Mechanics lecture notes mechanical engineering

ENGINEERING MECHANICS I
Mechanical I/I
2024
COURSE INTRODUCTION

• Know the engineering mechanics and its subdivisions and their
applications.
2
Course Objectives
• Learn the basic concepts and fundamental principles of
engineering mechanics.
• Develop model and solve problems involving forces acting on a
particle and its equilibrium.
• Develop model and solve problems involving forces acting on a
rigid body and its equilibrium.
• Solve problems involving friction.
• Determine the location of the centroid and the center of mass for
a system of discrete particles and for objects of arbitrary shape.
• Calculate moments of inertia for lines, areas, and volumes.

• Determine the internal forces in trusses and in general frame
structures.
3
• Determine the internal reactions in a beam, draw correct shear-
force and bending moment diagrams, and write equations for the
shear-force and bending moment as functions of position along
the beam.

4
Course Outlines
1. Forces Acting on Particle (6 hours)
Introduction to Mechanics; Scalars, Vectors and Vector
Operations; Definitions and Concept of Particle, Deformable and
Rigid Bodies; Fundamental concepts and principles of
mechanics: Newtonian Mechanics; Concept of Particles and Free
Body Diagram; Equation of Equilibrium in Two Dimension;
Force in Space

5
2. Forces Acting on Rigid Body (6 hours)
External and Internal Force; Principle of Transmissibility and
Equivalent Forces; Resolution of a Force into Forces and a
Couple; Resultant of Force and Moment for a System of Force;
Equilibrium of Two force and three force body; Equation of
Equilibrium in Three Dimension
3. Center of Gravity and Centroids (6 hours)
Centre of Gravity of Two Dimensional Body; Centroid of Areas
and Lines; First Moment of Area and Lines; Determination of
Centroids by Integration and Theorems of Pappus-Guldinus;
Centre of Gravity of Three Dimensional Body; Introduction to
Composite Plate and Wire

6
4. Moment of Inertia (7 hours)
Moment of Inertia of Area; Second moment of Area; Polar
Moment of Inertia; Radius of Gyration; Parallel axis Theorem
for area moment of Inertia; Moment of Inertia of Composite
Areas; Mass Moments of Inertia; Moment of Inertia of a Simple
Mass; Parallel-Axis Theorem for Mass Moments of Inertia;
Moments of Inertia of Thin Plates; Mass moment of inertia of
Composite Bodies

7
5. Friction (6 hours)
Laws of Friction, Static and Dynamic Coefficient of Friction,
Angle of Friction; Application of friction in engineering:
Example as High Tension Friction, Wedge and Screw, Belt
Friction, Friction on axles, disks, and wheels, Journal Bearings
and Axle Friction, Thrust Bearings and Disk Friction, Wheel
Friction and Rolling Resistance

8
6. Introduction to Structure: Beam, Frame (8 hours)
Reactions for a Two-Dimensional Structure: Rocker/Roller
support, Short Cable/Link; Hinge/Rough Surface, Frictionless
Pin; Fixed Support; Types of loading and supports; Examples
and Standard symbols; Point Load, Uniformly Distributed Load,
Uniformly Varying Load; Concept of Statically/Kinematically
Determinate and Indeterminate Beams and Frames; Relation of
Load, Axial Force, Shear Force and Bending Moment;
Determinate Beams; Determinate Frame; Axial Force, Shear
Force and Bending Moment Calculation and Diagram

9
7. Analysis of Trusses (6 hours)
Concept of Simple Truss and pin joints/joint loads in trusses;
Calculation of Member Forces of Truss by method of joints;
Calculation of Member Forces of Truss by method of joints
under Special Loading Conditions; Calculation of Member
Forces of Truss by method of sections; Introductory Concept of
Machines

10
Books
1. F.P. Beer and E.R. Johnston, “Vector Mechanics for
Engineers – Statics and Dynamics”, Mc Graw Hill.
2. R.C. Hibbler, “Engineering Mechanics”, Pearson, New
Delhi.
3. J.L. Meriam., “Engineering Mechanics – Statics and
Dynamics”, John Wiley and Sons.
4. J.C. Jong and B.G. Rogers, “Engineering Mechanics, Statics and
Dynamics”-Saunders College Publishing, International Edition
5. Bela I. Sandor, “Engineering Mechanics – Statics and
Dynamics”, Prentice Hall, Inc. Englewood Cliffs.
6. S. Neupane and H.R. Parajuli, “A Textbook of Applied
Mechanics for Engineers”, Heritage Publishers and Distributors.

11
7. Engineering Machanics, (Statics and Dynamics) Irving H.
Shames
8. Engineering Mechanics Statics, William F. Riley, Leroy D.
Sturges
9. Engineering Mechanics Statics, Andrew Pytel, Jaan Kiusalaas,
Ishan Sharma
10. Engineering Mechanics Statics, Bedford Fowler
11. Engineering Mechanics Statics Michael Plesha, Gary Gray,
Francesco Costanzo
12. Engineering Mechanics, Timoshenko
13. Schaum's Outline of Engineering Mechanics, McLean W.G.,
E.W.Nelson, C.L.Best
14. Introduction to Statics and Dynamics, Andy Ruina, Rudra Pratap
15. Engineering Mechanics, K. L. Kumar, Veenu Kumar

12
Assignments
Assignment 1: Statics of Particles (Resultant and Resolution)
Assignment 2: Statics of Particles (Equilibrium)
Assignment 3: Statics of Rigid Bodies (Moments about a point and an
axis)
Assignment 4: Statics of Rigid Bodies (Equivalent System)
Assignment 5: Statics of Rigid Bodies (Equilibrium)
Assignment 6: Friction
Assignment 7: Centroid, Center of Gravity, Center of Mass
Assignment 8: Moment of Inertia
Assignment 9: Analysis of Frames and Machines
Assignment 10: Shear Force and Bending Moments of a Beam
Assignment 11: Analysis of a Truss (Joint Method)
Assignment 12: Analysis of a Truss (Section Method)

13
Evaluation
S. No. Evaluation Tools Weight
1. Unit Tests 15
2. Assignments 20
3. Projects 15
4. Test 1 25
5. Test 2 25
TOTAL 100

14
What Next?
Engineering
Mechanics I
(Statics)
Engineering
Mechanics II
(Dynamics)
I/I I/II
Strength of
Materials
II/I
Mechanics of
Solids
II/II
Theory of
Machines and
Mechanisms
II/II
Machine
Dynamics
III/I
Machine
Design I
III/I
Machine
Design II
III/II
Finite
Element
Method
III/II

Mechanical I/I
2024
FORCES ACTING ON PARTICLES
Session 01: Definition and Scope of
Engineering Scope

• Define engineering mechanics.
Learning Objectives
• Know the scope of engineering mechanics.
• Know the divisions of engineering mechanics.
• Define each division of engineering mechanics.
• Know the subdivision of rigid body mechanics and its scope.
• Know the subdivision of deformable body mechanics and its
scope.
• Know subdivision of fluid mechanics and its scope.

1.1 Engineering Mechanics: Definition and Scope
The subject of mechanics occupies a unique position in the physical
sciences because it is fundamental to so many fields of study.
In its broadest sense, mechanics may be defined as the science,
which describes and predicts the conditions of rest or motion of
bodies under the action of forces.
A thorough understanding of this subject is required for the study
of structural engineering, machine design, fluid flow, electrical
devices, and even the molecular and atomic behavior of elements.

Division of Engineering Mechanics
ENGINEERING MECHANICS
RIGID BODY DEFORMABLE BODY FLUID
STATICS DYNAMICS INCOMPRESSIBLE COMPRESSIBLE
STRENGTH OF
MATERIALS
THEORY OF
ELASTICITY
THEORY OF
PLASTICITY
KINEMATICS KINETICS

The term rigid body means that the body under consideration does
not undergo significant deformation under loading.
Rigid body mechanics is divided into two areas: statics and
dynamics. Statics deals with the forces and their effects, while
acting upon the bodies at rest, whereas dynamics deals with the
forces and their effects while acting upon the bodies in motion.
Statics deserves separate treatment in engineering, since most
structures, or frameworks are designed with the attention that they
remain in equilibrium.

Actual structures and machines, however, are never absolutely rigid
and deform under the loads to which they are subjected. But these
deformations are usually small and do not appreciably affect the
conditions of equilibrium or motion of the structure under
consideration.
They are important, though, as far as the resistance of the structures
to failure is concerned and are studied in strength of materials;
which is a part of the deformable body mechanics. The deformable
body mechanics is further subdivided into strength of material,
theory of elasticity and theory of plasticity.
The third division of mechanics, fluid mechanics is subdivided into
the study of incompressible fluids and of compressible fluids. An
important subdivision of the study of incompressible fluids is
hydraulics, which deals with problems involving liquids.

Review Questions
1. Define engineering mechanics. What is its scope?
2. What are the divisions of engineering mechanics?
3. What is rigid body mechanics? What are its sub-divisions?
4. What is deformable body mechanics? What are its sub-
divisions?
5. What is fluid mechanics? What are its sub-divisions?

Mechanical I/I
2024
Session 02: Fundamental Concepts

• Define quantities: Space, Mass, Time and Force.
Learning Objectives
• Know idealization concept.
• Define particle, rigid body, deformable body and fluid.
• State fundamental principles: Parallelogram law of forces,
Principle of transmissibility, Newton’s laws of motion and
Newton’s law of gravitation.

1.2 Fundamental Concepts
Before we begin the study of engineering mechanics, it is important
to understand the meaning of certain fundamental concepts and
principles.
1.2.1 Basic Quantities
The basic concepts used in mechanics are space, time, mass, and
force. These concepts cannot be truly defined; they should be
accepted on the basis of our intuition and experience and used as a
mental frame of reference for our study of mechanics.

(a) Space
The concept of space is associated with the position of a point 𝑃.
We can define the position of 𝑃 by providing three lengths
measured from a certain reference point, or origin, in three given
directions. These lengths are known as the coordinates of 𝑃.
(b) Time
To define an event, it is insufficient to indicate its position in space.
We also need to specify the time of the event. Although the
principles of statics are time independent, this quantity plays an
important role in the study of dynamics.

(c) Mass
We use the concept of mass to characterize and compare bodies on
the basis of certain fundamental mechanical experiments. Two
bodies of the same mass, for example, are attracted by the earth in
the same manner; they also offer the same resistance to a change in
translational motion.
(d) Force
A force represents the action of one body on another. A force can be
exerted by actual contact, like a push or a pull, or at a distance, as in
the case of gravitational or magnetic forces. A force is characterized
by its point of application, its magnitude, and its direction; a force is
represented by a vector.

1.2.2 Idealizations
Models or idealizations are used in mechanics in order to simplify
application of the theory. Here, we will consider important
idealizations.
(a) Particle
A particle is a portion of matter which is indefinitely small in size,
or which, for the purpose of our investigations, is so small in the
distances between its different parts may be neglected.
A body may be regarded as an indefinitely large number of
indefinitely small portions, or as a conglomeration of particles.

(b) Rigid body
When the dimensions, linear and angular, of a body do not change
during the course of observation, the body is modeled as a rigid
body. A rigid body, in other words, is one in which the distance
between any two arbitrary points is invariant. A body, therefore,
qualifies to be represented as a rigid body if the deformation
between its parts is negligible in the course of its analysis.
(c) Deformable Body
When the dimensions, linear or angular, of a body change during its
analysis, the body is modeled as a deformable body. Deformation
may be brought about in a variety of ways: it may temporary or
permanent, instantaneous or continuous. A body is, therefore,
represented as a deformable body if the relative deformation
between its parts cannot be ignored in the course of its analysis.

(d) Fluid
A substance which deforms continuously under the application of
shear stress, however small, is called a fluid. The process of
continuous deformation is called flow. A fluid must, therefore, flow
when subjected to a shear stress. In the absence of shear stresses the
fluids behave as static masses or as rigid bodies in motion.

1.2.3 Fundamental Principles
The study of elementary mechanics rests on six fundamental
principles, based on experimental evidence.
(a) The Parallelogram Law for the Addition of Forces
Two forces acting on a particle may be replaced by a single force,
called their resultant, obtained by drawing the diagonal of the
parallelogram with sides equal to the given forces.
(b) The Principle of Transmissibility
The conditions of equilibrium or of motion of a rigid body remain
unchanged if a force acting at a given point of the rigid body is
replaced by a force of the same magnitude and same direction, but
acting at a different point, provided that the two forces have the
same line of action.

(c) Newton’s First Law of Motion
If the resultant force acting on a particle is zero, the particle remains
at rest (if originally at rest) or moves with constant speed in a
straight line (if originally in motion).
(d) Newton’s Second Law of Motion
If the resultant force acting on a particle is not zero, the particle has
an acceleration proportional to the magnitude of the resultant and in
the direction of this resultant force.
(e) Newton’s Third Law of Motion
The forces of action and reaction between bodies in contact have the
same magnitude, same line of action, and opposite sense.

(f) Newton’s Law of Gravitation
Two particles of mass 𝑀 and 𝑚 are mutually attracted with equal
and opposite forces 𝑭 and −𝑭 of magnitude 𝐹, given by the formula
𝐹 =
𝐺𝑀𝑚
𝑟2
….. (1.1)

Review Questions
1. Define Length, Time, Mass and Force.
2. Define Particle, Rigid body, Deformable body and Fluid.
3. State six fundamental principles of mechanics.

Mechanical I/I
2024
Session 03: General Procedure of Problem
Solving

• Know the general steps of problem analysis.
Learning Objectives
• Know the activities of each step.

1.3 General Procedure of Problem Solving
In general, you can usually solve problems in several different ways;
there is no one approach that works best for everybody. However, it will
be helpful to follow the general guidelines to solve the problems of
engineering mechanics.
1.3.1 Problem Identification
The statement of a problem should be clear and precise, and it should
contain the given data and indicate what information is required. The first
step in solving the problem is to decide what concepts you have learned
that apply to the given situation and to connect the data to the required
information. It is often useful to work backward from the information
you are trying to find: Ask yourself what quantities you need to know to
obtain the answer, and if some of these quantities are unknown, how can
you find them from the given data.

1.3.2 Modeling
The first step in modeling is to define the system; that is, clearly
define what you are setting aside for analysis. After you have
selected a system, draw a neat sketch showing all quantities
involved, with a separate diagram for each body in the problem. For
equilibrium problems, indicate clearly the forces acting on each
body along with any relevant geometrical data, such as lengths and
angles. These diagrams are known as free-body diagrams.

1.3.3 Analysis
After you have drawn the appropriate diagrams, use the
fundamental principles of mechanics listed in Section 1.2.3 to write
equations expressing the conditions of rest or motion of the bodies
considered. Each equation should be clearly related to one of the
free-body diagrams and should be numbered. If you do not have
enough equations to solve for the unknowns, try selecting another
system, or reexamine your strategy to see if you can apply other
principles to the problem. Once you have obtained enough
equations, you can find a numerical solution by following the usual
rules of algebra, neatly recording each step and the intermediate
results. Alternatively, you can solve the resulting equations with
your calculator or a computer. (For multipart problems, it is
sometimes convenient to present the Modeling and Analysis steps
together, but they are both essential parts of the overall process.)

1.3.4 Interpretation
After you have obtained the answer, check it carefully. Does it make
sense in the context of the original problem? For instance, the
problem may ask for the force at a given point of a structure. If your
answer is negative, what does that mean for the force at the point?

Review Questions
1. What are general steps of problem analysis?
2. What activities are done during each step?

Mechanical I/I
2024
Session 04: Addition of Planar Forces

• Familiar with planar forces.
Learning Objectives
• Define a force.
• Know parallelogram law of forces.
• Define scalars and vectors.
• Know the method to add two and several vectors.
• Know the meaning of a product of a scalar and a vector.
• Determine resultant of several concurrent forces.
• Resolve a force into components.

In this chapter, you will study the effect of forces acting on
particles. By the word “particle” we do not mean only tiny bits of
matter, like an atom or an electron. Instead, we mean that the sizes
and shapes of the bodies under consideration do not significantly
affect the solutions of the problems.
Another way of saying this is that we assume all forces acting on a
given body act at the same point. This does not mean the object
must be tiny- if you were modeling the mechanics of the Milky
Way galaxy, for example, you could treat the sun and the entire
Solar System as just a particle.
Our first step is to explain how to replace two or more forces acting
on a given particle by a single force having the same effect as the
original forces. This single equivalent force is called the resultant
of the original forces.

After this step, we will derive the relations among the various
forces acting on a particle in a state of equilibrium. We will use
these relations to determine some of the forces acting on the
particle.
1.4 Addition of Planar Forces
Many important practical situations in engineering involve forces
in the same plane. These include forces acting on a pulley,
projectile motion, and an object in equilibrium on a flat surface. We
will examine this situation first before looking at the added
complications of forces acting in three-dimensional space.

1.4.1 Force on a Particle: Resultant of Two Forces
A force represents the action of one body on another. It is generally
characterized by its point of application, its magnitude, and its
direction. Forces acting on a given particle, however, have the same
point of application. Thus, each force considered in this chapter is
completely defined by its magnitude and direction.
We define the direction of a force by its line of action and the sense
of the force. The line of action is the infinite straight line along
which the force acts; it is characterized by the angle it forms with
some fixed axis (Figure 2.1).

The force itself is represented by a segment of that line; through the
use of an appropriate scale, we can choose the length of this
segment to represent the magnitude of the force. We indicate the
sense of the force by an arrowhead. It is important in defining a
force to indicate its sense.
Two forces having the same magnitude and the same line of action
but a different sense, such as the forces shown in Figure 2.1(a) and
(b), have directly opposite effects on a particle.

Experimental evidence shows that two forces 𝑷 and 𝑸 acting on a
particle 𝐴 (Figure 2.2a) can be replaced by a single force 𝑹 that has
the same effect on the particle (Figure 2.2c). This force is called the
resultant of the forces 𝑷 and 𝑸. We can obtain 𝑹, as shown in
Figure 2.2(b), by constructing a parallelogram, using 𝑷 and 𝑸 as
two adjacent sides. The diagonal that passes through 𝐴 represents
the resultant.
This method for finding the resultant is known as the parallelogram
law for the addition of two forces.

1.4.2 Vectors
We have just seen that forces do not obey the rules of addition
defined in ordinary arithmetic or algebra. For example, two forces
acting at a right angle to each other, one of 40 𝑁 and the other of
30 𝑁, add up to a force of 50 𝑁 acting at an angle between them,
not to a force of 70 𝑁.
Forces are not the only quantities that follow the parallelogram law
of addition. As you will see later, displacements, velocities,
accelerations, and momenta are other physical quantities possessing
magnitude and direction that add according to the parallelogram
law. All of these quantities can be represented mathematically by
vectors.

Those physical quantities that have magnitude but not direction,
such as volume, mass, or energy, are represented by plain numbers
often called scalars to distinguish them from vectors.
Vectors are defined as mathematical expressions possessing
magnitude and direction, which add according to the parallelogram
law. Vectors are represented by arrows in diagrams and are
distinguished from scalar quantities in this text through the use of
boldface type (𝑷). In longhand writing, a vector may be denoted by
drawing a short arrow above the letter used to represent it (𝑃). The
magnitude of a vector defines the length of the arrow used to
represent it. In this text, we use italic type to denote the magnitude
of a vector. Thus, the magnitude of the vector 𝑷 is denoted by 𝑃.

A vector used to represent a force acting on a given particle has a
well-defined point of application-namely, the particle itself.
Other physical quantities, however, such as couples, are represented
by vectors that may be freely moved in space; these vectors are
called free vectors. Still other physical quantities, such as forces
acting on a rigid body, are represented by vectors that can be moved
along their lines of action; they are known as sliding vectors.
Two vectors that have the same magnitude and the same direction
are said to be equal, whether or not they also have the same point of
application (Figure 2.3); equal vectors may be denoted by the same
letter.

The negative vector of a given vector 𝑷 is defined as a vector
having the same magnitude as 𝑷 and a direction opposite to that of
𝑷 (Figure 2.4); the negative of the vector 𝑷 is denoted by −𝑷. The
vectors 𝑷 and −𝑷 are commonly referred to as equal and opposite
vectors. Thus, we have
𝑷 + −𝑷 = 0 ….. (1.2)
1.4.3 Addition of Vectors
By definition, vectors add according to the parallelogram law.
Thus, we obtain the sum of two vectors 𝑷 and 𝑸 by attaching the
two vectors to the same point 𝐴 and constructing a parallelogram,
using 𝑷 and 𝑸 as two adjacent sides (Figure 2.5). The diagonal that
passes through 𝐴 represents the sum of the vectors 𝑷 and 𝑸,
denoted by 𝑷 + 𝑸.

Because the parallelogram constructed on the vectors 𝑷 and 𝑸 does
not depend upon the order in which 𝑷 and 𝑸 are selected, we
conclude that the addition of two vectors is commutative, and we
write
𝑷 + 𝑸 = 𝑸 + 𝑷
From the parallelogram law, we can derive an alternative method
for determining the sum of two vectors, known as the triangle rule.
Consider Figure 2.5, where the sum of the vectors 𝑷 and 𝑸 has
been determined by the parallelogram law.
….. (1.3)

Because the side of the parallelogram opposite 𝑸 is equal to 𝑸 in
magnitude and direction, we could draw only half of the
parallelogram (Figure 2.6a). The sum of the two vectors thus can
be found by arranging 𝑷 and 𝑸 in tip-to-tail fashion and then
connecting the tail of 𝑷 with the tip of 𝑸. If we draw the other half
of the parallelogram, as in Figure 2.6(b), we obtain the same result,
confirming that vector addition is commutative.

We define subtraction of a vector as the addition of the
corresponding negative vector. Thus, we determine the vector
𝑷 − 𝑸, representing the difference between the vectors 𝑷 and 𝑸, by
adding to 𝑷 the negative vector −𝑸 (Figure 2.7). We write
𝑷 − 𝑸 = 𝑷 + −𝑸 ….. (1.4)
We now consider the sum of three or more vectors. The sum of
three vectors 𝑷, 𝑸 and 𝑺 is, by definition, obtained by first adding
the vectors 𝑷 and 𝑸 and then adding the vector 𝑺 to the vector
𝑷 + 𝑸. We write
𝑷 + 𝑸 + 𝑺 = 𝑷 + 𝑸 + 𝑺 ….. (1.5)

Similarly, we obtain the sum of four vectors by adding the fourth
vector to the sum of the first three. It follows that we can obtain the
sum of any number of vectors by applying the parallelogram law
repeatedly to successive pairs of vectors until all of the given
vectors are replaced by a single vector.

Product of a Scalar and a Vector
It is convenient to denote the sum 𝑷 + 𝑷 by 2𝑷, the sum 𝑷 + 𝑷 + 𝑷
by 3𝑷, and, in general, the sum of 2 equal vectors 𝑷 by the product
2𝑷. Therefore, we define the product 𝑛𝑷 of a positive integer 𝑛 and
a vector 𝑷 as a vector having the same direction as 𝑷 and the
magnitude 𝑛𝑃.
Extending this definition to include all scalars and recalling the
definition of a negative vector given earlier, we define the product
𝑘𝑷 of a scalar 𝑘 and a vector 𝑷 as a vector having the same
direction as 𝑷 (if 𝑘 is positive) or a direction opposite to that of 𝑷
(if 𝑘 is negative) and a magnitude equal to the product of 𝑷 and the
absolute value of 𝑘 (Figure 2.9).

1.4.4 Resultant of Several Concurrent Forces
Consider a particle 𝐴 acted upon by several coplanar forces, i.e., by
several forces contained in the same plane (Figure 2.10a). Because
the forces all pass through 𝐴, they are also said to be concurrent. We
can add the vectors representing the forces acting on 𝐴 by the
polygon rule (Figure 2.10b). Because the use of the polygon rule is
equivalent to the repeated application of the parallelogram law, the
vector 𝑹 obtained in this way represents the resultant of the given
concurrent forces. That is, the single force 𝑹 has the same effect on
the particle 𝐴 as the given forces. As before, the order in which we
add the vectors 𝑷, 𝑸 and 𝑺 representing the given forces is
immaterial.

1.4.5 Resolution of a Force into Components
We have seen that two or more forces acting on a particle may be
replaced by a single force that has the same effect on the particle.
Conversely, a single force 𝑭 acting on a particle may be replaced by
two or more forces that, together, have the same effect on the
particle. These forces are called components of the original force 𝑭,
and the process of substituting them for 𝑭 is called resolving the
force 𝑭 into components.
Each force 𝑭 can be resolved into an infinite number of possible
sets of components. Sets of two components 𝑷 and 𝑸 are the most
important as far as practical applications are concerned. However,
even then, the number of ways in which a given force 𝑭 may be
resolved into two components is unlimited (Figure 2.11).

