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Hydraulic Lift equilibrium

  1. Apr 9, 2008 #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 700 cm2. It is filled with oil of density 570 kg/m3.

    a) What mass must be placed on the small piston to support a car of mass 1300 kg at equal fluid levels? (answer: 46.4kg)

    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? (answer: 1.77m)

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

    3. The attempt at a solution

    I know that the fluid is incompressible therefore the volume is conserved. I also know that the height changed is proportional to the area of each of the pistons. I know that A1*d1=A2*d2 where d1 is the distance piston 1 is pushed down and the volume is A1 that flowed into the piston.

    Where I am confused is how do we relate the weight to the change in height.

    Any help would greatly be appreciated!
  2. jcsd
  3. Apr 9, 2008 #2
    If I say : the difference in fluid heights provides a weight that balance with that of the car.
    Does that help you in understanding their relations?:smile:
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