Question

The small piston of a hydraulic lift has an area of 0.25 m2. A car weighing 1.35 x 104 N sits on a rack mounted on the large piston. The large piston has an area of 0.85 m2. How large a force must be applied to the small piston to support the car?

Answer 1

We can find the force that must be applied to the small piston to support the car from the hydraulic press formula:

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

here, $F_1 = 1.35 \cdot 10^4 \ N$ is the force applied to the large piston, $F_2$ is the force applied to the small piston, $A_1 = 0.85 \ m^2$ is the area of the large piston, $A_2 = 0.25 \ m^2$ is the area of the small piston.

Then, we get:

$$F_2 = F_1 \dfrac{A_2}{A_1} = 1.35 \cdot 10^4 \ N \cdot \dfrac{0.25 \ m^2}{0.85 \ m^2} = 3970.6 \ N.$$

Answer:

$F_2 = 3970.6 \ N.$