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Hydraulic lift help

  1. Jan 23, 2004 #1
    A hydraulic lift has two connected pistons with cross-sectional areas 25 cm2 and 420 cm2. It is filled with oil of density 730 kg/m3.

    a) What mass must be placed on the small piston to support a car of mass 1000 kg at equal fluid levels?
    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?
    c) How much did the height of the car drop when the person got in the car?
    *************
    a) Simple. Did F1/A1=F2/A2. Solved for mass.
    m1*g/.0025m^2=m2*g/.0420m^2

    m2 = 1000kg
    solve for m1 i get 59.523kg

    b) Ok. With the use of bernoulii's equation, the velocity is zero, so I get p1+rho*g*h1=p2+rho*g*h2
    so (h1-h2)=(p2-p1)/(rho*g)

    where p = pressure
    A2=.0420m^2
    A1=.0025m^2
    p2= (mass car+guy)*g/A2
    p1= 59.523kg*g/A1

    sovle for h1-h2
    I get 2.2835m, which is right.

    c) Okay, I was thinking find the differnce in height when its only the car, then find the difference in heigh when guy is in car. Take the two answers, and subtract them from each other to get the answer, but its not working right.
     
  2. jcsd
  3. Jan 23, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    One side goes down (H1), one side goes up (H2). You know the difference in height, H = H1 + H2.

    Now make use of the fact that the volume of the fluid (oil) must not change. This will give you enough info to solve for the individual height changes.
     
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