave
equation.
I We begin with the proposed solution
/‘(L 1) =Ae“"‘ --* (1)
Dilierentiating with respect to x, we find that
3‘: _2A8(x_U')e<ati-I-1)? (2)
and
3‘ . . .
é=[-ZAB +4AB"(x -vi)-|r""""' (3)
Diflerentiating with respect to 1, we find that
g= 2ABv(x - UI)e "" "'*‘ (4)

23.44
23.45
23.46
23.47
23.48
23.49
23.50
WAVE MOTION I 375
support; at the other end. a tension of l00N is applied. lf a transverse blow IS
struck at one end of the tube.
how long does it take to reach the other end?
I u — J-LT1= \)%)=18.3m/s .r = U! so 5.0=18.31 and r=0.27s
Refer to Prob. 23.43. What frequency of vibration must be applied to the tube to
produce a standing wave
with four segments in the length of the tube? (This frequency is called the fourth
harmonic frequency.)
I We have 4(Al2) =S.0m. or A =2.5 m. Then v= v//1 = 18.3/2.5 =7.3 Hz.
.0
Standing waves are produced in a rubber tube 12m long. If the tube vibrates in five
segments and the velocity
of the wave is 20 mls. what is (a) the wavelength of the waves. (b) the frequency
of the waves?
I (a) = 12m or A =-_1_._8l
_20m/s__
lb) Y-4v8m—h
A string has a length of 0.4m and a mass of 0.16g. if the tension in the string is
70 N. what are the three
lowest frequencies it produces when plucked?
_ E- /L,/i_
| u-\/;- (0_ml6/M)- 1750(1)-418111/s.
The fundamental frequency v, corresponds to a wavelength of 2(0.4) = 0.8 m; thus,
418m/s
V-— 0_8m W
The second and third harmonic frequencies are then 2v, = 1046 Hz and 3v, = 1569 Hz.
The third overtone produced by a vibrating string 2 m long is 1200 Hz. What are the
frequencies of the lower
overtones and of the fundamental? What is the velocity of propagation?
I The third overtone is the fourth harmonic v,. v, = 12001-lz implies that v, =
l2(D/4 = 30] flz. Then
v, = 2v,=QQ!;l1, and v, = 3v. =900 fig. u = /Lv, = (1 m)(l200 Hz) = 12(1) tnls (or v
= /1,v, =
(4 m)(300 Hz) = 12(1) rn/s).
A 160-cm<long string has two adjacent resonances at frequencies of 85 and 102 Hz.
(a) What is the
fundamental frequency of the string? (b) What is the length of a segment at the 85-
Hz resonance?
(c) What is the speed of the waves on the string’!
I (a) The fundamental is f, = u/(2L), the nth harmonic frequency is nf, = 85 Hz
while the next harmonic
frequency is (n + 1))’, = IUZ Hz. From these we obtain], = 171-lz and n = 5. (b)
For n = 5. there are five
segments; the length of each is 160/5 =§2 cm. (e) Fromf, =17 = t1/|2(1.6)]. v =
§4.4 mls.
A vertically suspended 2m-cm length of string is given a tension equal to the
weight of an 800-g mass. The
string is found to resonate in three segments to a frequency of 480 Hz. What is the
mass per unit length of the
string?
I In the third harmonic. (3).)l2 = 2.00m; so that u =fi. = 480(1.33)= 640 mls. We
then use the relation
u‘= T/it to find it -= (mg)/v'= [0.800(9.80)]/640:= 1.91 >< 10‘ ’kg[m.
The equation for a particular standing wave on a string is y = 0.15 (sin Sx cos
3001) m. Find the (a) amplitude
of vibration at the antinode. (bl distance between nodes. (c) wavelength. (4)
frequency (e) speed of the wave.
I (er) By comparison with y,, I A sin [(2:rx)/A] oos (lzrfr) we have A =0.1S m. (b)
When the argument of
the sine = 0. II. 2.'r, . . . , we have nodes. Since x = 0 is a node we have for
the next node 5x = ll’ and
x = ll/5 =Q._§Z_8_m. (c) The wavelength is twice this. so A = (Zn)/5 = 1.26m. (4)
Since 2:11 = 3(1), I =1S0/n =
47.7 fig and (e) u = Af = [(2.1)/S](1S0/x) =60 mls.
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376 J CHAPTER 23
23.51 An organ pipe l ft long is open at both ends. ll the velocity of sound is
I100 ft/s. what are the frequencies of
the fundamental and of the first twu overtones?
I As can he seen from Fig. 23—8. n()./2) = I. holds for an open organ pipe as well
as a string. Thus. the
fundamental frequency is v, = v/2L =(l10(l ft/s)/2 ft = 550 Hz. The first Overtone
is the second harmonic. and
v, = 2v, = 2(SSl)) = lit!) Hz. Similarly v, = 3v, = 3(S50) = l650 llz (second
overtone).
|--—_1. - mil
It
Fuut|;iment;il
2., -*-' II.
I
|'ttst uwitone
,1, = 1, I-‘lg. as-a
73.52 A closed organ pipe 2.5 ft long is sounded. If the velocity of the sound is
lltll ft/s. what are the fundamental
frequency and the first two overtones‘?
I For a "closed" pipe (that is. open at one end) (2n - l)t)./4) = L. as can he seen
from Fig. 23-9. Hence
v,,, ,= (Zn - llv/-tL = (Zn — l)v,, so only the odd harmonies appear. For our case.
A, = 4L =4(2.5) = 10ft.
ti = v,l, or H00 = \',(1O) and v, = 110 Hz. The first overtone is the third
harmonic. so )._. =4LI3 and
v, =11/A. = 3v/4L = 3v, = 31110) = 330llz. Similarly. vt = 5v, = 550 Hz.
/
Z
K
i_ ;:.;
mi?
\_“\//\
Ft I It It
:31: ll
I ' l
t
fig. 23-9 4! . ta‘. 23-to
l—
23.53 A sounding tuning fork whose frequency is 256 Hz is held over an empty
measuring cylinder. See Fig. 23-IO.
The sound is faint. hut if just the right amount of water is poured into the
cylinder. it becomes loud. If the
optimal amount of water produces an air column of length 0.31 m. what is the speed
of sound in air to a first
approximation?
| The loudest sound will be heard at resonance. when the frequency of vibration of
the air column in the
cylinder is the same as that of the tuning fork. Sinec the air column is open at
one end and closed at the
other. we conclude that the wavelength of the vibration is four times the length of
the column:
it = 4L = (4l(0.3l m) = 1.24 m. Here we have assumed that the observed resonant
oscillation of the air column
is its fundnnmual oscillation. Since the frequency v = 256 H2. the sound speed ti =
v/1 = 317 mls. This is an
underestimate since the displacement antinode (or the pressure node) which is
located a distance of
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376 - CHAPTER 23
23.99‘ A uniform string (length L. linear density ii, and tension F) is vibrating
with amplitude A, in its nth mode.
Show that its total energy of oscillation is given by E = :r’vf,A,',uL.
I The displacement of the string is given by
y(x. I) = A, sin %xc0s (2:rv,,l + 6)
so the transverse velocity is
%:= —2:rv,,A,, sin %sin(2nv,,1+ b)
The string‘s total energy of oscillation is equal to the maximum kinetic energy.
(Note that all points on the
string achieve their maximum kinetic energy at the same time, when y = 0 for all
x.) Since dm = it dx, we
have
E = K =max = max [LM('3l):dx]
"“‘ 2 81 2 L, 6|
The maximum value occurs when sin’ (2.-rv,,r + 6) = l, so we find that
I.
E = g (Lw,/1_)=J; §in"'T""d.i
The average value of sin’ [(n.'Lx)/L] over any number of half cycles is given by §.
so the integral has the
I 2 Z Z
value L/2. Therefore E = 2n'uv,,A,,(L/2) =2! v,§fEL, as desired. [Compare with the
result for a traveling
wave, Prob. 23.29.]
73.60 A taut square membrane 85 cm on a side is fastened to a rigid frame on its
edges. When tapped lightly at its
center. it gives off a tone of about 200 Hz. Assuming it to be resonating in the
mode shown in Fig. 23-12,
what is the wave speed in the membrane?
Hg. 25-12
| The resonant frequencies of a rectangular drumhead are given by the formula
‘T
u In ’ n
’~~=5\/(Z) *(-)
where m and n are the respective numbers of half-wavelengths that fit into the
dimensions L, and L,. Figure
23-12 shows the mode m = n = 1; hence,
v 2
200 = — if i
2 (0.85):
P
giving v = 2-10 mls.
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CHAPTER 24
Sound
SOUND VELOCITY; BEATS; DOPPLER SHIFT
Helium is a monatomic gas that has a density of 0.179 kg/m’ at a pressure of 76 cm
of mercury and a
temperature of 0°C. Find the speed of compressional waves (sound) in helium at this
temperature and
pressure.
I u = VB/p. where B is the adiabatic bulk modulus. For an ideal gas. B = yp (see
Prob. 21.24), and
y = 1.67 for a monatomic gas. Then
1.671.013 10‘ NI ’
»~=v‘<...+.,...’.~"‘=9_-Lt='»§
Using the fact that hydrogen gas consists of diatomic molecules with M = 2
kg/ltmol. find the speed of sound
in hydrogen at 27 °C.
I For ideal gases v = l(YP)/Pl": = [(yRT)/MI“. so
U = [(].40)(B3l4J/ltmO| ' K)(300 K)/(2 kg/kmol)]"" = 1321 ml;
From the fact that the molecular weight of oxygen molecules is 32 kg/kmol, find the
speed of sound in oxygen
at 0 “C.
I As in Prob. 24.2. 2: = [(l.40)(83l4)(273)/32]':=315 mls
The velocity of sound in a container of hydrogen at -73 ‘C is approximately 4000
ft/st What would the
velocity be (in ft/s) it the temperature of the hydrogen were raised to 127°C
without a change in volume?
I Since v =[(yRT)/M1“ we have. assuming constant 7. v = t'.,VT/7.'.= (4000
ft/s)\/(400 K)/(200 K) =
5657 ftls.
What is the speed of sound in air when the temperature is 35 “C? The speed of sound
in air at 0°C is 331m/s.
I tr =< \/i'= V273 + I; so if 11,. = speed at t = 0°C. tt = u..(l +1/273)": =
331(1+ 35/273)” = 351.6 mls.
By how much must the temperature of air near 0°C he changed to cause the speed of
sound in it to change
by 1 percent’?
(273 +1)‘ 1 - Q73") _
I " — 0.0]
Noting [1 +1/273]": = l +1/546. we get (t/546)(l00) = 1.0. which leads to t = 5459C
A certain gas mixture is composed of two diatomic gases (molecular weights M, and
M,). The ratio of the
masses of the two gases in a given volume is m;/n1, = r. Show that the speed of
sound in the gas mixture is as
follows if the gases are ideal:
i‘l.4ORTM;+ VM,
[I = m-
\' .u..-4, 1 + I
I 1' = |(YP)/pl‘ 1. but p =n1/V = (m, + 111;)/V. and (Dallon‘s law) p =p, +p; = (nx
+ n,)[(RT)/V)|. with
n, = m,/M, and :1; = m,/My so that p/P = [(m,/M, + m,/MQRT]/(m, + m,) = [(1/M. +
r/M,)RT]/(1 + r).
with r = m,/m,. Substitute p/p and y = 1.40 into the expression for v to find the
required result.
Suppose (with Newton) that the compressional vibration in a gas were isothermal in
character rather than
adiabatic. Find the expression equivalent to (yp/p)' I for the speed of sound in
that case.
I For pV a constant. A[pV) =p AV + V Ap =0. or AV/Ap = —V/p. But AV/Ap = —V/B from
the
definition of B. Therefore. B=p and 1' = (B/p)‘ ’ = (p/p)’ F Thus t',,,/u,_,= y‘ :4
The Learning App For 379

