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Math Help - Hydrostatic Force Problem

  1. #1
    Junior Member Coco87's Avatar
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    Hydrostatic Force Problem

    Hey,
    I've done an even problem from my book, and am not 100% sure if it's correct. It asks to find the Hydrostatic Force acting against an Area.

    <br />
\setlength{\unitlength}{0.30cm}<br />
\begin{picture}(13,12)<br />
\put(1,0){\line(1,0){12}}<br />
\put(13,0){\line(0,1){12}}<br />
\put(13,12){\line(-1,0){12}}<br />
\put(1,12){\line(0,-1){12}}<br />
\qbezier(5,0)(5,2)(7,2)<br />
\qbezier(7,2)(9,2)(9,0)<br />
\put(1,10){\line(1,0){12}}<br />
\put(6.8,0.5){$i$}<br />
\put(0,5.5){$l$}<br />
\put(11.5,10.6){$w$}<br />
\end{picture}<br />

    w=2 \mbox{ m}

    i=4 \mbox{ m}

    l=12 \mbox{ m}

    This figure represents a Dam. The top rectangle represents... I guess air? The w is for height, at 2 \mbox{ m}. The lower rectangle is the water. The half circle at the bottom is the Object will act as the area. The i in the circle is the diameter of the half circle, which is 4 \mbox{ m}. And last, but not least, l represents the entire length of the Dam, which is 12 \mbox{ m}.

    To find the Hydrostatic Force, I did the following:

    F = pgAd

    p = 1000 \mbox{ }kg/m^3

    g = 9.8 \mbox{ }m/s^2

    d = 12-2 = 10

    x^2+y^2=(2)^2 \Rightarrow y=\sqrt{4-x^2}

    A=\int_{-2}^{2}\sqrt{4-x^2}dx

    \Rightarrow \frac{x}{2} \sqrt{4-x^2} + 2\sin^{-1}{\frac{x}{2}}=I

    A=I(2)-I(-2)=6.283185

    F=(1000)(9.8)(6.283185)(10)=615752.13 \mbox{ N}

    Did I correctly apply the concept? If so, is my Arithmetic correct also?

    Thanks!
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    Junior Member Coco87's Avatar
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