In many practical problems, we start with a given vector 𝑭 and want
to determine a useful set of components. Two cases are of particular
interest:
1. One of the two components, 𝑷, is known. We obtain the
second component, 𝑸, by applying the triangle rule and joining
the tip of 𝑷 to the tip of 𝑭 (Figure 2.12). We can determine the
magnitude and direction of 𝑸 graphically or by trigonometry.
Once we have determined 𝑸, both components 𝑷 and 𝑸 should
be applied at 𝐴.

2. The line of action of each component is known. We obtain the
magnitude and sense of the components by applying the
parallelogram law and drawing lines through the tip of 𝑭 that
are parallel to the given lines of action (Figure 2.13). This
process leads to two well-defined components, 𝑷 and 𝑸, which
can be determined graphically or computed trigonometrically by
applying the law of sines.

Review Questions
1. What are planar forces?
2. How is a force defined?
3. How can we add forces acting on a particle?
4. Differentiate between scalars and vectors.
5. How two vectors are added?
6. How several vectors are added?
7. What does a product of a scalar and a vector mean?
8. How can we determine the resultant of several concurrent
forces?
9. What are common rules to determine components of a force?

Mechanical I/I
2024
Session 05: Adding Forces by Components

• Define force as a vector using rectangular components and unit
vectors.
Learning Objectives
• Determine scalar components of a force.
• Determine direction and magnitude of a force from given
rectangular components of a force.
• Determine sum of forces by adding rectangular components of
forces.

1.5 Adding Forces by Components
In previous section, we described how to resolve a force into
components. Here we discuss how to add forces by using their
components, especially rectangular components. This method is
often the most convenient way to add forces and, in practice, is the
most common approach.
We can readily extend the properties of vectors established in this
section to the rectangular components of any vector quantity, such
as velocity or momentum.

1.5.1 Rectangular Components of a Force: Unit Vectors
In many problems, it is useful to resolve a force into two
components that are perpendicular to each other. Figure 2.14
shows a force 𝑭 resolved into a component 𝑭𝒙 along the 𝑥 axis and
a component 𝑭𝒚 along the 𝑦 axis. The parallelogram drawn to
obtain the two components is a rectangle, and 𝑭𝒙 and 𝑭𝒚 are called
rectangular components.
The 𝑥 and 𝑦 axes are usually chosen to be
horizontal and vertical, respectively, as in
Figure 2.14; they may, however, be chosen in
any two perpendicular directions, as shown in
Figure 2.15.

Force in Terms of Unit Vectors
To simplify working with rectangular components, we introduce
two vectors of unit magnitude, directed respectively along the
positive 𝑥 and y axes. These vectors are called unit vectors and are
denoted by 𝒊 and 𝒋, respectively (Figure 2.16).
Recalling the definition of the product of a scalar
and a vector, note that we can obtain the
rectangular components 𝑭𝒙 and 𝑭𝒚 of a force 𝑭
by multiplying respectively the unit vectors 𝒊 and
𝒋 by appropriate scalars (Figure 2.17). We have
𝑭𝒙 = 𝐹𝑥𝒊 𝑭𝒚 = 𝐹𝑦𝒋
𝑭 = 𝐹𝑥𝒊 + 𝐹𝑦𝒋 ….. (1.6)

Scalar Components
Denoting by 𝐹 the magnitude of the force 𝑭 and by 𝜃 the angle
between 𝑭 and the 𝑥 axis, which is measured counter-clockwise
from the positive 𝑥 axis (Figure 2.17), we may express the scalar
components of 𝑭 as 𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃 𝐹𝑦 = 𝐹𝑠𝑖𝑛𝜃
Direction of a Force
When a force 𝑭 is defined by its rectangular
components 𝐹𝑥 and 𝐹𝑦 (see Figure 2.17), we can
find the angle 𝜃 defining its direction from
𝑡𝑎𝑛𝜃 =
𝐹𝑦
𝐹𝑥
We can obtain the magnitude 𝐹 of the force by applying the
Pythagorean theorem,
𝐹 = 𝐹𝑥
2
+ 𝐹𝑦
2

1.5.2 Addition of Forces by Summing 𝒙 and 𝒚 Components
We described in the previous section, how to add forces according
to the parallelogram law.
From this law, we derived two other methods that are more readily
applicable to the graphical solution of problems: the triangle rule for
the addition of two forces and the polygon rule for the addition of
three or more forces. We also explained that the force triangle used
to define the resultant of two forces could be used to obtain a
trigonometric solution.
However, when we need to add three or more forces, we cannot
obtain any practical trigonometric solution from the force polygon
that defines the resultant of the forces.

In this case, the best approach is to obtain an analytic
solution of the problem by resolving each force into two
rectangular components. Consider, for instance, three
forces 𝑷, 𝑸 and 𝑺 acting on a particle 𝐴 (Figure 2.18a).
Their resultant 𝑹 is defined by the relation
𝑹 = 𝑷 + 𝑸 + 𝑺 ….. (1.7)
Resolving each force into its rectangular components,
we have
𝑅𝑥𝒊 + 𝑅𝑦𝒋 = 𝑃𝑥𝒊 + 𝑃𝑦𝒋 + 𝑄𝑥𝒊 + 𝑄𝑦𝒋 + 𝑆𝑥𝒊 + 𝑆𝑦𝒋
= 𝑃𝑥 + 𝑄𝑥 + 𝑆𝑥 𝒊 + 𝑃𝑦 + 𝑄𝑦 + 𝑆𝑦 𝒋
From this equation, we can see that
𝑅𝑥 = 𝑃𝑥 + 𝑄𝑥 + 𝑆𝑥 𝑅𝑦 = 𝑃𝑦 + 𝑄𝑦 + 𝑆𝑦

or, in short form
𝑅𝑥 = 𝐹𝑥 𝑅𝑦 = 𝐹𝑦
We thus conclude that when several forces are acting on a particle,
we obtain the scalar components 𝑅𝑥 and 𝑅𝑦 of the resultant 𝑹 by
adding algebraically the corresponding scalar components of the
given forces. (This result also applies to the addition of other vector
quantities, such as velocities, accelerations, or momenta.)

Review Questions
1. How is a force defined with rectangular components and unit
vectors ?
2. How the scalar components of a force are determined?
3. How the magnitude and direction of a force are determined
from its rectangular components ?
4. How the sum of forces can be determined by adding rectangular
components of forces?

Mechanical I/I
2024
Session 06: Adding Forces in Space

• Define a force in space as a vector when angles made by the
force are given.
Learning Objectives
• Define a force in space as a vector when two points on its line
of action are given.
• Determine the resultant of several concurrent forces in space.

1.6 Adding Forces in Space
The problems considered in the first part of this chapter involved
only two dimensions; they were formulated and solved in a single
plane. In the last part of this chapter, we discuss problems
involving the three dimensions of space.
1.6.1 Rectangular Components of a Force in Space
Consider a force 𝑭 acting at the origin 𝑂 of the
system of rectangular coordinates 𝑥, 𝑦 and 𝑧. To
define the direction of 𝑭, we draw the vertical plane
𝑂𝐵𝐴𝐶 containing 𝑭 (Figure a).
This plane passes through the vertical 𝑦 axis; its
orientation is defined by the angle 𝜙 it forms with
the 𝑥𝑦 plane.

The direction of 𝑭 within the plane is defined by the angle 𝜃𝑦 that
𝑭 forms with the 𝑦 axis. We can resolve the force 𝑭 into a vertical
component 𝑭𝒚 and a horizontal component 𝑭𝒉; this operation,
shown in Figure (b), is carried out in plane 𝑂𝐵𝐴𝐶 according to the
rules developed earlier. The corresponding scalar components are
𝐹𝑦 = 𝐹𝑐𝑜𝑠𝜃𝑦 and 𝐹ℎ = 𝐹𝑠𝑖𝑛𝜃𝑦 ….. (1.8)

However, we can also resolve 𝑭𝒉 into two rectangular components
𝑭𝒙 and 𝑭𝒛 along the 𝑥 and 𝑧 axes, respectively. This operation,
shown in Figure (c), is carried out in the 𝑥𝑧 plane. We obtain the
following expressions for the corresponding scalar components:
𝐹𝑥 = 𝐹ℎ𝑐𝑜𝑠𝜙 = 𝐹𝑠𝑖𝑛𝜃𝑦𝑐𝑜𝑠𝜙 and 𝐹𝑥 = 𝐹ℎ𝑠𝑖𝑛𝜙 = 𝐹𝑠𝑖𝑛𝜃𝑦𝑠𝑖𝑛𝜙
….. (1.9)
The given force 𝑭 thus has been resolved into three rectangular
vector components 𝑭𝒙, 𝑭𝒚, 𝑭𝒛, which are directed along the three
coordinate axes.

We can now apply the Pythagorean theorem to the triangles 𝑂𝐴𝐵
and 𝑂𝐶𝐷 of Figure:
𝐹2 = 𝑂𝐴 2 = 𝑂𝐵 2 + 𝐵𝐴 2 = 𝐹𝑦
2
+ 𝐹ℎ
2
𝐹ℎ
2
= 𝑂𝐶 2 = 𝑂𝐷 2 + 𝐷𝐶 2 = 𝐹𝑥
2
+ 𝐹𝑧
2
Eliminating 𝐹ℎ
2
from these two equations and solving for 𝐹, we
obtain the following relation between the magnitude of 𝐹 and its
rectangular scalar components:
𝐹 = 𝐹𝑥
2
+ 𝐹𝑦
2
+ 𝐹𝑧
2 ….. (1.10)

The relationship between the force 𝐹 and its three components 𝐹𝑥,
𝐹𝑦 and 𝐹𝑧 is more easily visualized if we draw a “box” having 𝐹𝑥, 𝐹𝑦
and 𝐹𝑧 for edges, as shown in Figure.
These triangles occupy positions in the box comparable with that of
triangle 𝑂𝐴𝐵. Denoting by 𝜃𝑥, 𝜃𝑦 and 𝜃𝑧, respectively, the angles
that 𝐹 forms with the 𝑥, 𝑦 and 𝑧 axes, we can thus write
𝐹𝑥 = 𝐹𝑐𝑜𝑠𝜃𝑥 𝐹𝑦 = 𝐹𝑐𝑜𝑠𝜃𝑦 and 𝐹𝑧 = 𝐹𝑠𝑖𝑛𝜃𝑧 ….. (1.11)

The three angles 𝜃𝑥 , 𝜃𝑦 and 𝜃𝑧 define the direction of the force 𝑭;
they are more commonly used for this purpose than the angles 𝜃𝑦
and 𝜙 introduced at the beginning of this section. The cosines of 𝜃𝑥
, 𝜃𝑦 and 𝜃𝑧 are known as the direction cosines of the force 𝑭.
Introducing the unit vectors 𝒊, 𝒋 and 𝒌, which are
directed respectively along the 𝑥, 𝑦 and 𝑧 axes
(Figure), we can express 𝑭 in the form
𝑭 = 𝐹𝑥𝒊 + 𝐹𝑦𝒋 + 𝐹𝑧𝒌 ….. (1.12)
where the scalar components 𝐹𝑥, 𝐹𝑦 and 𝐹𝑧 are
defined by the relations in Equation (1.11).
Substituting into Equation (1.12) the expressions obtained for 𝐹𝑥, 𝐹𝑦
and 𝐹𝑧 in Equation (1.11), we have
𝑭 = 𝐹 𝑐𝑜𝑠𝜃𝑥𝒊 + 𝑐𝑜𝑠𝜃𝑦𝒋 + 𝑐𝑜𝑠𝜃𝑧𝒌 ….. (1.13)

This equation shows that the force 𝑭 can be expressed as the
product of the scalar 𝐹 and the vector
𝝀 = 𝑐𝑜𝑠𝜃𝑥𝒊 + 𝑐𝑜𝑠𝜃𝑦𝒋 + 𝑐𝑜𝑠𝜃𝑧𝒌 ….. (1.14)
The vector 𝝀 is a vector whose magnitude is
equal to 1 and whose direction is the same
as that of 𝑭 (Figure). The vector 𝝀 is
referred to as the unit vector along the line
of action of 𝑭.
It follows from Equation (1.14) that the components of the unit
vector 𝝀 are respectively equal to the direction cosines of the line of
action of 𝑭:
𝜆𝑥 = 𝑐𝑜𝑠𝜃𝑥 𝜆𝑦 = 𝑐𝑜𝑠𝜃𝑦 𝜆𝑧 = 𝑐𝑜𝑠𝜃𝑧 ….. (1.15)

Note that the values of the three angles 𝜃𝑥, 𝜃𝑦 and 𝜃𝑧 are not
independent. Recalling that the sum of the squares of the
components of a vector is equal to the square of its magnitude, we
can write
𝜆𝑥
2
+ 𝜆𝑦
2
+ 𝜆𝑧
2
= 1
Substituting for 𝜆𝑥, 𝜆𝑦 and 𝜆𝑧 from Equation (1.15), we obtain
𝑐𝑜𝑠2𝜃𝑥 + 𝑐𝑜𝑠2𝜃𝑦 + 𝑐𝑜𝑠2𝜃𝑧 = 1 ….. (1.16)
When the components 𝐹𝑥, 𝐹𝑦 and 𝐹𝑧 of a force 𝑭 are given, we can
obtain the magnitude 𝑭 of the force from Equation (1.10). We can
then solve relations in Equation (1.11) for the direction cosines as
𝑐𝑜𝑠𝜃𝑥 =
𝐹𝑥
𝐹
; 𝑐𝑜𝑠𝜃𝑦 =
𝐹𝑦
𝐹
; 𝑐𝑜𝑠𝜃𝑧 =
𝐹𝑧
𝐹
….. (1.17)

1.6.2 Force Defined by Its Magnitude and Two Points on Its
Line of Action
In many applications, the direction of a force 𝑭 is defined by the
coordinates of two points, 𝑀 𝑥1, 𝑦1, 𝑧1 and 𝑁 𝑥2, 𝑦2, 𝑧2 , located
on its line of action (Figure).
Consider the vector 𝑀𝑁 joining
𝑀 and 𝑁 and of the same sense
as a force 𝑭. Denoting its scalar
components by 𝑑𝑥, 𝑑𝑦 and 𝑑𝑧
respectively, we write
𝑀𝑁 = 𝑑𝑥𝒊 + 𝑑𝑦𝒋 + 𝑑𝑧𝒌 ….. (1.18)

We can obtain a unit vector 𝝀 along the line of action of 𝑭 (i.e.,
along the line 𝑀𝑁) by dividing the vector 𝑀𝑁 by its magnitude
𝑀𝑁. Substituting for 𝑀𝑁 from Equation (1.18) and observing that
𝑀𝑁 is equal to the distance 𝑑 from 𝑀 to 𝑁, we have
𝝀 =
𝑀𝑁
𝑀𝑁
=
1
𝑑
𝑑𝑥𝒊 + 𝑑𝑦𝒋 + 𝑑𝑧𝒌 ….. (1.19)
Recalling that 𝑭 is equal to the product of 𝐹 and 𝝀, we have
𝑭 = 𝐹𝝀 =
𝑀𝑁
𝑀𝑁
=
𝐹
𝑑
𝑑𝑥𝒊 + 𝑑𝑦𝒋 + 𝑑𝑧𝒌 ….. (1.20)
It follows that the scalar components of 𝑭 are, respectively,
𝐹𝑥 = 𝐹
𝑑𝑥
𝑑
𝐹𝑦 = 𝐹
𝑑𝑦
𝑑
𝐹𝑧 = 𝐹
𝑑𝑦
𝑑
….. (1.21)

The relations in Equation (1.18) considerably simplify the
determination of the components of a force 𝑭 of given magnitude 𝐹
when the line of action of 𝑭 is defined by two points 𝑀 and 𝑁. The
calculation consists of first subtracting the coordinates of 𝑀 from
those of 𝑁, and then determining the components of the vector 𝑀𝑁
and the distance d from 𝑀 to 𝑁. Thus,
𝑑𝑥 = 𝑥2 − 𝑥1; 𝑑𝑦 = 𝑦2 − 𝑦1; 𝑑𝑧 = 𝑧2 − 𝑧1;
𝑑 = 𝑑𝑥
2
+ 𝑑𝑦
2
+ 𝑑𝑧
2
Substituting for 𝐹 and for 𝑑𝑥, 𝑑𝑦, 𝑑𝑧 and 𝑑 into the relations in
Equation (1.18), we obtain the components 𝐹𝑥, 𝐹𝑦 and 𝐹𝑧 of the
force.

We can then obtain the angles 𝜃𝑥, 𝜃𝑦 and 𝜃𝑧 that 𝑭 forms with the
coordinate axes from Equations (1.17). Comparing Equations (1.14)
and (1.19), we can write
𝑑𝑥
𝑑
𝑑𝑦
𝑑
𝑑𝑧
𝑑
….. (1.22)
In other words, we can determine the angles 𝜃𝑥, 𝜃𝑦 and 𝜃𝑧 directly
from the components and the magnitude of the vector 𝑀𝑁.

1.6.3 Addition of Concurrent Forces in Space
We can determine the resultant 𝑹 of two or more forces in space by
summing their rectangular components. Graphical or trigonometric
methods are generally not practical in the case of forces in space.
The method followed here is similar to that used in Section 1.5.2
with coplanar forces. Setting
𝑹 = 𝑭
we resolve each force into its rectangular components:
𝑅𝑥𝒊 + 𝑅𝑦𝒋 + 𝑅𝑧𝒌 = 𝐹𝑥𝒊 + 𝐹𝑦𝒋 + 𝐹𝑧𝒌
= 𝐹𝑥 𝒊 + 𝐹𝑦 𝒋 + 𝐹𝑧 𝒌

From this equation, it follows that
𝑅𝑥 = 𝐹𝑥 ; 𝑅𝑦 = 𝐹𝑦 ; 𝑅𝑧 = 𝐹𝑦 ….. (1.23)
The magnitude of the resultant and the angles 𝜃𝑥, 𝜃𝑦 and 𝜃𝑧 that the
resultant forms with the coordinate axes are obtained using the
method discussed earlier in this section. We end up with
𝑅 = 𝑅𝑥
2
+ 𝑅𝑦
2
+ 𝑅𝑧
2 ….. (1.24)
𝑅𝑥
𝑅
𝑅𝑦
𝑅
𝑅𝑧
𝑅
….. (1.25)

Review Questions
1. How is a force in space defined in a vector form when the
angles made by the force are given?
2. How is a force in space defined in a vector form when two
points on its line of actions are defined?
3. How can we determine the resultant of several concurrent
forces in space?

Mechanical I/I
2024
Session 07: Examples for Components of a
Force and Resultant of Forces

• Determine the components of a planar force and a three-
dimensional force.
Learning Objectives
• Determine the resultant of concurrent forces in a plane and space

Example 1.1
Determine the components of the 𝟐 𝒌𝑵 force shown in Figure
E1.1 along the oblique axes 𝒂 and 𝒃. Determine the projections
of 𝑭 onto the 𝒂- and 𝒃-axes.
Figure E1.1
𝟎. 𝟓𝟗𝟖 𝒌𝑵 𝟏. 𝟔𝟑𝟑 𝒌𝑵 𝟏. 𝟒𝟏𝟒 𝒌𝑵 𝟏. 𝟗𝟑𝟐 𝒌𝑵

Example 1.2
Resolve the force 𝑭𝟐 shown in Figure E1.2 into components
acting along the 𝒖 and 𝒗 axes and determine the magnitudes of
the components.
Figure E1.2
𝟔. 𝟎𝟎 𝒌𝑵 𝟑. 𝟏𝟏 𝒌𝑵

Example 1.3
A stake is being pulled out of the ground by means of two ropes
as shown in Figure E1.3. Knowing that 𝜶 = 𝟑𝟎𝟎
, determine by
trigonometry (a) the magnitude of the force 𝑷 so that the
resultant force exerted on the stake is vertical, (b) the
corresponding magnitude of the resultant.
Figure E1.3
𝟏𝟎𝟏. 𝟒 𝑵 𝟏𝟗𝟔. 𝟔 𝑵

Example 1.4
Determine the magnitude and direction 𝜽 of 𝑭𝑨 so that the
resultant force is directed along the positive 𝒙 axis and has a
magnitude of 𝟏𝟐𝟓𝟎 𝑵 [Figure E1.4].
Figure E1.4
𝟓𝟒. 𝟑𝟑𝟎 𝟔𝟖𝟓. 𝟖𝟗 𝑵

Example 1.5
Determine the magnitude of the resultant force acting on the
plate shown in Figure E1.5 and its direction, measured
counterclockwise from the positive 𝒙 axis.
Figure E1.5
𝟓𝟒𝟔 𝑵
𝟐𝟓𝟑𝟎

Example 1.6
Determine the magnitude of force 𝑭 shown in Figure E1.6 so
that the resultant 𝑭𝑹 of the three forces is as small as possible.
What is the minimum magnitude of 𝑭𝑹?
Figure E1.6
𝟗. 𝟗𝟑 𝒌𝑵
𝟏. 𝟐𝟎 𝒌𝑵

Example 1.7
Find the magnitude and direction of the resultant of the two
forces shown in Figure E1.7 knowing that 𝑷 = 𝟑𝟎𝟎 𝑵 and
𝑸 = 𝟒𝟎𝟎 𝑵.
Figure E1.7
𝟓𝟏𝟓 𝑵
𝟕𝟎. 𝟐𝟎, 𝟐𝟕. 𝟔𝟎, 𝟕𝟏. 𝟓𝟎

Example 1.8
Determine the magnitude and coordinate direction angles of
the resultant force of the two forces acting on the sign at point
𝑨 shown in Figure E1.8.
Figure E1.8
𝟕𝟓𝟔. 𝟕𝟐 𝑵
𝟏𝟒𝟗. 𝟎𝟒𝟎, 𝟗𝟎. 𝟎𝟎, 𝟓𝟗. 𝟎𝟒𝟎

Mechanical I/I
2024
Session 08: Equilibrium of a Particle

• Derive equations of equilibrium for a particle in a plane.
Learning Objectives
• Derive equations of equilibrium for a particle in space.
• Know space diagram and free body diagram.

1.7 Equilibrium of a Particle in a Plane
In the preceding sections, we discussed methods for determining
the resultant of several forces acting on a particle. Although it has
not occurred in any of the problems considered so far, it is quite
possible for the resultant to be zero.
In such a case, the net effect of the given forces is zero, and the
particle is said to be in equilibrium. We thus have the definition:
When the resultant of all the forces acting on a particle is zero, the
particle is in equilibrium.

To express algebraically the conditions for the equilibrium of a
particle, we write
𝑹 = 𝑭 = 𝟎 ….. (1.26)
Resolving each force 𝑭 into rectangular components, we have
𝐹𝑥𝒊 + 𝐹𝑦𝒋 = 0 or, 𝐹𝑥 𝒊 + 𝐹𝑦 𝒋 = 0
We conclude that the necessary and sufficient conditions for the
equilibrium of a particle are
𝐹𝑥 = 0 and 𝐹𝑦 = 0 ….. (1.27)

1.8 Equilibrium of a Particle in Space
According to the definition given in Section 1.7, a particle 𝐴 is in
equilibrium if the resultant of all the forces acting on 𝐴 is zero. The
components 𝑅𝑥, 𝑅𝑦 and 𝑅𝑧 of the resultant of forces in space are
given by Equations (1.23); when the components of the resultant
are zero, we have
𝐹𝑥 = 0; 𝐹𝑦 = 0; 𝐹𝑦 = 0 ….. (1.28)
Equations (1.28) represent the necessary and sufficient conditions
for the equilibrium of a particle in space. We can use them to solve
problems dealing with the equilibrium of a particle involving no
more than three unknowns.

Free-Body Diagrams
In practice, a problem in engineering mechanics is derived from an
actual physical situation. A sketch showing the physical conditions
of the problem is known as a space diagram.
The methods of analysis discussed in the
preceding sections apply to a system of
forces acting on a particle. A large number
of problems involving actual structures,
however, can be reduced to problems
concerning the equilibrium of a particle.
The method is to choose a significant
particle and draw a separate diagram
showing this particle and all the forces
acting on it. Such a diagram is called a
free-body diagram.

Review Questions
1. Write down the conditions for equilibrium of a particle in a
plane.
2. Write down the conditions for equilibrium of a particle in
space.
3. What are space diagram and free body diagram?

