CHAPTER 4 FORCE

CHAPTER 4
FORCES

[ 10 Hours ]

4.1 Basic of Forces
4.2 Newton’s Laws of Motion

4.0 Introduction

A force is something that is capable of changing an
object’s state of motion.

Push Force

Pull Force

A force is a push or pull upon an object resulting
from the object’s interaction with other object.

Force is vector quantity. SI unit : kg m s-2 @
Newton (N)

4.1 Basic of Forces and free body diagram

a) Identification Forces On The Moving Body

Forces acting on a body in different situations:
i) Weight, W
ii)Tension, T
iii)Normal force, N
iv)Friction, f
v)External force, F

Forces

Weight, W Tension, T Normal Friction, f External
Force, N @ R force

the force the pulling force that a reaction force as a force that resists A force exerted on a
exerted on a is directed away from that exerted by the motion of one system by other objects or
body under the surface to an surface relative to
gravitational the object and object interact sources
attempts to stretch or another with which it is
field elongate the object with it in contact

Free body diagram (FBD) Coefficient Pull Push
Simple diagram that shows the of friction, 
magnitude and direction of all
forces acting upon an object Static, s Kinetic, k

4.1 BASIC OF FORCES & FREE BODY
: DIAGRAM

1. The weight of a body is the gravitational force
acting on the body itself, W. The magnitude of the
weight is given by:
W = m g where g is the acceleration due to gravity.

W=mg

1.1 Weight, W .
- Weight

W mg

ex:
If the mass is 1kg, then
since g = 9.81 m s–2, the
weight W = 1 × 9.81 = 9.81 N

- Vector quantity; SI unit: Newton (N) @ kg m s-2

- varies slightly with altitude because weight
depends on the strength of the gravitational
field (g).

2. Tension, T is the force in a cord that pulls on a
body.
It is always directed away from a body and along the
cord. The tension pulls at either end of the attachment
with the same magnitude.

Tension (T) T T
Tension (T) T T 1 kg

4 kg

T

3. When a body is in contact with a surface, a force
that is perpendicular to the surface is exerted on the
body. This force is known as the normal force ( N ) @
reaction force (R).

N

N

N
N



3.1 Normal Reaction Force
-- is the force exists whenever 2 solid surface are

in contact.
-- acts perpendicular (90°) to the surface.

Normal reaction force is NOT
necessarily always equals to the
weight mg of the object.

Example : for Inclined Plane

N Reaction
θ
mg cos θ is an action pushes on the
surface. The surface pushes back with
a normal reaction force, N which is
perpendicularly upwards on the object.
N is the reaction to the action.

mg sin θ

mg cos θ Action
θ

mg

According to Newton’s 3rd law : N = mg cos θ

Example 
a, Fnet
The diagram shows the 
forces acting on a person in a, Fnet
a lift. The person has a mass
of 70 kg. Calculate the
normal reaction force N
when :
(a) the lift is at rest.
(b) the lift is accelerating

upwards at 1.0 m s–2.
(c) the lift is accelerating

downwards at 2.0 m s–2.
(d) the lift is ascending at a

steady speed.

Solution 
a
(a) At rest,  0  Fnet  0

N  mg  0

N  mg  709.81  686.7 N

(b) Accelerating upwards, applying Newton’s 2nd law

Fnet  ma

N  mg  ma

N  ma  mg

 70(1)  70(9.81)

 756.7 N We feel heavier

(c) Accelerating downwards,

Fnet  ma

mg  N  ma

N  mg  ma

 70(9.81)  70(2)

 616.7 N We feel lighter

(d) Constant speed  acceleration a = 0 m s–2 ,

Fnet  ma  0

N  mg

 686.7 N

4. Friction, Ff is the force that resists the motion
of a body as it slides over the rough surfaces.
Friction force always acts parallel to the surface
in contact and in the opposite direction of the
motion.

5. External force, F is a push or pull upon an object
resulting from the object’s interaction with other
object.

Push Force

Pull Force

b) Sketch Free body diagram

Free body diagram: a diagram that shows all the
forces acting on just one object.

Object is represented by a point or a
box.The force arrow is draw from the
center of the point ( or box ) outward
in the direction in which the force is
acting.

Direction of arrow reveals the direction in which
the force acts.

Example

Draw a free body diagram showing all
the forces acting on the object moving on
rough surface.

Solution N F

Ff .

W

This is free-body diagram.
We can find the value of force, normal force,
weight and friction force

Free body Diagram for Object in Motion
(a) An object is speeding up as it’s pulled to the

right, with no friction

N

F

W

Force x-comp y-comp

N FF 0
N0 N

FW0 W

W

(b) An object moving to the right slowing down
due to friction

Ff

W

N Force x-comp y-comp

Ff Ff 0
Ff N 0 N

W0 W

W

(c) An object accelerating to the right with a
force applied at an upward angle

N F

Fy Fx

Ff

W

(d) An object accelerating at inclined
without friction

N

θ

W W

W

θ

Exercise

Physical Situation Draw Free Body Diagram showing
all the forces acting on each of the
A book is at rest on top of objects in the system.
a table.

