Question

The areas of pistons in a hydraulic machine are 5 cm² and 625 cm². What force on the smaller piston will support a load of 1250 N on the larger piston? State any assumption which you make in your calculation.

Answer 1

Area of small piston A₁ = 5 cm²

Area of wider piston, A₂ = 625 cm²

Force on small piston be F₁

Force on wider piston or load, F₂ = 1250 N

By the principle of hydraulic machine,

Pressure on narrow piston = pressure on wider piston

or, ${\displaystyle \frac{F_1}{A_1}=\frac{F_2}{A_2}}$

or, ${\displaystyle \frac{F_1}{5}=\frac{1250}{625}}$

or, ${\displaystyle F_1=\frac{1250}{625}\times 5}$

or ${\displaystyle F_1=10}$ N

Assumption: No friction or leakage of liquid happens.