Study UnitEngineeringMechanics, Part 1ByAndrew Pyt.docx

Study Unit
Engineering
Mechanics, Part 1
By
Andrew Pytel, Ph.D.
Associate Professor, Engineering Mechanics
The Pennsylvania State University
About the Author
Dr. Andrew Pytel is well qualified to write about engineering
mechanics. He obtained his M.S. and Ph.D. degrees in
engineering
mechanics from The Pennsylvania State University. Since 1957
he has taught engineering mechanics at Rochester Institute of
Technology, Northeastern University, and The Pennsylvania
State
University. He has authored and coauthored several articles on
engineering mechanics.
Dr. Pytel is a member of the American Academy of Mechanics,
American Society of Engineering Education, American Society
of
Mechanical Engineers, and Society of Sigma Chi.
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When you complete this study unit, you’ll be
able to
• Explain some of the methods, laws, and procedures
used in engineering mechanics
• Describe the characteristics and classifications of forces
and their effects on bodies
• Apply the rules and formulas that pertain to collinear
forces
• Apply the analytic methods used to determine the
resultant of concurrent forces
• Calculate the resultant of a system of nonconcurrent
forces as they relate to couples and components
• Describe the principles pertaining to the center of
gravity of a body and the methods used to calculate the
distance to the centroid
INTRODUCTORY EXPLANATIONS 1
Scope of Engineering Mechanics 1
Applications of Engineering Mechanics 2
Branches of Engineering Mechanics 2
Mass and Weight of Body 3

Historical Introduction to Newtonian Mechanics 4
Statement of Newton’s Laws 5
Scalar Quantities and Vector Quantities 6
Graphic and Analytic Methods of Solving Problems 6
General Procedure in Solving a Problem 7
Suggestions for Studying Engineering Mechanics 7
FORCES 11
Representation of Forces 11
Combining Collinear Forces 20
Combining Concurrent Forces 26
Combining Nonconcurrent Forces 47
CENTER OF GRAVITY 65
Simple Body 65
Composite Body 78
SELF-CHECK ANSWERS 85
PRACTICE PROBLEMS ANSWERS 87
EXAMINATION 89
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1
INTRODUCTORY EXPLANATIONS
Scope of Engineering Mechanics
1 Engineering mechanics is that branch of engineering that
deals with problems involving the effects of forces
on bodies. There are many different kinds of actions that are
called forces. For example, a person exerts a force on a book
when he or she moves the book from one place to another.
Also, the earth exerts a force on every object, and this force
tends to keep the object in contact with the earth. Numerous
other types of forces could be mentioned. It is therefore impos-
sible to give a simple definition of the term force in the usual
way. The best general definition of a force is as follows: A
“force” is any action that causes or tends to cause a change
in the motion of a body. Various kinds of forces must be con-
sidered in engineering. Many general types will be discussed
in these texts on engineering mechanics.
A body may be composed entirely of one material, or it may
consist of parts made of two or more different materials. In
these texts, we shall assume that every body is composed of
material in the solid state (rather than in the liquid state or

the gaseous state) and that each body has a specific shape
and size. In a certain problem, it may be desirable to con-
sider the entire earth as a single body. In another problem,
an automobile may be treated as a single body. In still another
problem, the body under consideration may be a small block
having the shape of a rectangular prism.
Engineering Mechanics,
Part 1
Engineering Mechanics, Part 12
Applications of Engineering Mechanics
2 The principles of engineering mechanics are applied inall
kinds of engineering. A few examples will be men-
tioned here. A civil engineer is interested in the forces exerted
on the parts of a building frame by objects on the floors. A
mechanical engineer must consider the forces acting upon
and produced by rotating machinery. An electrical engineer
may have to consider the forces acting on cables for trans-
mission lines. An aeronautics engineer must deal with the
forces acting on aircraft. A nuclear engineer must protect a
reactor against the forces that would be produced by an
earthquake.
Branches of Engineering Mechanics
3 Every actual body is always under the influence of atleast two
forces. The effects of the forces acting on a par-
ticular body at any instant depend on the conditions at that
instant. Let us consider the simple case of a book resting on
the top of a desk. If you push vertically downward on the
book, the book will not move noticeably, because movement

will be prevented by the support provided by the desk. If you
push the book sideways and apply enough force, the book
will slide along the top of the desk in the direction in which
you push it. In either case, the shape and size of the book
will be changed, even though the change may be so small
that you will not be able to see any difference in the appear-
ance of the book. Any change in the dimensions of an object
is called a deformation.
Three types of problems involving systems of forces acting on
bodies are common in engineering. In one type of problem, it
is desired to keep a body at rest. In such a problem, the body
is subjected to the actions of certain forces that tend to cause
the body to move, but movement is prevented by the applica-
tion of other forces. In another type of problem, the body
moves under the actions of a system of forces, and it is nec-
essary to analyze the movement. In either of these cases, the
deformation of the body is ignored. In a problem of the third
type, the deformation of a body under the actions of the
forces is considered.
Engineering Mechanics, Part 1 3
Problems of the first type are studied in a branch of
engineering mechanics called statics. The branch that
includes problems of the second type is called dynamics. The
branch that pertains to problems of the third type is called
mechanics of materials or strength of materials. In these texts
on engineering mechanics, only the basic principles of statics
and dynamics will be covered. The deformation of a body will
be ignored, and it will be assumed that every body is rigid. In
other words, it will be assumed that the shape and size of the
body will not be changed by the forces acting on it.

