Category: Machine Kinematics Objective Questions

Machine Kinematics Multiple Choice Questions on “Advantages and Disadvantages of Chain Drive Over Belt or Rope Drive”.

1. The advantages of the V-belt drive over flat belt drive are
a) The V-belt drive gives compactness due to the small distance between the centres of pulleys.
b) The drive is positive, because the slip between the belt and the pulley groove is negligible.
c) Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth.
d) all of the mentioned
Answer: d
Clarification: Following are the advantages of the V-belt drive over flat belt drive:
a) The V-belt drive gives compactness due to the small distance between the centres of pulleys.
b) The drive is positive, because the slip between the belt and the pulley groove is negligible.
c) Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth.
d) It provides longer life, 3 to 5 years.
e) It can be easily installed and removed.
f) The operation of the belt and pulley is quiet.
g) The belts have the ability to cushion the shock when machines are started.
h) The high velocity ratio (maximum 10) may be obtained.
i) The wedging action of the belt in the groove gives high value of limiting ratio of tensions.

2. The disadvantages of the V-belt drive over flat belt drive are
a) The V-belt drive cannot be used with large centre distances.
b) The V-belts are not so durable as flat belts.
c) The construction of pulleys for V-belts is more complicated than pulleys for flat belts.
d) all of the mentioned
Answer: d
Clarification: Following are the disadvantages of the V-belt drive over flat belt drive :
a) The V-belt drive cannot be used with large centre distances.
b) The V-belts are not so durable as flat belts.
c) The construction of pulleys for V-belts is more complicated than pulleys for flat belts.
d) Since the V-belts are subjected to certain amount of creep, therefore these are not suitable for constant speed application such as synchronous machines, and timing devices.
e) The belt life is greatly influenced with temperature changes, improper belt tension and mismatching of belt lengths.
f) The centrifugal tension prevents the use of V-belts at speeds below 5 m/s and above 50m/s.

3. The advantages of the V-belt drive over flat belt drive are
a) It provides longer life, 3 to 5 years.
b) It can be easily installed and removed.
c) The operation of the belt and pulley is quiet.
d) all of the mentioned
Answer: d
Clarification: Following are the advantages of the V-belt drive over flat belt drive:
a) The V-belt drive gives compactness due to the small distance between the centres of pulleys.
b) The drive is positive, because the slip between the belt and the pulley groove is negligible.
c) Since the V-belts are made endless and there is no joint trouble, therefore the drive is smooth.
d) It provides longer life, 3 to 5 years.
e) It can be easily installed and removed.
f) The operation of the belt and pulley is quiet.
g) The belts have the ability to cushion the shock when machines are started.
h) The high velocity ratio (maximum 10) may be obtained.
i) The wedging action of the belt in the groove gives high value of limiting ratio of tensions.

4. The disadvantages of the V-belt drive over flat belt drive are
a) Since the V-belts are subjected to certain amount of creep, therefore these are not suitable for constant speed application such as synchronous machines, and timing devices.
b) The belt life is greatly influenced with temperature changes, improper belt tension and mismatching of belt lengths.
c) The centrifugal tension prevents the use of V-belts at speeds below 5 m/s and above 50m/s.
d) all of the mentioned
Answer: d
Clarification: Following are the disadvantages of the V-belt drive over flat belt drive :
a) The V-belt drive cannot be used with large centre distances.
b) The V-belts are not so durable as flat belts.
c) The construction of pulleys for V-belts is more complicated than pulleys for flat belts.
d) Since the V-belts are subjected to certain amount of creep, therefore these are not suitable for constant speed application such as synchronous machines, and timing devices.
e) The belt life is greatly influenced with temperature changes, improper belt tension and mismatching of belt lengths.
f) The centrifugal tension prevents the use of V-belts at speeds below 5 m/s and above 50m/s.

5. The distance between the hinge centre of a link and the corresponding hinge centre of the adjacent link is called
a) pitch of the chain
b) bush roller chain
c) block chain
d) none of the mentioned
Answer: a
Clarification: A bush roller chain, consists of outer plates or pin link plates, inner plates or roller link plates, pins, bushes and rollers.
The distance between the hinge centre of a link and the corresponding hinge centre of the adjacent link is called pitch of the chain.

