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# Question 48

A force of 50kgf is applied to the smaller piston of a hydraulic machine. Neglecting friction, find the force exerted on the large piston, if the diameters of the pistons are 5cm and 25cm respectively.

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Comparing the diameter of the narrow piston and broader piston = 5/25 or 5:25 = 25:625

Force exerted on narrow piston, F1 = 50kgf

Consider F2 to be the force exerted on the broader piston

We know from the principle of Hydraulic machine,

Pressure on narrow piston = pressure on the broader piston

\begin{array}{l} \frac{\mathrm{F}_{1}}{\mathrm{A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{A}_{2}} \\ \frac{\mathrm{F}_{1}}{\mathrm{F}_{2}}=\frac{\mathrm{A}_{1}}{\mathrm{A}_{2}} \\ \Rightarrow 50 / \mathrm{F}_{2}=25 / 625 \\ \mathrm{F}_{2}=1250 \mathrm{kgf} \end{array}

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