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Hydraulic lift

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A hydraulic lift has two connected pistons with cross-sectional areas 25 cm2 and 250 cm2. It is filled with oil of density 510 kg/m3.

    a) What mass must be placed on the small piston to support a car of mass 1200 kg at equal fluid levels? I got this answer - 120 kg

    b) With the lift in balance with equal fluid levels, a person of mass 90 kg gets into the car. What is the equilibrium height difference in the fluid levels in the pistons?
    i got this answer 7.0588m

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

    C is the part that i am very confused on. I thought i was doing it correct and apparently not.

    2. Relevant equations

    h1+h2 = 7.0588
    I thought this was the equation i used to figure out part C but i am just lost on it.

    3. The attempt at a solution
  2. jcsd
  3. Nov 29, 2009 #2


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    Homework Helper

    One cylinder goes down h1, the other up h2. Their total is 7.0588 m.
    The volume of oil pushed out of the first cylinder equals the volume of oil pushed into the other cylinder. If you write an equation for this fact, you'll have a second equation relating h1 and h2. Then you can solve the two equations as a system to find h1 and h2.
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