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Fluid Pressures on a Vertical Surface, Oh My!

  1. Dec 10, 2004 #1

    All right, so, in Physics, we've been learning about pressure, and now we're tying it in with elements of calculus. Today, we began covering fluid pressures on vertical surfaces, and though I've understood most of it (8 out of 10 on a 10 problem assignment), this one problem has me stuck.

    You're supposed to find the fluid force on the vertical side of a tank full of water, where the dimensions of the side are given in feet. The side has the shape of a parabola, y=x^2, with width 4 and height 4, and you're to assume that the "top" of the parabola is at the surface of the water.

    Now, I'm not looking for you guys to do this for me, but only for a hint. I've just spent three hours on this, and I've still got a Physics paper on time travel to work on. >_> Our teacher hasn't really stressed the "formula" for fluid force of a vertical side, but instead has us relying on geometric analysis. I just need a little idea on where to begin, please? ^_^

  2. jcsd
  3. Dec 11, 2004 #2

    Doc Al

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    Here's a hint: The water pressure is just a function of depth. [itex]P = \rho g h = \rho g (4 - y)[/itex]. The force exerted on the "vertical" wall will be perpendicular to the wall at any point, but you can calculate the x and y components of the net force on the side independently.
  4. Dec 11, 2004 #3


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    Imagine a narrow HORIZONTAL rectangle, of width "x" and thickness "dy" (since the depth is measured in the y direction). The water pressure at each depth y is the density of water (a constant) times the depth (y). The force on that rectangle is the pressure times the area of the rectangle which is itself xdy. Both the depth and the width of the rectangle vary as y changes so you do that for each depth y and add them up. In the limit that becomes the integral: [itex][\integral \rho g(4-y)x(y)dy[/itex].
    Notice that "x(y)" the width x depends upon y: in your example y= x2 so
    x= 2√(y). (Do you see where the "2" came from?)
  5. Dec 11, 2004 #4

    Thanks both of you guys. Just one of those things I didn't see for a bit, probably because I was trying to work it from the bottom up, instead of the top down (if that makes any sense).

    It also didn't help that I was going from 0 to 3 in my integral (3 was from an earlier problem. *-_-). Anyway, I got the right answer, so thanks. ^_^

    Was a great first-time Physics Forums experience!

  6. Dec 11, 2004 #5
    One More...

    Ok, just one more quick question. The next problem features a semiellipse, under the same conditions, with equation y=-.5squareroot(36-9x^2), with width 4, and height three.

    Using information garnered from the parabola problem, I've worked it down into F = 62.4 (integral from 0 to 3) (squareroot (4-4y/9))(3-y)dy

    (I know that's messy...) Now, I'm just having problems solving the integral.
    -_- Any suggestions? Or did I get the integral wrong, too? >_<

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