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, `F_1/A_1 = F_2/A_2`

or, `F_1/5 = 1250/625`

or, `F_1 = 1250/625 xx 5`

or `F_1 = 10` N

Assumption: No friction or leakage of liquid happens.