Question 8

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

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  = F / A

= 29400  /  425 x 10-4

= 6.917 × 105 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 × 105 Pa.

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