A 12,000-N car is raised using a hydraulic lift,which consists of a U-tube with arms of unequal areas,filled with 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 oil must exert on the smaller piston in order to lift the car after it has been raised 1.20 m? For purposes of this problem,neglect the mass of the pistons. A)0.954 kN B)1.88 kN C)1.96 kN D)3.67 kN E)1.20 kN

Question

Quizplus · Accepted Answer

We can find the pressure at the bottom of the U-tube using the formula P = ρgh, where ρ is the density of the oil, g is the acceleration due to gravity, and h is the height difference between the two pistons.

For the wider piston, the pressure is P = (800 kg/m³)(9.8 m/s²)(1.20 m) = 9,408 Pa.

Because the U-tube is filled with an incompressible fluid, the pressure at the bottom of the narrower arm must also be equal to 9,408 Pa in order for the system to be in equilibrium.

We can use the formula F = PA to find the force required to lift the car, where A is the area of the smaller piston.

The area of the smaller piston is A = πr² = π(0.050 m)² = 0.00785 m².

So the force required is F = (9,408 Pa)(0.00785 m²) = 73.7 N ≈ 1.96 kN.

Therefore, the correct answer is (C) 1.96 kN.

A 12,000-N Car Is Raised Using a Hydraulic Lift,which Consists