Mechanical I/I
2024
Session 09: Examples for Equilibrium of a
Particle

• Solve problems related to equilibrium of particles in a plane and
space.
Learning Objectives

Example 2.1
The cylinders shown in Figure E2.1 have the same diameter but
the cylinder ‘𝟏’ weighs 𝟐𝟎𝟎 𝑵 and cylinder ‘𝟐’ weighs 𝟏𝟓𝟎 𝑵.
Find the reactions at the supports.
Figure E2.1
𝟑𝟖. 𝟖𝟐 𝑵 𝟏𝟒𝟒. 𝟖𝟗 𝑵 𝟗𝟑. 𝟕𝟖 𝑵 𝟐𝟏𝟕. 𝟒𝟓 𝑵

Example 2.2
Two cables are tied together at 𝑪 and are loaded as shown in
Figure E2.2. Determine the range of values of 𝑷 for which both
cables remain taut.
Figure E2.2
𝟏𝟕𝟗. 𝟑 𝑵 < 𝑷 < 𝟔𝟔𝟗 𝑵

Example 2.3
The members of a truss are pin connected at joint 𝑶 as shown
in Figure E2.3. Determine the magnitudes of 𝑭𝟏 and 𝑭𝟐 for
equilibrium. Set 𝜽 = 𝟔𝟎𝟎
.
Figure E2.3
𝟏. 𝟖𝟑 𝒌𝑵 𝟗. 𝟔𝟎 𝒌𝑵

Example 2.4
The 𝟑𝟎 𝒌𝒈 pipe is supported at 𝑨 by a system of five cords as
shown in Figure E2.4. Determine the force in each cord for
equilibrium.
Figure E2.4
𝑻𝑯𝑨 = 𝟐𝟗𝟒 𝑵
𝑻𝑨𝑩 = 𝟑𝟒𝟎 𝑵
𝑻𝑨𝑬 = 𝟏𝟕𝟎 𝑵
𝑻𝑩𝑫 = 𝟒𝟗𝟎 𝑵
𝑻𝑩𝑪 = 𝟓𝟔𝟐 𝑵

Example 2.5
Determine the tensions in cables 𝑨𝑩, 𝑨𝑪, and 𝑨𝑫 shown in
Figure E2.5.
Figure E2.5
𝑻𝑨𝑩 = 𝟓𝟔𝟗 𝑵, 𝑻𝑨𝑪 = 𝟑𝟕𝟔 𝑵, 𝑻𝑨𝑫 = 𝟒𝟔𝟕 𝑵

Example 2.6
The 𝟏𝟎 𝒌𝒈 lamp in Figure E2.6 is suspended from the three
equal-length cords. Determine its smallest vertical distance 𝒔
from the ceiling if the force developed in any cord is not
allowed to exceed 𝟓𝟎 𝑵.
Figure E2.6
𝟓𝟏𝟗 𝒎𝒎

Mechanical I/I
2024
FORCES ACTING ON RIGID
BODIES
Session 10: Forces and Moments

• Know the concepts of rigid body statics.
Learning Objectives
• Differentiate between external and internal forces.
• State principle of transmissibility and know its application and
limitations.
• Define equivalent forces.
• Define a vector product and know its meaning and properties.
• Define moment of a force.
• Know necessary and sufficient conditions for equivalent forces.
• State Varignon’s theorem.
• Determine the moment of a force about a point.

In Chapter 1, we assumed that each of the bodies considered could
be treated as a single particle. Such a view, however, is not always
possible. In general, a body should be treated as a combination of a
large number of particles. In this case, we need to consider the size
of the body as well as the fact that forces act on different parts of
the body and thus have different points of application.
Most of the bodies considered in elementary mechanics are
assumed to be rigid. We define a rigid body as one that does not
deform. Actual structures and machines are never absolutely rigid
and deform under the loads to which they are subjected. However,
these deformations are usually small and do not appreciably affect
the conditions of equilibrium or the motion of the structure under
consideration.

In this chapter, you will study the effect of forces exerted on a rigid
body, and you will learn how to replace a given system of forces by
a simpler equivalent system. This analysis rests on the fundamental
assumption that the effect of a given force on a rigid body remains
unchanged if that force is moved along its line of action (principle
of transmissibility). It follows that forces acting on a rigid body can
be represented by sliding vectors.
Two important concepts associated with the effect of a force on a
rigid body are the moment of a force about a point and the moment
of a force about an axis.
The determination of these quantities involves computing vector
products and scalar products of two vectors, so in this chapter, we
introduce the fundamentals of vector algebra and apply them to the
solution of problems involving forces acting on rigid bodies.

Another concept introduced in this chapter is that of a couple, i.e.,
the combination of two forces that have the same magnitude,
parallel lines of action, and opposite sense.
As you will see, we can replace any system of forces acting on a
rigid body by an equivalent system consisting of one force acting at
a given point and one couple. This basic combination is called a
force-couple system.
In the case of concurrent, coplanar, or parallel forces, we can further
reduce the equivalent force-couple system to a single force, called
the resultant of the system, or to a single couple, called the resultant
couple of the system.

2.1 Forces and Moments
The basic definition of a force does not change if the force acts on a
point or on a rigid body. However, the effects of the force can be
very different, depending on factors such as the point of application
or line of action of that force. As a result, calculations involving
forces acting on a rigid body are generally more complicated than
situations involving forces acting on a point. We begin by
examining some general classifications of forces acting on rigid
bodies.

2.1.1 External and Internal Forces
Forces acting on rigid bodies can be separated into two groups:
external forces and internal forces.
(a) External forces
External forces are exerted by other bodies on the rigid body under
consideration. They are entirely responsible for the external
behavior of the rigid body, either causing it to move or ensuring
that it remains at rest.
(b) Internal forces
Internal forces hold together the particles forming the rigid body. If
the rigid body is structurally composed of several parts, the forces
holding the component parts together are also defined as internal
forces.

The people pulling on the rope exert the force 𝑭. The point of
application of 𝑭 is on the front bumper. The force 𝑭 tends to make
the truck move forward in a straight line and does actually make it
move, since no external force opposes this motion. (We are ignoring
rolling resistance here for simplicity.) This forward motion of the
truck, during which each straight line keeps its original orientation
(the floor of the truck remains horizontal, and the walls remain
vertical), is known as a translation

Other forces might cause the truck to move differently. For
example, the force exerted by a jack placed under the front axle
would cause the truck to pivot about its rear axle. Such a motion is a
rotation.
We conclude, therefore, that each external force acting on a rigid
body can, if unopposed, impart to the rigid body a motion of
translation or rotation, or both.

2.1.2 Principle of Transmissibility: Equivalent Forces
The principle of transmissibility states that the conditions of
equilibrium or motion of a rigid body remain unchanged if a force
𝑭 acting at a given point of the rigid body is replaced by a force 𝑭′
of the same magnitude and same direction, but acting at a different
point, provided that the two forces have the same line of action.
The two forces 𝑭 and 𝑭′ have the same effect on the rigid body and
are said to be equivalent forces.

We indicated in Chapter 1 that we could represent the forces acting
on a particle by vectors. These vectors had a well-defined point of
application––namely, the particle itself––and were therefore fixed,
or bound, vectors.
In the case of forces acting on a rigid body, however, the point of
application of the force does not matter, as long as the line of action
remains unchanged. Thus, forces acting on a rigid body must be
represented by a different kind of vector, known as a sliding vector,
since forces are allowed to slide along their lines of action.

We observe that the line of action of the force 𝑭 is a horizontal line
passing through both the front and rear bumpers of the truck
(Figure). Using the principle of transmissibility, we can therefore
replace 𝑭 by an equivalent force 𝑭′
acting on the rear bumper. In
other words, the conditions of motion are unaffected, and all of the
other external forces acting on the truck (𝑾, 𝑹𝟏, 𝑹𝟐) remain
unchanged if the people push on the rear bumper instead of pulling
on the front bumper.

If the bar is in tension and, if not absolutely rigid, increases in
length slightly; and similarly if the bar is in compression and, if not
absolutely rigid, decreases in length slightly. Thus, although we can
use the principle of transmissibility to determine the conditions of
motion or equilibrium of rigid bodies and to compute the external
forces acting on these bodies, it should be avoided, or at least used
with care, in determining internal forces and deformations.
Thus, although we can use the principle of transmissibility to
determine the conditions of motion or equilibrium of rigid bodies
and to compute the external forces acting on these bodies, it should
be avoided, or at least used with care, in determining internal forces
and deformations.

2.1.3 Vector Products
In order to gain a better understanding of the effect of a force on a
rigid body, we need to introduce a new concept, the moment of a
force about a point. However, this concept is more clearly
understood and is applied more effectively if we first add to the
mathematical tools at our disposal the vector product of two
vectors.
The vector product of two vectors 𝑷 and 𝑸 is defined as the vector
𝑽 that satisfies the following conditions:
a) The line of action of 𝑽 is perpendicular to the
plane containing 𝑷 and 𝑸.

b) The magnitude of 𝑽 is the product of the magnitudes of 𝑷 and
𝑸 and of the 𝑠𝑖𝑛𝑒 of the angle 𝜃 formed by 𝑷 and 𝑸 (the
measure of which is always 1800 or less). We thus have
𝑉 = 𝑃𝑄 𝑠𝑖𝑛𝜃 ….. (2.1)
c) The direction of V is obtained from the right-hand rule.
As stated previously, the vector 𝑽 satisfying these three conditions
(which define it uniquely) is referred to as the vector product of 𝑷
and 𝑸. It is represented by the mathematical expression
𝑽 = 𝑷 × 𝑸 ….. (2.2)

The magnitude 𝑽 of the vector product of 𝑷 and 𝑸 is equal to the
area of the parallelogram that has 𝑷 and 𝑸 for sides (Figure). The
vector product 𝑷 × 𝑸 is therefore unchanged if we replace 𝑸 by a
vector 𝑸′ that is coplanar with 𝑷 and 𝑸 such that the line joining the
tips of 𝑸 and 𝑸′ is parallel to 𝑷:
𝑷 × 𝑸 = 𝑷 × 𝑸′ ….. (2.3)
From the third condition used to define the vector product 𝑽 of 𝑷
and 𝑸 ––namely, that 𝑷, 𝑸, and 𝑽 must form a right-handed triad––
it follows that vector products are not commutative; i.e., 𝑸 × 𝑷 is
not equal to 𝑷 × 𝑸. Indeed, we can easily check that 𝑸 × 𝑷 is
represented by the vector −𝑽, which is equal and opposite to 𝑽:
𝑸 × 𝑷 = − 𝑷 × 𝑸 ′ ….. (2.4)

We saw that the commutative property does not apply to vector
products. However, it can be demonstrated that the distributive
property
𝑷 × 𝑸𝟏 + 𝑸𝟐 = 𝑷 × 𝑸𝟏 + 𝑷 × 𝑸𝟐 ….. (2.5)
does hold.
A third property, the associative property, does not apply to vector
products; we have in general
𝑷 × 𝑸 × 𝑺 ≠ 𝑷 × 𝑸 × 𝑺 ….. (2.6)

2.1.4 Rectangular Components of Vector Products
We can now easily express the vector product 𝑽 of two given
vectors 𝑷 and 𝑸 in terms of the rectangular components of these
vectors. Resolving 𝑷 and 𝑸 into components, we can write
𝑽 = 𝑷 × 𝑸 = 𝑃𝑥𝒊 + 𝑃𝑦𝒋 + 𝑃𝑧𝒌 × 𝑄𝑥𝒊 + 𝑄𝑦𝒋 + 𝑄𝑧𝒌
….. (2.7)
= 𝑃𝑦𝑄𝑧 − 𝑃𝑧𝑄𝑦 𝒊 + 𝑃𝑧𝑄𝑥 − 𝑃𝑥𝑄𝑧 𝒋 + 𝑃𝑥𝑄𝑦 − 𝑃𝑦𝑄𝑥 𝒌
Thus, the rectangular components of the vector product 𝑽 are
𝑉
𝑥 = 𝑃𝑦𝑄𝑧 − 𝑃𝑧𝑄𝑦
𝑉
𝑦 = 𝑃𝑧𝑄𝑥 − 𝑃𝑥𝑄𝑧
𝑉
𝑧 = 𝑃𝑥𝑄𝑦 − 𝑃𝑦𝑄𝑥
….. (2.8)

We can express the vector product 𝑽 in the following form, which
is more easily memorized:
𝑽 =
𝒊 𝒋 𝒌
𝑃𝑥 𝑃𝑦 𝑃𝑧
𝑄𝑥 𝑄𝑦 𝑄𝑧
….. (2.9)

2.1.5 Moment of a Force about a Point
As we know, the force 𝑭 is represented by a vector that defines its
magnitude and direction. However, the effect of the force on the
rigid body depends also upon its point of application 𝐴. The
position of 𝐴 can be conveniently defined by the vector 𝒓 that joins
the fixed reference point 𝑂 with 𝐴; this vector is known as the
position vector of 𝐴. The position vector 𝒓 and the force 𝑭 define
the plane shown in Figure.

We define the moment of 𝑭 about 𝑂 as the vector product of 𝒓 and
𝑭:
𝑴𝒐 = 𝒓 × 𝑭 ….. (2.10)
Denoting by 𝜃 the angle between the lines of action of the position
vector 𝒓 and the force 𝑭, we find that the magnitude of the moment
of 𝑭 about 𝑂 is
𝑀𝑜 = 𝑟𝐹 𝑠𝑖𝑛𝜃 = 𝐹𝑑 ….. (2.11)
where 𝑑 represents the perpendicular
distance from 𝑂 to the line of action of 𝑭
(see Figure).
The magnitude of 𝑴𝒐 measures the tendency of the force 𝑭 to
make the rigid body rotate about a fixed axis directed along 𝑴𝒐.

Mechanics lecture notes mechanical engineering

Recall from Section 2.1.2 that the principle of transmissibility states
that two forces 𝑭 and 𝑭′ are equivalent (i.e., have the same effect on
a rigid body) if they have the same magnitude, same direction, and
same line of action. We can now restate this principle:
Two forces 𝑭 and 𝑭′ are equivalent if, and only if, they are equal
(i.e., have the same magnitude and same direction) and have equal
moments about a given point 𝑂.
The necessary and sufficient conditions for two forces 𝑭 and 𝑭′ to
be equivalent are thus
𝑭 = 𝑭′ 𝑴𝒐 = 𝑴𝒐′ ….. (2.12)

Two-Dimensional Problems
Many applications in statics deal with two-dimensional structures.
Such structures have length and breadth but only negligible depth.
Often, they are subjected to forces contained in the plane of the
structure. We can easily represent two-dimensional structures and
the forces acting on them on a sheet of paper or on a blackboard.
Their analysis is therefore considerably simpler than that of three-
dimensional structures and forces.

The moment of 𝑭 about a point 𝑂, which is
chosen in the plane of the figure, is represented
by a vector 𝑴𝒐 perpendicular to that plane and of
magnitude 𝐹𝑑. In the case of Figure (a), the
vector 𝑴𝒐 points out of the page, whereas in the
case of Figure (b), it points into the page.
As we look at the figure, we observe in the first
case that 𝑭 tends to rotate the slab counter-
clockwise and in the second case that it tends to
rotate the slab clockwise. Therefore, it is natural
to refer to the sense of the moment of 𝑭 about 𝑂
in Figure (a) as counter-clockwise , and in
Figure (b) as clockwise .

Since the moment of a force 𝑭 acting in the plane of the figure must
be perpendicular to that plane, we need only specify the magnitude
and the sense of the moment of 𝑭 about 𝑂. We do this by assigning
to the magnitude 𝑴𝒐 of the moment a positive or negative sign
according to whether the vector 𝑴𝒐 points out of or into the page.

2.1.5 Rectangular Components of the Moment of a Force
We can use the distributive property of vector products to
determine the moment of the resultant of several concurrent forces.
If several forces 𝑭𝟏, 𝑭𝟐, . . . are applied at the same point 𝐴
(Figure) and if we denote by 𝒓 the position vector of 𝐴, it follows
immediately from Equation (2.5) that
𝒓 × 𝑭𝟏 + 𝑭𝟐 + ⋯ = 𝒓 × 𝑭𝟏 + 𝒓 × 𝑭𝟐 + ⋯
….. (2.13)
In words,
The moment about a given point 𝑂 of the
resultant of several concurrent forces is
equal to the sum of the moments of the
various forces about the same point 𝑂.
This property is known as Varignon’s theorem.

In general, determining the moment of a force in space is
considerably simplified if the force and the position vector of its
point of application are resolved into rectangular 𝑥, 𝑦, and 𝑧
components.
Consider, for example, the moment 𝑴𝒐 about 𝑂 of a force 𝑭 whose
components are 𝐹𝑥, 𝐹𝑦, and 𝐹𝑧 and that is applied at a point 𝐴 with
coordinates 𝑥, 𝑦, and 𝑧 (Figure).
Since the components of the position
vector 𝒓 are respectively equal to the
coordinates 𝑥, 𝑦, and 𝑧 of the point 𝐴, we
can write 𝒓 and 𝑭 as
𝒓 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 ….. (2.14)
𝑭 = 𝐹𝑥𝒊 + 𝐹𝑦𝒋 + 𝐹𝑧𝒌 ….. (2.15)

Substituting for 𝒓 and 𝑭 from Equations (2.14) and (2.15) into
Equation (2.10) and recalling Equations (2.7) and (2.8), we can
write the moment 𝑴𝒐 of 𝑭 about 𝑂 in the form
𝑴𝒐 = 𝑀𝑥𝒊 + 𝑀𝑦𝒋 + 𝑀𝑧𝒌 ….. (2.16)
where the components 𝑀𝑥, 𝑀𝑦, and 𝑀𝑧 are defined by the relations
𝑀𝑥 = 𝑦𝐹𝑧 − 𝑧𝐹𝑦 𝑀𝑦 = 𝑧𝐹𝑥 − 𝑥𝐹𝑧 𝑀𝑧 = 𝑥𝐹𝑦 − 𝑦𝐹𝑥
….. (2.17)
The scalar components 𝑀𝑥, 𝑀𝑦, and 𝑀𝑧 of the moment 𝑴𝒐 measure
the tendency of the force 𝑭 to impart to a rigid body a rotation about
the 𝑥, 𝑦, and 𝑧 axes, respectively.

Substituting from Equation (2.17) into Equation (2.16), we can also
write 𝑴𝒐 in the form of the determinant, as
𝑴𝒐 =
𝒊 𝒋 𝒌
𝑥 𝑦 𝑧
𝐹𝑥 𝐹𝑦 𝐹𝑧
….. (2.18)
To compute the moment 𝑴𝑩 about an arbitrary point 𝐵 of a force 𝑭
applied at 𝐴 (Figure), we must replace the position vector 𝒓 in
Equation (2.10) by a vector drawn from 𝐵 to 𝐴. This vector is the
position vector of 𝐴 relative to 𝐵, denoted by 𝒓𝑨/𝑩. Observing that
𝒓𝑨/𝑩 can be obtained by subtracting 𝒓𝑩 from 𝒓𝑨, we write
𝑴𝑩 = 𝒓𝑨/𝑩 × 𝑭 = 𝒓𝑨 − 𝒓𝑩 × 𝑭 ….. (2.19)

or using the determinant form,
𝑴𝒐 =
𝒊 𝒋 𝒌
𝑥𝐴/𝐵 𝑦𝐴/𝐵 𝑧𝐴/𝐵
….. (2.20)
where 𝑥𝐴/𝐵, 𝑦𝐴/𝐵, and 𝑧𝐴/𝐵 denote the components of the vector
𝒓𝑨/𝑩:
𝑥𝐴/𝐵 = 𝑥𝐴 − 𝑥𝐵
𝑦𝐴/𝐵 = 𝑦𝐴 − 𝑦𝐵 𝑧𝐴/𝐵 = 𝑧𝐴 − 𝑧𝐵 ….. (2.21)
In the case of two-dimensional problems, we can
assume without loss of generality that the force 𝑭
lies in the 𝑥𝑦 plane (Figure). Setting 𝑧 = 0 and
𝐹𝑧 = 0 in Equation (2.18), we obtain
𝑴𝒐 = 𝑥𝐹𝑦 − 𝑦𝐹𝑥 𝒌 ….. (2.22)

We can verify that the moment of 𝑭 about 𝑂 is perpendicular to the
plane of the figure and that it is completely defined by the scalar
𝑀0 = 𝑀𝑧 = 𝑥𝐹𝑦 − 𝑦𝐹𝑥 ….. (2.23)
To compute the moment about 𝐵 𝑥𝐵, 𝑦𝐵 of a force lying in the 𝑥𝑦
plane and applied at 𝐴 𝑥𝐴, 𝑦𝐴 (Figure), we set 𝑧𝐴/𝐵 = 0 and
𝐹𝑧 = 0 in Equation (2.20) and note that the vector 𝑴𝑩 is
perpendicular to the 𝑥𝑦 plane and is defined in magnitude and sense
by the scalar
𝑀𝐵 = 𝑥𝐴 − 𝑥𝐵 𝐹𝑦 − 𝑦𝐴 − 𝑦𝐵 𝐹𝑥
….. (2.24)

Review Questions
1. How do you analyse the effects of forces on a rigid body?
2. Differentiate between external and internal forces.
3. State principle of transmissibility. What are its application and
limitations?
4. What are equivalent forces?
5. How magnitude and direction of a vector product is
determined?
6. Which of the following properties are satisfied for a vector
product: commutative, distributive, associative?
7. How can a vector product be expressed as a determinant?

8. What is the moment of a force? How is it determined?
9. What are necessary and sufficient conditions for two forces to
be equivalent?
10. State Varignon’s theorem.
11. How can a moment of a force about a point be expressed as a
determinant?
12. How can a moment of a force about any arbitrary point be
expressed as a determinant?

Mechanical I/I
2024
BODIES
Session 11: Examples for Moment of a
Force about a Point

• Solve problems related to moment of a force about a point.
Learning Objectives

Example 3.1
A force of 𝟖𝟎𝟎 𝑵 acts on a bracket as shown in Figure E3.1.
Determine the moment of the force about 𝑩.
Figure E3.1
𝟐𝟎𝟐. 𝟔 𝑵. 𝒎, 

Example 3.2
A 𝟑𝟎 𝑵 force acts on the end of the 𝟏. 𝟓 𝒎 lever as shown in
Figure E3.2. Determine the moment of the force about 𝑶.
Figure E3.2
𝟏𝟓. 𝟒 𝑵. 𝒎, 

Example E3.3
The wire 𝑨𝑬 is stretched between the corners 𝑨 and 𝑬 of a bent
plate as shown in Figure E3.3. Knowing that the tension in the
wire is 𝟒𝟑𝟓 𝑵, determine the moment about 𝑶 of the force
exerted by the wire (a) on corner 𝑨, (b) on corner 𝑬.
Figure E3.3
𝟐𝟖. 𝟖𝒊 + 𝟏𝟔. 𝟐𝒋 − 𝟐𝟖. 𝟖𝒌 𝑵. 𝒎
−𝟐𝟖. 𝟖𝒊 − 𝟏𝟔. 𝟐𝒋 + 𝟐𝟖. 𝟖𝒌 𝑵. 𝒎
Also determine the perpendicular
distance from point 𝑶 to cable 𝑨𝑬.
𝟏𝟎𝟎. 𝟖 𝒎𝒎

Example 3.4
Determine the coordinate direction angles of force 𝑭 shown in
Figure E3.4, so that the moment of 𝑭 about 𝑶 is zero.
Figure E3.4
𝟓𝟓. 𝟔𝟎, 𝟒𝟓𝟎, 𝟏𝟏𝟓𝟎
or
𝟏𝟐𝟒𝟎, 𝟏𝟑𝟓𝟎, 𝟔𝟒. 𝟗𝟎

Mechanical I/I
2024
BODIES
Session 12: Moment of a Force about an
Axis

• Define a scalar product and know its meaning and properties.
Learning Objectives
• Know applications of a scalar product.
• Define a scalar triple product and know its meaning and
properties.
• Define moment of a force about an axis.

2.2 Moment of a Force about an Axis
We want to extend the idea of the moment about a point to the
often useful concept of the moment about an axis.
However, first we need to introduce another tool of vector
mathematics. We have seen that the vector product multiplies two
vectors together and produces a new vector. Here we examine the
scalar product, which multiplies two vectors together and produces
a scalar quantity.