6. Industrial rotors will not have uniform diameter throughout their lengths.
a) True
b) False
Answer: a
Clarification: Industrial rotors are not uniform, because of fitment of pulleys, gears, belt etc.

7. For cutting multi-start threads, the speed ratio is expressed in terms of the lead of the job thread and lead of the lead screw threads.
a) True
b) False
Answer: a
Clarification: During multi-start thread cutting operation, speed is reduced to one-third to one-fourth of that in turning operation.

8. Which one of the following is a positive drive?
a) Crossed flat belt drive
b) Rope drive
c) V-belt drive
d) Chain drive
Answer: d
Clarification: The chains are mostly used to transmit motion and power from one shaft to another, when the centre distance between their shafts is short such as in bicycles, motor cycles, agricultural machinery, conveyors, rolling mills, road rollers etc.

9. The chain drive transmits ____________ power as compared to belt drive.
a) more
b) less
c) equal
d) none of the mentioned
Answer: a
Clarification: Following are the advantages and disadvantages of chain drive over belt or rope drive:
1. As no slip takes place during chain drive, hence perfect velocity ratio is obtained.
2. Since the chains are made of metal, therefore they occupy less space in width than a belt or rope drive.
3. It may be used for both long as well as short distances.
4. It gives a high transmission efficiency (upto 98 percent).
5. It gives less load on the shafts.
6. It has the ability to transmit motion to several shafts by one chain only.
7. It transmits more power than belts.
8. It permits high speed ratio of 8 to 10 in one step.
9. It can be operated under adverse temperature and atmospheric conditions.

10. The relation between the pitch of the chain (p) and pitch circle diameter of the sprocket (D) is given by
a) p = D sin(90⁰/T)
b) p = D sin(120⁰/T)
c) p = D sin(180⁰/T)
d) p = D sin(360⁰/T)
Answer: c
Clarification: None.

11. In order to have smooth operation, the minimum number of teeth on the smaller sprocket, for moderate speeds, should be
a) 15
b) 17
c) 21
d) 25
Answer: b
Clarification: In order to have smooth operation, the minimum number of teeth on the smaller sprocket or pinion may be taken as 17 for moderate speeds and 21 for high speeds.

12. The speed of the sprocket reduces as the chain pitch _____________ for a given number of teeth.
a) increases
b) decreases
c) remains same
d) none of the mentioned
Answer: a
Clarification: The r.p.m. of the sprocket reduces as the chain pitch increases for a given number of teeth.

13. The greater angle of articulation will increase the life of the chain.
a) True
b) False
Answer: b
Clarification: Greater angle of articulation will lead to breaking of chain & reduction in life of the chain.

Tricky Machine Kinematics Questions and Answers on “Differential Equation of Simple Harmonic Motion”.

1. By how much angle in degrees does the velocity leads the displacement in a body undergoing SIMPLE HARMONIC MOTION?
a) 90
b) 45
c) 180
d) 0
Answer: a
Clarification: For a body undergoing Simple Harmonic Motion, the velocity leads the displacement by an angle of 90 degrees as shown by the differential equation of the motion.

2. For a body undergoing SIMPLE HARMONIC MOTION, the acceleration is always in the direction of the displacement.
a) True
b) False
Answer: b
Clarification: For a body undergoing SIMPLE HARMONIC MOTION, the acceleration is always in the opposite direction of the displacement. This is indicated by a negative sign in the equation.

3. The maximum displacement of the body under Simple Harmonic Motion from its mean position is known as_______
a) Amplitude
b) Frequency
c) Time period
d) Range
Answer: a
Clarification: The maximum displacement of a body undergoing Simple Harmonic Motion is known as Amplitude, it is generally denoted by ‘A’.

4. The piston of an engine moves with SIMPLE HARMONIC MOTION. The crank rotates at a speed of 120 r.p.m. with a stroke of 2 metres. Find the velocity of the piston in m/s, when it is at a distance of 0.75 metre from the centre.
a) 8.31
b) 7.33
c) 8.41
d) 9.02
Answer: a
Clarification: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
(v = omega sqrt{(A^2-x^2)})
substituting the values we get
v = 8.31 m/s.

