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Help please f= PxA

  1. Oct 13, 2011 #1
    hey everyone, can someone please help me out with this one, im a bit stuck with this
    using the law F = PXA and F = mxa

    calculate the pressure in a hydraulic cylinder, when the load of 375kg, is applied to the piston
    piston diameter it 95mm

    so far what i have worked out is as follows
    force on the piston is:
    F = M x 9.81
    = 250x9.81 = 2452.5N

    now for the area
    pi x rxr
    = 3.1416 x 37.5x37.5
    =4417.86 mm2

    now if im correct i need to change the mm2 to m2?
    so...
    4417.86mm2 /1,000,000
    =0.00441786m2

    therefore....
    f=2452.5N
    A=0.00441786m2
    P = F/A ?
    555,133.028p
    or 555.133kp

    is this correct guys or am i way out?
    if anyone could help me out with this it would be greatly appreciated
    thansk





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  2. jcsd
  3. Oct 13, 2011 #2

    gneill

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    Staff: Mentor

    You might want to check the value that you used for the radius of the cylinder.

    Other than that your method looks fine.
     
  4. Oct 13, 2011 #3

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    If the load is 375 kg, why did you use 250 kg in your calculations?
     
  5. Oct 13, 2011 #4

    gneill

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    Staff: Mentor

    Good catch SteamKing. I missed that second error!
     
  6. Oct 17, 2011 #5
    Actually , all your work is messed up , the pressure inside the cylinder would be the summation of the external force and the weight of the piston divided by the area of the piston..

    So:

    P= ( Fex + W ) / A

    A= 3.14 * d^2 / 4 = 3.14 * (0.095)^2 / 4 = 0.0071 m^2

    Fex = 375 * 9.81 = 3.44 k N

    ..and since you didn't mention the mass of the piston I'll assume its negligible

    so now:

    P = 3.44 / 0.0071 = 484.5 kPa
     
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