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



Ah. Apologies. I just caught up on the thread, and it appears "net" force is
being used as [perhaps] pressures integrated over the surface. Sorry for the
noise.


Still, what a strange use of the term "net force"!







________________________________
From: curtis osterhoudt <flutzpah@yahoo.com>
To: Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu>
Sent: Fri, November 5, 2010 3:19:07 PM
Subject: Re: [Phys-l] A Crude Attempt at Analysis


I think I'm confused as to what is being called the "net force". If I make it a
cube of water (same density as the surroundings), it's not going to accelerate
anywhere. But if it's a cube of iridium, it's definitely going somewhere. How
can the net forces be the same on the two cubes?

/**************************************
"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
unknowable." ~~Roger Zelazny, in "Lord of Light"
***************************************/




________________________________
From: Chuck Britton <cvbritton@embarqmail.com>
To: Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu>
Sent: Fri, November 5, 2010 3:08:47 PM
Subject: Re: [Phys-l] A Crude Attempt at Analysis

The net force on a (submerged) cubical volume doesn't depend on what
the volume is filled with.
Could be lead, could be extremely light plastic, could even be water
itself. The net force is the same.

The net force on a (submerged) cubical volume doesn't depend on where
that volume is positioned.
Could be a bit off the glass or could be smack intimately up against
the glass. The net force is the same.

Help me with my math here, please. ( and uh, assuming static equilibrium)

Let's stick with cubes, cuz my mind does better with simple things.

At 3:22 PM -0400 11/5/10, 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.

ok so far? (I think I saw this a physics text - so it MUST be right.)

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.

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.

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.

Go ahead - shoot me down.
I ain't no theoretician and I CAN take a hit when it's called for.
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l

_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l