CHAPTER 3: FORCES
3.1 DEFINITION OF A FORCE

Force may be defined as that which causes, or tends to cause, a change in the state of an
object.

When a force acts on an object, the effects can be
i.
ii.
iii.
iv.
v.
change of the size of the object,
change of the shape,
change of the stationary state,
change of the speed and
change of direction of the object.

Because both the magnitude and direction of a force are significant, forces are vector
quantities.

Types of forces in mechanics include weight, pushes and pulls, friction, normal reaction,
tension and thrust.

A force is characterized by:
i)
ii)
iii)
iv)

its magnitude
its direction
its point of action
the type of force
The unit of force is Newton or kgms-2. One newton is the force required to give a mass of 1
kilogram an acceleration of 1 metre per second square ( 1 m/s2 ).
1 N = 1 kgm/s2
3.2 MASS AND WEIGHT
3.2.1 DEFINITION OF MASS

Mass is defined as the amount of matter contained in an object.
3.2.2 DEFINITION OF WEIGHT

Weight is a measure of how strongly gravity pulls on that matter and depends on where it is.
Where
W = mg

W = Weight
m = mass
g = gravity,
where g = 9.81 m/s2 (N/kg)
Well, mass is a measurement of how much matter is in an object; weight is a measurement of
how hard gravity is pulling on that object.
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CHAPTER 3: FORCES

Your mass is the same wherever you are--on Earth, on the moon, floating in space--because
the amount of stuff you're made of doesn't change. But your weight depends on how much
gravity is acting on you at the moment; you'd weigh less on the moon than on Earth, and in
interstellar space you'd weigh almost nothing at all.

Thus if you were to travel to the moon your weight would change because the pull of gravity
is weaker there than on Earth but, your mass would stay the same because you are still
made up of the same amount of matter.

Imagine yourself out is space away from any gravitational field, with a bowling ball in your
hands. Let it go and it just floats in front of you. Without gravity, it has no weight.
3.2.3 DIFFERENCES BETWEEN MASS AND WEIGHT
The following table will cover a few differences between mass and weight
Mass
Weight
Mass is the quantity of matter in a
body.
Weight is the force with which a body is attracted
towards the center of the earth by the gravity.
The mass of an object is constant on
Earth and even in space.
The weight of an object can vary from place to place
and becomes zero at the center of the earth. It is also
zero in places that are far away from earth. It has no
weight without gravity,
m = F/a is the mass of a moving body.
Where a is an acceleration
W = mg, is the weight of a body, where g = 9.81 (N/kg)
Mass is a scalar quantity.
Weight is a vector quantity.
Mass is a base quantity.
Weight is a derived quantity.
The unit of mass in the S.I system is
Kilogram (kg).
The unit of weight in S.I system is Newton (N)
1. Mass is an intrinsic property of a
body.
2. It is independent of any external
factor.
1.
2.
Weight of an object depends on the mass of an
object that is attracting it.
Weight is also dependent on the force with which
it is attracted.
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CHAPTER 3: FORCES
3.3 NEWTON’S LAW
3.3.1 NEWTON'S FIRST LAW

Newton's first law of motion is often stated as an object at rest stays at rest and an object in
motion stays in motion with the same speed and in the same direction unless an unbalanced
force acted upon it

There are two parts to this statement - one that predicts the behavior of stationary objects
and the other that predicts the behavior of moving objects. The two parts are summarized in
the following diagram.
FORCES ARE BALANCED
Object at rest
(v = 0 m/s )
Object in
Motion
(v ≠ 0 m/s )
a= 0 m/s2
a= 0 m/s2
Stay at rest

Stay in motion at same
speed and direction
Have you ever experienced inertia in an automobile while it is braking to a stop? The force of
the road on the locked wheels provides the unbalanced force to change the car's state of
motion, yet there is no unbalanced force to change your own state of motion. Thus, you
continue in motion, sliding along the seat in forward motion. A person in motion stays in
motion with the same speed and in the same direction.
3.3.2 NEWTON’S SECOND LAW

Newton’s second law of motion states that the rate of change of momentum of a moving
body is proportional to the resultant external force acting on the body and takes place in the
direction of the force.

It pertains to the behavior of objects for which all existing forces are not balanced.

The law states that the acceleration of an object is dependent upon two variables:
1. the net force acting upon the object
2. the mass of the object.
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CHAPTER 3: FORCES


The acceleration of an object depends directly proportional to the net force acting upon the
object (a ∞ F), and inversely proportional to the mass of the object (a ∞ 1/m). As the force
acting upon an object is increased, the acceleration of the object is increased. As the mass
of an object is increased, the acceleration of the object is decreased.
Therefore :
Force (F) = mass (m) x acceleration (a)
F = ma
Where :
F = Force (measured in kg/ms-2 @ Newton (N)).
m = mass (measured in kg)
a = acceleration (measured in ms-2)

Example 1:
A force of 10 N acts on an object of mass 5 kg on a smooth floor. Find its acceleration.
a
10N
𝐹 = 10 𝑁 ,
𝑚 = 5 𝑘𝑔
𝐹=𝑚𝑥𝑎
10 = 5𝑥𝑎
𝑎=
10
5
= 2 𝑚𝑠 −2
3.3.3 NEWTON’S THIRD LAW

Newton’s third law of motion states that action and reaction forces are equal in magnitude
and opposite direction.
FA
FB
Action Force (FA) = Reaction Force (FB )
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CHAPTER 3: FORCES
3.5
FORCE IN EQUILIBRIUM

When all the forces that act upon an object are balanced, then the object is said to be in a
state of equilibrium. The forces are considered to be balanced if the rightward forces are
balanced by the leftward forces and the upward forces are balanced by the downward forces.

