An effort of 21 Kgf is applied on a machine for the movement of a load of 120 Kgf. Find the mechanical advantage of the machine.

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Solution

Given that,

Load = 120 kgf

Effort = 21 kgf

We know that,

Mechanical Advantage = load/ effort

$M . A = \frac{L}{E}$

$M . A = \frac{120}{21}$

$M . A = 5.7$

Hence, the mechanical advantage of machine is $5.7$

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