Lesson 1

Engineering Mechanics: Introduction

Engineering mechanics as its term implies is basically attributed to engineers or people

whose line of work somehow needs the assistance of the expertise of an engineer. In this regard
Architects, who one way or the other consult Engineers, should learn the course or somehow have
knowledge about engineering mechanics for as far as the former’s line of works are concern they are
generally engineering in nature.
It is basically the role of an engineer to have knowledge of the forces, known or unknown,
acting on the object, and the motions, if any, of the body’s various related parts. To do this
successfully he must have a thorough knowledge and understanding of the principles of mechanics,
and their applications to particular problems.
As for the architects and others whose line of works involves these forces in nature, they
should also have an understanding of the works of the engineer so that they both could come up with
the best design that satisfies the needs of the project concern.

Lesson 1 of this manual focuses on the following as for its discussions:

Main Objective:
Introduce to the students the different fields of engineering mechanics.
Specific Goals:
1. Know the different fields of engineering mechanics and differentiate one from the other.
2. Branch of engineering mechanics that applies to their chosen career.
3. Realize the significance of the subject to their chosen field.

Static
Rigid s
Bodie Kinetics
s Dynamics

Strength of
Mechanic
Materials
s of Solids
 Kinematics

Fig. 1.1 Outline of Engineering Mechanics

Engineering Mechanics is a branch of physical science that treats of motion, forces, and the effects
of forces on the bodies upon which they act. Two major aspects of which as far as the bodies on
which forces acts are mechanics of solids and mechanics of fluids. Of these two major fields of
engineering mechanics, this manual limits its focus on mechanics of solids specifically statics of
rigid bodies.

The following fields of engineering mechanics are so defined to have an understanding as to

how one differ from the other.

Mechanics of Solids is that branch of engineering mechanics that treats of effects of forces on
objects which are solid in nature i.e. materials that have strong molecular structure that they retain
definite shapes unless acted upon by external forces.

Mechanics of Fluids treats of forces that act on materials that flows by nature such as liquid, or
gases. These materials differ from solids in that they apparently take the shape of the container and
readily change shape when exposed to external forces.
Rigid Bodies is a branch mechanics of solids that deals with objects that is treated to have no
deformation however forces acts upon them while on Deformable Bodies, solids are treated to have
deformations (change in size and/or shape) and stresses when acted upon by forces.
Statics of Rigid Bodies deals with force and the effects of forces upon rigid bodies which are and
remain at rest.

Dynamics is the other branch of mechanics of solids that deals with motion and with the effects of
forces acting on bodies in motion. This is further divided into two branches namely a) Kinematics and
b) Kinetics.

Kinematics is the geometry of motion that essentially treats of relations between displacement,
velocity, and acceleration without regard to the forces causing the motion.

Kinetics is another branch of dynamics that relates the forces acting on a body to its mass and
acceleration

Lesson 2
General Concepts and Definitions
 Lesson 2 deals with the basic concepts and definitions of terms not new to college students
for they are somehow taken in earlier lessons in high school Physics. Terms like force, vector
quantities, scalar quantities, etc. are but common terms that are taken here. The general concepts
and principles discussed here are very fundamental that students are advised to learn them well for
these are recurrent i.e. they are to be use and re-used for the entirety of this manual.

Main Objective:

Introduce to students the analysis of work problems in statics of rigid bodies.

Specific Goals:

1. Identify force and/or force systems as to type and effects on bodies on which they act.

2. Construct a vector diagram showing the relative position and action of the force or system of
forces on bodies on which they act.

3. Interpret such vector diagram constructed as to its relative effect on the bodies on which
they act.

Definition of Terms. The following are basic definition of terms used in this manual,

Rigid Body is a definite amount of matter the parts of which are fixed in position relative to each
other.
Vector quantity is a quantity whose measurement is specified by both magnitude and direction.
Those which have only magnitude are called scalar quantities.

Force is arbitrarily defined as any influence capable of producing a change in the state of motion of
an object. Force is a vector quantity; as such it can be represented graphically by a directed line
segment whose length drawn to scale indicates its magnitude and its inclination or rotation from an
axis its direction.

External Effect of a Force is that which changes or tends to change the state of motion of a body.

Internal Effect of a Force is that which produces stresses and deformations on the body to which it
acts.

Force is characterize or completely described as having (a) a magnitude, (b) position of line of
action and (c) sense or direction of its line of action

Force System is any arrangement where two or more forces act on a body or group of related bodies.

Co-Planar force systems are when the lines of action of the forces lie on the same plane and/or
straight line.

Concurrent force system is when the lines of action of the forces pass or meet at a common point.

Parallel force systems are when the lines of action of the forces are parallel.

Non-Concurrent force systems are when the lines of action of the forces do not lie on the same

plane and/or do not meet at a common point. This is also considered as general force system.