2.2.1 Scalar Products
The scalar product of two vectors 𝑷 and 𝑸 is defined as the product
of the magnitudes of 𝑷 and 𝑸 and of the cosine of the angle 𝜃
formed between them (Figure). The scalar product of 𝑷 and 𝑸 is
denoted by 𝑷. 𝑸 and is given by
𝑷. 𝑸 = 𝑃𝑄 𝑐𝑜𝑠𝜃
It follows from its very definition that the scalar product
of two vectors is commutative, i.e., that
….. (2.25)
𝑷. 𝑸 = 𝑸. 𝑷 ….. (2.26)
It can also be proven that the scalar product is distributive, as
shown by
𝑷. 𝑸𝟏 + 𝑸𝟐 = 𝑷. 𝑸𝟏 + 𝑷. 𝑸𝟐 ….. (2.27)

As far as the associative property is concerned, this property cannot
apply to scalar products. Indeed, 𝑷. 𝑸 . 𝑺 has no meaning, because
𝑷. 𝑸 is not a vector but a scalar.
We can also express the scalar product of two vectors 𝑷. 𝑸 in terms
of their rectangular components. Resolving 𝑷 and 𝑸 into
components, we first write
𝑷. 𝑸 = 𝑃𝑥𝒊 + 𝑃𝑦𝒋 + 𝑃𝑧𝒌 . 𝑄𝑥𝒊 + 𝑄𝑦𝒋 + 𝑄𝑧𝒌 = 𝑃𝑥𝑄𝑥 + 𝑃𝑦𝑄𝑦 + 𝑃𝑧𝑄𝑧
….. (2.28)

Applications of the Scalar Product
(a) Angle formed by two given vectors
Let two vectors be given in terms of their components:
𝑷 = 𝑃𝑥𝒊 + 𝑃𝑦𝒋 + 𝑃𝑧𝒌 𝑸 = 𝑄𝑥𝒊 + 𝑄𝑦𝒋 + 𝑄𝑧𝒌
Then, the angle formed by the two vectors can be determined as
𝑃𝑄 𝑐𝑜𝑠𝜃 = 𝑃𝑥𝒊 + 𝑃𝑦𝒋 + 𝑃𝑧𝒌 . 𝑄𝑥𝒊 + 𝑄𝑦𝒋 + 𝑄𝑧𝒌
𝑐𝑜𝑠𝜃 =
𝑃𝑥𝑄𝑥 + 𝑃𝑦𝑄𝑦 + 𝑃𝑧𝑄𝑧
𝑃𝑄
….. (2.29)

(b) Projection of a vector on a given axis
Consider a vector 𝑷 forming an angle 𝜃 with an
axis, or directed line, 𝑂𝐿 (Figure). We define the
projection of 𝑷 on the axis 𝑂𝐿 as the scalar
Now consider a vector 𝑸 directed along 𝑂𝐿 and
of the same sense as 𝑂𝐿 (Figure). We can
express the scalar product of 𝑷 and 𝑸 as
𝑃𝑂𝐿 = 𝑃 𝑐𝑜𝑠𝜃 ….. (2.30)
𝑷. 𝑸 = 𝑃𝑄 𝑐𝑜𝑠𝜃 = 𝑃𝑂𝐿𝑄 ….. (2.31)
from which it follows that
𝑃𝑂𝐿 =
𝑷. 𝑸
𝑄
=
𝑃𝑥𝑄𝑥 + 𝑃𝑦𝑄𝑦 + 𝑃𝑧𝑄𝑧
𝑄
….. (2.32)

In the particular case when the vector selected along 𝑂𝐿 is the unit
vector 𝝀 (Figure), we have
𝑃𝑂𝐿 = 𝑷. 𝝀 ….. (2.33)
The components of 𝝀 along the coordinate
axes are respectively equal to the direction
𝑐𝑜𝑠𝑖𝑛𝑒𝑠 of 𝑂𝐿. Resolving 𝑷 and 𝝀 into
rectangular components, we can express the
projection of 𝑷 on 𝑂𝐿 as
𝑃𝑂𝐿 = 𝑃𝑥𝑐𝑜𝑠𝜃𝑥 + 𝑃𝑦𝑐𝑜𝑠𝜃𝑦 + 𝑃𝑧𝑐𝑜𝑠𝜃𝑧 ….. (2.34)
where 𝜃𝑥, 𝜃𝑦, and 𝜃𝑧 denote the angles that the axis 𝑂𝐿 forms with
the coordinate axes.

2.2.2 Mixed Triple Products
We have now seen both forms of multiplying two vectors together: the
vector product and the scalar product. Here we define the mixed triple
product of the three vectors 𝑺, 𝑷, and 𝑸 as the scalar expression
𝑺. 𝑷 × 𝑸 . This is obtained by forming the scalar product of 𝑺 with
the vector product of 𝑷 and 𝑸.
The mixed triple product is thus equal, in absolute
value, to the volume of the parallelepiped having the
vectors 𝑺, 𝑷, and 𝑸 for sides (Figure).
The sign of the mixed triple product is positive if 𝑺, 𝑷,
and 𝑸 form a right-handed triad and negative if they
form a left-handed triad. [That is, 𝑺. 𝑷 × 𝑸 is
negative if the rotation that brings 𝑷 into line with 𝑸
is observed as clockwise from the tip of 𝑺].

Since the parallelepiped defined in this way is independent of the
order in which the three vectors are taken, the six mixed triple
products that can be formed with 𝑺, 𝑷, and 𝑸 all have the same
absolute value, although not the same sign. It is easily shown that
𝑺. 𝑷 × 𝑸 = 𝑷. 𝑸 × 𝑺 = 𝑸. 𝑺 × 𝑷
= −𝑺. 𝑸 × 𝑷 = −𝑷. 𝑺 × 𝑸 = −𝑸. 𝑷 × 𝑺 ….. (2.35)
We can also express the mixed triple product
𝑺. 𝑷 × 𝑸 = 𝑆𝑥 𝑃𝑦𝑄𝑧 − 𝑃𝑧𝑄𝑦 + 𝑆𝑦 𝑃𝑧𝑄𝑥 − 𝑃𝑥𝑄𝑧 + 𝑆𝑧 𝑃𝑥𝑄𝑦 − 𝑃𝑦𝑄𝑥
….. (2.36)
We can also express the mixed triple product
𝑺. 𝑷 × 𝑸 =
𝑆𝑥 𝑆𝑦 𝑆𝑧
𝑃𝑥 𝑃𝑦 𝑃𝑧
𝑄𝑥 𝑄𝑦 𝑄𝑧
….. (2.38)

2.2.3 Moment of a Force about a Given Axis
The moment 𝑴𝑶𝑳 of 𝑭 about 𝑂𝐿 can be defined as the projection
𝑂𝐶 of the moment 𝑴𝒐 onto the axis 𝑂𝐿.
Suppose we denote the unit vector along
𝑂𝐿 by 𝝀, we can express 𝑀𝑂𝐿 as
𝑀𝑂𝐿 = 𝝀. 𝑴𝒐 = 𝝀. 𝒓 × 𝑭 ….. (2.39)
This shows that the moment 𝑀𝑂𝐿 of 𝑭
about the axis 𝑂𝐿 is the scalar obtained by
forming the mixed triple product of 𝝀, 𝒓,
and 𝑭. We can also express 𝑀𝑂𝐿 in the form
of a determinant,
𝑀𝑂𝐿 =
𝜆𝑥 𝜆𝑦 𝜆𝑧
𝑥 𝑦 𝑧
….. (2.40)

The physical significance of the moment 𝑀𝑂𝐿 of a force 𝑭 about a
fixed axis 𝑂𝐿 becomes more apparent if we resolve 𝑭 into two
rectangular components 𝑭𝟏 and 𝑭𝟐, with 𝑭𝟏 parallel to 𝑂𝐿 and 𝑭𝟐
lying in a plane 𝑃 perpendicular to 𝑂𝐿 (Figure). Resolving 𝒓
similarly into two components 𝒓𝟏 and 𝒓𝟐 ,we get
𝑀𝑂𝐿 = 𝝀. 𝒓𝟏 + 𝒓𝟐 × 𝑭𝟏 + 𝑭𝟐
= 𝝀. 𝒓𝟏 × 𝑭𝟏 + 𝝀. 𝒓𝟏 × 𝑭𝟐
+ 𝝀. 𝒓𝟐 × 𝑭𝟏 + 𝝀. 𝒓𝟐 × 𝑭𝟐
….. (2.41)
Note that all of the mixed triple products except the last one are
equal to zero because they involve vectors that are coplanar when
drawn from a common origin. Therefore, this expression reduces to
𝑀𝑂𝐿 = 𝝀. 𝒓𝟐 × 𝑭𝟐 ….. (2.42)

The vector product 𝒓𝟐 × 𝑭𝟐 is perpendicular to the plane 𝑃 and
represents the moment of the component 𝑭𝟐 of 𝑭 about the point 𝑄
where 𝑂𝐿 intersects 𝑃. Therefore, the scalar 𝑀𝑂𝐿, which is positive
if 𝒓𝟐 × 𝑭𝟐 and 𝑂𝐿 have the same sense and is negative otherwise,
measures the tendency of 𝑭𝟐 to make the rigid body rotate about
the fixed axis 𝑂𝐿. The other component 𝑭𝟏 of 𝑭 does not tend to
make the body rotate about 𝑂𝐿, because 𝑭𝟏 and 𝑂𝐿 are parallel.
Therefore, we conclude that
The moment 𝑴𝑶𝑳 of 𝑭 about 𝑂𝐿 measures the
tendency of the force 𝑭 to impart to the rigid
body a rotation about the fixed axis 𝑂𝐿.

From the definition of the moment of a force about an axis, it
follows that the moment of 𝑭 about a coordinate axis is equal to the
component of 𝑴𝑶 along that axis. If we substitute each of the unit
vectors 𝒊, 𝒋, and 𝒌 for 𝝀 in Equation (2.38), we obtain expressions
for the moments of 𝑭 about the coordinate axes. These expressions
are respectively equal to those obtained earlier for the components
of the moment 𝑴𝑶 of 𝑭 about 𝑂:
𝑀𝑥 = 𝑦𝐹𝑧 − 𝑧𝐹𝑦 𝑀𝑦 = 𝑧𝐹𝑥 − 𝑥𝐹𝑧 𝑀𝑧 = 𝑥𝐹𝑦 − 𝑦𝐹𝑥
….. (2.43)
Just as the components 𝐹𝑥, 𝐹𝑦, and 𝐹𝑧 of a force 𝑭 acting on a rigid
body measure, respectively, the tendency of 𝑭 to move the rigid
body in the 𝑥, 𝑦, and 𝑧 directions, the moments 𝑀𝑥, 𝑀𝑦, and 𝑀𝑧 of
𝑭 about the coordinate axes measure the tendency of 𝑭 to impart to
the rigid body a rotation about the 𝑥, 𝑦, and 𝑧 axes, respectively.

More generally, we can obtain the moment of a force 𝑭 applied at 𝐴
about an axis that does not pass through the origin by choosing an
arbitrary point 𝐵 on the axis (Figure) and determining the
projection on the axis 𝐵𝐿 of the moment 𝑴𝑩 of 𝑭 about 𝐵. The
equation for this projection is given here
𝑀𝐵𝐿 = 𝝀. 𝑴𝑩 = 𝝀. 𝒓𝑨/𝑩 × 𝑭 ….. (2.44)
where 𝒓𝑨/𝑩 = 𝒓𝑨 − 𝒓𝑩 represents the vector
drawn from 𝐵 to 𝐴. Expressing 𝑀𝐵𝐿 in the
form of a determinant, we have
𝑀𝐵𝐿 =
𝜆𝑥 𝜆𝑦 𝜆𝑧
𝑥𝐴 − 𝑥𝐵 𝑦𝐴
− 𝑦𝐵
𝑧𝐴 − 𝑧𝐵
….. (2.45)

Note that this result is independent of the choice of the point 𝐵 on
the given axis. Indeed, denoting by 𝑀𝐶𝐿 the moment obtained with a
different point 𝐶, we have
𝑀𝐶𝐿 = 𝝀. 𝒓𝑨 − 𝒓𝑪 × 𝑭
= 𝝀. 𝒓𝑨 − 𝒓𝑩 × 𝑭 + 𝝀. 𝒓𝑩 − 𝒓𝑪 × 𝑭
….. (2.46)
However, since the vectors 𝝀 and 𝒓𝑩 − 𝒓𝑪 lie along the same line,
the volume of the parallelepiped having the vectors 𝝀, 𝒓𝑩 − 𝒓𝑪, and
𝑭 for sides is zero, as is the mixed triple product of these three
vectors. The expression obtained for 𝑀𝐶𝐿 thus reduces to its first
term, which is the expression used earlier to define 𝑀𝐵𝐿 . In
addition, it follows that, when computing the moment of 𝑭 about
the given axis, 𝐴 can be any point on the line of action of 𝑭.

Review Questions
1. Define a scalar product.
2. Which of the following properties are satisfied for a scalar
product: commutative, distributive, associative?
3. What are the applications of a scalar product?
4. Define a scalar triple product.
5. What is the meaning of a scalar triple product?
6. How can a scalar triple product be expressed as a determinant?
7. Define moment of a force about an axis.
8. How can a moment of a force about an axis be expressed as a
determinant?

9. How can a moment of a force about any arbitary axis be
expressed as a determinant?

Mechanical I/I
2024
BODIES
Session 14: Couples and Force-Couple
Systems

• Know the necessity of couple.
Learning Objectives
• Define moment of a couple and know its main characteristics.
• Define equivalent couples.
• Know the operations that can be used to transform one system of
forces into another.
• Determine resultant couple.
• Specify couple vectors.
• Resolve a given force into a force at any arbitrary point and a
couple.
• Transfer a force and a couple at a given point into an equivalent
force and couple at another point.

2.3 Couples and Force-Couple Systems
Now that we have studied the effects of forces and moments on a
rigid body, we can ask if it is possible to simplify a system of
forces and moments without changing these effects.
It turns out that we can replace a system of forces and moments
with a simpler and equivalent system.
One of the key ideas used in such a transformation is called a
couple.

2.3.1 Moment of a Couple
Two forces 𝑭 and −𝑭, having the same magnitude, parallel lines of
action, and opposite sense, are said to form a couple (Figure).
Clearly, the sum of the components of the two forces in any
direction is zero. The sum of the moments of the two forces about a
given point, however, is not zero. The two forces do not cause the
body on which they act to move along a line (translation), but they
do tend to make it rotate.

Let us denote the position vectors of the points of application of 𝑭
and −𝑭 by 𝒓𝑨 and 𝒓𝑩, respectively (Figure). The sum of the
moments of the two forces about 𝑂 is
𝒓𝑨 × 𝑭 + 𝒓𝑩 × −𝑭 = 𝒓𝑨 − 𝒓𝑩 × 𝑭
Setting 𝒓𝑨 − 𝒓𝑩 = 𝒓, where 𝒓 is the
vector joining the points of application
of the two forces, we conclude that the
sum of the moments of 𝑭 and −𝑭 about
𝑂 is represented by the vector
𝑴 = 𝒓 × 𝑭 ….. (2.47)
The vector 𝑴 is called the moment of the couple. It is perpendicular
to the plane containing the two forces, and its magnitude is
𝑀 = 𝑟𝐹 𝑠𝑖𝑛𝜃 = 𝐹𝑑 ….. (2.48)

Note that the vector 𝒓 in Equation (2.47) is independent of the
choice of the origin 𝑂 of the coordinate axes. Therefore, we would
obtain the same result if the moments of 𝑭 and −𝑭 had been
computed about a different point 𝑂′. Thus, the moment 𝑴 of a
couple is a free vector, which can be applied at any point (Figure).
From the definition of the moment of a couple, it
also follows that two couples––one consisting of
the forces 𝑭𝟏 and −𝑭𝟏, the other of the forces 𝑭𝟐
and −𝑭𝟐 (Figure)––have equal moments if
𝐹1𝑑1 = 𝐹2𝑑2 ….. (2.49)
provided that the two couples lie in parallel
planes (or in the same plane) and have the
same sense (i.e., clockwise or counter-
clockwise).

2.3.2 Equivalent Couples
Imagine that three couples act successively on the same rectangular
box (Figure). As we have just seen, the only motion a couple can
impart to a rigid body is a rotation. Since each of the three couples
shown has the same moment 𝑴 (same direction and same
magnitude 𝑀 = 120 𝑙𝑏. 𝑖𝑛), we can expect each couple to have the
same effect on the box.

We state that two systems of forces are equivalent (i.e., they have
the same effect on a rigid body) if we can transform one of them
into the other by means of one or several of the following
operations:
• replacing two forces acting on the same particle by their
resultant;
• resolving a force into two components;
• cancelling two equal and opposite forces acting on the same
particle;
• attaching to the same particle two equal and opposite forces; and
• moving a force along its line of action.
Each of these operations is easily justified on the basis of the
parallelogram law or the principle of transmissibility.

The property we have just established is very important for the correct understanding
of the mechanics of rigid bodies. It indicates that when a couple acts on a rigid body,
it does not matter where the two forces forming the couple act or what magnitude and
direction they have. The only thing that counts is the moment of the couple
(magnitude and direction). Couples with the same moment have the same effect on
the rigid body.

2.3.3 Addition of Couples
Consider two intersecting planes 𝑃1 and 𝑃2 and
two couples acting respectively in 𝑃1 and 𝑃2.
Recall that each couple is a free vector in its respective plane and can
be represented within this plane by any combination of equal,
opposite, and parallel forces and of perpendicular distance of
separation that provides the same sense and magnitude for this couple.
Thus, we can assume, without any loss of generality, that the couple in
𝑃1 consists of two forces 𝑭𝟏 and −𝑭𝟏 perpendicular to the line of
intersection of the two planes and acting respectively at 𝐴 and 𝐵
(Figure). Similarly, we can assume that the couple in 𝑃2 consists of
two forces 𝑭𝟐 and −𝑭𝟐 perpendicular to 𝐴𝐵 and acting respectively at
𝐴 and 𝐵.

It is clear that the resultant 𝑹 of 𝑭𝟏 and 𝑭𝟐 and
the resultant −𝑹 of −𝑭𝟏 and −𝑭𝟐 form a
couple.
Denoting the vector joining 𝐵 to 𝐴 by 𝒓 and recalling the definition
of the moment of a couple, we express the moment 𝑴 of the
resulting couple as
𝑴 = 𝒓 × 𝑹 = 𝒓 × 𝑭𝟏 + 𝑭𝟐 ….. (2.50)
By Varignon’s theorem, we can expand this expression as
𝑴 = 𝒓 × 𝑭𝟏 + 𝒓 × 𝑭𝟐 ….. (2.51)
The first term in this expression represents the moment 𝑴𝟏 of the
couple in 𝑃1, and the second term represents the moment 𝑴𝟐 of the
couple in 𝑃2. Therefore, we have
𝑴 = 𝑴𝟏 + 𝑴𝟐 ….. (2.52)

We conclude that the sum of two couples of moments 𝑴𝟏 and 𝑴𝟐
is a couple of moment 𝑴 equal to the vector sum of 𝑴𝟏 and 𝑴𝟐
(Figure). We can extend this conclusion to state that any number of
couples can be added to produce one resultant couple, as
𝑴 = 𝑴 = 𝒓 × 𝑭 ….. (2.53)

2.3.4 Couple Vectors
We have seen that couples with the same moment, whether they act
in the same plane or in parallel planes, are equivalent. Therefore,
we have no need to draw the actual forces forming a given couple
in order to define its effect on a rigid body (Figure a).
It is sufficient to draw an arrow equal in magnitude and direction to
the moment 𝑴 of the couple (Figure b).

We have also seen that the sum of two couples is itself a couple and
that we can obtain the moment 𝑴 of the resultant couple by forming
the vector sum of the moments 𝑴𝟏 and 𝑴𝟐 of the given couples.
Thus, couples obey the law of addition of vectors, so the arrow used
in Figure (b) to represent the couple defined in Figure (a) truly can
be considered a vector.
The vector representing a couple is called a couple vector. Note that
we added the symbol  to the red arrow to avoid any confusion
with vectors representing forces. A couple vector, like the moment
of a couple, is a free vector.

Therefore, we can choose its point of application at the origin of the
system of coordinates, if so desired (Figure c). Furthermore, we can
resolve the couple vector 𝑴 into component vectors 𝑴𝒙, 𝑴𝒚, and
𝑴𝒛 that are directed along the coordinate axes (Figure d). These
component vectors represent couples acting, respectively, in the 𝑦𝑧,
𝑧𝑥, and 𝑥𝑦 planes.

2.3.5 Resolution of a Given Force into a Force at a Point 𝑶 and
a Couple
Consider a force 𝑭 acting on a rigid body at a point 𝐴 defined by
the position vector 𝒓 (Figure a). Suppose that for some reason it
would simplify the analysis to have the force act at point 𝑂 instead.
Although we can move 𝑭 along its line of action (principle of
transmissibility), we cannot move it to a point 𝑂 that does not lie
on the original line of action without modifying the action of 𝑭 on
the rigid body.

We can, however, attach two forces at point 𝑂, one equal to 𝑭 and
the other equal to −𝑭, without modifying the action of the original
force on the rigid body (Figure b). As a result of this
transformation, we now have a force 𝑭 applied at 𝑂; the other two
forces form a couple of moment 𝑴𝑶 = 𝒓 × 𝑭. Thus,
Any force 𝑭 acting on a rigid body can be moved to an arbitrary
point 𝑂 provided that we add a couple whose moment is equal to
the moment of 𝑭 about 𝑂.

If we move force 𝑭 from 𝐴 to a different point 𝑂′ (Figure a and
Figure c), we have to compute the moment 𝑴𝑶′ = 𝒓′ × 𝑭 of 𝑭
about 𝑂′ and add a new force-couple system consisting of 𝑭 and the
couple vector 𝑴𝑶′ at 𝑂′.
We can obtain the relation between the moments of 𝑭 about 𝑂 and
𝑂′ as
𝑴𝑶′ = 𝒓′ × 𝑭 = 𝒓 + 𝒔 × 𝑭 = 𝒓 × 𝑭 + 𝒔 × 𝑭
𝑴𝑶′ = 𝑴𝑶 + 𝒔 × 𝑭 ….. (2.54)

Review Questions
1. What is the necessity of a couple?
2. Define moment of a couple. What is its main characteristics?
3. What are equivalent couples?
4. What are the operations that can be used to transform one
system of forces into another?
5. How is a resultant couple determined?
6. How couple vectors specified?
7. How a given force is resolved into a force at any arbitrary point
and a couple.
8. How a force and a couple at a given point can be transferred
into an equivalent force and couple at another point.

Mechanical I/I
2024
BODIES
Session 15: Simplifying Systems of Forces

• Know common methods of simplifying a system of forces.
Learning Objectives
• Reduce a system of forces into a force and a couple.
• Define equivalent system of forces and know its physical
significance.
• Simplify concurrent forces, coplanar forces and parallel forces.
• Reduce a system of forces into a wrench.

2.4 Simplifying Systems of Forces
We saw in the preceding section that we can replace a force acting
on a rigid body with a force-couple system that may be easier to
analyze.
However, the true value of a force-couple system is that we can use
it to replace not just one force but a system of forces to simplify
analysis and calculations.

2.4.1 Reducing a System of Forces to a Force-Couple System
Consider a system of forces 𝑭𝟏, 𝑭𝟐, 𝑭𝟑, . . . , acting on a rigid body
at the points 𝐴1, 𝐴2, 𝐴3, . . . , defined by the position vectors 𝒓𝟏,
𝒓𝟐, 𝒓𝟑, etc. (Figure a). As seen in the preceding section, we can
move 𝑭𝟏 from 𝐴1 to a given point 𝑂 if we add a couple of moment
𝑴𝟏 equal to the moment 𝒓𝟏 × 𝑭𝟏 of 𝑭𝟏 about 𝑂.
Repeating this procedure with 𝑭𝟐, 𝑭𝟑,
. . . , we obtain the system shown in
Figure (b), which consists of the
original forces, now acting at 𝑂, and
the added couple vectors.

Since the forces are now concurrent, they can be added vectorially
and replaced by their resultant 𝑹. Similarly, the couple vectors 𝑴𝟏,
𝑴𝟐, 𝑴𝟑, . . . , can be added vectorially and replaced by a single
couple vector 𝑴𝑶
𝑹
.
Thus,
We can reduce any system of forces, however complex, to an
equivalent force-couple system acting at a given point 𝑂.

The equivalent force-couple system is defined by
𝑹 = 𝑭 ….. (2.55)
𝑴𝑶
𝑹
= 𝑴𝑶 = 𝒓 × 𝑭
These equations state that we obtain force 𝑹 by adding all of the
forces of the system, whereas we obtain the moment of the resultant
couple vector 𝑴𝑶
𝑹
, called the moment resultant of the system, by
adding the moments about 𝑂 of all the forces of the system.
Once we have reduced a given system of forces to a force and a
couple at a point 𝑂, we can replace it with a force and a couple at
another point 𝑂′. The resultant force 𝑹 will remain unchanged,
whereas the new moment resultant 𝑴𝑶′
𝑹
will be equal to the sum of
𝑴𝑶
𝑹
and the moment about 𝑂′ of force 𝑹 attached at 𝑂 (Figure).

