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 cm2. What
maximum pressure would the smaller piston have to bear ?
Solution
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
=
= 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.
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
=
= 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.
