# Hydrostatic Force Problem In Calc. 2

by coco87
Tags: calc, force, hydrostatic
 P: 15 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. 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. $$w=2 \mbox{ m}$$ $$i=4 \mbox{ m}$$ $$l=12 \mbox{ m}$$ 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 $$12 \mbox{ m}$$. 2. Relevant equations $$\int\sqrt{a^2-u^2}du \Rightarrow \frac{u}{2} \sqrt{a^2-u^2} + \frac{a^2}{2}\sin^{-1}{\frac{u}{a}}+C$$ $$F = pgAd$$ $$p = 1000 \mbox{ }kg/m^3$$ $$g = 9.8 \mbox{ }m/s^2$$ Now, I am not 100% sure I am applying these formulas correctly. $$F$$ 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 $$p$$ as the density of water, and of course $$g$$ as gravity. 3. The attempt at a solution $$d = 12-2 = 10$$ $$x^2+y^2=(2)^2 \Rightarrow y=\sqrt{4-x^2}$$ $$A=\int_{-2}^{2}\sqrt{4-x^2}dx$$ $$\Rightarrow \frac{x}{2} \sqrt{4-x^2} + 2\sin^{-1}{\frac{x}{2}}=I$$ $$A=I(2)-I(-2)=6.283185$$ $$F=(1000)(9.8)(6.283185)(10)=615752.13 \mbox{ N}$$ Did I correctly apply the concept? If so, is my Arithmetic correct also? Thanks!

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