Question

A hydraulic automobile lift is designed to lift cars with a maximum mass of `3000Kg`. The area of cross section of the piston carrying the load is `425cm^(2)`. What maximum pressures would the smaller piston have to bear?

Answer 1

`A=425cm^2=425xx10^-4m^2`,`F=3000kgwt=3000kgwt=3000xx10N=30000N`
pressure on the larger piston, `P=(F)/(A)=(30000N)/(425xx10^-4m^2)=705882Pa`
Since the pressure is transmitted equally in all direction, pressure on the smaller piston is also 705882 Pa