Note Chap 4 SP015 (Forces) Nun

Chapter 04 Physics

CHAPTER 4:
Forces

(F2F:4 Hours)

1

Overview: Chapter 04 Physics

Forces

Basic forces and free Newton’s law of
body diagram motion

Identification Free body Static and kinetic

of forces diagram frictions

Newton’s Newton’s Newton’s Applications of Applications of
1st law 2nd law 3rd law Newton’s 1st law Newton’s laws

for equilibrium for linear motion
of particle

2

Chapter 04 Physics

4.0 Force

 is something capable of changing state of motion or size
or dimension of a body .

 There are four types of fundamental forces in nature:
⚫ Gravitational forces
⚫ Electromagnetic forces
⚫ Strong nuclear forces
⚫ Weak nuclear forces

 is a vector quantity.
 S.I. unit of force, F:

F = ma

( )unit of F : (kg) m s−2

: kg m s−2 OR newton (N)

3

Chapter 04 Physics

Learning Outcome:

4.1 Basic of forces and free body diagram (1 hour)
At the end of this chapter, students should be able to:
 Identify the forces acting on a body in different

situations:
⚫ Weight
⚫ Tension
⚫ Normal Force
⚫ Frictional force
⚫ External force (pull or push)
 Determine static friction and kinetic friction.
 Sketch free body diagram.

4

Chapter 04 Physics

4.1 Basic of forces and free body diagram

4.1.1 Basic of forces

At the end of this lesson, students should be able to:

✓ Identify the forces acting on a body in different situations:
Weight, tension, normal force, frictional force and external

force (pull or push).
Weight, W

 is a force exerted on a body under gravitational field.

 a vector quantity.

 dependant on where it is measured, because the value of g
varies at different localities on the earth’s surface.

 always directed toward the centre of the earth or same
direction of acceleration due to gravity, g.

 S.I. unit : kg m s-2 or new ton (N).
 Equation: W mg
=
5

Chapter 04 Physics

Case 1: Horizontal surface
 An object lies at rest on a flat horizontal surface.

Figure 4.6

Case 2 : Inclined plane W = mg

 An object lies at rest on a rough inclined.

Figure 4.7 

W = mg 6

Chapter 04 Physics

Case 3: Hanging object

 An object is hang by using a light string.

Figure 4.8

W = mg


Tension in the string/rope/cabel , T

 is a pulling force that is directed away from the object and
attempts to stretch or elongate the object.

 a vector quantity.
 S.I. unit : kg m s-2 or newton (N).

7

Chapter 04 Physics

Case 1: Horizontal surface
 A box of mass m is pulled by using a light string along a

horizontal surface.

T

Case 2 : Pulley Figure 4.9

T T
T T 1 kg
Figure 4.10 8
4 kg

Figure 4.11

Chapter 04 Physics

Normal force (reaction force), N or R

 is a reaction force that exerted by the surface to an object
interact with it and the direction always perpendicular to
the surface.

 a vector quantity.
 S.I. unit: kg m s-2 or newton (N).
Case 1: Horizontal surface

N N

Figure 4.12 Figure 4.13

9

Case 2: Inclined plane Chapter 04 Physics

N



Figure 4.14

Case 3: Wall

N 10

Figure 4.15

Frictional force, f Chapter 04 Physics

At the end of this lesson, students should be able to:

✓ Determine static friction and kinetic friction.

 is a force that resists the motion of one surface relative to
another with which it is in contact.

 is independent of the area of contact between the two
surfaces.

 is directly proportional to the reaction force.

OR f  N

f = N where f : frictional force

μ : coefficient of friction

N : reaction force

 The direction of the frictional force is always in the opposite
direction of the motion.

 The frictional and the reaction forces are always 11
perpendicular.

 Coefficient of friction,  Chapter 04 Physics

⚫ is a ratio between frictional force to reaction force.

