ENGINEERING SCIENCE

QUICK CHECK

Calculate the weight of each at planet if the body of mass is 40 kg

Planet Gravitational Mass,m Weight,w

Mercury Field Strength,g
Venus
Earth 3.78 N/kg X 40 kg =
Moon 40 kg =
Mars 8.94 N/kg X 40 kg =
Jupiter 40 kg =
Saturn 9.81 N/kg X 40 kg =
Uranus 40 kg =
Neptune 1.7 N/kg X 40 kg =
Pluto 40 kg =
3.79 N/kg X 40 kg =
40 kg =
25.4 N/kg X

10.7 N/kg X

9.2 N/kg X

12 N/kg X

0.3 N/kg X

Which planet has the most of gravitational field strength? Explain more

**Remember
Unit of gravitational is N/kg or −2

Chapter 3 – Force/ Page 94

Newton’s Law of Motion

1. Newton’s First Law said that the body at rest tends to remain at rest, and a body in motion
continues to move at constant velocity unless acted upon by a net force. It also called a
law of inertia.

2. Newton’s Second Law said that the acceleration of an object depends directly upon the net
force acting upon the object, and inversely upon the mass of the object. It also called a law
of acceleration.

3. Newton’s Third Law said that for every action there is an equal and opposite reaction. It
also called a law of attraction.

Newton’s Second Law

1. The acceleration of an object is directly proportional to the net force acting on it and
inversely proportional to its mass.

F = ma

F = ( )

F = ( )

ma = ( −2)

Figure 3.1: Triangle of Newton’s 2nd Law

2. The acceleration of an object increases with increased force, decreases with increased
mass, and is in the same direction as the force.

Figure 3.2a: Situation of Newton’s 2nd Law Figure 3.2b: Situation of Newton’s 2nd Law
Chapter 3 – Force/ Page 95

Example 2
How much force is needed to accelerate a 30 kg scooter at a rate of 2 −2 ?

Solution:
Given: mass, m = 30 kg , acceleration, a = 2 ms−2 , force, F ?

= ma
= 30 × 2
=

QUICK CHECK 2) A net force of 16 N causes a mass to
accelerate at a rate of 5 − .
1) What acceleration will result when a
12 N net force applied to a 3 kg object? Determine the mass?

Answer: = − Answer: = 3.2

3) What is the force on a 1000 kg elevator 4) A worker pushes a box with mass 80 kg
that is falling freely at 9.81 − on a horizontal floor with constant
acceleration 2.0 −2.Calculate

a) The force applied by worker
b) The acceleration if 70 kg is released

from the box

Answer: F = 9810 N Answer: a) F = 160 N b) a = 16 ms−2

Chapter 3 – Force/ Page 96

Inclined Plane

1. An inclined planed is any slope or ramp shown in Figure 3.3
2. The force of gravity (also known as weight) acts in a downward direction which is the

normal force acts in a direction perpendicular to the surface.

mgsinθ θ W mgcosθ

Figure 3.3: Inclined planed
3. Component of weight parallel to the plane = mg sin θ
4. Component of weight normal to the plane = mg cos θ

Example 3

A box with a mass of 10 kg moves down a smooth plane inclined at an angle of 300.
Calculate the acceleration of the object. (Use g= 9.81 −2)

300

Solution:
Given: mass, m = 10 kg , , ?
F = mgsin θ
F = 10 × 9.81 × sin 30
F = 49.05 N

F = m
49.05 = 10

49.05
= 10
= . −

Chapter 3 – Force/ Page 97

QUICK CHECK

1) Determine the acceleration of the moving down the plane (g = 9.81 ms−2)

400

Answer: 6.31 −2

2) A friction force of magnitude 35 N acts on an object of mass 16 kg which is placed
on a rough inclined plane as shown in figure. Determine the acceleration of the
object moving down the plane ( = 9.81 −2)

300

Answer: 2.72 −2

Video description: Newton’s 2nd Law

Chapter 3 – Force/ Page 98

Net/Resultant Force for Individual Forces

1. The net force is the combination all of the forces acting on the object
2. Whenever more than one force act on an object, always sum the forces into a single net

force
3. Forces in the same direction :

• When forces act in the same direction, add the forces together
• The net force will be in the same direction as the two original forces

