Question

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.

Answer 1

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

Answer 2

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: