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

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

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?

Bill


William C. Robertson, Ph.D.


On Nov 3, 2010, at 6:41 PM, Chuck Britton wrote:

> So what it boils down to is that you don't want to use word
> 'Buoyancy' to describe this upward force exerted by the empty box on
> the bottom of the aquarium. That's your prerogative.
>
>
> At 5:26 PM -0700 11/3/10, John Mallinckrodt wrote:
> > What Philip said...  But in direct response to your question:  As
> > long as the box had an average density different from that of water,
> > it would, of course, affect the flexion of the base of the aquarium
> > relative to what it would be with only water of the same depth.  If
> > the box had a GREATER average density than water, than the base of
> > the aquarium would be subject to greater downward force (for the
> > simple reason that it would have to support more weight) and would
> > tend to flex MORE in the downward direction and vice versa.
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