Content standard Learning standard

1.1 Resultant force

• Define resultant force Youtube
• Determining resultant force for;
• Parallel forces in the same direction
• Parallel forces in the opposite direction
• Forces perpendicular to each other
• Non-parallel forces (using scale diagram – triangle method and
parallelogram method)
• Communicate the resultant force when object is;
• At rest, F = 0 N
• Moving with constant velocity, F = 0 N
• Moving with constant acceleration, F≠ 0 N
• Solving problems involving resultant force, mass and acceleration of
an object
• Solving problems involving;
• Object moving horizontally or vertically
• Passenger in an elevator
• Object pulled with a pulley
1.2 Resolution of forces • Describe resolution of forces
• Solving problems involving resultant forces and resolution of forces
• Resolving forces into two components for objects moving in non-
parallel forces such as;
• Object pulled or pushed along an inclined plane
• Object sliding along inclined plane due to its weight
1.3 Forces in equilibrium • Explaining forces in equilibrium
• Sketching triangle of forces for three forces in equilibrium
• Sketching triangle of forces involving forces in equilibrium such as;
• Object at rest on an inclined plane
• Picture frame hung using string
• Ship pulled by tugboat with constant velocity
• Solving problems involving forces in equilibrium
• Solving problems involving forces in equilibrium by;
• Resolution of forces
• Drawing scale diagram of triangle of forces
1.4 Elasticity • Describing elasticity
• Experimenting to find the relationship between F and x
• Communicating the law related to F and x
• Solving problems involving F and x

Most of the image, vector or diagram in this module are either original content or available from Freepik.com
1.1 Resultant force
A single force that represents the effect of combination of two or more forces by magnitude and direction.

1. Same direction
2. Opposite direction Youtube
3. Perpendicular force
4. Non-perpendicular forces

Parallel forces - same direction ( Assumption: Force towards the right is +, towards the left is - )
1. Calculate the resultant force.

a. 8N b.
5N
2 kg 2 kg
15 N 10 N

2. If the resultant force is 12N, calculate F.

a. 5N b. F
2 kg 2 kg
F 10 N

Parallel forces in opposite direction ( Assumption: Force towards the right is +, towards the left is - )

3. Calculate the resultant force.

a. b.
8N 15 N 7N 5N
2 kg 2 kg

4. If the resultant force is 12N, calculate F.

a. b.
8N F F 5N
2 kg 2 kg

2
 Perpendicular forces (Pythagoras’ theorem)

5N
3N
3N 3N
37°

4N 4N 4N

CAH TOA
SOH c = 𝑎2 + 𝑏 2
𝑂 𝐴 𝑂
sin Ɵ = cos Ɵ = tan Ɵ =
𝐻 𝐻 𝐴

Non-perpendicular forces (scale drawing)

10 N
1. Choose a scale.
50°
2. Draw the lines with correct length and angle.
8N Use ruler and protractor.
3. Draw triangle (a) or parallelogram (b).
4. Measure length and calculate force using scale
chosen in (1).
1 Scale → 1 cm : 2 N

2 3 a.
5 cm 5 cm

50°

4 cm 4 cm

3 b. 5 cm
4

4 cm (Length in cm) x 2 N = F

3
Exercise
1. Calculate the resultant force.

a. 2N
8N
5N

b.
2N 8N
1N

2. Calculate the resultant force.

15 N
a. b.

8N
12 N

5N

3. Find the resultant force.

250 N

50°
200 N

4
Free body diagram

• Diagram that shows all the forces acting on the object only.

R = normal reaction from ground Engine thrust

Normal reaction

Air resistance
Engine thrust

Air resistance
Frictional force
W = weight of ball Weight
W = weight of rocket

• F = ma where F is the resultant force acting on an object.

- When object at rest or in uniform velocity, acceleration is 0 ms-2 → F = 0
- When object is accelerating, a is not 0ms-2 → F ≠ 0

R = normal reaction // W = weight // T = engine thrust // FR = frictional force

R R R

T T
FR FR
W W W
Car at rest Car at uniform velocity Car accelerating
v=0 ms-1 v is uniform v increases
a = 0 ms-2 a = 0 ms-2 a ≠ 0 ms-2
F=0N F=0N F≠0N

R=W R=W R=W

T = FR T > FR
F = T - FR

5
 Exercise
1. The box is pulled on a rough surface with a string. It moves to the
right at constant speed. The surface of the ground is rough. string
a. Sketch the free body diagram of the box.
b. Compare the normal reaction and weight of the box.
c. Compare the tension in string and the frictional force.

