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

[William Robertson <wrobert9@ix.netcom.com>]

I agree that the bottom will flex up under the box, but for reasons  
outlined by others. The pressure exerted on the bottom of the aquarium  
just below the box is less than the pressure exerted on the aquarium  
bottom all around it. Not because of a buoyant force on the box,  
because you have removed the mechanism for that force, but because  
there is "less water" above that part of the bottom of the aquarium.  
And yes, it will be equal to the weight of the water displaced, but  
that doesn't mean it's a buoyant force pushing up on the box.

Bill


William C. Robertson, Ph.D.
Bill Robertson Science, Inc.
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On Nov 3, 2010, at 7:14 PM, Chuck Britton wrote:

> How about two identical aquaria - each filled to the rim.
> One has an empty box glued to its bottom.
>
> My claim is that the box will cause that piece of glass to flex less.
>
> Maybe it will, maybe t won't.
> My expectation is that the 'boxed' bottom will flex less.
> The 'unboxed' bottom could be made to closely match the 'boxed'
> bottom if one would push up on its center with a force equal to the
> 'weight of water that the box is displacing' in the other aquarium.
>
>
> At 6:56 PM -0600 11/3/10, William Robertson wrote:
> > I've been watching from the sidelines on this. Here's how I see it.
> > The buoyant force is exerted by a fluid on objects partially or
> > entirely submerged in them. You can calculate this by using the  
> > weight
> > of the fluid displaced, but the mechanism for the force is the
> > pressure difference between the bottom and top of the object. No
> > gravity--no buoyant force, because no pressure difference. Now if  
> > your
> > box does not allow any fluid beneath it, then you have removed the
> > mechanism for the buoyant force. There is no buoyant force, and the
> > weight of the displaced fluid means nothing.
> >
> > Am I missing something?
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