Question

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

Answer 1

Answer:

the force applied to the smaller piston is 600 N

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;

$P = \frac{F}{A} = \frac{f}{a} \\\\\frac{F}{R^2} = \frac{f}{r^2} \\\\f = \frac{F\times r^2}{R^2} = \frac{15,000 \times (0.04)^2}{(0.2)^2} = 600 \ N$

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