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.
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.
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)
Pman+car - P = rho g h
h = Pman+car - p / rho g
h = -10.39m = wrong! please help me!