1.2 Resolution of Force (Force & Motion II)

PHYSICS F5
Theme 1 Newtonian Mechanics

CHAPTER 7: FORCE & MOTION II

7.2 Resolution of Forces
Leraian Daya

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

by CHU MP

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

CHAPTER 7: FORCE & MOTION II

You will learn:

1.1 Resultant Force (Daya Paduan)
1.2 Resolution of Forces (Leraian Daya)
1.3 Forces in Equilibrium (Keseimbangan Daya)
1.4 Elasticity (Kekenyalan)

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

1.2 Resolution of Forces

Direction of Motion

Why is the motion of the friend to the right while the
direction of the applied pulling force is inclined upwards?

Every force has a horizontal component Vertical component
and a vertical component.

Resolution of forces is the process of Horizontal component
resolving a force into two components.

Normal FORCE & MOTION II
Reaction 7.2 RESOLUTION OF FORCES

Frictional Vertical
Force component of
the pulling force
Weight of
Object Horizontal
component of
the pulling force

Vertical FORCE & MOTION II
component of 7.2 RESOLUTION OF FORCES
the pulling force
The pulling force can be
Normal resolved into two
Reaction perpendicular components.

Frictional Horizontal
Force component of
the pulling force
Weight of
Object

The vertical component is to balance the weight of the boy
while the horizontal component overcomes friction and
moves the boy to the right.

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

How do we determine the magnitudes
of the two components?

1. Draw the horizontal component, Fx and
the vertical component, Fy to form a
rectangle as shown in the diagram.

2. Calculate the magnitude of the
components. Fx = F cos θ

Fy = F sin θ

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

Fy

Triangle ACD: Triangle ABC:

cos θ = sin θ =


Fx = F cos θ Fy = F sin θ

Example 1: FORCE & MOTION II
7.2 RESOLUTION OF FORCES

Figure below shows a wooden block being pulled by force, T that inclines
at an angle of 30° above the horizontal surface. Table below shows the
magnitudes of the forces acting on the block.

T
Fy

Fx

(a) Calculate the magnitudes of the horizontal component and vertical

component of the pull, T.

Fx = F cos θ Fy = F sin θ

Fx = T cos 30 o Fy = T sin 30 o
= 36 cos 30 o = 36 sin 30 o

= 31.18 N = 18 N

Example 1: FORCE & MOTION II
7.2 RESOLUTION OF FORCES

(b) Determine the magnitude and direction of the resultant force
acting on the block.

Vertical: Horizontal:

F = W – (R + Fy) F = Fx - FR
W = 24 N – (6 N + 18 N) F = 31.18 N - 20 N

=0N = 11.18 N moving to the right.

Example 1: FORCE & MOTION II
7.2 RESOLUTION OF FORCES

(c) What is the acceleration of the block if its mass is 2.4 kg?

Resultant force, F = 11.18 N F=ma
11.18 N = (2.4kg) a

a = 4.66 ms-2

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

Example 2:

Figure below shows the free body diagram of a block, 24 N sliding
down a smooth inclined plane.
(a) Sketch the component of the weight of the block parallel to

the inclined plane and the component of the weight of the
block perpendicular to the inclined plane.

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

Example 2:

Figure below shows the free body diagram of a block, 24 N sliding
down a smooth inclined plane.
(b) Determine the resultant force acting on the block.

- WR

Example 2: FORCE & MOTION II
7.2 RESOLUTION OF FORCES
(b) Determine the resultant force acting
on the block.

60°

Wy
Wx

Wx = W- WsRin 60°
= 24 sin 60° = 20.78 N

Wy = W cos 60°
= 24 cos 60° = 12 N

Resultant of the forces perpendicular to the
inclined plane = 12 + (–12) = 0 N

Resultant force on the block = 20.78 N

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

Example 2:

(c) Calculate the acceleration of the block if its mass is 2.4 kg.

Resultant force,
F = 20.78 N
Mass of block, m = 2.4 kg

F = ma

Acceleration of block,


a =

= 20.78
2.4

= 8.66 m s–2

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

Force can be resolved to two components at any direction as
long as the two components are perpendicular to each other.

Component Y Component Y

90o Component X

Component X

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

Formative Practice 1.2

1. Resolve the following forces into horizontal component
and vertical component.

40° 70 cos 42° Fy = 90 sin 64o N 90 sin 64°
70 sin 42° = 80.89 N
Fy = 70 sin 42o N
= 46.84 N Fy = 90 cos 64o N
= 39.45 N
Fx = 70 cos 42o N
= 52.02 N 70 cos 40° 64°

90 cos 64°

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

2. Figure below shows a man pushing a lawn mower with a
force of 90 N.

(a) Resolve the pushing force into
its horizontal component and
vertical component.

Fx = 90 sin 60o N

F =90 N Fy = 90 cos 60o N
60°

Fy = 90 cos 60o N Fx = 90 sin 60o N
= 45 N = 77.94 N

FORCE & MOTION II
7.2 RESOLUTION OF FORCES

2. Figure below shows a man pushing a lawn mower
with a force of 90 N.

(b) State the function of the
horizontal component and vertical
component of the pushing force
when the lawn mower is being
pushed.

The horizontal component forces move the lawn
mower forward.
The vertical component forces press the lawn
mower on the surface of the field.

Solving Problems Involving Resultant Force and
Resolution of Forces - Google Drive