# Hydraulic Lift of a car

1. Apr 7, 2009

### mrnastytime

1. The problem statement, all variables and given/known data
A hydraulic lift has two connected pistons with cross-sectional areas 15 cm2 and 460 cm2. It is filled with oil of density 560 kg/m3.

2. Relevant equations
a)What mass must be placed on the small piston to support a car of mass 1100 kg at equal fluid levels?
A:35.86 kg
b)With the lift in balance with equal fluid levels, a person of mass 70 kg gets into the car. What is the equilibrium height difference in the fluid levels in the pistons?
A:2.72 m
c) How much did the height of the car drop when the person got in the car?

3. The attempt at a solution
Take the difference in height when its the only the car, then find difference in height of car + guy. Subtract those two. But its not working.

2. Apr 8, 2009

### Carid

It seems to me that, when the guy got in the car, the car went down but the 38.5 kg weight on the other piston went up. So you have to share out that 2.72 metres between how much the car went down and how much the other piston went up.