F1 F2
F2
F1

F3 F3
 (a) Co-Planar Forces

F2
F1 F3 F4 F5
F2
F1 F2 F3 F4 F5 F1

F3

(b) Con-Current Forces (c) Parallel Forces

F1 F1
F2 F2

F1

F3
Forces in Plane Forces in Space

(d) Non-Concurrent Forces

Fig. 2.1 Illustrations of the Different Force Systems

Principle of transmissibility of a force states that the external effects of a force on a body does
not change along its line of action.

F F

(a)

F F
 (b)

F F
(c)

F F
(d)

F F
(e)

F F
(f)

Fig. 2.2 (a) and (b) Shows that whether force F is applied at the right end or left end of the block its
motion is the same, while for the frame at (c), & (d), its motion would only be the same if a third
member as in (e) & (f) is added to consider it as a rigid body.
Free Body Diagram is a sketch of the isolated body which shows only the forces of the removed
elements acting upon it.

D Dh

T
B T B
C C
A P W

A A

E Eh E
 Ev

(a) Derrick (b) Shown are all the force effecting the
Derrick

(c) Free-Body Diagram of Pin at C

(d) Free-Body Diagram of Derrick

Fig. 2.3 Illustration of how free-body diagrams are formed

Parallelogram Law states that the resultant of two forces is the diagonal of the parallelogram
formed by these vectors.

F1 F1 R
F1

F2 F2 F2

(a) Forces F1, F2 acting (b) Wrong diagonal for Resultant, R (c) Right diagonal for Resultant, R
on an object
 Fig. 2.4 Resultant of Two Forces Obtained by Parallelogram

Triangle Law is a corollary of the parallelogram law i.e. the resultant of two forces is the closing
side of the triangle formed by the vectors connected tip-to-tail.

F1 R
F1

F2 F2

(a) Forces F1, F2 acting (c) F1 and F2 connected tip-to-tail forming a

on an object triangle with Resultant R as closing side

Fig. 2.5 Resultant of Two Forces Obtained by Triangle Method

(Note: If Graphical method is applied force vectors must be drawn to scale and as directed)

Two Forces are in Equilibrium only when equal in magnitude, opposite in direction and collinear in
action.
A Set of Forces in Equilibrium may be added to any system of forces without changing the effect
of the original system.
Action and Reaction are equal but oppositely directed.

Component of a force is that effective value of a force in a given direction.

Free Vector is a vector arrow that may be moved anywhere in space provided it maintains the
same direction and magnitude.
Y
Free vector of Fx

Fy F Free vector of Fy

θx
X
O Fx
 Fig. 2.6 Rectangular Components

Sin θx = Fy/F or Fy = F (Sin θx)

Cos θx = Fx/F or Fx = F (Cos θx)
F = √Fx2 + Fy2
Tan θx = Fy / Fx or θx = tan-1 Fy/Fx

Table 2.1 Sign Conventions of Rectangular Components

X-
General Direction of Force Y-component
Diagram component
(F) (Fy)
(Fx)

Up to Right + +

Down to Right + ―

Up to Left ― +

Down to Left ― ―

Illustrative Problems

1. Determine the x- and y- components of each of the forces shown in Fig. 2.7.

Ty
Y
Y
T=722 Tx Px P
T
kN

O X
 P=200 kN √13
3
√13 3 Py
60 o
2 60 o

2
O X

45o 45o Fx
F=500 kN
Fy F

(a) (b)
Fig. 2.7

Solution:

From Figure 2.7b, with the rectangular components drawn for each of the forces,
for force P = 200 kN,
Px = P (Cos 60o) = 200 kN (Cos 60o) = 100 kN, to the left or P = -100 kN
Py = P (Sin 60o) = 200 kN (Sin 60o ), downward or P = -173.2 kN

for force T = 722 kN, with hypotenuse of its slope equal to √22 + 32 = √13
Tx = T (2/√13 ) = 722 kN (2/√13) = 400 kN, to the left or T = -400 kN
Ty = T (3/√ 13) = 722 kN (3/√13) = 600 kN, upward or T = 600 kN

for force F = 500 kN

Fx = F(Sin 45o) = 500 kN(Sin 45o) = 353 kN, to the right or Fx = 353 kN
Fy = F(Cos 60o) = 500 kN(Cos 45o)= 353 kN, downward or Fy = -353 kN

2. In Fig. 2.8, the body on the inclined surface is acted upon by a force P inclined at 20o with the
horizontal. If P is resolved into components parallel and perpendicular to the incline and the
value of the parallel component is 400 kN, compute the value of the perpendicular
component and that of P. Y
X
P Py
P
20 o

