Question

What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2500-kg car (a large car) resting on the slave cylinder? The master cylinder has a 4.00-cm diameter and the slave has a 25.0-cm diameter.

Answer 1

Given:

$m=2500\:\rm kg$

$d_1=\rm 4.00\: cm$

$d_2=\rm 25.0\: cm$

The hydraulic press equation says

$$\frac{F_1}{A_1}=\frac{F_2}{A_2}$$

Hence, the required force

$$F=mg\frac{A_1}{A_2}=2500*9.8*\frac{(4.00)^2}{(25.0)^2}=627\: \rm N$$