Example 4

Find the net force Solution:
Same direction: 8N + 3 N = 11 N
8N So that net force is:
3N
11 N

The direction moves to the right

4. Forces in the opposite direction
• When two forces act in opposite directions, subtract the smaller force from the larger
force
• The net force will be in the direction of the larger force

Example 5 Solution:
Different direction: 8N - 3 N = 5 N
Find the net force
8N So that net force is:
5N

3 N The direction moves to the upward
Chapter 3 – Force/ Page 99

QUICK CHECK

Calculate the net force of each individual force :

a) 1N b) 10 N
4N 2N 30 N

Answer: Fx = 7 N Answer: Fx = 20 N

c) d)
15 N 7N
5N

25 N

Answer: Fy =10 N 11 N
6N
e) f)
Answer: Fy = 3 N
30 N 40 N
20 N 10 N 12 N 4N

5N

4N

Answer: Fx = 0 N Answer: R = 17 N

Chapter 3 – Force/ Page 100

Forces in Equilibrium

1. The forces acting on an object are in equilibrium when the resultant force acting on the
object is equal to zero. An object at rest is in equilibrium and move with constant velocity.
=
Acceleration = (stationary or uniform)
velocity)

2. If the forces on an object are balanced (or if there are no forces acting on it) this is what
happens:
• an object that is not moving stays still.
• an object that is moving continues to move at the same speed and in the same
direction.

3. Balanced force is forces in opposite directions produce no motion.
=

Figure 3.4a: Balanced forces
4. Unbalanced force is unequal opposing forces produce an unbalanced force causing

motion.
≠

Figure 3.4b: Unbalanced forces

Chapter 3 – Force/ Page 101

Component of Forces

1.  is an angle between the force,F to the horizontal line or vertical line.
2. The sign of the force depends on the quadrant where the force, F is placed.
3. Single force can be resolved into two perpendicular components to Fx and Fy


θ F

θ θθ


Figure 3.5 a: Angle between F to x-axis Figure 3.5 b: Angle between F toθ y-axis

Horizontal component, = HorizontalFcomponent, = F
Vertical component, = Vertical component, =

Example 6 Example 7
Calculate horizontal component, and Determine horizontal component,
vertical component, and vertical component,


125 N 30 N

400 550


Solution: Solution:
Horizontal component, = 125 cos 40 Horizontal component, = 30 sin 55

= . = 24.57 N
Vertical component, = 125 sin 40 Vertical component, = 30 cos 55

= 80.35 N = 17.21 N

Chapter 3 – Force/ Page 102

Example 8
A man pushes a lawnmower with a force of 85 N at an angle of 450 with the horizontal

85 N
45°

Calculate:
a) the horizontal component of the force that causes the lawnmower to move forward
b) the vertical component of the force that presses on the lawn

Solution: 450

85 N


a) Horizontal component, = 85 cos 45
= .

b) Vertical component, = 85 sin 45
= .

Chapter 3 – Force/ Page 103

QUICK CHECK

Calculate horizontal component, and vertical component,
a) b)

15 N
300

400
70 N

Answer: = . Answer: =
= . = .

c) d) 500
43 N 6 kN

210

Answer: = . Answer: = .
= . = .

Chapter 3 – Force/ Page 104

e) f)

75 N 43 N

18 N

250

29 N

10 N

g) Answer: = . Answer: =
150 N = . =

400 h) 30 N

8N 550
16 N
370
60 N 65 N

Answer: = . Answer: = .
= . = .

Chapter 3 – Force/ Page 105

Resultant Force Using Resolution Method

Fy
R


Fx

Figure 3.6 : Diagram of resultant force

Resultant force, = √∑( ) + ∑( )

Direction, = − ( )



Example 9
Find the resultant and direction of force in a body applied

40 N

12 N

Solution:

, = √∑(Fx)2 + ∑(Fy)2 Direction, = tan−1 ( )

R = √(40)2 + (12)2
R = .
= tan−1 (12)

40

= °

Fx

41.76 N
Fy

Chapter 3 – Force/ Page 106

Example 10
Find the resultant and direction of force in a body applied.