For a system of forces to be in equilibrium, when :
a)
No resultant force must act. If a resultant force acts then the system
would accelerate and would therefore not be in equilibrium.
Summation of force in the direction of x-axis must be zero.
Summation of force in the direction of y-axis must be zero.
∑ Fx = 0, ∑ Fy = 0
b)
No resultant turning effect must act.
∑M = 0

This however does not necessarily mean that all the forces are equal to each other. Consider
the two objects pictured in the force diagram shown below. Note that the two objects are at
equilibrium because the forces that act upon them are balanced; however, the individual
forces are not equal to each other.

Examples1:
8N
8N
∑ F = 8 – 8 = 0 ( state of equilibrium)
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
Example 2 :
F3 = 2N
F1 = 4N
F2 = 4N
F4 = 2N
∑ F x = F 1 – F2 = 4 – 4 = 0 N
∑ Fy = F 3 – F4 = 2 – 2 = 0 N
( state of equilibrium)
Equal in magnitude but opposite in direction
3.5.1

LAMI'S THEOREM
In statics, the Lami's theorem is an equation relating the magnitudes of three coplanar,
concurrent and non-collinear forces, which keeps an object in static equilibrium, with the
angles directly opposite to the corresponding forces.
b
c
FORCE
OPPOSITE
ANGLE
a
Aº
b
Bº
c
Cº
Aº
Bº
Cº
a

According to the law,
𝐚
𝐛
𝐜
=
=
𝐬𝐢𝐧 𝐀° 𝐬𝐢𝐧 𝐁° 𝐬𝐢𝐧 𝐂°

where a, b, and c are the magnitudes of three coplanar, concurrent and non-collinear forces,
which keeps the object in static equilibrium, and A, B and C are the angles directly opposite
to the forces a, b and c respectively
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CHAPTER 3: FORCES
3.5.2 PROOF OF LAMI'S THEOREM

Suppose there are three coplanar, concurrent and non-collinear forces, which keeps the
object in static equilibrium.

By the triangle law, we can re-construct the diagram as follow:
 Draw a closed triangle of force.
 Draw the straight line, a with arrow
 Then draw the parallel line b and c
 Determine position of the angle
(180-Cº)
b
c
Cº
b
Aº
Aº
B
Cº
a
(180 - Aº )
c
(180 - Bº)
Bº
a

After draw the triangle, use simple trigonometry to solve the problem

By the law of sines,
𝐬𝐢𝐧 (𝟏𝟖𝟎°−𝐀°)
𝐚
=
𝐬𝐢𝐧 (𝟏𝟖𝟎−𝐁°)
𝐛
=
𝐬𝐢𝐧 (𝟏𝟖𝟎°−𝐂°)
𝐜
Where
sin (180º – Aº) = sin Aº
sin (180º – Bº) = sin Bº
sin (180º – Cº) = sin Cº
𝐬𝐢𝐧 𝐀° 𝐬𝐢𝐧 𝐁° 𝐬𝐢𝐧 𝐂°
=
=
𝐚
𝐛
𝐜
𝐚
𝐛
𝐜
=
=
𝐬𝐢𝐧 𝐀° 𝐬𝐢𝐧 𝐁° 𝐬𝐢𝐧 𝐂°
Sinus Rules
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Lami’s theorem
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CHAPTER 3: FORCES

Example 1:
A 20 N weight is suspended by two string as shown in figure. Find the tensions in both strings
.
1200
P
1050
Q
1350
20 N
Solution:
FORCE
OPPOSITE
ANGLE
P
135º
Q
105
20N
120º
According to the lami’s theorem
P
Q
20
=
=
sin 135 sin 105 sin 120
So that;
P
20
=
sin 135° sin 120°
20
P=
x sin 135°
sin 120°
20
=
x 0.707 = 16.33N
0.866
Q
20
=
sin 105° sin 120°
20
Q=
x sin 105°
sin 120°
20
=
x0.966 = 22.31N
0.866
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CHAPTER 3: FORCES

Example 2:
A 10 N weight is suspended by two string. Find the tensions in both strings.
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
1000
P
1100
Q
1500
10 N
Solution:
FORCE
OPPOSITE
ANGLE
A
150º
B
110º
10N
100º
According to the lami’s theorem
A
B
10
=
=
sin 150 sin 110 sin 100
So that;
A
10
=
sin 150° sin 100°
10
A=
x sin 150 °
sin 100°
10
=
x 0.5 = 5.07 N
0.985
B
10
=
sin 110° sin 100°
10
x sin 110°
sin 100°
10
=
x 0.94 = 9.54 N
0.985
B=
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CHAPTER 3: FORCES

