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 board is 425 ~cm ^2 . What maximum pressure would smaller piston have to bear?

Step-by-step solution

The maximum force, which the bigger piston can bear, F=30000 kgf =3000 9.8 ~N Area of piston, A=425 ~cm ^2=425 10⁻⁴ ~m ^2 Maximum pressure on the bigger piston, p= F A = 300 9.8 425 10⁻⁴ =6.92 10^4 ~Pa Since, the liquid transmits pressure equally, therefore the maximum pressure the smaller piston can bear is 6.92 10^4 ~Pa