Chapter 1

Introduction

Prepared by: Engr. Lucia V. Ortega 8/28/20 Statics of Rigid Bodies

Statics of Rigid Bodies Chapter 1: Introduction to Statics

Objectives:

 To introduce the basic quantities and idealizations of mechanics.

 To introduce free-body diagram and its importance in problem solving.
 To give a statement of Newton’s Laws of Motion and Gravitation.
 To review the principles for applying SI systems of units.
 To examine the standard procedures for performing numerical calculations.
 To present a general guide for solving problems.

Introduction:

Mechanics is a branch of the physical sciences that is concerned with the state of rest or motion of bodies that are
subjected to the action of forces. In general, this subject can be subdivided into three branches: rigid-body
mechanics, deformable body mechanics and fluid mechanics

Subdivision of Rigid Body Mechanics (Engineering Mechanics)

1. Statics – branch of mechanics, which concerns the equilibrium of bodies under the action of forces.
2. Dynamics – branch of mechanics, which concerns the motion of bodies.

Subdivision of Dynamics

a. Kinematics – deals with pure motion of rigid bodies.

b. Kinetics - relates the motion to the applied forces.

Historical Development:

 287 – 212 B.C. – Principle of lever and the principle of buoyancy by Archimedes

 1548 – 1620 – Laws of vector combination of forces and the principles of statics by Stevinus

 1564 – 1620 – the first investigation of dynamics by the experiments with falling stones by Galileo

 1642 – 1727 – the accurate formulation of the laws of motion and the law of gravitation by Isaac Newton

 Substantial contribution to the development of mechanics were also made by da Vinci, Varignon, Euler,
D’Alembert, Lagrange, Laplace and others

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Statics of Rigid Bodies Chapter 1: Introduction to Statics

Fundamental Concepts

a) Basic Concepts:

 Space – is the geometric region occupied by bodies whose position are described by the linear and angular
measurement relative to a coordinate system
 Time – is the measure of the succession of events and is basic quantity in dynamics.
 Mass – is a measure of the inertia of a body, which is its resistance to a change of velocity. Mass can also
be thought as the quantity of matter in a body. The mass of a body affects the gravitational attraction
force between it and other bodies.
 Force – is the action of one body on another. A force tends to move a body in the direction of its action.
The action of a force is characterized by its magnitude, by the direction of its action, and by its point of
application.

b) Idealizations:

 Particle – is a body of negligible dimensions. In the mathematical sense, a particle is a body whose
dimensions are considered to be near zero so that we may analyze it as a mass concentrated at a point.
 Rigid body – is a definite amount of matter the parts of which are fixed in position relative to each other.
 Concentrated force - represents the effect of a load, which is assumed to act at a point on a body.

Characteristics of a Force
1) It has magnitude
2) The position of its line of action (point of application)
3) The direction (or sense) in which the force is acting along its line of action

Force System

A force system is any arrangement where two or more forces acts on a body or a group of related bodies.
 Coplanar force system – the lines of action of all the forces lie in one plane
 Non-coplanar force system - the lines of action of all the forces lie in more than one plane
 Concurrent force system - the lines of action of all the forces pass through a common point
 Non-concurrent force system - the lines of action of all the forces do not pass through a common point
 Parallel force system - the lines of action of all the forces are parallel.

Principle of Transmissibility of a Force

The principle of transmissibility states that the point of application of a force can be moved anywhere along its line
of action without changing the external reaction forces on a rigid body.

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Statics of Rigid Bodies Chapter 1: Introduction to Statics

Axioms of Mechanics

1. The parallelogram law: The resultant of two forces is the diagonal of the parallelogram formed on the vectors
of these forces.
2. Two forces are in equilibrium only when equal in magnitude, opposite in direction, and collinear in action.
3. A set of forces in equilibrium maybe added to any system of forces without changing the effect of the original
system.
4. Action and reaction forces are equal but oppositely directed.

Free-Body Diagram

A free-body diagram is a sketch of the isolated body, which shows only the forces acting upon the body. The forces
acting on the free-body are the section forces, also called the applied forces. The reaction forces are those exerted
by the free-body upon other bodies.

The free-body diagram method is the key to the understanding of mechanics. This is so because the isolation of a
body is the tool by which cause and effect are clearly separated, and by which our attention is clearly focused on the
literal application of a principle of mechanics.

Scalars and Vector Quantities

Scalar quantities – are those with which only a magnitude is associated. Examples are time, volume, density, speed,
energy, and mass.

Vector quantities – possess direction as well as magnitude, and must obey the parallelogram law of addition.
Examples of vector quantities are displacement, velocity, acceleration, force, moment, and momentum.