Mass and Weight of Body
4 Every material of which bodies are composed may beincluded
in the general term matter. What is called the
mass of a body indicates the amount of matter contained in
the body. However, it is not practicable to attempt to deter-
mine the mass of a body by direct measurement. The usual
practice is to measure what is called the weight of the body
and to compute the mass from the weight.
It is a well-established fact that every body exerts a force of
attraction on every other body. At this time, we shall refer
only to the force of attraction between the earth and any other
body. Such a force is called the force of gravity, and the mag-
nitude of this force for a particular body is known as the
weight of the body. The weight of a body that is of reasonable
size can be measured directly by means of a suitable type of
scale.
In the English system of measures, the unit most commonly
used for expressing weights is the pound. Small weights may
be expressed in ounces, and large weights may be expressed
in tons. In the metric system, the basic unit of weight is the
gram, but the unit most commonly used is the kilogram, which
is equal to 1000 grams. Since a weight is a force, the units
used for measuring weights are also used for forces of all
kinds.
The relation between the weight of a body and its mass need
not be considered at this time. In Engineering Mechanics,
Parts 1, 2, and 3, we shall discuss only the effects of forces
on bodies, and the mass of a body is not a factor in the types

of problems covered in these texts.
Historical Introduction to
Newtonian Mechanics
5 In the year 1687, Sir Isaac Newton published a bookentitled
Mathematical Principles of Natural Philosophy, in
which he presented his now famous three laws of motion.
These laws form the basis of the principles of engineering
mechanics that are covered in these texts. However, before
we discuss these laws themselves, we shall mention some of
the outstanding scientific developments that led up to
Newton’s conclusions.
Even in very ancient times, human beings were interested in
the motions of bodies. In particular, the movements of the
heavenly bodies across the sky have had an extraordinary
influence on the philosophy, religions, and daily life of people.
We can readily understand why some of the first scientific
studies included attempts to explain and describe the motions
of the planets with respect to the sun and the stars. Early
philosophers, among them Aristotle and Ptolemy, based their
theories on the concept that the earth was a stationary body
at the center of the universe, and that the sun and the other
planets revolved around the earth along complex orbits. This
concept was accepted for many centuries. However, in 1543,
the Polish astronomer Copernicus published his theory that
the planets moved in circular orbits around the sun and that
each planet also rotated on its own axis. This theory stimu-
lated astronomers in the sixteenth century to keep more
careful records pertaining to the motions of the planets. The
most significant information was obtained by Brahe, and his
records were studied extensively by one of his assistants,
named Kepler.
In addition to stating his three laws of motion for mechanical

systems, Newton was able to apply these laws to the motions
of the planets and to derive results that agreed with the data
obtained by Brahe and Kepler. For more than 200 years,
scientists believed that Newton’s principles could be used to
solve all problems pertaining to the motions of mechanical
systems. However, exceptions to the universal validity of
Newton’s laws were discovered by Einstein in 1905 and by
Schroedinger and others in 1925. Now scientists realize that
Einstein’s theory of relativity must replace Newton’s laws in
the analysis of a system that moves with a speed approaching
that of light (186,000 miles per second). Also, Schroedinger’s
wave equation, or quantum mechanics, must be used in the
analysis of the motion of a body that moves through only an
extremely small distance. Although the theories of Einstein
and Schroedinger replace Newton’s laws in certain special
cases, Newtonian mechanics has a very high degree of accu-
racy for the speeds and distances involved in ordinary
engineering problems. Since Newton’s theory can be applied
to such problems with relatively little difficulty, Newton’s laws
are generally used in the analysis of these problems.
Statement of Newton’s Laws
6 Newton’s three laws of motion are usually stated in
thefollowing way:
First Law. A particle continues in a state of rest, or of
uniform motion along a straight line, unless acted upon by a
force.
Second Law. If a resultant force acts upon a particle,
the particle will be accelerated in the direction of the force.