380 I CHAPTER 24
241.9
24.10
24.11
24.12
24.13
74.14
A I000-Hz sound wave in air strikes the surface of ll lake and penetrates into the
water. What are the
frequency and wavelength of the wave in water’? Assume that the speed of sound in
water is 1500 m/s.
I The number of complete waves passing any point in air and in water in unit time
is the same. so
I = l000 Hz for both media. Therefore.
' l5(X)m/s
,\‘=‘_"=i=‘
~ I mos. Hi
An underwater sonar source operating at a frequency of 60 kHz directs its beam
toward the surface. What is
the wavelength of the beam in the air above? What frequency sound due to the sonar
source does a bird
flying above the water hear? Assume that v in air = 330 m/s.
I As indicated in Prob. 24.9. I rcmains constant--in this case at 60 kHz. The
wavelength = v/f =
330/(6.0 x 10‘) = 5.5 mm.
Define pitch. loudness. quality. decibel. reverberation lime, inlzrfzrence. beats.
Doppler effect. supersonic
velocity, shock wave, and Mach number.
I Pitch is a sound characteristic that depends on the frequency of the fundamental.
Higher pitch means
higher frequency.
Loudness refers to the strength of the auditory sensation produced by a sound. lt
depends on the intensity
and frequency of thc sound.
The quality (or timbre) of a sound depends on the number and intensities of the
overtones.
Decibel (dB) is the unit of intensity-level, n, of sound. One dB is ten times the
log of an intensity ratio of
1.26:1. The equation for n is n = l0log(I/In). where I.,= 10'" W/m’.
The reverberation lime of a room is the time required for the sound level to fall
60dB after the sound
source is shut off; that is, for the intensity to decrease by a factor of a
million.
Interference is the superposition of sound waves to produce either constructive or
destructive addition.
Bears are fluctuations in sound intensity that occur when there is interference
between two sound waves of
equal intensity and slightly different frequency.
The Doppler ejfecl is the apparent change in frequency as the source of a sound and
an observer of the
sound move relative to each other.
A supersonic velocity is one that is greater than the local velocity of sound.
A shock wave is the wave motion accompanying objects traveling at supersonic
speeds.
Mach number is the ratio of the velocity of an object. or of a shock front. to the
local velocity of sound.
Two closed organ pipes sounded simultaneously give five beats per second between the
fundamentals. lf the
shorter pipe is l.l m long. find the length L of the longer pipe. Assume that v in
air = 340m/s.
I Each pipe is occupied by a quarter-wave (refer to Prob. 23.52); hence,
340 / 340 I
5“’=”'_"=fi_%
Solving. L =1.18 m.
Two open organ pipes. Ont: 2.5 ft and one 2.4 ft in length. are sounded
simultaneously. How many beats per
second will he produced between the fundamental tones if the velocity of the sound
is ll0(I ft/s?
I Each pipe is occupied by a half-wave (see Prob. 23.51). so
g l1lI0ftls 1100 ft/s__
"' " ' can 5.0ri ‘gm
Four beats per seoond are heard when two tuning forks are sounded simultaneously.
After attaching a small
piece of tape to one prong of the second tuning fork, the two tuning forks are
sounded again and two beats
per second are heard. lf the first fork has a frequency of l8!) Hz, what must the
original frequency of the
second fork have been?
I The frequency of the second fork must he higher than that of the first fork or
adding the tape would have
increased the number of beats. Therefore. v, - 180 = 4 or v, = 184 Hz.
24.15‘ Some of the low keys of the piano have two strings. On a particular key one
of the strings is tuned correctly