A block is being pushed
to the right across a
frictionless table surface
by a force, F

Physical Situation Draw Free Body Diagram showing all the
forces acting on each of the objects in the
system.

A man stand in an
upward moving
elevator .

A block of mass m =
10 kg hanging from
a cord.

Physical Situation Draw Free Body Diagram showing all
the forces acting on each of the objects
in the system.

A block of mass m
hanging from
three wires.

Mass m rest on an
inclined plane.

c) Determine static, fs and kinetic friction, fk

The origin of friction. No surface is perfectly smooth. When
viewed on the atomic level, even the “smoothest” surface is
actually rough and jagged. This type of roughness contributes
to friction. Force is required to overcome the microscopic hills
when surfaces slide across each other.

-- Factors affecting friction force :

(1) Type of surfaces in contact

(2) How hard the surfaces are pressed together
-- friction force is directly proportional to
the normal reaction force.

Friction between solids is generally classified
into 2 types :

(1) Static friction, fs

(2) Kinetic friction, fk

Static friction, fs

-- is the friction force exist between 2 object when
there is no motion.

May have different values up to some maximum.

fs  s N

Just before an object starts to slide, fs is maximum.

fs max  s N

where
µs – coefficient of static friction
N – normal reaction force

Kinetic friction, fk
-- is the friction force exists between 2 object

when there is relative motion at the interface of
the surfaces in contact.

fk acts in the direction opposite to the direction of
motion and has a magnitude of

fk  k N

where
µk – coefficient of kinetic friction
N – normal reaction force

Values of µs and µk depends on the nature of the
surfaces in contact.
In general, µs > µk  fs > fk



Example
Consider a block on a rough surface. An external
force, F is applied on the block.
If F < fs (max) , the block won’t move.

As F increases, fs will also increase until its
reach its maximum value.

When F = fs (max) , the block start to move which is
called the point of slipping.

Once an object starts to move, the force of
friction change to kinetic friction ( fk ).

The plot of the frictional force vs. the applied force

Box begin to move
when F applied = fs(max)

As applied force
increases, fs
increases to
keep box at rest

Once box is in
motion, fs is
replaced by fk

4.2 Newton’s Law of Motion

Newton’s 1st Law of Motion
States:

If there is no net force acting on an object:
• if it is at rest, it will stay at rest
• if it is moving, it keeps on moving at a

constant velocity.

-- also known as the law of inertia.

-- ‘no net force’ means
1. no force acts on object, or
2. the summation of all forces acting on the

object is equal to zero.
F  0

Inertia
-- is the tendency of an object to resist any

changes in its state of rest or motion.

-- Inertia depends on mass.

-- A bigger mass needs a bigger force to overcome
its inertia and change its motion.

Easy to start moving
Hard to start moving

Mass, m

- is an inherent property of a body.

- is the amount of matter you
contain.

- The value of mass is independent
of location.

- Scalar quantity ; SI unit is kilogram (kg ).

Newton’s 2nd Law of Motion

States:
the rate of change of momentum of an
object is directly proportional to the net
force acting on it.

According to this law:

  dp
Fnet
dt

 d (mv)
dt

  v dm 0 dv =a
Fnet
m
dt dt

If the mass is constant, Newton’s 2nd law becomes:
Fnet  ma

Force is directly proportional to acceleration.

The greater the net force acting, the greater the
acceleration of an object.

The acceleration is in the same direction as the
net force.

One newton (1 N) is the force which produces a
linear acceleration of 1 ms–2 in the direction of
the force when it acts on a body of mass 1 kg .

The net force is the vector sum of all forces acting
on the object or system.

If no net force acts on an object,
Fnet  0  a  0m s1

This links back to Newton’s 1st law.
With no net force the object must
continue at the same velocity or
remain at rest.

Example

A force of 15 N is applied at an angle of 30° to
the horizontal on a 0.750 kg block at rest on a
frictionless surface as shown in figure.

F a
30°

a) What is the magnitude of the resulting
acceleration of the block?

b) What is the magnitude of the normal force?

Solution

Step 1 : Sketch an appropriate free body diagram
Step 2 : List out the information given
Step 3 : Choose related equation
Step 4 : Substitute correct value and calculate
Step 5 : Write final answer with correct unit

Solution

N

Fx Fy

30°

F

W  mg

Note: the forces act in same directions as
in space diagram

Given,
F = 15 N, m = 0.7250 kg, θ = 30°, a = ?,N=?

a)From body diagram,

Fx = m a
F cos 30° = m a
(15)(0.866)= (0.75) a

a = 17.3 m s-2

b) The forces in y direction with upward
as positive,

N – Fy – W = 0
N – F sin 30 ° – mg = 0

N= 14.75 N

Example

Two blocks with masses of m1 = 2 kg, m2 = 5 kg are
connected by a string that passes over a
frictionless pulley as shown in figure below.

The coefficient of friction between the table and
m1 is 0.03. Find the acceleration of the system.

Solution

a

fk

a

For m1 : F  ma

T  fk  m1a (1)

T  kN  m1a
T  k(m1g)  m1a

T  2a  0.5886

For m2 : F  ma

m2g T  m2a

T  49.05  5a (2)

Equating (1) & (2) we get : a  6.92 m s2