Furthermore, the magnitude of the acceleration will be directly
proportional to the magnitude of the resultant force and
inversely proportional to the mass of the particle.
Third Law. For every action, there is an equal and opposite
reaction.
The meaning of each of these laws will be clarified when
applications of that law are discussed.
Scalar Quantities and
Vector Quantities
7 Some quantities can be described completely by a num-ber
and a unit of measurement. For example, it may be
said that a person’s age is 40 years or that the cost of an
automobile is 4,200 dollars. No additional information is
needed to clarify such a measurement. In these texts on engi-
neering mechanics, however, it is often necessary to consider
a quantity that involves not only a number of units but also a
direction. Three of the quantities that are included in this
class are forces, distances through which bodies are moved,
and rates of movement of bodies.
A quantity that can be described completely by its magnitude,
or by a number of units, is called a scalar quantity, or scalar.
A quantity that involves both magnitude and direction is called
a vector quantity, or vector. The characteristics of scalars and
the methods of performing mathematical calculations with
scalars should require no additional explanations. However,
the characteristics of vectors and the methods of combining
vectors will be explained in detail.

In Engineering Mechanics, Parts 1 and 2, the only vectors that
will be considered are forces. Hence, the general principles
relating to vectors will be explained in the discussion of
forces.
Graphic and Analytic Methods of
Solving Problems
8 In some cases, it is convenient to solve a problem completely
by constructing a suitable diagram. Such a
solution is called a graphic solution. In other cases, it is
advisable to solve a problem by performing the proper calcu-
lations. Such a solution is called an analytic solution.
An advantage of a graphic solution is that the use of a
diagram makes it easier for the person to visualize the condi-
tions and to understand the solution. However, the accuracy
obtainable by using a graphic solution is limited by the
degree of precision possible with the instruments used for
measuring distances and angles. When a problem is solved
by an analytic method, it is possible to determine results to
any desirable degree of precision. On the other hand, the
mathematical procedure that is applied does not always indi-
cate clearly the meaning of the calculations. For example, it
may be easy for a person to sum numbers in a formula and
compute a result without understanding the significance of
the result.
In these texts on engineering mechanics, computed values
will usually be expressed to three significant figures when the

first significant figure is 2 or more, and to four figures when
the first figure is 1. The number of figures used in a result
based on a graphic solution will depend on the conditions in
the particular problem.
General Procedure in Solving
a Problem
9 Since it is usually easier for a person to visualize
theconditions in a problem when a graphic solution is used,
the procedure in this text for presenting a general principle
will be to explain the graphic method before the analytic
method. Also, when a problem is to be solved by an analytic
method, the first step will be to prepare a suitable diagram to
indicate the conditions. This diagram need not picture the
conditions accurately, but it should be complete enough to
show all the known information and also the relation of the
required quantities to the known quantities.
Suggestions for Studying
Engineering Mechanics
10 In the texts on Engineering Mechanics, the primaryemphasis
will be on the techniques for solving prob-
lems. Nevertheless, we shall include a discussion of the
principle applied in the solution of each problem and enough
of the theory underlying the principle to enable you to under-
stand the significance of the principle and its limitations.
To solve some of the problems, you will have to apply princi-
ples of arithmetic, algebra, geometry, and trigonometry.
Therefore, if you have forgotten some of the mathematical

procedures that will be used, you should review the portions
of previous texts in which these procedures are explained.
Also, many of the principles described in the texts on engi-
neering mechanics are applied in other texts in your course.
Hence, you should study each principle in these texts very
carefully and be sure that you understand it completely before
you go on to the next one. Moreover, if you come to some
procedure that requires a knowledge of a principle that was
explained previously and you find that you cannot recall that
principle, you are expected to review the appropriate text
material. Where we have reason to believe that you may
encounter difficulty because of insufficient knowledge of a
preceding explanation, we shall try to help you by including a
reference to the place in which the necessary basic informa-
tion is given. Remember that every explanation in these texts
is important.
Self-Check 1
At the end of each section of Engineering Mechanics, Part 1,
you’ll be asked to pause and
check your understanding of what you’ve just read by
completing a “Self-Check” exercise.
Answering these questions will help you review what you’ve
studied so far. Please
complete Self-Check 1 now.
1. What general types of problems are dealt with in engineering
mechanics?
_____________________________________________________
_____

2. How would you define the term force in a general way?
_____________________________________________________
_____
3. What are the names commonly given to the three main
branches of engineering mechanics?
_____________________________________________________
_____
4. When a rigid body is acted upon by a number of forces, is it
necessary to consider the
deformation of the body?
_____________________________________________________
_____
5. What is meant by the “weight” of a body?
_____________________________________________________
_____
6. Name the unit commonly used to measure force in the a)
English system and b) metric
system.
_____________________________________________________
_____
7. How many separate laws are included in what are called
Newton's laws of motion?
_____________________________________________________
_____

(Continued)
Self-Check 1
8. What are the two essential characteristics of a vector
quantity?
_____________________________________________________
_____
9. What is the main advantage of a graphic solution of a
problem over an analytic solution?
_____________________________________________________
_____
10. In a problem in engineering mechanics, when a value is
computed, how many significant
figures are generally used to express the result?
_____________________________________________________
_____
Check your answers with those on page 85.
FORCES
Representation of Forces