OR  = f

N

⚫ is a quantity without unit and depends on the nature of the
surfaces.

 There are two types of frictional force :

⚫ Static, fs (frictional force act on the object before its move)
⚫ Kinetic, fk (frictional force act on the object when its move)

fs = sN where fk  fs
fk = kN
thus k  s

12

Chapter 04 Physics

Case 1: Horizontal surface

 A box of mass m is pulled along a rough horizontal surface by a

horizontal force, F.  
a

F

f

Figure 4.16

Case 2: Inclined plane

 A box of mass m is pulled up along a rough inclined plane by a
force, F.  F

a

f  Figure 4.17 13

Chapter 04 Physics

Push or Pull (External Force), F OR Fext

 Another force which may act on an object could be any physical

push or pull.

Case 1: Push

 object
i) F

Figure 4.18 (a)

ii) 
F

object

Figure 4.18 (b)

14

Chapter 04 Physics

Case 2: Pull 
F
i)
object

Figure 4.19 (a)


F

ii)
object

Figure 4.19 (b)

Note:
External force: For a force to accelerate an object it must
come from outside it.

15

Chapter 04 Physics

4.1.2 Free body diagram

At the end of this lesson, students should be able to:

✓ Sketch free body diagram.

 is a diagram showing the chosen body by itself, with
vectors drawn to show the magnitude and directions of all
the forces applied to the body by the other bodies that
interact with it.

Example 4.1 :
Sketch free body diagrams for each case.
Case 1: Horizontal surface
a. An object lies at rest on a flat horizontal surface.

N

m

Figure 4.18

W = mg 16

Chapter 04 Physics

b. A box is pulled along a rough horizontal surface by a horizontal

force, F.  N
a a
m
F fF

Figure 4.19

Case 2: Inclined plane W = mg

A box is pulled up along a rough inclined plane by a force, F.
a aF
F N

f

 Figure 4.20 W = mg
17

Chapter 04 Physics

Case 3: Hanging object using the cable/string/rope
a.

T

m

Figure 4.21 W = mg

b. T2

m T1

Figure 4.22

W = mg 18

Chapter 04 Physics

Case 4: Pulley m1: m2:
a.
T T

m1 a a
m2 Figure 4.23 m1 g m2 g

b. m1: m2:
m1
aT
smooth NT a

m2 m1g m2 g

Figure 4.24 19

Chapter 04 Physics

Learning Outcome:

4.2 Newton’s laws of motion (3 hours)
At the end of this chapter, students should be able to:
 State Newton’s laws of motion.
 Apply Newton’s laws of motion.

20

Chapter 04 Physics
4.2 Newton’s laws of motion

4.2.1 Newton’s first law of motion

At the end of this lesson, students should be able to:

✓ State Newton’s First Law.

 states “an object at rest will remain at rest, or continues to
move with uniform velocity in a straight line unless it is
acted upon by a external forces”

OR



Fnett = F = 0

Inertia

 is a tendency of an object to resist any change in its state
of rest or motion.

 a scalar quantity.

21

Chapter 04 Physics

 Examples of real experience of inertia.

Figure 4.25a
Figure 4.25b

22

Chapter 04 Physics

Mass, m

At the end of this lesson, students should be able to:

✓ Define mass as a measure of inertia.

 is a measure of a body’s inertia.
 a scalar quantity.
 S.I. unit: kilogram (kg).
 independent of location.
 If the mass of a body increases then its inertia will increase.

mass  inertia

23

Chapter 04 Physics

4.2.2 Equilibrium of a particle

At the end of this lesson, students should be able to:

✓ Define the equilibrium of a particle.

 is a vector sum of all forces acting on a particle (point)
must be zero.

 The equilibrium of a particle ensures the body in translational

equilibrium.  Newton’s first law
F = Fnett = 0 of motion
 Condition :

  Fx = 0 , Fy = 0 , Fz = 0

 Two types of equilibrium of a particle:

⚫ Static equilibrium (v=0) body remains at rest (stationary).