10 N

35 N 20 N

45 N

Solution:

Force, F(N) Horizontal force, Fx(N) Vertical force,Fy(N)

20 20 -
10 - 10
35 35 -
45 - 45
ΣF 55
15

Resultant force, = √∑( )2 + ∑( )2
R = √(15)2 + (55)2
R = √3250
R =

57 N Fy

Direction , = Tan−1 (Fy)

Fx

= Tan−1 (55) .
Fx
15

= .

Chapter 3 – Force/ Page 107

Example 11

Find the resultant and direction of force in a body applied.

10 N

35 N

300

150
20 N

Solution:

Force, F(N) Horizontal force, Fx(N) Vertical force,Fy(N)
10 - 10
35
20 35 cos 30 = 30.31 35 sin 30 = 17.5
20 sin 15 = 5.18 20 cos 15 = -19.32

ΣF 25.13 8.18

Resultant force, = √∑( )2 + ∑( )2 Fy
= √(25.13)2 + (8.18)2
= √698.43 26.43 N
= .

Direction , = −1 ( )



= −1 ( 8.18 ) .
Fx
25.13

= 18.030

Chapter 3 – Force/ Page 108

QUICK CHECK

1) Find the resultant and direction of force in a body applied
60 N

37 N 13 N
45 N

Answer: R = 52.20 N, θ = 16.7 0

Chapter 3 – Force/ Page 109

2) Find the resultant and direction of force in a body applied
10 N 5 N

600 5N
100
4N

Answer: R = 12.42 N, θ = 56.760

Chapter 3 – Force/ Page 110

3) Calculate the magnitude of resultant force of the figure below:

25 kN 80 kN
800 1400

62 kN

Answer: R = 65.65 kN

Chapter 3 – Force/ Page 111

4) A load of mass 2 kg is suspended by strings T, Q and R attached to retort
stands as shown in Figure below. By using the resolving method, calculate
the tension of string Q, R and T.
R
T

400

Q
2 kg

Answer: Q = 19.62 N, R = 30.52 N, T = 23.38 N

Chapter 3 – Force/ Page 112

3.2 CONCEPT OF MOMENT

Definition The turning effect of a force is called a moment
Formula Moment = Force (N) × perpendicular distance from the force to the pivot (m)

M
Fd
Figure 3.7: Triangle of moment formula
Unit Newton meter (Nm)
Example This is an anticlockwise turn. The turning effect of a force is called a moment.

Figure 3.8: The left-hand side of the see-saw moves downwards
when a force is applied to it.

Figure 3.9: Turning effect of moment
Chapter 3 – Force/ Page 113

Example 12

Gina weighs 500 N and stands on one end of a seesaw. She is 0.5 m from the pivot.
What moment does she exert?

0.5 m

Pivot

500 N

Solution :

500 N
0.5 m

Given: Force, F = 500 N, Distance, d = 0.5 m, Moment, M ?

Moment, M = F × d
= 500 × 0.5
=

Chapter 3 – Force/ Page 114

QUICK CHECK 2) What moments of vase exert?

1) Find the moment

Pivot

Pivot

m = 4 kg

Answer: M = 60 Nm Answer: M = 196.2 Nm

3) Find the moment of spanner exert? 4) Determine the moment of beam

Pivot 50 N

45°

10 m

Pivot

Answer: M = 13 Nm Answer: M = 353.55 Nm
Nm

Chapter 3 – Force/ Page 115

Principle of Moments

1. The principle of moments states that when in equilibrium the total sum of the
anticlockwise moment is equal to the total sum of the clockwise moment.

2. When a system is stable or balance it is said to be in equilibrium as all the forces acting
on the system cancel each other out.
Total anticlockwise moment = Total clockwise moment

Resultant force of vertical = N
assume R = reaction and location of center of gravity

↺= ↻
=

Pivot

Figure 3.10a: Two balance forces acting on the bar

Pivot ↺= ↻
= +

Figure 3.10b: Three balance forces acting on the bar
Chapter 3 – Force/ Page 116

3. The principle can be explained by considering two people on a seesaw.

Figure 3.11: Two people on a seesaw
Both people exert a downward force on the seesaw due to their weights.
Person A’s weight is trying to turn the seesaw anticlockwise while person B’s weight is
trying to turn the seesaw clockwise.
Person A’s Moment = Person B’s Moment

ΣM ↺ = ΣM ↻
1 1 = 2 2
1000 × 1 = 500 × 2
1000 Nm = 1000 Nm
Therefore seesaw is in equilibrium because the value of moment clockwise and moment
anticlockwise are same.

