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Homework Help: Force and Continuity?

  1. Mar 30, 2008 #1
    1. The problem statement, all variables and given/known data

    The small piston of a hydraulic lift has a cross sectional area of (2.8 cm^2), and the large piston connected to it has an area of (17 cm^2). What force of F must be applied to the smaller piston to maintain a load of (28000 N)?


    2. Relevant equations

    A_1 * V_1 = A_2 * V_2


    3. The attempt at a solution

    Since the volumes are the same, I set the A * F to be set equal to each other on both sides and solved....but I'm not seeing something in the unit conversion.....Idk what to do.
     
  2. jcsd
  3. Mar 30, 2008 #2
    Maybe it would help to think of it as Force/Area. You have the right idea setting it up as a proportion.
     
  4. Mar 31, 2008 #3

    rock.freak667

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    You need to convert cm^2 to m^2

    Remember that [itex]1cm=10^{-2}m[/itex] square both sides of that and you will get your conversion.
     
  5. Mar 31, 2008 #4
    OH it'd be F1/A1 = F2/A2, and ya I would have to convert to m^2, by X100 but squaring them??? you mean make it into flat out (m)?
     
  6. Mar 31, 2008 #5

    rock.freak667

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    [itex]1cm=10^{-2}m \Rightarrow 1cm^{2}=10^{-4}m^2[/itex]..squaring like that to get the conversion.
     
  7. Mar 31, 2008 #6
    Why convert, rock.freak667? The units in the area will cancel. OP will be left with just newtons.
     
  8. Apr 1, 2008 #7

    rock.freak667

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    oh both are in cm^2...my bad, I read it wrong, I thought one was in m^2 and the other was in cm^2..No need to convert then.
     
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