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BB101
ENGINEERING SCIENCE
CHAPTER 3:
FORCES
LEARNING OUTCOME


CLO 1 Identify the basic concept of force.
(C1)
CLO 3 Apply concept of force in real basic
engineering problems. (C2, A1)
2
UNDERSTANDING OF
FORCE

Apply the concept of force

Define force and its units.

Differentiate between weight and mass

Define Newton’s Second Law.

Define forces in equilibrium.

Apply the concept of force in solving problems.

Calculate Resultant Force.
Understand the concept of Moment Force

Define Moment Force

Describe principle of moment

Apply the concept and formula of moment force in solving the
related problems
3
CONCEPTUAL MAP
INTRODUCTION
FORCE
Definition
Type of force
Effect of force
Newton
Second Law
Resultant force
using resolution
force
Weight and
mass
Force in
equilibrium
Definition
MOMENT
Resultant Moment
Formula of moment
Force Moment
Concept
4
Application
Force in our life…
http://lmspsp.cidos.edu.my/course/vie
w.php?id=333
5
INTRODUCTION

Definition of Force and its units.
 Force is the action of pushing or pulling on an objects.
 The SI unit of force is Newton (N) or kgms-2.

The effect of force on an object:
 A stationary object to move.
 A moving object to change its speed.
 A moving object to change its direction of motion.
 An object to change its size and shape.

Application of force
 Using in mechanical system
 Transferred using mechanical instrument, eg: gear, pulley, screw
and piston.

How to measure the force?
 using a spring balancing together with Newton scale (N).
6
THE TYPES OF FORCES

The balance force is:
When two or more external forces acting on a body produce no net
force,


Pulling forces = pushing forces
The following are some of the situation where forces are balanced on
a body :



a pile of book resting on a hard surface
a car moving at constant velocity along a straight road
an airplane is flying horizontally at a constant height with a
constant velocity
10 N
10 N
7
THE TYPE OF FORCES

Unbalanced force is :
When two or more forces acting on a body are not balanced, there
must be a net force on it. This net force is known as the
unbalanced force or the resultant force.


Pulling forces ≠ pushing forces
The effects of unbalanced forces acting on an object are shown in
the following examples



Golfer hits a stationary golf ball
A footballer kicks a fast moving ball towards him
When the engine of a moving car is shut down
8N
12 N
8
WEIGHT & MASS

The weight of an object is defined as the gravitational force acting on the
object.
Weight (N) = mass (kg) x gravity (ms-2)
W = mg
where,
W = weight, m = mass, g = gravitational field (9.81ms-2)
The SI unit for weight is Newton (N) and it is a vector quantity

The mass of an object is a measure of its inertia

Mass is constant quantity and it is a scalar quantity. It is the same
irrespective of where the object is.
Mass (kg) ≠ weight (N)
9
WEIGHT & MASS
• The difference between weight and mass is summarized in below
table
Weight
Mass
Dependent on the acceleration
due to gravity
Is a constant quantity
Is a vector quantity
Is a scalar quantity
Is measured in Newton (N)
Is measured in kilogram (kg)
10
Please think of ……..

Why we say, we weigh our body, while
the given unit is in ‘kg’. Remember “kg”
is unit of mass, not the unit of weight.
Therefore, why we didn’t say we scale
the mass. So, please think of …..
Unit in ‘kg’
11
NEWTON’S LAW

Definition :

When net force acting on an object is not zero, the object will
accelerate at the direction of exerted force
F  a
F = ma
Where,
F = Force
m = mass
a = acceleration (9.81ms-2)
12
GROUP ACTIVITY
 Gallery
walk
13
Exercise 1
1)
a)
Determine the direction of situations below and give the explanation.
6N
15 N
b)
c)
4N
8N
10 N
4N
4N
2) A certain force is applied to a 2.0 kg mass. The mass is accelerated at
1.5 ms-2. if the same force is applied to a 5.0 kg mass, what is the
acceleration og the mass?
3) A car of mass 700 kg accelerates from rest to 105 km h-1 in 10 s. what
is the accelerating force developed by the car engine?
5N
4) A toy car of mass 800 g is pulled along a level runway with a constant
speed by a force of 2 N.
a) what is the friction on the toy car?
b) when the force is increased to 6 N, what is
i) the unbalanced force acting on it?
ii) the acceleration of the toy car?
14
EXERCISE 2