Characteristics of a Force
11 When any force is applied to a body, the force actsalong a
certain straight line, which is called the line
of action of the force. The line of action of a force may be hor-
izontal or vertical, or it may be inclined at any angle to the
horizontal or vertical. Also, a force may act in either of two
directions along the line of action. A force really involves two
distinct directions. One is the direction of the line of action in
space; the other is the direction of the force along its line of
action. To avoid confusion in regard to the meaning of the
direction of a force, we shall use the expressions “inclination
of the line of action” and “direction of the force along the line
of action.” Of course, the magnitude of a force must be
considered. Therefore, to describe a force completely, it is
necessary to specify the inclination of the line of action of the
force, the direction of the force along that line of action, and
the magnitude of the force. In these texts on engineering
mechanics, the magnitude of every force will be expressed in
pounds. The inclination of the line of action of a force and
the direction of the force along its line of action may be
described or indicated in more than one way.
The usual method of indicating the inclination of the line of
action of a force is by means of the angle between that line of
action and either a horizontal or a vertical reference line. This
method will be used in all problems in this text. The direction
of a force along its line of action is indicated in most cases by
placing an arrowhead on the line of action. Other methods
will be described where they are used.
In the analysis of some problems in engineering mechanics, it
is necessary to assume that a force is applied to a body at
some particular point on the body. In other problems, it may
be assumed that a force is applied to a body at any point on
the line of action of the force.

Graphic Representation of a Force
12 Although it is possible to describe a force by usingwords
alone, the easiest and clearest method of
describing a force is by means of a diagram, as indicated in
Figure 1. For each of the forces represented, the straight line
1 indicates the position of the line of action of the force with
respect to the body 2 on which the force acts, and the arrow-
head 3 indicates the direction of the force along its line of
action. The arrowhead may be placed at one end of the line of
action or at some convenient point on the line of action. The
magnitude of each force is the number of pounds shown, and
the length of the line of action is chosen arbitrarily. In some
diagrams constructed for solving problems, the relative mag-
nitudes of two or more forces are indicated by the lengths of
specific segments of lines that are drawn parallel to the lines
of action of the forces. The use of such segments will be dis-
cussed a little later in this text. For simplicity, each body in
this illustration is represented by a small rectangle, but the
body in a practical problem may have any shape and size.
FIGURE 1—Graphic
Representation of Forces
1. Line of action of force
2. Body acted upon by force
3. Arrowhead showing direction of force along its line of action

In Figure 1A, the line of action of the force is horizontal and
the force acts toward the right along the line of action.
Because of the relative positions of the force and the body,
the force acts toward the body and, therefore, the force tends
to push the body horizontally toward the right. In Figure 1B,
the line of action of the force is vertical, and the force acts
downward along its line of action. This force acts away from
the body and tends to pull the body vertically downward. The
line of action of the force in Figure 1C is inclined to the hori-
zontal at an angle equal to 30° (degrees) and the force tends
to push the body. In Figure 1D, the line of action of the force
is inclined at the indicated angle with the horizontal, and the
force tends to pull the body. For the conditions in Figures 1C
or 1D, the actual effect of the force on the body will depend
on the effects of other forces that will also act on the body.
For the problems in these texts on engineering mechanics, it
is not necessary to consider the manner in which a force is
applied to a body. In other words, we shall only consider the
direction in which a force acts along its line of action, and it
will not be necessary to specify whether the force tends to
push a body or tends to pull the body.
Description of a Force
13 Now that you have seen how a force can be repre-sented
graphically, you should be able to understand
a description of a force in words. Typical descriptions of the
four forces represented in Figure 1 are as follows: Figure 1A
represents a force acting horizontally toward the right and
having a magnitude equal to 200 lb (pounds); Figure 1B
represents a force acting vertically downward and having a
magnitude equal to 50 lb; Figure 1C represents a force acting

downward and toward the right at an angle with the horizon-
tal equal to 30° and having a magnitude equal to 115 lb;
Figure 1D represents a force acting upward and toward the
right at an angle with the horizontal equal to 45° and having
a magnitude equal to 78 lb.
In the solution of a problem or in an explanation of a proce-
dure, the common practice is to use a symbolic notation for
describing a force. The notations for the forces in Figure 1
would be as shown in Figure 2. In every case, the magnitude
of the force is given as if no other information were needed,
and the inclination of the line of action and the direction of
the force along its line of action are indicated diagrammati-
cally. This symbolic method of describing a force will be used
frequently in these texts.
Vector Representing Force
14 When a problem involving forces is to be solved com-pletely
by a graphic method, each force must be
represented by a line that has both the correct inclination
and the correct length. Such a line is, of course, a vector. A
vector must usually be parallel to the actual line of action of
the force represented by the vector, and the length of the vec-
tor must represent the magnitude of the force to some suitable
scale. For example, if the magnitudes of two forces that act
on a single body at the same time are 200 lb and 50 lb, the
relative lengths of the vectors representing the forces may be
determined as follows: It is assumed that each inch of length
of a vector corresponds to some convenient number of pounds
of force, such as 100 lb. Then the length of the vector repre-
senting the 200-lb force should be 200/100 or 2.00 in. (inches)