ELECTRIC POTENTIAL AND CAPACITANCE .7 431
V = q/C gives
3 I1’ . .
°' "-=°-’5"-
Substitution of this in the previous equation gives
U.75qj+q§=6uC or q§=3.43 EC
Then qt = 0.7sq; - 2.57 gc.
16.107 Two capacitors, C . = 3 14F and C, = 6 uF. are connected in series and
charged by connecting a battery of
voltage V = 10V in series with them. They are then disconnected from the battery,
and the loose wires are
connected together. What is the final charge on each?
I The capacitors are charged in series so they originally have equal charges. When
the loose wires are
reconnected. they neutralize each other giving zero final charge.
Ct
I Z CI C2
C1 Ill 3'4
cl cl CI C2
‘llz j l7!2 4'71‘
3 4
,,
V
Before After Hg. 76-33 Before After Hg. 26-30
16.113 Repeat Proh. 26.107 if after being disconnected from the battery. the
capacitors are disconnected from each
other. They are now reconnected as shown in Fig. 26-33. What is the final charge on
each?
I The original charge on each is Q = C,,V = 20 uC. After being connected as shown.
Q, + Q; = 2Q = 40uC.
Also V, = l/, and so Q./C, = Q,/C,. Solving these two equations simultaneously
gives for the two charges
26.7 EC on the 6;4F and 13.3 EC on the 3;4F.
26.109 lf two capacitors C, =4uF and C; = 6;4F are originally connected to a
battery V = 12 V. as shown in Fig.
Z6-34. and then disconnected and reconnected as shown. what is the final charge on
each capacitor?
I Originally, Q. = 4s ac. Q. = 12 MC; hence. Q; + Q; = 12 - as = 24 tic. Also. v; =
vg gives Q;/c. =
Q;/C1. Solving simultaneously gives E and 14.4 EC.
AAJ K/-\ TOPPER Ii'EeE§°§i'§i|'§gJ'E'épR':I’§
L“E‘*E#°Z.1?l2“r’é‘é"R'i{’é
CHAPTER 27 /T
Q O I I I
Simple Electric C1rcuttsJ
27 1 OHM‘S LAW. CURRENT, RESISTANCE
TABLE 27-l Resistivities (p) at 20°C and
Temperature Coeflklents (tr).
material p.!2~ tn u.'C"
1
Silver l.6 X l(l " 3.8X HI
Copper 1.7 X 10 " 3.9 X ll) ‘
Aluminum 2.8x Ill “ 3.9 X l0 ‘
Tungsten 5.6 >< I0 " 4.5 x ltl ‘
lrtvn t0><10 ‘ 5.l)x to ‘
Graphite (carbon) 3SO0X l0 " -0.5 X I0 ‘
What is the relation between resistance and resistivity’?
I The resistance R of a wire of length L and cross-sectional area A is
'*="(,'i)~
where p is a constant called the resit/ivily and is a characteristic of the
material from which the wire is made.
For L in m. A in m’. and R in Q. the units of p are Q - m.
How does the resistance of a conductor vary with temperature?
I If a wire has a resistance Rr, at a temperature 11,. then its resistance R at a
temperature Tis
R = R.. + crR.,(T — 7;). where a is the Iemprralure corflicienl of rrsislance of the
material of the wire. Usually
a varies with temperature and so a linear relation is applicable only over a small
temperature range. The units
ofa are K ' or “C '.
A similar relation applies to the variation of resistivity with temperature. ll p.,
and p are the resistivities at
‘II, and T respectively. then p = p., + nflr-( T - T“).
Table 27~l lists the resistivities of a number of conductors for 7I.= 20°C. as well
as temperature coelficients
of resistance.
How are current and current density related’!
I The rate ol flow of electric charge across a given area (within a conductor) is
defined as the electric
current I through that area. Thus.
_"_'I
1- dl (A)
The electric current density J at a point (within a conductor) is a vector whose
direction is the direction of flow
of charge at that point and whose magnitude is the current through a unit urea
perpendicular to the /low
direction at that point. Thus. the current through an element of area dS.
arbitrarily oriented with respect to
the flow direction, is given by (see Fig. 27-l) dl =1 -11$ = J dA. where dA = dS cos
6 is the projection of d8
perpendicular to the flow direction. The total current through a surface S (e.g.. a
cross section of the
conductor) is then
1=[.|-4s=f/aa
\ §
AAJ K/-\ TOPPER Ir'E°Ei°§i'§i'§gJ'E'EpR':I§
AAJ K/-\ TQPPER it'Eiz'¥e§|i'ii§gJ'§EpR'i§’§
SIMPLE ELECTRIC CIRCUITS J’ 433
§
\ 4»
‘T ‘\
-g —>
as n as T’
n Flow
J 0 l|nc\
—>
\\_ / t‘/ Z
\\ \ Fig. 27-1
Starting from the standard form of Ohm‘s law, V = IR, find the relation between J.
the current density. and
E. the electric field in a eurrent~carrying conductor.
| We consider a conductor of uniform cross-sectional area A and length L. The
resistance R = p(L/A).
where p is the resistivity. The current can be expressed as I =JA. and the
potential drop across the resistor is
related to the average electric field by V = EL. Then V = IR becomes EL
=JA[p(L//1)], or E = pl. Often
one specifies the conductivity. 0. instead of the resistivity. where a= I/p. Then J
= 0E. This result can be
generalized to an arbitrary conductor in the vector form: J = 0E. which holds at
each point in the conductor.
How many electrons per second pass through a section of wire carrying a current of
0.7 A?
I l= 0.7 A means 0.7 C/s. Dividing by e = 1.6>< 10'" C. the magnitude of charge on
a single electron. we
get number of electrons per second = 0.7l(1.6 X 10‘ "') = 4.4 X 10"‘.
A current of 7.5 A is maintained in a wire for 45 s. In this time la) how much
charge and (b) how many
electrons flow through the wire?
I (4) q = ll = (7.5 A)(45 s) = 337.SC (bl The number of electrons N is given by
_ 5 _ 331.5 c _ .,
N3" |.r>>< 10 "c'2i‘1X'°
where e = 1.6 X l0 "C is the charge of an electron.
If 0.6 mol of electrons flow through a wire in 45 min, what are (a) the total charge
that passes through the
wire, and (bl the magnitude of the current?
I (a) The number N of electrons in 0.6 mol is
N = (0.6 mol)(6.02 X 10*’ electrons/mol) = 3.6 X 10*‘ electrons
q = Ne = (3.6 X l0“)(l.6 X 10"“ C) =5.78 X l0'C
(b) t = (45 min)(60 s/min) = 2.7 X 10" s
_g_s.1s><10‘c_
'3“ 2.7Xl0‘s T
An electron gun in a TV set shoots out a beam of electrons. The beam current is 10
uA. How many electrons
strike the TV screen each second? How much charge strikes the screen in a minute?
I Let n, = number of electrons per second. n, = I/e = (1.0 X 10 5 C/s)/(l.6 X 10""
C) =
6.3 X 10"‘ electrons Er second. The charge Q striking the screen obeys |Ql = IT =
(l0pC/s)(60 s) = 600pC.
Since the charges are electrons, the actual charge is Q = —600;4C.
in the Bohr model, the electron of a hydrogen atom moves in a circular orbit of
radius 5.3 X 10 " m with a
speed of 2.2 X 10“ mls. Determine its frequency f and the wrrent I in the orbit.
AAJ K/-\ TOPPER Il'EeE§°§|i?i'§gJ'E'épR':I’§
AAJ K/-\ TQPPER Ir'EE'ie§|i'm§gJ'§EpRi§’§
466 1'
27.142
27.143
27.!“
27.145
27.146
CHAPTER 27
E3
switch is closed the capacitor will discharge through resistance R. Find an
expression for qtl). the charge on
the capacitor at any time I.
Fig. 27-50
I With the switch closed we must have for the loop equation q/C + IR =0. Since I=
rate of discharge.
I= dq/dl. Then. substituting in the loop equation. we get rlq/rll = —(l/RClq, or
dq/q = —(l/RC) (II.
Integrating.
"t1q'_ 'l , q_ 1
I..T"I.,Rcd' ""d '“Q"nc
_ rkr _
Then q - Qt’ . The quantity R(' - r is called the time constant of the circuit; the
larger r. the longer the
discharge time.
ln the laboratory. a student charges a 2-44F capacitor by placing it across a l.5-V
battery. While disconnecting
it. the student holds its two lead wires in two hands. Assuming that the resistance
of the body between the
hands is 60 M2. what is the time constant of the series circuit composed of the
capacitor and the student‘s
body? llow long does it take for the charge on the capacitor to drop to l/e of its
orgiinal value? To l/100?
(Hint: In l(ll== 2.30 log ND.)
I r = RC = (6 X l0‘)[2 x l0 ") = 0.12s. Now Q = Q“ exp (—rlRC) = Qr|exp(—r/0.12).
If exp (—!/0. 12) =
expt-I). then l=0.l2s. If exp(—1/0.l2) =0.tll. then exp(i/0.l2)= 1(1). or t/0.l2 =
In 100==2.30log l(0=
4.60; so I = 0.55 5.
ln a certain electronic device, a 10-uF capacitor is charged to 2000 V. When the
device is shut off. the
capacitor is discharged for safety reasons by a so-called hlcedcr resistor of l MQ
placed across its temtinals.
How long does it take for the charge on the capacitor to decrease to 0.01 oi its
original value?
I We recall that lnx *1 2.30 logx; r = RC = 10 s. We wish that Q/Q4,= exp(—I/Y)
=0.0l. From this.
t/l0 = 2.30 log 100. which gives t= 4_6_s.
A 400-uF capacitor is connected through a resistor to a battery. Find (a) the
resistance R and (b) the emf of
the battery 1 if the time constant of the circuit is 0.5s and the maximum charge on
the capacitor is 0.024 C.
_1-.i_ -_z_ °-014° _
' R'c'400><1o~r=‘l “"0"-tm><i0*r='m
A 50-;rF capacitor initially uncharged is connected through a 300-Q resistor to a
l2-V battery. (0) What is the
magnitude of the final charge qr. on the capacitor’? tb) How long after the
capacitor is connected to the
battery will it be charged to §q.,'.‘ (cl How long will it take for the capacitor
to be charged to 0.9(lq.,‘?
I ln charging we have q =q.,(l — e"""). with r1., = CV. tn) q..= CV = (S0 X l0 “ F)
(l2 V) = 6(Xl EC.
(bl r = RC = (3(I) Q)(50 X l0 ’ F) = 15 X l0"‘s = 1§__rr_i§. From the formula the
charge reaches §q., in the time I
such that e ""‘ = Q. or I/(RC) =1/r = ln2=0.7. Thus: r.,.==t).7r=lU.5ms. (cl
Similarly, e“"'=0.l and
I/r = In l0 =- 2.3. Thus In» == 2.31’ = 34.5 ms.
A 150-;iF capacitor is connected through a 5009 resistor to a 40-V battery. ta)
What is the final charge qt. on
a capacitor plate’! (b) What is the time constant of the circuit? (cl How long does
it take the charge on a
capacitor plate to reach 0.3q,,'.’
I (ll) q.,= VC = (40V)(l5(l X10 " F) =6.0mC. (D) f = RC = (5(l) Q)(l5() X 10 “F)
=75 ms.
(e) Following the procedure in Prob. 27.145: -I/1 = ln 0.2 or I/1' = ln 5 = L6.
Then I = (l.6)(7S ms) = 120 ms.
AAJ K/-\ TOPPER It'EeEi°§i'ii'§gJ'E'é°R'§I’§
AAJ K/-\ TQPPER ii'EE'ie§ii'm§gi'§EpR'j§’§

28.13
B.“
28.15
28.16
28.17
28.18
28.19
THE MAGNETIC FIELD I 469
_ v’_ (3 X l0‘m/s)‘ _
l‘) " II '1.s1><1o"m/s"h
If it is assumed that a circular path around the earth (radius = 6400 km) can be
found upon which the earth's
field is horizontal and constant at 0.50 G. how fast must a proton be shot in order
to circle the earth’? In what
direction?
I The centripetal force (mur)/r is fumished by the force euB,,. Then v = (¢B,,r)/m
— [(1.6 X 10"“) x
(S x l0 ‘)(6.4 x l0")]/(1.67 x l0 ”) = 3.1 >< 10'" mls. lt must be shot westward,
since the field points north.
Why is the solution to Prob. 28.13 faulty’?
I The speed of the proton cannot exceed the speed of light. Evidently. the
relativistic equation of motion
should have been employed.
Alpha particles (m =6.68 x I0 ” kg. q = +20). accelerated through a potential
difference V to 2-lteV. enter a
magnetic field B = 0.2 T perpendicular to their direction of motion. Calculate the
radius of their path.
I The KE of a particle is conserved in the magnetic field:
1 . “T
Emu" = Vq or u = \/27‘!
They follow a circular path in which
_»L1/_1 !2Vq_l ZVm_ 1 2(1000v)(ts.as><1o ”k;)_
"qs'qa\/m ‘a\/ q ‘0.2'r‘/ 3.2><10-"c ‘3i2“"“
After being accelerated through a potential difference of 5(ll0 V, a singly charged
carbon ion moves in a circle
of radius 2l cm in the magnetic field of a mass spectrometer. What is the magnitude
of the field‘?
I Following the approach of Prob. 28. l5.
,, , 2v ~ 2(s000v)(19.9><i0=’1t)
m.»=(12u)(l.o6><l0 kg/u)=l9.9X10 kg a=T?=
" "' 'kZ.82>< ll) ’T’ 'B='0.l§§T
A particle with charge q and mass m is shot with kinetic energy K into the region
between two plates as
shown in Fig. 28-2. If the magnetic field between the plates is B and as shown, how
large must B be if the
particle is to miss collision with the opposite plate’!
XXX
XXX|
qQ—>
X
X
X
X 3-xi I-1g.28-2
I To just miss the opposite plate. the particle must move in a circular path with
radius d so from Bqd = mo;
and using K = (muz)/2. we have B = (2mK)"’/fqd).
A cathode-ray beam (an electron beam; m = 9.l X 10'“ kg. q = -2) is bent in a
circle of radius 2 cm by a
uniform field with B = 4.5 mT. What is the speed of the electrons?
I To describe a circle like this. the pat ticle must be moving perpendicular to B.
Then
_,. _.
U=5£;=(0.02m)(|.t>><10 Vc|)(4.5><10 T)=1. ax 0.ms
m 9.l><l0 kg
ln Fig. 28-3(a). a proton (q = +e. m = l.67 x I0 3’ kg) is shot with speed 8 X 10“
m/s at an angle of 30° to an
x-directed field B = 0.15 T. Describe the path followed by the pl'Ot0n.
AAJ K/-\ TOPPER ii'EeEi°§i'ii'§gJ§'épR':I’§

211.28
18.29
THE MAGNETIC FIELD 5 473
I Here,
V
E=;k=(5><l0')k V/m
and the Lorentz equation gives
I-‘,,,. I (3 x lO'°)|(5 X 10‘)|r + (4 X 10‘)(0.766j + 0.643!) X (1.5Ir)] = (3 x 10'
")[(S x l0‘)l 1-(4 X l0‘)(0.766)(l.5)l]
=0.l397i + 0.l5k N
The value of elm, can be obtained by using a specially designed vacuum tube
illustrated in Fig. 28-7. lt
contains a heated filament F and an anode A which is maintained at a positive
potential relative to the
filament by a battery of known voltage V. Electrons evaporate from the heated
filament and are accelerated
to the anode. which has a small hole in the center for the electrons to pas through
into a region of constant
magnetic field B. which points into the paper. The electrons then move in a
semicircle of diameter d. hitting
the detector as shown. Prove that elm, = 8V/(Bd)‘.
l-—d*-l
A
to
F
Li. .. Mal
II
O
v fig. za-1
I Refer to Fig. 28-7. Newton's second law implies that (m,u’)/r = evB. where r = Qd
is the radius of the
electron's orbit. Solving for the speed. we find that
B
U = '78: = 4 (1)
Since the electrons have been accelerated essentially from rest through a potential
dilferenoe V. each has
kinetic energy im,u’ = ev, so that
, 2eV
u -= -——
"L
e 8V
Z’ B:dz (3)
(2)
Combining Eqs. (l) and (2). we obtain
Solving for the charge-to-mass ratio. we have
which was to be shown.
If in a mass spectrograph carbon ions move in a circle of radius r(- = 9.0 cm and
oxygen ions move in a circle
of radius r‘, = I0.-4cm. what is the mass of an oxygen ion?
I We assume that V and 8 are the same for both ions. so that from Eq. (3) of Prob.
28.28.
m = (8111:)/(ZV) and
mo If, (l0.4 cm)’
_ = _ = i_._- = 133
mc rf- (9.0cm)'
That is. the mass of an oxygen ion is 1.33 times the mas of a carbon ion. Since the
mass of carbon is defined
to he exactly 12 u. the mass of oxygen is nr‘, = 1.33m; = (l.33)(l2 u) = liq.
AAJ K/-\ TOPPER ir'EeE§°§i?i'§.gi'E'épR':I’§

