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



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?

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

This would be a fun experiment to try, but I don't have a scale sensitive enough to try this at home.

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 6:40 PM
Subject: Re: [Phys-l] Definition of upthrust or buoyancy


On 10/19/2010 08:19 PM, M. Horton wrote:

Where does the buoyant force come from? Or to ask it in a slightly
different way, what causes the buoyant force on an object and how? I've
seen answers such as "The difference in density between the liquid at the
bottom of the object versus the top," "The differnce between the pressure at
the bottom of the object and the top," "the now-elevated water level pushing
back," "the displaced water trying to get back in," and other equally
non-convincing arguments.

As Jeffrey Schnick hinted: It helps to think about the microscopic
physics. At the end of the day, the only thing that is going on is
molecules of the fluid colliding with the surface of the object.

1) When we consider a large number of microscopic collisions and
summarize them in macroscopic terms, this is what we call "pressure".
So at the end of the day, the buoyant force on the object is due to
differences in pressure.

We are making a number of conventional assumptions, notably that
neither the object nor the fluid is in motion.

Also: It is not just the topmost top and the bottommost bottom that
are acted upon by the pressure; the sides count also, if there is
any slope to the sides. There are scads of classic exam problems
that revolve around odd-shaped buoyant objects.

2) Differences in density have got nothing to do with any of this.
An incompressible fluid would exhibit no differences in density, but
would still produce buoyancy in the usual way.

3) The /total/ density matters indirectly, because it is what produces
the differences in pressure required by item (1). This is easy to
understand in terms of force balance (aka momentum budget). The
gravitational force on each parcel of fluid has to be balanced by
something ... and in fact it is balanced by delta pressure.

It is an interesting exercise to calculate the pressure profile of
the entire atmosphere by integrating the weight (per unit area) of
the air column.
http://www.av8n.com/physics/altimetry.htm
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