Classification of Vectors

1. Free vector – is one whose action is not confined to or associated with a unique line in space. For example, if a
body moves without rotation, then the movement or displacement of any point in the body may be taken as a
vector. This vector describes equally well in the direction and magnitude of the displacement of every point in
the body. Thus, we may represent the displacement of such a body as a free vector.

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Statics of Rigid Bodies Chapter 1: Introduction to Statics

2. A sliding vector has a unique line of action in space but not a unique point of application. For example, when
an external force acts on a rigid body, the force can be applied at any point along its line of action without
changing its effect on the body as a whole.

3. A fixed vector is one for which a unique point of application is specified. The action of a force on a deformable
or non-rigid body must be specified by a fixed vector at the point of application of the force. In this instance
the forces and the deformations within the body depend on the point of application of the force, as well as on
its magnitude and line of action.

Newton’s Laws of Motion

First Law: A particle originally at rest, or moving in a straight line with constant
velocity, tends to remain in this state provided the particle is subjected t0 an
unbalanced force.

Second Law: A particle acted upon by an unbalanced force 𝐹 experiences an

acceleration 𝑎 that has the same direction as the force and a magnitude that is
directly proportional t0 the force. If 𝐹 is applied to a particle of mass 𝑚, this law
may be expressed as 𝐹 = 𝑚𝑎.

Third Law: The mutual forces of action and reaction between two particles are
equal, opposite and collinear.

Newton’s Law of Gravitational Attraction

Shortly after formulating his three laws of motion, Newton postulated a law governing the gravitational attraction
between any two particles. Stated mathematically.
m1 m2
F=G
r2

Where:

F = the mutual force of attraction between two particles

G = a universal constant known as the constant of gravitation
m1, m2 = the masses of the two particles
r = the distance between the centers of the particles
G = 66.73 𝑥 10−12 𝑚3 /(𝑘𝑔 ∗ 𝑠 2 )

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Statics of Rigid Bodies Chapter 1: Introduction to Statics

Weight

In the case of a particle located at or near the surface of the earth, however,
the only gravitational force having any sizable magnitude is that between the
earth and the particle. Consequently, this force, termed the weight, will be the
only gravitational force considered in our study of mechanics.
𝑚𝑀𝑒
𝐹=𝐺  Where: 𝑚 = mass of the particle; 𝑀𝑒 = mass of the
𝑟2
earth; 𝑟 = the distance between the center of the
earth and the center of the particle
𝐺𝑀𝑒
Letting 𝑔 =  𝑊 = 𝑚𝑔 ; where: 𝑔 = acceleration due to gravity
𝑟2

Units of Measurements

In SI Units, the base units for mass, length, time and force are kilogram, meter, second and Newton respectively
while in U.S. Customary Units, the base unit for length, time and force are foot, second and pound respectively.

Table 1-1: Systems of Units

Dimensional SI Units U. S. Customary Units
Quantity
Symbol Unit Symbol Unit Symbol

Mass M kilogram kg slug -

Length L meter m foot ft
Time T Second s second sec
Force F newton N pound lb

Conversion of Units

Table 1-2: Conversion of Units

Force 9.81 𝑁 = 2.20 𝑙𝑏𝑓 1 𝑙𝑏𝑓 = 4.4482 𝑁 32.2 𝑙𝑏𝑓 = 143.2 𝑁

Mass 1 𝑘𝑔 = 2.20 𝑙𝑏𝑚 1 𝑙𝑏𝑚 = 0.45359 𝑘𝑔 1 𝑠𝑙𝑢𝑔 or 32.2 𝑙𝑏𝑚 = 14.5938 𝑘𝑔

Length 1 𝑓𝑡 = 0.3048 𝑚 1 𝑙𝑏𝑚 = 3.28 𝑓𝑡

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Statics of Rigid Bodies Chapter 1: Introduction to Statics

The International Systems of Units

Table 1-3: Prefixes

Multiple Exponential Form Prefix SI Symbols

1 000 000 000 109 giga G

1 000 000 106 mega M
3
1 000 10 kilo k
Submultiple
0.001 10−3 milli m
−6
0.000 001 10 micro 
−9
0.000 000 001 10 nano n

Rules in Using SI Symbols

 Quantities defined by several units which are multiples of one another are separated by a dot to avoid
𝑚
confusion with prefix notation, as indicated by 𝑁 = 𝑘𝑔 ∙ 2 = 𝑘𝑔 ∙ 𝑚 ∙ 𝑠 −2 . Also, 𝑚 ∙ 𝑠 (𝑚𝑒𝑡𝑒𝑟 ∙ 𝑠𝑒𝑐𝑜𝑛𝑑),
𝑠
whereas 𝑚𝑠 (𝑚𝑖𝑙𝑙𝑖𝑠𝑒𝑐𝑜𝑛𝑑).

