Note Chap 4 SP015 (Forces) Nun

Chapter 04 Physics

Exercise 4.2 :

1. A block is dragged by forces, F1 and F2 of the magnitude

20 N and 30 N respectively as shown in Figure 4.36. The

frictional force f exerted on the block is 5 N. If the weight of

the block is 200 N and it is move horizontally, determine the

acceleration of the block.  
F1 F2
a(Given g = 9.81 m s−2)
50 

f 20 

Figure 4.36

ANS. : 1.77 m s−2

51

Chapter 04 Physics

2. One 3.5 kg paint bucket is hanging by a massless cord from
another 3.5 kg paint bucket, also hanging by a massless cord
as shown in Figure 4.37. If the two buckets are pulled upward
with an acceleration of 1.60 m s−2 by the upper cord, calculate
the tension in each cord.

(Given g = 9.81 m s−2)

ANS. : 39.9 N; 79.8 N

Figure 4.37

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Chapter 04 Physics

4.2.7 Newton’s third law of motion

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

✓ State and apply Newton’s Third Law.
 states “every action force has a reaction force that is equal

in magnitude but opposite in direction”.

 For example :

⚫ When the student push on the wall it will push back with the
same force.

A (hand) B (wall) 
FBA FAB = − FBA

FAB

Figure 4.38 53
Where FAB is a force by the hand on the wall (action)

FBA is a force by the wall on the hand (reaction)

Chapter 04 Physics

⚫ When a book is placed on the table.

Force by the table on the book (reaction)

Figure 4.39

Force by the book on the table (action)

⚫ If a car is accelerating forward, it is because its tyres are
pushing backward on the road and the road is pushing
forward on the tyres.

⚫ A rocket moves forward as a result of the push exerted on it
by the exhaust gases which the rocket has pushed out.

 In all cases when two bodies interact, the action and reaction
forces act on different bodies.

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Chapter 04 Physics

Case 1: Horizontal surface

 An object lies at rest on a flat horizontal surface.
N

Figure 4.40 
W = mg

Action: weight of an object is exerted on the horizontal
surface

Reaction: surface is exerted a force, N on the object .

 Fy = N − mg = 0

N = mg

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Chapter 04 Physics

Case 2 : Inclined plane

 An object lies at rest on a rough inclined plane.

 y Component of the weight :
N
x W x = mg sin θ
W y = mg cosθ

Wx Action: y-component of the object’s
weight is exerted on the inclined
  Wy surface.

 Reaction: surface is exerted a force, N on
W
the object.

Figure 4.41 = 
mg

 Fy = N −Wy = 0

N = mg cos

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Chapter 04 Physics

Case 3 : Motion of a lift

a. Lift moving upward at a uniform velocity

 Since the lift moving at a
N uniform velocity, thus

ay = 0

 Fy = 0

N − mg = 0

N = mg

Figure 4.42a  
W = mg

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Chapter 04 Physics

b. Lift moving upwards at a constant acceleration, a

  Fy = may
N
N − mg = ma

a N = m(a + g )

Figure 4.42b  
W = mg

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Chapter 04 Physics

c. Lift moving downwards at a constant acceleration, a

  Fy = may
N
mg − N = ma
a
N = m(g − a)

Figure 4.42c  
W = mg

 Caution : N is also known as apparent weight and W is true

weight.

59

Chapter 04 Physics

4.2.8 Application of Newton’s laws of motion
Case 1 : Horizontal surface

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

horizontal force, F. N
a a

F fF

Figure 4.43 W = mg

⚫ x-component : Fx = Fnett = ma

F − f = ma

⚫ y-component : Fy = 0
N = mg

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Chapter 04 Physics

Case 2 : Inclined plane

 A box of mass m is pulled along a rough inclined plane by a
force, F xF
F.    y
N a N a

f Wx

W = mg   Wy
f 
Figure 4.44
W = mg
⚫ x-component ⚫ y-component

(parallel to the inclined (perpendicular to the inclined

plane) : Fx = ma plane: Fy = 0

FF=−mWax+−mfg=simn aθ + f N − Wy =0 θ 61
N = mg cos

Example 4.7 : Chapter 04 Physics


F

30

Figure 4.45

A box of mass 20 kg is on a rough horizontal plane. The box is

pulled by a force, F which is applied at an angle of 30 above

horizontal as shown in Figure 3.28. If the coefficient of static
friction between the box and the plane is 0.3 and the box moves
at a constant speed, calculate

a. the normal reaction force,

b. the applied force F,

c. the static friction force. (Given g = 9.81 m s-2) 62

Chapter 04 Physics
 F
Solution : m = 20 kg; μ s = 0.3 N

a.  30

fs a=0

Force  y-component
mg

x-component

F F cos 30 F sin 30
fs −s N 0
N0
N
mg 0 −mg

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Chapter 04 Physics

Solution :

 Fx = 0

F cos 30 − μs N = 0
F = 0.3N
cos30 (1)

 Fy = 0

N + F sin 30 − mg = 0

N + F sin 30 = (20)(9.81)
N + F sin 30 = 196
(2)

By substituting eq. (1) into eq. (2), hence

N +  0.3N  sin 30 = 196
 cos30 N = 167
N

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Chapter 04 Physics

Solution :
b. Therefore the applied force is given by

0.3(167)

