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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.

1. The problem statement, all variables and given/known data

A vertical dam has a semicircular gate as shown in the figure. Find the Hydrostatic force against the gate.

[tex]w=2 \mbox{ m}[/tex]

[tex]i=4 \mbox{ m}[/tex]

[tex]l=12 \mbox{ m}[/tex]

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

2. Relevant equations

[tex]\int\sqrt{a^2-u^2}du \Rightarrow \frac{u}{2} \sqrt{a^2-u^2} + \frac{a^2}{2}\sin^{-1}{\frac{u}{a}}+C[/tex]

[tex]F = pgAd[/tex]

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

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

Now, I am not 100% sure I am applying these formulas correctly. [tex]F[/tex] is what I'm assuming to be the Hydrostatic Force since the book states that it is, "The force exerted by the fluid on an area". And the book gives [tex]p[/tex] as the density of water, and of course [tex]g[/tex] as gravity.

3. The attempt at a solution

[tex]d = 12-2 = 10[/tex]

[tex]x^2+y^2=(2)^2 \Rightarrow y=\sqrt{4-x^2}[/tex]

[tex]A=\int_{-2}^{2}\sqrt{4-x^2}dx[/tex]

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

[tex]A=I(2)-I(-2)=6.283185[/tex]

[tex]F=(1000)(9.8)(6.283185)(10)=615752.13 \mbox{ N}[/tex]

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

Thanks!

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# Hydrostatic Force Problem In Calc. 2

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