An effort of $50$ kgf is applied at the end of a lever of the second order which supportsb a load of $750$ kgf. Find the mechanical advantage.

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Solution

REF.Image.
Length of lever = Effort arm = AE
Load arm AB = 0.1 m
Effort E = 50 kgf
Load L = 750 kgf
$A E \times 50 = 750 \times 0.1$
$A E = \frac{750}{50} \times \frac{1}{10} = 1.5 m$

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