Video description: Moment of force

Chapter 3 – Force/ Page 117

Example 13

Two girls are standing on opposite sides of on a see-saw. Mimi weighs 200 N and is 1.5 m
from the pivot. Where must Lisa 150 N stand if the seesaw is to balance?

Mimi Lisa

1.5 m d=?

200 N Pivot 150 N

Solution : 150 N
200 N

1.5m d
Pivot =
?

When the see-saw is balanced:

Mimi’s moment = Lisa’s moment

ΣM ↺ = ΣM ↻

F1d1 = F2d2

200 × 1.5 = 150 × d

300 = 150d
= 300

150

= 2 meters

The distance of Lisa from pivot is 2 meters

Chapter 3 – Force/ Page 118

Example 14

If a 10,000 N counterweight is 3 m from the tower, what weight can be lifted when the
loading platform is 6 m from the tower?

Using the principle of moments, the crane is balanced.

Solution: 10 000N
F =?

6m 3 m

Pivot

moment of load = moment of counterweight

ΣM ↺ = ΣM ↻

F1d1 = F2d2
× 6 = 10000 × 3

6 = 30000

= 30000

6

=

The weight of load is 5000 N

Chapter 3 – Force/ Page 119

Example 15

Calculate the amount of force,F and R if the beam is in equilibrium. Assume R = reaction
and location of center of gravity.

20 N F

R 5N
4m
4m 12 m

Solution :

20 N F

R 5N

4 m Pivot 12 m 4m

Total Anticlockwise moment = Total Clockwise moment

ΣM ↺ = ΣM ↻
F1d1 + F2d2 = F3d3
(20 × 4) + (5 × 12) = (16)

80 + 60 = 16
140 = 16
= 140

16

= .

Chapter 3 – Force/ Page 120

∑ = (Equilibrium)
Assume direction of force downward is negative and direction of force upward is
positive

−20 + + 5 − 8.75 = 0

−23.75 + = 0

= .

QUICK CHECK

1) Two girls are sitting on opposite sides of a see-saw. One girl weighs 200 N and is
1.5 m from the pivot. How far from the pivot must her 250 N friend sit if the see-
saw is to balance?

250 N 200 N

Pivot

d=? 1.5 m
m

Answer: d = 1.2 m

Chapter 3 – Force/ Page 121

2) The ruler is balanced state, calculate the value of weight, W

20 cm 30 cm 50 cm

Pivot

100N 20 N W=?

Answer: w = 17.5 N

Chapter 3 – Force/ Page 122

Centre of Gravity

If a body is hanging freely, its center of gravity will always be vertically below the pivot.
To find distance pivot from origin, we can use two methods:
i) Moment Force Method (MFM)

F1 F2



1 Pivot
2

Figure 3.12a: The balance bar

↺ = ↻
( − ) = ( − )

ii) Moment Resultant Method (MRM) F2
F1



1 Pivot
2

Figure 3.12b: The balance bar


=

= +
+

Chapter 3 – Force/ Page 123

Example 16
Determine the gravitational point, x to balance the bar.

10 20

20
=?
2 Pivot 6

Solution:

i) Moment force method :
10


=?

2 Pivot

8

ΣM ↺ = ΣM ↻
10( − 2) = 20(8 − )

10 − 20 = 160 − 20

30 = 180
= from the left

ii) Moment resultant method :
Σ

= Σ
= (10 ×2) + (20 ×8)

10 + 20

= 180

30

= from the left

Chapter 3 – Force/ Page 124

Example 17
Determine the gravitational point, x to balance the bar.

15 N 20 25

= ?

3 Pivot 4 25

Solution :
i) Moment force method:

15 20
= ?

3 Pivot
7

ΣM ↺ = ΣM ↻
15 ( − 0) + 20( − 3) = 25(7 − )

15 + 20 − 60 = 175 − 25
60 = 235
= . from the left

ii) Moment resultant method :
Σ

= Σ
(15 × 0) + (20 × 3) + (25 × 7)

= 15 + 20 + 25
235

= 60
= . from the left

Chapter 3 – Force/ Page 125

Example 18

Determine the gravitational point, x to balance the bar.


=?

