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Force in Varying Places in a Swimming Pool

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data
    A swimming pool has dimensions 25 m x 10 m x 3m (length x width x height.) When it is filled with water, what is the force on the bottom (Fb) of the pool? On the long side (Fl)? On the short sides (Fw)? (note that integrals are required.) If you are concerned with whether or not the concrete walls and floor will collapse, is it appropriate to take the atmospheric pressure into account?

    2. Relevant equations

    d = m/v
    F = ρA

    3. The attempt at a solution
    I'm having a lot of trouble setting up the integral. I legitimately am kind of lost on how to do it. Would you just do

    F = ∫ ρA (from the top of the pool to the bottom). And would you use the density of the entire pool for ρ????
     
  2. jcsd
  3. Apr 20, 2015 #2

    SteamKing

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    Hint: Use Pascal's Law for this problem.

    And how is the density of the entire pool different from the density of any random unit volume of water in the pool?
     
  4. Apr 20, 2015 #3

    haruspex

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    Are you using ρ for pressure, density, or mixing the two?
     
  5. Apr 20, 2015 #4
    Ok, so I tried again:

    Fbottom = ∫ (p0 + ρdensity*l*w)dh
    =p0 * l * w * h + ρdensity*l*w*h
    for the bottom of the pool, the integral doesn't matter so much.

    ...but...I'm confused. How do you find p0 (which is pressure at the surface of the liquid due to the atmosphere) and even if you do find it, the units won't add up?
     
  6. Apr 20, 2015 #5

    haruspex

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    That's because this term is wrong:
    ρdensity*l*w.dh gives a mass.
    What equation do you know for pressure at a depth in a liquid?
     
  7. Apr 20, 2015 #6
    only p = p0 + ρdensity*g*h. It's literally the only equation my book gives.
     
  8. Apr 20, 2015 #7

    haruspex

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    That's right, but it's not what you posted before.
    p0 also features wrongly in these two equations:
    The first one would give p0's contribution to the force at the bottom as p0h, a pressure multiplied by a distance, while in the second you have a pressure multiplied by a volume.
    The force at the bottom should be easy, no need for integrals. Take the whole body of water as a unit. What are the vertical forces acting on it?
     
  9. Apr 20, 2015 #8
    The vertical forces would just be the atmosphere, and the weight of water above it?

    But what about the sides?
     
  10. Apr 20, 2015 #9

    haruspex

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    One more... what stops it descending?
    Those would be horizontal forces.
     
  11. Apr 20, 2015 #10
    What do you mean by "what stops it descending"?

    How do I solve for that?
     
  12. Apr 21, 2015 #11

    haruspex

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    I asked about the forces acting on the body of water. You mentioned atmosphere and its own weight (gravity). Both of those act downwards.
    However, I guess you instead tried to answer what forces act on the floor of the pool. The short answer to that is the pressure of the water. (The pool floor doesn't know anything about the atmosphere, and feels no direct force from it. The force is transferred by the water.)
    But in the end, yes, the force on the floor is the sum of the force from atmospheric pressure and the weight of the water. So what do those add up to?

    Note that this shortcut works because the pool has vertical sides. The more correct approach is to figure out the total pressure at the floor of the pool (using the equation you posted in post#6) and multiply by the pool floor area. That works for any pool profile.
     
  13. Apr 21, 2015 #12
    But
    Fbottom = 32349779.5 N
     
  14. Apr 21, 2015 #13

    Does that mean that
    Fl = ∫ (p0 + ρ*g*h)A*dl
    = p0 ∫A * dl + ρ*g*h∫A dl
    ...None of those units add up?
     
  15. Apr 21, 2015 #14

    haruspex

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    Are you saying that's the given answer or what you calculated? Looks about right to me. Please post your working.
     
  16. Apr 21, 2015 #15

    haruspex

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    p0 + ρ*g*h is the pressure at the bottom.
    (p0 + ρ*g*h)A is the force on the bottom.
    What does it mean to integrate that dl (whatever l is)?
     
  17. Apr 21, 2015 #16
    F = (p0 + ρ*g*h)A
    = A * (1.0 x 105 kg/(m*s2) + 999.97 kg/m3 * 9.8 m/s2 * 3 m)
    = 250 m2 * 129399.118 kg/(m*s2)
    = 32349779.5 N
     
  18. Apr 21, 2015 #17
    I don't know. I figured I would integrate with respect to length here.
     
  19. Apr 21, 2015 #18

    haruspex

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    Yes, except that the precision makes no sense. The atmospheric pressure is only accurate to two or three decimal places since it varies, so an answer should look more like 3.23*108N.
    By what logic? And what do you mean by length here... the length of the pool or its depth? And what are you trying to calculate with this integral (i.e. what do you mean by Fl)?
     
  20. Apr 21, 2015 #19
    Well, part of the question asks to find the Force on the long side of the pool (Fl). I only really guessed here.

    And the precision thing makes sense. It's just, the integrals don't.
     
  21. Apr 21, 2015 #20

    haruspex

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    For the forces on the sides you will need integrals, but you can't use the force on the floor as the integrand. The force on the side won't depend on the pool area.
    What is the pressure at depth x? What force does that exert on a wall over a horizontal strip of height dx?
     
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