Question

What force is aplied on a piston of area of cross section $2c{m}^2$ to obtain a force $150$N on the piston of area cross section $12c{m}^2$ in a hydraulic machine?

Answer 1

Here let force on piston 1 be $F_1$

cross sectional area of piston 1, $A_1=2c{m}^2$

force on piston 2 be $F_2=150N$

cross sectional area of piston 2, $A_2=12c{m}^2$

we know that the pressure on smaller piston(1) is equal to the pressure on wider piston(2) in a hydraulic machine.

$P_1=P_2$

So, $\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$\Rightarrow \frac{F_1}{2c{m}^2}=\frac{150N}{12c{m}^2}$

$\Rightarrow F_1=25N$

Hence, the force on smaller piston is $25N$.