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Re: [Phys-l] A Crude Attempt at Analysis

Confusion - what confusion might you be referring to? ;-)

Any mathematically defined volume that is 'under water' (perhaps also in contact with the bottom my aquarium) must experience a NET upward force that is equal to the weight of the displaced water.

Sounds simple enough to me.


At 1:12 PM -0700 11/5/10, John Mallinckrodt wrote:

uh, I was thinking that the total SIX sided integral would have to be the same for the same volume.
(regardless of what was filling up that volume.)

I guess I AM still confused.

I was thinking that he mathematical integral didn't depend on silly details like what was contained within the volume. Or whether it was a micron or so above the glass.

Straighten me out here please.



--------------------

Going forward ....

Now if you replace the cubical volume of water with a physical cube of identical shape but smaller mass:

1. The gravitational force (which is, of course, downward) gets smaller;

2. the integral over the five exposed sides is still the same as it was above (and also obviously downward), and thus;

3. the integrated force over the remaining (bottom) side of the cube (which is obviously upward since the others are ALL downward) need not be as large as it used to be due to the reduction of the gravitational force.

I still don't see why there should be any confusion.

Just a thought that might help. Imagine that you place the physical cube JUST in contact with the bottom so that all fluid is excluded, but the bottom exerts NO significant force on the cube. Hopefully you see that the cube will be DRIVEN downward into MORE significant contact with the bottom. It is the resulting compression of the bottom that results in it exerting a greater--and ultimately balancing--upward force. The same mechanism explains how horizontal surfaces always manage to exert JUST the right amount of upward force on whatever sits on them.