Question

What force (N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2300.0 kg car resting on the slave cylinder? The master cylinder has a 1,70 cm diameter, while the slave's is 24.0 cm.

Answer 1

Since the pressure in the master and the slave cylinders should be equal so

$P_1=P_2\\ \implies \frac{F_1}{A_1}=\frac{F_2}{A_2}\\ \implies \frac{F_1}{\pi (\frac{d_1}{2})^2}=\frac{F_2}{\pi (\frac{d_2}{2})^2}\\ \implies \frac{F_1}{{d_1}^2}=\frac{F_2}{{d_2}^2}\\ \implies F_1=\frac{F_2}{{d_2}^2}\times d_1^2\\ \implies F_1=\frac{2300\times 9.8}{0.24^2}\times 0.017^2\\ \implies F_1=113.09N$

$F_1$ is force applied on master cylinder.

$F_2$ is force applied on slave cylinder.

$A_2$ is area of slave cylinder

$A_1$ is area of master cylinder.