600

3 3.5


600
Solution :
i) Moment force method:




=?

3

6.5
5

ΣM ↺ = ΣM ↻
35( − 3) = 5 ( − 0) + 10 sin 60 (6.5 − )

35( − 3) = 5 ( − 0) + 8.66(6.5 − )
35 − 105 = 5 + 56.29 − 8.66

38.66 = 161.29
= . from the left

ii) Moment resultant method :

ΣM
= ΣF

= (0 ×−5) + (35 ×3) + (10 60×6.5)
−5 + 35 + 10 60

161.29
= 38.66

= 4.17 from the left

Chapter 3 – Force/ Page 126

QUICK CHECK

1) Determine the gravitational point, x to balance the bar.

10 N 20 N 5N

2m 6m


=?

Answer: x = 2.29 m from left

Chapter 3 – Force/ Page 127

2) Determine the gravitational point, x to balance the bar.
5N

450

5 m 5m 3m

20 N 10 N

Answer: x = 5.82 m from left

Chapter 3 – Force/ Page 128

3) Find the reaction of and its distance,x from point A to keep the beam in
equilibrium.

54 N 31 N 47 N

400 5
A m

10 m



Answer: R = 112.71 N, x = 9 m from point A

Chapter 3 – Force/ Page 129

ACTIVITY CHAPTER 3

1) Define and state the SI units of following terms below:
a) Force
…………………………………………………………………………………….
b) Moment
…………………………………………………………………………………….

2) List two effects of force with examples
a) …………………………………………………………………………………….
…………………………………………………………………………………….
b) …………………………………………………………………………………….
…………………………………………………………………………………….

3) Define forces in equilibrium
………………………………………………………………………………………

4) A beam is in equilibrium with the pivot in the centre at O. A weight is hung at W,
with a distance of d based on figure below :
d

O O
o

0

W0
0

a) Describe the reaction of beam 0
0

……………………………………………o………………………………………

……………………………………………………………………………………

b) Explain briefly how to ensure the beam is in equilibrium
……………………………………………………………………………………
……………………………………………………………………………………
……………………………………………………………………………………

Chapter 3 – Force/ Page 130

5) A box of mass 70 kg is places on horizontal floor with smooth surface. The velocity
of the box changes from stationary to 50 m/s in 10 seconds when it is acted by a force.
Find the magnitude of the force.

6) Find the resultant force of the following: b) Answer: F = 350 N
a)
3N
36 N 100 N
70 4N
N

Answer: a) 6 N b) 5 N

Chapter 3 – Force/ Page 131

7) A contractor uses a scraper to scrape old paint away from a wall. He uses a force of
12 N at an angle of 360 with the wall. Calculate :
a) The vertical component
b) The horizontal component

360
12 N

Answer: a) Fy = 9.71 N b) Fx = 7.05
N

8) A traffic light weighing 200 N being held by two identical cables at an angle of 600 with
the horizontal beam. Calculate the tension of each cable.

600 600

Answer: F = 115.5 N

Chapter 3 – Force/ Page 132

9) Calculate the magnitude and angle of the resultant force for the Figure below :

50 N

300

80 N

400

80 N

Answer: R = 112.3 N, = 50. 200

Chapter 3 – Force/ Page 133

10) Some children are playing on a see-saw as shown in the Figure below. The see-saw
is a plank of wood 3.0 m long, with a pivot exactly in the middle.

d=?

a) What are the weights of the children?

Child A Child B Child C

Weight, W

Answer: 343.35N Answer: 245.25 N Answer: 392.4N
m

b) Child B is sitting 0.4 m from the pivot. Where should child C sit so that the see-
saw remains level?

Answer: d = 1.06 m

Chapter 3 – Force/ Page 134

11) Figure shows a loaded beam. Find the reaction of R and its distance, x from point
A to keep the beam in equilibrium.
7N

450 5 kg

xR 3N
2m 2m 3m

Answer: R = 57 N, x = 3.98 m from left

Chapter 3 – Force/ Page 135

Concept Map

FORCE Is or Newton’s third
a law
Of an object is PUSH PULL
given by States that
for every
F = ma
Action
Which is Reaction
the
There is an
Newton’s second Law equal and
opposite

Weight Is given mg
by

Chapter 3 – Force/ Page 136

4 Work, Energy
and Power

4.1 Concept of work
4.2 Renewable energy
4.3 Concept of energy
4.4 Concept of power

Renewable energy supplies will never run
out. The supplies of coal, oil, and natural
gas are limited but sunshine, wind, biomass
and water power are considered almost
limitless resources.