Example 3:
Sarah and Mimi carried a heavy bag as shown in a diagram below.
Find the tensions in both strings (X and Y).
Y
X
500
700
600
Bag
50 N
Solution
FORCE
OPPOSITE
ANGLE
X
160º
Y
150º
50N
50º
According to Lami’s theorem
X
Y
50
=
=
sin 160 sin 150 sin 50
So that;
X
50
=
sin 160° sin 50°
50
X=
x sin 160°
sin 50°
50
=
x 0.342 = 22.32 N
0.766
Y
50
=
sin 150° sin 50°
50
Y=
x sin 150°
sin 50°
50
=
x 0.5 = 32.63 N
0.766
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CHAPTER 3: FORCES

Exercise 1:
A 5 kg weight is supported by two ropes as shown in the figure below. Find the tensions in
both ropes (A and B).
600
300
A
B
5 kg
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CHAPTER 3: FORCES

Exercise 2:
Use Lami’s theorem to find the values of F1 and F2
F1
15 N
1300
1200
F2
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CHAPTER 3: FORCES

Exercise 3:
An object with a mass of 10 kg is suspended by a string that is tied to a hook which is
attached to the ceiling as shown in the figure below. A horizontal force F p is pulling the string
at the middle, so that it makes an angle of 60º from the horizontal plane. Find the value of Fp .
( assume g = 10 N/kg)
60º
Fp
10 kg
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CHAPTER 3: FORCES
3.6

RESULTANT FORCE
The net force that acts on an object when two or more forces act on it is known as the
resultant force. If the forces act in the same straight line the resultant is found by simple
addition or subtraction as shown in Figure below
a. Different magnitude
and equal direction
b. Different magnitude and
differrent direction
a.
b.
Dif
5N
3N
Dif
4N
R=5+3=8N
7N
R=7-4=3N
5N
d.
c.
7N
3N
4N
4N
R=7-4=3N
R=5+3–4=4N
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CHAPTER 3: FORCES
3.6.1
RESULTANT FOR TWO FORCES

The resultant of forces that do not act in the same straight line can be determined by using
the parallelogram law.

The parallelogram law states that if two forces acting at a point are represented in size and
direction by the sides of a parallelogram drawn from the point, their resultant is represented in
size and direction by the diagonal of the parallogram drawn from the point.

Using the parallelogram law, the resultant of the forces P and Q in Figure (a) is thus
represented by R in Figure (b) below.
P
P
θ
θ
O
O
θ
Q
Q
(a)

R
180° - θ
(b)
The resultant force, R can be calculated by the cosinus rule,
𝑹 = √𝑷𝟐 + 𝑸𝟐 − 𝟐𝑷𝑸 𝒄𝒐𝒔 (𝟏𝟖𝟎° – 𝜽)
since
cos (180° – θ) = - cos θ
so that,
𝑹 = √𝑷𝟐 + 𝑸𝟐 + 𝟐𝑷𝑸 𝒄𝒐𝒔 𝜽

The direction of the resultant R is given by the sinus rule
Sinus rule
C
R
θ
sin α =
(180º - θ)
α
O
sin (180° − θ) sin α
=
OC
AC
θ
A
AC
OC
sin (180° − θ)
α = sin −1 (
AC
sin (180° − θ))
OC
since
sin (180° - θ) = sin θ
so
AC
α = sin −1 ( sin θ)
OC
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CHAPTER 3: FORCES

Example 1:
Two concurrent forces of 3 N and 4 N act at a body at right angles to one another. Determine
the magnitude and direction of the resultant force.
Solution:
A
C
R
3N
900
α
B
4N
O
Using Cosinus Rule
Y = 3 N, X = 4 N, θ = 900
sin α =
𝑅 = √𝑋 2 + 𝑌 2 + 2𝑋𝑌𝑐𝑜𝑠𝜃
𝑅 = √[
32
+
42
+
OA
sin θ
OC
3
= (sin 90° ) = 0.6
5
2(3)(4)cos900
α = sin−1 0.6
= 5N
= 370
= sin-1 0.6 = 370

Example 2:
Compute the resultant forces produce by the following system of forces:
b)
a)
1.5 kN
8N
600
O
800
12 N
O
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1 kN
74
CHAPTER 3: FORCES
Solution:
a)
C
A
8N
R
600
α
B
O
12 N
Direction:
Magnitude:
sin α
sin (180° − 60°)
=
𝐵𝐶
𝑅
R =  ( P2 + Q2 + 2PQcos θ
sin α
sin 60°
=
= 0.398
8
17.4
=  [ 82 + 122 + 2(8)(12)cos 600 ]
α = sin-1 0.398 = 23.5º
= 17.4 N
b)
A
C
Magnitude:
1.5 kN
R =  [ 1.52 + 12 + (1.5)(1)cos 800 ]
R
= 1.9 kN
800
Direction:
α
O
1 kN
B
sin α = (1.5/1.9) sin 800 = 0.777
α = sin-1 0.777 = 510

