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, $P =\frac{ F }{A}$

$= \frac{29400 }{ 425 × 10 ^{- 4}}$

= 6.917 × 10 5 Pa
Pressure is transmitted equally in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 × 10 5 Pa.