Question

A hydraulic lift consists of two pistons that connect to each other by an incompressible fluid. If one piston has an area of 0.15 m² and the other an area of 6.0 m², how large a mass can be raised by a force of 130 N exerted on the smaller piston?

Answer 1

The presure exerted on the smaller piston is equal to the preasure provided by the big one:

$$\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$$

where $F_1 = 130N$ is the foce exerted on the smaller piston, $A_1 = 0.15m^2$ is the area of the smaller piston, $F_2$ is the force provided by the big piston, and $A_2 = 6m^2$ is the area of the big piston. Expressing $F_2$, get:

$$F_2 = F_1\dfrac{A_2}{A_1} =130\times \dfrac{6}{0.15} = 5200N$$

A mass that can be raised by this force is:

$$m = \dfrac{F_2}{g} = \dfrac{5200N}{9.81m/s^2} \approx 530.07kg$$

Answer. 530.07 kg.