5. The piston of an engine moves with SIMPLE HARMONIC MOTION. The crank rotates at a speed of 120 r.p.m. with a stroke of 2 metres. Find the velocity of the piston in m/s², when it is at a distance of 0.75 metre from the centre.
a) 118.46
b) 117.33
c) 128.41
d) 119.02
Answer: a
Clarification: When a body is undergoing SIMPLE HARMONIC MOTION, its acceleration is given by the equation
a=ω²x
substituting the values we get
a = 118.46 m/s²

6. A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its angular velocity in rad/s.
a) 5.7
b) 7.5
c) 6.7
d) 7.6
Answer: a
Clarification: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
v = (omega sqrt{r^2 – x^2})
Given values:
when point (x) = 0.75 m, then its velocity (v) = 11 m/s
when point (x) = 2 m, then its velocity (v) = 3 m/s
When point is 0.75 m away from the mid path (v) is,
v = (omega sqrt{r^2 – x^2})
11 = (omega sqrt{r^2 – (0.75)^2}) —>Eq(1)
Similarly, When point is 2 m away from the centre (v) is,
v = (omega sqrt{r^2 – x^2})
3 = (omega sqrt{r^2 – 2^2}) —>Eq(2)
Solving Eq(1) & Eq(2) we get,
(frac{11}{3} frac{omega sqrt{r^2 – (0.75)^2}}{omega sqrt{r^2 – 2^2}} = frac{sqrt{r^2 – (0.75)^2}}{sqrt{r^2 – 2^2}} )
Squaring on both sides, we get
(frac{121}{9} = frac{r^2 – 0.5625}{r^2 – 4})
121r² – 484 = 9r² – 5.0625
121r² – 9r² = 484 – 5.0625
112r² = 478.9375
r² = (frac{478.9375}{112})
r² = 4.27622
r = 2.07m
Substituting the value of r in Eq(1) we get,
11 = ( omega sqrt{(2.07)^2 – (0.75)^2})
11 = ( omega sqrt{4.2849 – 0.5625})
11 = ( omega sqrt{3.7224})
11 = 1.9293 ω
ω = (frac{11}{1.9293}) = 5.7 rad/s.

7. A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its time period in s.
a) 1.1
b) 1.2
c) 1.3
d) 1.4
Answer: a
Clarification: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
v = (omega sqrt{r^2 – x^2})
Given values:
when point (x) = 0.75 m, then its velocity (v) = 11 m/s
when point (x) = 2 m, then its velocity (v) = 3 m/s
When point is 0.75 m away from the mid path (v) is,
v = (omega sqrt{r^2 – x^2})
11 = (omega sqrt{r^2 – (0.75)^2}) —>Eq(1)
Similarly, When point is 2 m away from the centre (v) is,
v = (omega sqrt{r^2 – x^2})
3 = (omega sqrt{r^2 – 2^2}) —>Eq(2)
Solving Eq(1) & Eq(2) we get,
(frac{11}{3} frac{omega sqrt{r^2 – (0.75)^2}}{omega sqrt{r^2 – 2^2}} = frac{sqrt{r^2 – (0.75)^2}}{sqrt{r^2 – 2^2}} )
Squaring on both sides, we get
(frac{121}{9} = frac{r^2 – 0.5625}{r^2 – 4})
121r² – 484 = 9r² – 5.0625
121r² – 9r² = 484 – 5.0625
112r² = 478.9375
r² = (frac{478.9375}{112})
r² = 4.27622
r = 2.07m
Substituting the value of r in Eq(1) we get,
11 = ( omega sqrt{(2.07)^2 – (0.75)^2})
11 = ( omega sqrt{4.2849 – 0.5625})
11 = ( omega sqrt{3.7224})
11 = 1.9293 ω
ω = (frac{11}{1.9293}) = 5.7 rad/s.
We know periodic time is,
T_p = (frac{2 pi}{omega} = frac{2 pi}{5.7} = frac{2 times 3.14}{5.7} = frac{6.28}{5.7}) = 1.1 s.