and the length of the vector representing the 50-lb force
should be 50/100 = 0.50 in. The number of pounds correspon-
ding to 1 in. of length of a line in a diagram is called the
scale of the diagram. In the example just mentioned, the
scale would be 1 in. = 100 lb.
The person graphically solving a problem involving forces
must also consider which end of the vector representing a
force is the termination of the vector, or its tail, and which
FIGURE 2—Representation
of Forces by Symbolic
Notation
end is the beginning of the vector, or its tip, or head. In
general, the direction of a vector from its tail to its tip must
correspond to the direction of the arrowhead on the line of
action of the force represented by the vector. In Figure 3 are
shown vectors representing each of the forces in Figure 1 to a
scale of 1 in. = 100 lb. In each case, the end of the vector
marked A is its tail, and the end marked B is the tip of the
vector. The arrowhead is here placed on each vector at its tip,
but the arrowhead may be placed anywhere on the vector. It
is often convenient to describe a vector representing a force
by means of two letters that are placed at the ends of the
vector. For instance, each vector in Figure 3 may be called
the vector AB. When such a method of describing a vector is
used, the first letter should always be the letter at the tail.
The order of the letters then indicates the direction of the
force along its line of action.
Classification of Systems of Forces

15 In a practical problem involving forces, it is oftennecessary
to consider the combined effects of two or
more forces. A group of forces that act on a single body at the
same time is called a system of forces. Systems of forces can
be divided into four general classes:
1. Collinear forces
2. Concurrent forces
3. Parallel forces
4. Nonconcurrent, non-parallel forces
FIGURE 3—Vectors
Representing Forces
A. Tail of vector
B. Tip of vector
SCALE: 1 in. = 100 lb
Bodies or groups of bodies that are acted upon by each of
the four types of systems of forces will be mentioned in prob-
lems that are considered in these texts. The bodies to which
references are made are only a few of those that could be
mentioned.
In a system of collinear forces, or a collinear system, the lines
of action of all the forces coincide. That is, all the forces have

a common line of action. For example, in Figure 4A the forces
F1 and F2 are collinear. The two forces F1 and F2 in Figure 4B
are also collinear. In each case, it is assumed that the two
forces of the system act on a small body (rectangular prism).
As indicated by the arrowheads, the two forces in Figure 4A
act in the same direction, and the two forces in Figure 4B act
in opposite directions. A system of collinear forces may con-
sist of more than two forces. All the forces in such a system
may act in the same direction, or some forces may act in one
direction while the other forces act in the opposite direction.
In a system of concurrent forces, or a concurrent system, the
lines of action of all the forces pass through a common point.
A collinear system is a special type of concurrent system.
Typical systems of concurrent forces that are not collinear
are represented in Figure 5. Here the body acted upon by the
forces is represented by a small circle so that you can easily
see that the forces of each system pass through a single
point. There may be any number of forces in a concurrent
system; the lines of action of the forces may have any inclina-
tions; and each force may act in either direction along its line
of action.
FIGURE 4—Systems of
Two Collinear Forces
A. Forces acting in same direction
B. Forces acting in opposite directions
In a system of parallel forces, or parallel system, the lines of
action of all the forces are parallel, as indicated in Figure 6.

In each of these cases, it is assumed that the body on which
the forces act is a long, slender bar in a horizontal position.
There may be any number of forces; the lines of action of the
forces may be located in any positions; and each force may
act in either direction along the line of action.
In a system of nonconcurrent, nonparallel forces, the lines of
action of the forces do not pass through a common point and
are not parallel. A typical system of forces of this type is rep-
resented in Figure 7, in which the body acted upon by the
forces is shown with an irregular shape. There may be any
number of forces in the system; the lines of action of the
forces may have any positions and any inclinations; and each
force may act in either direction along its line of action.
FIGURE 5—Typical
Systems of Concurrent
Forces
FIGURE 6—Typical
Systems of Parallel Forces
In most problems in ordinary engineering work, the lines of
action of all the forces in a system lie in one plane. In other
words, the forces are coplanar. However, in some problems
the lines of action of all the forces of a system do not lie in
the same plane. We shall assume that all the forces in any
system considered in Engineering Mechanics, Part 1, are
coplanar.
Resultant of System of Forces

