Engineering Questions with Answers - Multiple Choice Questions

# Machine Kinematics Questions and Answers – Relative Velocity of Two Bodies Moving in Straight Lines

1 - Question

A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate mechanical advantage.
a) 6
b) 7
c) 8
d) 9

Explanation: Given, load raised, W = 360 N
Effort applied, P = 60 N
Distance moved by the effort, y = 1.8 m
Distance moved by the load, x = 200 mm = 0.2 m
Mechanical advantage, M.A. = W/P = 360/60 = 6.

2 - Question

A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate velocity ratio.
a) 6
b) 7
c) 8
d) 9

Explanation: Given, load raised, W = 360 N
Effort applied, P = 60 N
Distance moved by the effort, y = 1.8 m
Distance moved by the load, x = 200 mm = 0.2 m
Velocity ratio, V.R. = y/x = 1.8/0.2 = 9.

3 - Question

A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate efficiency at this load.
a) 44.44%
b) 55.55%
c) 66.66%
d) 77.77%

Explanation: Given, load raised, W = 360 N
Effort applied, P = 60 N
Distance moved by the effort, y = 1.8 m
Distance moved by the load, x = 200 mm = 0.2 m
Mechanical advantage, M.A. = W/P = 360/60 = 6
Velocity ratio, V.R. = y/x = 1.8/0.2 = 9
Efficiency at the load, ȵ = M.A./V.R. x 100 = 6/9 x 100 = 66.66%.

4 - Question

A machine raised a load of 360 N through a distance of 200 mm. The effort, a force of 60 N moved 1.8 m during the process. Calculate effect of friction.
a) 10 N
b) 20 N
c) 30 N
d) 40 N

Explanation: Given, load raised, W = 360 N
Effort applied, P = 60 N
Distance moved by the effort, y = 1.8 m
Distance moved by the load, x = 200 mm = 0.2 m
Mechanical advantage, M.A. = W/P = 360/60 = 6
Velocity ratio, V.R. = y/x = 1.8/0.2 = 9
Effort lost in friction, FP = P – W/V.R. = 60 – 360/9 = 20 N.

5 - Question

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 law of the machine.
a) P = 1/10W +30
b) P = 1/20W +30
c) P = 1/30W +30
d) P = 1/40W +30

Explanation: Let the law of machine be P = mW + C
where P = effort applied, W = load lifted and m and C being constants.
when W = 200 N P = 50 N
when W = 300 N P = 60 N
Putting these values in the law of machine.
50 = 200m + C …………(i)
60 = 300m + C …………(ii)
Subtracting (i) and (ii), we get
1 = 10 m
or, m = 1/10
Putting this value in equation (i), we get
50 = 200 x 1/10 + C
C = 30

Hence, the machine follows the laws
P = 1/10W +30.

6 - Question

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 efficiency to load of 200 N.
a) 10 %
b) 15 %
c) 20 %
d) 25 %

Explanation: When W = 200 N, P = 50 N
M.A. = W/P = 200/50 = 4
V.R. = 20
Efficiency at this load ȵ = M.A./V.R. x 100 = 4/20 x 100 = 20 %.

7 - Question

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 efficiency to load of 300 N.
a) 10 %
b) 15 %
c) 20 %
d) 25 %

Explanation: When W = 300 N, P = 60 N
M.A. = W/P = 300/60 = 5
V.R. = 20
Efficiency at this load ȵ = M.A./V.R. x 100 = 5/20 x 100 = 25 %.

8 - Question

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