8. A point moves with SIMPLE HARMONIC MOTION. When this point is 0.75 m away from the mid path, it has a velocity of 11 m/s and when 2 m from the centre of its path its velocity is 3 m/s. Find its maximum acceleration in m/s².
a) 61.1
b) 67.2
c) 51.3
d) 41.4
Answer: b
Clarification: When a body is undergoing SIMPLE HARMONIC MOTION, its velocity is given by the equation
v = (omega sqrt{r^2 – x^2})
Given values:
when point (x) = 0.75 m, then its velocity (v) = 11 m/s
when point (x) = 2 m, then its velocity (v) = 3 m/s
When point is 0.75 m away from the mid path (v) is,
v = (omega sqrt{r^2 – x^2})
11 = (omega sqrt{r^2 – (0.75)^2}) —>Eq(1)
Similarly, When point is 2 m away from the centre (v) is,
v = (omega sqrt{r^2 – x^2})
3 = (omega sqrt{r^2 – 2^2}) —>Eq(2)
Solving Eq(1) & Eq(2) we get,
(frac{11}{3} frac{omega sqrt{r^2 – (0.75)^2}}{omega sqrt{r^2 – 2^2}} = frac{sqrt{r^2 – (0.75)^2}}{sqrt{r^2 – 2^2}} )
Squaring on both sides, we get
(frac{121}{9} = frac{r^2 – 0.5625}{r^2 – 4})
121r² – 484 = 9r² – 5.0625
121r² – 9r² = 484 – 5.0625
112r² = 478.9375
r² = (frac{478.9375}{112})
r² = 4.27622
r = 2.07m
Substituting the value of r in Eq(1) we get,
11 = ( omega sqrt{(2.07)^2 – (0.75)^2})
11 = ( omega sqrt{4.2849 – 0.5625})
11 = ( omega sqrt{3.7224})
11 = 1.9293 ω
ω = (frac{11}{1.9293}) = 5.7 rad/s.
Maximum acceleration is
A_max = ω²r = (5.7)² × 2.07 = 32.49 × 2.07
= 67.25 m/s².

9. If V is the maximum velocity of a body undergoing SIMPLE HARMONIC MOTION, then what is the average velocity of motion from one extreme to other extreme is?
a) 2V/π
b) 4V/π
c) V/2π
d) 2V/3π
Answer: a
Clarification: V = 2πA/T
V av = 2A/T÷2 = 4A/T
A/T = V/2π
Vav = 2V/π

10. If V is the maximum velocity of a body undergoing SIMPLE HARMONIC MOTION, then what is the average velocity of motion?
a) 2V/π
b) 4V/π
c) V/2π
d) 2V/3π
Answer: a
Clarification: V = Aω
= 4A/T
= 2aω/π
= 2V/π

11. Which of the following is the correct differential equation of the SIMPLE HARMONIC MOTION?
a) d²x/dt² + ω²x = 0
b) d²x/dt² – ω²x = 0
c) d²x/dt + ω²x = 0
d) d²x/dt – ω²x = 0
Answer: a
Clarification: For body undergoing Simple Harmonic Motion, it’s motion can be represented as projected uniform circular motion with radius equal to the amplitude of motion.
Therefore x =Acosωt
dx/dt = -Aωsinωt
d²x/dt² = -aω²cosωt
therefore
d²x/dt² + ω²x = 0

12. For a body undergoing Simple Harmonic Motion, the acceleration is maximum at the extreme.
a) True
b) False
Answer: a
Clarification: For a body undergoing SIMPLE HARMONIC MOTION, the acceleration is always in the opposite direction of the displacement. This is indicated by a negative sign in the equation and at an extreme position, the acceleration attains a maximum value.