 The exponential power on a unit having a prefix refers to both the unit and its prefix. For example, 𝑁 2 =
(𝜇𝑁)2 = 𝜇𝑁 ∙ 𝜇𝑁 . Likewise, 𝑚𝑚2 represents (𝑚𝑚)2 = 𝑚𝑚 ∙ 𝑚𝑚.

 With the exception of the base unit the kilogram, in general avoid the use of a prefix in the denominator of
composite units. For example, do not write 𝑁/𝑚𝑚, but rather 𝑘𝑁/𝑚; also, 𝑚/𝑚𝑔 should be written as
𝑀𝑚/𝑘𝑔.

 When performing calculations, represent the numbers in terms of their base or derived units by converting all
prefixes to powers of 10. The final result should then be expressed using a single prefix. Also, after calculation,
it is best to keep numerical values between 0.1 and 1000; otherwise, a suitable prefix should be chosen. For
example,
(50 𝑘𝑁)(60𝑛𝑚) = [50(103 )𝑁][60(10−9 )𝑚]
= 3000(10−6 )𝑁 ∙ 𝑚 = 3(10−3 )𝑁 ∙ 𝑚 = 3𝑚𝑁 ∙ 𝑚
Numerical Calculations

 Dimensional Homogeneity: The terms of any equation used to describe a physical process must be expressed
in the same units.

 Significant Figures: the number of significant figures in any number determines the accuracy of the number.
For large number, it is better to present the result in scientific notation.

 Rounding Off Numbers: rounding off number is necessary so that the accuracy of the result will be the same
as that of the problem data. As a general rule, any numerical figure ending in five or greater is founded up and
a number less than five is rounded down.

 Calculations: when a sequence of calculations is performed, it is best to store intermediate results in the
calculator. In other words, do not round off calculations until the final result.

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Statics of Rigid Bodies Chapter 1: Introduction to Statics

General Procedure for Analysis

 Read the problem carefully and try to correlate the actual physical situation with the theory studied.

 Tabulate the problem data and draw any necessary diagrams.

 Apply the relevant principles, generally in mathematical form. When writing any equations, be sure they are
dimensionally homogeneous.

 Solve the necessary equations, and report the answer with no more than three significant figures.

 Study the answer with technical judgment and common sense to determine whether or not it seems
reasonable.

Important Points

 Statics is the study of bodies that are at rest or move with constant velocity.

 A particle has a mass but a size that can be neglected.

 A rigid body does not deform under load.

 Concentrated forces are assumed to act at a point on a body.

 Newton's three laws of motion should be memorized.

 Mass is measure of a quantity, of matter that does not change from one location to another.

 Weight refers to the gravitational attraction of the earth on a body or quantity of mass. Its magnitude depends
upon the elevation at which the mass is located.

 In the SI system the unit of force, the newton. is a derived unit. The meter, second, and kilogram are base
units..

 Prefixes G, M, k, m, , and n are used to represent large and small numerical quantities. Their exponential
size should be known, along with the rules for using the SI units.

 Perform numerical calculations with several significant figures, and then report the final answer to three
significant figures.

 Algebraic manipulations of an equation can be checked in part by verifying that the equation remains
dimensionally homogeneous.

 Know the rules for rounding off numbers.

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Statics of Rigid Bodies Chapter 1: Introduction to Statics

Example 1.1

Convert 2 𝑘𝑚/ℎ to 𝑚/𝑠? How many 𝑓𝑡/𝑠 is this?

Solution:
Since 1 𝑘𝑁 = 1000 𝑚 and 1 ℎ𝑟 = 3600 𝑠, the factors of conversion are arranged in the following order so that
a cancellation of the c units can be applied:
𝑘𝑚 𝑘𝑚 1000 𝑚 1ℎ
2 =2 ( )( )
ℎ ℎ 1 𝑘𝑚 3600 𝑠

𝑘𝑚 2600 𝑚 𝑚
2 = = 0.556 Answer
ℎ 3600 𝑠 𝑠

From Table 1-2, 1 𝑓𝑡 = 0.3048 𝑚. Thus

𝑚 0.556 𝑚 1 𝑓𝑡
0.556 =( )( )
𝑠 𝑠 0.3048 𝑚

𝑚 𝑓𝑡
0.556 = 1.82 Answer
𝑠 𝑠

Note: Remember to round of the final answer to three significant figures.

Example 1.2

𝑠𝑙𝑢𝑔
Convert the quantities 300 𝑙𝑏 ∙ 𝑠 and 52 to appropriate SI units.
𝑓𝑡 3
Solution:

Using Table 1-2, 1 𝑙𝑏 = 4.4482 𝑁

4.4482 𝑁
300 𝑙𝑏 ∙ 𝑠 = 300 𝑙𝑏 ∙ 𝑠 ( )
1 𝑙𝑏

300 𝑙𝑏 = 1334.5 𝑁 ∙ 𝑠 = 1.33 𝑘𝑁 ∙ 𝑠 Answer

Since 1 𝑠𝑙𝑢𝑔 = 14.5938 𝑘𝑔 and 1 𝑓𝑡 = 0.3048 𝑚, then

𝑠𝑙𝑢𝑔 𝑠𝑙𝑢𝑔 1 𝑓𝑡 3
52 = 52 (1 𝑠𝑙𝑢𝑔) ( )
𝑓𝑡 3 𝑓𝑡 3 0.3048 𝑚

𝑠𝑙𝑢𝑔 26.8(103 )𝑘𝑔 𝑀𝑔

52 = = 26.8 Answer
𝑓𝑡 3 𝑚3 𝑚3

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Statics of Rigid Bodies Chapter 1: Introduction to Statics

Example 1.3

Evaluate each of the following and express with SI units having an appropriate prefix: (a) (50 𝑚𝑁)(6 𝐺𝑁), (b)
(400 𝑚𝑚)(0.6 𝑀𝑁)2 , (c) (45 𝑀𝑁 3 )/(900 𝐺𝑔).

Solution:

First convert each number to basic units, perform the indicated operations, then choose an appropriate prefix.

Part a: (50 𝑚𝑁)(6 𝐺𝑁) = [50(10−3 )𝑁][6(109 )𝑁] = 300(106 )𝑁 2

1 𝑘𝑁 1 𝑘𝑁
(50 𝑚𝑁)(6 𝐺𝑁) = 300(106 )𝑁 2 ( )( ) = 300 𝑘𝑁 2 Answer
103 𝑁 103 𝑁

Note: Keep in mind the convention 𝑘𝑁 2 = (𝑘𝑁)2 = 106 𝑁 2

Part b: (400 𝑚𝑚)(0.6 𝑀𝑁)2 = [400(10−3 )𝑚][0.6(106 )𝑁]2 = [400(10−3 )𝑚][0.36(1012 )𝑁 2 ]

(400 𝑚𝑚)(0.6 𝑀𝑁)2 = 144(109 )𝑚 ∙ 𝑁 2 = 144 𝐺𝑚 ∙ 𝑁 2 Answer

We can also write

1 𝑀𝑁 1 𝑀𝑁
144(109 )𝑚 ∙ 𝑁 2 = 144(109 )𝑚 ∙ 𝑁 2 ( )( ) = 0.144 𝑚 ∙ 𝑀𝑁 2 Answer
106 𝑁 106 𝑁
3 3 1
(45 𝑀𝑁3 ) 45(106 𝑁) 𝑁3 1 𝑘𝑁 𝑘𝑁3
Part c: = = 50(109 ) ( ) = 50 Answer
(900 𝐺𝑔) 900(106 )𝑘𝑔 𝑘𝑔 103 𝑁 𝑘𝑔 𝑘𝑔

Example 1.4

Determine the weight in newtons of a car whose mass is 1400 kg. Convert the mass of the car to slugs then
determine its weight in pounds.

Solution:
a) From relationship 1/3, we have
𝑊 = 𝑚𝑔 = 1400(9.81) = 13,734 𝑁 Answer
b) From the table of conversion factors, we see that 1 slug = 14.594 kg.
Thus the mass of the car in slugs is,
1 𝑠𝑙𝑢𝑔
𝑚 = 1400 𝑘𝑔 [ ] = 95.93 𝑠𝑙𝑢𝑔𝑠 Answer
14.594 𝑘𝑔
c) Finally, its weight in pounds is,
𝑊 = 𝑚𝑔 = (95.93)(32.2) = 3088.94 𝑙𝑏 Answer
d) As another rule to the last result, we convert from kg to 𝑙𝑏𝑚. Again
using the table, we have
1 𝑙𝑏𝑚
𝑚 = 1400 𝑘𝑔 [ ] = 3086.49 𝑙𝑏𝑚 Answer
0.45359 𝑘𝑔

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Statics of Rigid Bodies Chapter 1: Introduction to Statics

Example 1.5

Use Newton’s law of universal gravitation to calculate the weight of a 70-kg person standing on a surface of the
earth. Then repeat the calculation by using W = mg and compare your two results.

Solution:

The two results are

𝐺𝑚1 𝑚2 (6.673 ∙ 10−11 )(5.976 ∙ 1024 )(70)
𝑊= =
𝑅2 (6371 ∙ 103 )2
𝑊 = 687.72 𝑁 Answer
or
𝑊 = 𝑚𝑔 = (70)(9.81) = 686.7 𝑁 Answer

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