A hydraulic lift in a garage has two pistons: a small one of cross-sectional area 3.70cm2 and a large one of cross-sectional area 210cm2 .

Question

Part A
If this lift is designed to raise a 3100-kg car, what minimum force must be applied to the small piston?

Part B
If the force is applied through compressed air, what must be the minimum air pressure applied to the small piston?

Answer 1

F₁/A₁=F₂/A₂
F₁=F₂A₁/A₂=mg A₁/A₂=
=3100 •9.8•3.7/210 = 535.3 N
p=F₁/A₁=535.3/3.7•10⁻⁴=1.447•10⁶ Pa