Balancing a Hydraulic Lift with Two Pistons

[mujadeo]

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!

 

[Shooting Star]

The pressure due to the extra height of the fluid must be supporting the pressure due to weight of the man. Think afresh.

 

[Moetasim]

Your calculations and approach seem to be correct. It is possible that there is a small error in your calculations or that the values given in the problem are not exact. I would suggest double-checking your calculations and rounding to the appropriate number of significant figures. Additionally, make sure to use consistent units throughout your calculations. If you are still having trouble, it may be helpful to consult with a classmate or instructor for further assistance.