(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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!

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Hydralic lift

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