Hydrostatic Force Problem In Calc. 2

[coco87]

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.

Homework Statement

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

http://sites.google.com/site/lcphr3ak/Home/prbm14.png

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

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

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

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 $$12 \mbox{ m}$$.

Homework 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.

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!

 

[mighty2000]

Thank you for your post. I am happy to help you with your problem. From what I can see, you have correctly applied the concepts of hydrostatic force and density to find the force acting against the semicircular gate. Your arithmetic also seems correct.

However, I would like to suggest a slightly different approach to solving this problem. Instead of using integrals, you can use the formula for the area of a semicircle (A = πr^2/2) to find the area of the gate. Then, you can use the formula for hydrostatic force (F = pgA) to calculate the force. This may be a simpler and more direct method.

Additionally, I noticed that you have used a value of 10 for the depth (d) in your calculation. However, according to the diagram, the depth should be 12 meters (d = 12 - 2 = 10). This may just be a typo, but I wanted to point it out to ensure the accuracy of your calculation.

Overall, it seems like you have a good understanding of the concept and have applied it correctly. Keep up the good work! If you have any further questions, please don't hesitate to ask.

Best of luck with your studies,