
The force that act on the object are
balanced in all direction.

The force cancel each other, so that the
resultant force or net force is zero.

Newton’s Third Law of motion states that
to every action there is an equal but
opposite reaction.

Is a single force that represents the
combined effect of 2 or more forces in
magnitude and direction.
1.
2 forces that act along the same direction.
F1 = 8 N
F2 = 3 N

2.
F = 11 N
=
Resultant force ,F = F1 + F2
2 forces that act in opposite direction.
F1 = 8 N
F2 = 3 N
F=5N
=

Resultant force, F = F1 – F2
3.
2 forces acting at a point at an angle to
each other.
a) 2 non-parallel force
F2
θ
F1
i.
Parallelogram law
F2
F2
θ
θ

F
θ
F1
F1
F1
ii.
F2
The tails of 3 forces F , F1 and F2 must originate
from the same point.
The triangle method (tail to tip method)
F
F2
F2
θ
F1
θ
F1
θ
F1
The tail of
F2 F1connecting to
the tip of F2
b) 2 perpendicular forces
F2
F1
i.
Using parallelogram law
F2
F2
θ
F1
i.
F
F1
Using Pythagoras Theorem
› Resultant force, F = √(F1)² + (F2)²
› tan θ = F2
F1
1.
Find the resultant force for the two
forces as shown in figure 1.
8N
120°
12 N
Figure 1
Answer : F = 10.6 N at angle of 41°
2.
Samy and Heng Gee pull a crate with
force of 70 N and 90 N respectively. Find
the resultant force on the crate due to
these two forces.
Samy
F2 = 70 Ns
70 N
F
θ
Heng Gee
F1 = 90 N
Answer :
 Resultant force, F = 114 N
 tan θ = 0.7778
 θ = 37.9°
90 N

A single force can be resolved into 2
perpendicular components.
F
Fy
θ
Fy = F sin θ
θ
Fx
Fx = F cos θ ( Fx / F = cos θ)
 Fy = F sin θ
( Fy / F = sin θ )

Fx = F cos θ
1.



A tourist is pulling a bag with a force of
12 N at an angle 60° to the horizontal
floor. What is the horizontal and vertical
components of the force?
Answer :
12 N
Fy
Fx = 6 N
Fy = 10.4 N
Fx
Santhiran pulls a 5 kg crate on the floor
with a force of 35 N.
 Find the horizontal component of the
force.
 If the crate is moving with constant
velocity, what is the friction against the
crate.
 If the friction against the box is 8 N, what is
the acceleration of the crate?
2.
35 N
25°
Answer :

›
›
›
Fx = 31.72 N
Friction : 31.72 N
a = 4.7 ms-2
ANALYSING FORCES IN
EQUILIBRIUM
2 types:
Object in equilibrium on an inclined
plane.

A.
I.
II.
B.
If the object is at rest.
If the object is moving on a smooth
inclined plane.
Three forces in equilibrium.
ANALYSING FORCES IN EQUILIBRIUM
Object in equilibrium on an inclined plane.
A.
If the object is at rest.
I.
Normal reaction
friction
mg sin θ
mg
mg cos θ
The net force perpendicular to the plane = 0
Normal reaction – mg cos = 0



F normal = mg cos
The net force parallel to the plane = 0
Frictional force – mg sin = 0



F friction = mg sin
If the object is moving on a smooth
inclined plane, vertical component of
the forces are balanced but the force
down the plane is not balanced.
II.
Normal reaction
Motion
friction
a
mg sinθ
mg



F net = ma
mg sin = ma
a = g sin
mg cos θ
Example

A carton of mass 5 kg is at rest on an inclined plane
making an angle of 15° with the horizontal. Find the
frictional force and the normal force acting on the
carton.
Normal reaction
friction
mg sin θ
15°
mg mg cos θ
Answer :
 F friction = 12.9 N
 Normal force = 48.3 N

Three forces in equilibrium
B.
B
A
C
Problem involving 3 forces in equilibrium can
be solving using :

i.
Resolution of forces
 Total force to the left = total force to the right
 Total force upward = total force downward
ii.
Drawing a closed tail- to –tip ( triangle method)
Example

A 12 kg mass is suspended from a hook in the
ceiling. The object is pulled aside by a horizontal
string and makes an angle of 45° with the
horizontal. Find the tension in both string.
T1 sin θ
T1
45°
T1 cos θ 45°
12 kg



Answer :
T1 = 169.7 N
T2 = 120 N
T2
mg
Problem involving F = ma and
W = mg
1.
2.
2)


Lift ( refer the notes)
Pulley system
Pulley system
A frictionless pulley serves to change the direction of a
force
Have 2 types :
i. A force pulling a mass over a pulley
 In this situation, the tension, T is equal to the pulling
force, F even if the rope is slanting.
ii. A pulley with 2 masses.
 The heavier mass will accelerate downwards while the
lighter one will accelerate upwards at the same rate.
 The tension is not equal to the weight of either mass.
Example pulley system

A 5 kg mass is used to accelerate a 3 kg block
along a table as shown in figure1. The friction
between the table and the block is 10 N. Assuming
that the pulley is smooth and the string is of
negligible mass, find
› The acceleration of the system
› The tension in the string
a
3kg



Answer :
a = 5 ms-2
T = 25 N
a
5 kg