Question

Two pistons of a hydraulic lift have radii of 2.2cm and 10.8cm. A mass of 1.71×103kg is placed on the larger piston. Calculate the minimum downward force needed to be exerted on the smaller piston to hold the larger piston level with the smaller piston.

Answer 1

radius of the smaller piston,r₁= 2.2 cm

radius of the larger piston,r2= 10.8 cm

Area of the smaller piston, A₁ = pi* r₁² = (3.14 * 2.2 * 2.2) = 15.1976 cm²

Area of the larger piston, A2 = pi* r₂² = (3.14 * 10.8* 10.8) = 366.2496 cm²

Force on the larger piston, F = Mg = ( 1.71 * 10³ * 9.81 ) = 16775.1 N

Pressure on the larger Piston , P = (F/A₂) = (16775.1 N / 366.2496 cm² ) = 45.8024 N /cm² .

For balancing of the two pistons, Pressure on the smaller piston = Pressure on the larger Piston = P = 45.8024 N /cm²

If the minimum weight to be put on the smaller piston is m, then P = (mg / A₁ )

m = (P*A₁ / g ) = (45.8024* 15.1976 ) / ( 9.81) = 70.9568 kg (ans)