30o 30o
 o
50o 20
30o

Px=
400kN

(a) (b)
Fig. 2.8

Solution:
Consider x- and y- axes drawn parallel and perpendicular to the inclined plane respectively
(also considered as rotated axes) as shown on Fig. 2.8b. Then draw the rectangular components of
P with respect to these axes and take note of the angles formed by P with the inclined. Hence,
for the perpendicular component of P,
Tan (20o + 30o) = Py/Px or Tan 50o = Py/400 kN
= Tan 50o (400 kN)
Py = 477 kN
for the force P,
Sin (20o + 30o) = Px/P or Tan 50o = 400 kN/P
= 400 / Sin 50o)
P = 622 kN
Checking:
by definition
P = √Px2 + Py2
= √160 000 + 227 529
= √387 529
P = 622 kN (Checked)

Exercises 1

1. In Fig. E-2.1, the X component of a force P is 140 kN to the left. Determine P and its vertical
component.
Y

4
 7 X

Fig. E-2.1

2. The triangular block shown in Fig. E-2.2 is subjected to loads P=1600 N and F=600 N. If
AB = 80 cm and BC = 60 cm, resolve or find the components of each force normal and
tangential to AC.

F = 600
N
 B C

P = 1600 N

Fig. E-2.2

3. Determine the x- and y- components of each of the forces shown in Fig. E-2.3.

60o
Q=200 kN
P = 400 kN X

3 60o F=300 kN
R=300 kN
 Fig. E-2.3

Lesson 3

Resultants of Con-Current Force Systems

In this part of the manual, applications of the basic concepts and definitions learned in
Lesson 2 will be discussed. The illustrative problems discussed here hopes to bring out a more
comprehensive meaning of the definitions previously taken.

Main Objective:

Enhance students to problem solving through applications.

Specific Goals:

1. Identify what method is easy to apply.

2. Develop skills in drafting by if graphical method is applied.

3. Enhance mathematical computation learned previous years of studies.

Resultant of a System of Forces means to take the sum or add up the force vectors to come up with
a single force that would have the same effect as the individual forces on the system taken together.

Determination of the resultant of a system of forces can be done through (a) Graphical
method and (b) Analytical method. If graphical method all force vector arrows must be drawn to
scale and directions properly.

Resultant Force by Parallelogram Method is done by forming a series of parallelogram

simultaneously until all forces are considered or taken. That is two force vectors are added to get a
resultant which is then added to a third force vector, etc., until all the vectors have been added
together to come up with an overall resultant.

Consider the system of three concurrent forces F, P, and Q shown in Fig. 3.1 application of
parallelogram method of vector addition is used, with first forming a parallelogram from F and P to
yield to resultant R1 which is the diagonal drawn from the tails of F and P. The obtained resultant R1,
which replaced F and P as a single force is then added to Q by forming the next parallelogram. Hence
the diagonal projected form the tails of R1 and Q then combines F, P and Q to an overall resultant R.

Y
F F R1
P P R
θx
X X
O O

Q Q

(a) (b)
Fig. 3.1 Resultant Determined Graphically by Parallelogram Method

Resultant Force by Polygon Method is done by connecting the vector forces tip-to-tail in any order
and form a close polygon with the closing side as the resultant.

This is a more direct solution of arriving at a resultant of system of forces. With the definition
of a free vector and the triangle law, consider the same forces F, P and Q above. First the free vector
of P was moved and its tail connected to the tip of force F, then the free vector of Q was moved and
its tail connected to the tip of P. For the overall resultant, the tail of the first vector, F in this
illustration is connected to the tip of the last vector which is Q. Notice that in this graphical method
of finding the overall resultant R1 need not be drawn.

Y
Free vector of P P
Q
Y
F R1
R
θx
 Free Vector of Q

F
P

(a) Q (b)

Fig. 3.2 Resultant Determined Graphically by Polygon Method

For the two graphical solutions of obtaining the resultant of a system of forces, the
magnitude of the resultant is the measure of the length of its vector arrow while the direction is the
measure of its rotation or angle from the positive x-axis.

Triangle Method of Solving for the Resultant is done by connecting vector forces tip-to-tail in any
order and form a close polygon with the resultant as closing side then dividing the polygon with series
of triangles. This is one analytical way of determining resultant where Law of Sine and Law of Cosine
are applied.

Again consider the forces F, P and Q in Fig. 3.1, with angles 1, 2 and 3 drawn for the
respective forces. Once a close polygon is derived by connecting the vectors tip-to-tail as in Fig. 3.2b,
the polygon is divided into two triangles. (For n number of forces the number of triangles that must
be formed is equal n ― 2).The triangle with two known forces F and P, with their subtended angle σ
is first analyzed to get R1, the first resultant. The angle σ can be obtained from angle 1 and 2
considering the horizontal line projected from their point of intersection, i.e. σ = 180o – 1 + 2. Applying
Law of Cosine will solve the magnitude of R1, i.e. R1 = √ F2 + P2 – 2FP Cos σ and the angles α and δ
can be obtained by Law of Sine i.e. R1/Sin σ = F/Sin α = P/Sin δ. With R1, α, and δ known, consider
next triangle formed by this resultant and the next force connected to it, in this case Q. The angle
formed by R1 and Q is ρ and is equal to 180o – (α + 2 + 3). The magnitude of the final resultant then
can be obtained by the Law of Cosine, i.e. R = √ R12 + Q2 – 2R1Q Cos ρ. The direction θx by this
resultant as seen from Fig. 3.3b is θx = 1– (β + δ ), where β can be obtained from Law of Sin, i.e. R/Sin
ρ = Q/Sin β.