Thus,
….. (2.56)
𝑴𝑶′
𝑹
= 𝑴𝑶
𝑹
+ 𝒔 × 𝑹
In practice, the reduction of a given system of
forces to a single force 𝑹 at 𝑂 and a couple
vector 𝑴𝑶
𝑹
is carried out in terms of components.
Resolving each position vector 𝒓 and each force
𝑭 of the system into rectangular components, we
have
….. (2.57)
𝒓 = 𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌 ….. (2.58)
𝑭 = 𝐹𝑥𝒊 + 𝐹𝑦𝒋 + 𝐹𝑧𝒌
Substituting for 𝒓 and 𝑭 in Equation (2.55) and factoring out the
unit vectors 𝒊, 𝒋, and 𝒌, we obtain 𝑹 and 𝑴𝑶
𝑹
in the form
….. (2.59)
𝑹 = 𝑅𝑥𝒊 + 𝑅𝑦𝒋 + 𝑅𝑧𝒌
….. (2.60)
𝑴𝑶
𝑹
= 𝑀𝑥
𝑅𝒊 + 𝑀𝑦
𝑅𝒋 + 𝑀𝑧
𝑅𝒌

The components 𝑅𝑥, 𝑅𝑦, and 𝑅𝑧 represent, respectively, the sums of
the 𝑥, 𝑦, and 𝑧 components of the given forces and measure the
tendency of the system to impart to the rigid body a translation in
the 𝑥, 𝑦, or 𝑧 direction.
Similarly, the components 𝑀𝑥
𝑅
, 𝑀𝑦
𝑅
, and 𝑀𝑧
𝑅
represent, respectively,
the sum of the moments of the given forces about the 𝑥, 𝑦, and 𝑧
axes and measure the tendency of the system to impart to the rigid
body a rotation about the 𝑥, 𝑦, or 𝑧 axis.
….. (2.59)
𝑹 = 𝑅𝑥𝒊 + 𝑅𝑦𝒋 + 𝑅𝑧𝒌
….. (2.60)
𝑴𝑶
𝑹
= 𝑀𝑥
𝑅𝒊 + 𝑀𝑦
𝑅𝒋 + 𝑀𝑧
𝑅𝒌

2.4.2 Equivalent Systems of Forces
We have just seen that any system of forces acting on a rigid body
can be reduced to a force-couple system at a given point 𝑂. This
equivalent force-couple system characterizes completely the effect
of the given force system on the rigid body.
Two systems of forces are equivalent if they can be reduced to the
same force-couple system at a given point 𝑂.
Recall that the force-couple system at 𝑂 is defined by the relations
in Equation (2.55). Therefore, we can state that
Two systems of forces, 𝑭𝟏, 𝑭𝟐, 𝑭𝟑, . . . , and 𝑭𝟏′, 𝑭𝟐′, 𝑭𝟑′, . . . ,
that act on the same rigid body are equivalent if, and only if, the
sums of the forces and the sums of the moments about a given
point 𝑂 of the forces of the two systems are, respectively, equal.

Mathematically, the necessary and sufficient conditions for the two
systems of forces to be equivalent are
Resolving the forces and moments in Equations (2.61) into their
rectangular components, we can express the necessary and
sufficient conditions for the equivalence of two systems of forces
acting on a rigid body as
𝑭 = 𝑭 ′ ….. (2.61)
𝑴𝑶 = 𝑴𝑶 ′
𝐹𝑥 = 𝐹𝑥 ′ 𝐹𝑦 = 𝐹𝑦 ′ 𝐹𝑧 = 𝐹𝑧 ′ ….. (2.62)
𝑀𝑥 = 𝑀𝑥 ′ 𝑀𝑦 = 𝑀𝑦 ′ 𝑀𝑧 = 𝑀𝑧 ′ ….. (2.63)

These equations have a simple physical significance. They express
that
Two systems of forces are equivalent if they tend to impart to the
rigid body (i) the same translation in the 𝑥, 𝑦, and 𝑧 directions,
respectively, and (ii) the same rotation about the 𝑥, 𝑦, and 𝑧 axes,
respectively.

2.4.3 Further Reduction of a System of Forces
We have now seen that any given system of forces acting on a rigid body
can be reduced to an equivalent force-couple system at 𝑂, consisting of a
force 𝑹 equal to the sum of the forces of the system, and a couple vector
𝑴𝑶
𝑹
of moment equal to the moment resultant of the system.
When 𝑹 = 0, the force-couple system reduces to the couple vector 𝑴𝑶
𝑹
.
The given system of forces then can be reduced to a single couple called
the resultant couple of the system.
It follows from the preceding section that we can replace the force-couple
system at 𝑂 by a single force 𝑹 acting along a new line of action if 𝑹 and
𝑴𝑶
𝑹
are mutually perpendicular. The systems of forces that can be
reduced to a single force, or resultant, are therefore the systems for which
force 𝑹 and the couple vector 𝑴𝑶
𝑹
are mutually perpendicular.

This condition is generally not satisfied by systems of forces in
space, but it is satisfied by systems consisting of concurrent forces,
coplanar forces, or parallel forces. Let’s look at each case
separately.
(a) Concurrent Forces
Concurrent forces act at the same point; therefore, we can add them
directly to obtain their resultant 𝑹. Thus, they always reduce to a
single force. Concurrent forces were discussed in detail in Chapter
1.

(b) Coplanar Forces
Coplanar forces act in the same plane, which we assume to be the
plane of the figure (Figure a). The sum 𝑹 of the forces of the
system also lies in the plane of the figure, whereas the moment of
each force about 𝑂 and thus the moment resultant 𝑴𝑶
𝑹
are
perpendicular to that plane. The force-couple system at 𝑂 consists,
therefore, of a force 𝑹 and a couple vector𝑴𝑶
𝑹
that are mutually
perpendicular (Figure b).

We can reduce them to a single force 𝑹 by moving 𝑹 in the plane
of the figure until its moment about 𝑂 becomes equal to 𝑴𝑶
𝑹
. The
distance from 𝑂 to the line of action of 𝑹 is 𝑑 = 𝑀𝑂
𝑅
/𝑅 (Figure c).
As noted earlier, the reduction of a system of forces is considerably
simplified if we resolve the forces into rectangular components.

As noted earlier, the reduction of a system of forces is considerably
simplified if we resolve the forces into rectangular components.
The force-couple system at 𝑂 is then characterized by the
components (Figure a)
𝑅𝑥 = 𝐹𝑥 𝑅𝑦 = 𝐹𝑦 𝑀𝑧
𝑅 = 𝑀𝑂
𝑅
= 𝑀𝑂 ….. (2.64)

To reduce the system to a single force 𝑹, the moment of 𝑹 about 𝑂
must be equal to 𝑴𝑶
𝑹
. If we denote the coordinates of the point of
application of the resultant by 𝑥 and 𝑦, we get
𝑥𝑅𝑦 − 𝑦𝑅𝑥 = 𝑀𝑂
𝑅
….. (2.65)
This represents the equation of the line of action of 𝑹. We can also
determine the 𝑥 and 𝑦 intercepts of the line of action of the resultant
directly by noting that 𝑴𝑶
𝑹
must be equal to the moment about 𝑂 of
the 𝑦 component of 𝑹 when 𝑹 is attached at 𝐵 (Figure b) and to the
moment of its 𝑥 component when 𝑹 is attached at 𝐶 (Figure c).

(c) Parallel Forces
Parallel forces have parallel lines of action and may or may not have
the same sense. Assuming here that the forces are parallel to the 𝑦 axis
(Figure a), we note that their sum 𝑹 is also parallel to the 𝑦 axis.
On the other hand, since the moment of a given force must be
perpendicular to that force, the moment about 𝑂 of each force of the
system and thus the moment resultant 𝑴𝑶
𝑹
lie in the 𝑧𝑥 plane. The
force-couple system at 𝑂 consists, therefore, of a force 𝑹 and a couple
vector 𝑴𝑶
𝑹
that are mutually perpendicular (Figure b).

We can reduce them to a single force 𝑹 (Figure c) or, if 𝑹 = 0, to a
single couple of moment 𝑴𝑶
𝑹
.
In practice, the force-couple system at 𝑂 is characterized by the
components
𝑅𝑦 = 𝐹𝑦 𝑀𝑥
𝑅 = 𝑀𝑥 𝑀𝑧
𝑅 = 𝑀𝑧 ….. (2.66)

The reduction of the system to a single force can be carried out by
moving 𝑹 to a new point of application 𝐴 𝑥, 0, 𝑧 , which is chosen
so that the moment of 𝑹 about 𝑂 is equal to 𝑴𝑶
𝑹
.
𝒓 × 𝑹 = 𝑴𝑶
𝑹
𝑥𝒊 + 𝑧𝒌 × 𝑅𝑦𝒋 = 𝑀𝑥
𝑅𝒊 + 𝑀𝑧
𝑅𝒌
By computing the vector products and equating the coefficients of
the corresponding unit vectors in both sides of the equation, we
obtain two scalar equations that define the coordinates of 𝐴:
−𝑧𝑅𝑦 = 𝑀𝑥
𝑅
𝑥𝑅𝑦 = 𝑀𝑧
𝑅

2.4.4 Reduction of a System of Forces to a Wrench
In the general case of a system of forces in space, the equivalent
force-couple system at 𝑂 consists of a force 𝑹 and a couple vector
𝑴𝑶
𝑹
that are not perpendicular and where neither is zero (Figure a).
This system of forces cannot be reduced to a single force or to a
single couple. However, we still have a way of simplifying this
system further.

The simplification method consists of first replacing the couple
vector by two other couple vectors that are obtained by resolving
𝑴𝑶
𝑹
into a component 𝑴𝟏 along 𝑹 and a component 𝑴𝟐 in a plane
perpendicular to 𝑹 (Figure b).
Then we can replace the couple vector 𝑴𝟐 and force 𝑹 by a single
force 𝑹 acting along a new line of action.

The original system of forces thus reduces to 𝑹 and to the couple
vector 𝑴𝟏 (Figure c), i.e., to 𝑹 and a couple acting in the plane
perpendicular to 𝑹.
This particular force-couple system is called a wrench because the
resulting combination of push and twist is the same as that caused
by an actual wrench. The line of action of 𝑹 is known as the axis of
the wrench, and the ratio 𝑝 = 𝑀1/𝑅 is called the pitch of the
wrench.

A wrench therefore consists of two collinear vectors: a force 𝑹 and
a couple vector
𝑴𝟏 = 𝑝𝑹 ….. (2.67)
We note that the projection of 𝑴𝑶
𝑹
on the line of action of 𝑹 is
𝑀1 =
𝑹. 𝑴𝑶
𝑹
𝑅
Thus, we can express the pitch of the wrench as
𝑝 =
𝑴𝟏
𝑹
=
𝑹. 𝑴𝑶
𝑹
𝑅2
….. (2.68)

To define the axis of the wrench, we can write a relation involving
the position vector 𝒓 of an arbitrary point 𝑃 located on that axis. We
first attach the resultant force 𝑹 and couple vector 𝑴𝟏 at 𝑃
(Figure).
Then, since the moment about 𝑂 of this
force-couple system must be equal to the
moment resultant 𝑴𝑶
𝑹
of the original
force system, we have
𝑴𝟏 + 𝒓 × 𝑹 = 𝑴𝑶
𝑹
….. (2.69)
Alternatively, using Equation (2.67), we have
𝑝𝑹 + 𝒓 × 𝑹 = 𝑴𝑶
𝑹
….. (2.70)

Review Questions
1. What are the common methods to simplify a system of forces?
2. How a system of forces can be reduced to a force and a couple?
3. How a force-couple system defined for a point can be
transferred to another point?
4. Define equivalent system of forces. What is its physical
significance?
5. Under what condition a force-couple system reduce to a single
couple?
6. How can concurrent, coplanar and parallel forces be simplified?
7. Under what condition a system of forces can be reduced to a
wrench? How is it reduced?

Mechanical I/I
2024
BODIES
Session 16: Examples for Simplification of
System of Forces

• Simplify systems of forces.
Learning Objectives

Example 4.1
If 𝛉 = 𝟑𝟎𝟎
, determine the magnitude of force 𝐅 shown in
Figure E4.1 so that the resultant couple moment is 𝟏𝟎𝟎 𝐍. 𝐦,
clockwise. The disk has a radius of 𝟑𝟎𝟎 𝒎𝒎.
Figure E4.1
𝟏𝟏𝟏 𝑵

Example 4.2
Determine the resultant couple moment of the two couples that
act on the pipe assembly shown in Figure E4.2. The distance
from 𝑨 to 𝑩 is 𝒅 = 𝟒𝟎𝟎 𝒎𝒎. Express the result as a Cartesian
vector.
Figure E4.2
−𝟏𝟐. 𝟏𝒊 − 𝟏𝟎. 𝟎𝒋 − 𝟏𝟕. 𝟑𝒌 𝑵. 𝒎

Example 4.3
The bracket shown in Figure E4.3 is fastened to a wall by
anchor bolts at 𝑨 and 𝑩, is loaded by the force 𝑷 = 𝟏𝟒𝟎 𝑵 and
the couple 𝑪 = 𝟏𝟖𝟎 𝑵. 𝒎 . Replace 𝑷 and 𝑪 with (a) an
equivalent force-couple system, the force of which acts at 𝑨;
and (b) two vertical forces, one acting at 𝑨 and the other at 𝑩.
Figure E4.3
𝟏𝟒𝟎 𝑵 , 𝟖𝟐. 𝟎 𝑵. 𝒎 ; 𝟒𝟎𝟕 𝑵 , 𝟓𝟒𝟕 𝑵 

Example 4.4
A 𝟑 𝒎 long beam shown in Figure E4.4 is subjected to a variety
of loadings. Replace each loading with an equivalent force-
couple system at end 𝑨 of the beam.
Figure E4.4
𝟓𝟎𝟎 𝑵 , 𝟐𝟎𝟎 𝑵. 𝒎 

Example 4.5
The two couples shown in Figure E4.5 are to be replaced with a
single equivalent couple. Determine (a) the couple vector
representing the equivalent couple, (b) the two forces acting at
𝑩 and 𝑪 that can be used to form that couple.
Figure E4.5
𝟏𝟑. 𝟔𝟑 𝑵. 𝒎; 𝜽𝒙 = 𝟐𝟕. 𝟖𝟎
, 𝜽𝒚 = 𝟔𝟐. 𝟐𝟎
; 𝜽𝒛 = 𝟗𝟎𝟎
;
𝟏𝟖. 𝟏𝟕 𝑵 𝟔𝟐. 𝟐𝟎
𝒂𝒕 𝑩 and 𝟏𝟖. 𝟏𝟕 𝑵𝟔𝟐. 𝟐𝟎
𝒂𝒕 𝑪

Example 4.6
Replace the force and couple moment system acting on the
beam shown in Figure E4.6 by an equivalent resultant force,
and find where its line of action intersects the beam, measured
from point 𝑶.
Figure E4.6
𝟓. 𝟑𝟕 𝒌𝑵, 𝟐𝟔. 𝟔𝟎
𝒘𝒊𝒕𝒉 𝒙 − 𝒂𝒙𝒊𝒔, 𝟐. 𝟐𝟓 𝒎 to the right of 𝑶

Example 4.7
Replace the loading on the frame shown in Figure E4.7 by a
single resultant force. Specify where its line of action intersects
member 𝑪𝑫, measured from 𝑪.
Figure E4.7
𝟗𝟗𝟏 𝑵, 𝟔𝟑. 𝟎𝟎
, 𝟐. 𝟔𝟒 𝒎

Example 4.8
A machine component is subjected to the forces and couples
shown in Figure E4.8. The component is to be held in place by a
single rivet that can resist a force but not a couple. For 𝑷 = 𝟎,
determine the location of the rivet hole if it is to be located (a)
on line 𝑭𝑮, (b) on line 𝑮𝑯.
Figure E4.8 𝟎. 𝟑𝟔𝟓 𝒎 above 𝑮, 𝟎. 𝟐𝟐𝟕 𝒎 to the right of 𝑮

Example 4.9
Replace the parallel force system acting on the plate shown in
Figure E4.9 by a resultant force and specify its location on the
𝒙𝒛 plane.
Figure E4.9 𝟏𝟎 𝒌𝑵, , 𝟐. 𝟏𝟕 𝒎

Example 4.10
Replace the force system by a wrench and specify the
magnitude of the force and couple moment of the wrench and
the point where the wrench intersects the 𝒙𝒛 plane [Figure
E4.10].
Figure E4.10
𝟑𝟕𝟗 𝑵, 𝟓𝟗𝟎 𝑵. 𝒎 , 𝒙 = −𝟐. 𝟕𝟔 𝒎, 𝒛 = 𝟐. 𝟔𝟖 𝒎

Mechanical I/I
2024
BODIES
Session 17: Equilibrium of Rigid Bodies

• Know equations of equilibrium for rigid bodies.
Learning Objectives
• Know the step of solving problems of equilibrium of rigid
bodies in two-dimension.
• Know the reactions for common two-dimensional structures.
• Reduce equations of equilibrium for rigid bodies in two-
dimension.
• Define statically determinate, statically indeterminate, partially
constrained and improperly constrained structures.
• Know the conditions of forces in two-force and three-force
members.
• Know the reactions for common three-dimensional structures.

• Know equations of equilibrium for rigid bodies.
• Know the step of solving problems of equilibrium of rigid
bodies in two-dimension.
• Know the reactions for common two-dimensional structures.
• Reduce equations of equilibrium for rigid bodies in two-
dimension.
• Define statically determinate, statically indeterminate, partially
constrained and improperly constrained structures.
• Know the conditions of forces in two-force and three-force
members.

2.5 Equilibrium of Rigid Bodies
We saw in previous sections how to reduce the external forces
acting on a rigid body to a force-couple system at some arbitrary
point 𝑂. When the force and the couple are both equal to zero, the
external forces form a system equivalent to zero, and the rigid body
is said to be in equilibrium.
We can obtain the necessary and sufficient conditions for the
equilibrium of a rigid body by setting 𝑹 and 𝑴𝑶
𝑹
equal to zero as:
𝑭 = 𝟎 𝑴𝑶 = 𝒓 × 𝑭 = 𝟎 ….. (2.71)

Resolving each force and each moment into its rectangular
components, we can replace these vector equations for the
equilibrium of a rigid body with the following six scalar equations:
𝐹𝑥 = 0 𝐹𝑦 = 0 𝐹𝑧 = 0 ….. (2.72)
𝑀𝑥 = 0 𝑀𝑦 = 0 𝑀𝑧 = 0 ….. (2.73)
Note that Equations (2.72) express the fact that the components of
the external forces in the 𝑥, 𝑦 and 𝑧 directions are balanced;
Equations (2.73) express the fact that the moments of the external
forces about the 𝑥, 𝑦 and 𝑧 axes are balanced.

Therefore, for a rigid body in equilibrium, the system of external forces
imparts no translational or rotational motion to the body.
We can use these equations to determine unknown forces applied to the
rigid body or unknown reactions exerted on it by its supports.
In order to write the equations of equilibrium for a rigid body, we must
first identify all of the forces acting on that body and then draw the
corresponding free-body diagram.
We first consider the equilibrium of two-dimensional structures subjected
to forces contained in their planes and study how to draw their free-body
diagrams. In addition to the forces applied to a structure, we must also
consider the reactions exerted on the structure by its supports. A specific
reaction is associated with each type of support. You will see how to
determine whether the structure is properly supported, so that you can
know in advance whether you can solve the equations of equilibrium for
the unknown forces and reactions.

Free-Body Diagrams
In solving a problem concerning a rigid body in equilibrium, it is
essential to consider all of the forces acting on the body. It is equally
important to exclude any force that is not directly applied to the
body. Omitting a force or adding an extraneous one would destroy
the conditions of equilibrium.
Therefore, the first step in solving the problem is to draw a free-
body diagram of the rigid body under consideration. We have
already used free-body diagrams on many occasions in the previous
Chapter. However, in view of their importance to the solution of
equilibrium problems, we summarize here the steps you must
follow in drawing a correct free-body diagram.

1. Start with a clear decision regarding the choice of the free body to
be analyzed. Mentally, you need to detach this body from the
ground and separate it from all other bodies. Then you can sketch
the contour of this isolated body.
2. Indicate all external forces on the free-body diagram. These forces
represent the actions exerted on the free body by the ground and by
the bodies that have been detached. In the diagram, apply these
forces at the various points where the free body was supported by
the ground or was connected to the other bodies.
3. Clearly mark the magnitudes and directions of the known external
forces on the free-body diagram. Recall that when indicating the
directions of these forces, the forces are those exerted on, and not
by, the free body. Known external forces generally include the
weight of the free body and forces applied for a given purpose.

4. Unknown external forces usually consist of the reactions
through which the ground and other bodies oppose a possible
motion of the free body. The reactions constrain the free body to
remain in the same position; for that reason, they are sometimes
called constraining forces. Reactions are exerted at the points
where the free body is supported by or connected to other
bodies; you should clearly indicate these points.
5. The free-body diagram should also include dimensions, since
these may be needed for computing moments of forces. Any
other detail, however, should be omitted.

2.6 Equilibrium of Rigid Bodies in Two-dimensions
First, we consider the equilibrium of two-dimensional structures;
i.e., we assume that the structure being analyzed and the forces
applied to it are contained in the same plane. Clearly, the reactions
needed to maintain the structure in the same position are also
contained in this plane.
2.6.1 Reactions for a Two-Dimensional Structure
The reactions exerted on a two-dimensional structure fall into three
categories that correspond to three types of supports or
connections.

a) Reactions Equivalent to a Force with a Known Line of
Action
Each of these supports and connections can prevent motion in one
direction only.

b) Reactions Equivalent to a Force of Unknown Direction
and Magnitude
Supports and connections causing reactions of this type include
frictionless pins in fitted holes, hinges, and rough surfaces. They can
prevent translation of the free body in all directions, but they cannot
prevent the body from rotating about the connection.

c) Reactions Equivalent to a Force and a Couple
These reactions are caused by fixed supports that oppose any motion
of the free body and thus constrain it completely. Fixed supports
actually produce forces over the entire surface of contact; these
forces, however, form a system that can be reduced to a force and a
couple.

2.6.2 Rigid-Body Equilibrium in Two Dimensions
The conditions stated in Section 2.5 for the equilibrium of a rigid
body become considerably simpler for the case of a two-
dimensional structure. Choosing the 𝑥 and 𝑦 axes to be in the plane
of the structure, we have
𝐹𝑧 = 0 𝑀𝑥 = 𝑀𝑦 = 0 𝑀𝑧 = 𝑀𝑂 ….. (2.74`)
for each of the forces applied to the structure. Thus, the six
equations of equilibrium stated at the beginning of this section
reduce to three equations:
𝐹𝑥 = 0 𝐹𝑦 = 0 𝑀𝑂 = 0 ….. (2.75)

Since 𝑀𝑂 must be satisfied regardless of the choice of the origin
𝑂, we can write the equations of equilibrium for a two-dimensional
structure in the more general form
𝐹𝑥 = 0 𝐹𝑦 = 0 𝑀𝐴 = 0 ….. (2.76)
where 𝐴 is any point in the plane of the structure. These three
equations can be solved for no more than three unknowns.
The unknown forces include reactions and that the number of
unknowns corresponding to a given reaction depends upon the type
of support or connection causing that reaction.

𝑀𝐴 = 0 gives the magnitude 𝐵
𝐹𝑥 = 0 gives the magnitude 𝐴𝑥
𝐹𝑦 = 0 gives the magnitude 𝐴𝑦
𝑀𝐵 = 0 can be used to checking the solution

An alternative system of equations for equilibrium is
𝐹𝑥 = 0 𝑀𝐴 = 0 𝑀𝐵 = 0 ….. (2.77)
A third possible set of equilibrium equations is
𝑀𝐴 = 0 𝑀𝐵 = 0 𝑀𝐶 = 0 ….. (2.78)

𝐹𝑥 = 0 𝑀𝐶 = 0 𝑀𝐷 = 0 ….. (2.79)

2.6.2 Statically Indeterminate Reactions and Partial Constraints
In the two examples considered, the types of supports used were
such that the rigid body could not possibly move under the given
loads or under any other loading conditions. In such cases, the rigid
body is said to be completely constrained.
Recall that the reactions corresponding to these supports involved
three unknowns and could be determined by solving the three
equations of equilibrium. When such a situation exists, the
reactions are said to be statically determinate.

