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 with a density of 800 kg/m3 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 force that must be applied to the smaller piston in order to lift the car after it has been raised 1.20 m? You may neglect the mass of the pistons.

Answer 1

Answer:

F = 1958.4 N

Explanation:

By volume conservation of the fluid on both sides we can say that volume of fluid displaced on the side of the car must be equal to the volume of fluid on the other side

so we have

$L_1A_1 = L_2A_2$

$1.20(\pi 18^2) = L_2(\pi 5^2)$

$L_2 = 15.55 m$

so the car will lift upwards by distance 1.2 m and the other side will go down by distance 15.55 m

So here the net pressure on the smaller area is given as

$P = P_{atm} + \frac{12,000}{\pi (0.18)^2} + \rho g (1.2 + 15.55)$

excess pressure exerted on the smaller area is given as

$P_{ex} = \frac{12000}{\pi (0.18)^2} + 800(9.81)(16.75)$

$P_{ex} = 2.49\times 10^5 Pascal$

now the force required on the other side is given as

$F = P_ex (area)$

$F = (2.49 \times 10^5)(\pi (0.05)^2)$

$F = 1958.4 N$