29.33
29.2
29.34
29.35
29.36
29.37
MAGNETIC PROPERTIES OF MATTER .7 519
over the volume of the wire. The magnetic energy dU,_ contained in a cylindrical
shell of length I. inner radius
r, and outer radius r + dr is
B’ '1!
dU,, = 5.. av =@zm1ar=i,r‘ar
2;t., Mm
Hence the magnetic energy per unit length contained in the wire is
l/.._1 _M-it“ “ . _;t<.|" L‘ _£
1'1idU""4;m'i, ' d,_4:'m 4 .,)_l61r
The external magnetic field of a spherical object of radius R. surrounded by vacuum
and carrying an
(idealized) central point magnetic dipole. contains magnetic energy U = (It/2)
[(B§R")/(2uti)]_ The quantity B,
is the maximum magnetic field strength at the surface of the object—that is, the
value of B at the objects
magnetic poles.
ta) The earth's external magnetic field is not that of a pure centered point dipole.
nor is the earth
surrounded by a perfect vacuum. Nevertheless. a reasonable estimate of the energy
content U in the external
field is U = (Bf,R’)/(2440). Evaluate this. using B, = 6.0 >< 10“ T and R = 6.4 x
10” rn.
(b) Compare the energy U to the total annual usage of electric energy in the United
States. lt was
1.7 x 10" kW ~ h in 1972.
I (a) With B, = 6.0 X l0"T. R = 6.4 X 10° m. and it" =4.~t X10" N/A’. we have
2 .\ <1 .\
U_%§= =;_..,0-r,
(b) The electric energy usage in the Unites States for 1972 was
E = l.7 X i0"kW ' h = (1.7 X l0")(l0‘.l/s)(36(I) s) =6.l x 10"‘J
MAGNETS; POLE STRENGTII
Describe how the torque on a bar magnet placed in a magnetic field can be expressed
in terms of pole
strength.
I When a bar magnet is placed with its axis perpendicular to a uniform magnetic
field of intensity B. it
experiences a torque r = Rm = Bpl, where m is the magnetic moment. I is the length
of the magnet. and p is
the pole strength of the magnet. Fundamentally, this torque arises from the
interaction between the magnetic
induction and the infinitesimal current loops associated with the orbits and spin of
certain of the magnet's
electrons. An alternative approach. frequently convenient. is to endow the magnet
with a pair of equal and
oppositely signed pole: at its two ends. and to attribute the torque to the equal
and opposite forces exerted by
the field B on these poles. When a pole of strength p is placed in a magnetic field
of intensity B, the pole
experiences a force F = Bp. ln general. 1 = mB sin 0. where 6 = angle between In
and ll.
An iron bar magnet of length l0cm and cross section l.0 cm’ has a magnetization of
10' A/m. Calculate the
magnet‘s magnetic pnle strength.
I The magnetic moment of a bar magnet of length 2d has magnitude m = 2 |p| d. where
lpl is the pole
strength. ln tenns of the magnetization M. the magnetic moment of the bat magnet is
II = M(2ad), where a is
the cross-sectional area. Hence |p| =m/2d =aM. With a = l.0 cm’ = 1.00 >< lll ‘ m‘
and M = 1.00 X 10’ A/m.
we obtain lpl = (l.[l)>< 10 ‘)(1.00 >4 I0‘) = l.00 X10 IA - m.
The field of a bar magnet can be considered to be caused by a surface current flowing
on the surface of the
magnet. If a bar magnet is to act like a solenoid whose interior field is 0.3 T. how
large a surface current must
flow on each centimeter length of the bar’!
I ln Fig. 29-7 take segment length l= 1.0 cm. Since B is axial and ~0 outside the
solenoid. the circuital law
yields B1 equal to u,.l,,,. so I... = 0.3(10 ’)/(4:-r X 10") = 2390A[m = 23.9 A/cm.
What is the relationship between the magnetization (or Amperean) surface current
I,,, in a long har magnet
and the pole strength?
I From Prob. 29.35, p = aM. Applying the circuital law for B to the magnet (Fig.
29.7). and noting that
B = a.,(kI + M) and that H cannot contribute since there are no true currents. we
get M1= I,,: Then p =
aI,.,/I.
AAJ K/-\ TOPPER 1-ltIIEeE|Te:lTliIiggJll?épR|:fSr

680 J CHAPTER 36
36.46‘
36.47
36.48
Reconsider Prob. 36.40 when the source emits a continuous mixture of wavelengths.
The resolving power, at
wavelength A. of the grating is defined as R I A/6. where 6 is the smallest
wavelength diiferenoe for which
the spectral lines A — §b and A + £6 are resolvable. Find R... the resolving power
in the mth order.
I According to Rayleigh‘: criterion, two peaks are just resolvable when their
angular separation is half the
angular width of either peak. This minimal separation is given by (1) of Prob.
36.40 as
A A
(A9)|nin mg W d 00$ 9(A9)..... — Fl
Ditferentiation of the grating equation, d sin 6 - ml, gives
docs 0 A9 =m Al whence duos 9(A9),,,;_ =m6
Comparing the two expressions for d cos 6(A8),,,,,, gives R, = mN.
Note that the resolving power is the same for all wavelengths.
A transmission grating is used with light incident normal to its plane. The width
of each slit is one-third the
spacing between slits. By considering single-slit diffraction. show that the third-
order (i = 3) multislil
diflraction maxima are mising from the dilfraetion pattern of the grating.
I As discussed in Prob. 36.35, the intensity profile in the overall diffraction
pattern due to N slits of finite
width is the product of the "ideal" N-slit intensity profile and the single-slit
profile. According to Eq. (l) of
Prob. 36.35. the jth maximum oi the ideal N-slit profile occurs at an angle 6, such
that
, A
$tnQ=,E (1)
where d is the slit spacing. However, if each slit has width w =d/3, then Fig. 36-
14b [or (I) of Prob. 36.33]
implies that the single slit profile has zeros at angles given by
_ ml 3mA\
sin 6,, - W - d (2)
for m =0. il. 12. 13. Equations (1) and (Z) show that the overall profile must have
a zero at every third
maximum of the ideal N-slit pattem. In particular, the third-order maxima (i = 13)
are removed by the first
minima (m = ii) of the single-slit pattem.
A beam oi light of wavelength i falls on a difiraction grating of line spacing D at
angle of incidence o
measured from the normal to the plane of the grating. Show that the maxima in the
diflraction pattern occur
at angles 0 which are determined by the equation [/1 = D (sin 6 — sin ¢), where j =
0. tl, 12, 13. . . . .
I The situation is indicated in Fig. 36-l7. The initial beam direction is assumed
to be perpendieular to the
ex‘ 1 /
,ai'-Dino
A.r'=Dsin¢ I
— _ _¢
I)
I ._+_”_._._
Flg.36-17
AAJ K/-\ TQPPER I|'E°Ei°§i'ii'§gJ'E'EpR':I’§
AAJ KA TOPPER I.“;E$Z.1?.‘;*‘.‘;’;".-t'i{’s'

686;!
36.70
36.71
36.72
36.73
36.74
CHAPTER 36
now polarized at 30° to the axis of the analyzer, so the intensity of the light
passing through the analyzer is
I’ = I cos: 8 = 0.'I5I,, cos‘ 30° = 0.56%.
Polarized light of intensity L, is incident on a pair of Polaroid sheets. Let 0,
and 6, be the angles between the
incident amplitude and the axes of the first and second sheets, respectively. Show
that the intensity of the
transmitted light is I = I, eos’ 9, cos’ (6, - 9;).
I The intensity of the light after passing through the first polarizer is I = L,cos’
9,. This light is polarized in
the direction of the axis of the first sheet, and so its axis makes the angle 9, -
9, with the axis of the second
sheet (see Fig. 36-25). Consequently. the intensity of the light after passing
through the second polarizer is
I’ =Icos=(8,— 9\)=L,cos’8,cos2(9;—0,)
%/t.
/\
9:
_-- Tut‘
.\.. ,1
#0,
I
/Ans
' Fig. ans
Two Polaroids are aligned with their axes of transmission making an angle of 45°.
They are followed by a
third Polaroid whose axis of transmission makes an angle of 90° with the first
Polaroid. If all three are ideal,
what traction of the maximum possible light (if all Polaroids were at the same
angle) passes through all three?
I The reduction factor for either of the last two Polaroids is cos‘ 45° = 1/2; so
the overall factor is (E): =
The axes of a polarizer and an analyzer are oriented at right angles to each other.
A third Polaroid sheet is
placed between them with its axis at 45° to the axes of the polarizer and analyzer.
(a) li unpolarized light of
intensity lo is incident on this system. what is the intensity of the transmitted
light? (b) What is the intensity
of the transmitted light when the middle Polaroid sheet is removed?
I (a) The light that passes through the polarizer has an intensity oi M, and is
polarized at 45" to the middle
sheet. Thus the light that passes through the middle sheet has the intensity I =
ll‘, cos’ 45" = 0.251,, and is
polarized 45° to the axis of the analyzer. Thus the intensity of the light passing
through the analyzer is
I ' = I cos’ 45" = 0.1251,. (b) lf the middle Polaroid sheet is removed, we have a
crossed polarizer-analyzer and
no light gets through. (Compare Prob. 36.68.)
Four perfect polarizing plates are stacked so that the axis of each is tumed 30°
clockwise with respect to the
preceding plate; the last plate is crossed with the first. How much of the intensity
of an incident unpolarized
beam oi light is transmitted by the stack?
I The first plate transmits one half oi the incident intensity. Each succeeding
plate makes a vector resolution
at angle 30°, transmitting the fraction cos 30° of the amplitude. or cos’ 30° = §
of the intensity. leaving the
preceding plate. The fraction of the initial intensity transmitted by the stack is
then QC)’ =0.2l1.
A quarter-wave plate is made from a material whose indices of refraction for light
of free-space wavelength
/in = 589 nm are nl = 1.732 and n" = 1.456. What is the minimum necessary thickness
of the plate tor this
wavelength?
I The optical path length (Prob. 34.53) of the ordinary wave in a plate of
thickness I is n,I and the optical
path length of the extraordinary ray is n I. Since the two rays must emerge from
the plate with a 90° phase
diiference. the optical paths must differ by (k + 1)/1", k = 0, 1. 2, . . . . The
minimum thickness thus satisfies
A.,_ _ _ A0 _ 589nm _
4"("‘ "") °' 1'40‘.-n,,)’4(1.732-1.4ss)"534“"'
AA-I K/-\ TOPPER Lt°E#°:.1?l2“.’;'.;"Ri(’;

