Question

You must design a hydraulic jack (see picture). Its large piston should be able to lift 3 tons when the force on the small piston is F = 200N. The diameter of the large piston is determined by the customer to 7cm. Determine the diameter of the small piston.

Answer 1

By the hydraulic lift formula, we get:

$$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2},$$ $$\dfrac{F_1}{\dfrac{\pi d_1^2}{4}}=\dfrac{F_2}{\dfrac{\pi d_2^2}{4}},$$ $$\dfrac{F_1}{d_1^2}=\dfrac{F_2}{d_2^2},$$ $$d_2=\sqrt{\dfrac{F_2\cdot d_1^2}{F_1}}=\sqrt{\dfrac{200\ N\cdot(0.07\ m)^2}{3000\ kg\cdot9.8\ \dfrac{m}{s^2}}}=0.0058\ m=0.58\ cm.$$