2. The box is pulled on a rough surface with a string. It accelerates

to the right. The surface of the ground is rough. string
a. Sketch the free body diagram of the box.
b. Compare the normal reaction and weight of the box.
c. Compare the tension in string and the frictional force.

3. A car with engine thrust 1500 N is accelerating at 2ms-2. If the air resistance acting on the car is 500 N,
what is the mass of the car?

1500 N 500 N

4. An engine causes a model rocket to propel upwards with 2000 N force. If the mass of the rocket is
150kg, calculate its acceleration. Ignore air resistance.

Elevator
R = normal reaction or reading on balance // m = mass // g = gravitational acceleration // a = acceleration of elevator // W = weight

R R R
Downwards acceleration
At rest / constant velocity

“Feels lighter”
Upwards acceleration

R=mg-ma
“Feels normal”
R=mg
“Feels heavier”
R=mg+ma

W W W

Youtube

6
Exercise

1. Diagram shows a boy standing in an elevator on a weighing scale. The mass

of the boy is 70kg. Calculate the reading on the scale if the elevator;
a. is at rest.
b. is accelerating upwards at 3ms-2.
c. is accelerating downwards at 2ms-2.

2. Diagram shows a boy standing in an elevator on a weighing scale. The mass

of the boy is 50kg. Calculate the reading on the scale if the elevator;
a. is moving upwards with constant speed 5ms-1.
b. is accelerating downwards at 1.5ms-2.

Pulley system
Youtube
• Tension in string, T and acceleration, a in all parts in the system are the same
- T1 = T2
- a1 = a2
• Forces in the system;
- When accelerating upwards, T – W = ma
- When accelerating downwards, W – T = ma
- When at rest / constant speed, T = W

a2
T2
T1
FR T2 T1
T2
T1
a2

a1

W2 a1
W
W
W1
7
Exercise

1. A boy pulls on a rope attached to a pulley system. Calculate the tension

in the string if the object has mass of 10 kg and is;
a. at rest.
b. accelerating 1.5 ms-2 upwards.

2. A boy pulls on a rope attached to a pulley system. Calculate the tension in

the string if the object has mass 5 kg and is;
a. moving upwards with constant velocity of 2 ms-1.
b. accelerating 3 ms-2 downwards.

3. When the weight is released, it accelerates 3kg

downwards. Given that the frictional force is 5N. Friction

a. Calculate the resultant force of the trolley.

b. Calculate the tension in the string.
c. Calculate the mass of the weight. W

1.5ms-2

4. Two masses are attached to each of a string in a pulley system. Calculate the
tension of the string and the acceleration when the masses are released.

5 kg

10 kg

8
1.2 Resolution of forces
(vertical component)
y-component

F
Fy

x-component FX
(horizontal component)

𝐹𝑦 𝐹𝑥
sin Ɵ = FY = F sin Ɵ cos Ɵ = FX = F cos Ɵ
𝐹 𝐹 Youtube

Exercise
1. A box is pulled with a string at 40° angle from the smooth
surface with a force 5N. If the mass is 2kg, calculate the
acceleration.

40°
2 kg

2. A box is pulled with a string at 45° angle from the ground with
a force 15N. Given that the mass is 3kg and the friction with
the surface is 8N, calculate the acceleration.

45°
3 kg

9
Object on inclined plane
R

W
θ

𝐴
cos θ =
𝐻
A = H cos θ

W cos θ θ
A
H
θ θ

θ W sin θ
O
W 𝑂
W sin θ =
𝐻
O = H sin θ

Exercise
1. A 5kg box is placed at rest on an inclined plane. The angle of the plane is 40°.
a. Calculate its normal reaction force.
b. Calculate the minimum frictional force acting on the plane.

2. A 5kg box is on a smooth inclined plane (no friction). The angle of the inclined plane is 30°.
a. Calculate its normal reaction force.
b. Calculate the acceleration of the box.

10
1.3 Forces in equilibrium
• An object is said to be in equilibrium of forces when the forces acting on it produce a zero resultant force.
• Triangle of forces can be used to show the equilibrium of three forces acting on an object.
• If drawing to scale, magnitude of forces are represented by the length of the sides of the triangle i.e. the
larger the force, the longer the length of the side.

Tip-to-tail method
Arrow “main kejar-kejar”

1 2 1 2
1

3
3 3 Rearrange
2
Triangle of forces

Free body diagram Youtube

Original condition

Exercise
Sketch the triangle of forces for the situations below.