F=
cos 30

F = 57.9 N

c. The static friction force is

fs = μs N

= 0.3(167)

fs = 50.1 N

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Chapter 04 Physics

Example 4.8 : F

20 

30 

Figure 4.46

A block of mass 200 kg is pulled along an inclined plane of 30 by

a force, F = 2 kN as shown in Figure 4.46. The coefficient of

kinetic friction of the plane is 0.4. Determine

a. the normal force,

b. the nett force,

c. the acceleration of the block,

d. the time taken for the block to travel 30 m from rest. 66

Chapter 04 Physics

Solution :

m = 200 kg; F = 2000 N; μk = 0.4
 y
a. a

F x

N 20

 x-component y-component
f Force
k

F F cos 20 F sin 20

30 fk −k N 0
mg

N0 N

mg −mg sin 30 −mg cos 30

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Chapter 04 Physics

Solution :  Fy = 0

F sin 20 + N − mg cos 30 = 0

2000sin 20 + N − (200)(9.81)cos 30 = 0

N = 1015 N

b. The nett force is directed along the inclined plane surface.

Fnett = Fx

= F cos 20 − μk N − mg sin 30

= 2000 cos 20 − (0.4)(1015) − (200)(9.81)sin 30

Fnett = 492 N

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Chapter 04 Physics

Solution :

c. Fnett = ma

492 = 200a
a = 2.46 m s−2

d. Given s = 30 m ; u = 0
s = ut + 1 at2
2

30 = 0 + 1 (2.46)t2

2
t = 4.94 s

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Chapter 04 Physics

Example 4.9 :

Two blocks, A of mass 10 kg and B of mass 30 kg, are side by
side and in contact with each another. They are pushed along a

smooth floor under the action of a constant force F of magnitude

200 N applied to A as shown in Figure 4.47. Determine

a. the acceleration of the blocks,  AB
b. the force exerted by A on B. F

Figure 4.47

Solution : mA = 10 kg; mB = 30 kg; F = 200 N 70

a. Fx = (mA + mB )a

F = (mA + mB )a

200 = (10 + 30)a

a = 5.0 m s−2

Chapter 04 Physics

Solution :  Fx = F − FBA = mAa

b. For the object A,   200 − FBA = 10(5.0)
NA a
FBA = 150 N

F  FBA

mA g

From the Newton’s 3rd law, thus F A B = FB A = 150 N

OR

For the object B,   Fx = FAB = mBa

NB FAB = 30(5.0)
a
FAB  FAB = 150 N
mB g

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Chapter 04 Physics

Exercise 4.3 :

1.

Figure 4.48

A 5.00 kg object placed on a frictionless horizontal table is
connected to a string that passes over a pulley and then is
fastened to a hanging 9.00 kg object as in Figure 4.48.

a. Sketch free body diagrams of both objects,

b. Calculate the acceleration of the two objects and the

tension in the string.

(Given g = 9.81 m s−2)

ANS. : 6.30 m s−2; 31.5 N 72

Chapter 04 Physics

2. Two object are connected by a light Figure 4.49
string that passes over a frictionless
pulley as in Figure 4.49. 73

The coefficient of kinetic friction of

the plane is 0.3 and m1 = 2.00 kg,
m2 = 6.00 kg and  = 55.

a. Sketch free body diagrams of

both objects.

b. Determine

i. the accelerations of the objects,

ii. the tension in the string

iii. the speed of each object 2.00 s
after being released from rest.

(Given g = 9.81 m s−2)

ANS. : 2.31 m s−2; 24.2 N; 4.62 m s−1

Chapter 04 Physics

3. A 5.00 g bullet is fired horizontally into a 1.20 kg wooden
block resting on a horizontal surface. The coefficient of
kinetic friction between block and surface is 0.20. The bullet
remains embedded in the block, which is observed to slide
0.230 m along the surface before stopping. Calculate the
initial speed of the bullet.

(Given g = 9.81 m s−2)

Tips : Use
⚫ Newton’s second law of motion involving
acceleration.
⚫ Principle of conservation of linear momentum.
⚫ Equation of motion for linear motion.

ANS. : 229 m s−1

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Chapter 04 Physics

4. The block shown in Figure 4.50, Figure 4.50

has mass, m =7.0 kg and lies on 75

a smooth frictionless plane tilted

at an angle,  = 22.0 to the

horizontal.
a. Determine the acceleration of

the block as it slides down the
plane.
b. If the block starts from rest
12.0 m up the plane from its
base, calculate the block’s
speed when it reaches the
bottom of the incline plane.

(Given g = 9.81 m s−2)

ANS. : 3.68 m s−2; 9.40 m s−1

Chapter 04 Physics

Solution : m = 7.0 kg;  = 22.0  ;

4. y

N

xa

m g sin  76

mg cos   
mg

a. x-component : Fx = ma

mg sin  = ma
a = g sin 

= (9.81)sin 22.0

a = 3.68 m s−2

Solution : Chapter 04 Physics

b. u = 0 s = 12 m

v=?

By applying the following equation of linear motion, thus

v2 = u2 + 2as

v2 = 0 + 2(3.68)(12)

v = 9.40 m s−1

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Chapter 04 Physics

THE END.

Next Chapter…

CHAPTER 5 :
Work, Energy and Power

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