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A car weighing 12,000N12,000N is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressible oil with a density 800 kg/m^3800kg/m 3 and capped at both ends with tight-fitting pistons. The car rests on the piston on the wider arm of the U-tube that has a radius of 18.0cm18.0cm, and the narrower arm has a radius of 5.00cm5.00cm. The pistons are initially at the same level. 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

Respuesta :

LammettHash

A force of magnitude F applied to the smaller piston exerts a pressure of

P = F / (π (0.0500 m)²)

while the 12 kN car exerts the same pressure at the other piston of

P = (12,000 N) / (π (0.180 m)²)

Solve for F :

F / (π (0.0500 m)²) = (12,000 N) / (π (0.180 m)²)

F = (12,000 N) (0.0500 m)² / (0.180 m)²

F = 925.926 N ≈ 930 N

Then a minimum force of 930 N must be applied in order to start lifting the car.