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Re: buoyancy puzzle (long!)



Carl Mungan wrote:

Question: What is the maximum depth H before the cubes break apart
(resulting in cube 1 rising to the surface and cube 2 sinking to the
bottom)?

This is an amazingly odd question. Suppose the two cubes weren't
glued. Then the only forces on the cubes would be weight (which is
obviously independent of depth) an buoyancy (which is obviously
independent of depth). The glue has to overcome the difference in
weight -- so the burden on the glue is so obviously independent of
depth that I can't imagine why anybody would ask the question. I
can't imagine why any calculation is necessary.

provided fluid does not seep into the glued interface.

That proviso is not even necessary. Assume the glued face is only
90% covered with glue, and the remaining area is a channel whereby
fluid can come and go. Result is unchanged.

You can perfectly well go to the extreme of replacing the glue with
a string made of a tiny filament of glue-stuff.

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

The total load borne by the glue is:

F = (W2-W1)/2

Yes. To derive this, all you need to know is that the objects are
the same size, and the pair is in equilibrium, and the fluid is
incompressible.

F1 := Net downward force on cube 1
F2 := Net downward force on cube 2
B := buoyancy of object of stated size
F1 = W1 - B + F (from force diagram) [a]
F2 = W2 - B - F (ditto) [b]
F1 = 0 (object is in equilibrium)
F2 = 0 (ditto)

Subtract [b] from [a].

I agree with A that the blocks can be torn apart by the fluid.

I wouldn't say that. The expression
F = (W2-W1)/2
has no dependence on fluid properties, so if/when the blocks
are ripped apart, it's hard to interpret it as being done "by
the fluid".