SPECIAL FIELATIVITY J 689
Equations (ll) and (l2) agree with Eqs. (5) and (8) of Prob. 37.1; the inversion of
the y’ and z’ equations l5
trivial.
37.3 For the situation of Prob. 37.1, suppose that at the instant origin 0'
coincides with 0 (at I = 1' =0). a
llashbulb is exploded at this common origin. According to observers in .2‘, a
spherical wavefront expands
outward from O at speed c. Show that, even though 2‘ is moving relative to 2’ with
velocity u, observers in
1” note an exactly similar wavefront expanding outward from 0'.
I The equation of the wavefront in 2’ is
Jr’-9-y’+z’=c’l’ (I)
The Lorentz transformation of (1) is (Eqs. (S) to (8). Prob. 37.1)
x'+nr' * iz '2_ 1 1'+(vx’/c’) ‘
it/‘T-<»=/¢=>‘i *‘" *‘** "m
which reduces to
(-V)’+(y')’+(1'l’=¢’(I')’ (2)
Equation (2) represents a spherical wavefront expanding outward from 0' at speed c.
Actually. the reasoning goes the other way round. The 1’ and 2" observers must see
exactly the same kind
of wave. by the postulates of special relativity. This (almost) forces the
relations between the coordinates in
the two frames to have a particular forrn—that of the Lorentz transformations. See
Prob. 37.5.
37.4 When inertial reference frames 9’ and ft" coincide. let a flash of light be
produced at the common origin.
Each observer is justified in considering himself at the center of an expanding
sphere of light. Experiment has
revealed that each obtains the same value c for the speed of light. The galilean
transformation. x‘ = x — vi,
does not give this result. Therefore try a modification. x‘ = y(x — vl). where y is
to be determined. The
principle of equivalence requires that this equation hold for the inverse
transformation. x - y(x' — v't') =
y(.r‘ + ut'). ln this equation. we use the assertion that v‘ = -11. But for
generality. the possibility has been
allowed that I' may be difierent from I.
If x and x’ are the intersections of the sphere with the axis at times I and I’,
respectively: (a) To what is
x’/I’ equal? (b) To what is x/I equal? (e) Use the results of pans (a) and (b) to
eliminate x and x‘ in the
transfonnation equations and thus to detemtine y.
I We note that the problem statement includes the tacit assumption that y’ = y when
u’ = —u.
(n) At time r‘, the light sphere observed by 1" intersects the positive .r' axis at
a coordinate x’ = ct’.
Therefore x'¢t‘ = c. (b) At time I, the light sphere observed by 2’ intersects the
positive z axis at a coordinate
x =1 ct. Therefore xlr = c. (e) Since the equations obtained in parts (a) and (la)
refer to the same point event.
we use x’ = cl’ and x = cl in our equations to get
ct‘ = y(ct — U!) (1)
and er = y(cl' + ul') (2)
Multiplying Eqs. (1) and (2) we have
fin’ = y‘(c’ — r.-’)n'
Solving this for 7. we find
l
= m (3)
Y V1-(u'/c‘)
37.5 Continuing with the results of Prob. 37.4. use the relation for Y. express I’
in terms of I and x.
I When Eq. (3) is used in the transfomtation equations (Prob. 37.4). we obtain
~=i'._ _ .
1 (x ll) (4)
and X =ao'+ w‘) <5)
These equations relate arbitrary event-coordinate observations of observers O and
O‘. Using Eq. (4) to
Th L ' A F
A/-\J K/-\ TOPPER NEeETe:lll:I|lggJEEpR:Sr

690 J CHAPTER 37
eliminate x’ from Eq. (5). we find that
X = (1 - U!) +
1- t-1’/H) \/W/7’)
Solving this for r’. we find (after some algebra) that
t’=--—l-;—(:—94:)
\/1——‘(=T/T) ¢
which is the standard Lorentz transformation equation for I’.
37.6 As measured by 1’ a firrshbulb goes off at x = 1(1) km. y = 10 km, z = l km at
t= 0.5 ms. What are the
coordinates x’. y’. Z’. and !' of this event as determined by a second observer.
if‘. moving relative to Y at
—().8c along the common xx’ axis?
I From the Lorentz transformations,
_ _ _ K I
I, = x or ~ = l00km ( 0.8X3>< 10 ‘km/s)(5>< 10 s)=367km
\/1-(w/t~>> \/1 - us‘ —‘
, r - (u/c‘): 5 X 10' ‘ s — [(—U.8)(l00 km)]/(3 X10’ km/s) ,
t = _ , = 1 = l.-8 ms.
\/1 — (v‘/r‘) \/1 — (0.8)
y‘=y=l0km z’=z=lkm
37.7 At time I’ = 0.4ms. as measured in 1”, a particle is at the point x’ = 10 m.
y’ = 4 m. :' = 6 m. (Note that
this constitutes an event.) Compute the corresponding values of X. y, 2. t, as
measured in fl’. for
(3) u = +500 m/s. (b) tr = -500 m/s, and (ct v = 2 X HY‘ m/s.
I From the inverse Lorentz transformation. Prob. 37.1.
4
(.1) x= ~i0+(sw)(4x10‘)=10.2m y=m :=@
,= -4X10--as
Vi —— (500/c‘)
tr» x= =i0-(sw)(4x10“)=9.sm y=m z=@
(4 x 10 ‘) — [(S00)(l0)]/c: ,
= ii =4 10
‘ \/isms m
10 Z 10' 4 10" 3.U0l X 10'
“’ ‘= =m=% -"H1 "E
(4 X 10") + [(2 X l0")(l0)]c’ _,
= L = _37
I ail _ (U:/C2) 5 X 10 s
37.8 At I = l ms (in ft’), an explosion occurs at x = S km. What is the time of the
event for the Y’ observer. if for
him it occurs at x’ = 35.354 km?
I From the Lorentz transformation for x’,
1 _ -.\
35.354 >< 10-‘ = or U = -3 >< 10’ l"l'I/S
Then. from the Lorentz transformation for 1',
10‘~‘ + [(3 x l0’)(5 x 10’)]/c‘
'=i=|.nt 7 .
I ,il_(U2/cl) )6 ms
37.9 A spaceship of (rest) length I00 m rakes Mrs to pass an observer on earth.
What is its velocity relative to the
earth?
I The observer measures the length of the spaceship to be l= l00Vl — (iv:/czl (m).
where v is the velocity of
AA‘-I R NEET AIIMS JEE RAS
37.51
37.52
37.53
37.54
37.55
37.56
SPECIAL HELATIVITY U 709
I The rest mass lost in the formation of a deutron is
(m, + m,,) - m, =(1.67261+1.67492 - 3.34357) X 10*" = 3.96 x10"°kg
which is equivalent to a rest-energy loss
A5,, = 3.96 >< 10*"? = 3.56 >< 10"’! = 2.22 MeV
Hence, by energy conservation, the surroundings must gain exactly Z_.22_M5_\_{ of
energy. This same amount of
energy, the binding energy of the deuteron. would have to be supplied to the
deuteron to tear it apart into an
infinitely separated proton and neutron.
Detennine the mas and speed of an eleetmn having kinetic energy of 1001teV (1.6 X
10'“J).
I The kinetic energy is K = mc’—m,,c’. or 1.6 X 10‘"J = (In —m,)(3 X 10' m/s)’. and
m —m,= 1.78 X
10"‘ kg. ln Prob. 37.49 we are given m =9.l1 X 10'“ kg; so m = (9.11 + 1.78) x 10"‘
kg= 1.089 ><10"°kg.
To get velocity we note that m = mo/V1 — (vi/c’), or 1 — (v‘/c’) = (mo/m)’ =0.7lX].
Then vlc = 0.548. and
v =-|.64x1Qm[s.
Find the speed that a proton rnust be given if its mass is to be twioe its rest
mass of 1.67 X 10'" kg. What
energy must be given the proton to achieve this speed?
I Use the mass-increase formula.
v: u‘ u’ v’ 3
m =m,,/t/1-F zm.,=m.,/\/1-c—, 4(1-;;)=1; C-,-5 v -0_.5g§£
AW = (m - m.,)¢’ = (2m., - m..)¢* = (1.61 ><10'” kg)(3 x 10'm/5)’ = 1.50 x 10-"'1 =
93; M;v
How much energy must be given to an electron to accelerate it to 0.951;?
I Use the mass-increase formula.
m,, 9.1 X 10”‘ kg _,,
=i-—=ii=29.l X 10 ' k
'” W-(U/C) \/1‘-055’ ‘
AW = (m — m.,)c’ =[(29.1- 9.1)(10’”)](9 X 10"’) I 1.8 X 10- "1 = 1.lZ§ MgV
A 2-kg object is lifted from the floor to a tabletop 30cm above the floor. By how
much did the mass of the
object increase because of its increased PE?
I We use AE = (Am)c’. with AE =mgh. Therefore
_ LE_flLI_ (2 ltg)(9.8 in/s=)(0.3 m)_ _,,
Am- C, - C, -asxlym/S), 6.Sx10 kg
The rest mass of a proton is mu = 1.672614 X 10“" kg. lts laboratory speed after
having fallen through a high
diflerence in potential, AV. is u = 2 X 10' nt/s. (n) Evaluate AV. (la) Find the
total energy of the proton.
I ta) The kinetic energy of the proton is
K =(m —m,,)r:*=[ —1]m,,c’=eAV
from which Av = [(1/0.744943) — 1](1.6726l4 x 10’")(2.99792S X10’): = 321 My
1.602 X l0"" T
(b) The rest energy of the proton is m.,c’ = 938.3 MeV. and so the total energy is
E = 938.3 + 321 = 1252.} Mgv
Show that KE = (m — mule: reduces 10 KE. = §m,,u’ when u is very much smaller than
c.
I KE = (m — m.,)c’ = - m,,)c' = m¢,c*[(1 - vz/c=)_"’ - 1]
bet b = -v*lc‘ and expand (1 + b)"" by the binomial theorem. Then
A
>4»-
V
!~jA
.-: T '3) 1 I 3 l
(1+b)'=l+(-§)b+i-b’+---=t+E%+§%+‘~
AAJ K/-\ TOPPER Ii'EeE§°§|T§i'§gJ'E'épR':I’§

