Question

A force of 150 kgf is applied to the smaller piston of hydraulic machine. Neglecting the friction force, find the force exerted on the larger piston. The diameter of the pistons is 10 cm and 25cm​

Answer 1

Answer:

force/area = force/area

=> 150/π(25) = 4force/π(625)

=> force = (625×150)/100 = 937.5 kgf

Answer 2

Ratio of diameter of smaller piston to bigger piston = 5:25

.•. Ratio of area of smaller piston to bigger piston = 25:625

Force applied on smaller piston, F1 = 50kgf

Let F2 be the force on the bigger piston.

By the principle of hydraulic machine,

Pressure on narrow piston = Pressure on wider piston

$\frac{F1}{A1} = \frac{F2}{A2} \\ \\ \implies \: \frac{F1}{F2} = \frac{A1}{A2} \\ \\ \implies \: \frac{50}{F2} = \frac{25}{625} \\ \\ \implies \: F2 \: = \: \cancel{50} \: \: ²\times \frac{625}{ \cancel{25} } \\ \\ \ = 1250 \: kgf \: \: \: \: \: (ans)$