Example 3
A resultant force of 130 N was produced by the action of two concurrent forces of 70 N and
80 N. Find the angle between the two forces.
Solution:
R2 = P2 + Q2 + 2PQcos θ
130 N
P=70 N
θ
𝑐𝑜𝑠𝜃 =
𝑅 2 − 𝑃2 − 𝑄2
2𝑃𝑄
𝑐𝑜𝑠𝜃 =
1302 − 702 − 802
= 0.5
2(70)(80)
Q=80 N
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θ = cos −1 0.5 = 60°
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CHAPTER 3: FORCES
3.6.2
RESOLUTION OF FORCES

Any single force may be split into two components. This is, in fact, the reverse of combining
two forces into a resultant force. In this case the single force is resolved into two other forces.
This process is known as resolution of forces.

The most useful way to resolve a single force into two forces is to produce components which
are perpendicular to each other.

Figure below shows single force, F. Applying the parallelogram rule, the force F is resolved
into a pair of perpendicular components: the vertical force, Fy (Y- components) and
horizontal force, Fx (X- components)
Y (axis)
(90º)
F
Fy
F
θ

Resolved into two
components ,
Fx and Fy
θ
0
Fx
X(axis)
(0º or 360º)
From the geometry of the rectangle in the figure above, the magnitude of each component
are :
Horizontal component:
Vertical component:
Fx = F cos θ
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Fy = F sin θ
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CHAPTER 3: FORCES

Example 1:
Find the horizontal and vertical components for the force below.
80 N
60º
Solution:
The force F is resolved into a pair of perpendicular components:
i. the horizontal force, Fx (X- components)
ii. the vertical force, Fy (Y- components)
Y (axis)
(90º)
80 N
80 sin 60º
60º
0
80 cos 60º
X(axis)
(0º or 360º)
Horizontal component:
Vertical component:
Fx = F cos θ
Fy = F sin θ
= 80 x cos 600
= 80 x sin 600
= 40 N
= 69.3 N
= 40 N to right
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CHAPTER 3: FORCES

Example 2:
Find the horizontal and vertical components for the force.
40 N
60º
Solution
Y (axis)
(90º)
40 sin 120º
40N
120º
60º
X(axis)
(180º)
40 cos 120º
0
X(axis)
(0º 360º)
Horizontal component:
Vertical component:
Fx = F cos (180º - 600)
Fy = F sin (180º - 600)
= 40 x sin (1200)
= 40 x cos
(1200)
= 34.6 N
= - 20N
= 40 N to right

Example 3:
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CHAPTER 3: FORCES
Find the horizontal and vertical components for the force.
30º
50N
Solution
90º
(360º - 30º)
Fx =50 cos (360º- 30º)
180º
0º or 360º
30º
Fy = 50 sin (360º-30º)
50N
270º
Horizontal component:
Vertical component:
Fx = F cos (360 - 300)
Fy = F sin (360 - 300)
= 50 x cos (3300)
= 50 x sin (3300)
= 43.3N
= - 25 N
= 40 N to right

Exercise 1:
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CHAPTER 3: FORCES
Find the horizontal and vertical components for the forces in the diagram below
Y axis
F1=8N
10º
60º
X axis
F1=8N
Solution
Horizontal
component
3.6.3
Vertical
component
SOLVING PROBLEM USING RESOLUTION OF FORCE
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CHAPTER 3: FORCES


Problem involving three and more than three forces in equilibrium can be solved by
decomposition of force to find the magnitude and the direction of the resultant force.
There are purely mathematical, analytical, techniques for obtaining the resultant of force.

Use formulas to calculate the magnitude and direction of the resultant force:
Magnitude:
Let:
R is a resultant of force
∑ Fx is tthe total force in direction of x- axis ( x- component)
∑ Fy is tthe total force in direction of y- axis (y – component)
so
2
R2 = ∑ 𝐹𝑥 + ∑ 𝐹𝑦
2
R = √∑ 𝐹𝑥 + ∑ 𝐹𝑦
2
2
Direction:
𝐭𝐚𝐧 𝛉 =
∑ 𝐹𝑦
∑ 𝐹𝑥
𝜃 = 𝑡𝑎𝑛−1
∑ 𝐹𝑦
∑ 𝐹𝑥
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CHAPTER 3: FORCES

The method of components being the most convenient in handling multiple vectors addition
or subtraction is discussed below.
Y axis
(90º)
F
1
F1
θ1θ1
X axis
(180º)
θ2
X axis
(0º 0r 360º)
θ3
F2
F3
Y axis
(270º)
Figure 3.8.2

The equilibrant of the vectors F1, F2 and F3, which was measured graphically in Figure 3.8.2,
can be broken into components x and y :

Assume that the +ve X-axis as the reference direction or 0º then the angles are measured
initially from the +ve X-axis.
Y axis
(90º)
F
1
F1
(180º+θ2)
θ1
X axis
(180º)
θ2
θ3
F2
F3
Y axis
(270º)