Y
2
3
P α ρ
Q
Y 2
F 1 σ R1 φ
F R
1 P δ β
2 1
θ
X X
O O x
3

(a) (b)
Fig. 3.3 Resultant Determined by Triangle Method

Component Method of Determining the Resultant is the most effective way of finding the resultant
of any force systems wherein each of the forces is resolved into their rectangular components,
usually taken as the x- and y- axes.

Consider again the set of forces F, P and Q of the previous methods of finding the resultant.
Also, consider the polygon formed by both the Polygon Method and the Triangle Method. With the
components of the forces drawn as shown in Fig. 3.4b,

Rx = ∑x-components Rx = ∑x-components R = √ Rx2 + Ry2

Rx = Fx + Px + Qx Ry = Fy + Py + (-Qy) Tan θx = Ry/Rx
or or
Ry = Fy + Py – Qy θx = Tan-1 Ry/Rx

Y
Y
Fy Py P
Qy
Py
F
Fx Px
1 P R
2
X
O O
3
 Ry Fy
θx
X

Fx Px Qx
Qx
Rx
Q
Qy
(a) (b)

Fig. 3.4 Resultant Determined by Components Method vs Polygon Method

Illustrative Problems

1. Find the resultant of the concurrent force system shown in Fig. 3.5.

300
kN

y
30o

100 kN
1
O 2 X
45o 30
o

400 kN
200 kN

Fig. 3.5

Solution:
a) By Triangle Method
 Connect the force vectors tip-to-tail and form a close polygon with resultant R as the
closing side and divide it with series of triangles as follows (Fig. 3.5a)

100 kN 1ω 30o
δ
φ2
R1
ρ 400 kN
R

α
α R1

β 45o
300 ε
kN 30o 200 kN
Fig. 3.5a

Consider the triangle with two known sides and solve for the angle formed between them,
i.e. 200 kN and 300 kN, & ε,

α R1

β 45o
300 ε=75o
kN 30o 200 kN

Fig. 3.5b

ε = 30o + 45o = 75o

Solve for R1 and interior angles of Cosine law and Sine law respectively, that is

R1 = √3002 + 2002 – 2(300)(200)Cos 75o = 314.55 kN

R1/Sin 75o = 200 kN/Sin α = 300 kN/Sin β

Sin α = 200 kN(Sin 75o)/R1

= 200 kN(Sin 75o)/314.55 kN

 = 0.614163

α = sin-1 0.614163 = 37.89o

Sin β = 300 kN(Sin 75o)/R1

= 300 kN(Sin 75o)/314.55 kN

= 0.921245
β = Sin-1 0.922859 = 67.11o

Consider the next triangle with R1, R2 & 400 kN load, and angles δ, ρ, σ and solve for angle
σ from the geometry of the figure
σ = 180o – (β + 45o + 60o) = 180o – (67.11o + 105o) = 7.89o

30o
δ
R2
ρ 400 kN

σ 60o
Fig. 3.5c R1= 314.55kN

β=67.11o 45
o

Solve for R2 by cosine law

200 kN
R2 = √400 + 314.55 – 2(400)(314.55)Cos 7.89 = 98.41 kN
2 2 o

Solve for δ & ρ, by the law of sine, that is

R2/Sin 7.89o = 400 kN/Sin ρ = 314.55 kN/Sin δ,

Hence

Sin δ = 314.55 kN(Sin 7.89o)/R2

 = 314.55 kN(Sin 7.89o)/98.41 kN

=0.438764

δ = Sin-1 0.438764 = 26.02o

Sin ρ = 400 kN(Sin 7.89o)/R2

= 400 kN(Sin 7.89o)/98.41 kN

= 0.557958

ρ = Sin-1 0.557958 = 146.09o

Consider the last triangle with resultant R & angles ω & φ and solve for ω from the geometry
of the figure,

100 kN 1ω
30o
2
δ
φ
Fig.3.5d R2=98.41
R kN
400 kN

ω = 180o – (δ + 30o + 26.57o) = 180o – (26.02o + 56.57o) = 97.41o

Where: Tan-1 (1/2) = 26.57o
Solve for R and interior angles of last triangle by cosine law and sine law respectively, that is