37.63
37.8
37-65'
37.66‘
SPECIAL RELATIVITY U 705
Thus the desired relation is
E'=E§+(P'-‘)2 (1)
Because of the torm of (I) momenta are often specified in MeV/c, on the atomic
level, where energies are
given in eV or Me‘/. (See Prob. 37.63.) (b) The total energy of the electron is its
rest energy, 0.Sll MeV.
plus its kinetic energy, lMeV. Then (1) gives 1.511‘ =0.5l1‘ + (pc)’. from which pc
= 1.42 MeV, or
p =1.42 MeVl;.
Compute the conversion factor between the ordinary units of momentum and MeV/c.
I 1MeV/c = (1.6 X 10-“ J)/(3 >< 10' Ill/S) =s.sa >< 10-" ti; - m/s.
Show that rest mass is not conserved in a symmetrical, perfectly inelastic
collision.
l Let two identical bodies. each with rest mass mo, approach each other at equal
speeds u, collide, and stick
together. By momentum conservation, the conglomerate body must be at rest, with
energy Mac‘. The initial
energy of the system was 2[m,c'/Vl - (ll!/05)]. Hence.
..J.'i"i_ = M061 or Mo = L
V1—(u,/c!) V1 — (M:/c7)
It is seen that the final rest mass, Mo, exceeds the initial rest mass, Zmo.
ln newtonian rnechania the relation dE/dt = I‘ - v is valid. where E is the total
energy oi a particle that is
moving with velocity v and is acted on by a net force F. Show that this relation is
also valid in relativistic
mechanics. (Assume that Newton‘: second law is valid under special relativity.)
I As shown in Prob. 37.62, the particle's instantaneous momentum p(t) and
instantaneous energy E(|) are
related by
E’=¢'v-|>+t»-..¢*>’ <1)
Difierentiating Eq. (1) with respect to time. we find that
dE _ , Q
2E dr _ 2c P dz
dE Q dp
°' at E at (2)
But p- mv = ymov and E I ms’ -I yrmrc‘, so (c']t)/E = v. Using this in Eq. (2), we
obtain
“-9. _ .
dt-dr v-F v (3)
since I-‘ = dp/dt.
Show that the components of the velocity of a particle of energy E and momentum p
are given by
v =2 u I BE v = E
’ fir. ' an ‘ 8:».
These relations apply in both the relativistic and newtonian domains.
I The energy is given in terms of the momentum by
E = W!!!‘ + (mi): = V¢‘(P.' + P3 + 0.’) + (mUcE)5
Difierentiating with respect to p,. we llnd that
E _ 1 2¢’P. = Q = ¢’>'m¢v. =
911» 2V¢'i(P.§+Pi*P-i)+("I»¢§)2 E 1"":-C’ U‘
The proofs for the y and z components follow the same procedures.
Since the newtonian result is just the low-velocity limit of the relativistic
result, the above relations hold in
tne newtonian domain. They may also be easily derived from E I p’/(2m,,) -pf/(2m,,)
+ pf/(Zm,,) + pf/(2m,,).
Then 8E/6p, =p,,/m,= u,, etc.
AAJ K/-\ TOPPER iiE°Ei°§i?i'§gJ'EEpR':i’§

SPECIAL HELATIVITY U 707
so the magnetic field magnitude at distance I" is
_ _M _ M_ (2 x in-’)(1.0s x 10"‘)_
B -21!/'— 21!! ' 1.00 X 10"‘ 'T2'12°"T
Finally. F, = 01:8‘-(1.6 X l0"“')(5.294 X l07)(Z.l2 X 10 ”) = 1.79 X IO “N tmvard
the beam,
31.70 A large cyclotron is designed to accelerate deuterons to 450 MeV of kinetic
energy. This means that their
speed will become a substantial fraction ol c. Hence their mass will become
substantially larger titan the rest
mass. Ii the magnetic field is everywhere of the same value. this requires that the
frequency of the oscillating
potential difference applied between the Dees be decreased during the acceleration
of a group of deuterons.
What is the ratio of the final to the initial frequency? The rest mass of a deuteron
is 3.3 X 10"’ kg.
I For a particle circling at relativistic speed in a magnetic field_ the
relativistic force. I-‘ = dpldl. is still
correctly given by the borentz formula:
F I ‘gr = qv X B (l)
, .
The momentum is given by p = ymnv. where y- [1 -— (v’/c )] "i If the particle is
circling with angular
velocity u,, we have, since the magnitude of p is constant,
55 = w, X ll = vm»(w.i X v) (Z)
Equations (1) and (Z) imply that, in magnitude.
wt =£,fl (3,
"I
Ymo
where m E ynn, is the relativistic mass. Since the deuterons are initially
nonrelativistic. Eq. (3) implies that
tu,, y, 1
= (4)
(".1 Yr Y1
In temts of the final kinetic energy K,. we have
K +m cl
Y1 = (5)
The rest energy of the deuteron is
:_(3.3>< 10 "kg)(3.uu>< |0'm/s)=_ ,
'""' ‘ 1.60 >< l0' "J/Mev ' 1'8“ '0 M”
Then with K, = 450MeV. Eqs. (4) and (5) yield
uI./ I i860
J,,=;,=m=@
AAJ K/-\ TOPPER ii'i§Ei°§m'§gJ'E'épR':i’§

PARTICLES OF LIGHT AND WAVES OF MATTER U 709
38.9 A desk lamp illuminates a desk top with violet light of wavelength 412 nm. The
amplitude of this
electromagnetic wave is 63,2 V/ m. Find the number N of photons striking the desk
per second per unit area.
assuming that the illumination is normal.
I From $8 = (he)/A and the energy density formula, (s,,E,§)/2. tor a wave with
electric field amplitude E‘,
(Prob. 33.64)
N=%= =1.1ox10‘°9notons/s-m’
38.10 A sensor is exposed for 0.1 s to a Z00-W lamp 10 in away. The sensor has an
opening that is 20 mm in
diameter. How many photons enter the sensor if the wavelength of the light is 6(1)
nm? Assume that all the
energy of the lamp is g'ven ofi as light.
I The energy of a photon of the light is
_5_ (6.63 X10"‘)(3 ><1o')_ _‘,
E-A-——-—-————600xm_, -3.3><t0 J
The lamp us/es ZIDW of power. The number of photons emitted per second is therefore
Z00
n -W - 6.1 X 10'” photons/s
Since the radiation is spherically symmetrical, the number of photons entering the
sensor per second is n
multiplied by the ratio of the aperture area to the area of a sphere of radius 10
m:
(6.1 x 10*") %;:),°;) = 1.53 >< 10" photons/s
and the number of photons that enter the sensor in 0.1 s is (0.l)(ll53 X 10") =
1.53 x 10".
38.11 What is the momentum-energy relation for photons?
I From Prob. 37.62. $’= (pc)’ + (m,c‘)’. But a photon has zero rest mass, so Z=pc.
38.12 What is the momentum of a single photon of red light (v = 4(X) X I0“ Hz)
moving through free space?
I The momentum is given by p = It/A = (Irv)/c. Hence
(6.6 X 10"‘)(-100 x 10") _,,
= _ . y
p Jxwl 88><l0 kg m/s
38.13 What wavelength must electromagnetic radiation have if a photon in the beam
is to have the same momentum
as an electron moving with a speed 2 >< 10’ mls?
I The requirement is that (mv),,,¢,,,, - (h /A),,,,,,,,,. From this,
A _ L - - , 64 ,,,,,
mu (9.1 x 10"‘ kg)(2 X 10‘ mls) '7
This wavelength is in the x-ray region.
38.14 A stream of photons impinging normally on a completely absorbing screen in
vacuum exerts a pressure 9.
Show that 9* - I/c, where I is the irradiance.
I The time rate ot change of momentum equals the force, that is, Ap/At = F. The
force per unit area. F/A.
is the pressure, and so
_lfl
914m
But 8 = cp for photons, which means that A3 = c Ap and
1 A8
Q = — —-
Ac Ar
Since irradiance by definition is energy per unit area per unit time. we obtain 9 =
I (c.
AAJ K/-\ TQPPER ii'E°Ei°§i'ii'§gJ'E'EpR'ii§