Chapter 4–Work, Energy and Power/ Page 137

Work,

CHAPTER

4 Energy and

Power

Learning Outcomes

A student should be able to:
❖ Define work
❖ Calculate work by using formulas:
❖ Apply the concept and formula of work in solving the related problems
❖ Define renewable energy and non-renewable energy
❖ Describe the primary renewable energy resources and technologies
❖ Explain the advantages of renewable energy to mankind
❖ Define Kinetic Energy and Potential Energy
❖ Calculate form of energy by using formulas
❖ State principle of conservation of energy
❖ Describe conversion from one form of energy to another
❖ Apply the concept and formula of energy in solving the related problems
❖ Define power
❖ Calculate power by using formula
❖ Define energy efficiency
❖ Calculate the efficiency of mechanical system

Chapter 4–Work, Energy and Power/ Page 138

MIND MAP

WORK, ENERGY AND POWER

Work Energy Power
Efficiency

s of the s of the object is Gravitational Kinetic
object is not parallel of , F potential Energy
parallel of, F energy

Conservation
energy

KEYWORDS

Work – Kerja
Energy – Tenaga
Renewable energy – Tenaga yang boleh diperbaharui
Gravitational potential energy – Tenaga keupayaan graviti
Kinetic energy – Tenaga kinetik
Principle of conservation of energy– Prinsip keabadian tenaga
Power – Kuasa
Efficiency – Kecekapan

Chapter 4–Work, Energy and Power/ Page 139

4.1 CONCEPT OF WORK

Definition Work, W is the product of an applied force and the
displacement of an object in the direction of the applied force.
Formula
Unit For work to be done, two
Type of quantity things must occur:
Examples of work 1.We must apply a force
to an object.
2. The object must move
in the same direction as
the force we apply.

Figure 4.1: A man push a box in the direction of the force

= ×

F = force (N)
s = displacement (m)

Newton meter (Nm) or Joule (J)

Scalar quantity

Figure 4.2a: Push a trolley move on Figure 4.2b: Pull a box
using a rope

Figure 4.2c: Climb the stairs Figure 4.2d: Throw a ball

Chapter 4–Work, Energy and Power/ Page 140

Calculation of Work

a) The displacement, s of the
object is in the direction of
the force, F

= × Figure 4.3a: The displacement,s is parallel with the
direction force to horizontal line
b) The displacement, s of
the object is in the m =mass (kg)
direction of the force, F
g = gravitational
= × = acceleration (9.81 −2)

or h = height (m)

=

c) The displacement, s of the Figure 4.3b: The displacement is parallel with the
object is not in the direction force to vertical line
direction of the force, F

θ

= ×

Figure 4.3c: The displacement is parallel with the
direction force to horizontal line (angle)

Chapter 4–Work, Energy and Power/ Page 141

Example 1
A boy pushes his bicycle with a force of 25 N through a distance of 3 m. Calculate the work
done by the boy.

Solution:
Given: = 25 , = 3 , , ?
= ×
= 25 × 3
=

Example 2

A girl is lifting up a 3 kg flower pot steadily to a height of 0.4 m. What is the work done by
the girl? (using = 9.81 −2)

3 kg Solution:
Given: = 3 , ℎ = 0.4 , = 9.81 −2

, ?
= ℎ
= 3 × 9.81 × 0.4
= .

Chapter 4–Work, Energy and Power/ Page 142

Example 3

A man is pulling a crate of fish along the floor with a force of 40 N through distance of 6 m
What is the work done in pulling the crate?

Solution:
Given: = 40 , = 6 , , ?
= cos ×
= 40 cos 50 × 6
= .

QUICK CHECK

1) An object is pulled 4 m on a horizontal 2) A boy pushes a box for 50 m on a
surface by applying a force of 60 N. frictionless surface and he does a work of
Calculate the work done on the object. 10 000 J. Find the force that he applies to
the box.

Answer: W = 240 J Answer: F = 200 N

Chapter 4–Work, Energy and Power/ Page 143