Question

What force must be exerted on the pedal cylinder of a hydraulic lift to support the weight of a 2411-kg car (a large car) resting on the wheel cylinder? The master cylinder has a 2.1-cm diameter and the second cylinder has a 29.2-cm diameter.

Round your answer to 2 decimal places

Answer 1

The main equation is the following (see https://en.wikipedia.org/wiki/Hydraulic_press):

$$\dfrac{F_m}{A_m} = \dfrac{F_2}{A_2}$$

where $F_m$ is the unknown force applied to the master cylinder, $A_m$ is the area of the master cylinder, $F_2 = mg$ is the force applied to the second cylinder ($m = 2411kg$ is the mass of the car), and $A_2$ is the area of the second cylinder.

The areas are:

$$A_m = \pi d_m^2/4\\ A_2 = \pi d_2^2/4$$

where $d_m = 2.1cm = 0.021m,\space d_2 = 0.292m$ are the diameters of the master and second cylinders respectively. Expressing $F_m$, obtain:

$$F_m = \dfrac{F_2A_m}{A_2}\\ F_m = \dfrac{mg\cdot \pi d_m^2\cdot 4}{4\cdot \pi d_2^2} =\dfrac{mg\cdot d_m^2}{d_2^2}\\ F_m = \dfrac{2411\cdot 9.8\cdot0.021^2}{0.292^2} \approx 122N$$

Answer. 122 N.