How Does a Sphere Act as a Valve in Fluid Dynamics?

[Feynmanfan]

Fluid mechanics problem.

In the picture below we have a sphere (consider it massless) acting as a valve in a water tank. At the bottom, there's a hole (which magnitude is determined by the angle theta.

I am asked to calculate the total vertical force acting on the sphere. And after that, when theta is (3Pi/4) what must the height h be so that the water doesn't flow through the hole.

I can calculate the pressure on any point of the sphere but how do I calculate this vertical force? Is it archimedes i have to use?

Thanks for your help.

 

[Dr.Brain]

Where is the picture?

 

[Feynmanfan]

sorry. I forgot the picture

 

[Feynmanfan]

do I have to integrate over the sphere? just give me a hint please.

 

[OlderDan]

Fluid mechanics problem.

In the picture below we have a sphere (consider it massless) acting as a valve in a water tank. At the bottom, there's a hole (which magnitude is determined by the angle theta.

I am asked to calculate the total vertical force acting on the sphere. And after that, when theta is (3Pi/4) what must the height h be so that the water doesn't flow through the hole.

I can calculate the pressure on any point of the sphere but how do I calculate this vertical force? Is it archimedes i have to use?

Thanks for your help.

Archimedes principle is equivalent to a pressure calculation over the surface of a submerged object, assuming pressure at any depth is uniform and increases with depth due only to the weight of the fluid. A floating object displaces its own weight of fluid because pressure at any depth exceeds the pressure at a lesser depth by the weight of a column of fluid of cross-sectional area A divided by that area. Under any other conditions, Archimedes principle is not going to hold.

If you know the pressure at all points on the surface of your sphere, you should be able to calculate the vertical force. It should have the form

[tex]\overrightarrow F = -\oint P \widehat n dA}[/tex]

if [tex]\widehat n[/tex] is the outward normal

The net force on the upper hemisphere will be vertically downward. The net force on the lower hemisphere will be vertically upward. If there were no drain, the lower pressure would exceed the upper pressure and the upward force would exceed the downward force by the weight of the displaced water. With the drain, the upward force will be reduced. If you don't add atmospheric pressure to the pressure at all levels, you can take the pressure over the region spanned by the drain opening as zero. This region results in lower upward force, and when it is just the right size the downward force and the upward forces will cancel. At that point the ball will still be displacing a lot of water, so what does that say about Archimedes principle?

do I have to integrate over the sphere?

Yes