Consider Figures (a), in which the truss shown is held by pins at 𝐴
and 𝐵. These supports provide more constraints than are necessary
to keep the truss from moving under the given loads or under any
other loading conditions.
Note from the free-body diagram of Figures (b) that the
corresponding reactions involve four unknowns. We pointed out in
the last section that only three independent equilibrium equations
are available; therefore, in this case, we have more unknowns than
equations.

As a result, we cannot determine all of the unknowns. The
equations 𝑀𝐴 = 0 and 𝑀𝐵 = 0 yield the vertical components
𝐵𝑦 and 𝐴𝑦, respectively, but the equation 𝐹𝑥 = 0 gives only the
sum 𝐴𝑥 + 𝐵𝑥 of the horizontal components of the reactions at 𝐴
and 𝐵. The components 𝐴𝑥 and 𝐵𝑥 are statically indeterminate.
We could determine their magnitudes by considering the
deformations produced in the truss by the given loading, but this
method is beyond the scope of statics and belongs to the study of
mechanics of materials.

Let’s consider the opposite situation. The supports holding the truss
shown in Figures (a) consist of rollers at 𝐴 and 𝐵. Clearly, the
constraints provided by these supports are not sufficient to keep the
truss from moving. Although they prevent any vertical motion, the
truss is free to move horizontally. The truss is said to be partially
constrained.

From the free-body diagram in Figure (b), note that the reactions at
𝐴 and 𝐵 involve only two unknowns. Since three equations of
equilibrium must still be satisfied, we have fewer unknowns than
equations. In such a case, one of the equilibrium equations will not
be satisfied in general.
The equations 𝑀𝐴 = 0 and 𝑀𝐵 = 0 can be satisfied by a proper
choice of reactions at 𝐴 and 𝐵, but the equation 𝐹𝑥 = 0 is not
satisfied unless the sum of the horizontal components of the applied
forces happens to be zero. We thus observe that the equilibrium of
the truss cannot be maintained under general loading conditions.

From these examples, it would appear that, if a rigid body is to be
completely constrained and if the reactions at its supports are to be
statically determinate, there must be as many unknowns as there
are equations of equilibrium.
When this condition is not satisfied, we can be certain that either
the rigid body is not completely constrained or that the reactions at
its supports are not statically determinate. It is also possible that the
rigid body is not completely constrained and that the reactions are
statically indeterminate.
However, the fact that the number of unknowns is equal to the
number of equations is no guarantee that a body is completely
constrained or that the reactions at its supports are statically
determinate.

Consider Figure (a), which shows a truss held by rollers at 𝐴, 𝐵,
and 𝐸. We have three unknown reactions of 𝑨, 𝑩, and 𝑬 (Figure b),
but the equation 𝐹𝑥 = 0 is not satisfied unless the sum of the
horizontal components of the applied forces happens to be zero.
Although there are a sufficient number of constraints, these
constraints are not properly arranged, so the truss is free to move
horizontally. We say that the truss is improperly constrained.

Since only two equilibrium equations are left for determining three
unknowns, the reactions are statically indeterminate. Thus,
improper constraints also produce static indeterminacy.
The truss shown in Figure is another example of improper
constraints and of static indeterminacy. This truss is held by a pin at
𝐴 and by rollers at 𝐵 and 𝐶, which altogether involve four
unknowns.

Since only three independent equilibrium equations are available,
the reactions at the supports are statically indeterminate.
On the other hand, we note that the equation 𝑀𝐴 = 0 cannot be
satisfied under general loading conditions, since the lines of action
of the reactions 𝐵 and 𝐶 pass through 𝐴. We conclude that the truss
can rotate about 𝐴 and that it is improperly constrained.
From the last two examples, we can conclude
A rigid body is improperly constrained whenever the supports
(even though they may provide a sufficient number of reactions)
are arranged in such a way that the reactions must be either
concurrent or parallel.

2.6.3 Two Special Cases
In practice, some simple cases of equilibrium occur quite often,
either as part of a more complicated analysis or as the complete
models of a situation. By understanding the characteristics of these
cases, you can often simplify the overall analysis.
(a) Two-Force Body
A particular case of equilibrium of considerable interest in practical
applications is that of a rigid body subjected to two forces. Such a
body is commonly called a two-force body. We show here that, if a
two-force body is in equilibrium, the two forces must have the
same magnitude, the same line of action, and opposite sense

Consider a corner plate subjected to two forces 𝑭𝟏 and 𝑭𝟐 acting at
𝐴 and 𝐵, respectively (Figure a). If the plate is in equilibrium, the
sum of the moments of 𝑭𝟏 and 𝑭𝟐 about any axis must be zero.
First, we sum moments about A. Since the moment of 𝑭𝟏 is
obviously zero, the moment of 𝑭𝟐 also must be zero and the line of
action of 𝑭𝟐 must pass through A (Figure b).
Similarly, summing moments about 𝐵, we can show that the line of
action of 𝑭𝟏 must pass through 𝐵 (Figure c).

(b) Three-Force Body
Another case of equilibrium that is of great practical interest is that
of a three-force body, i.e., a rigid body subjected to three forces or,
more generally, a rigid body subjected to forces acting at only three
points.
Therefore, both forces have the same line of action (line 𝐴𝐵). You
can see from either of the equations 𝐹𝑥 = 0 and 𝐹𝑦 = 0 that
they must also have the same magnitude but opposite sense.
Consider a rigid body subjected to a system of
forces that can be reduced to three forces 𝑭𝟏, 𝑭𝟐,
and 𝑭𝟑 acting at 𝐴, 𝐵, and 𝐶 respectively. We show
that if the body is in equilibrium, the lines of action
of the three forces must be either concurrent or
parallel.

Since the rigid body is in equilibrium, the sum of
the moments of 𝑭𝟏, 𝑭𝟐, and 𝑭𝟑 about any axis
must be zero.
Assuming that the lines of action of 𝑭𝟏 and 𝑭𝟐
intersect and denoting their point of intersection
by 𝐷, we sum moments about 𝐷 (Figure b).
Because the moments of 𝑭𝟏 and 𝑭𝟐 about 𝐷 are
zero, the moment of 𝑭𝟑 about 𝐷 also must be zero,
and the line of action of 𝑭𝟑 must pass through 𝐷
(Figure c).
Therefore, the three lines of action are concurrent.
The only exception occurs when none of the lines
intersect; in this case, the lines of action are
parallel.

2.7 Equilibrium of Rigid Bodies in Three-dimensions
The most general situation of rigid-body equilibrium occurs in
three dimensions. The approach to modeling and analyzing these
situations is the same as in two dimensions: Draw a free-body
diagram and then write and solve the equilibrium equations.
However, you now have more equations and more variables to deal
with. In addition, reactions at supports and connections can be
more varied, having as many as three force components and three
couples acting at one support.

2.7.1 Rigid-Body Equilibrium in Three Dimensions
As explained earlier, six scalar equations are required to express the
conditions for the equilibrium of a rigid body in the general three-
dimensional case:
𝐹𝑥 = 0 𝐹𝑦 = 0 𝐹𝑧 = 0 ….. (2.72)
𝑀𝑥 = 0 𝑀𝑦 = 0 𝑀𝑧 = 0 ….. (2.73)
We can solve these equations for no more than six unknowns,
which generally represent reactions at supports or connections.

2.7.2 Reactions for a Three-Dimensional Structure
The reactions on a three-dimensional structure range from a single
force of known direction exerted by a frictionless surface to a
force-couple system exerted by a fixed support. Consequently, in
problems involving the equilibrium of a three-dimensional
structure, between one and six unknowns may be associated with
the reaction at each support or connection.

Review Questions
1. What are equations of equilibrium for rigid bodies?
2. Explain the step of solving problems of equilibrium of rigid
bodies in two-dimension
3. Show the reactions for common two-dimensional structures
with sketches.
4. Write down the equations of equilibrium for rigid bodies in
two-dimension.
5. What are statically determinate, statically indeterminate,
partially constrained and improperly constrained structures?
6. How can concurrent, coplanar and parallel forces be simplified?

7. What are the conditions of forces in two-force and three-force
members?
8. Show the reactions for common two-dimensional structures
with sketches.

Mechanical I/I
2024
BODIES
Session 18: Examples for Equilibrium of
Rigid Bodies

• Solve problems related to equilibrium of rigid bodies in two-
dimension and three-dimension.
Learning Objectives

Example 5.1
Determine the horizontal and vertical components of reaction
at the pin 𝑨 and the reaction of the rocker 𝑩 on the beam
shown in Figure E5.1.
Figure E5.1
𝑨𝒙 = 𝟏. 𝟕𝟑 𝒌𝑵 𝑨𝒚 = 𝟏. 𝟎𝟎 𝒌𝑵 𝑩 = 𝟑. 𝟒𝟔 𝒌𝑵

Example 5.2
The bracket 𝑩𝑪𝑫 is hinged at 𝑪 and attached to a control cable
at 𝑩 as shown in Figure E5.2. For the loading shown, determine
(a) the tension in the cable, (b) the reaction at 𝑪.
Figure E5.2
𝟐 𝒌𝑵 𝟐. 𝟑𝟐 𝒌𝑵 𝟒𝟔. 𝟒𝟎

Example 5.3
The boom supports the two vertical loads as shown in Figure
E5.3. Neglect the size of the collars at 𝑫 and 𝑩 and the
thickness of the boom, and compute the horizontal and vertical
components of force at the pin 𝑨 and the force in cable 𝑪𝑩. Set
𝑭𝟏 = 𝟖𝟎𝟎 𝑵 and 𝑭𝟐 = 𝟑𝟓𝟎 𝑵.
Figure E5.3
𝑨𝒙 = 𝟔𝟐𝟓 𝑵 𝑨𝒚 = 𝟔𝟖𝟏 𝑵
𝑭𝑪𝑩 = 𝟕𝟖𝟐 𝑵

Example 5.4
Bar 𝑨𝑪 supports two 𝟒𝟎𝟎 𝑵 loads as shown in Figure E5.4.
Rollers at 𝑨 and 𝑪 rest against frictionless surfaces and a cable
𝑩𝑫 is attached at 𝑩. Determine (a) the tension in cable 𝑩𝑫, (b)
the reaction at 𝑨, (c) the reaction at 𝑪.
Figure E5.4
𝟏𝟒𝟑𝟐 𝑵
𝟏𝟏𝟎𝟎 𝑵 
𝟏𝟒𝟎𝟎 𝑵 

Example 5.5
A man raises a 𝟏𝟎 𝒌𝒈 joist with a length of 𝟒 𝒎 by pulling on a
rope as shown in Figure E5.5. Find the tension 𝑻 in the rope
and the reaction at 𝑨.
Figure E5.5
𝟓𝟖. 𝟔𝟎
𝟖𝟏. 𝟗 𝑵 𝟏𝟒𝟕. 𝟖 𝑵 𝟓𝟖. 𝟔𝟎

Example 5.6
Determine the reactions at 𝑨 and 𝑩 when 𝒂 = 𝟏𝟓𝟎 𝒎𝒎 [Figure
E5.6].
Figure E5.6
𝟔𝟎𝟎 𝑵 , 𝟔𝟖𝟎𝑵 𝟐𝟖. 𝟏𝟎

Example 5.7
A 𝟐. 𝟒 𝐦 boom is held by a ball-and-socket joint at 𝐂 and by
two cables 𝐀𝐃 and 𝐀𝐄 as shown in Figure E5.7. Determine the
tension in each cable and the reaction at 𝐂.
Figure E5.7
𝟐. 𝟖𝟎 𝒌𝑵 𝟐. 𝟔𝟎 𝒌𝑵
𝟏. 𝟖𝟎𝒋 + 𝟒. 𝟖𝟎𝒌 𝒌𝑵

Example 4.8
A machine component is subjected to the forces and couples
shown in Figure E4.8. The component is to be held in place by a
single rivet that can resist a force but not a couple. For 𝑷 = 𝟎,
determine the location of the rivet hole if it is to be located (a)
on line 𝑭𝑮, (b) on line 𝑮𝑯.
Figure E4.8
𝟎. 𝟑𝟔𝟓 𝒎 above 𝑮, 𝟎. 𝟐𝟐𝟕 𝒎 to the
right of 𝑮

Mechanical I/I
2024
CENTER OF GRAVITY AND
CENTROIDS
Session 19: Planar Centers of Gravity and
Centroids

• Define center of gravity and centroids.
Learning Objectives
• Derive expressions for coordinates of center of gravity of a two-
dimensional surface (plate).
• Derive expressions for coordinates of center of gravity of a two-
dimensional line.
• Derive expressions for coordinates of centroids of a two-
dimensional surface (plate).
• Derive expressions for coordinates of centroids of a two-
dimensional line.
• Define first moment or areas and lines.
• Derive expressions for coordinates of center of gravity and centroid
of a composite surface and a composite line.

3.1 Planar Centers of Gravity and Centroids
In the previous chapter, we showed how the locations of the lines
of action of forces affects the replacement of a system of forces
with an equivalent system of forces and couples.
In this section, we extend this idea to show how a distributed
system of forces (in particular, the elements of an object’s weight)
can be replaced by a single resultant force acting at a specific point
on an object. The specific point is called the object’s center of
gravity.

3.1.1 Center of Gravity of a Two-Dimensional Body
𝑾 = ∆𝑾𝟏 + ∆𝑾𝟐 + ⋯ . . +∆𝑾𝒏
𝐹𝑧 : ….. (3.1)
𝑥𝑾 = 𝑥1∆𝑾𝟏 + 𝑥2∆𝑾𝟐 + ⋯ . . +𝑥𝑛∆𝑾𝒏
𝑀𝑦 : ….. (3.2)
𝑦𝑾 = 𝑦1∆𝑾𝟏 + 𝑦2∆𝑾𝟐 + ⋯ . . +𝑦𝑛∆𝑾𝒏
𝑀𝑥 : ….. (3.3)

Solving these equations for 𝑥 and 𝑦
If we now increase the number of elements into which we divide the
plate and simultaneously decrease the size of each element, in the
limit of infinitely many elements of infinitesimal size, we obtain the
expressions
𝑥 =
𝑥1∆𝑾𝟏 + 𝑥2∆𝑾𝟐 + ⋯ . . +𝑥𝑛∆𝑾𝒏
𝑾
….. (3.4)
𝑦 =
𝑦1∆𝑾𝟏 + 𝑦2∆𝑾𝟐 + ⋯ . . +𝑦𝑛∆𝑾𝒏
𝑾
….. (3.5)
𝑥𝑊 = 𝑥𝑑𝑊
𝑊 = 𝑑𝑊 𝑦𝑊 = 𝑦𝑑𝑊 ….. (3.6)
Solving for 𝑥 and 𝑦
𝑥 =
𝑥𝑑𝑊
𝑊
𝑊 = 𝑑𝑊 𝑦 =
𝑦𝑑𝑊
𝑊
….. (3.7)

The same equations can be derived for a wire lying in the 𝑥𝑦 plane
(Figure). Note that the center of gravity 𝐺 of a wire is usually not
located on the wire.

3.1.2 Centroids of Areas and Lines
In the case of a flat homogeneous plate of uniform thickness, we
can express the magnitude ∆𝑊 of the weight of an element of the
plate as
∆𝑊 = 𝛾𝑡∆𝐴 ….. (3.8)
where,
𝛾 = specific weight (weight per unit volume) of the material
𝑡 = thickness of the plate
∆𝐴 = area of the element
Similarly, we can express the magnitude 𝑊 of the weight of the
entire plate as
𝑊 = 𝛾𝑡𝐴 ….. (3.9)
where 𝐴 is the total area of the plate.

Substituting for ∆𝑊 and 𝑊 in the moment Equations (3.2) and (3.3)
and dividing throughout by 𝛾𝑡, we obtain
𝑥𝐴 = 𝑥1∆𝐴1 + 𝑥2∆𝐴2 + ⋯ . . +𝑥𝑛∆𝐴𝑛
𝑀𝑦 : ….. (3.10)
𝑦𝐴 = 𝑦1∆𝐴1 + 𝑦2∆𝐴2 + ⋯ . . +𝑦𝑛∆𝐴𝑛
𝑀𝑥 : ….. (3.11)
If we increase the number of elements into which the area 𝐴 is
divided and simultaneously decrease the size of each element, in the
limit we obtain
𝑥𝐴 = 𝑥𝑑𝐴
𝐴 = 𝑑𝐴 𝑦𝐴 = 𝑦𝑑𝐴 ….. (3.12)
𝑥 =
𝑥𝑑𝐴
𝐴
𝐴 = 𝑑𝐴 𝑦 =
𝑦𝑑𝐴
𝐴
….. (3.13)

In the case of a homogeneous wire of uniform cross section, we can
express the magnitude ∆𝑊 of the weight of an element of wire as
∆𝑊 = 𝛾𝑎∆𝐿 ….. (3.14)
where,
𝛾 = specific weight of the material
𝑎 = cross-sectional area of the wire
∆𝐿 = length of the element
We can obtain the coordinates 𝑥 and 𝑦 of the centroid of line 𝐿 from
the equations
𝑥𝐿 = 𝑥𝑑𝐿 𝑦𝐿 = 𝑦𝑑𝐿 ….. (3.15)
𝑥 =
𝑥𝑑𝐿
𝐿
𝑦 =
𝑦𝑑𝐿
𝐿
….. (3.16)

3.1.3 First Moments of Areas and Lines
The integral 𝑥𝑑𝐴 in Equations (3.12) is known as the first
moment of the area 𝐴 with respect to the 𝑦 axis and is denoted by
𝑄𝑦. Similarly, the integral 𝑦𝑑𝐴 defines the first moment of 𝐴 with
respect to the 𝑥 axis and is denoted by 𝑄𝑥. That is,
𝑄𝑦 = 𝑥𝑑𝐴 𝑄𝑥 = 𝑦𝑑𝐴 ….. (3.17)
Comparing Equations (3.12) with Equations (3.17), we note that we
can express the first moments of the area 𝐴 as the products of the
area and the coordinates of its centroid:
𝑄𝑦 = 𝑥𝐴 𝑄𝑥 = 𝑦𝐴 ….. (3.18)

3.1.4 Composite Plates and Wires
𝑋 𝑊1 + 𝑊2 + ⋯ . . 𝑊𝑛 = 𝑥1𝑊1 + 𝑥2𝑊2 + ⋯ . . +𝑥𝑛𝑊𝑛
𝑀𝑦 :
….. (3.19)
𝑌 𝑊1 + 𝑊2 + ⋯ . . 𝑊𝑛 = 𝑦1
𝑊1 + 𝑦2
𝑊2 + ⋯ . . +𝑦𝑛
𝑊𝑛
𝑀𝑥 :
….. (3.20)
𝑋 =
𝑥𝑊
𝑊
𝑌 =
𝑦𝑊
𝑊
….. (3.21)

If the plate is homogeneous and of uniform thickness, the center of
gravity coincides with the centroid 𝐶 of its area.
𝑄𝑦 = 𝑋 𝐴1 + 𝐴2 + ⋯ . . + 𝐴𝑛 = 𝑥1𝐴1 + 𝑥2𝐴2 + ⋯ . . +𝑥𝑛𝐴𝑛
𝑀𝑦 :
….. (3.22)
𝑄𝑥 = 𝑌 𝐴1 + 𝐴2 + ⋯ . . + 𝐴𝑛 = 𝑦1
𝐴1 + 𝑦2
𝐴2 + ⋯ . . +𝑦𝑛
𝐴𝑛
𝑀𝑥 :
….. (3.23)
𝑋 =
𝑥𝐴
𝐴
𝑌 =
𝑦𝐴
𝐴
….. (3.24)
Again, in shorter form,
𝑄𝑦 = 𝑋 𝐴 = 𝑥𝐴 𝑄𝑥 = 𝑌 𝐴 = 𝑦𝐴 ….. (3.25)

Similarly, it is possible in many cases to determine the center of
gravity of a composite wire or the centroid of a composite line by
dividing the wire or line into simpler elements.

Review Questions
1. Define center of gravity
2. Write down the expressions for coordinates of center of gravity
of a two-dimensional surface (plate).
of a two-dimensional line.
4. Write down the expressions for coordinates of centroid of a
two-dimensional surface (plate).
two-dimensional line.
6. Define first moment or areas and lines.

of a composite surface (plate).
of a composite line.
composite surface (plate).
composite line.

Mechanical I/I
2024
CENTROIDS
Session 20: Examples for Planar Centers of
Gravity and Centroids

Learning Objectives
• Determine centroid of two-dimensional surfaces and lines.

Example 6.1
Locate the centroid of the plane area shown in Figure E6.1.
Figure E6.1
𝟏𝟗. 𝟐𝟕 𝒎𝒎, 𝟐𝟔. 𝟔 𝒎𝒎

Example 6.2
Determine the centroid of the area shown in Figure E6.2.
Figure E6.2
𝟏𝟐𝟐 𝒎𝒎, 𝟎

Example 6.3
The shape shown in Figure E6.3 is made from a piece of thin,
homogeneous wire. Determine the location of its center of
gravity.
Figure E6.3
𝟏𝟎 𝒎𝒎, 𝟑 𝒎𝒎

Example 6.4
The frame shown in Figure E6.4 for a sign is fabricated from
thin, flat steel bar stock of mass per unit length 𝟒. 𝟕𝟑 𝒌𝒈/𝒎.
The frame is supported by a pin at 𝑪 and by a cable 𝑨𝑩.
Determine (a) the tension in the cable, (b) the reaction at 𝑪.
Figure E6.4 𝟏𝟐𝟓. 𝟑 𝑵, 𝟏𝟑𝟕. 𝟎𝟑 𝟓𝟔. 𝟕𝟎

Mechanical I/I
2024
CENTROIDS
Session 21: Determination of Centroids by
Integration

• Determine centroid of a two-dimensional surface by direct
integration.
Learning Objectives
• Determine centroid of a two-dimensional line by direct integration.

3.2 Determination of Centroids by Integration
For an area bounded by analytical curves (i.e., curves defined by
algebraic equations), we usually determine the centroid by
evaluating the integrals in Equations (3.13):
𝑥 =
𝑥𝑑𝐴
𝐴
𝑦 =
𝑦𝑑𝐴
𝐴
….. (3.13)
If the element of area 𝑑𝐴 is a small rectangle of sides 𝑑𝑥 and 𝑑𝑦,
evaluating each of these integrals requires a double integration with
respect to 𝑥 and 𝑦. A double integration is also necessary if we use
polar coordinates for which 𝑑𝐴 is a small element with sides 𝑑𝑟
and 𝑟𝑑𝜃.

In most cases, however, it is possible to determine the coordinates
of the centroid of an area by performing a single integration. We can
achieve this by choosing 𝑑𝐴 to be a thin rectangle or strip, or it can
be a thin sector or pie-shaped element (Figure).

Then we obtain the coordinates of the centroid of the area under
consideration by setting the first moment of the entire area with
respect to each of the coordinate axes equal to the sum (or integral)
of the corresponding moments of the elements of the area. Denoting
the coordinates of the centroid of the element 𝑑𝐴 by 𝑥𝑒𝑙 and 𝑦𝑒𝑙
, we
have
𝑄𝑦 = 𝑥𝐴 = 𝑥𝑒𝑙𝑑𝐴 𝑄𝑥 = 𝑦𝐴 = 𝑦𝑒𝑙
𝑑𝐴 ….. (3.26)
When a line is defined by an algebraic equation, you can determine
its centroid by evaluating the integrals in Equations (3.16):
𝑥 =
𝑥𝑑𝐿
𝐿
𝑦 =
𝑦𝑑𝐿
𝐿
….. (3.16)

You can replace the differential length 𝑑𝐿 with one of the following
expressions, depending upon which coordinate, 𝑥, 𝑦, or 𝜃, is chosen
as the independent variable in the equation used to define the line
(these expressions can be derived using the Pythagorean theorem):
𝑑𝐿 = 1 +
𝑑𝑥
𝑑𝑦
2
𝑑𝑦
𝑑𝐿 = 1 +
𝑑𝑦
𝑑𝑥
2
𝑑𝑥 𝑑𝐿 = 𝑟2 +
𝑑𝑟
𝑑𝜃
2
𝑑𝜃
….. (3.27)

Review Questions
1. How can we determine centroid of a two-dimensional surface
by direct integration?
2. How can we determine centroid of a two-dimensional line by
direct integration?