An astronaut has mass of 70 kg. what is his weight if
(a) he is on the surface of the Earth where the gravitational field strength
is 9.8 N kg-1?
(b) he is on the surface of the moon where the gravitational field strength
is 1/6 of that on the surface of the Earth?
A spacecraft of mass 800 kg is orbiting above the Earth’s surface at a height
where its gravitational field strength is 2.4 N kg-1.
(a) what is meant by gravitational field strength at a point in the
gravitational field?
(b) Calculate the gravitational force experienced by the spacecraft
A rock has a mass of 20.0 kg and weight of 90.0 N on the surface of a planet.
(a) What is the gravitational field strength on the surface of the planet?
(b) what are the mass and the weight of the rock on the surface of the
Earth where its gravitational field strength is 9.8 N kg-1?
15
PROPERTIES OF VECTOR

Showed by symbol and arrow
magnitude
direction
A
B
Write as vector AB
16
VECTOR OF FORCES


Showed by arrow.
The length of this diagonal represents the magnitude resultant force,
F and its direction.
magnitude
direction
F = 15 N
F = 30 N
30°
Vertical vector F
Force at 30 °
17
FORCES IN EQUILIBRIUM

An object at rest is in equilibrium. This is because the forces acting
on it are balanced and the resultant force is zero.
Fy 1
Horizontal force
Therefore
Fx 2
Fx 1
Fy 2
Vertical force
Therefore
Fx1 = Fx2
Fx1 – Fx2 = 0
Fy1 = Fy2
Fy1 – Fy2 = 0
Since the resultant force on an object in equilibrium is zero, if the forces
are resolved into horizontal and vertical components, then
(a) The sum of all the horizontal components of the forces = 0
(b) The sum of all the vertical components of the forces = 0
18
ADDITION FORCES :
RESULTANT FORCES

A resultant force is a single force that represents the combined
effect of two or more forces with magnitude and direction

Vertical force = The forces acting at y-axial
Horizontal force = The forces acting at x-axial


The effects of force are depends on:
1. The magnitude – the value of forces in Newton’s unit
2. The directions – left, right, up and down
C
AB + BC = AC
AC is resultant force
A
B
19
ADDITION FORCES:
RESOLUTION FORCE
A resolution forces is a single force that can be resolved into
two perpendicular components

Fy
Figure shows a force F is resolved into two
perpendicular components Fx and Fy.
F
θ
Fx
With that,
Fx = F cos 
Fy = F sin 
Fx is the Vertical component of force whereas Fy is the Horizontal
component of force
20
PROBLEM SOLVING

Example 1:
The table being pulled by two forces with the magnitude of 6N and 8N
respectively. The angle between the two forces is 60°
Solution
Method 1 : can be determined by using the parallelogram of
forces
Method 2 : can be resolved into two perpendicular
components (using formula)
21
Method 1 : Parallelogram

Steps:

Choose the scale . Eg: 1 cm = 1m.

Using the graph paper and set the point

Draw the forces F1 and F2 from a point with an angle of  with
each other.
F1
F

F2


Draw another two lines to complete the parallelogram
Draw the diagonal of the parallelogram. The length of this
diagonal represents the magnitude resultant force, F and its
direction, α can be determined by measuring the angle
between the diagonal with either one side of the parallelogram
F1
F

F2
22
Method 1 : Parallelogram

Steps:
6.0 N
Fx = ______ N
Fy = ______ N
60°
8.0 N
Fx and Fy are the vertical and horizontal components of the force :
Magnitude of Fx = 6.0 cos 60° = 3.0 N
Magnitude of Fy = 8.0 sin 60° = 7.0 N
23
METHOD 2 : FORMULA

Using formula :
R
= √ FY 2 + FX 2
= √ 62 + 82
= √ 36 + 64
= √ 100
= 10 m
24
Exercise 3
1.
Who will win?
15 N
45 N
2.
20 N
Calculate the total net force between the following interaction:
a)
c)
25 N
5N
8N
32 N
25 N
10 N
b)
30 N
d)
45 N
25 N
55 N
25 N
25 N
25
THE TOTAL OF FORCE
1.
The Total of force between two or more interaction
Example 1:
Total net force
F1
F2
F1
F = F1 + F2
Total net force
F2
F3
F = (F2 + F3 ) - F1
26
THE TOTAL OF FORCE
1.
The Total of force between two or more interaction
Example 2:
Total net force
15 N
5N
10 N
= ( 5 + 10 ) – 15 N
= 0 (equilibrium state)
Total net force
24 N
28 N
= 28 – 24 N
= 4 N (move to right side)
27
THE TOTAL OF FORCE
2.
The total of force acting at on angle 
10 N
FX
30°
Method 2: Scale drawing :
• Using paper graph to get accurate reading / value
•Scale 1cm = 1N
•Use the protractor to measure the angle of 30°
10 cm
Fx = _________ cm
= _________ N
30°
Fx
28
THE TOTAL OF FORCE
2. The total of force acting at on angle 

Example 3:
Method 1 : Analysis
10 N
FX
30°
FX
= 10 cos θ
= 10 cos 30°
= 8.67 N
29
THE TOTAL OF FORCE
The total of force acting at on angle 
2.