16 To solve a problem in engineering, it is often desir-able to
replace two or more forces acting on a body
by a single force that would have the same effect on the body
as the given forces it replaces. A single force that can replace
two or more given forces is called the resultant of the given
forces. For example, the motion of either body in Figure 4
theoretically would not be affected if the two forces F1 and F2
were replaced by a single horizontal force having the proper
magnitude and the proper direction along its line of action.
The method of determining the magnitude and the direction
of the resultant will be explained in the next article. Also, the
motion of either body in Figure 5 theoretically would not be
affected if the three given forces acting on the body were
replaced by an equivalent single force whose line of action
passes through the common point of intersection of the lines
of action of the given forces. Procedures for determining the
inclination of the line of action of this resultant force, the
magnitude of the resultant, and the direction of the resultant
along its line of action will be described a little later in this
text. The characteristics of the resultant of a system of paral-
lel forces and of a system of nonconcurrent, nonparallel
forces will also be discussed in this text.
FIGURE 7—System of
Nonconcurrent,
Nonparallel Forces
Self-Check 2
1. A force is described by the following notation: . What is a)
the magnitude of the force,
b) the angle between a horizontal reference line and the line of

action of the force, and
c) the direction of the force along its line of action?
_____________________________________________________
_____
2. A force acts upward and toward the left at an angle with the
horizontal equal to 20°, and its
magnitude is 120 lb. How would you show the characteristics of
this force by the symbolic
method?
_____________________________________________________
_____
3. Draw a vector representing the force in question 2 so that 1
in. corresponds to 40 lb.
_____________________________________________________
_____
4. Place the words tail and tip at the proper ends of the vector in
question 3.
_____________________________________________________
_____
5. Describe the relative positions of the lines of action of a
group of forces that form a collinear
system.
_____________________________________________________
_____
6. If the forces forming a system are concurrent, what
requirement must be satisfied by the

lines of action of the forces?
_____________________________________________________
_____
7. The lines of action of the forces of a system are not parallel
and do not pass through a single
point. How would you describe such a system of forces in the
simplest terms?
_____________________________________________________
_____
8. A certain body is acted upon by two concurrent forces. If
these two forces are replaced by a
single force that would have the same effect on the body as the
given forces, what is this
single force commonly called?
_____________________________________________________
_____
200
60�
Combining Collinear Forces
Resultant of Collinear Forces

17 To determine the magnitude of the resultant of twocollinear
forces that act in the same direction, it is
simply necessary to add the magnitudes of the two given
forces. Obviously, the direction of the resultant along its line
of action is the same as the direction of each given force. For
instance, let us consider a freight train that is pulled along a
railroad track by one locomotive at the front of the train with
a force whose magnitude is 20,000 lb, and is also pushed by
a second locomotive at the rear of the train with a force whose
magnitude is 15,000 lb. The conditions are really similar to
those represented in Figure 4A, where the body represents
the train, the force F1 represents the pull exerted by the front
locomotive, and the force F2 represents the push exerted by
the rear locomotive. In this case, the pull and the push tend
to move the train along the track in the same direction. Hence,
the magnitude of the resultant of these forces—the total force
tending to move the train—is 20,000 + 15,000 = 35,000 lb.
Also, the direction of the resultant, or the direction of move-
ment, would be the same as the direction of each actual force
applied to the train.
If two collinear forces act in opposite directions, the magni-
tude of the resultant of the given forces may be found by
subtracting the magnitude of the smaller given force from the
magnitude of the larger given force; and the direction of the
resultant along its line of action is the same as the direction
of the larger force. For example, let us suppose that two men
are having a tug-of-war. The conditions are similar to those
represented in Figure 4B, where the body represents the
rope, the force F1 represents the pull exerted by the man at
the right, and the force F2 represents the pull exerted by the
man at the left. If the magnitude of the first force F1 is 150 lb
and the magnitude of the second force F2 is 120 lb, then the
magnitude of the resultant of these forces, or the force tend-
ing to move the rope, would be 150 – 120 = 30 lb; and the
direction of the resultant, or the direction of movement, would

be the same as that of the larger force F1, toward the right.
When a system of collinear forces consists of more than two
forces and all the given forces act in the same direction, the
magnitude of the resultant of the system is equal to the sum
of the magnitudes of all the given forces, and the direction of
the resultant along its line of action is the same as the direc-
tion of each given force.
If a system of collinear forces consists of more than two
forces and some of the given forces act in one direction while
the other given forces act in the opposite direction, the proce-
dure for determining the magnitude and the direction of the
resultant of the system may be outlined as follows: First, the
magnitude and direction of the partial resultant of all the
forces acting in one direction are determined by applying the
method described in the preceding paragraph, and the mag-
nitude and direction of the partial resultant of all the forces
acting in the other direction are also determined in a similar
manner. Then, the magnitude and direction of the resultant
of the entire system are determined by combining the two
partial resultants, which are treated as two collinear forces
acting in opposite directions.
Example Problem
At intervals throughout this text you will find one or more
example problems solved to illustrate clearly the application
of a principle, rule, or formula. Read each problem carefully,
and study the solution until you understand it thoroughly.
Problem: In Figure 8 is represented a system consisting of

