Question

What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2200 kg car (a large car) resting on the slave cylinder? The master cylinder has a 2.50 cm diameter, while the slave has a 24.0 cm diameter.

Answer 1

Given that

Mass of car m = 2200 kg

Diameter of master cylinder d1 = 2.50 cm

Radius of master cylinder r1 = 2.50 / 2 = 1.25 cm = 0.0125 m

Diameter of slave  cylinder d2 = 24 cm

Radius of slave cylinder r2 = 24 / 2 = 12 cm = 0.12 m

to lift the weight at the same height pressure should be same at both ends

P1 = P2

F1 / A1 = F2 / A2

F1 = (A1 / A2) * F2

where

F2 = Force on car = mg

F1 = force exerted on the master cylinder

F1 = ( r1^2 /  r2^2 )* mg

F1 = (0.0125^2 / 0.12^2 )*2200*9.8

F1 = 233.94 N

Good Luck !