Question

For the hydraulic lift shown in Fig. 12.7, what must be the ratio of the diameter

of the vessel at the car to the diameter of the vessel where the force F1 is

applied so that a 1520-kg car can be lifted with a force F1 of just 125 N?

Answer 1

Explanations & Calculations

• The question sounds to the basic theory of hydraulic jack.

• The theory is the hydraulic pressure within a liquid at a given level is the same anywhere at that level.
• Then,

$\qquad\qquad \begin{aligned} \small \frac{F}{A}&=\small \frac{f}{a}\\ \small \frac{1250kg\times9.8ms^{-2}}{\pi R^2}&=\small \frac{125\,N}{\pi r^2}\\ \small \frac{R^2}{r^2}&=\small \frac{1250\times9.8}{125}\\ \small \frac{R}{r}&=\small \sqrt{98}\\ \small \frac{\frac{D}{2}}{\frac{d}{2}}&=\small 9.9\\ \small \frac{D}{d}&\approx\small 10 \end{aligned}$

• The required ratio would be $\small D:d \approx 10:1$