a fixed pulley is driven by a 100 kg mass falling at rate of 8.0m in 4.0sec.It lifts a load of 500kgf.Calculate the power input to the pulley taking the force due to gravity on 1 kg as 10 N.if the efficiency of pulley is 75% find the to which load raised in 4.0 sec.
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Answered by
5
The efficiency of a pulley will be n = output / input = load x height / effort x distance
or
height = n x effort x distance / load
Here
n = 0.75
load = 500 kg x 10 m/s2 = 5000 N
distance = 8 m
effort = 100kg x 10 m/s2 = 1000 N
thus, the height moved will be
h = 0.75 x 1000 x 8 / 5000
or
h = 6000/5000 = 1.2m
or
height = n x effort x distance / load
Here
n = 0.75
load = 500 kg x 10 m/s2 = 5000 N
distance = 8 m
effort = 100kg x 10 m/s2 = 1000 N
thus, the height moved will be
h = 0.75 x 1000 x 8 / 5000
or
h = 6000/5000 = 1.2m
Answered by
7
Answer:
Effort=E=100kg=100×10=1000N
d=8m, t=4sec
Load=L=75kgf=75×10=750N
(a) Power = E×d/t= 1000×8/4=2000W
(b)MA=L/E =750/1000 = 75/100 = 0.75
VR = d/d =8/8 = 1
Efficiency =MA/VR =0.75/1 = 0.75
(c) Effort goes down in 4 sec = 8 m
Hence the load raised in 4 sec = 8 m
Explanation:
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