Mechanical I/I
2024
CENTROIDS
Session 22: Examples for Determination of
Centroids by Integration

Example 6.5
Locate the centroid of the area shown in Figure E6.5.
𝟎. 𝟕𝟓 𝒎, 𝟎. 𝟑 𝒎
Figure E6.5

Example 6.6
Locate the centroid for the area of a quarter circle shown in
Figure E6.6.
𝟒𝑹/𝟑𝝅, 𝟒𝑹/𝟑𝝅
Figure E6.6

Example 6.7
Locate the centroid of the shaded area shown in Figure E6.7.
𝟏. 𝟓 𝒎, 𝟐. 𝟒 𝒎
Figure E6.7

Example 6.8
Determine by direct integration the centroid of the area shown
in Figure E6.8. Express your answer in terms of 𝒂 and 𝒉.
𝒂/𝟐, 𝟐𝒉/𝟓
Figure E6.8

Example 6.9
Determine the location of the centroid of the circular arc shown
in Figure E6.9.
𝒓𝒔𝒊𝒏𝜶/𝜶
Figure E6.9

Example 6.10
Locate the centroid of the rod bent into the shape of a parabolic
arc as shown in Figure E6.10.
𝟎. 𝟒𝟏𝟎 𝒎, 𝟎. 𝟓𝟕𝟒 𝒎
Figure E6.10

Mechanical I/I
2024
CENTROIDS
Session 23: Theorems of Pappus-Guldinus

• State and prove Theorems of Pappus-Guldinus.
Learning Objectives
• Know the applications of Theorems of Pappus-Guldinus.

3.3 Theorems of Pappus-Guldinus
A surface of revolution is a surface that can be generated by
rotating a plane curve about a fixed axis.
A body of revolution is a body that can be generated by rotating a
plane area about a fixed axis.

Theorem I. The area of a surface of revolution is equal to the length
of the generating curve times the distance travelled by the centroid
of the curve while the surface is being generated.
Proof
Consider an element 𝑑𝐿 of the line 𝐿 (Figure)
that is revolved about the 𝑥 axis.
The circular strip generated by the element 𝑑𝐿
has an area 𝑑𝐴 equal to 2𝜋𝑦 𝑑𝐿.
Thus, the entire area generated by 𝐿 is 𝐴 = 2𝜋𝑦 𝑑𝐿.
Recall our earlier result that the integral 𝑦 𝑑𝐿 is equal to 𝑦𝐿.
Therefore, we have
𝐴 = 2𝜋𝑦𝐿 ….. (3.28)

Theorem II. The volume of a body of revolution is equal to the
generating area times the distance travelled by the centroid of the
area while the body is being generated.
Proof
Consider an element 𝑑𝐴 of the area 𝐴 that
is revolved about the 𝑥 axis (Figure).
The circular ring generated by the element
𝑑𝐴 has a volume 𝑑𝑉 equal to 2𝜋𝑦 𝑑𝐴.
Thus, the entire volume generated by 𝐴 is 𝑉 = 2𝜋𝑦 𝑑𝐴 and since
we showed earlier that the integral 𝑦 𝑑𝐴 is equal to 𝑦𝐴, we have
𝑉 = 2𝜋𝑦𝐴 ….. (3.29)

The theorems of Pappus-Guldinus offer a simple way to compute
the areas of surfaces of revolution and the volumes of bodies of
revolution.
Conversely, they also can be used to determine the centroid of a
plane curve if you know the area of the surface generated by the
curve or to determine the centroid of a plane area if you know the
volume of the body generated by the area.

Review Questions
1. State and prove Theorems of Pappus-Guldinus.
2. What are applications of Theorems of Pappus-Guldinus?

Mechanical I/I
2024
CENTROIDS
Session 24: Examples for Theorems of
Pappus-Guldinus

Learning Objectives
• Apply Theorems of Pappus-Guldinus.

Example 7.1
Determine the surface area and volume of the full solid in
Figure E7.1.
Figure E7.1
𝟏𝟒𝟐𝟗𝟎 𝒎𝒎𝟐, 𝟒𝟕𝟔𝟒𝟖 𝒎𝒎𝟑

Mechanical I/I
2024
CENTROIDS
Session 25: Centers of Gravity and
Centroids of Volumes

Learning Objectives
• Derive expressions for coordinates of center of gravity and
centroid of a three-dimensional body.
centroid of a composite three-dimensional body.
centroid of a three-dimensional body by direct integration.

3.4 Centers of Gravity and Centroids of Volumes
So far in this chapter, we have dealt with finding centres of gravity
and centroids of two-dimensional areas and objects such as flat
plates and plane surfaces.
However, the same ideas apply to three-dimensional objects as
well. The most general situations require the use of multiple
integration for analysis, but we can often use symmetry
considerations to simplify the calculations.

3.4.1 Three-Dimensional Centres of Gravity and Centroids
−𝑊𝒋 = −∆𝑊𝒋
𝒓 × −𝑊𝒋 = 𝒓 × −∆𝑊𝒋
….. (3.30)
….. (3.31)
𝑭 :
𝑴𝑶 :
𝒓𝑊 × −𝒋 = 𝒓∆𝑊 × −𝒋 ….. (3.32)

From these equations, we can see that the weight 𝑾 of the body is
equivalent to the system of the elemental weights ∆𝑾 if the
following conditions are satisfied:
𝒓𝑊 = 𝒓∆𝑊 ….. (3.33)
𝑊 = ∆𝑊
Increasing the number of elements and simultaneously decreasing
the size of each element, we obtain in the limit as
𝒓𝑊 = 𝒓𝑑𝑊 ….. (3.34)
𝑊 = 𝑑𝑊
Resolving the vectors 𝒓 and 𝒓 into rectangular components, we
note that the second of the relations in Equations (3.34) is
equivalent to the three scalar equations
𝑦𝑊 = 𝑦𝑑𝑊 ….. (3.35)
𝑥𝑊 = 𝑥𝑑𝑊 𝑧𝑊 = 𝑧𝑑𝑊

Or,
𝑦 =
𝑦𝑑𝑊
𝑊
….. (3.36)
𝑥 =
𝑥𝑑𝑊
𝑊 𝑧 =
𝑧𝑑𝑊
𝑊
If the body is made of a homogeneous material of specific weight
𝛾, We obtain
𝑦 =
𝑦𝑑𝑉
𝑉
….. (3.37)
𝑥 =
𝑥𝑑𝑉
𝑉 𝑧 =
𝑧𝑑𝑉
𝑉
The integral 𝑥𝑑𝑉 is known as the first moment of the volume
with respect to the 𝑦𝑧 plane. Similarly, the integrals 𝑦𝑑𝑉 and
𝑧𝑑𝑉 define the first moments of the volume with respect to the 𝑧𝑥
plane and 𝑥𝑦 plane.

3.4.2 Composite Bodies
𝑋 =
𝑥𝑊
𝑊
….. (3.38)
𝑌 =
𝑦𝑊
𝑊
𝑍 =
𝑧𝑊
𝑊
If the body is made of a homogeneous material, its center of gravity
coincides with the centroid of its volume, and we obtain
𝑌 =
𝑦𝑉
𝑉
….. (3.39)
𝑋 =
𝑥𝑉
𝑉
𝑍 =
𝑧𝑉
𝑉

3.4.3 Determination of Centroids of Volumes by Integration
We can determine the centroid of a volume bounded by analytical
surfaces by evaluating the integrals given earlier in this section:
𝑦𝑉 = 𝑦𝑑𝑉 ….. (3.40)
𝑥𝑉 = 𝑥𝑑𝑉 𝑧𝑉 = 𝑧𝑑𝑉
If we choose the element of volume 𝑑𝑉 to
be equal to a small cube with sides 𝑑𝑥, 𝑑𝑦,
and 𝑑𝑧, the evaluation of each of these
integrals requires a triple integration.
However, it is possible to determine the
coordinates of the centroid of most volumes
by double integration if we choose 𝑑𝑉 to be
equal to the volume of a thin filament
(Figure).

We then obtain the coordinates of the centroid of the volume by
rewriting Equations (3.40) as
𝑦𝑉 = 𝑦𝑒𝑙
𝑑𝑉 ….. (3.41)
𝑥𝑉 = 𝑥𝑒𝑙𝑑𝑉 𝑧𝑉 = 𝑧𝑒𝑙𝑑𝑉
Then we substitute the expressions given in
Figure for the volume 𝑑𝑉 and the
coordinates 𝑥𝑒𝑙 , 𝑦𝑒𝑙
, 𝑧𝑒𝑙 . By using the
equation of the surface to express 𝑧 in terms
of 𝑥 and 𝑦, we reduce the integration to a
double integration in 𝑥 and 𝑦.

If the volume under consideration possesses two planes of
symmetry, its centroid must be located on the line of intersection of
the two planes. Choosing the 𝑥 axis to lie along this line, we have
𝑦 = 𝑧 = 0
and the only coordinate to determine is 𝑥. This can be done with a
single integration by dividing the given volume into thin slabs
parallel to the 𝑦𝑧 plane and expressing 𝑑𝑉 in terms of 𝑥 and 𝑑𝑥 in
the equation
….. (3.42)
𝑥𝑉 = 𝑥𝑒𝑙𝑑𝑉

Review Questions
and centroid of a three-dimensional body.
and centroid of a composite three-dimensional body.
3. How can we determine centroid of a three-dimensional body by
direct integration?

Mechanical I/I
2024
CENTROIDS
Session 26: Examples for Centers of
Gravity and Centroids of Volumes

Learning Objectives
• Determine center of gravity and centroid of a composite three-
dimensional body.
• Determine center of gravity and centroid of a three-dimensional
body by direct integration.

Example 7.2
Determine the distance 𝒙 to the centroid of the solid which
consists of a cylinder with a hole of length 𝒉 = 𝟓𝟎 𝒎𝒎 bored
into its base as shown in Figure E7.2.
Figure E7.2
𝟔𝟒. 𝟏 𝒎𝒎

Example 7.3
For the machine element shown in Figure E7.3, locate the 𝒙 and
𝒚 coordinates of the center of gravity.
Figure E7.3
𝟔𝟐. 𝟏 𝒎𝒎, 𝟏𝟗. 𝟏𝟑 𝒎𝒎

Example 7.4
Locate the center of gravity of the volume shown in Figure
E7.4. The material is homogeneous.
Figure E7.4
𝟎, 𝟎, 𝟏. 𝟑𝟑 𝒎

Example 7.5
Locate the centroid of the volume obtained by rotating the
shaded area shown in Figure E7.5 about the 𝒙 axis.
Figure E7.5
𝟐. 𝟑𝟒 𝒎, 𝟎, 𝟎

Example 7.6
Locate the centroid of the wire shown in Figure E7.6.
Figure E7.6
𝟒𝟓. 𝟓 𝒎𝒎, − 𝟐𝟐. 𝟓, − 𝟎. 𝟖𝟎𝟓 𝒎𝒎

Mechanical I/I
2024
MOMENT OF INERTIA
Session 27: Moments of Inertia of Areas

Learning Objectives
• Know the need of moment of inertia of an area.
• Derive expressions for moment of inertias of an area with
respect to rectangular coordinate axes.
• Derive expression for polar moment of inertia of an area.
• Derive expressions for radii of gyration of an area.

In previous Chapter, we analyzed various systems of forces
distributed over an area or volume. The three main types of forces
considered were (a) weights of homogeneous plates of uniform
thickness and (b) weights of homogeneous three-dimensional
bodies.
In all of these cases, the distributed forces were proportional to the
elemental areas or volumes associated with them. Therefore, we
could obtain the resultant of these forces by summing the
corresponding areas or volumes, and we determined the moment of
the resultant about any given axis by computing the first moments
of the areas or volumes about that axis.

In the first part of this chapter, we consider distributed forces ∆𝑭
where the magnitudes depend not only upon the elements of area
∆𝐴 on which these forces act but also upon the distance from ∆𝐴 to
some given axis. More precisely, we assume the magnitude of the
force per unit area ∆𝑭/∆𝐴 varies linearly with the distance to the
axis. Forces of this type arise in the study of the bending of beams
and in problems involving submerged nonrectangular surfaces.
Starting with the assumption that the elemental forces involved are
distributed over an area 𝐴 and vary linearly with the distance 𝑦 to
the 𝑥 axis, we will show that the magnitude of their resultant 𝑹
depends upon the first moment 𝑄𝑥 of the area 𝐴. However, the
location of the point where 𝑹 is applied depends upon the second
moment, or moment of inertia, 𝐼𝑥 of the same area with respect to
the 𝑥 axis.

You will see how to compute the moments of inertia of various
areas with respect to given 𝑥 and 𝑦 axes. We also introduce the
polar moment of inertia 𝐽0 of an area. To facilitate these
computations, we establish a relation between the moment of
inertia 𝐼𝑥 of an area 𝐴 with respect to a given x axis and the
moment of inertia 𝐼𝑥′ of the same area with respect to the parallel
centroidal 𝑥′ axis (a relation known as the parallel-axis theorem).
In the second part of this chapter, we will explain how to determine
the moments of inertia of various masses with respect to a given
axis. Moments of inertia of masses are common in dynamics
problems involving the rotation of a rigid body about an axis. To
facilitate the computation of mass moments of inertia, we introduce
another version of the parallel-axis theorem.

4.1 Moments of Inertia of Areas
In the first part of this chapter, we consider distributed forces ∆𝑭
whose magnitudes ∆𝑭 are proportional to the elements of area ∆𝐴
on which the forces act and, at the same time, vary linearly with the
distance from ∆𝐴 to a given axis.
4.1.1 Second Moment or Moment of Inertia of an Area
The magnitude of the resultant 𝑹 of the
elemental forces ∆𝑭 that act over the
entire section is
𝑅 = 𝑘𝑦𝑑𝐴 = 𝑘 𝑦𝑑𝐴 ….. (4.1)

You might recognize this last integral as
the first moment 𝑄𝑥 of the section about
the 𝑥 axis; it is equal to 𝑦𝐴 and is thus
equal to zero, since the centroid of the
section is located on the 𝑥 axis. The
system of forces ∆𝑭 thus reduces to a
couple.
𝑀 = 𝑘𝑦2𝑑𝐴 = 𝑘 𝑦2𝑑𝐴 ….. (4.2)
The magnitude 𝑀 of this couple (bending moment) must be equal
to the sum of the moments ∆𝑀𝑥 = 𝑦∆𝐹 = 𝑘𝑦2∆𝐴 of the elemental
forces. Integrating over the entire section, we obtain

This last integral is known as the second moment, or moment of
inertia, of the beam section with respect to the 𝑥 axis and is denoted
by 𝐼𝑥. We obtain it by multiplying each element of area 𝑑𝐴 by the
square of its distance from the 𝑥 axis and integrating over the beam
section. Since each product 𝑦2𝑑𝐴 is positive, regardless of the sign
of 𝑦, or zero (if 𝑦 is zero), the integral 𝐼𝑥 is always positive.

We just defined the second moment, or moment of inertia, 𝐼𝑥 of an
area 𝐴 with respect to the 𝑥 axis. In a similar way, we can also
define the moment of inertia 𝐼𝑦 of the area 𝐴 with respect to the 𝑦
axis (Figure):
𝐼𝑦 = 𝑥2𝑑𝐴 ….. (4.3)
𝐼𝑥 = 𝑦2𝑑𝐴
4.1.2 Determining the Moment of Inertia of an Area by
Integration

Moment of Inertia of a Rectangular Area
𝑑𝐼𝑥 = 𝑦2𝑑𝐴 = 𝑦2𝑏𝑑𝑦
….. (4.4)
𝑑𝐴 = 𝑏𝑑𝑦
𝐼𝑥 = 𝑦2𝑏𝑑𝑦
ℎ
0
=
1
3
𝑏ℎ3
Computing 𝑰𝒙 and 𝑰𝒚 Using the Same Elemental Strips
….. (4.5)
𝑑𝐼𝑥 =
1
3
𝑦3𝑑𝑥
….. (4.6)
𝑑𝐼𝑦 = 𝑥2𝑑𝐴 = 𝑥2𝑦𝑑𝑥
Setting 𝑏 = 𝑑𝑥 and ℎ = 𝑦 in formula (4.4), we
obtain

4.1.3 Polar Moment of Inertia
An integral of great importance in problems
concerning the torsion of cylindrical shafts and
in problems dealing with the rotation of slabs is
….. (4.7)
𝐽𝑂 = 𝑟2𝑑𝐴
where 𝑟 is the distance from 𝑂 to the element of area 𝑑𝐴 (Figure).
This integral is called the polar moment of inertia of the area 𝐴 with
respect to the “pole” 𝑂.
Noting that 𝑟2
= 𝑥2
+ 𝑦2
, we have
….. (4.8)
𝐽𝑂 = 𝑟2
𝑑𝐴 = 𝑥2
+ 𝑦2
𝑑𝐴 = 𝑦2
𝑑𝐴 + 𝑥2
𝑑𝐴 = 𝐼𝑥 + 𝐼𝑦

4.1.4 Radius of Gyration of an Area
Consider an area 𝐴 that has a moment of inertia 𝐼𝑥 with respect to
the 𝑥 axis (Figure a). Imagine that we concentrate this area into a
thin strip parallel to the 𝑥 axis (Figure b).
….. (4.9)
𝐼𝑥 = 𝑘𝑥
2
𝐴
If the concentrated area 𝐴 is to have the same moment of inertia
with respect to the 𝑥 axis, the strip should be placed at a distance 𝑘𝑥
from the 𝑥 axis, where 𝑘𝑥 is defined by the relation
Figure (a) Figure (b)

Solving for 𝑘𝑥, we have
….. (4.10)
𝑘𝑥 =
𝐼𝑥
𝐴
The distance 𝑘𝑥 is referred to as the radius of
gyration of the area with respect to the 𝑥 axis.
Figure (a)
Figure (c)
Figure (d)
In a similar way, we can define the radii of
gyration 𝑘𝑦 and 𝑘𝑂 (Figure c and d); we have
𝐼𝑦 = 𝑘𝑦
2
𝐴 ….. (4.11)
𝑘𝑦 =
𝐼𝑦
𝐴
𝐽𝑂 = 𝑘𝑂
2
𝐴 ….. (4.12)
𝑘𝑂 =
𝐽𝑂
𝐴

Review Questions
1. Why moment of inertia of area is necessary?
2. Write down the expressions for moment of inertias of an area
with respect to rectangular coordinate axes.
3. How can we determine moments of inertia of an area with
respect to rectangular coordinate axes by using the same strip?
4. Write down the expression for the polar moment of inertia of an
area.
5. Write down the expressions for radii of gyration of an area.
6. What is the meaning of radius of gyration of an area?

Mechanical I/I
2024
MOMENT OF INERTIA
Session 28: Parallel-Axis Theorem and
Composite Areas

Learning Objectives
• State and derive parallel axis theorem.
• Determine moments of inertia of composite areas.

4.2 Parallel-Axis Theorem and Composite Areas
In practice, we often need to determine the moment of inertia of a
complicated area that can be broken down into a sum of simple
areas. However, in doing these calculations, we have to determine
the moment of inertia of each simple area with respect to the same
axis.
In this section, we first derive a formula for computing the moment
of inertia of an area with respect to a centroidal axis parallel to a
given axis. Then we show how you can use this formula for finding
the moment of inertia of a composite area.

4.2.4 The Parallel-Axis Theorem
Consider the moment of inertia 𝐼 of an area 𝐴 with respect to an
axis 𝐴𝐴′ (Figure). We denote the distance from an element of area
𝑑𝐴 to 𝐴𝐴′ by 𝑦. This gives us
….. (4.9)
𝐼 = 𝑦2𝑑𝐴
Let us now draw through the centroid 𝐶 of
the area an axis 𝐵𝐵′ parallel to 𝐴𝐴′; this
axis is called a centroidal axis.
Denoting the distance from the element 𝑑𝐴 to 𝐵𝐵′ by 𝑦′, we have
𝑦 = 𝑦′ + 𝑑, where 𝑑 is the distance between the axes 𝐴𝐴′ and 𝐵𝐵′.
Substituting for 𝑦 in the previous integral, we obtain
….. (4.10)
𝐼 = 𝑦2
𝑑𝐴 = 𝑦′
+ 𝑑 2
𝑑𝐴 = 𝑦′2
𝑑𝐴 + 2𝑑 𝑦′𝑑𝐴 + 𝑑2
𝑑𝐴

The first integral represents the moment of inertia 𝐼 of the area with
respect to the centroidal axis 𝐵𝐵′. The second integral represents
the first moment of the area with respect to 𝐵𝐵′, but since the
centroid 𝐶 of the area is located on this axis, the second integral
must be zero. The last integral is equal to the total area 𝐴.
Therefore, we have
….. (4.11)
𝐼 = 𝐼 + 𝐴𝑑2
This formula states that the moment of inertia 𝐼 of an area with
respect to any given axis 𝐴𝐴′ is equal to the moment of inertia 𝐼 of
the area with respect to a centroidal axis 𝐵𝐵′ parallel to 𝐴𝐴′ plus
the product of the area 𝐴 and the square of the distance 𝑑 between
the two axes. This theorem is known as the parallel-axis theorem.

Substituting 𝑘2𝐴 for 𝐼 and 𝑘2𝐴 for 𝐼 , we can also express this
theorem as
….. (4.12)
𝑘2 = 𝑘2 + 𝑑2
A similar theorem relates the polar moment of inertia 𝐽𝑂 of an area
about a point 𝑂 to the polar moment of inertia 𝐽𝐶 of the same area
about its centroid 𝐶. Denoting the distance between 𝑂 and 𝐶 by 𝑑,
we have
….. (4.13)
𝐽𝑂 = 𝐽𝐶 + 𝐴𝑑2
….. (4.14)
𝑘𝑂
2
= 𝑘𝐶
2
+ 𝑑2

Consider a composite area 𝐴 made of several component areas 𝐴1,
𝐴2, 𝐴3, . . . . The integral representing the moment of inertia of 𝐴
can be subdivided into integrals evaluated over 𝐴1, 𝐴2, 𝐴3, . . . .
Therefore, we can obtain the moment of inertia of 𝐴 with respect to
a given axis by adding the moments of inertia of the areas 𝐴1, 𝐴2,
𝐴3, . . . with respect to the same axis.
4.2.2 Moments of Inertia of Composite Areas

Review Questions
1. State and derive parallel axis theorem.
2. How can we determine moments of inertia of composite areas.

Mechanical I/I
2024
MOMENT OF INERTIA
Session 29: Mass Moments of Inertia

Learning Objectives
• Define mass moment of inertia. Know the necessity of mass
moment of inertia
• Determine mass moments of inertia of thin plates.
• State and derive parallel axis theorem for mass moment of
inertia.
• Determine mass moment of inertia of a three-dimensional body
by direct integration.
• Determine mass moments of inertia of composite areas.

4.3 Mass Moments of Inertia
So far in this chapter, we have examined moments of inertia of
areas. In the rest of this chapter, we consider moments of inertia
associated with the masses of bodies. This will be an important
concept in dynamics when studying the rotational motion of a rigid
body about an axis.
4.3.1 Moment of Inertia of a Simple Mass
Consider a small mass ∆𝑚 mounted on a rod of
negligible mass that can rotate freely about an
axis 𝐴𝐴′ (Figure a). If we apply a couple to the
system, the rod and mass (assumed to be initially
at rest) start rotating about 𝐴𝐴′.

At present, we wish to indicate only that the time required for the
system to reach a given speed of rotation is proportional to the mass
∆𝑚 and to the square of the distance 𝑟. The product 𝑟2∆𝑚 thus
provides a measure of the inertia of the system; i.e., a measure of
the resistance the system offers when we try to set it in motion. For
this reason, the product 𝑟2∆𝑚 is called the moment of inertia of the
mass ∆𝑚 with respect to axis 𝐴𝐴′.

Now suppose a body of mass m is to be rotated about an axis 𝐴𝐴′
(Figure b). Dividing the body into elements of mass ∆𝑚1, ∆𝑚2,
etc., we find that the body’s resistance to being rotated is measured
by the sum 𝑟1
2
∆𝑚1 + 𝑟2
2
∆𝑚2 + ⋯. This sum defines the moment
of inertia of the body with respect to axis 𝐴𝐴′. Increasing the
number of elements, we find that the moment of inertia is equal, in
the limit, to the integral
….. (4.15)
𝐼 = 𝑟2𝑑𝑚
We define the radius of gyration 𝑘 of the body with respect to axis
𝐴𝐴′ by the relation
𝐼 = 𝑘2
𝑚 ….. (4.16)
𝑘 =
𝐼
𝑚

The radius of gyration 𝑘 represents the distance at which the entire
mass of the body should be concentrated if its moment of inertia
with respect to 𝐴𝐴′ is to remain unchanged (Figure c). Whether it
stays in its original shape (Figure b) or is concentrated as shown in
Figure (c), the mass 𝑚 reacts in the same way to a rotation (or
gyration) about 𝐴𝐴′.

