In hydraulic machine, the two pistons are of area of cross section in the ratio $1 : 10$ . What force is needed on the narrow piston to overcome a force of $100 N$ on the wider piston?

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Solution

$F_{1} = 100 N A_{1} = 10 m^{2}$
$F_{2} = x N A_{2} = 1 m^{2}$
Since pressure = constant
hence, $\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \Rightarrow \frac{100}{10} = \frac{x}{1}$
$x = 10 N$

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