[Phys-L] Re: Help with Fluids....

[Rick Swanson <swansonr@SANDHILLS.EDU>]

Basically, as I see it, static fluids put forces on surfaces that are perpendicular to the surface because a fluid cannot sustain shear forces.  This is a great logical thought exercise for engaging students; similar to "you can't push on a rope."

Solids can sustain shear forces.  So ... the force vector due to the fluid pressure on the angled sides of the container has sideways and downwards components.  Do the free body diagram and sum the forces to find the normal force.

Tipler and Mosca's fifth edition of Physics for Scientists and Engineers has this as a chapter 13 conceptual problem with a solution in the student solution manual.  I have the book but not the manual :-)

Cheers,

Rick

Richard E. Swanson, Ph.D.
Dean of Instruction
Physics Professor
Sandhills Community College, Pinehurst, NC 28374
swansonr@sandhills.edu (910) 695-3715

> > > betwys1@SBCGLOBAL.NET 12/2/2005 7:42:33 AM >>>
At 12:09 AM 12/2/2005, Ralph v. Philp,  you wrote:
> Okay....
> Imagine 3 vessels filled with water to the same height resting on a table.
> One vessel is a vertical cylinder, one is shaped like a "V", and one is
> shaped like an "A".  They all have the same area for the base.  You've see
> the situation before....Pascal's Vases?
>
> Now, I understand that the pressure on each base would be the same, and
> therefore the total force on each base would be the same.
>
> I also understand that the "V" shape would contain more total water, and
> therefore would weight more, and therefore would exert more total force on
> the table.
>
> What I don't understand is how the force on each base is the same, but the
> force on the table is not.  Is the additional weight of the water
> transmitted through the walls of the container to the table via the "edges"
> of the base, and that does not count as the force on the base due to the
> fluid pressure?
>
> What would be an effective way to explain this apparent paradox to students?
>
> Thanks for the help!
> ~ Ralph von Philp
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