Hydrostatic force, half circle

1. Dec 3, 2008

kevtimc

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

Find the hydrostatic force on any side of a bottom half-circle with a 6 m diameter with the top 1 m above water level.

http://img377.imageshack.us/my.php?image=45095885io3.png"

2. Relevant equations

P = 1000 * 9.8 (mass density of water) * displacement
A = l * w
F = P * A

3. The attempt at a solution

Pressure = density * gravity * displacement
displacement = (2 - y)

x = $$2\sqrt{9 - y^2}$$
A = x * delta y (for any rectangular strip)

I'm concerned with my displacement; would it be 2 - y?
Would I be integrating from -1 to -3?

EDIT: I looked over the problem again, I think what we really have is a semi circle with a radius of 2 at sea level.
The area would be 2 * sqrt(4 - y^2) and the displacement just y, I think :)

Last edited by a moderator: Apr 24, 2017
2. Dec 3, 2008

HallsofIvy

Staff Emeritus
No, you do not have a semi-circle with a radius of 2: the problem specifically says a "half-circle with a 6 m diameter" so you have a semi-circle with a radius of 3 m, not two. The fact that the water only comes up to 1 m below the top means that the depth for any y (with $x= \sqrt{9- y^2}$) is y-1. You will want to integrate from y= 1 to y= 3.
(Note: I've taken my positive y axis downward to avoid negatives.)

3. Dec 4, 2008

kevtimc

Thanks, but I'm having trouble with the second integral in the problem:

$$2pg \int (y-1) * \sqrt{9-y^2}$$

For: $$2pg\int \sqrt{9-y^2}$$ from 1 - 3 (when integrals are seperated)

doesn't that evaluate to $$2pg \int9*cos^2\theta$$ ?

I'm not getting the correct answer when I evaluate this integral.

Last edited: Dec 4, 2008
4. Feb 13, 2011

lakeccrunner

I'm having a problem with the exact same problem. The answer should be 6.7x104N, but I have yet to get this answer.

I evaluate $$2pg\int(y-1)\sqrt{9-y^{2}}dy$$ from 1 to 3

And get: $$2pg\inty\sqrt{9-y^{2}dy - 2pg\int\sqrt{9-y^{2}}dy$$ from 1 to 3.

I then do u substitution for the first integral using $$u=9-y^{2}, du=-2y, -\frac{1}{2}du=ydy$$

For the second integral, since the are is a semicircle, I use $$2pg\pi3^{2}$$

So I have: $$-pg\inty\sqrt{u}du - 2pg\pi3^{2}$$ from 1 to 3.

Then I take the antiderive of the first integral to get

$$-pg\frac{2}{3}(9-y^{2})^{\frac{3}{2}} - 2pg\pi3^{2}$$

From here I plug in the numbers and get:

$$-147832-277088$$ which give me $$-424920$$ I then divide it by 2, since it says to find the Hydrostatic Force on one side of the semicircle to get $$2.1x10^{5}$$

I think I'm making a little mistake somewhere, but I can't seem to figure out where. Any help would be appreciated.