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

[curtis osterhoudt <flutzpah@yahoo.com>]

OK. We *all* understand that there is pressure from the water (or atmosphere, or 
any other fluid).  Perhaps my use of "compression" was tremendously misguided in 
this instance. I apologize for that.

Jeffrey's original question about the normal stress in the piling *seems* to 
remain. Or does it?

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________________________________
From: John Denker <jsd@av8n.com>
To: Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu>
Sent: Thu, October 21, 2010 9:11:16 AM
Subject: Re: [Phys-l] Definition of upthrust or buoyancy

On 10/21/2010 08:37 AM, curtis osterhoudt wrote:

> I haven't found, yet, anything that convinces me that the piling is under 
> compression at any point.

But it is under compression.  

As I wrote previously, 
-- Buoyancy concerns dE/dh at constant V.
-- Compression concerns dE/dV at constant h.

Changing the volume of an underwater object involves a large amount
of energy.  This can be verrry significant for objects underwater.
  Example:  A submarine's operations are limited by the "crush depth" 
   of the hull.
  Example:  Scuba divers would be crushed except for the fact that
   they put high-pressure air into their lungs ... which leads to 
   other problems, notably nitrogen narcosis and decompression 
   sickness.
  Example: If you lower a piece of open-cell foam into the water, 
   it won't get crushed, but it will get soaked.  It resists the
   compression sorta the way the scuba diver does, but using water
   on the inside rather than air.
  Example: If you lower a piece of closed-cell foam into the water, 
   it will get crushed.
  Example: A water-logged wooden piling will not get crushed, for 
   the same reason that the open-cell foam did not get crushed.
   It is under compression, but it resists the compression.  If
   you do NMR on the cellulose molecules, you will observe a 
   "pressure shift".
  Example: A hollow steel piling (with air on the inside) *will* get 
   crushed, just like the submarine gets crushed.

Again:
-- Buoyancy concerns dE/dh at constant V.
-- Compression concerns dE/dV at constant h.

Applying buoyancy formulas to compression problems is guaranteed to
produce wrong results.
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