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?

#### Solution 1

The maximum mass of a car that can be lifted, *m* = 3000 kg

Area of cross-section of the load-carrying piston, *A* = 425 cm^{2 }= 425 × 10^{–4} m^{2}

The maximum force exerted by the load, *F = m*g

= 3000 × 9.8

= 29400 N

The maximum pressure exerted on the load-carrying piston, `P = F/A`

`= 29400/(425xx10^(-4))`

= 6.917 × 10^{5} 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 × 10^{5} Pa.

#### Solution 2

Pressure on the piston due to car = `"Weight of car"/"Area of piston"`

`P = (3000xx9.8)/(425xx10^(-4)) Nm^(-2) = 6.92 xx 10^5 "Pa"`

This is also the maximum pressure that the smaller piston would have to bear.