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Find the volume of the piston

  1. May 19, 2005 #1
    Hello! I was wondering if you could help me with a homework problem:

    The piston of a hydraulic automobile lift is X meters in diameter.

    What gauge pressure, in pascals, is required to lift a car with a mass of Y kg?


    I know that a piston is a cylindrical shaped apparatus - however I believe that does not affect my calculations.

    I also know this about pressure: p = F/A

    F = mass x a so in my case that would be: Y kg x 9.8m/s

    A (of a circle) = 2 x pi x r^2 r being half my diameter so X/2


    From using this formula, the pressure is calculated in Pascals. The answer I get however is wrong. Am i using the wrong formula? should find the volume of the piston? I'm not sure how that would help... hope someone can give me some hints! =)
     
  2. jcsd
  3. May 19, 2005 #2
    If you are considering the pressure on just the bottom of the piston (or top) then you only need to consider the area of one circle, not two.
     
  4. May 19, 2005 #3
    that's right, i should only consider the area of one circle. that "2" in front of pi r squared was a typo... =) i've been trying to calculate it assuming it's one circle but i'm doing something wrong. =/
     
  5. May 19, 2005 #4
    Is your final result

    [tex] \frac{4(Y kg)(9.8 m/s^2)}{\pi x^2}[/tex]?
     
  6. May 19, 2005 #5
    Why is the equation being multiplied by 4?
     
  7. May 19, 2005 #6
    The 4 comes from the denominator, since you are squaring X/2, you get x^2/4, and instead of having stacked fractions, you can multiply top and bottom by 4 to get what I gave.
     
  8. May 19, 2005 #7
    My final equation looks like this:


    (Y kg)*(9.8 m/s^2) / (pi)*(r^2) = p (pascals)
     
  9. May 19, 2005 #8
    regarding the equation that I am using (p = F/A), does that even look like the correct equation I need to solve this type of problem?
     
  10. May 19, 2005 #9
    Express it in terms of x and y, I think thats what you need to do. Is this online homework? Is that the complete problem?
     
  11. May 19, 2005 #10

    OlderDan

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    Yes it is the right equation. Your first calculation of area was wrong. If you get the area right, I think you will have it.
     
  12. May 19, 2005 #11
    yes, it is online hw and that is the complete problem except "x" and "y" there are real numbers. so instead of multiplying by 4 i just divided x by two before I plugged it into the equation.
     
    Last edited: May 19, 2005
  13. May 19, 2005 #12
    Did you divide correctly? ;)
    Lets see the numbers.
     
  14. May 19, 2005 #13
    got it.. i WAS getting the area wrong.. i was imputing it incorrectly into the calc. thanks so much for your time and patience!!
     
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