Re: [Phys-l] buoyancy on a submerged pole

[Hugh Haskell <haskellh@verizon.net>]

At 14:52 -0400 11/05/2010, Chuck Britton wrote:
> At 1:41 PM -0500 11/5/10, Rauber, Joel wrote:
> > This will be a repeat somewhat of John M's comment: but I would ask
> > Chuck B. two questions
> >
> > 1) This may sound silly, but what is your definition of submerged?
> > As it may have bearing on how I evaluate your statement below.
>
> Underwater
>
> > 2) What is the agent of the UPWARD force to which you refer?
> >     (I'm assuming we are analyzing the situation in a "Newtonian"
> > inertial reference frame.)
>
> Pseudo Forces don't have agents, they are free-agents!
So far, more experiments have been proposed or thought about or 
carried out than were needed to establish the theory of relativity, 
but let's try one more thought experiment (but one that could, in 
principle, be carried out):

Construct an aquarium out of a plastic material that is less dense 
than water, and then provide internal support for the bottom so that 
when the aquarium is suspended any sagging of the bottom due to its 
own weight is at least very difficult to detect. Now fill the 
aquarium with water. if Chuck's hypothesis is correct, then we should 
see the bottom of the aquarium bulge upward into the water, or at 
least bulge downward less than it would if an equal weight of some 
solid material denser than water was placed in the aquarium (to avoid 
adding to the complexity of the description, we will assume that its 
bouyancy in air is negligible compared to that in water).

Then restructure the aquarium so that the bottom has a cylindrical 
block in the center of the bottom (made of the same material as the 
rest of the aquarium and as an integral part of the bottom, so that 
there is no seam between the bottom and the cylindrical block). The 
internal structure is again installed so that the bottom of the empty 
aquarium has no measurable sag. Again, we fill the aquarium with with 
water and measure any sag, upward or downward.

Finally, cut the cylinder away from the bottom of the aquarium and 
put it in the original aquarium but now simply tie it to the bottom, 
so that it can try to float, and repeat the sagging measurements on 
the bottom. I will now bet that the sagging of the bottom will be 
considerably less, if earlier measurements found downward sagging, or 
more, if earlier measurements found upward sagging. than what was 
measured in either of the first two cases.

I assert that this will show that any buoyant force on an object 
immersed in a fluid arises purely from the difference in pressure 
between the upward-facing surfaces of the object and the lower-facing 
surfaces of the object, when integrated over the entire surface of 
the object, where the pressure forms a vector field at the surface of 
the object that is pointing inward and is everywhere perpendicular to 
the surface and proportional to its depth beneath the surface of the 
fluid, and the integration is done in the same manner as one would 
carry out a Gauss's Law calculation on an arbitrarily-shaped object 
in and electromagnetic or gravitational field (the difference in this 
case being that the integration will never yield a null result, 
unless the density of the object is equal to the density of the water.

If part of the surface of the object is in contact with the bottom of 
the container in such a way that there is no contact between that 
part of the surface of the object and the water, then the net flux of 
the water pressure will be changed accordingly--reduced if this part 
of the surface is downward-facing, increased if it is upward-facing.

If there is no downward-facing surface of the object in contact with 
the water, then there can be no buoyant force on the object--the 
agent of the buoyancy of the object being the integrated water 
pressure at the surface of the object.

If the object is porous or its surface is rough, then the problem 
becomes much more complicated and the results become difficult to 
predict (e.g., will a saturated sponge float or sink?).

Since surface tension effects also complicate the picture, I have 
assume the object in question is always totally immersed.

For what its worth, this thread may have set a new record for 
activity on this list. And it's all due to the obstinacy of one 
friend of mine who shall remain nameless, but who should know better, 
considering how much physics he has taught me over the years >:-}.

Hugh
--

Hugh Haskell
mailto:hugh@ieer.org
mailto:haskellh@verizon.net

It isn't easy being green.

		--Kermit Lagrenouille