⚫ Dynamic equilibrium (a=0) body moving at a uniform
(constant) velocity.

24

Chapter 04 Physics

4.2.3 Polygon of forces

Case 1:

 A particle in equilibrium as a result of two forces acting on it.
F2 F1
Figure 4.26

 They are equal in magnitude but opposite in the direction.

  F = F1 − F2 = 0 i.e. Fx = 0 OR Fy = 0

Case 2: 
 A particle in equilibrium as a result of F3

three forces acting on it. 
F1
Figure 4.27
25

F2

Chapter 04 Physics

 They are form a closed triangle of forces.
F1   
 
F3  F = F1 + F2 + F3 = 0 F3

 F2 i.e. Fx = 0 and Fy = 0

Case 3:

 A particle in equilibrium as a result of  
four forces acting on it. F4 F2

 26
Figure 4.28 F1

 They will form a closed polygon of forces.
   
  F = F1 + F2 + F3 + F4 = 0
 F4
F1  F3
 i.e. Fx = 0 and Fy = 0

F2

Chapter 04 Physics

4.2.4 Problem solving strategies for equilibrium of
a particle

At the end of this lesson, students should be able to:

✓ Apply Newton’s First Law in equilibrium of forces.

 Procedures to solve problem of equilibrium of a particle:
⚫ Sketch a simple diagram.
⚫ Sketch a separate free body diagram for each body.
⚫ Choose a convenient coordinate axes for each body and
construct a table to resolve the forces into their
components.
⚫ Apply the condition for equilibrium of a particle in
component form :

 Fx = 0 and Fy = 0

⚫ Solve the component equations for the unknowns.

27

Chapter 04 Physics

Example 4.2 :
A load of 250 kg is hung by a crane’s cable. The load is pulled by
a horizontal force such that the cable makes a 30 angle to the
vertical plane. If the load is in the equilibrium, calculate

a. the magnitude of the tension in the cable,

b. the magnitude of the horizontal force. (Given g =9.81 m s−2)

Solution : m = 250 kg Free body diagram of the load :

30  
F 30 T Ty 
60 F
Tx


mg

28

Chapter 04 Physics

Solution : m = 250 kg y-component (N)

1st method : − mg = −(250)(9.81)

a. Force x-component (N) = −2453
0

mg 0 T sin 60

 F
F − T cos 60
T

Since the load is in the equilibrium, then

F =0

Thus Fx = 0 F − T cos 60 = 0 (1)
(2)
 Fy = 0 T sin 60  − 2453 = 0

T = 2833 N
b. By substituting eq. (2) into eq. (1), therefore

F − (2833 )cos 60 = 0 F = 1417 N 29

Solution : m = 250 kg Chapter 04 Physics

2nd method :

a. Since the load is in the equilibrium, then a closed triangle of

forces can be sketched as shown below.

 30   mg = cos30 (250 )(9.81) = cos30
mg T T
T T = 2833 N

F F = sin 30
2833
b. F = sin 30
T F = 1417 N

30

Chapter 04 Physics

Example 4.3 : F1 = 12 N

F2 = 20 N 55.0

30.0 A

45.0

F3 = 30 N

Figure 4.29

Calculate the magnitude and direction of a force that balance the
three forces acted at point A as shown in Figure 4.29.

