# Hydraulic Lift (Fluids)

1. Jun 19, 2008

### Pat2666

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.

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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 [Broken]

Last edited by a moderator: May 3, 2017
2. Jun 20, 2008

### RTW69

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