six forces, all of which have a common horizontal line of
action. The forces are shown separately for easier identifica-
tion of the magnitude and direction of each. Determine the
magnitude and the direction of the resultant of the entire
system.
FIGURE 8—Forces in Example Problem, Article 17
Solution
: The magnitude of the partial resultant of the three
forces acting toward the right is
60 + 110 + 80 = 250 lb
and the magnitude of the partial resultant of the three forces
acting toward the left is
90 + 100 + 40 = 230 lb
Therefore, the magnitude of the resultant of the entire system
is 250 – 230 = 20 lb, and the direction of this resultant along

the common line of action of the given forces is toward the
right.
Symbolic Notation for Combination
of Collinear Forces
18 Since forces are vectors, a convenient symbolic wayto
indicate the procedure for determining the result-
ant of a system of collinear forces is to represent each force
in the manner indicated in Figure 2 and to place the symbol
between each two forces. For instance, the resultant of
the 20,000-lb force and the 15,000-lb force referred to in the
first paragraph of Article 17 may be indicated as follows:
20,000 lb 15,000 lb = 35,000 lb
Likewise, the resultant of the 120-lb force and the 150-lb
force referred to in the second paragraph of Article 17 would
be
120 lb 150 lb = 30 lb
Similarly, the steps in the solution of the preceding example

problem could be indicated as follows:
60 lb 110 lb 80 lb = 250 lb
90 lb 100 lb 40 lb = 230 lb
250 lb 230 lb = 20 lb
The symbol is intended to be a combination of the symbol
+ and an arrow. It is used to distinguish between the addition
of scalar quantities and the “addition” of vector quantities. You
should note that the symbol is used between two forces
regardless of whether the forces act in the same direction or
in opposite directions, and regardless of whether a force acts
toward the right or toward the left. However, if two forces act
in the same direction, their magnitudes are added; whereas,
if two forces act in opposite directions, the magnitude of one
force is subtracted from the magnitude of the other force.
Simplified Formula for Resultant
of Collinear Forces

19 Since it is so easy to determine analytically the magnitude
and the direction of the resultant of any
system of collinear forces, the graphic method is rarely, if
ever, used for solving a problem of collinear forces. The sev-
eral rules given in Article 17 for determining the magnitude
of the resultant may be replaced by a single simple formula,
which may be stated as follows:
R = �F
In this formula, R represents the resultant of a system, and
the combination of the Greek letter � (sigma) and F is used to
denote the vector sum of the magnitudes of all the given
forces in a system.
To determine the vector sum �F for any system of collinear
forces, it is simply necessary to consider as positive quantities
all forces acting in one direction along the common line of
action, and to consider as negative quantities all forces acting
in the opposite direction. The usual practice is to assume
that horizontal forces acting toward the right are positive and
that horizontal forces acting toward the left are negative.
Also, vertical forces acting upward are usually considered as
positive and vertical forces acting downward are considered

as negative. The magnitude of the resultant is the same as
the magnitude of the vector sum. If the vector sum for a sys-
tem of horizontal forces is a positive quantity, the resultant of
the system acts toward the right; and if the vector sum is
negative, the resultant acts toward the left. Similarly, if the
vector sum for a system of vertical forces is positive, the
resultant acts upward; and if the vector sum is negative, the
resultant acts downward.
One method of applying the formula R = �F to a system of
forces that act along a common inclined line is to imagine
that the inclined line of action is horizontal and to treat the
system as if it were a system of horizontal forces.
When the preceding formula is applied to solve the example
problem in Article 17, the vector sum may be found as
follows:
The sum of the positive forces is

+(60 + 110 + 80) = +250 lb
The sum of the negative forces is
–(90 + 100 + 40) = –230 lb
The vector sum of the forces for the entire system is
+250 + (–230) = +20 lb
Thus, the magnitude of the resultant is 20 lb. Also, since the
vector sum is positive, the resultant acts toward the right.
Practice Problems 1
Practice problems are included in this text to test your ability to
apply a rule or a formula.
Work each problem carefully and check your answer against the
answer given after the
statement of the problem.
In each of the following problems, assume that the forces are

collinear. Determine the
magnitude of the resultant and the direction of the resultant
along the common line of
action of the given forces.
1. F1 = 25 lb toward the right and F2 = 75 lb toward the right
_____________________________________________________
_____
2. F1 = 150 lb upward and F2 = 60 lb upward
_____________________________________________________
_____
3. F1 = 80 lb toward the left and F2 = 60 lb toward the right
_____________________________________________________
_____
4. F1 = 1500 lb upward and F2 = 800 lb downward
_____________________________________________________
_____

