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# Re: [Phys-l] Definition of upthrust or buoyancy

• From: John Denker <jsd@av8n.com>
• Date: Thu, 21 Oct 2010 19:04:46 -0400

Starting from the buoyant object held down by a string:

1) If you replace the string by a thin hollow empty tube (such
as a soda straw) it will be in compression in the X and Y
directions even while it is in tension in the Z direction.
Indeed, under a wide range of conditions, it could get
crushed in the radial direction even while it is under
tension in the Z direction.

Remark: As the situations get more intricate, we lose the
ability to describe the whole situation using simple one-word
expressions such as "tension" and "compression".

2) This directionality is completely nontrivial, as we can
see in the case of a buoyant object held down by _putty_.
I suspect this is getting rather close to the intent of the
original question.

In the case where the putty extends over a large area and
the buoyancy is not very large, the putty will be in
compression in the Z direction, with perhaps thousands of
pounds of compressive force.

However ..... This compression in the Z direction is for
practical purposes nullified by the compression in the X
and Y directions, in the following sense: The putty, by
definition, is free to creep. If it creeps inward by 1%
in the X and Y directions and outward by 2% in the Z
direction, there is no change in volume of the putty, and
therefore no "P dV" work. Meanwhile, there is a negative
amount of "F dh" work, where F is the buoyant force and
h is the height of the buoyant object. So this creep is
energetically favored. It is favored even though it moves
against the multi-thousand-pound compressive force!

3) This gets even more relevant if we replace the putty
with plain water. If you push down on the buoyant object
you can (maybe!) push the water out from between the
object and the solid thing beneath it. This (theoretically)
requires only a modest force to overcome the net buoyancy.
However, if you stop pushing, the water will come back in
and the object will float away. It will float away even
though there was a multi-thousand-pound solid/solid contact
force pushing the object against the other thing while the
two were in contact.

This solid/solid contact force is entirely real, but you may
not be able to count on it, because of the strong tendency
of the water to weasel its way into cracks. You are going
to need some good glue (not putty) or a good O-ring seal
or something like that, if you want to keep the water out.

On 10/21/2010 06:53 PM, Philip Keller wrote:

I have a wooden beam, circular cross-section. I install it into a
concrete base inside a caisson (is that the big pipe they use for
in-water projects?). The concrete sets and we allow water to fill in
and completely cover the beam so that it is well below the surface of
the water. I believe that the beam will shrink due to the pressure x
area at the top and that if the water was high enough above the top
of the beam, you could even get a measurable change in length.

And I guess if I believe the above then I also have to say that if
the water is not above the top of the beam, then the beam will stay
the same length as it was when it was vertical in dry air.

First of all, I would prefer to consider an impermeable plastic
beam rather than a wooden beam, because wood is notoriously
permeable to water.

Secondly I assume the density of the beam is close to the density
of water so that buoyancy effects are small compared to the
pressure effects.

The beam will shrink when pressurized, but it will shrink less
than you would have predicted based only on the pressure x area
of the top end-face of the beam. That's because of the inward
pressure on the sides, in the X and Y directions.

And for a beam sticking out of the water, it might well stick
out *more* when the sides are pressurized, sorta like squeezing
toothpaste out of a tube.