# Hydraulic Lift.( U tube) and oil

1. Jan 29, 2008

### dr3vil704

1. The problem statement, all variables and given/known data
A 12,000 N car is raised using a hydraulic lift, which consist of a U - tube with arm of unequal areas, filled with oil with a density of 800 kg/m^3 and capped at both ends with tight-fitting pistons. The wider arm of the U tube had a radius of 18. cm and the narrower arm has a radius of 5 cm. The car rest on the piston on the wider arm of the U-tube, The piston are initially at the same level. What is the force that must applied to the smaller piston in order to lift the car after it has been raised 1.2 m. Neglect the weight of the pistons

Ok, so I converted the radius to Meter, just for future reference. 5 cm = .005 m, 18 cm = .018 m.

I'm thinking of the Hyrdralift lift formula. A1 X d1 = A2 X d2. A= area, and D= vertical distance The sub 1 represent the smaller arm, and sub 2 represent the wider arm. and since it a circular surface, I should use $$\pi$$r^2 for the Area.

So I could do d1=(A2/A1) X d2...
But then, I have no idea what to do with the density of the oil and use that distance to find the force (Mass X Gravity) of the smaller arm.

2. Jan 29, 2008