Question

In a car lift compressed air exerts a force F, on a small piston having a radius of 5 cm. This pressure is transmitted to a second piston of radius 15 cm. If the mass of the car to be lifted is 1500 kg, What is $F_1$? What is the pressure required to do this task?

Answer 1

Given, $r_1=5cm=5\times {10}^{-2}m;F_1=?$

$r_2=1cm=1\times {10}^{-2}m;F_1=1500\times 9.8N$

$A_1=\pi r_1^2=\pi \times {(5\times {10}^{-2})}^2=\pi \times 25\times {10}^{-4}{m}^2$

$A_2=\pi r_2^2=\pi \times {(15\times {10}^{-2})}^2=\pi \times 225\times {10}^{-4}{m}^2$

From Pascal's law

$P_1=P_2\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2}$

$\therefore F_1=\frac{A_1}{A_2}\times F_2=\frac{\pi \times 25\times {10}^{-4}}{\pi \times 225\times {10}^{-4}}\times 1500\times 9.8$

$=\frac{1500\times 9.8}{9}=\frac{500\times 9.8}{3}$

$=163.33N$

Pressure $P_1=\frac{F_1}{A_1}=\frac{500\times 9.8}{3\times \pi \times 25\times {10}^{-4}}$

$=\frac{20\times 9.8\times {10}^4}{3\times 3.14}$

$=20.8\times {10}^4$

or $P_1=2.08\times {10}^{5}N{m}^{-2}$