# Buoyancy of a sealed hemisphere underwater

Nathan B

## Homework Statement

You have a hemisphere, like half a ping pong ball for example, sitting in a cup under a few centimeters of water. The hemisphere is sealed to the bottom so that no water can get underneath it. What is the buoyant force the hemisphere experiences?

P = P0 + ρgh

## The Attempt at a Solution

So the hemisphere displaces water, which would lead me to believe that there would be an upward buoyant force on the object, but as I understand it force vectors due to pressure are always perpendicular to the surface, leading me to think there would be a net downward force. Which is it?

haruspex
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## Homework Statement

You have a hemisphere, like half a ping pong ball for example, sitting in a cup under a few centimeters of water. The hemisphere is sealed to the bottom so that no water can get underneath it. What is the buoyant force the hemisphere experiences?

P = P0 + ρgh

## The Attempt at a Solution

So the hemisphere displaces water, which would lead me to believe that there would be an upward buoyant force on the object, but as I understand it force vectors due to pressure are always perpendicular to the surface, leading me to think there would be a net downward force. Which is it?
Archimedes' principle only applies to bodies for which the portion below the surface of the fluid is completely surrounded by it. As you note, if the fluid is unable to reach some of that then the buoyant force may be different.
In the present case, can you think of a way to calculate how much less?

conscience
Hello haruspex ,

As you note, if the fluid is unable to reach some of that then the buoyant force is less.

Is there any buoyant force at all ?

Assuming it is a thin spherical shell and water is absent on the concave side , I believe water pushes the shell down .Even if air at atmospheric pressure is present beneath the shell , and assuming the fluid is exposed to atmosphere , water still exerts a net downward force on the shell .

You can calculate how much less by considering what the pressure of the fluid on the base would be if it could reach it.

For that we need radius of the concave side .

haruspex
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Is there any buoyant force at all ?
The net force will be downward in this case, but for, say, a sphere with a cap removed there may still be a net upward force. Regarding the second as buoyant and the first as not conforms with everyday usage, but in a scientific context I feel it is more useful to allow that a "buoyant" force can be net downward.
For that we need radius of the concave side .
No. Would the net force be any different if we were to close off the hemisphere with a disc?

No. Would the net force be any different if we were to close off the hemisphere with a disc?

If hemisphere is closed with a circular disc such that water is beneath the disk , there would be a net upward force on the disk given by P(πr2) .P is pressure at the bottom of water .

In case disk is not present and water is allowed to be present on the concave side , we need to integrate . For calculating the force exerted by water on the concave side we would require it's radius .

haruspex
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If hemisphere is closed with a circular disc such that water is beneath the disk , there would be a net upward force on the disk given by P(πr2) .P is pressure at the bottom of water .
There would be that force, but that would not be the net force. What would the net force be?
In case disk is not present and water is allowed to be present on the concave side , we need to integrate
No, just apply Archimedes' principle.

No, just apply Archimedes' principle.

If there is a hemispherical shell completely submerged in water then the net upward force due to water will be equal to the weight of volume of water displaced by it . Isn't volume of shell (2/3)(πR13 - πR23 ) . R2 is inner radius .

haruspex
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If there is a hemispherical shell completely submerged in water then the net upward force due to water will be equal to the weight of volume of water displaced by it . Isn't volume of shell (2/3)(πR13 - πR23 ) . R2 is inner radius .
Right.

Right.

But this is what I said earlier to which you objected
For that we need radius of the concave side .

No

haruspex
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But this is what I said earlier to which you objected
Look at my post #6 carefully. What, exactly, did I say no to?

conscience
Look at my post #6 carefully. What, exactly, did I say no to?

Oops ! I misunderstood you . You objected to the integration . I thought you said inner radius was not required .

Nathan B
Follow up question from reading the responses, if you did have a submerged ball with variable pressure due to depth what would the double integral of pressure and area look like? The area part would be ∫2πrdr, but how about pressure varying with depth?

haruspex
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Follow up question from reading the responses, if you did have a submerged ball with variable pressure due to depth what would the double integral of pressure and area look like? The area part would be ∫2πrdr, but how about pressure varying with depth?
You would be integrating force, a vector, so you have to take direction into account. Archimedes' Principle is a very effective shortcut when it can be applied.
Can you answer my question in post #2?

Nathan B
In answer to your question, the difference could be calculated by subtracting the for force due to pressure on the underside if it were submerged. In other words, there is no buoyant force in the the traditional sense, there's just the force due to gravity (causing pressure) above it.

haruspex