Question

In a single movable pulley system ,a load of 125 kgf is lifted by an effort of 75kgf. find the percentage efficiency of system? ​

Answer 1

Load=125 kgf

Effort =75kgf

M.A= load / effort

=125/75

=1.66

V.R= 1

Efficiency = M.A / V.R

= 1.66/1 × 100

= 166%

I think there is a mistake in the question

Answer 2

Explanation:

Load=125 kgf

Effort =75kgf

M.A= load / effort

=125/75

=1.66

V.R=2

Efficiency = M.A / V.R

= 1.66/2 x 100

=83%

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