Machine Kinematics Questions and Answers – Pulleys

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

1. In a wheel and differential axle, the diameter of the effort wheel is 400 mm. The radii of the load axles are 150mm and 100 mm respectively. The diameter of the rope is 10 mm. Find the load which can be lifted by an effort of 100 N, assuming an efficiency of the machine to be 75%.
a) 800 N
b) 725 N
c) 615 N
d) none of the mentioned
View Answer

Answer: c
Explanation: Diameter of effort wheel, D = 400 mm
Diameter of longer axle, d1 = 2 x 150 = 300 mm
Diameter of the smaller axle, d2 = 2 x 100 = 200 mm
Diameter of the rope, dr = 10 mm

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therefore, V.R. = 2(D + dr )/(d1 + dr ) – (d2 + dr )
= 2(400 + 10)/ (300 + 10) – (200 + 10)
= 820/100 = 8.2
Effort, P = 100 N
ȵ = 75%
Let W = load which can be lifted by the machine

ȵ = M.A./V.R.
0.75 = W/P x 8.2
W = 0.75 x 100 x 8.2 = 615 N.

2. Four movable pulleys are arranged as in the first system. If the weight of each pulley is 5 N, calculate the effort which can lift a load of 10 kN.
a) 629.7 N
b) 615 N
c) 625 N
d) none of the mentioned
View Answer

Answer: a
Explanation: We know that M.A. = 2nW/W + w(2n – 1)
where W = load to be lifted
w = weight of each pulley
n = no. of movable pulleys

therefore, M.A. = 24 x 10000/10000 + 5(24 – 1) = 10000/P
P = 10000 + 5(24 – 1)/ 24 = 629.7 N.

3. A person weighing 600 N platform attached to the lower block of a system of 5 pulleys arranged in the second system. The platform and the lower block weigh 100N. The man himself supports by exerting a downward pull at the free end of the rope. Neglecting friction, the minimum pull of the man will be
a) 1000 N
b) 200 N
c) 116.7 N
d) none of the mentioned
View Answer

Answer: c
Explanation: n = total number of pulleys in the system = 5
W = 600 N
Weight of the lower block and platform = 100 N
Total weight = 600 + 100 = 700 N
Let the pull exerted by the man be P newton.
Due to this pull the effective load on the lower platform will reduce to (700 – P)
nP = effective load = 700 – P
therefore, 5P = 700 -P
5P = 700
P = 116.7 N.
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4. Five pulleys are arranged in the second system of pulleys. When not loaded the effort required to raise the movable block is 35N. Further wastage in friction increases the pull at the rate of 3% of the load lifted. What is the effort required to raise a load of 2kN?
a) 500 N
b) 400 N
c) 495 N
d) none of the mentioned
View Answer

Answer: c
Explanation: n = no. of pulleys = 5
Frictional effort at zero loading = 35N
Frictional effort at 2 kN loading = 35 + 2000 x 3/100 = 95 N
When the system is considered frictionless nP = W
5P = 2000
P = 400 N
Hence total effort = 400 + 95 = 495 N.

5. Five pulleys are arranged in the second system of pulleys. When not loadwd the effort required to raise the movable block is 35N. Further wastage in friction increases the pull at the rate of 3% of the load lifted. What is the efficiency of the system at 2kN?
a) 80%
b) 80.81%
c) 80.50%
d) none of the mentioned
View Answer

Answer: b
Explanation: n = no. of pulleys = 5
Frictional effort at zero loading = 35N
Frictional effort at 2 kN loading = 35 + 2000 x 3/100 = 95 N
When the system is considered frictionless nP = W
5P = 2000
P = 400 N
Hence total effort = 400 + 95 = 495 N
Efficiency at this load = effort without friction/effort with friction
= 400/495 x 100
= 80.81%.

6. In a weston differential pulley block, the number of recesses in the smaller wheel is 9/10 of that of the larger wheel. If the efficiency of the machine is 50%, find the load lifted by an effort of 300N.
a) 2000N
b) 3000N
c) 4000N
d) none of the mentioned
View Answer

Answer: b
Explanation: Let the recesses in the larger wheel, n1 = 10
Recesses in the smaller wheel, n2 = 9/10 x 10 = 9

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V.R. = 2n1 /n1 – n2 = 2×10/10 – 9 = 20
and mechanical advantage M.A. = W/P
= W/300
efficiency = M.A./V.R.
0.5 = W/300×20
W = 3000N.

7. If the velocity ratio for an open belt drive is 8 and the speed of driving pulley is 800 r.p.m, then considering an elastic creep of 2% the speed of the driven pulley is
a) 104.04 r.p.m
b) 102.04 r.p.m
c) 100.04 r.p.m
d) 98.04 r.p.m
View Answer

Answer: d
Explanation: Velocity Ratio = Velocity of belt on driver/Velocity of belt on driven
Velocity of belt on driven = 800/8 = 100 r.p.m
Elastic creep = velocity of belt at driven pulley – Velocity of driven pulley
0.02 × Vp = [100-Vp] Vp = 100/1.02 = 98.04r.p.m.

8. If the angle of wrap on smaller pulley of diameter 250 mm is 1200 and diameter of larger pulley is twice the diameter of smaller pulley, then the centre distance between the pulleys for an open belt drive is
a) 1000 mm
b) 750 mm
c) 500 mm
d) 250 mm
View Answer

Answer: d
Explanation: sin α = (D -d)/2c
Angle of wrap on smaller pulley = п – 2α
2п/3 = п – 2sin-1(D -d)/2c
c = 250 mm.

9. In short open-belt drives, an idler pulley is used in order to decrease the angle of contact on the smaller pulley for higher power transmission.
a) True
b) False
View Answer

Answer: b
Explanation: In short open-belt drives, an idler pulley is used in order to increase the angle of contact on the smaller pulley for higher power transmission.
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10. In design of arms of a pulley, in belt drive, the cross-section of the arm is elliptical with minor axis placed along the plane of rotation.
a) True
b) False
View Answer

Answer: b
Explanation: Arms of a pulley in belt drive are subjected to complete reversal of stresses and is designed for bending in the plane of rotation.

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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