One semantic point. The word "base" seems to have been used in three =
different ways,=20
(1) the horizontal section through only the fluid at the bottom of t=
he vessel,
(2) the horizontal section of the glass vessel at its bottom, and
(3) the container holding the vessels resting on the table.
Al Bachman
----- Original Message -----=20
From: von Philp<mailto:vonphilp@MYACTV.NET>=20
To: PHYS-L@LISTS.NAU.EDU<mailto:PHYS-L@LISTS.NAU.EDU>=20
Sent: Saturday, December 03, 2005 8:34 PM
Subject: Help with Fluids (Thanks!)....
Many thanks to John (and many others) who helped me sort this out.
Turns out I was on the right track. The additional weight of the f=
luid
pushing on the "V" shaped vessel walls is transmitted through the w=
alls to
the base, therefore the total force on the V-shaped base would natu=
rally be
greater than the other shapes, even though the force due to the flu=
id
pressure would indeed be the SAME on each base (given the equal wat=
er
heights and base area).
One thing that helped was to imagine an *empty* V-shaped vessel. I=
f I were
to put my hands inside and push normally on the sloped walls (as th=
e fluid
would do) that force would increase the force the base exerts on th=
e table,
but is independent of fluid pressures on the base, which is only du=
e to the
weight of the fluid above the base.
The total force on the table naturally equals the total weight of t=
he vessel
plus the fluid, which is different for each vessel. The force due =
to the
fluid pressure on the base would, in this problem, be the same. Wh=
ere I got
hung up was thinking that the force on the base due to the fluid pr=
essure
was then the same as the force the base exerted on the table.
I think I got it now. Thanks!
~ Ralph
----- Original Message -----
From: "John Denker" <jsd@AV8N.COM<mailto:jsd@AV8N.COM>>
To: <PHYS-L@LISTS.NAU.EDU<mailto:PHYS-L@LISTS.NAU.EDU>>
Sent: Friday, December 02, 2005 3:57 AM
Subject: Re: Help with Fluids....
> von Philp wrote:
>> Okay....
>> Imagine 3 vessles filled with water to the same height resting o=
n a
>> table.
>> One vessle 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 sa=
me,
>
> True, if given a favorable interpretation.
>
>> and
>> therefore the total force on each base would be the same.
>
> False.
>
> We need to be careful about "the" pressure and "the" total force.
> -- Pressure*area tells us the force of the fluid acting agains t=
he
> _inside_
> surface of the base; call this the fluid/base force.
> -- The total base/table force is something else entirely. It in=
cludes
> contributions carried by the stiffness of the walls of the vess=
el.
> (The fluid _per se_ has no stiffness, but the vessel does.)
>
>> I also understand that the "V" shape would contain more total wa=
ter, and
>> therefore would weight more, and therefore would exert more tota=
l force
>> on
>> the table.
>
> Yes, the V has more total force.
>
>> What I don't understand is how the force on each base is the sam=
e, but
>> the
>> force on the table is not.
>
> It is good to not understand that, because it's not true.
>
>> Is the additional weight of the water
>> transmitted through the walls of the container to the table via =
the
>> "edges"
>> of the base,
>
> It is.
>
>> and that does not count as the force on the base due to the
>> fluid pressure?
>
> It does count. We know it *must* count. We know this because of=
Newton's
> third law, aka conservation of momentum.
>
>> What would be an effective way to explain this apparent paradox =
to
>> students?
>
> 1) I would start with Newton's third law. As always, when in dou=
bt about
> how
> to interpret the third law, it is helpful to recast it as a cons=
ervation
> of
> momentum proposition. Downward momentum flows from the fluid in=
to the
> walls;
> it has to flow out somewhere.
>
> 2) At the next level of detail, draw the force diagram. For stud=
ents who
> can't
> handle calculus or even trigonometry, approximate the V-shaped v=
essel by
> a
> jaggy V with (say) 5 steps:
>
> |_ _|
> |_ _|
> |_ _|
> |_ _|
> |_|
>
> Then it's just a simple sum with 5 terms.
>
> Show that there is a "coincidence" that the 5-term expression for=
the
> total force
> looks very similar to the 5-term expression for the total mass. =
Of course
> it's
> not really a coincidence; at each level, part of the fluid is su=
pported
> by
> lower-lying fluid (making pressure) and the rest is supported by =
the wall
> (making force in the wall). Each bit of fluid has to be supporte=
d by
> *something*.
> It's just conservation of momentum, aka third law.
>
>
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