# Machine Kinematics Questions and Answers – Laws of Solid Friction and Limiting Friction

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This set of Machine Kinematics Multiple Choice Questions & Answers (MCQs) focuses on “Laws of Solid Friction and Limiting Friction”.

1. The force of friction always acts in a direction, ___________ to that in which the body tends to move.
a) same
b) opposite
c) both of the mentioned
d) none of the mentioned

Explanation: The force of friction always acts in a direction, opposite to that in which the body tends to move.

2. The magnitude of the force of friction is ____________ to the force, which tends the body to move.
a) equal
b) different
c) both of the mentioned
d) none of the mentioned

Explanation: The magnitude of the force of friction is exactly equal to the force, which tends the body to move.

3. The magnitude of the limiting friction (F ) bears a constant ratio to the normal reaction (RN) between the two surfaces.
a) True
b) False

Explanation: The magnitude of the limiting friction (F ) bears a constant ratio to the normal reaction (RRN) between the two surfaces. Mathematically
F/RRN = constant.
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4. The force of friction is _____________ of the area of contact, between the two surfaces.
a) dependent
b) independent
c) both of the mentioned
d) none of the mentioned

Explanation: The force of friction is independent of the area of contact, between the two surfaces.

5. The force of friction does not depends upon the roughness of the surfaces.
a) True
b) False

Explanation: The force of friction depends upon the roughness of the surfaces.

6. The ratio of magnitude of the kinetic friction to the normal reaction between the two surfaces is_____________ than that in case of limiting friction.
a) greater
b) less
c) equal
d) none of the mentioned

Explanation: The magnitude of the kinetic friction bears a constant ratio to the normal reaction between the two surfaces. But this ratio is slightly less than that in case of limiting friction.

7. For moderate speeds, the force of friction
a) increases
b) decreases
c) remains constant
d) none of the mentioned

Explanation: For moderate speeds, the force of friction remains constant. But it decreases slightly with the increase of speed.

8. A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find the angle made by the contact force on the body with the vertical.
a) 350
b) 360
c) 370
d) 380

Explanation: Let the contact force on the block by the surface be F which makes an angle ϴ with the vertical.
The component of F perpendicular to the contact surface is the normal force N and the component of F parallel to the surface is the friction f As the surface is horizontal, N is vertically upward. For vertical equilibrium,
N = mg = (0.400 kg) (10 m/s2) = 4.0 N.
The frictional force is f = 3.0 N.
tan ϴ = f/N = 3/4
or, ϴ = tan-1 (3/4) = 370.

9. A body of mass 400 g slides on a rough horizontal surface. If the frictional force is 3.0 N, find the magnitude of the contact force.
a) 5 N
b) 10 N
c) 15 N
d) 20 N

Explanation: Let the contact force on the block by the surface be F which makes an angle ϴ with the vertical.
The component of F perpendicular to the contact surface is the normal force N and the component of F parallel to the surface is the friction f As the surface is horizontal, N is vertically upward. For vertical equilibrium,
N = mg = (0.400 kg) (10 m/s2) = 4.0 N.
The frictional force is f = 3.0 N.
F = √N2
= √42 + 32 = 5 N.

10. A heavy box of mass 20 kg is pulled on a horizontal surface by applying a horizontal force. If the coefficient of kinetic friction between the box and the horizontal surface is 0.25, find the force of friction exerted by the horizontal surface on the box.
a) 29 N
b) 39 N
c) 49 N
d) 59 N

Explanation: As the box slides on the horizontal surface, the surface exerts kinetic friction on the box. The magnitude of the kinetic friction is
f = μN
= μmg
= 0.25 x (20 kg) x (9.8 m/s 2) = 49 N.
This force acts in the direction opposite to the pull.

11. A boy (30 kg) sitting on his horse whips it. The horse speeds up at an average acceleration of 2.0 m/s 2.If the boy does not slide back, what is the force of friction exerted by the horse on the boy ?
a) 20 N
b) 30 N
c) 40 N
d) 60 N

Explanation: The forces acting on the boy are
(i) the weight Mg.
(ii) the normal contact force N and
(iii) the static friction fs

As the boy does not slide back, its acceleration a is equal to the acceleration of the horse. As friction is the only horizontal force, it must act along the acceleration and its magnitude is given by Newton’s second law
fs = Ma = (30 kg) (2.0 m/s2) = 60 N.

12. A boy (30 kg) sitting on his horse whips it. The horse speeds up at an average acceleration of 2.0 m/s 2.If the boy slides back during the acceleration, what can be said about the coefficient of static friction between the horse and the boy.
a) 0.10
b) 0.20
c) 0.30
d) 0.40

Explanation: If the boy slides back, the horse could not exert a
friction of 60 N on the boy. The maximum force of static
friction that the horse may exert on the boy is
fs = μs = μsMg

μs(30 kg) (10m/s2) = μs 300 N
where μs is the coefficient of static friction. Thus,
μs(300 N) <60 N
or, μs <60/300 = 0.20.

13. A wooden block is kept on a polished wooden plank and the inclination of the plank is gradually increased. It is found that the block starts slipping when the plank makes an angle of 18° with the horizontal. However, once started the block can continue with uniform speed if the inclination is reduced to 15°. Find the coefficient of static friction between the block and the plank.
a) tan 180
b) tan 150
c) tan 330
d) tan 30

Explanation: The coefficient of static friction is
μs = tan 180.

14. A wooden block is kept on a polished wooden plank and the inclination of the plank is gradually increased. It is found that the block starts slipping when the plank makes an angle of 18° with the horizontal. However, once started the block can continue with uniform speed if the inclination is reduced to 15°. Find the coefficient of kinetic friction between the block and the plank.
a) tan 180
b) tan 150
c) tan 330
d) tan 30

Explanation: The coefficient of kinetic friction is
μk = tan 150.

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