31

Chapter 04 Physics

Solution : F1 = 12 N ; F2 = 20 N ; F3 = 30 N

Force x-component (N) y-component (N)

 12cos 55.0 12sin 55.0
F1 = 6.88 = 9.83
 − 20cos 30.0 20sin 30.0
F2 = −17.3 = 10.0
 − 30cos 45.0 − 30sin 45.0
F3 = −21.2 = −21.2

 Fx Fy
F

To find a force to balance the three forces means the system must

be in equilibrium hence

 Fx = 0

6.88 −17.3 − 21.2 + Fx = 0 32
Fx = 31.6 N

Chapter 04 Physics

Solution :

 Fy = 0

9.83 +10.0 − 21.2 + Fy = 0
Fy = 1.37 N

The magnitude of the force,

( )F = (Fx )2 + Fy 2 = (31.6)2 + (1.37)2

F = 31.6 N

and its direction, tan −1  Fy 
Fx
θ = y

θ = tan−1 1.37   2.48
 31.6  F
x
θ = 2.48

33

Example 4.4 : Chapter 04 Physics


F

50.0

Figure 4.30

A window washer pushes his scrub brush up a vertical window at
constant speed by applying a force F as shown in Figure 4.30.
The brush weighs 10.0 N and the coefficient of kinetic friction is

k= 0.125. Calculate

a. the magnitude of the force F ,

b. the normal force exerted by the window on the brush. 34

Chapter 04 Physics

Solution : W = 10.0 N ; μ k = 0.125 y-component (N)

a. The free body diagram of the brush : F sin 50.0
− 10.0
constant  Force x-component (N) 0
F F − μk N
speed W F cos 50.0 = −0.125N
N 50.0 N 0
 fk
W  −N

fk 0

The brush moves up at cFon=smtaant=s0peed (a=0) so that

Thus Fx = 0 N = F cos 50.0 (1)

 Fy = 0 F sin 50.0 − 0.125N = 10.0 (2)

35

Chapter 04 Physics

Solution :

a. By substituting eq. (1) into eq. (2), thus

( )F sin 50.0 − 0.125 F cos 50.0 = 10.0

F = 14.6 N

b. Therefore the normal force exerted by the window on the brush

is given by N = F cos 50.0

N = (14.6)cos 50.0

N = 9.39 N

36

Chapter 04 Physics

Exercise 4.1 :
Use gravitational acceleration, g = 9.81 m s−2

1.

Figure 4.31

The system in Figure 4.31 is in equilibrium, with the string at the
centre exactly horizontal. Calculate

a. the tensions T1, T2 and T3.

b. the angle .

ANS. : 49 N, 28 N, 57 N; 29

37

Chapter 04 Physics

2.

Figure 4.32

A 20 kg ball is supported from the ceiling by a rope A. Rope B
pulls downward and to the side on the ball. If the angle of A to
the vertical is 20 and if B makes an angle of 50 to the vertical
as shown in Figure 4.32, Determine the tension in ropes A and
B.
ANS. : 134 N; 300 N

38

Chapter 04 Physics

3.

Figure 4.33

A block of mass 3.00 kg is pushed up against a wall by a force

P that makes a 50.0  angle with the horizontal as show in

Figure 4.33. The coefficient of static friction between the block
and the wall is 0.250. Determine the possible values for the

magnitude of P that allow the block to remain stationary.

ANS. : 31.8 N; 48.6 N

39

Chapter 04 Physics

4.2.5 Newton’s second law of motion

At the end of this lesson, students should be able to:

✓ State and apply Newton’s Second Law.

 states “the rate of change of linear momentum of a moving

body is proportional to the resultant force and is in the
same direction as the force acting on it”

OR 
dp
   dt
F


where F : resultant force
dp : change in linear momentum

dt : time interval

40

Chapter 04 Physics

From the NFew=todnp’sa2ndnd law of motion, it also can be written as
 d t(mv
p = mv
d
=  )  =  dm + m 
F v dv
 F
dt dt dt

Case 1:

 Object at rest or in motion with constant velocity but with

changing mass. For example : Rocket
 
 =  dm + m dv and dv = 0
F v
dt dt dt

  =  dm
F v dt

41

Chapter 04 Physics

Case 2:

 Object at rest or in motion with constant velocity and

constantmass. dm + m  where dm =0 and  =0
F =v dv dv

 dt dt dt dt

F =0 Newton’s 1st law of motion

 =  =0
dp
F dt


p = constant

42

Chapter 04 Physics

Case 3:

 Object with constant mass but changing velocity.
m dv and dm = 0
 dm +
F =v
 d t dt dt
F m dv  
a dv
 dt dt= and =

  
F = ma
where F : resultant force
m : mass of an object

a : accelerati on

 The direction of the resultant force always in the same
direction of the acceleration.

