A hydraulic lift has a small piston with a diameter 5.0 cm piston and the large piston with a diameter of 25 cm. What force must be applied to the small piston in order to lift a car on the large piston that weighs 13,000 N?

Question

Quizplus · Accepted Answer

The force required can be calculated using the principle of hydraulic systems, which states that the force divided by the area is constant for both pistons. Thus, $F_1/A_1 = F_2/A_2$, where $F_1$ is the force applied on the small piston, $A_1$ is the area of the small piston, $F_2$ is the force applied by the large piston (13,000 N), and $A_2$ is the area of the large piston. The areas can be calculated using the formula for the area of a circle, $A = \pi r^2$, where $r$ is the radius of the piston. The radius is half of the diameter, so the radius of the small piston is 2.5 cm (or 0.025 m) and the radius of the large piston is 12.5 cm (or 0.125 m). Plugging in the values, we get:$$F_1 = \frac{F_2 \cdot A_1}{A_2} = \frac{13,000 \cdot \pi \cdot (0.025)^2}{\pi \cdot (0.125)^2} = \frac{13,000 \cdot 0.001963}{0.049087} \approx 520 N$$Thus, the correct answer is 520 N.

A Hydraulic Lift Has a Small Piston with a Diameter