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Homework Help: Fluid force in a pool

  1. Dec 7, 2004 #1
    Let us define a pool. Viewed from top, we see a rectangle of sides 16 ft and 10 ft. When we look from the side (so the 16 ft side is perpendicular to our view), it is a trapezoid with width 16 ft and 4 ft, 16 ft being the height. It is roughly like this:

    From top:

    Code (Text):

              16 ft
        ------------------
        |                |
        |                |
        |                |  10 ft
        ------------------
     
    From one side (and the other opposite side):

    Code (Text):

              16 ft
        ------------------
    4ft |                |
         ....            |
             ....        |  8 ft
                 ....    |
                     ....|
     
    The two other sides are rectangular (4 ft x 10 ft and 8 ft x 10 ft). The floor (OK, it's also a "side") is a rectangle with dimension 10 ft x [itex]\sqrt{16^2+4^2}[/itex] ft.

    Let the pool be filled completely with water with weight density [itex]\pi[/itex] = 62.4 lb/ft^3.

    The question is: what is the fluid force on the bottom of the pool?

    Let us use the second image for this discussion. I define an vertical x axis with x = 0 located at 4 ft below the water surface and the positive x is below x = 0. I define h(x) = 4 + x as the height of water surface at a particular x location and w(x) = 10 as the width of the pool.

    To find the fluid force F, I use the equation

    [tex]
    F = \int_0^4\pi h(x)w(x)\sec t\;dx
    [/tex]

    Where t is the angle between the x axis and the inclined pool floor.

    Since sec t = [itex]\sqrt{16^2+4^2}/4[/itex],

    [tex]
    F = \int_0^4 (62.4) (4 + x) (10) \frac{\sqrt{16^2+4^2}}{4} dx
    [/tex]

    However the answer given by my book isn't the same as mine (the book's answer is bigger). Since I ruled out the possibility of miscalculation the integral, it must be my formulation of the integral (or the book's answer is wrong).

    However I can't see what's wrong with my integral... I tried doing it "low level", formulating the Riemann Sums first, but it results in the integral I use.

    What's wrong?

    Thanks a lot.
     
  2. jcsd
  3. Dec 8, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I don't understand your choice of variable. I would pick x as measured along a horizontal axis, from x = 0 to x = 16. Then the water depth is h(x) = 4 + x/4.

    If you rewrite your integral using the variable x as I have defined it, you should be OK.
     
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