X axis
(0º 0r 360º)
(360º- θ3º)
The direction of each forces are refer to the +ve X - axis
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CHAPTER 3: FORCES


So that;
The direction of F1 is θ1
the direction of F2 is (180° + θ2)
the direction of F3 is (360° - θ3)
The resolution of the three forces are;
(resolve into X axis and Y axis)
F1x = F1 cos θ1°
F2x = F2 cos (180° + θ2)
F3x = F3 cos (360° - θ3)
and F1y = F1 sin θ1
and F2y = F2 sin (180° + θ2)
and F3y = F3 sin (360° - θ3)
Angle
Horizontal component,
(refer to +ve X axis)
Fx
F1
θ1
F1 cos θ1 =
F1 sin θ1 =
F2
(180 + θ2°)
F2 cos (180 + θ2°) =
F2 sin (180 + θ2°) =
F3
(360-θ3°)
F3 cos (360-θ3°) =
Σ Fx =
F3 sin (360-θ3°) =
Σ Fy =
Force , F (N)

{starting from 0° direction }
Vertical component, Fy
Next
sum all the x-components and y-components.
Σ Fx = F1 cos θ1 + F2 cos (180 + θ2) + F3 cos (360-θ3)
Σ Fy = F1 sin θ1 + F2 sin (180 + θ2) + F3 sin (360-θ3)

The magnitude and angle of the resultant are then found by
2
R2 = ∑ 𝐹𝑥 + ∑ 𝐹𝑦
2
R = √∑ 𝐹𝑥 + ∑ 𝐹𝑦
2
2
Direction:
𝐭𝐚𝐧 𝛉 =
∑ 𝐹𝑦
∑ 𝐹𝑥
𝜃 = 𝑡𝑎𝑛−1
∑ 𝐹𝑦
∑ 𝐹𝑥
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CHAPTER 3: FORCES

Example 1:
Using the method of force decomposition, compute the magnitude and direction for the
resultant force produced by the system of forces below.
8N
F1
θ145º
20º
30º
10N
12N
Solution
We take the +ve x axis as the 0 direction then the angle, is measured initially from the + x
axis.
Y axis
(90º)
F
1 =8N
F1
(180º+20º)
45º
X axis
(180º)
20º
X axis
(0º 0r 360º)
30º
F3=12N
F2=10N
Y axis
(270º)
(360º- 30º)
Angle
Horizontal
Vertical component,
refer to +ve X axis
component, Fx
Fy
F1 = 8N
45°
8 cos 45 = 5.66
8sin 45 = 5.66
F2 =10N
(180+20°)= 200°
10cos (200°) = -9.40
10sin (200°) = -3.42
F3 = 12N
(360-30°)= 330°
12cos (330°) = 10.39
Σ Fx = 6.65
12sin (330°) = -6.00
Σ Fy = - 3.76
Force , F (N)
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The magnitude of resultant
force is;
The angle of resultant force
Angle = tan-1 (ΣFy / ΣFx)
FR = √(Σ
=
Fx)2
√(6.65)2
+ (Σ
Fy)2
= tan-1 (-3.76 / 6.65)
+
(-3.76)2
= tan-1 (-0.57)
= √(44.22) + (14.14)
= - 29.48 atau (360 -29.48)
= 330.52º
= √ 58.36
= 7.64 N

Example 2:
Using the decomposition of force, determine the magnitude and direction for the resultant
forces produced by the system of forces below..
15N
60º
45º
70º
25N
30N
Solution:
Y axis
(90º)
(180º - 60º)
F1=15N
X axis
(180º)
60º
(180º+ 45º)
45º
X axis
( 0º 0r 360º)
70º
F2 = 25N
F3 = 30N
(270º+70º)
Y axis
(270º)
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Angle
Horizontal component,
(refer to +ve X axis)
Fx
F1 = 15N
(180-60°)= 120°
5 cos 120 = -7.5
15 sin 120 = 13.00
F2 =25N
(180+45°)= 225°
25 cos 225° = -17.70
25 sin 225° = -17.70
F3 = 30N
(270+70°)= 340°
30 cos 340° = 28.20
Σ Fx = 3.00
30 sin 340° = -26.00
Σ Fy = -30.7
Force , F (N)
The magnitude of resultant
force is;
FR = √(Σ Fx)2 + (Σ Fy)2
= √(3)2 + (-30.7 )2
= √(9) + (942.5)
Vertical component, Fy
The angle of resultant force
Angle = tan-1 (ΣFy / ΣFx)
= tan-1 (-30.7 / 3.0)
= tan-1 (-10.234)
= - 84.4º atau 360 - 84.4
= √ 951.5
= 275.6º
= 30.84 N
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CHAPTER 3: FORCES

Exercise 1:
Find the resultant force produced from system of forces below.
8N
5N
30º
60º
4N
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CHAPTER 3: FORCES

Exercise 2:
Determine the resultant force produced by the action of forces below.
10N
12N
45º
60º
30º
15N
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
Exercise 3:
Using the decomposition of force, compute the resultant force for the system of forces below.
400N
500N
45º
30º
300N
60º
700N
100N
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CHAPTER 3: FORCES
3.7
MOMENT OF FORCE

Definition - The turning effect produced by a force on a rigid body about a point, pivot
or fulcrum

The moment of a force about a point is the product of the force and perpendicular distance of
its line of action from the point.