Example 4
Method 1: Analysis
10 N
35°
40°
FX
12 N
Fx
= FX1 + FX2
FX1
= 10 cos 35° = 8.19 N
Fx2
= 12 cos 40° = 9.19 N
Fx
= 8.19 + 9.19
= 17.38 N
30
THE TOTAL OF FORCE
2.
The total of force acting at on angle 
Method 2: Scale drawing
10 N
• Using paper graph to get accurate reading/value
•Scale 1cm = 1N
35°
•Use the protractor to measure the angle of 35°
40°
FX
12 N
40°
10 cm
12 cm
Fx = _________ cm
= _________ N
35°
Fx
31
MOMENT

The moment of a force can be worked out using the
formula: moment = force applied × perpendicular distance
from the pivot. If the magnitude of the force is F newtons
and the perpendicular distance is d metres then:
MOMENT = Force X Perpendicular distance (arm)
=FXd
= (Newton) x (meter)
 SI unit  = Nm

Force (F)
Distance (d)
32
MOMENT OF FORCES

For an object to be in static equilibrium,
1.
the sum of the forces must be zero, but also the sum of the
torques (moments) about any point. For a two-dimensional
situation with horizontal and vertical forces, the sum of the
forces requirement is two equations:
The total of anti-clockwise moment = the total of clockwise moment
F1
F2
F1 d1 = F2 d2 Nm
d1 m
d2 m
RF
2. The total of normal force = the total of interaction force
F1 + F2 = RF N
33
MOMENT OF FORCES
For an object to be in static equilibrium,
The centre of gravity can be determine using moment resultant
method
Resultant moment = the total of force moment


Given, the centre of gravity at x is A
F1
F2
x1
F3
x2
Then, label the reference point of moment at A
Resultant moment = ( F1 + F2 + F3 ) x
B The total of force moment = F1 (0) + F2 (x1) + F3 (x1 + x2)
A
x
So, ( F1 + F2 + F3 ) x = F1 (0) + F2 (x1) + F3 (x1 + x2)
X=
F1 (0) + F2 (X1) + F3 (X1 + X2)
( F 1 + F2 + F3 )
34
MOMENT OF FORCES

Example 5:
Determine the centre of gravity for force action, so that the bar
remains in horizontal equilibrium
20 N
50 N
4m
A
B
x
35
MOMENT OF FORCES

Example 5 :
Solution
Method 2:
Method 1:
20 ( 0 ) + 50 ( 4 )
Given the centre of gravity at x form A is,
The total of anti-clockwise moment = the
total of clockwise moment
20 ( x )
= 50 ( 4 - x )
20 ( x )
= 200 – 50 x
20 x + 50 x = 200 x
70 x
= 200
x =
=
( 20 + 50 )
200
70
x = 2.86 m
x = 2.86 m
36
MOMENT OF FORCES

Example 6:

Determine the centre of gravity for force action, so that the bar
remains in horizontal equilibrium
50 N
25 N
1m
100 N
4m
P
Q
x
37
MOMENT OF FORCES
Example 5 :

Solution
Method 2:
Method 1:
Given the centre of gravity at x
form P is,
x =
50 ( 0 ) + 25 ( 1 ) + 100 ( 5 )
50 + 25 + 100
525
The total of anti-clockwise moment =
the total of clockwise moment
=
50 x + 25 ( x - 1 )
= 3.00 m
50 x
+ 25 x
= 100 ( 5 - x )
175
- 25 = 500 – 100 x
175 x
= 525
x = 3.00 m
38
EXERCISE 4
1.
Determine the centre of gravity for force action, so that the bar
remains in horizontal equilibrium
a)
c)
5N
8N
15 N
1m 1m
6m
2m
10 N 15 N 12 N
8m
b)
8N
15 N
12 N
10 N
d)
16 N
55 N
25 N
45°
1m
4m
2m
12 m
39
EXERCISE 5
1.
a)
Find the interaction for both of following points, RA and RB :
0.5m
b)
RA
1m
RA
0.5m
1m
2N
3N
1kg
1kg
1m
RB
1m
1m
2kg
RB
40
THE END