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

[curtis osterhoudt <flutzpah@yahoo.com>]

Let me extend the Gedankenexperiment a bit.

Consider the glass-bottomed aquarium, with water. 

1) Lower into it a stiff-sided, flat-bottomed box, filled with water. If the 
walls of the box are of the same density as the water, this box should be 
neutrally-buoyant in the water, and float (albeit unstably) at just about any 
depth you'd like (neglecting the minor compressibility of water). Of course, the 
water level in the aquarium will go up a smidgen.
2) Glue this box securely to the bottom of the aquarium, sealing the edges well. 
Presumably, there is no change to the net force on the aquarium bottom due to 
this process.
3) Pump out the water in the box, and replace it with air. 
4) Pump out the air in the box, and replace it with He.
5) Pump out the He in the box -- it's "replaced" with vacuum.

Suppose now we somehow dissolve the bottom of the smaller box, so that its 
contents are fully in contact with the aquarium bottom. We've blended the box 
bottom and the aquarium bottom.

I think that:

* The water-filled box will have no change to the net forces exerted on it.
* The air-filled box will have no change to the net forces exerted on it.
* The vacuum-filled box will have no change to the net forces exerted on it.

This last one -- the vacuum-filled box -- is effectively a suction cup at this 
point, firmly stuck to the bottom of the aquarium, and held there by the weight 
of the water. Yet if we only consider Archimedes' "rule", it should have the 
_greatest_ buoyant force on it, as it displaces the same amount of surrounding 
water as all the other boxes, yet contains  the least mass to be acted upon by 
gravity. 


And if the vacuum-filled box is firmly stuck to the bottom of the aquarium due 
to overpressure, wouldn't the original box-with-a-bottom be the same?


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"The four points of the compass be logic, knowledge, wisdom and the unknown. 
Some do bow in that final direction. Others advance upon it. To bow before the 
one is to lose sight of the three. I may submit to the unknown, but never to the 
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On Nov 3, 2010, at 5:41 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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