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?

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;

Where;

A₂ is the cross-sectional area of the larger piston

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

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