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Homework Help: Force on the sides of a swimming pool, differeniation

  1. Feb 16, 2008 #1

    TFM

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    [SOLVED] Force on the sides of a swimming pool, differeniation

    1. The problem statement, all variables and given/known data

    A swimming pool measures a length of 5.7m , width 3.8m , and depth 2.9m. Compute the force exerted by the water against either end. Do not include the force due to air pressure.

    I have already calculated the force on the bottom as 6.2*10^5N

    Hint: Calculate the force on a thin, horizontal strip at a depth , and integrate this over the end of the pool.

    2. Relevant equations

    f = pgya

    3. The attempt at a solution

    pretend the swimming pool is on its end, using formula
    f = intergration,/' pgdya
    from 0 to 5.7

    getting F = 1/2 pg(y^2)a

    But this doesn't seem to work?

    Any help/suggestions/ideas

    TFM
     
    Last edited: Feb 16, 2008
  2. jcsd
  3. Feb 16, 2008 #2

    Mapes

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    Why doesn't it seem to work?

    Edit: I think you need to integrate more carefully.
     
    Last edited: Feb 16, 2008
  4. Feb 16, 2008 #3

    TFM

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    I get an answer of 1754395.02, which Mastering physics says is wrong. Any idea where in my integration I have gone wrong?

    TFM
     
  5. Feb 16, 2008 #4

    Mapes

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    I think I see the problem. Don't turn the pool on its end; that's changing the problem. We want to find the force on either end in its orginal configuration. This will affect a couple of your variables.
     
    Last edited: Feb 16, 2008
  6. Feb 16, 2008 #5

    TFM

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    Keeping the Pool in the same way, I get F = 892586.94
    However, if i ignore the 1/2 from the integration, I get 1785173.88

    Is this the answer, because there are two sides?

    TFM
     
  7. Feb 16, 2008 #6

    Mapes

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    I don't think so. Can you show how you integrated and what variables you multiplied together to get this result?
     
  8. Feb 16, 2008 #7

    TFM

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    I integrated:

    f = intergration,/' p*g*y*a dy
    and got
    F = 1/2 pg(y^2)a

    I then used
    p = 1000 (density of water)
    g = 9.8
    y = 2.9
    a = 5.7*3.8

    Giving me 892586.94

    TFM
     
  9. Feb 16, 2008 #8

    Mapes

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    Right away I can tell that the units are off: your units are N-m for a force answer, which is incorrect of course. That's why checking units is so useful is troubleshooting a problem.

    I agree with setting force equal to (density)(g)(depth)(area). However, it seems to me that your area should be a rectangular differential element at the end of the pool. This differential element would have a width of w and a height of dy. Does this make sense?
     
  10. Feb 16, 2008 #9

    TFM

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    I'm not sure...? Would it be:
    Area = width*length
    Area = ,/' width*length dlength
    giving area = 1/2 * width*length^2
    putting bvack into the equation,

    F = pgya

    F = pgy(1/2 * width*length^2)

    If not, then I'm don't understand, sorry:blushing:

    TFM
     
  11. Feb 16, 2008 #10

    Mapes

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    How about [itex]dF=pgy\,da=pgy(w\,dy)[/itex], [itex]F=\int_0^{\mathrm{depth}}dF=\frac{1}{2}pgw(\mathrm{depth})^2[/itex]?
     
  12. Feb 16, 2008 #11

    TFM

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    That makes sense - I was still using the area as being the floor. I now get an answer of 15979, alkthough this is ten times smaller than the force on the floor? (and apparantly isn't the answer)
    TFM
     
    Last edited: Feb 16, 2008
  13. Feb 16, 2008 #12

    Mapes

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    x g?
     
  14. Feb 16, 2008 #13

    TFM

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    D'oh

    With g in, gives right answer:

    156594.2

    Although Mastering Physics only accepts 160000

    Thanks, Mapes

    TFM
     
  15. Feb 16, 2008 #14

    Mapes

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    You're welcome, good luck.
     
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