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



I mean the sum total of all forces on all six sides of the cube.
Five sides will experience force from the pressure and the sixth side is the interesting one.
I'm claiming that the sixth surface MUST have the force contribution - either from pressure or from contact.

The VOLUME is the same - the force integral is also (maybe?)

At 4:18 PM -0500 11/5/10, Rauber, Joel wrote:
By net force in your comments below, do you mean

1) sum of all forces acting on the object (this would include gravitational forces, integrated forces due to pressure from the fluid in which the object is immersed, and contact forces from entities other than the fluid that surfaces of the cube may be in contact with while it is immersed in the fluid?

Or

2) Only the sum of the integrated pressure from the fluid in which the object is immersed

Or

3) something else (in which case please specify).


PS - I'm following your suggestion to restrict my share of this discussion to the simplified example that you wanted me/us to consider and I'm happy to stick with a cube in order to KISS. And in an equilibrium situation.


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