Question

In a hydraulic garage lift, the small piston has a radius of 5.0 cm and the large piston has a radius of 15 cm. what force must be applied on the small piston in order to lift a car weighing 20,000 n on the large piston?

Answer 1

For Pascal's principle, the pressure exerted on the first piston is equal to the pressure exerted on the second piston:
$p_1 = p_2$
which means
$\frac{F_1}{A_1}= \frac{F_2}{A_2}$
where F1, F2 are the two forces and A1, A2 the areas of the two pistons.

The areas of the pistons are:
$A_1 =\pi r_1^2 = \pi (5.0 cm)^2=25\pi cm^2$
$A_2 = \pi r_2^2 = \pi (15 cm)^2 =225 \pi cm^2$
And the force on the second piston is $F_2 = 20000 N$, so the force that should be applied on the first piston is
$F_1 = F_2 \frac{A_1}{A_2}=(20000 N) \frac{25 \pi cm^2}{225 \pi cm^2}=2222 N$