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Huge problem on hydraulic lifts

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data
    the hydraulic lift at a car repair shop is filled with oil. the car rests on a 25cm diameter piston. to lift the car, compressed air is used to push down on a 6cm diameterpiston. By how much must the air pressure force b eincreased to life the car 2.0m.


    2. Relevant equations
    DeltaF = pg(A1+A2)h where p is the density (900 for oil) and h is the height


    3. The attempt at a solution
    please read this: I didnt use the above equation because it was just stated in the text book with no logic, so i used another equation derived from the logic of previous pages. We know F1/A1 = F2/A2 to keep the car up without lifting it a certain height, and to lift it a certain height, the equation is F1/A1 = F2/A2 + pgh where p is density and h is height. these equation are given in the textbook as well and they make sense. so to find the change in pressure, dont you just arrange the two equations for F1 and subtract them? Why is this wrong!!!!!
     
  2. jcsd
  3. Dec 9, 2008 #2

    mgb_phys

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    You have effectively increased the load by the weight of the column of oil.
    You know that the pressure in both pistons must be the same
     
  4. Dec 9, 2008 #3
    I have not effectively done this because my answer is incorrect!!!! the answer using their equation of DeltaF=pg(A1 + A2)h yeilds 920N while mine yields a MUCH smaller number. can you tell me how my approach is effective?
     
  5. Dec 9, 2008 #4

    mgb_phys

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    Isn't it just the pressure at the bottom of the extra oil column?
    Pressure in a column of incompressible liquid = [tex]\rho[/tex] g h
    = 900 kg/m^3 * 9.8 m/s^2 * 2m = 17 640 kg/m/s^2 = 17.6kPa
     
  6. Dec 9, 2008 #5
    yea that is what i got and i multiplied it by the area of the force pushed to get the change in the force. its not working and i am yielding a change of 49.9212N as you would be as well. the answer given is 920N!!! please help! my exam in in a couple hours!
     
  7. Dec 9, 2008 #6

    mgb_phys

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    Not sure what that means - "air pressure force" isn't very clear.
    The force to lift the liquid is, mass * g = rho g h * area
    900 kg/m^3 9.8 * 2 * 0.125m^2 pi = 865.9N

    If you used the sum of the areas
    900 kg/m^3 9.8 * 2 * (0.125m^2 pi + 0.03m^2 pi) = 917N but I don't see how this is justified.
     
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