Question

What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2100 kg car (a large car) resting on the slave cylinder

Answer 1

Complete Question:

What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2100 kg car (a large car) resting on the slave cylinder ? The master cylinder has a 2.00-cm diameter and the slave has a 24.0-cm diameter

Answer:

$F_1 = 142.92N$

Explanation:

Given

$m = 2100kg$ --- mass

$D_1 = 2.00\ cm$ --- diameter of the large cylinder

$D_2 = 24.0\ cm$ --- diameter of the slave cylinder

To do this, we apply Archimedes' principle of buoyancy which implies that:

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

Where

$F_1 = Force\ on\ the\ master\ cylinder$

$F_2 = Force\ on\ the\ slave\ cylinder$

$A_1 = Area\ of\ the\ master\ cylinder$

$F_2 = Area\ of\ the\ small\ cylinder$

Calculating the area of the master cylinder.

$A_1 = \pi r_1^2$

$r_1 = \frac{1}{2}D_1 = \frac{1}{2} * 2.00cm = 1.00cm$

$A_1 = \pi* 1^2$

$A_1 = \pi * 1$

$A_1 = \pi$

Calculating the area of the slave cylinder.

$A_2 = \pi r_2^2$

$r_2 = \frac{1}{2}D_2 = \frac{1}{2} * 24.00cm = 12.00cm$

$A_2 = \pi* 12^2$

$A_2 = \pi* 144$

$A_2 = 144\pi$

Substitute these values in:

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

$\frac{F_1}{\pi} = \frac{F_2}{144\pi}$

Multiply both sides by $\pi$

$\pi * \frac{F_1}{\pi} = \frac{F_2}{144\pi} * \pi$

$F_1 = \frac{F_2}{144}$

The force exerted on the slave cylinder (F2) is calculated as:

$F_2 = mg$

$F_2 = 2100 * 9.8$

$F_2 = 20580$

Substitute 20580 for F2 in $F_1 = \frac{F_2}{144}$

$F_1 = \frac{20580}{144}$

$F_1 = 142.92N$

Hence, the force exerted on the master cylinder is approximately 142.92N