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 cm${}^2$. What maximum pressure would the smaller piston have to bear ?

Answer 1

Pressure on the piston $P=\frac{F}{A}$

Force $F=m\times a$

$=3000\times 9.8$

$=29400N$

Area of cross section $A=425\times {10}^{-4}sqm$

Therefore the pressure $P=\frac{3000\times 9.8}{425\times {10}^{-4}}=6.92\times {10}^{5}Pa$.