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

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

I'm not saying the two POV are equal. The fact that the force you  
remove from the downward force is equal to the weight of the water  
displaced is due to similarities in how the buoyant force and this  
force are computed. I'd have to think about it more, but I'm pretty  
sure the force you remove will be equal to the weight of water  
displaced no matter what the shape of the box, provided there isn't  
any surface area on the bottom of the box that is not in contact with  
the bottom of the aquarium. For example, if the box is a pyramid, then  
the volume displaced is less than a cubic box of the same height, but  
you also are reducing the weight of the water above less than with a  
cubic box all the way down the sides of the pyramid--more water above  
the farther you go down the sides.

Bill


William C. Robertson, Ph.D.
Bill Robertson Science, Inc.
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On Nov 3, 2010, at 8:13 PM, Josh Gates wrote:

> Wait - I have a question:
>  _____
>
>
> > At 7:24 PM -0600 11/3/10, William Robertson wrote:
> > > 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. OK - that's fine by me.  This force is  
> > > dependent upon the surface area of the bottom of the box.  Bigger  
> > > box -> less downward force on the bottom.  Except for the edges of  
> > > the box, right?  Those push down the same as the water above  
> > > would've (less the pressure differential (times A) from the height  
> > > of the box, I think).
> 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.Now  
> > > this "force" - which I think you're arguing is equal to the force  
> > > discussed above (the microscopic view, I guess) - is dependent on  
> > > the volume of the box, which doesn't necessarily have any  
> > > relationship to its bottom surface area.  Is there a reason to  
> > > think that these two POV really are equal?
>
> Thanks,
> Josh
>
>
> > > Bill
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