How Does a Hydraulic Lift Work with Different Piston Sizes?

AI Thread Summary
A hydraulic lift operates based on Pascal's principle, allowing force to be transferred through incompressible fluid. In this scenario, the smaller piston has an area of 15 cm² and supports a car weighing 1600 kg with a required mass of 55.8 kg on the smaller piston. When an additional 80 kg person enters the car, the equilibrium height difference between the pistons is calculated to be 3.02 m. The challenge arises in determining how much the car drops when the person enters, with initial calculations yielding an incorrect height of 0.109 m. The correct approach involves applying the principle of work conservation, leading to a calculated drop of approximately 0.1003 m for the car.
Pat2666
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Okay, so I'm stuck on the last part of the problem :

A hydraulic lift has two connected pistons with cross-sectional areas 15 cm2 and 430 cm2. It is filled with oil of density 620 kg/m3.

--------------------------------------------------------------------------------

a) What mass must be placed on the small piston to support a car of mass 1600 kg at equal fluid levels?

kg *
55.8 OK

b) With the lift in balance with equal fluid levels, a person of mass 80 kg gets into the car. What is the equilibrium height difference in the fluid levels in the pistons?

m *
3.02 OK

c) How much did the height of the car drop when the person got in the car?

m

HELP: The fluid is incompressible, so volume is conserved.
HELP: Remember, one side will go up and one side will go down. The difference you calculated in part (b) was the sum of those two changes.


Okay, so I managed to find out the first two parts of the problem but I'm having trouble with Part C. Any way I attempt either comes out to the answer for B, or is just wrong. Since it mentioned that the volume is the same I thought that maybe the volumes of the cylindrical area displaced for each piston would be equal to one another (as you will see below). However, I came up with 0.109m as the height that the car moved, but it was wrong. I don't know what I'm doing wrong or what I might be missing completely. Any help would be greatly apprciated :)

My work :

http://img255.imageshack.us/img255/7595/workns5.jpg
 
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Your result looks fine to me. Another approach is Work-in=Work-out.

F1d1=F2d2

(3.02)(55.8)(9.80)= X(1680)(9.8)

X= .1003 m
 
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