# Hydrostatic Force Problem

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• Mar 22nd 2009, 04:18 PM
Coco87
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.

$
\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}
$

$w=2 \mbox{ m}$

$i=4 \mbox{ m}$

$l=12 \mbox{ m}$

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!
• Mar 24th 2009, 03:30 AM
Coco87
Bump (Itwasntme)