If F be a force and d the perpendicular distance of its line of action from the fixed point O,
then moment of F about O = F × d.

Formula for Moment:
d
O
pivoted
point
F
Moment of a force = Force x Perpendicular distance between force
and the pivoted point
Mo = F x d

The unit of moment of the force is newton metre (N m)

There are many examples around us where we use the principle of moments. A son can
balance a much heavier father on a see-saw by sitting at a greater distance from the fulcrum.

Here the distance at father’s side being less, you will have to apply more force to cause the
same turning effect.
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CHAPTER 3: FORCES

The fact that the ground reaction normally acts 5-6 cm in front of the ankle has big
implications. Imagine two people on a see-saw, each of weight m.g Newtons:

In order to balance the see-saw, the weight of each person (m.g) has to equal.

What happens when one person moves closer to the fulcrum or pivot?

Clearly, the see-saw will tip down at the left hand end. The reason for this is that the turning
effect of the force is dependent on the distance of the force from the pivot. This turning effect
is called the moment of force and the distance is called the moment arm of the force.

Example 7:
Find the magnitude and sense of the moment of the given force about O?
8N
O
2m
MO = F x d = 8 x 2 = 16 Nm
Sense of the moment  the body turn in anticlockwise direction
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
Exercise 8:
A spanner that is used to tighten a nut is 300 mm long. A perpendicular force, 100 N is
exerted at the end of the spanner. Calculate the moment produced at the nut.
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3.7 PRINCIPLE OF MOMENTS

If a body is in equilibrium under the action of a number of forces, then the algebraic sum of
the moments of the forces about any point is equal to zero.
3.7.1 The Sign of a Moment



Earlier, when we were collecting components of forces, we chose a positive direction ;
components in that direction had a positive sign while components in the opposite direction
took a negative sign.
In the same way, we choose a positive sense of rotation when dealing with a system of
moments.
If, for example, we decide to make clockwise the' positive sense, an clockwise moment
has a positive sign while a anticlockwise moment has a negative sign. The resultant
moment of a number of forces is then the algebraic sum of the separate moments.
Sum of the clockwise moments = Sum of the anti clockwise moments
∑ 𝐜𝐥𝐨𝐜𝐤𝐰𝐢𝐬𝐞 𝐦𝐨𝐦𝐞𝐧𝐭 = ∑ 𝐚𝐧𝐭𝐢 𝐜𝐥𝐨𝐜𝐤𝐰𝐢𝐬𝐞 𝐦𝐨𝐦𝐞𝐧𝐭

The positive sense does not always have to be anticlockwise; an individual choice can be
made for each problem.
3.7.2 Zero Moment


When a force passes through the axis of rotation, its distance from that axis is zero.
Therefore the moment of the force about that axis is zero.
Example 1:
F1
F2
d1
C
d2
A
B
R
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
If a beam is in equilibrium under the action of the forces, then the algebraic sum of the
moments of the forces about any point on the beam is equal to zero.
∑ M (at any point on the beam) = 0
i.
If we choose point A as a reference point (moment point)
F1
F2
d1
C
d2
MA
A
B
R
∑MA
=0
F1 as a reference point ( moment = 0)
R rotate anticlockwise, -ve sign
F2 rotate clockwise, +ve sign
Assume that an anticlockwise moment has a +ve sign
while a clockwise moment has a -ve sign.
Moment at A = F1 x (0)
Moment at C = - R x (d1)
Moment at B = + F2 x (d1 + d2)
MA = F1( 0 ) + F2( d1 + d2 ) - R( d1 ) = 0
Therefore
F2( d1 + d2 ) = R( d1 )
ii.
If we choose point C as a reference point (moment point)
F1
F2
d1
C
d2
A
MC
B
R
∑MC
=0
F1 rotate anticlockwise, -ve sign
R as a reference point ( moment = 0)
F2 rotate clockwise, +ve sign
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CHAPTER 3: FORCES
Therefore
-F1d1 + R(0) + F2 d2 = 0
F1d1 = F2d2

Sum of the clockwise moments equals sum of the anticlockwise moments when the
body is in equilibrium.
Sum of the anticlockwise moments = Sum of the clockwise moments

Newton’s third law of motion states that action and reaction forces are equal in magnitude
and opposite direction.
Action Forces = Reaction Forces or Downward Forces = Upward Forces

