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# Re: [Phys-l] Definition of upthrust or buoyancy

I got it now, thanks. I didn't take from the previous message that it is a pressure effect AND that it depends on the differential pressure on the object. Makes sense.

Thx,
Mike

----- Original Message ----- From: "John Denker" <jsd@av8n.com>
To: "Forum for Physics Educators" <phys-l@carnot.physics.buffalo.edu>
Sent: Tuesday, October 19, 2010 9:59 PM
Subject: Re: [Phys-l] Definition of upthrust or buoyancy

On 10/19/2010 10:28 PM, M. Horton wrote:
I don't actually know the answer to this question. Let's say that you put a
solid object (a rock) in water and measured its effective weight*. Then,
you heat up the water. Will the effective weight decrease by a proportion
equal to the ratio of the before and after temperatures (or the square
thereof bc of the relationship between temperature and molecular velocity)?
If it is the pressure of the water molecules colliding with the object that
cause the buoyant force, then the answer should be "yes." But gut instinct
tells me that the answer is "no."

I cannot justify the discrepancy in my thought experiment and have never
tried the experiment. In fact, since the mass of water displaced would be
less (due to the decreased density of the water at higher temperatures), I
would expect that the effective weight of the rock might increase slightly.
This seems to contradict the molecular collision model. What am I missing?
Is my thought experiment all wrong?

The thought experiment leaves out some critical things.

For one thing, water is a liquid, so a simple kinetic-theory
analysis -- leaving out the binding energy -- will be very
misleading. In particular, consider the free surface. If
you neglect the binding energy, the molecules would instantly
fly out of the surface and the whole thing would explode.

In fact what happens when you heat the water is that it expands.
The molar volume is to a good approximation a linear function
of temperature, with no squares or square roots. It's linear
because there is a balance between the kinetic energy and the
binding energy, and the added temperature shifts the balance
a little bit.

This expansion may or may not make the water deeper. This may
or may not increase the overall pressure on the rock, depending
on where the rock sits in the water column. (Ten feet up from
the bottom? Or ten feet down from the surface?) Most importantly,
we don't care! We don't care about the overall pressure, because
the overall pressure acts on the rock in all directions, and has
no effect on the effective weight.

This is a really useful general result: The force of the fluid
on objects (and the force of the fluid on itself) never depends
on the overall pressure; it only depends on _differences_ in
pressure. This point is worked out in almost any physics text.
It is needed in the section where they derive the wave equation
for sound, among other places.

In fact, since the mass of water displaced would be
less (due to the decreased density of the water at higher temperatures), I
would expect that the effective weight of the rock might increase slightly.

Not just might increase ... it must increase, because the
buoyancy decreases. (We are assuming that the water expands
more than the rock does, which is a pretty safe assumption.)

The pressure profile (i.e. pressure as a function of depth)
is what gives rise to the _delta_ pressure that gives rise to
buoyancy. The dP/dh is just gravity times density, and that
is what matters.

* I just made up the term "effective weight" to mean "weight as measured
when the object is under water" to simplify the discussion.

Works for me.

============================

Also, a clarification: My previous two notes must be read together.
In the first note I said that the "buoyancy" of an object sitting
on the floor of the ocean is a pathological case, and in the second
note I talked about the microscopics of fluid particles hitting the
object. The second note applies in the non-pathological case;
the floor case is still pathological.

Let me now explain what happens when something sits on the floor,
or, better, sits on a scale at the bottom of the ocean. Suppose
we have an object that weighs 10 N on dry land. Its displacement
is 9 N so it must have an effective weight of 1 N.

We have a scale deep in the ocean, all properly calibrated, and
when we set the object on the scale, sure enough it reads 1 N.
So far so good.

We do *not* need to ask whether there is any water between the
object and the scale. Maybe there is a lot; maybe there is
only a little. Microscopically and theoretically this makes a
very big difference, because my object is very thin and flat,
so that there is 1000 N of downward hydrostatic force on its
upper surface. That's just pressure times area equals force.
There is of course 1009 N of upward hydrostatic force on its
lower surface when it is submerged in the normal way.

If we imagine that the object/scale interface gets "dewetted"
to some degree, the hydrostatic force gets replaced by a
solid/solid contact force. The crucial observation is that
whenever and wherever dewetting occurs, it removes a certain
amount of upward force from the object, *and* it removes the
exact same amount of downward force on the scale! So it's
a break-even proposition. The effective weight, as measured
operationally by the scale, is independent of how much
dewetting occurs. Macroscopically and operationally speaking,
we don't know and don't care how much dewetting there is.

If you insist on a pedantic microscopic notion of buoyancy,
you might think that under some conditions the object would
be subject to -1000 N of buoyancy, if the lower surface is
fully dewetted. But this is insanely impractical. We should
define a notion of "generalized buoyancy" that is connected
to the effective weight, and is invariant with respect to
dewetting. The generalized buoyancy includes the "normal"
wet buoyancy, but also includes a credit for any dewetted
areas when the object is loosely resting on something else.

If it is not _loosely_ resting, e.g. if there are suction
cups or bolts or chemical bonds involved, then all bets
are off, because the system is now too complicated to be
described in terms of simple concepts such as buoyancy and
effective weight.

So ... the generalized buoyancy strictly upholds Archimedes's
principle: it is equal to the weight of the displaced fluid,
whether or not the object is resting on the bottom. This is
very useful in practice, and I reckon it is what most people
mean by "buoyancy".

If anybody wants to define a stricter notion of buoyancy,
they are of course free to do so, but they run the risk of
being misunderstood, unless they are super-careful to explain
what they mean.
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