Question

A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressible oil and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the initial force that must be applied to the smaller piston in order to start lifting the car? (For purposes of this problem, you can neglect the weight of the pistons.)

Answer 1

Answer:

926 N

Explanation:

Metric unit conversion:

R = 18 cm = 0.18 m

r = 5 cm = 0.05 m

The pressure exerted by the F = 12000N car on the wider arm would be ratio of the gravity over area

$P = F/A = \frac{F}\pi R^2} = \frac{12000}{2*\pi*0.18^2} = 117892 Pa$

The pressure must be the same on the smaller pressure for it to be able to start lifting the car. We can calculate the force f acting on it:

$f = Pa = P\pi r^2 = 117892 * \pi * 0.05^2 = 926 N$