710 U
36.15
38.16
3G.l7
38.18
38.19
3l.N
38.21
38.22
CHAPTER 38
A collimated beam of light of flux density 30 kWlm’ is incident normally on a 100-
mm’ completely absorbing
screen. Using the results of Prob. 38.14. determine both the pressure exerted on
and the momentum
transferred to the screen during a IIXX)-s interval.
I ‘Hie presure is simply
I 3 10‘ F A /A _
9= ;=3-:3, =10 ‘ lja -7% and Ap = 9,4 A: = (to ')(t0-‘)(1o’)=1o"g-m/5
(One could actually build an interplanetary sailboat using solar pressure.)
What is the largest momentum we can expect for a microwave photon?
I Microwave frequencies go up to 3 x 10" Hz. 'l1terefote. since p = It/A = (Irv)/c,
—a4 ||
P = =6_6X 10~"ks.m.s~1
How many red photons (A = 663 nm) must strike a totally reflecting screen per
second, at nonnal incidence. if
the exerted force is to be 0.22$lb?
I We know that F - Ap/At and, in this ease, Ap is twite the incident momentum.
Thus, if N is the number
of incoming photons per second. F = NI(2II)/A]_ Since 0.225 lb= 1 N. we have F = l.
and
I. 663 X 10"
N-5- _5x 10” photons/s
What potential difference must be applied to stop the fastest photoelectrons
emitted by a nickel surface under
the action of ultraviolet light of wavelength 20(1) A? The work function of nickel
is 51]] eV.
I energy of photon = If = = 6.20 eV
Then, from the photoelectric equation. the energy of the fastest emitted electron
is 6.20 eV - 5.00 eV =
l.20eV. Hence a retarding potential of LZQV is required,
The work function of sodium metal is 2.3 eV. What is the longest-wavelength light
that can cause
photoelectron emission from sodium?
I At threshold, the photon energy just equals the energy required to tear the
electron loose from the metal,
namely. the work function W,,,,,.
he I240 V~
W....=T Or 1= =o!1
What is the work function of sodium metal if the photoelectric threshold wavelength
is 680 nm7
. h 1240 v-
I work funetton= w,,,,_=:w,,__=fi-fi=1.ss;v
Will photoelectrons be emitted by a copper surface, of work function 4.4 eV, when
illuminated by visible
light?
I As in mt». 33.19,
V.
threshold/t=%n= =2s2nm
Hence visible light (400 to 700 nm) cannot eject photoelectrons from oogpg.
Light of wavelength 6(1) nm falls on a metal having photoelectric work function 2
eV. Find (u) the energy of a
photon, (bl the kinetic energy of the most energetic photoelectron. and (c) the
stopping potential.
hc 1240 eV - nm
I no E-7=p-2.m=v
Th L i A F
AAJ KA TOPPER NETETe:lrlr|illggJEEpR:Sr
PARTICLES o1= uem /mo wnves or MATTER 17 111
(b) Kn,-E—W,.,,,=2.07—2=0.D7eV
(c) eV,,-K,,,,,=0.07 eV or V,=0.07V
38.23 lt takes 4.2eV to remove one of the least tightly bound electrons from a
metal surface. When ultraviolet
photons of a single frequency strike a metal, electrons with kinetic energies from
zero to 2.6 eV are ejected.
What is the energy of the incident photons?
I w1==4.z¢v 1tE.__,=2.s=v and £,_,,,_=wr+K5__=4.2+z.e=o.aev
38.24 A photon of energy 4.0 eV imparts all its energy to an electron that leaves a
metal surface with 1.leV of
kinetic energy. What is the work function W,,,,,. of the metal?
I W,,,,,=hv— x,___=4.0-1.1=2.9ev
38.25 Detemiine the maximum KE of photoelectrom ejected from a potassium surface by
ultraviolet light of
wavelength Ztlll A. What retarding potential dilterence is required to stop the
emission of electrons? The
photoelectric threshold wavelength for potassium is 44(1) A.
I Work function = w_,=A%= = 2.82 =v
11¢ 1. 2400
then T—}i;=KE,_,,=%i-—2.82I6.20—2.82=3.38eV
The retarding potential must be just large enough to stop electrons with KEN, so V
=1.
38.26 A surface has light of wavelength A, = 550 nm incident on it, musing the
ejection of photoelectrons for which
the stopping potential is V,, = 0.19 V. Suppose that radiation of wavelength A, =
190 nm were incident on the
surface. Calculate (0) the stopping potential I/.1, (ll) the work function of the
surface, and (e) the threshold
frequency for the surface.
I (0) Since eV = KE,_,, = hv — Wm, with W,“ a constant of the surface, we have
e(Vj, — P1,) = Ii(v, — v,) or
l
1| 11¢ 1 1 1
v,,=v,,+;(v,-V.)-v,,+7(Z-z)=o.19+1240(5)-§)=4.41v
h 1240
(5) w,_,_=f-=v,.=fi-o.19=g.o7=v
Wm 2.07) 1.602 X10"
(cl lIv.=W..... or v,=T= =i
38.2‘! In the pholoionization of atomic hydrogen. what will be the maximum kinetic
energy of the ejected electron
when a 60-nm photon is absorbed by the atom? The ionization energy of H is 13.6 eV.
I The ionization energy is minimum energy to remove the electron from the atom.
Then
he l240eV ' nm
K_,,- A l3.6eV-Tr?-— l3.6eV -7.1 eV
38.2 COMPTON SCATTERING; X-RAYS: PAIR PRODUCTION AND ANNIHILATION
36.28 Suppose that a 3.64 nm photon going in the +x direction collides head-on with
a 2 X 10’ m/s electron moving
in the —x direction. lf the collision is perfectly elastic. what are the conditions
after collision.
I From the law of conservation of momentum.
momentum before = momentum after £— mu, = 2 — mu
But, from Prob. 38.13. It//1q= mu, in this case. Hence, ll/A = mu. Also, for a
perfectly elastic collision,
KE before = KE after '51- 1 mu’ = '£+1mu’
A, 2 " A 2
A/-\J KA TOPPER NEETAIIMSJEE RAS

712 -'1 CHAPTER 38
$.29
38.30
38.31
38.32
38.33
Using the facts that It/An =mvn and It//1 = mu. we find that
v,,(c + iv») = v(c + iv)
Therefore u = on and the electron rebounds with its original speed. Because It/A =
mu = mu», the photon also
rebounds and with its original wavelength.
The Compton equation can be written as A‘ — A = [hI(m°c)](1 - cos ¢). The factor
It/(mqc). called the
Compton wavelength, is the wavelength shift that occurs when ¢ = 90". Evaluate it
for scattering from
(a) electrons and (b) protons. ln the case of electrons, what percentage change is
this it the incident radiation
has a wavelength (cl 500 nm (visible light) and (d) 0.050 nm (0.5-A x-rays)?
I Inserting the rest masses of the electron (9.1 X 10*" kg) and proton (1.67 X 10
"ltg) into h/me yields
Compton wavelengths of 0.024 A and 1.32 X 10"’ m. respectively. For the electron
(0.024/5000)(llX)) =
4.8 x 10 ‘ Ereent, while for the 0.5-A x-ray. (0.024/0.500)(100) =4.8 Ercent.
A photon (I. = 04(1) nm) strikes an electron at rest and rebounds at an angle ol
150’ to its original direction.
Find the speed and wavelength of the photon after the collision.
I The speed of a photon is always the speed of light in vacuum. c. To obtain the
wavelength after collision,
we use the equation for the Compton effect:
,_ h _ H, 6.63X10‘“.l-s
A -}.+m£(l cosp)-4 >< l0 m-+01xw_,,kg)(3xw,ms)(l cosl50°)
= 4 X10 '“rn + (2.43 X 10 '2 m)(1+0.866)=0.4045 nm
Suppose that a beam of 0.2-MeV photons is scattered by the electrons in a carbon
target. (a) What is the
wavelength associated with these photons? (bl What is the wavelength of those
photons scattered through an
angle of 90°? (r) What ts the energy of the scattered photons that emerge at an
angle ol 60° relative to the
incident direction?
hr l240MeV-tm
' "" "E-"tfinzt/-'-‘MM
h
(b) A’=A+m(l-cos8)=6200+(2430)(l-0)=8630tm
hr: Itc I240
“t 7' 'o2w+(24so)(1-§)'°" M”
Verity Compton‘s equation (Prob. 38.29) when the photon is back-scattered (4: =
190°).
I Assuming the electron to be at rest initially. we have from conservation of
momentum
h h 1 1
1-"7 °' "-"(Fri
and from conservation of relativistic energy
1| h 1 l
T‘+s(.=7f+£ or E=s,,+/"(I-1-,)
Substituting these expressions for the clectron's final momentum and final energy
into the momentum-energy
relation for the electron. we obtain
l 1 ‘ 1 l ’ l l . l 1
[£.,+ hrs -31)] = 125+ [1-¢(; + 2£,.h¢(Z -7) - 2/fic Z1-,= 21865 5.41‘ - A) = 211:
whence A‘ — A = (Z116)/E‘,=(2lt)/(m,c). which is Compton's equation with cos 0 = —-
1.
A photon strikes a free electron at rest and is scattered straight backward. lt the
electron's speed after
collision is ac. where or << l, show that the electron's kinetic energy is a
fraction tr of the photon‘s initial
energy.
I Since v = ac and or (< l, the electron is nonrelativistic. Conservation of
momentum yields It/A + h/it‘ =
mu; multiplying by c, (hc)/A + (hc)/1' = mvc = (mu')/a. The energy equation is
(hc)/A - (Irc)//l’ = (mt?)/2.
AAJ K/-\ TOPPER Ii'EeEi°§i'§i'§gJ'E'épR':I’§
AAJ K/-\ TQPPER ii'EE'ie§ii'ii§gi§EpR'i§’§

