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

[John Mallinckrodt <ajm@csupomona.edu>]

Chuck Britton wrote:

> Assume a cubical volume of water. If we integrate the net force over 
> the entire surface of that cube, we come up with a total force that 
> is the weight of the fluid in that cubical volume.

Right, but no need to restrict yourself to a cube.  The logic works for any shape.  This is the basis for Archimedes' principle.  The net force on any imagined blob of fluid completely surrounded by more fluid all of which is in hydrostatic equilibrium must be zero.  Then, because there are only two forces on the blob:

1. Gravitational force equal to its weight and 

2. Integrated effect of the external pressure acting on the blob's surface

force 2 must be upward and equal in magnitude to force 1.

> Let's position that cubical, mathematical volume directly ojn the 
> glass at the bottom of my aquarium, positioned exACTLY such that 
> there is not one single molecule of water below this mathematical 
> volume.

Okay.

> It seems to me (naively) that we better come up with exACTLY the same 
> total force when we integrate force around all SIX sides of this 
> mathematical volume as we did the first time.

Right, because the integration must still yield a force that balances the weight of the same cubical volume of water. 

> The pseudoContact force of the bottom is supplying exACTLY the same 
> contribution to the total force as did the layer of water that used 
> to be there.

There's nothing "pseudo" about it.  There is a very real contact force and it is equal to the force that used to be supplied by contact with water at the lower boundary.

> Go ahead - shoot me down.
> I ain't no theoretician and I CAN take a hit when it's called for.

I don't see anything to shoot down here other than the needless term "pseudoContact force."

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

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.

John Mallinckrodt
Cal Poly Pomona