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A 15 000-N car on a hydraulic lift rests on a cylinder with a piston of radius 0.20 m. If a connecting cylinder with a piston of 0.040-m radius is driven by compressed air, what force must be applied to this smaller piston in order to lift the car

User Jozzy
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1 Answer

19 votes
19 votes

Answer:

the force applied to the smaller piston is 600 N

Step-by-step explanation:

Given;

weight of the car, F = 15,000 N

radius of the lager piston, R = 0.2 m

radius of the smaller piston, r = 0.04 m

let the force applied to the smaller piston = f

The pressure applied on both piston is constant;


P = (F)/(A) = (f)/(a) \\\\(F)/(R^2) = (f)/(r^2) \\\\f = (F* r^2)/(R^2) = (15,000 * (0.04)^2)/((0.2)^2) = 600 \ N

Therefore, the force applied to the smaller piston is 600 N

User Adi H
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