Question

A fixed pulley is driven by a 100 kg mass falling at a rate of 8 m in 4 s. It lifts a load of 75 kgf. Calculate-
1) power input to the pulley taking the force of gravity on 1 kg as 10 N.
2) efficiency of pulley
3) height to which load is raised in 4 s.

Answer 1

Given,

Effort,E=100kgf=100❌10N

=1000N

load,L=75kgf

=750N

Height,h orD.E=8m where h or D.E is the distance to which effort is moved.

time,t= 4s

1) power input to the pulley means work done by effort on the machines i.e pulley

P= W/t

=Effort❌D.E/t

=1000❌8/4

=2000 watt (W).

2)M.A= load/ effort

=750/1000

=0.75

Velocity ratio,V=1

Efficiency,n=M.A/ V.R

=0.75/1

=0.75 (or, 75%).

3) power supplied by effort is spent by pulley in doing work against load.

power output=work input on pulley by effort❌efficiency

=load❌load arm/time

So,

2000 ❌0.75 = 750❌ load arm/4

load arm=2000❌0.75❌4/ 750

=7.95 metre(m)

=8 m .