Example

Question:

In a car lift compressed air exerts a force \(F_1\) on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm. If the mass of the car to be lifted is 1350 kg, calculate \(F_1\). What is the pressure necessary to accomplish this task? (\(g = 9.8\,\mathrm{ms}^{-2}\))

Solution:

Since pressure is transmitted undiminished throughout the fluid, \[ F_1 = \frac{A_1}{A_2} F_2 = \frac{\pi (5 \times 10^{-2}\,\mathrm{m})^2}{\pi (15 \times 10^{-2}\,\mathrm{m})^2} \cdot (1350\,\mathrm{kg} \times 9.8\,\mathrm{ms}^{-2}) \] \[ = 1470\,\mathrm{N} \approx 1.5 \times 10^3\,\mathrm{N} \]
The air pressure that will produce this force is \[ P = \frac{F_1}{A_1} = \frac{1.5 \times 10^3\,\mathrm{N}}{\pi (5 \times 10^{-2})^2\,\mathrm{m}^2} = 1.9 \times 10^5\,\mathrm{Pa} \]
This is almost double atmospheric pressure.
Hydraulic brakes in automobiles also work on the same principle.