13. Which of the following is the solution of the differential equation of the SIMPLE HARMONIC MOTION?
a) x = Acosωt + B sinωt
b) x = (A+B)cosωt
c) x = (A+B)sinωt
d) x = Atanωt + B sinωt
Answer: a
Clarification: We know that dx/dt = -Aωsinωt
d²x/dt² = -aω²cosωt
therefore
d²x/dt² + ω²x = 0
is the standard differential equation of Simple Harmonic Motion
It’s solution is/are:
x = Acosωt + B sinωt

To practice Tricky questions and answers on all areas of Machine Kinematics,

Machine Kinematics Multiple Choice Questions on “Number of Degrees of Freedom for Plane Mechanisms”.

1. The total number of instantaneous centres for a mechanism consisting of n links are
a) n/2
b) n
c) n-1
d) n(n-1)/2
Answer: d
Clarification: The number of pairs of links or the number of instantaneous centres is the number of combinations of n links taken two at a time. Mathematically, number of instantaneous centres, n(n-1)/2.

2. According to Kennedy’s theorem, if three bodies move relatively to each other, their instantaneous centres will lie on
a) straight line
b) parabolic curve
c) triangle
d) rectangle
Answer: a
Clarification: The Aronhold Kennedy’s theorem states that if three bodies move relatively to each other,
they have three instantaneous centres and lie on a straight line.

3. Which of the following property of the instantaneous center is correct?
a) A rigid link rotates instantaneously relative to another link at the instantaneous centre for the configuration of the mechanism considered.
b) The two rigid links have no linear velocity relative to each other at the instantaneous centre.
c) The velocity of the instantaneous centre relative to any third link is same whether the instantaneous centre is regarded as a point on the first link or on the second rigid link.
d) all of the mentioned
Answer: d
Clarification: None.

4. The magnitude of velocities of the points on a rigid link is
a) directly proportional to the distance from the points to the instantaneous centre and is parallel to the line joining the point to the instantaneous centre.
b) directly proportional to the distance from the points to the instantaneous centre and is perpendicular to the line joining the point to the instantaneous centre.
c) inversely proportional to the distance from the points to the instantaneous centre and is parallel to the line joining the point to the instantaneous centre.
d) inversely proportional to the distance from the points to the instantaneous centre and is perpendicular to the line joining the point to the instantaneous centre.
Answer: d
Clarification: None.

5. In a mechanism, the fixed instantaneous centres are those centres which
a) remain in the same place for all configurations of the mechanism
b) vary with the configuration of the mechanism
c) moves as the mechanism moves, but joints are of permanent nature
d) none of the mentioned
Answer: a
Clarification: Fixed instantaneous centres remain in the same place for all configurations of the mechanism.

6. The instantaneous centres, which moves as the mechanism moves but joints are of permanent nature, are called permanent instantaneous centres.
a) True
b) False
Answer: a
Clarification: The permanent instantaneous centres move when the mechanism moves, but the joints are of permanent nature.
Fixed instantaneous centres remain in the same place for all configurations of the mechanism.

7. The instantaneous centres which vary with the configuration of mechanism, are called
a) permanent instantaneous centres
b) fixed instantaneous centres
c) neither fixed nor permanent instantaneous centres
d) none of the mentioned
Answer: c
Clarification: Neither fixed nor permanent instantaneous centres vary with the configuration of the mechanism.

8. When two links are connected by a pin joint, their instantaneous centre lies
a) on their point of contact
b) at the centre of curvature
c) at the centre of circle
d) at the pin joint
Answer: d
Clarification: None.

9. The two links are said to have a pure rolling contact, when their instantaneous centre __________ on their point of contact.
a) lies
b) does not lie
Answer: a
Clarification: None

10. When a slider moves on a fixed link having ____________ their instantaneous center lies at infinity.
a) straight surface
b) curved surface
c) oval surface
d) none of the mentioned
Answer: a
Clarification: When a slider moves on a fixed link having curved surface, their instantaneous centre lies at the centre of curvature.
When a slider moves on a fixed link having straight surface their instantaneous center lies at infinity.

11. When a slider moves on a fixed link having curved surface, their instantaneous centre lies
a) on their point of contact
b) at the centre of curvature
c) at the centre of circle
d) at the pin joint
Answer: b
Clarification: When a slider moves on a fixed link having curved surface, their instantaneous centre lies at the centre of curvature.
When a slider moves on a fixed link having straight surface their instantaneous center lies at infinity.

12. A slider moving on a fixed link having constant radius of curvature will have its instantaneous centre at the center of the circle.
a) True
b) False
Answer: a
Clarification: None.

13. The instantaneous center of a rigid thin disc rolling on a plane rigid surface is located at
a) the centre of the disc
b) the point of contact
c) an infinite distance on the plane surface
d) the point on the circumference situated vertically opposite to the contact point
Answer: b
Clarification: None.

Machine Kinematics Multiple Choice Questions on “Motion of a Link”.

1. What is the direction of velocity of a point in a link relative to another point on the same link rotating in a specific direction.
a) Perpendicular to line joining both the links
b) Parallel to line joining both the links
c) Perpendicular to the surface of the link
d) Parallel to the surface of the link
Answer: a
Clarification: Velocity of any point on a link with respect to another point on the same link is always in the direction perpendicular to the line joining these points on the space diagram.

2. The direction of velocity is parallel if the rotation is anticlockwise and perpendicular to the line joining links if the rotation is clockwise.
a) True
b) False
Answer: b
Clarification: Velocity of any point on a link with respect to another point on the same link is always in the direction perpendicular to the line joining these points on the space diagram both during anticlockwise and clockwise rotations.

3. What is the correct representation of velocity of point A with respect to B in a link?
a) Vab
b) Vba
c) Va-b
d) Vb-a
Answer: a
Clarification: When there are two points on a link, the velocity of a point A wrt to other point on the same link is represented by Vab.

4. What is the velocity of point C with respect to A in the given figure?

a) Perpendicular to line joining BC
b) Perpendicular to line joining AC
c) Parallel to line joining BC
d) Parallel to line joining AC
Answer: b
Clarification: In the above figure we can assume an imaginary link between A and C, as per the rule, the velocity of a point with respect to any other point on the same link is perpendicular to the line joining the links.

5. Which type of pair formed by two elements which are so connected that one is constrained to turn or revolve about a fixed axis of another element?
a) Turning pair
b) Rolling pair
c) Sliding pair
d) Higher pair
Answer: a
Clarification: The type of pair formed by two elements which are so connected that one is constrained to turn or revolve about a fixed axis of another element is known as turning pair.

6. Which of the following is a lower pair?
a) Pulleys in belt drive
b) Cam and follower
c) Belt drive
d) Ball and socket joint
Answer: d
Clarification: If two moving elements do not have surface contact in motion, then the pair is known as lower pair, in the given question ball and socket joint is a lower pair.

7. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine effort loss in friction at 200 N.
a) 30 N
b) 35 N
c) 40 N
d) 45 N
Answer: d
Clarification: When W = 300 N, P = 60 N
Effort lost in friction, F_P = P – W/V.R. = 60 – 300/20 = 45 N.

8. In a lifting machine, the effort required to lift loads of 200N and 300N were 50N and 60N respectively. If the velocity ratio of the machine is 20 determine maximum efficiency which can be expected from this machine.
a) 30 %
b) 40 %
c) 50 %
d) 60 %
Answer: c
Clarification: Maximum possible efficiency of any machine = 1/m x V.R. = 1/1/10 x 20 = 0.5 = 50 %.

Machine Kinematics Multiple Choice Questions 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
Answer: b
Clarification: 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
Answer: a
Clarification: 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 (R_N) between the two surfaces.
a) True
b) False
Answer: a
Clarification: The magnitude of the limiting friction (F ) bears a constant ratio to the normal reaction (RR_N) between the two surfaces. Mathematically
F/RR_N = constant.

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
Answer: b
Clarification: 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
Answer: b
Clarification: 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
Answer: b
Clarification: 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
Answer: c
Clarification: 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
Answer: c
Clarification: 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
Answer: a
Clarification: 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
Answer: c
Clarification: 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
Answer: d
Clarification: The forces acting on the boy are
(i) the weight Mg.
(ii) the normal contact force N and
(iii) the static friction f_s

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
f_s = Ma = (30 kg) (2.0 m/s²) = 60 N.