R = √R22 + 1002 – 2(R2)(100)Cos 97.41o

R = √98.412 + 1002 – 2(98.41)(100)Cos 97.41o

R = 149.07 kN

R/Sin ω = R2/ Sin Ф

Sin φ = R2(Sin ω)/R
= 98.41 kN(sin 97.41o)/149.07 kN
= 0.654646
Ф = Sin-1 0.654646 = 40.89o
θx = Ф – 26.57o
= 40.89o – 26.57o
θx = 14.32o

b) By Components Method

F=300 kN
Fy
y Py
Fx 30 o

P=100
Px
kN
Fig. 3.5e 1
O 2 X
Tx
45o 30o
Qx
Ty
Q=400 kN
T=200 kN
Qy

By definition R = √Rx2 + Ry2 , and θx = Tan-1 Ry/Rx

where: Rx = ∑ x- components
Ry = ∑ y - components

Assign a letter to each of the forces, say P = 100 kN, F = 300 kN, T = 200 kN and Q = 400 kN
and project the rectangular components of each force anywhere from their line of action (see
illustration). From the definition of components of any force F,

Fx = F(Cos θx) while Fy = F(Sin θx)

make a table of the force with their corresponding value of components and take their
summation
 Force (kN) X-Component (kN) Y-component (kN)

P = 100 100 (2/√5) = +89.44 100 (1/√5) = +44.72

F = 300 300 ( Sin 30o) = -150 300 (Cos 30o) = +259.81

T = 200 200(Cos 45o) = -141.42 200(Sin 45o) = -141.42

Q = 400 400(Cos 30o) = +200 400(Sin 30o) = 150

∑ Rx = +144.43 Ry = -37.28

Compute value of resultant R from,

R = √Rx2 + Ry2

R = √144.432 + (-37.28)2

R = 149.16 kN

Finally for the direction of resultant R

θx = tan-1 Ry/Rx

= tan-1 -37.28/144.43

θx = 14.47o, from the +x-axis clockwise

 2. The block shown in Fig. 3.6a is acted upon by its own weight W = 200 kN, a horizontal
force Q = 600 kN, and the pressure P exerted by the inclined plane. The resultant R of these
forces is up and parallel to the inclined thereby sliding the block up it. Determine P and R.

W = 200 kN WyY R = Rx
Qy 30 W =Wx
200 kN X
o
β Px
Q=600 kN Q=600 kN

P 30o Qx
Py 15o P
30 o
30o Β=45o
(a) 15 (b)30
o o

Fig. 3.6

Solution:

In Fig. 3.6(b) an x- and y- axes was drawn parallel and perpendicular to the inclined. This axes
is rotated 30o with the horizontal, hence, this axes shows that forces Q and W were at an angle of 30o
with the x- and y- axes respectively while the resultant R (arbitrarily drawn) coincides with the x- axis
meaning R must be equal to Rx or the summation of x- components. That is, with components of
the forces resolved on the x- axis,
R = Rx = ∑X

R = Rx = Qx + (-Px) + (-Wx)
R = Rx = Q(Cos 30o) – P(Sin 15o) – 200(Sin 30o)
 R = Rx = 600 kN(Cos 30o) – P(Sin 15o) – 200 kN(Sin 30o)
R = Rx = 419.61 kN – P(Sin 15o) (1)

Since R = Rx, then Ry must be equal to zero, that is,

Ry = ∑Y = 0
Ry = Py + (-Qy) + (-Wy) = 0
Py – Q(Sin 30o) – W(Cos 30o) = 0
Py – 600 kN(Sin 30o) – 200 kN(Cos 30o) = 0
Py = 473.20 kN
With the value of Py = 473.20 kN obtained and by definition of components,
Py = P(Cos 15o) = 473.20 kN
From which the value of force P is obtained as,
P = Py/ Cos 15o
P = 473.20 kN/ Cos 15o
P = 489.90 kN
Going back with equation (1) with P = 489.90 kN, value of resultant R is,
R = Rx = 419.61 kN – P(Sin 15o)
= 419.61 kN – 489.90 kN (Sin 15o)
= 419.61 kN – 126.79kN
R = 292. 81 kN
 Exercises 2

1. Determine the resultant of the concurrent forces shown in Fig. E-2.1 by a) Graphical method
and b) by Components.
Y
P = 1 kN
20o
P = 2 kN

F = 3 kN X
30o
3 Q = 4 kN
T = 6 kN
4

Fig. E-2.1
2. The resultant of the concurrent forces shown in Fig. E-2.2 is 3000 N pointing up along the Y
axis. Compute the values of F and θ required to give this resultant. Verify also the direction
of the line of action of F as drawn.

Y
F

θ
Q = 5000 X
N 30 o

P=2400 N

Fig. E-2.2
 Lesson 4
Principle of Moments and Couples

The previous lesson as far as the external effects of forces is concern with their action to a
body has something to do with its linear motion or movement. This is to say, whether the forces tend
to push (or pull) up or down; to the right or left; or in a sloping manner the body acted upon by these
forces. This somehow limits the total effects of a force or system of forces on a body in that the
tendency of the object to spin or rotate about an axis was excluded. Hence, Lesson 4 deals with the
discussions of rotational or twisting effect of forces on rigid bodies.

Main Objective:

Expand the knowledge of students in effects of forces on rigid bodies.

Specific Goals:

1. How to define moments of a force or system of forces.

2. Locate center of action of the resultant of system of forces.

3. Define the resultant fully as to the three basic characteristics of a force (i.e. magnitude,
direction, and point of application).
Moment of a Force about an axis or line is the measure of its ability to produce turning or twisting
about the axis. The magnitude of the moment of a force about an axis is defined as

Moment of Force about a moment center O = Force x Distance

or

MFO = F x d

Axis of rotation

d O Moment center
F
Moment arm

Fig. 4.1 Moment of a Force

Moment Arm is the perpendicular distance between the axis of rotation and the line of action of the
force.

Center of Moments (or Moment Center) is the intersection of the axis of rotation with the plane of
the force and its moment arm.
Varignon’s Theorem states that at any point, the algebraic sum of the moments of the components
of a force equals the moment of the force. (See Fig. 4.2 for illustrations)

Y
Fy F
x θx
A Fx
Fy

B Fx
y

d iy
Fy

C Fx X
O
ix

Fig. 4.2 Illustration of Varignon’s Theorem

Figure 4.2 illustrates the application of the principle of moments also known as Varignon’s
Theorem. As seen from the figure, consider a force F, inclined at an angle θ x with the x- axis passing
through a point A having the coordinates (x, y). Defining the moment of this force with O as its
moment center, its action line is extended until it passes O then a line is drawn from O perpendicular
to this line of action which serves as the moment arm. By definition then, the moment of F about O
is given by,

+ MFO = F • d, clockwise

This means of determining the magnitude of MFO is rather inconvenient as it is sometimes

hard to get the measure of the moment arm d. Considering resolving the components of force F at A,
with its coordinates given as (x, y), the moment of F about O can be defined then as,

+ MFO = F • d = ±Fx • Y ± Fy • X
 As seen from Fig. 4.2, the extended line of action of force F intercepts the X- axis at C and the
Y- axis at B, from which ix and iy were the measures of these intercepts, respectively. The moment of
force F about O, can also be defined from these intercepts by considering the components F x at point
B with moment arm iy and Fy at point C with moment arm ix, as follows;
+ MFO = F • d = Fx • iy (where Fy passes thru O so its moment is zero)
+ MFO = F • d = Fy • ix (where Fx passes thru O so its moment is zero)

The forces of a system can be regarded as the components of their resultant. Thus, about
any point, the moment of the resultant of a force system equals the algebraic sum of the
moments of each force taken separately. Consider the diagram below with R as the resultant of
forces T and P. This diagram will prove that the moment of the resultant R about M is equal to the
algebraic sum of moments of forces T and P about M.
Y

R
A C

T t r P D

β p
α θ
E X
•
O M

Fig. 4.3
From the figure,
BE = CE + BC
or R Sin β = T Sin α + P Sin θ
Multiplying each term of the equation by OM, yields
R (OM Sin β) = T (OM Sin α) + P (OM Sin θ)
Where the following can be defined from the geometry of the figure
r = OM Sin β (perpendicular distance or moment arm of R with respect to M)
t = OM Sin α (perpendicular distance or moment arm of T with respect to M)
p = OM Sin θ (perpendicular distance or moment arm of P with respect to M)
Hence,
R
• r = •T t + •P p
 Illustrative Problems

1) In Fig. 4.4a, a force F passing through C causes a clockwise moment of 1200 N-m about A
and a clockwise moment of 700 N-m about B. Determine the force F and its x intercept ix.

Y Y
A A
• •
F Fy
2m 2m

•C •C
Fx
3m 3m Fy

X X
O • O • •
B Fx D B
ix
5m 5m

(a) (b)
Fig. 4.4
Solution:
From point C in Fig. 4.4a, draw a force vector F pointing up to the left (Fig. 4.4b) thereby
producing a clockwise moment about A. Also, its line of action when extended to point D should
produce a clockwise moment as described in the problem. Hence, resolving x- and y- components
of F at point C and taking moment with A as moment center gives
+ MFA = F•x 2m + F•y 0
Or
1 200 N-m = F•x 2m
Fx = 600 N
Also, taking moment at B as the moment center gives
+ • x 3m + F•y
MFB = F 5m
Or
700 N-m = -(600•N 3m) + •Fy 5m
700 N + 1 800 N = Fy(5)
Fy = 500 N , upward
With Fx = 600 N and Fy = 500 N, known then force F by definition of components of a force gives
 F = √Fx2 + Fy2 θx = Tan-1 Fy/Fx
= √(600 N)2 + (500 N)2 = Tan-1 500 N/600 N
F = 781.02 N θx = 39.8o
To solve for the x intercept ix , consider taking moment at B with the x- and y- components resolved
at point D
+ MFB = F•x 0 + F•y d
or
700 N-m = 500•N d
d = 1.4 m
Hence,
Ix = 5 m – d or ix = 5 m – 1.4 m
Ix = 3.6 m