1. Portrait of Isaac Newton hanging on the wall 2. A fancy lamp held by two strings

3. Two tugboats pulling on a ship

Water drag

11
Solving problems involving forces in equilibrium
(refer to page 21 – 23 of the textbook for more examples)

Cosine rule
C
𝑎2 = 𝑏 2 + 𝑐 2 - 2bc cos A

𝑏 2 = 𝑎2 + 𝑐 2 - 2ac cos B

b a
𝑐 2 = 𝑎2 + 𝑏 2 - 2ab cos C

Sine rule

𝑎 𝑏 𝑐
= =
sin 𝐴 sin 𝐵 sin 𝐶
A c B

Example using cosine rule

Given that the weight of the giant bulb is 50N, calculate the tension of the rope.

50° 50° 50°

T T 40° T

50 N

40° T
50 N 50°

𝑎2 = 𝑏2 + 𝑐 2 - 2bc cos A

40° T 502 = 𝑇 2 + 𝑇 2 - 2 (T)(T) cos 100

502 = 2𝑇 2 - 2 T2 cos 100

100° 2500 = 2𝑇 2 - T2 (-0.347)

50 N

a 2500 = 2𝑇 2 - T2 (-0.347)
40° T 2500 = 2.347 𝑇 2

T = 32.6 N

12
Using sine rule
Given that the weight of the giant bulb is 50N, calculate the tension of the rope.

50° 50° 50°

T T 40° T

50 N

40° T
50 N 50°
40° T A
A B
B =
100° b sin a sin b
50 N
T 50
=
a sin 40 sin 100 Youtube
40° T
T = 32.6N

Using resolution of forces

Given that the weight of the giant bulb is 50N, calculate the tension of the rope.

50° 50° a
T T

1. Since it is at rest, the weight of the bulb is equal to the

vertical force (a).
2. Total vertical force (a) is equal to the total of FY of each rope.
3. Each vertical component of the rope is 25 N (50 N weight is
held by two ropes at the same angle so 50 ÷ 2 = 25 N). 50 N
25
sin 50 =
50° 50° 𝑇
25
25 T=
25 sin 50

T = 32.6 N

13
Exercise

1. Given that the tension of the rope is 3.11 N. Calculate the

weight of the object.

40° 40°

2. Two tugboats A and B are pulling on a ship. Calculate the tension in each rope if boat is moving at
constant velocity.

Water drag 50°

1500N 30°

14
1.4 Elasticity
The property of material that enables an object to return to its original shape and size after the force
applied on it is removed.

Which of the following materials are elastic material?

1 2 3 4

Hooke’s Law

Encyclopædia Britannica, Inc.

Extension of spring is directly proportional to the force applied to the

spring provided that the elastic limit of the spring has not been
exceeded.

F = force Youtube
k = spring constant
x = extension of spring

F
Gradient : Spring constant

Area under the graph: Elastic potential energy

1
EP = Fx
2
1
EP = kx2
2
x

15
Exercise

1. Given that the length of spring without any weight attached on it is 15cm. When added a 5N weight,
the spring stretches and the length increases by 3cm.
a) What is the value of spring constant, k of the spring?
b) What is the length of the spring if 15N weight is used to replace the original weight?

2. Given that the length of spring without any weight placed on it is 10cm. When a 2N weight is placed
on it, it compresses and the length of the spring becomes 6cm.
a) What is the value of spring constant, k of the spring?
b) What is the length of the spring if 4N weight is used to replace the original weight?

3. Graph below represents the changes in the extension of a spring when different weights are attached
to it. Given that the initial length of a spring is 10cm.

F/N
a. What is the length of the spring if 1.5N of
weight is added to it?
b. Calculate the spring constant.
3 c. Calculate the elastic potential energy of the
spring when 3N weight is added to the
spring.
2

x/cm
0 2 4 6 8

16
Factors affecting spring constant

a. material of spring: spring constant b. length of spring: spring constant is

changes with different materials higher when spring is shorter

steel copper

>

c. diameter of spring: spring constant is d. thickness of spring wire: spring constant

higher when diameter is smaller is higher when wire is thicker

> >

Arrangement of spring

Parallel Series

F
Tension =
Number of spring

Extension =
x F
Number of spring

Tension = F
Extension = x . number of spring

Youtube

17
Exercise
(A) (B)
1. Diagram shows a spring without any weight
attached to it (A) and the spring when a weight is
attached to it (B). Based on diagram (C) and (D),
calculate; 10cm

a. tension in each spring. 15cm

b. extension of each spring.
c. total length of the system.

2N

(C) (D)

2N

2N

18