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Thread: Hydrostatic Force Problem

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    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.

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

    $\displaystyle w=2 \mbox{ m}$

    $\displaystyle i=4 \mbox{ m}$

    $\displaystyle l=12 \mbox{ m}$

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

    To find the Hydrostatic Force, I did the following:

    $\displaystyle F = pgAd$

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

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

    $\displaystyle d = 12-2 = 10$

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

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

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

    $\displaystyle A=I(2)-I(-2)=6.283185$

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

    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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