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Hydraulic lift problem

  1. Apr 8, 2008 #1
    i'm having difficulty with part C of this problem.

    A hydraulic lift has two connected pistons with cross-sectional
    areas 25 cm2 and 430 cm2. It is filled with oil of density 500 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: 75.58kg)

    b) With the lift in balance with equal fluid levels, a person of mass 60 kg gets into the car. What is the equilibrium height difference in the fluid levels in the pistons? (Answer= 2.8m)

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

    please help. my rationale thus far is that volume is conserved. could anyone direct me in applying the right concept? thanks.
  2. jcsd
  3. Apr 8, 2008 #2
    If your rationale is correct how was the mass accomodated? What changes are there to the density of the oil or the amount of force at the opposite piston?
  4. Apr 9, 2008 #3
    there are no changes to the density of the oil. the force is what i'm confused about. since on the opposite end, no additional mass is added, is the force acting upon that piston equivalent to part a? or do i have to find the new force, F2=F(A1/A2); where F=1360kg (mass of the car +person). I've tried this problem a million different ways (feels like), perhaps i'm rounding incorrectly.

    i know that if piston 1 drops by a certain amount, piston 2 will rise proportionally in volume.... the answer i keep getting is ~0.17 meters. which isn't correct....
  5. Apr 10, 2008 #4
    Let's call the mass acting on the smaller piston M1. (M1=75.58 kg, if it's correct)
    Let's call the mass of the person M2.

    You have to consider that the total volume of oil remains constant.
    So, if a volume of oil in the big cylinder (car side) goes down because of the added mass of the person, the same volume of oil will go up in the other cylinder. You can calculate the relation between the two volumes because you know the areas of the two pistons.
    The new equlibrium requires that (eliminating "g" constant on both sides):

    M2 + mass of the volume of oil in the big cylinder =
    = M1 + mass of the volume of oil in the small cylinder.

    Take care of the signs (one of the oil levels goes down, the other up).
  6. Apr 11, 2008 #5

    Shooting Star

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    (The density of oil does not change.)

    Do you remember on what principle the barometer works? Remember Pascal's law?
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