# Quick hydraulic jack question

1. Nov 1, 2005

### frigid

A 2000 kg car is raised 1.5 meters high on a hydraulic jack, which has a large piston with a diameter of 30cm. The small piston has a diameter of 5cm (btw the reservoir is filled with oil)

d) When the car is raised by 1.5 m, what distance will the small piston move?

Is it simply that you have to take off 1.5 m on the side of the small piston in order for the big piston to raise by 1.5 m? Or is there something more complicated about this?

Last edited: Nov 1, 2005
2. Nov 1, 2005

### FredGarvin

You need to apply the idea of work. The work on one side will be the same as the other piston. Calculate the work for the large side and then use that amount for what the small piston must do.

3. Nov 1, 2005

### lightgrav

Alternatively, you can remove Volume fom one side,
to put it on the other side.
(you get the same answer as using Work ...
W = F dh = P A dh = P dV)

4. Jul 1, 2009

### bsodmike

is d = 0.25 m ?

I got this from the assumption 1/6th the work done at the small piston (from the pressure relationship).