38.34
38.36
38.36
3.37
38.38
PARTICLES OF LIGHT AND WAVES OF MATTER J 713
Adding we find that (hr)/A = E, = (mt?)/4 + (mu‘)/(Zn). Since a<< l. (mul)/4 is
neglected and (mt/’)/2 =
aE,.
A photon with A = 0.5 nm strikes a free electron head-on and is scattered straight
backward. If the electron is
initially at rest, what is its speed after the collision?
I Momentum conservation gives h/A = "lull, — hilt’. assuming that the scattered
electron is nonrelativistic. A
is given as 0.5 nm and 1‘ for 180° scattering is 0.5 nm + (2h)/(mac) = 0.5048 nm,
so 1', = (h/mt.)/(1/A +1//1')=
[(6.63 >< 10"‘)/(9.1 X l0‘“)](3.98 X l0°) = 2.9 X 10° mls.
A photon with A = 0.5 nm is moving along the x axis when it strikes a tree electron
(initially at rest) and is
scattered so as to move along the y axis. What are the x and y components of the
electron‘s velocity after
collision‘!
I In the x direction. conservation of momentum yields hll = mntt, and in y
direction It/J.’ + m,,t-, = 0. From
the Compton scattering relation for o = 90°. A’ = A + 0.024 A = 5.0 + 0.024 = 5.024
A. Solving for I1, and u,
from above, with ml, the electron rest mass. gives v, = +1.46 X 10" m/s and ti, =
-1.46 x l0“m/s; the electron
is nonrelatitnstic.
Derive the Compton equation. using the relativistic expressions for the energy and
momentum oi the
electron.
I.
If
J\f\¢> -- -- ----- -- -t
I .‘,
It
I-1g.38-1
| For notation, see Fig. 38-l.
conservation of x momentum: g= % cos ¢ + P cos 0 (l)
conservation of y momentum: 0 = % sin ¢ - P sin 6 (2)
energy conservation: hf + mac: = hf’ + (P‘c’ + m§c‘)"" (3)
(i) Solve (1) for P cos 9 and (2) for P sin 9. Square and add; Iff’ — 2h’f}" cos 4:
+ Diff" =c’P’.
(ii) Square (3) to obtain: h’f* + Hf" — 2I|‘fl" + 2Irm,,c‘(/ —f') = c’P’.
(m) Subtract: (I —f')/1T = [It/(mt,c’)](l — cos 4)).
(iv) Butf=c/A and so (f —f')/())")= ()3 —/1)/c.
(U) Thtl’¢f01’¢, 1' - A =-|h/(m..c)](1— wt 4»).
In what amounts to the inverse of the photoelectric efiect. x-ray photons are
produced when a tungsten target
is bombarded by accelerated electrons. ll an x-ray machine has an accelerating
potential of 60 ltV. what is the
shortest wavelength present in its radiation?
I Since the work function of tungsten is very much smaller than the accelerating
potential. we may suppose
that the entire kinetic energy of an electron is lost to create a single photon of
maximum energy:
hc hc_1240keV-pm_
‘V Am *"““=ZV' 60keV ‘Em
An electron is shot down an it-ray tube. lts energy just before it strikes the
target oi the tube is 30keV. If it
loses all this energy in a single collision with a very massive atom. what is the
wavelength of the single x-ray
photon that is emitted?
I £='%‘=1fl—k?’-1' l=%=U.Nl3nm=O.4l3A
AAJ K/-\ TOPPER It'EeEi°§i'§i'§gJ'E'épR':I’§
38.47
38.48
38.49
38.50
38.51
PARTICLES OF LIGHT AND WAVES OF MATTER U 715
I For Bragg reflection the scattering angle is given by sin B = (ml)/(24) = [(2)
(l.2)]/4.4. and 0 = Q3.
The attenuation coetficient of aluminum tor soft X-rays is L73/cm. Compute the
traction of these X-rays
transmitted by an aluminum sheet l.l56cm thick.
I The exponential law of attenuation is
_, I _ _ 1 1 .
!=L,e “ E=¢ '”“'”’=e ‘=e—,=F87-0.135 or 13.5 d
One-half the intensity of a homogeneous x-ray beam is removed by an aluminum filter
5 mm thick. What is
the fraction of this homogeneous beam that would be removed by 15 mm of aluminum?
I 5 = e "' 1= 2"" L’= e‘ "“ = (e"")’ = G)’ =5 (fraction remaining) 1- 1 =2
(traction removed)
L, 2 In 2 8 8 8
Show that when a positron and an electron (both essentially at rest) annihilate,
Creating two photons, either
photon has the Compton wavelength (Prob. 38.29).
I Total energy before annihilation is 2m,c’; and after, 2[(hc)lA] (momentum
conservation requires that the
photon energies be equal). Then. by conservation of energy.
2m,c’=ZE or A=i=2430fm
A m,c
Show that the threshold wavelength for the production of a positron-electron pair
is half the Compton
wavelength.
I The incident photon must have at least enough energy to create a pair having zero
kinetic energy (we
ignore any kinetic energy oi the recoil nucleus). Thus,
hc 2 It
lz2m,c or Aszmic L
Alter pair annihilation. two 1-MeV photons move olf in opposite directions. Find
the kinetic energy oi the
electron and positron.
I The two photons have the same energy, and hence the same momentum, which we are
told are in
opposite directions. Thus by momentum conservation the electron and positron must
have had equal and
opposite momentum. Hence their kinetic energies are equal. By energy conservation:
2 MeV I Ztmoc‘) +
2 KE = 2(0.51)MeV + 2 KE and KE of each = 0.49 MeV.
38.3 DE BROGLIE WAVES AND THE UNCEIYTAINTY PRINCIPLE
3852
18.53
38.54
A proton is accelerated through a potential difference of l000V. What is its de
Broglie wavelength?
I We have A = It/(mu). But m = mo (nonrelativistic), so u may be found from (mv')/2
= llllle. Substitution
gives A = 9.1x 10"’ m.
Find the de Broglie wavelength of a thermal neutron of mass 1.67 X 10'" kg
traveling at a speed of 22(1) m/s.
I Use the dc Broglie wave equation. . . . - .
It 6.63Xl0 ‘u
a.=—=i-=0.
mv 1.67><l0""(22(l)) W
What is the dc Broglie wavelength for a particle moving with speed 2 >< 10° m/s if
the particle is (a) an
electron, (b) a proton, and (c) a 0.2-kg ball?
I We make use of the definition of the de Bmglie wavelength:
It 6.63x10"‘J~ s 3.3 X l0"°m - kg
*=.:=;@WF>=-m
AAJ K/-\ TOPPER il'E°$°§i?i'§gJ'EEpR':i§

PARTICLES OF LIGHT AND WAVES OF MATTER 5' 717
0 0 0 u
0 I e 0
t‘ 0 0 0 0
‘- gap
0 1 0 n
e 0 0 e
e I e 0
‘I.
I-'1g.38-4
substitution. T = 4.25 X 10 "‘/m. Using the electron mass and the helium mass (4 X
L67 X ll) 2" kg). we have
71 = Mil and Tn. L-3 K-
3860 Consider a beam of electrons shot toward a crystal. as shown in Fig. 38-4‘ The
crystal spacing is b. as
indicated. For what de Broglie wavelengths will the electron beam be strongly
reflected straight back upon
itself? For what electron kinetic energies? lt is found by expenment that electrons
having these energies are
unable to move through such a crystal in the direction shovm. Evaluate the energies
in eV for b = 2 X l0""m.
| Strong reflection will occur if 2b = ml. because the beams reflected from
successive planes Will then
reinforce. Therefore. A = (2b)/n. For nonrelativistic electrons E, = (pi)/(Zm) =
h’/(2mA§) = (n’h*)/(8b‘m).
which gives E, =9.4n= eV.
38.61 The nuclei of atoms have radii of order 10' '5 m. Consider a hypothetical
situation of a proton confined to a
narrow tube with length 2 X 10"’ m. What will he the de Broglie wavelengths which
will resonate in the tube?
To what momentum does the longest wavelength correspond? ll relativistic effects
are assumed negligible. to
what energy (in eV) does this correspond?
I This is a simple standing-wave problem. For symmetric boundary conditions, the
allowed A‘: are
/1,. = (ZL)/n = [(4 X 10' ")/n] m; since /1,, = h/p.., p. = (uh)/2L = n(1.66 >< l0
"'). Nonrelativistic kinetic
energy = E, =p,‘,/(Zm) = n‘(8.2 >< 10'" J) or 1:151 Mev}. n = 1 corresponds to
longest wavelength.
38.62 At what energy will the nonrclativistic calculation of the dc Broglic
wavelength of an electron be in error by 5
percent?
| For the nonrelativistic case. the de Broglie wavelength is
A _ if _ hc
"' pc
For the relativistic case.
1 Z Z ‘ 1 I K ll
(K +m..c) =(pc‘) + (m..r‘) or pC= l2m.,e K(1
and the de Broglie wavelength is
A _ E _ i"°___
' pc (2m(,c’K[1 + K/(Zm¢.c2)])""
For our case. 1., — K, = 0.051,; /1,,/1, = 1.05.
).,,, f K 1 K
A, _ \‘l +2m.,== 1'05" \ 1 +2(u.sn McV)
Solving. K = 0.105 Mg.
38.63 Show that the de Broglie wavelength of a particle is approximately the same
as that of a photon with the
same energy. when the energy of the particle is much greater than its rest energy.
. . E '—E.’ E _ it it
I E-=p=<~+£5 or p=?vll—(—l)=; it s>>£... So i.=;=€
Pi
For a photon. E =hv = (ht)//\,. so A, = (he)/E =- A.
AA‘-I R NEET AIIMS JEE RAS

718 U CHAPTER 38
38.60
38.65
38.66
38.6‘!
\- = <1 v = mt-\~'
w
Hg. 38-5
An electron is confined to a tube of length L. 'l'he electron‘s potential energy in
one half oi the tube is zero,
while the potential energy in the other half is l0 eV. If the electron has a total
energy E = 15 eV. what will be
the ratio of the de Broglie wavelength of the electron in the 10-eV region of the
tube to that in the other half
of the tube? Sketch the wave function w for the electron along the tube.
I The wave function is sketched in Fig, 38-5. A =It/p =h/(2mK)‘°; with K the
kinetic energy of the
electron. A,/A, I (K,lK,)'” = (1$lS)"’ = i
A particle of mass m is confined to a one-dimensional line of length L. From
arguments based on the wave
interpretation of matter. show that the energy of the particle can have only
discrete values and determine
these values.
I If the particle is confined to a line segment. say from x = 0 to x = L, the
probability of finding the particle
outside this region must be zero. Therefore. the wave function w must be zero for r
< 0 or x 2- L. since the
square of w gives the probability for finding the particle at a certain location.
lnside the limited region the
wavelength of qt must be such that w vanishes at the boundaries x = 0 and x = L, so
that it can vary
continuously to the outside region. Hence only those wavelengths will be possible
for which an integral
number of half wavelengths fit between x I 0 and x = L. that is. L = (ml)/2, where
n is an integer. called the
quantum number, with values n = l, 2. 3. . . . . From the de Broglie relationship A
= It/p we then find that the
particle's momentum can have only discrete values given by
_ E _ 1':
" ‘ A ‘ 2L
Since the particle is not acted upon by any forces inside the region. its potential
energy will be a constant
which we set equal to zero. Therefore the energy ol the body is entirely kinetic
and will have the discrete
values obtained from
1 * [tn/tzul’ . ts
£=K=5mu==%=”—2'—"— lhflllS, E,,=n28W n=l.2.3...,
This very simple problem illustrates one of the basic features of the probability
interpretation of matter;
namely. that the energy of a bound system can take on only discrete values, with
zero energy not being a
possible value.
Assume that the uncertainty in the position of a particle is equal to its de
Broglie wavelength. Show that the
uncertainty in its velocity is equal to or greater than 1/(4n') times its velocity.
I Use the Heisenberg uncertainty principle with Ax I »\ = It/(mu,,).
Ax A(mu,) 2% "Illa A(mu,) 2%
Since m is constant.
h h h It 1
mbn Alma; Eran, an Au, zau,
lf the uncertainty in the time during which an electron remains in an excited state
is l0"’ s. what is the least
uncertainty (in J) in the energy of the excited state?
I Let W be the energy of the excited state.
-11 ~17
aw A1 2% (aw)(10"') = 6ii'°3:n1° aw 26-——%0 2 0.523 x to " .l
AAJ K/-\ TQPPER it'E°Ei°§i'ii'§gi'E'EpR':I’§

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