Machine Kinematics Questions and Answers – Limiting Friction

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

1. What is the direction in which the limiting friction acts?
a) Opposite to the direction in which the body tends to move
b) Opposite to the direction in which the body moves
c) In the direction in which the body tends to move
d) In the direction in which the body moves
View Answer

Answer: a
Explanation: Limiting friction acts in the opposite direction in which the body tends to move. It has a maximum value beyond which it cannot increase.
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2. The force of friction is independent of the roughness of the surface.
a) True
b) False
View Answer

Answer: b
Explanation: The force of friction is dependent on the roughness of the surface. There is a coefficient of friction on which the frictional force depends.

3. If the area of contact between the two surfaces is increased by two times, what will be the effect of it on the force of friction?
a) Increases by two times
b) Decreases by two times
c) Remains same
d) Depends on the surface
View Answer

Answer: c
Explanation: If the area of contact between the two surfaces is increased by two times then there will be no effect on the force of friction as it is independent of the area of contact surface.

4. The ratio of frictional force to normal reaction is known as _______
a) Coefficient of friction
b) Coefficient of restitution
c) Coefficient of inertia
d) Coefficient of force
View Answer

Answer: a
Explanation: The ratio of frictional force to the normal reaction at the surface of contact bears a constant ratio, this ratio is known as the coefficient of friction.

5. Which of the following is true regarding the limiting friction?
a) Its value is equal to the force applied in the opposite direction which tends to move the body
b) Its value is equal to the force applied in the opposite direction which moves the body
c) Its value is equal to μ.N
d) It is equal to the normal reaction
View Answer

Answer: a
Explanation: For the limiting friction, the maximum value it can attain is μ.N, until then it has the same value as the force applied in the opposite direction which tends to move the body.
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6. The angle at which the body just begins to slide down an incline is known as ________
a) Angle of inclination
b) Angle of motion
c) Angle of repose
d) Angle of stability
View Answer

Answer: c
Explanation: When a body is kept on an incline, upto a certain angle it remains at rest but at a specific angle the body begins to slide down if the angle is increased further, this angle is known as the angle of repose.

7. For a body on an inclined plane, the value of coefficient of friction is equal to _____ (θ is the angle of friction)
a) Tanθ
b) Cotθ
c) Sinθ
d) Cosθ
View Answer

Answer: a
Explanation: For a body on an inclined plane, the coefficient of friction can also be expressed in the form of the inclination as μ = Tanθ.

8. Static frictional force is equal to the coefficient of friction if the normal reaction is unity.
a) True
b) False
View Answer

Answer: b
Explanation: The above statement talks about static frictional force, this force is always equal to the force which is applied on the body and tends to move away.

9. What is the maximum value of static friction?
a) Limiting friction
b) 0
c) Rolling friction
d) Kinetic friction
View Answer

Answer: a
Explanation: When the applied force is less than the limiting friction, the body remains at rest, and the friction into play is called static friction which may have any value between zero and limiting friction.
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10. A body of mass 1 kg is kept on a rough surface having coefficient of friction = 0.25 being pulled by a force of 2N, what will be the value of friction force in N?
a) 2
b) 2.5
c) 0
d) 4
View Answer

Answer: a
Explanation: Since the applied force of 2N has a value less than the limiting friction 0.25×9.81
The acting friction will be static having a value of 2N.

11. A body of mass 1 kg is kept on a rough surface having coefficient of friction = 0.25 being pulled by a force of 2N, which frictional force will be acting on the body?
a) Limiting friction
b) Static friction
c) Kinetic friction
d) Rolling friction
View Answer

Answer: b
Explanation: Since the applied force of 2N has a value less than the limiting friction 0.25×9.81
The acting friction will be static friction and the body will stay at rest.

12. A body of mass 1 kg is kept on a rough surface having coefficient of friction = 0.25 being pulled by a force of 2N, after how long will the body come to rest again?
a) 2s
b) 2.5s
c) 0s
d) 4s
View Answer

Answer: c
Explanation: Since the applied force of 2N has a value less than the limiting friction 0.25×9.81
The acting friction will be static friction and the body will stay at rest and will not undergo any motion until the applied force exceeds the value of limiting friction.

13. A body, resting on a rough horizontal plane required a pull of 181 N inclined at 30° to the plane just to move it. It was recorded that a push of 220 N inclined at 30° to the plane just moved the body. Estimate the weight of the body in N.
a) 2000N
b) 2500N
c) 1000N
d) 4000N
View Answer

Answer: c
Explanation: For vertical equilibrium we have reaction R = W – 181sin30
horizontally,
f = 181cos30
Now considering a push of 220N
horizontally
f =220cos30
vertically
R = W + 220sin30
solving the equations we get
W = 1000N.
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14. A body, resting on a rough horizontal plane required a pull of 181 N inclined at 30° to the plane just to move it. It was recorded that a push of 220 N inclined at 30°° to the plane just moved the body. Estimate the coefficient of friction.
a) 0.17
b) 0.2
c) 0.5
d) 0.75
View Answer

Answer: a
Explanation: For vertical equilibrium we have reaction R = W – 181sin30
horizontally,
f = 181cos30
Now considering a push of 220N
horizontally
f = 220cos30
vertically
R = W + 220sin30
solving the equations we get
μ = 0.171.

15. A body of mass 1 kg is kept on a rough surface is being pulled by a force of 2N, what will be the value of coefficient of friction if it just begins to move. Take g = 10m/s2.
a) 0.2
b) 0.25
c) 0
d) 0.4
View Answer

Answer: a
Explanation: The acting friction will be limiting having a value of 2N.
Since the body just begins to move, we have
f = μN = 2N
therefore
μx10 = 2N
μ = 0.2.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn