Question

A hydraulic lift raises a 4000 kg automobile when a 500N force is applied to the smaller piston. If the smaller piston has an area of 10 cm2, what is the cross-sectional area of the larger piston?

Answer 1

Answer:

The cross-sectional area of the larger piston is 800 cm².

Explanation:

Given;

mass of the automobile, m = 4000 kg

force applied on the small piston, F₁ = 500 N

area of the smaller piston, A₁ = 10 cm²

load lifted by the larger piston, F₂ = 4000 x 10 = 40,000 N

Pressure experienced by each piston is given as;

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

Where;

A₂ is the cross-sectional area of the larger piston

$\frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\A_2 = \frac{F_2A_1}{F_1} \\\\A_2 = \frac{40,000 \ \times\ 10}{500} \\\\A_2=800 \ cm^2$

Therefore, the cross-sectional area of the larger piston is 800 cm².