Question

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm² . What maximum pressure would the smaller piston have to bear?

Answer 1

The maximum mass of a car that can be lifted, m = 3000 kg

Area of cross-section of the load-carrying piston, A = 425 cm² = 425 × 10^–4 m² 

The maximum force exerted by the load, 

F = mg = 3000 × 9.8 = 29400 N

The maximum pressure exerted on the load-carrying piston, P = F/A

 = 29400/425 x 10^-4 = 6.917 x 10⁵Pa

Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 × 10⁵ Pa.