43

Chapter 04 Physics
 Newton’s 2nd law of motion restates that “The acceleration of

an object is directly proportional to the nett force acting on

it and proportional to its mass”.
F
OR
inversely


a
m

 One newton (1 N) is an amount of nett force that gives an
acceleration of one metre per second squared to a body
with a mass of one kilogramme.

OR 1 N = 1 kg m s-2

 Notes: 

⚫ F is a nett force or effective force or resultant force.
⚫ If the forces act on an object and the object moving at

euqnuifiolirbmriuamcc)ehleernacteionFn(netto=t atrest or not in the


F = ma
44

Chapter 04 Physics

4.2.6 Applications of Newton’s 2nd law of motion

 F = Fnett = ma

 Five steps to solve problems in mechanics:

⚫ Identify the object whose motion is considered.

⚫ Determine the forces exerted on the object.

⚫ Draw a free body diagram for each object.

⚫ Choose a system of coordinates so that calculations may be
simplified.

⚫ Apply the equation above for

 along x-axis: Fx = ma
x

 along y-axis: Fy = may

45

Chapter 04 Physics

Example 4.5 :

Three wooden blocks connected by a rope of negligible mass are
being dragged by a horizontal force, F in Figure 4.34.

  m3
F m1 T1 m2 T2

Figure 4.34

Suppose that F = 1000 N, m1 = 3 kg, m2 = 15 kg and m3 = 30 kg.
Determine
a. the acceleration of blocks system.
b. the tension of the rope, T1 and T2.
Neglect the friction between the floor and the wooden blocks.

46

Chapter 04 Physics

Solution :

a.Forthe block, m1 = 3 kg
a N1
  Fx = F − T1 = m1a

F 
m1 g
T1 Fx = 1000 − T1 = 3a (1)
T1 + 3a = 1000

For the block,am2N=2 15 kg − Fx = T1 − T2 = m2a
the T1  = T1 − T2 = 15a
 For  Fx = 15a
T2 T2
Fx = T2 = m3a
m2 g T1 (2)
N3
block, m3 = 30 kg
a
T2  T2 = 30a
(3)

m3 g 47

Chapter 04 Physics

Solution :
a. By substituting eq. (3) into eq. (2) thus

T1 − 45a = 0 (4)

Eq. (1)−(4) : a = 1000
48

a = 20.8 m s−2

b. By substituting the value of acceleration into equations (4) and

(3), therefore

T1 = 936 N
T2 = 624 N

48

Chapter 04 Physics

Example 4.6 :

Two objects of masses m1 = 10 kg and m2 = 15 kg are connected
by a light string which passes over a smooth pulley as shown in

Figure 4.35. Calculate

a. the acceleration of the object of mass 10 kg.

b. the tension in the each string.

(Given g = 9.81 m s−2) m1
m2
Solution :
Figure 4.35
For the object m1= 10 kg,
a. T1 (1)
 Fy = T1 − m1g = m1a
a T1 = T2 = T 49
where

T −10 g = 10a

W1 = m1g

Chapter 04 Physics

Solution : (2)

a. For the object m2= 15 kg,

  Fy = m2 g − T2 = m2a
a
T2 Fy = 15g − T = 15a

− T +15g = 15a

Eq. (1) + (2) : a = 5g = 5(9.81)
25 25

W2 = m2g a = 1.96 m s−2

b. Substitute the value of acceleration into equation (1) thus

T −10(9.81) = 10(1.96)

T = 118N

Therefore T1 = T2 = T = 118 N

50