Example 2:
A horizontal beam is supported at a point 20 cm from A. It is balanced by the weights 40 N
and 20 N at A and B respectively. Find the length of the beam
FA = 40 N
FB = 20 N
MR
A
B
d
20 cm
R
Solution:
FA = 40 N, FB = 20 N, distance AR = 20 cm and distance BR = d cm
Choose R as a reference point (moment point)
∑MR
=0
FA rotate anticlockwise, -ve sign
R as a reference point ( moment = 0)
FB rotate clockwise, +ve sign
Therefore
-40 x 20 + R(0) + 20 x d = 0
-800 + 20d = 0
20d = 800
d = 800/20 = 40 m
The length of the meter scale is 20m + 40 m = 60 m
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CHAPTER 3: FORCES

Example 3:
From the diagram below, determine the point must the beam be supported from A so that it
may rest horizontally?
5N
3N
5N
9N
B
A
200 m
150 m
250 m
Solution:
Choose point A as a reference point (moment point)
F1=5 N
F2=3 N
F3=5 N
A
MA
F4=9 N
B
2.0m
2.5m
x
1.5m
R
Let;
F1 = 5N , F2 = 3N, F3 = 5N , F4 = 9N and R as a support force, X meter from A
To be easier in calculation, choose the end of the beam as a reference point or the moment
point..
Method 1
F1 = 5N with distance of 0 m from A
F2= 3N with distance of 2.0 m from A (rotate clockwise , +ve sign)
F3 = 5N with distance of 4.5 m from A (rotate clockwise , +ve sign)
F4 = 9N with distance of 6.0 m from A (rotate clockwise , +ve sign)
R = ? with distance of x m from A
(rotate anticlockwise , -ve sign)
∑MA
=0
5(0) + 3(2.0) + 5(4.5) + 9(6.0) – R(X) = 0
0 + 6.0 + 22.5 + 54.0 – R(X) = 0
R(X) = 82.5
X = 82.5/R
.............................. (1)
Newton’s third law of motion states that
action forces = reaction forces
R= 5N + 3N + 5N + 9N = 22 N ...............................(2)
Subtitute (2) into (1)
X = 82.5/22 = 3.75 m
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Method 2
Choose point A as a reference point (moment point)
Force
Value of
Force
Distance from
reference point
A
Rotation ( cw
or acw)
Moment (F x d)
F1
5N
0
cw ( +ve)
0
F2
3N
2.0
cw ( +ve)
+ 3 x 2.0 = 6.0
F3
5N
4.5
cw ( +ve)
+ 5 x 4.5 = 22.5
F4
9N
6.0
cw ( +ve)
+ 9 x 6.0 = 54.0
R
R (unknown)
X (unknown)
Acw ( -ve)
- R x X = RX
∑MA
=0
6.0 + 22.5 + 54.0 – R(X) = 0
82.5 – R(X) = 0
R(X) = 82.5
X = 82.5/R
............................. (1 )
Newton’s third law of motion states that
action forces = reaction forces
R = 5N + 3N + 5N + 9N = 22 N ...............................(2)
Subtitute (2) into (1)
X = 82.5/22 = 3.75 m
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CHAPTER 3: FORCES