5. F1 = 1200 lb toward the left, F2 = 1000 lb toward the right,
F3 = 2000 lb toward the right,
and F4 = 2400 lb toward the left
_____________________________________________________
_____
Combining Concurrent Forces
Parallelogram of Forces
20 So that you may clearly understand the relationshipbetween
two given concurrent forces and their
resultant, we shall present first a graphic method for deter-
mining the characteristics of the resultant, and later an
analytic method. Figure 9A shows a body that is acted upon
by two forces F1 and F2. For example, one of these forces may
represent the effect of the propeller in moving a small airplane,
while the other force represents the effect of the wind on the

movement of the airplane. Numerous other bodies are acted
upon by two concurrent forces at the same time. The diagram
in Figure 9A shows the relative positions of the lines of action
of the forces, the direction of each force along its line of
action, and the magnitude of each force. Since the lines of
action of these two forces must intersect at some point, the
forces are concurrent. Figure 9B shows a graphic method for
determining directly the magnitude of the resultant, the incli-
nation of its line of action, and its direction along its line of
action.
The first step in constructing the diagram in Figure 9B is to
select a point O in some convenient position and to draw
through this point the vectors OA and OB, which represent
the given forces. It is assumed that the point O is the tail of
each of these vectors. To draw the vector OA, an inclined line
of unlimited length is drawn through the point O so that the
angle between a horizontal reference line OH through O and
this inclined line is 18°. Then the distance OA along this line
is laid off to represent the magnitude of the force F1 to some
convenient scale. In this case, the selected scale is 1 in. =
50 lb, and the distance OA represents 100 lb to that scale.
The vector OB is drawn in a similar manner. An inclined line
of unlimited length is drawn through the point O at the
proper angle with the horizontal reference line OH, and the

distance OB along this inclined line is laid off to represent
the magnitude of the force F2 to the selected scale. Thus, the
point A is the tip of the vector OA and the point B is the tip
of the vector OB. The distances OA and OB must be measured
in the proper directions from the point O. These directions
are indicated by the arrows on the lines of action of the
forces F1 and F2 in Figure 9A.
After the lines OA and OB have been drawn in Figure 9B, the
next step is to complete the parallelogram OACB in the fol-
lowing manner: From the point A, a line of unlimited length
is drawn parallel to the vector OB, and from the point B a
line of unlimited length is drawn parallel to the vector OA.
FIGURE 9—Resultant of
Two Concurrent Forces by
Parallelogram
A. Representation of forces

B. Parallelogram of forces
These two new lines, which are the broken lines in the
illustration, intersect at the point C. Finally, the vector repre-
senting the resultant R of the given forces F1 and F2 is drawn
from the initial point O to the point C. The magnitude of the
resultant is indicated by the length of the line OC expressed
in pounds by the relation 1 in. = 50 lb. The inclination of the
line of action of the resultant with respect to the horizontal
can be determined by measuring the angle between the lines
OH and OC. The direction of the resultant along its line of
action is determined by assuming that the tail of the vector
OC is at O and the tip of this vector is at C. For the given
characteristics of the forces F1 and F2 in Figure 9A, the mag-
nitude of the resultant would be about 225 lb, and the angle
between the horizontal and the line of action of the resultant
would be about 48°.
The actual line of action of the resultant of two concurrent
forces must pass through the point at which the lines of
action of the two given forces intersect.

A parallelogram like OACB in Figure 9B is called a
parallelogram of forces. Such a parallelogram can be con-
structed for determining the characteristics of the resultant
of any two concurrent forces. The procedure may be outlined
as follows:
Step 1. Select any convenient initial point, and through
this point draw vectors that represent the two given
forces. Be careful to draw these vectors in the proper
directions and to lay off their lengths correctly to a
selected scale.
Step 2. Complete the parallelogram by drawing a line that
passes through the tip of each vector located in
Step 1 and is parallel to the other vector.
Step 3. Draw the vector representing the required resultant
by connecting the initial point and the point of
intersection of the two lines drawn in Step 2.

Step 4. Determine the magnitude of the resultant by meas-
uring the length of the vector representing the
resultant; determine the inclination of the line of
action of the resultant by measuring the angle
between the vector representing it and some con-
venient reference line; and determine the direction
of the resultant along its line of action by consider-
ing the direction from the tail to the tip of the vector
representing the resultant.
In Figure 9B, the angle between the vectors representing the
given forces F1 and F2 is less than 90°. In the following exam-
ple problem, the angle between the vectors representing the
given forces in the parallelogram of forces is greater than 90°.
Example Problem
Problem: The characteristics of two forces acting on a body
are indicated in Figure 10A. Determine the characteristics of
the resultant of the given forces by constructing a parallelo-
gram of forces.

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