We can express the moment of inertia of a body with respect to a
coordinate axis in terms of the coordinates 𝑥, 𝑦, 𝑧 of the element of
mass 𝑑𝑚 (Figure). Noting, for example, that the square of the
distance 𝑟 from the element 𝑑𝑚 to the 𝑦 axis is 𝑧2
+ 𝑥2
, the
moment of inertia of the body with respect to the 𝑦 axis is
….. (4.17)
𝐼𝑦 = 𝑟2𝑑𝑚 = (𝑧2 + 𝑥2)𝑑𝑚
We obtain similar expressions for the
moments of inertia with respect to the 𝑥 and
𝑧 axes.
….. (4.18)
𝐼𝑥 = (𝑦2 + 𝑧2)𝑑𝑚
….. (4.19)
𝐼𝑧 = (𝑥2 + 𝑦2)𝑑𝑚

Consider again a body of mass 𝑚 and let 𝑂𝑥𝑦𝑧 be a system of
rectangular coordinates whose origin is at the arbitrary point 𝑂. Let
𝐺𝑥′𝑦′𝑧′ be a system of parallel centroidal axes; i.e., a system whose
origin is at the center of gravity 𝐺 of the body and whose axes 𝑥′,
𝑦′, 𝑧′ are parallel to the 𝑥, 𝑦, and 𝑧 axes, respectively (Figure).
4.3.2 Parallel-Axis Theorem for Mass Moments of Inertia
We denote by 𝑥, 𝑦, and 𝑧 the coordinates of 𝐺
with respect to 𝑂𝑥𝑦𝑧 . Then we have the
following relations between the coordinates 𝑥,
𝑦, 𝑧 of the element 𝑑𝑚 with respect to 𝑂𝑥𝑦𝑧
and its coordinates 𝑥′, 𝑦′, 𝑧′ with respect to the
centroidal axes 𝐺𝑥′𝑦′𝑧′:
𝑥 = 𝑥′ + 𝑥 𝑦 = 𝑦′ + 𝑦 𝑧 = 𝑧′ + 𝑧
….. (4.20)

Referring to Equation (4.18), we can express the moment of inertia
of the body with respect to the 𝑥 axis as
𝐼𝑥 = (𝑦2
+ 𝑧2
)𝑑𝑚 = 𝑦′
+ 𝑦
2
+ 𝑧′
+ 𝑧
2
𝑑𝑚
= (𝑦′
2
+ 𝑧′
2
)𝑑𝑚 + 2𝑦 𝑦′𝑑𝑚 + 2𝑧 𝑧′𝑑𝑚 + (𝑦2
+ 𝑧2
) 𝑑𝑚
he first integral in this expression represents the moment of inertia
𝐼𝑥′ of the body with respect to the centroidal axis 𝑥′. The second
and third integrals represent the first moment of the body with
respect to the 𝑧′𝑥′ and 𝑥′𝑦′ planes, respectively, and since both
planes contain 𝐺, these two integrals are zero. The last integral is
equal to the total mass 𝑚 of the body. Therefore, we have
….. (4.21)
𝐼𝑥 = 𝐼𝑥′ + 𝑚(𝑦2
+ 𝑧2
)

Similarly,
….. (4.22)
𝐼𝑦 = 𝐼𝑦′ + 𝑚(𝑧2
+ 𝑥2
)
….. (4.23)
𝐼𝑧 = 𝐼𝑧′ + 𝑚(𝑥2
+ 𝑦2
)
We easily verify from Figure that the sum
𝑧2
+ 𝑥2
represents the square of the
distance 𝑂𝐵 between the 𝑦 and 𝑦′ axes.
Similarly, 𝑦2
+ 𝑧2
and 𝑥2
+ 𝑦2
represent
the squares of the distance between the 𝑥
and 𝑥′ axes and the 𝑧 and 𝑧′ axes,
respectively.

We denote the distance between an arbitrary axis 𝐴𝐴′ and a parallel
centroidal axis 𝐵𝐵′ by 𝑑 (Figure). Then the general relation
between the moment of inertia 𝐼 of the body with respect to 𝐴𝐴′
and its moment of inertia 𝐼 with respect to 𝐵𝐵′, known as the
parallel-axis theorem for mass moments of inertia, is
….. (4.24)
𝐼 = 𝐼 + 𝑚𝑑2
Expressing the moments of inertia in terms of
the corresponding radii of gyration, we can also
write
….. (4.25)
𝑘2
= 𝑘
2
+ 𝑑2
where 𝑘 and 𝑘 represent the radii of gyration of the body about 𝐴𝐴′
and 𝐵𝐵′, respectively.

Now imagine a thin plate of uniform thickness 𝑡, made of a
homogeneous material of density 𝜌 (density = mass per unit
volume). The mass moment of inertia of the plate with respect to an
axis 𝐴𝐴′ contained in the plane of the plate (Figure a) is
4.3.3 Moments of Inertia of Thin Plates
….. (4.25)
𝐼𝐴𝐴′
,𝑚𝑎𝑠𝑠
= 𝑟2𝑑𝑚

Since 𝑑𝑚 = 𝜌𝑡𝑑𝐴, we have
….. (4.26)
𝐼𝐴𝐴′
,𝑚𝑎𝑠𝑠
= 𝜌𝑡 𝑟2𝑑𝐴
However, 𝑟 represents the distance of the element of area 𝑑𝐴 to the
axis 𝐴𝐴′. Therefore, the integral is equal to the moment of inertia of
the area of the plate with respect to 𝐴𝐴′.
….. (4.27)
𝐼𝐴𝐴′
,𝑚𝑎𝑠𝑠
= 𝜌𝑡 𝐼𝐴𝐴′
,𝑎𝑟𝑒𝑎
Similarly, for an axis 𝐵𝐵′ that is contained in the plane of the plate
and is perpendicular to 𝐴𝐴′ (Figure b), we have
….. (4.28)
𝐼𝐵𝐵′
,𝑚𝑎𝑠𝑠
= 𝜌𝑡 𝐼𝐵𝐵′,𝑎𝑟𝑒𝑎
Consider now the axis 𝐶𝐶′, which is perpendicular to the plate and
passes through the point of intersection 𝐶 of 𝐴𝐴′ and 𝐵𝐵′ (Figure c).
This time we have
….. (4.29)
𝐼𝐶𝐶′
,𝑚𝑎𝑠𝑠
= 𝜌𝑡 𝐽𝐶,𝑎𝑟𝑒𝑎

Recall the relation 𝐽𝐶 = 𝐼𝐴𝐴′ + 𝐼𝐵𝐵′ between the polar and
rectangular moments of inertia of an area. We can use this to write
the relation between the mass moments of inertia of a thin plate as
….. (4.30)
𝐼𝐶𝐶′ = 𝐼𝐴𝐴′ + 𝐼𝐵𝐵′
Rectangular Plate
In the case of a rectangular plate of sides 𝑎
and 𝑏 (Figure), we obtain the mass
moments of inertia with respect to axes
through the center of gravity of the plate as
….. (4.31)
𝐼𝐴𝐴′
,𝑚𝑎𝑠𝑠
,𝑎𝑟𝑒𝑎
= 𝜌𝑡
1
12
𝑎3𝑏
….. (4.32)
𝐼𝐵𝐵′
,𝑚𝑎𝑠𝑠
= 𝜌𝑡 𝐼𝐵𝐵′,𝑎𝑟𝑒𝑎 = 𝜌𝑡
1
12
𝑎𝑏3

Since the product 𝜌𝑎𝑏𝑡 is equal to the mass 𝑚 of the plate, we can
also write the mass moments of inertia of a thin rectangular plate as
….. (4.33)
𝐼𝐴𝐴′
,𝑚𝑎𝑠𝑠
=
1
12
𝑚𝑎2
….. (4.34)
𝐼𝐵𝐵′
,𝑚𝑎𝑠𝑠
=
1
12
𝑚𝑏2
….. (4.35)
𝐼𝐶𝐶′ = 𝐼𝐴𝐴′ + 𝐼𝐵𝐵′ =
1
12
𝑚 𝑎2 + 𝑏2
Circular Plate
In the case of a circular plate, or disk, of radius 𝑟
(Figure), Equation (4.27) becomes
….. (4.36)
𝐼𝐴𝐴′
,𝑚𝑎𝑠𝑠
,𝑎𝑟𝑒𝑎
= 𝜌𝑡
𝜋
4
𝑟4

In this case, the product 𝜌𝜋𝑟2𝑡 is equal to the mass 𝑚 of the plate,
and 𝐼𝐴𝐴′ = 𝐼𝐵𝐵′. Therefore, we can write the mass moments of
inertia of a circular plate as
….. (4.37)
𝐼𝐴𝐴′
,𝑚𝑎𝑠𝑠
= 𝐼𝐵𝐵′
,𝑚𝑎𝑠𝑠
=
1
4
𝑚𝑟2
….. (4.38)
𝐼𝐶𝐶′ = 𝐼𝐴𝐴′ + 𝐼𝐵𝐵′ =
1
2
𝑚𝑟2

We obtain the moment of inertia of a three-dimensional body by
evaluating the integral 𝐼 = 𝑟2𝑑𝑚. If the body is made of a
homogeneous material with a density 𝜌, the element of mass 𝑑𝑚 is
equal to 𝜌𝑑𝑉, and we have 𝐼 = 𝜌 𝑟2
𝑑𝑉. This integral depends
only upon the shape of the body. Thus, in order to compute the
moment of inertia of a three-dimensional body, it is generally
necessary to perform a triple, or at least a double, integration.
4.3.4 Determining the Moment of Inertia of a Three-
Dimensional Body by Integration
However, if the body possesses two planes of symmetry, it is
usually possible to determine the body’s moment of inertia with a
single integration. We do this by choosing as the element of mass
𝑑𝑚 a thin slab that is perpendicular to the planes of symmetry.

In the case of bodies of revolution, for example, the element of
mass is a thin disk (Figure). Using formula (4.38), we can express
the moment of inertia of the disk with respect to the axis of
revolution as indicated in Figure. Its moment of inertia with
respect to each of the other two coordinate axes is obtained by
using formula (4.37) and the parallel-axis theorem. Integration of
these expressions yields the desired moment of inertia of the body.

For a body consisting of several of these simple shapes in
combination, you can obtain the moment of inertia of the body with
respect to a given axis by first computing the moments of inertia of
its component parts about the desired axis and then adding them
together.
4.3.5 Moments of Inertia of Composite Bodies

Review Questions
1. Define mass moment of inertia. Why it is necessary?
2. State and derive parallel axis theorem for mass moment of
inertia.
3. Write down a general expression for the mass moment of
inertia of a thin plate. Use it to determine mass moment of
inertia of (a) a rectangular plate and (b) a circular plate.
4. How can we determine mass moment of inertia of a three-
dimensional body by direct integration?
5. How can we determine mass moment of inertia of a composite
three-dimensional body?

Mechanical I/I
2024
MOMENT OF INERTIA
Session 30: Examples for Moment of
Inertia Mass Moment of Inertia

Learning Objectives
• Determine moment of inertia of an area and a mass moment of
inertia of a body.

Example 8.1
Determine the moment of inertia of a triangle shown in Figure
E8.1 with respect to its base.
𝒃𝒉𝟑
/𝟏𝟐

Example 8.2
Determine by direct integration the moments of inertia of the
shaded area shown in Figure E8.2 with respect to the 𝒙 and 𝒚
axes.
𝟒𝟓𝟕 × 𝟏𝟎𝟔
𝒎𝒎𝟒
, 𝟓𝟑. 𝟑 × 𝟏𝟎𝟔
𝒎𝒎𝟒

Example 8.3
Determine by direct integration the moment of inertia of the
shaded area shown in Figure E8.3 with respect to the 𝒚 axis.
𝒓𝟎
𝟒
𝒔𝒊𝒏𝜶 + 𝜶 /𝟖

Example 8.4
Determine the centroidal polar moment of inertia of a circular
area shown in Figure E8.4 by direct integration. Using the
result obtained, determine the moment of inertia of a circular
area with respect to a diameter.
𝝅𝒓𝟒
/𝟐, 𝝅𝒓𝟒
/𝟒

Example 8.5
Determine the polar moment of inertia and the polar radius of
gyration of the shaded area shown in Figure E8.5 with respect
to point 𝑷.
𝟐𝟎𝒂𝟒
, 𝟏. 𝟖𝟐𝟔𝒂

Example 8.6
Determine the moments of inertia and the radii of gyration of
the shaded area shown in Figure E8.6 with respect to the 𝒙 and
𝒚 axes.
𝟑 𝒂𝒃𝟑
/𝟑𝟓, 𝒃 𝟗/𝟑𝟓; 𝟑𝒂𝟑𝒃/𝟑𝟓, 𝒂 𝟗/𝟑𝟓

Example 8.7
Determine the moments of inertia of the composite area shown
in Figure E8.7 about the 𝒙 and 𝒚 axes.
𝟕𝟗𝟖 × 𝟏𝟎𝟔
𝒎𝒎𝟒
, 𝟏𝟎. 𝟑 × 𝟏𝟎𝟗
𝒎𝒎𝟒

Example 8.8
Determine the shaded area and its moment of inertia with
respect to the centroidal axis parallel to 𝑨𝑨′ shown in Figure
E8.8 knowing that 𝒅𝟏 = 𝟐𝟓 𝒎𝒎 and 𝒅𝟐 = 𝟏𝟎 𝒎𝒎 and that its
moments of inertia with respect to 𝑨𝑨′ and 𝑩𝑩′ are 𝟐. 𝟐 ×
𝟏𝟎𝟔
𝒎𝒎𝟒 and 𝟒 × 𝟏𝟎𝟔
𝒎𝒎𝟒, respectively.
𝟑𝟎𝟎𝟎 𝒎𝒎𝟐
, 𝟑𝟐𝟓 × 𝟏𝟎𝟑
𝒎𝒎𝟒

Example 8.9
Determine the moments of inertia 𝑰𝒙 and 𝑰𝒚 of the area shown
in Figure E8.9 with respect to centroidal axes respectively
parallel and perpendicular to side 𝑨𝑩.
𝟏𝟑. 𝟖𝟗 × 𝟏𝟎𝟔
𝒎𝒎𝟒
, 𝟐𝟎. 𝟗 × 𝟏𝟎𝟔
𝒎𝒎𝟒

Example 8.10
A thin plate with a mass 𝒎 is cut in the shape of an equilateral
triangle of side 𝒂 shown in Figure E8.10. Determine the mass
moment of inertia of the plate with respect to (a) the centroidal
axes 𝑨𝑨′ and 𝑩𝑩′ , (b) the centroidal axis 𝑪𝑪′ that is
perpendicular to the plate.
𝒎𝒂𝟐/𝟐𝟒, 𝒎𝒂𝟐/𝟐𝟒, 𝒎𝒂𝟐/𝟏𝟐

Example 8.11
Determine the mass moment of inertia of the cylinder shown in
Figure E8.11 about the 𝒛 axis. The density of the material, 𝝆, is
constant.
𝒎𝑹𝟐
/𝟐

Example 8.12
Determine the mass moment of inertia 𝑰𝒚 of the solid formed by
revolving the shaded area shown in Figure E8.12 around the 𝒚
axis. The total mass of the solid is 𝟏𝟓𝟎𝟎 𝒌𝒈.
𝟏. 𝟕𝟏 × 𝟏𝟎𝟑
𝒌𝒈. 𝒎𝟐

Mechanical I/I
2024
FRICTION
Session 31: Laws of Friction

Learning Objectives
• Know the effects of friction in common engineering
applications.
• State and explain laws of friction.
• Define coefficients of friction.
• Define angles of friction.

In the previous chapters, we assumed that surfaces in contact are
either frictionless or rough. If they are frictionless, the force each
surface exerts on the other is normal to the surfaces, and the two
surfaces can move freely with respect to each other. If they are
rough, tangential forces can develop that prevent the motion of one
surface with respect to the other.
This view is a simplified one. Actually, no perfectly frictionless
surface exists. When two surfaces are in contact, tangential forces,
called friction forces, always develop if you attempt to move one
surface with respect to the other. However, these friction forces are
limited in magnitude and do not prevent motion if you apply
sufficiently large forces.

Thus, the distinction between frictionless and rough surfaces is a
matter of degree. You will see this more clearly in this chapter,
which is devoted to the study of friction and its applications to
common engineering situations.
There are two types of friction: dry friction, sometimes called
Coulomb friction, and fluid friction or viscosity. Fluid friction
develops between layers of fluid moving at different velocities.
This is of great importance in analyzing problems involving the
flow of fluids through pipes and orifices or dealing with bodies
immersed in moving fluids. It is also basic for the analysis of the
motion of lubricated mechanisms. Such problems are considered in
texts on fluid mechanics. The present study is limited to dry
friction, i.e., to situations involving rigid bodies that are in contact
along unlubricated surfaces.

5.1 The Laws of Dry Friction
Place a block of weight 𝑾 on a horizontal plane surface (Figure a).
The forces acting on the block are its weight 𝑾 and the reaction of
the surface. Since the weight has no horizontal component, the
reaction of the surface also has no horizontal component; the
reaction is therefore normal to the surface and is represented by 𝑵
in Figure (a).

Now suppose that you apply a horizontal force 𝑷 to the block
(Figure b). If 𝑷 is small, the block does not move; some other
horizontal force must therefore exist, which balances 𝑷. This other
force is the static-friction force 𝑭, which is actually the resultant of a
great number of forces acting over the entire surface of contact
between the block and the plane. The nature of these forces is not
known exactly, but we generally assume that these forces are due to
the irregularities of the surfaces in contact and, to a certain extent, to
molecular attraction.

If you increase the force 𝑷, the friction force 𝑭 also increases,
continuing to oppose 𝑷, until its magnitude reaches a certain
maximum value 𝐹𝑚 (Figure c). If 𝑷 is further increased, the friction
force cannot balance it anymore, and the block starts sliding. As soon
as the block has started in motion, the magnitude of 𝑭 drops from 𝐹𝑚
to a lower value 𝐹𝑘.

This happens because less interpenetration occurs between the
irregularities of the surfaces in contact when these surfaces move
with respect to each other. From then on, the block keeps sliding
with increasing velocity while the friction force, denoted by 𝑭𝒌 and
called the kinetic-friction force, remains approximately constant.
5.1.1 Coefficients of Friction
Experimental evidence shows that the maximum value 𝐹𝑚 of the
static friction force is proportional to the normal component 𝑁 of
the reaction of the surface. We have
….. (5.1)
𝐹𝑚 = 𝜇𝑠𝑁
where 𝜇𝑠 is a constant called the coefficient of static friction.

Similarly, we can express the magnitude 𝐹𝑘 of the kinetic-friction
force in the form
….. (5.2)
𝐹𝑘 = 𝜇𝑘𝑁
where 𝜇𝑘 is a constant called the coefficient of kinetic friction.
It appears that four different situations can occur when a rigid body
is in contact with a horizontal surface:
1. The forces applied to the body do not tend to move
it along the surface of contact; there is no friction
force (Figure a).

2. The applied forces tend to move the body along
the surface of contact but are not large enough to
set it in motion. We can find the static-friction
force 𝑭 that has developed by solving the
equations of equilibrium for the body. Since there
is no evidence that 𝑭 has reached its maximum
value, the equation 𝐹𝑚 = 𝜇𝑠𝑁 cannot be used to
determine the friction force (Figure b).

3. The applied forces are such that the body is
just about to slide. We say that motion is
impending. The friction force 𝑭 has reached
its maximum value 𝐹𝑚 and, together with the
normal force 𝑵, balances the applied forces.
Both the equations of equilibrium and the
equation 𝐹𝑚 = 𝜇𝑠𝑁 can be used. Note that the
friction force has a sense opposite to the sense
of impending motion (Figure c).
4. The body is sliding under the action of the applied forces, and
the equations of 𝐹𝑘 = 𝜇𝑘𝑁. The sense of 𝑭𝒌 is opposite to the
sense of motion (Figure d).

5.1.2 Angles of Friction
It is sometimes convenient to replace the normal force 𝑵 and the
friction force 𝑭 by their resultant 𝑹.
Consider again a block of weight 𝑾 resting on a
horizontal plane surface. If no horizontal force is
applied to the block, the resultant 𝑹 reduces to the
normal force 𝑵 (Figure a).
However, if the applied force 𝑷 has a horizontal
component 𝑷𝒙 that tends to move the block, force 𝑹
has a horizontal component 𝑭 and, thus, forms an
angle 𝜙 with the normal to the surface (Figure b).

If you increase 𝑷𝒙 until motion becomes impending,
the angle between 𝑹 and the vertical grows and
reaches a maximum value (Figure c).
This value is called the angle of static friction and is
denoted by 𝜙𝑠. From the geometry of Figure (c), we
note that
….. (5.3)
𝑡𝑎𝑛𝜙𝑠 =
𝐹𝑚
𝑁
=
𝜇𝑠𝑁
𝑁
= 𝜇𝑠
If motion actually takes place, the magnitude of the
friction force drops to 𝑭𝒌; similarly, the angle between
𝑹 and 𝑵 drops to a lower value 𝜙𝑘, which is called the
angle of kinetic friction (Figure d).

From the geometry of Figure (d), we have
….. (5.4)
𝑡𝑎𝑛𝜙𝑘 =
𝐹𝑘
𝑁
=
𝜇𝑘𝑁
𝑁
= 𝜇𝑘

Review Questions
1. How friction affects common engineering applications?
2. State and explain laws of friction
3. Define coefficient of static friction and coefficient of kinetic
friction.
4. Define angle of static friction and angle of kinetic friction. Also
write down the relations between (a) angle of static friction and
coefficient of static friction, and (b) angle of kinetic friction and
coefficient of kinetic friction.

Mechanical I/I
2024
FRICTION
Session 32: Examples for Laws of Friction

Learning Objectives
• Solve problems involving friction.

Example 9.1
Determine whether the block shown in Figure E9.1 is in
equilibrium, and find the magnitude and direction of the
friction force when 𝑷 = 𝟏𝟓𝟎 𝑵.
Figure E9.1
Block is in equilibrium. 𝟑𝟎. 𝟏 𝑵 𝟐𝟎𝟎

Example 9.2
A support block is acted upon by two forces as shown in Figure
E9.2. Knowing that the coefficients of friction between the
block and the incline are 𝝁𝒔 = 𝟎. 𝟑𝟓 and 𝝁𝒌 = 𝟎. 𝟐𝟓, determine
the force 𝑷 required to (a) start the block moving up the
incline; (b) keep it moving up; (c) prevent it from sliding down.
Figure E9.2
𝟕𝟖𝟎 𝑵 , 𝟔𝟒𝟗 𝑵 , 𝟖𝟎 𝑵 

Example 9.3
The 𝟖𝟎 𝑵 block is attached to link 𝑨𝑩 and rests on a moving
belt as shown in Figure E9.3. Knowing that𝝁𝒔 = 𝟎. 𝟐𝟓 and
𝝁𝒌 = 𝟎. 𝟐𝟎, determine the magnitude of the horizontal force 𝑷
that should be applied to the belt to maintain its motion (a) to
the right, (b) to the left.
Figure E9.3
𝟏𝟖. 𝟎𝟗 𝑵 , 𝟏𝟒. 𝟑𝟒𝑵 

Example 9.4
The coefficients of friction are 𝝁𝒔 = 𝟎. 𝟒𝟎 and 𝝁𝒌 = 𝟎. 𝟑𝟎
between all surfaces of contact [Figure E9.4]. Determine the
smallest force 𝑷 required to start the 𝟑𝟎 𝒌𝒈 block moving if
cable 𝑨𝑩 (a) is attached as shown, (b) is removed.
Figure E9.4
𝟑𝟓𝟑 𝑵 , 𝟏𝟗𝟔. 𝟐 𝑵 

Example 9.5
The 𝟓𝟎 𝑵 block 𝑨 and the 𝟐𝟓 𝑵 block 𝑩 are supported by an
incline that is held in the position shown in Figure E9.5.
Knowing that the coefficient of static friction is 𝟎. 𝟏𝟓 between
all surfaces of contact, determine the value of 𝜽 for which
motion is impending.
Figure E9.5
𝟒𝟔. 𝟒𝟎

Example 9.6
A 𝟔. 𝟓 𝒎 ladder 𝑨𝑩 leans against a wall as shown in Figure
E9.6. Assuming that the coefficient of static friction 𝝁𝒔 is zero
at 𝑩, determine the smallest value of 𝝁𝒔 at 𝑨 for which
equilibrium is maintained.
Figure E9.6
𝟎. 𝟐𝟎𝟖

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