2) Assuming clockwise moments positive, determine the moments of force F=4 500 N and force
P=3 610 N about points A, B, C and D.
A•
F
 •C

1m
P
• •
D B
1m

Fig. 4.5
Solution:
For force F = 4 500 N F
Fx
A• y

F = 4 500
d 5 3 N

Fy 4
Fig.4.5a •C
Fx

1m
• •
D B
1m

Resolving the components of F from its given inclination gives

3/Fy = F/5 = 4/Fx
Or
Fy = 3/5(F) Fx = 4/5(F)
= 3/5(4 500 N) = 4/5(4 500 N)
Fy = 2 700 N, to the left Fx = 3 600 N, upward
Defining the moments of F with respect to any points in general gives
+ MO = F x d = ∑M-components
Hence
+ MFA = Fx(0) + Fy(5m) + MFB = Fx(6m) + Fy(0)
 =―(2 700 N x 5m) =(3 600 N x 6m)
+ MFA =― 13 500 N-m, counterclockwise + MFB = 21 600 N-m, clockwise

+ MFC = Fx(0) + Fy(5m) + MF = Fx(3m) + Fy(1m)

D

=―(2 700 N x 5m) =(3 600 N x 5m) + [-(2 700 N x 1m)]

+ MFC =13 500 N-m, clockwise =108 000N-m ―2 700 N-m
+ MFD = 8 100 N-m, clockwise

Moments of force P=3 610 N with respect to points A, B, C and D are left for the students as
an exercise.

3) Two forces P and Q passes through a point A which is 4m to the right and 3m above a moment
center O. Force P is 2 000 N directed up to the right at 30o with the horizontal and force Q is 1
000 N directed up to the left at 60o with the horizontal. Determine the moment of the resultant
of the two forces with respect to O.
Solution:
 Draw a sketch of the problem for analysis

R
Q=1
000N P=2 000N
60 A 30
o
• o
Qx Px

Qy Py
3
m

O
•

4
m

Fig. 4.6
Considering Fig.4.6, by parallelogram law the resultant R of P and Q would have a moment
with respect to O equal to

+ MRO = R x d (this is with the line of action extended until it passed O)

Somehow this is an inconvenient approach and requires more analysis. Applying Varignon’s theorem
that the moment of resultant of system of forces equal the summation of the moment of the forces
taken separately gives

+ MRO = R • d = P • p + Q• q

This is with p and q as moment arm of forces P and Q with respect to O (not drawn in the
figure) which is still an inconvenient approach.

By considering the components of forces P and Q at point A and applying the principle that
moment of a force is equal to summation of the moment of its components yields to

+ MRO = R x d = Px(3m) + Py(4m) + Qx(3m) + Qy(4m)

Where:
 Px = P(Cos 30o) Py = P(Sin 30o)
= 2 000 N (Cos 30o) = 2 000 N (Sin 30o)
Px = 1 732 N (to the right) Py = 1 000 N (upward)

Qx = P(Cos 60o) Qy = P(Sin 60o)

= 1 000 N (Cos 60o) = 1 000 N (Sin 60o)
Qx = 500 N (to the left) Qy = 866 N (upward)
Hence

+ MRO = R x d = [1 732 N(3m)] + [-1000 N(4m)] + [-500 N(3m)] +[-866 N(4m)]

= 5 196 N-m – 4 000 N-m – 1 500 N-m – 3 464 N-m

+ MRO = – 3 768 N-m, (counterclockwise)

Exercises 3

Consider the truss loaded as shown in Fig. E-3.1 for the next set of problems.

26 kN

20 kN
12
3

4
B D
 5

3m

A
AX

AY

Fig. E-3.1

1. What is the moment of forces at joints (a) B, (b) C, (c) D, and (e) F with joint A as their moment
center?
2. Considering summation of moments of all the forces at joint A equal to zero, what should be
the magnitude of the reaction at the roller support at joint E? Take clockwise moment
positive.
3. What if the summation of the moments of the forces at joint E is equal to zero, what is the
magnitude of the vertical reaction at the hinge support at A?

A Couple consists of two equal, opposite and parallel forces having separate lines of action. Couple
is characterized by the following:
a) The resultant force of a couple is zero
b) Its moment is the product of one of the force and the perpendicular distance between their
action line
c) The moment of a couple is the same for all points in the plane of the couple, and
d) A couple can only be balanced by an equal and opposite couple in the same or parallel plane.

x a
C A B
• • •

F
 F

Fig. 4.8 Moment of a couple

To prove statement (c) for properties of couple, consider Fig. 4.8,

MFA = F(a), (clockwise) or MFB = F(a), (clockwise)
Hence for any arbitrary point C at a distance x from A in the same plane of the couple
MFC = F (x + a) – F(x)
= F(x) + F (a) – F(x)
+ MFC = F(a), clockwise

e) The effect of a couple is unchanged

1. If it is rotated through any angle in its plane.
2. If it is shifted to any other position in its plane.
3. If it is shifted to a parallel plane.