Example 4:
Determine the point the beam should be supported from A so that it remain in equilibrium
25N
35N
2m
40N
5m
15N
4m
60º
A
B
Solution:
Vertical
component
25 sine 60º
= 21.65 N
25N
35N
2m
5m
40N
15N
4m
60º
A
B
MA
Xm
R
Let say;
F1 = 35N ,
F2 = 25 sin 60 = 21.65N, ( resolved into vertical component )
F3 = 40N ,
F4 = 15N
R as a support force
Choose point A as a reference point (moment point)
F1 = 35N with distance of 0 m from A
F2= 21.65N with distance of 2 m from A
F3 = 40N with distance of 7 m from A
F4 = 15N with distance of 11 m from A
R = ? with distance of X m from A
(rotate clockwise , +ve sign)
(rotate clockwise , +ve sign)
(rotate clockwise , +ve sign)
(rotate anticlockwise , -ve sign)
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CHAPTER 3: FORCES
∑MA
Force
Value of
Force
Distance from
reference point
A
Rotation ( cw
or acw)
Moment (F x d)
F1
35N
0
cw ( +ve)
0
F2
21.65N
2
cw ( +ve)
+ 21.65 x 2 = 43.3
F3
40N
7
cw ( +ve)
+ 40 x 7 = 280
F4
15N
11
cw ( +ve)
+ 15 x 11 = 165
R
R (unknown)
X (unknown)
Acw ( -ve)
- R x X = R(X)
=0
43.4 + 280 + 165 – R(X) = 0
488.3 – R(X) = 0
R(X) = 488.3
X = 488.3/R
...................... (1 )
Newton’s third law of motion states that
action forces = reaction forces
R = 35N + 21.65N + 40N + 15N = 111.65 N .........................(2)
Subtitute (2) in (1)
X = 488.3/111.65
= 4.37 m
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CHAPTER 3: FORCES
Example 5:
Calculate the reaction forces at R1 and R2 if the bar is in equilibrium.
80N
20N
30N
4m
5m
14m
3m
R1
R2
Solution:
Let;
F1 = 80N ,
F2 = 30N
F3 = 20N ,
R1 and R2 to be calculated
F1=80N
8m
F2=30N
4m
F3=20N
5m
MR1
14m
3m
R1
R2
Choose point R1 as a reference point (moment point)
F1 = 80N with distance of 8m from R1
F2 = 30N with distance of 12m from R1
F3 = 20N with distance of 17m from R1
R1 = ?
with distance of 0m from R1
R2 = ?
with distance of 14m from R1
(rotate clockwise , +ve sign)
(rotate clockwise , +ve sign)
(rotate clockwise , +ve sign)
(reference point)
(rotate anticlockwise , -ve sign)
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CHAPTER 3: FORCES
Value of Force
Distance from
reference point A
Rotation ( cw
or acw)
Moment (F x d)
F1
80N
8
cw ( +ve)
+80 x 8 = 560
F2
30N
12
cw ( +ve)
+ 30 x 12 = 360
F3
20N
17
cw ( +ve)
+ 20 x 17 = 340
R1
R1
0
R2
R2 (unknown)
14
Force
∑MA
0
Acw ( -ve)
- R2 x 14 = -14R2
=0
560 + 360 + 340 – 14R2 = 0
1260 – 14R2 = 0
R2 = 1260/14
= 90N
............................ (1 )
Newton’s third law of motion states that
action forces = reaction forces
R1 + R2 = 80N + 30N + 20N = 130N
R1 + R2 = 130N
R1 = 130N - R2
.................................(2)
Subtitude (1) into (2)
R1 = 130 – 90 = 40 N
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Exercise 1:
Find the location to place a single support under the beam, so it will remain in equilibrium.
40N
50N
80m
30N
40m
A
60N
50m
B
Solution:
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CHAPTER 3: FORCES
Exercise 2:
Calculate the reaction forces at RA and RB if the bar in equilibrium.
4N
10N
80m
8N
40m
6N
50m
A
B
RB
RA
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CHAPTER 3: FORCES
Exercise 3:
Determine the point should be the beam being supported from A for that beam stay in
equilibrium
45N
35N
2m
60N
5m
25N
4m
50º
A
B
Solution:
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CHAPTER 3: FORCES
Exercise 4:
A uniform beam AB have a 6m long. There are 50N, 25N and 100N forces are acted on that
beam from A and B respectively. At what point must the beam be supported from B so that it may
in equilibrium
50N
25N
1m
100N
5m
A
B
Solution
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CHAPTER 3: FORCES
Exercise 5:
Calculate the reaction forces for R1 and R2 if the bar is in equilibrium.
5N
8N
4N
4m
3m
10m
R1
2m
R2
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CHAPTER 3: FORCES
SELF ASSESSMENT
Question 1
Determine the weight of an astronaut who has a mass of 100 kg (gravity=10 N/kg).
a) on earth?
b) on the Moon, where the gravitational acceleration is
c)
1
of the Earth?
6
in outer space?
Question 2
Two concurrent forces F1 and F2 act at an object with right angles to one another.
Determine the magnitude and direction of the resultant force.
a)
b)
F1 = 80 N, F2 = 100 N, θ = 500
F1 = 0.8 kN, F2 = 300 N, θ = 1200
Question 3
An object of weight 30 N is suspended to a horizontal beam by two chains. The
chains are attached to the same point on the object with the angle of 300 and 400 to the
vertical respectively. Determine the tensions in both chains.
Question 4
A 4 kg object is suspended by a single string at the ceiling as shown in figure below. A
horizontal force, Fp is pulling the string at the middle so that it will make an angle of 500 from
the horizontal plane. Determine the value of Fp.( g = 10 N/kg)
50º
Fp
4 kg
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CHAPTER 3: FORCES
Question 5
Calculate the magnitude and direction of resultant force for the following system of forces.
a)
b)
100N
20kN
120º
30º
200N
150º
40º
20º
30kN
300N
40kN
Question 6
Imran is using a 30 cm spanner to detach a wheel of an automobile. The wheel has 4 screws
(bolts). If he apply 50 N force to open it, calculate the moment of the force to open one
wheel.
Question 7
The beam is balanced horizontally by a weight W. Calculate the value of the weight.
30 N
70 N
30 m
W
50 m
30 m
Fulcrum
Question 8
Refer to the diagram below:
i) Calculate the magnitude of resultant force
Y
F1=15N
40º
X
F2=12N
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CHAPTER 3: FORCES
ii) Define the reaction forces at A and B
5N
4N
3N
A
B
2m
1.5 m
1m
Question 9
Determine the point must the beam be supported from A so that it remain in equilibrium
10 N
2m
25 N
2m
55 N
30 N
4m
60°
A
B
A
Question 10
Refer to the diagram below, calculate the resultant force
3N
60º
5N
2N
Question 11
Refer to the diagram below, determine the point must the beam be supported so that it stay
in equilibrium.
i.
12N
1m
10N
2m
8N
2m
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CHAPTER 3: FORCES
ii.
8N
20N
12N
4m
8m
iii
15N
10N
30N
4m
8m
30º
Question 12
Refer to the diagram below, calculate the tension of the string P and Q.
P
Q
45º
25º
50N
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