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

On 10/20/2010 12:02 PM, LaMontagne, Bob wrote:

If you want to imagine the buoyancy on an object, first imagine a
blob of water with exactly the same shape as the object. That water
is obviously in equilibrium. Integrate the pressure over the surface
area of the blob. If in equilibrium, this must give a force that is
equal and opposite to the weight of the water. Now replace the blob
with the object. The surfaces forces have not changed so they still
add to the weight of the blob. Hence, the buoyant force is the weight
of the water displaced. This definition of buoyancy does not require
water surrounding all surfaces of the blob - the blob could be on the
bottom of a glass beaker.

That's a very nice contribution to the discussion.
Let's call is the "integral d(volume)" approach : ∫dV.

I have not much to add except for the following relatively
minor points:

1a) The ∫dV analysis can be easily proved correct using the
principle of virtual work. When the object moves up, an
object-shaped slug of water must move down to take its

1b) Also, it can easily be proved correct using an appeal
to the third law of motion ... which is (as always) the
same thing as an appeal to conservation of momentum. When
you replace a slug of water of mass m with an object of
mass M, the net change to the momentum budget is (M-m) g dt
and therefore when we weigh the object under water, the
scale must exert a force of (M-m) g ... quite independent
of how much contact-area there is between object and scale.

2) Previously I described buoyancy in terms of the forces
i.e. the pressure acting on the surfaces of the object.
We can call this the "integral d(surface)" approach : ∫dS.

The two approaches are related by Stokes's theorem.

3) One should not be too quick ask whether one approach is
better than the other. Some students may find it easier to
start with one approach, while some may find it easier to
start with the other approach. More importantly: Mastery
of the subject means being able to use both approaches!

You can prove Archimedes's principle either way. Even peculiar
things like suction cups, pistons, and Cartesian divers can be
analyzed either way ... and it is instructive to do it both
wans and compare.

On the other hand, there are some situations where one
approach clearly wins over the other. For example, consider
fluid dynamics. We have fluid pushing against fluid (rather
than fluid pushing against a distinct object), so we have to
do everything as a function of position : velocity fields,
pressure fields, et cetera. In this situation, the ∫dS
approach would be a nightmare, because for each patch of
surface there is no good way to keep track of which side
of the surface you are talking about (A pushing on B, or B
pushing back on A). The conventional and sensible practice
(in most cases) is to use the ∫dV approach, keeping track
of the momentum in each parcel of fluid.


On 10/21/2010 07:19 AM, Jeffrey Schnick wrote:
I think that a negative effect of defining the buoyant force to be an
upward force equal in magnitude to the weight of the water dispaced, is
that I would expect that students who have been taught that definition
would have a greater chance of getting the following problem so wrong
that they come to the conclusion that the piling is in tension rather
than compression ....

We can agree on the maxim that no matter what the students are
doing, they can always do it wrong.

I do not agree that this reflects negatively on the ∫dV approach.

Let the height of the center-of-mass of the piling be h.
Let the volume of the piling be V.

Buoyancy concerns dE/dh at constant V.
Compression concerns dE/dV at constant h.

A compression question is not a buoyancy question. If a student
blithely applies a buoyancy formula to a non-buoyancy problem,
bad things are going to happen. Nobody should be surprised by

Any competent application of the ∫dV approach will give the right
answer for the dE/dV i.e. for the compression.