Does Buoyancy Apply When Not Fully Submerged?

[anuttarasammyak]

TL;DR

Is buoyancy still applicable in the case that surface of the body is not fully covered with fluids ? By buoyancy I would say is buoyant force = weight of displaced fluid .

I am discussing with friends whether concept of buoyancy is applicable even if the full surface of body is not covered by fluid(s).

As an example a clay ball is dropping in the pool water. Buoyancy is working on the ball. Then it reaches the pool flat and sticks with it to make a mountain shape. Does buoyancy still work on the clay mountain though upward water pressure does not work on its bottom surface ? Without the bottom upward pressure is the mountain is pressed by water pressure than when it was dropping in the water ?

I am skeptical to apply buoyancy to such a not-full-surface is covered by water case. We should rely on simple area integration of pressure to get force on the body. Your teaching will be appreciated.

EDIT By buoyancy I would say is buoyant force = weight of displaced fluid .

 

[Vanadium 50]

If you are swimming underwater and your pinky leaves the water do you immediately sink link a stone?

For that matter, what keeps ships afloat?

 

[Halc]

Say the (approx 1KG) clay wad weighs 10N dry, but 2N in water. It sinks in a meter deep pail.
At the bottom the pressure is 1N/cm². There's a 10cm * 10cm surface down there with a scale measuring it. It reads 100N with just water. When the clay wad just touches it, it's going to weigh 102N, most of that being water pressure. Now we spread it out into a square that size so it's all clay now.
The scale will still read 102. Any other figure and I can make a perpetual motion machine.

You can interpret this as 2 for the (still buoyant) clay and 100 for the water pressure, or as 10 for the now unsupported clay and 92 for the water pressing on the top of the wad. Either way you get the same answer, but one interprets it as buoyancy and the other does not.

Personally I'd vote for the non-buoyant interpretation, else it is hard to explain a cup of air inverted over an open-air drain. It will not float upward despite it weighing less than the water it displaces.

 

[Dale]

a cup of air inverted over an open-air drain. It will not float upward despite it weighing less than the water it displaces

This is the example i was thinking of also

 

[anuttarasammyak]

You can interpret this as 2 for the (still buoyant) clay and 100 for the water pressure, or as 10 for the now unsupported clay and 92 for the water pressing on the top of the wad. Either way you get the same answer, but one interprets it as buoyancy and the other does not.

Thanks. I also prefer non buoyant interpretation.

Is the buoyancy interpretation compatible with the mathematics of buoyancy
\int \mathbf{p}\cdot\mathbf{ds}=\rho Vg
though we are not sure about pressure working on bottom of the mountain ? I do not think so.

For confirmation of understanding, say density of the clay is less than that of water, we push down the clay ball against the thrust to touch the pool floor, deform it a mountain shape and leave our hands. Does buoyant force separate the clay mountain from the floor ?

 

[Halc]

For confirmation of understanding, say density of the clay is less than that of water, we push down the clay ball against the thrust to touch the pool floor, deform it a mountain shape and leave our hands. Does buoyant force separate the clay mountain from the floor ?

I can throw a small blob of clay onto the ceiling and it will stick there. Clay may stick to the pool floor for the same reason, but assuming a sort of no-stick surface, the low-density clay will buoy up (float) to the surface in short order. The pressure exerted by the water trying to get under the clay is greater than the pressure downward by the clay keeping the water out.

If it's a big enough clay wad (hill), the clay even if stuck to the bottom will simply deform upward like a semi-solid lava lamp.

 

[anuttarasammyak]

I agree with you in case water get under the clay buoyant force of
\int p \ \mathbf{dS} works accordingly.
If we can keep dry touch with no penetration of water ( though I am not sure how it can be done ), no upward but only downward force would be applied from water, we cannot apply buoyancy to the mountain clay block, density of which does not matter.

Thank you so much.

cf
I found in web the paper titled Steady State Vapor Bubble in Pool Boiling
https://www.nature.com/articles/srep20240#MOESM1
with interest how generated vapor bubble get buoyancy in leaving the hot plate.
Even in wet contact case bubble is stable. Other factors, e.g. surface tension, should be considered.

 

[A.T.]

If we can keep dry touch with no penetration of water ( though I am not sure how it can be done ), no upward but only downward force would be applied from water, we cannot apply buoyancy to the mountain clay block, density of which does not matter.

Yes, if you can seal the contact to the wall, then you will have a large net force towards the wall from the water, depending on the absolute value of the hydro-static pressure. Usually far greater than the buoyant force without the seal, which depends on the gradient of hydro-static pressure.