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Finding the force of water in a pool on a vertical wall

  1. Dec 14, 2009 #1
    1. The problem statement, all variables and given/known data
    A swimming pool has the dimensions 24 m multiplied by 9.0 m multiplied by 2.5 m.

    (a) When it is filled with water, what is the force (resulting from the water alone) on the bottom, on the short sides, and on the long sides?


    2. Relevant equations

    F = P*A

    P = rho * g * h

    3. The attempt at a solution

    Ok, I have no idea what I am doing wrong, but I am trying to find the force of water on the short vertical side of the pool.

    So, since pressure varies by height, I have to intergrate the pressure by height (h).

    So P = [tex]\int\rho g dh[/tex]

    and A = h*w where w = width of the pool

    So F = [tex]\int\rho g dh h*w [/tex]

    then you end up with [tex]\rho*w*g*(h^2)/2[/tex] from 0 to 9

    and get 992250 Pa

    I'm pretty sure what I did was correct, but im not getting the right answer.

    Can anyone help?
     
  2. jcsd
  3. Dec 14, 2009 #2

    ideasrule

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    Homework Helper

    Seems good to me. Are you sure you didn't get your dimensions mixed up? That is, are you sure you used the right values for w and h?
     
  4. Dec 14, 2009 #3

    cepheid

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    Gold Member

    Well, for one thing, 2.5 m is the only sensible choice for the depth of the pool! If either of the other two is the depth, then that is one strange pool.
     
  5. Dec 14, 2009 #4
    Why would 2.5 be the sensible height? i thought it would be the width of the pool.
     
  6. Dec 14, 2009 #5
    Who has swum in a pool of 30' depth?

    Edit: one other thing bothers me about the solution, the pressure at 0' depth is not 0 pressure but atmospheric pressure which I think should be accounted for if we are interested in the total pressure exerted on any of the surfaces, or am I confused?
     
  7. Dec 14, 2009 #6
    Oh, I guess that is rather large.......

    But yes, you are right, atmospheric pressure is the pressure at the surface, but the problem states to not take that into account.
     
  8. Dec 14, 2009 #7
    very well, I would try using 2.5 as the depth and 9 as the width as Cepheid suggests. See if you get the right answer
     
  9. Dec 14, 2009 #8
    I did and came out with the right answer. Thanks a lot!!!!
     
  10. Dec 14, 2009 #9
    Thank Cepheid, he caught the error, but you're welcome.
     
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