Balancing a Hydraulic Lift with Two Pistons

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The discussion revolves around a hydraulic lift with two pistons, where the smaller piston has an area of 5 cm² and the larger one 550 cm², filled with oil of density 520 kg/m³. The first part of the problem is solved, determining that a mass of 13.61 kg must be placed on the small piston to support a car weighing 1500 kg. The second part involves calculating the equilibrium height difference in fluid levels after a 70 kg person enters the car, which has proven challenging for the user. They attempted to calculate pressures on both pistons but arrived at an incorrect height difference of -10.39 m. The user seeks assistance in correctly applying the principles of fluid mechanics to find the accurate height difference.
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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 I am 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!
 
Physics news on Phys.org
The pressure due to the extra height of the fluid must be supporting the pressure due to weight of the man. Think afresh.
 
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