Resolution of a Force into a Force and a Couple is illustrated as follows: Given a force F acting at
point A (Fig. 4.9). By adding forces F’ and F” with the same magnitude as that of F and at the same
time are collinear and parallel to F at any point B, the effect of F does not changed.

•A
C = Fd
F” A•
F d
d
•B
•B

F’
F’

Fig 4.9
 Illustrative Problems

1. From Fig. 4.10a, transform the couple shown into an equivalent couple whose forces are
horizontal and act through points CD.

9 kN

9 kN 9 kN
A 5m 9 kN F =15
4m 9 kN
C B kN C
C
E 3m 3m
5m
D E
D
15 D
kN
F F 9 kN

(a) (b) (c) (d)

Fig. 4.10 Transformation of a couple

Solution:
From property of a couple, its effect does not changed if shifted form a parallel plane. Thus
in Fig. 4.10b, the couple was shifted from plane joining AB to parallel plane CDEF. Next, it is rotated
to plane CDEF as in Fig. 4.10c, still without changing its effect on the block. Considering now the
constant effect of the couple force F can be found as follows
M (Couple) = 9 kN x 5 m = F x 3m
Or F = 45 kN-m/3 m
F = 15 kN, acting horizontally at C and D

2. Determine the resultant moment about A of the system of forces shown in Fig. 4.11. Each
square is 1 m on a side.

800 N 800 N 1000 N

1000 N
1000 N

2 √5
1
2 √1
3
3
2000 N

A
Fig. 4.11
Solution:

From the principle of moment by Varignon’s, moment of resultant is equal to summation of

moments of each forces of the system. Hence,

+ MRA = ∑M of forces about A

In terms of the forces
+ MRA = (2000)X(3 ) + (2000)Y(2 ) + [-(1000)X(2 )] + [-(1000)Y (1 )] +(800)(2 ) + [-(1000)(√2 )]

Where:
(2 000)X = (2 000 N)(3/√13 ) = 1 664 N X-component of 2 000 N force
(2 000)Y = (2 000 N)(2/√13 ) = 1 109 N Y-component of 2 000 N force
(1 000)X = (1 000 N)(1/√5 ) = 447 N X-component of 1 000 N force
(1 000)Y = (1 000 N)(2/√5 ) = 894 N Y-component of 1 000 N force
√2 m = √(12 + 12) m = 1.414 m -- distance between the 1 000 N couple force
Then,
+ MRA = (1664N)(3m) + (1109N)(2m) - (447N)(2 m) - (893N)(1m) + (800N)(2 m) - (1000N)(1.414m)
+ MRA = 5 608 N-m, clockwise

3. Replace the system of forces shown in Fig. 4.12 by an equivalent force through O and a
couple acting through A and B. Solve if the forces of the couple are (a) horizontal and (b)
vertical.

1414 N
A•
1
1

√1 3
3 2

1000 N •B
√5
1
• 2
O 2000 N

Fig. 4.12
Solution:
P =1414 N
A•
1 PX
1

√1 3
3 2

T =1000 N •B
√5
1
• 2
O Q =2000 N

For the resultant of the forces, by component method

R = √RX2 + RY2 θX = Tan-1 RY/RX
Where:
RX = ∑x- components
= PX + QX + TX
= P(1/√2) + Q(2/√5) – T(2/√13)
= 1414(1/√2) + 2000(2/√5) – 1000(2/√13)
RX = 2 234 N

RY = ∑y- components
= PY + QY + TY
= P(1/√2) – Q(3/√5) – T(3/√13)
= 1414(1/√2) – 2000(1/√5) – 1000(3/√13)
RY = 727 N

R = √( 2 234 N)2 + ( 727 N)2 θX = Tan-1 (727/2 234)

= 2 349 N θX = 18.03o
The moment produced by the resultant R with respect to point O is
MRO = ∑moments of forces about O
= Px(5m) – Py(4m) + Qx(0) + Qy(4m) – Tx(2m) – Ty(1m)
= 999.85 N (5m) – 999.85 N (4m) + 1 788.85 N (4m) – 554.70 N (2m) – 832.05 N (1m)
 MRO = 6213.8 N-m, clockwise
F = 1242.76 N
A P =1414 N
F = 1553.45 N 1 PX
1

√1 3
3 2

T =1000 N B F = 1553.45 N

√5 F = 1242.76 N
1
• 2
O
R = 2 349 N
To replace the moment produced by R with a couple at A and B with forces acting horizontally
Mc = 6213.8 N-m = F x 4 m; Hence F = 6213.8 N-m/ 4m; F = 1553.45 N
Problem (b) is left for students as part of their exercises.
Mc = 6213.8 N-m = F x 5 m; Hence F = 6213.8 N-m/ 5m; F = 1242.76 N