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## Homework Statement

A hydraulic lift has two connected pistons with cross-sectional areas 5 cm2 and 550 cm2. It is filled with oil of density 520 kg/m3.

a) What mass must be placed on the small piston to support a car of mass 1500 kg at equal fluid levels?

this was no prob.

=13.61 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?

cant figure this out.

## Homework Equations

## The Attempt at a Solution

heres what im doing...

1500kg + 70kg (9.81) = 15401.7N = weight of man + car

15401.7N/5.5m^2 = 509.147Pa (pressure on large piston)

then i took answer from part I and did the following

13.64kg(9.81)/(.05^2m) = 53523.360Pa (Pressure on small piston)

Finally..

Pman+car - P = rho g h

rearrange eq..

h = Pman+car - p / rho g

h = -10.39m = wrong! please help me!