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Hydraulic Lift

  1. Jun 4, 2008 #1
    Anyone know how you get the formula F = rho*g*(A1+A2)d2 which is used to find the additional force needed to be applied to a piston to lift a car a distance of d2?

    Thanks.
     
  2. jcsd
  3. Jun 4, 2008 #2
    Well, although I don't know anything about pistons, with a little looking at the dimensions, you can notice that

    [tex]\frac{F}{A_1 + A_2} = \rho g d_2 = P[/tex]

    where [tex]P[/tex] is pressure.

    Edit: I just looked up Pascal's law, which states that

    [tex]\Delta P = \rho g \Delta h[/tex]

    where [tex]\Delta P[/tex] is hydrostatic pressure and [tex]\Delta h[/tex] is the height in the fluid. Your question probably has some relation to this, although, as I said, I don't know anything about pistons.
     
    Last edited: Jun 4, 2008
  4. Jun 4, 2008 #3
    OK here is the question that relates to the formula above:

    "The hydraulic lift at a car repair shop is filled with oil. The car rests on a 25-cm-diameter piston. To lift the car, compressed air is used to push down on a 6.0-cm-diameter piston. By how much must the air-pressure force be increased to lift the car 2.0 m?

    This is what I'm thinking: essentially, by Pascal's law, the pressure you give to the air-pressure force will be transmitted throughout the entire liquid. Thus, F1/A1 = F2/A2. The work done on both sides must also be the same F1d1 = F2d2 and finally because the liquid is incompressible, the volume's must be the same A1d1 = A2d2. I just don't know how to put everything together to get the formula.
     
  5. Jun 4, 2008 #4
    From what I gather, I don't think your formula is actually applicable here. In order to lift the car a distance [tex]\Delta h=2.0 m[/tex], the larger piston must do work equal to the weight of the car times [tex]\Delta h[/tex], and via conservation of energy, the smaller piston must do opposite work. In other words,

    [tex]F_g \Delta h_1 = - F \Delta h_2.[/tex]​

    where [tex]F_g[/tex] is the weight of the car. Unfortunately, the question doesn't seem to provide you with this information, so either there's another way to do it, I'm wrong, or they forgot to put it in.

    Edit: I just realized that h1 and h2 aren't the same, so I'll get back to you in a moment with a revised version of the above equation.
     
    Last edited: Jun 4, 2008
  6. Jun 5, 2008 #5
    Hi,

    I've actually figured it out, I can type up the solution later if you're curious. Thanks for your help anyways.
     
  7. Jun 5, 2008 #6
    I actually am curious.
     
  8. Jun 5, 2008 #7
    The extra force you exert is going to have a pressure of: F1/A1.

    Once the system establishes equilibrium again, the other side will match that pressure with one of: mg/A2 which is simply the total weight of the fluid above where the smaller piston is on the other side.

    mg/A2 = rho*A2*(d1+d2)*g/A2.

    Thus, F1 = rho*A1*(d1+d2)*g

    Because of conservation of mass, A1d1 = A2d2 (since the volume pushed down on one side equals the volume pushed up on the other) and

    d1 = A2*d2/A1 - sub this into the last equation and you get

    F1 = rho*A1*(A2*d2/A1 +d2)*g which you can then simply to rho*g(A1 + A2)d2.
     
  9. Jun 5, 2008 #8
    I'm not exactly sure how we could get that equation with the work argument, if you come up with something please explain.
     
  10. Feb 15, 2010 #9
    where did you get this from

    What equation originally is that? d1+d2?

    rho*A2*(d1+d2)*g/A2.
     
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