Question

A force of $50kgf$ is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, the diameters of the piston being $5cm$ and $24cm$ respectively.

Answer 1

$Area1=\frac{\Pi 5^2}{4}{F}_1=50kgf$

$Area2=\frac{\Pi {24}^2}{4}{F}_2=xkgf.$

Since Pressure is constant.

So, $\frac{F_1}{A_1}=\frac{F_2}{A_2}\Rightarrow \frac{x}{\frac{\Pi {24}^2}{4}}=\frac{50}{\frac{\Pi 5^2}{4}}$

$x=\frac{{50}^2\times {24}^2}{25}=576\times 2$

$x=1152kgf$