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 $425c{m}^2$ . 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, $A=425c{m}^2=425\times {10}^{–4}{m}^2$
The maximum force exerted by the load,
$F=mg=3000\times 9.8=29400N$
The maximum pressure exerted on the load-carrying piston, $P=\frac{F}{A}$
$=\frac{29400}{425\times {10}^{-4}}=6.917\times {10}^{5}Pa$
Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is $6.917\times {10}^{5}Pa.$