Re: buoyancy puzzle (long!)

[Brian Whatcott <inet@INTELLISYS.NET>]

> Two uniform cubes have sides of length L. Cube 1 has volume mass
> density d1, and cube 2 has d2 > d1. Their average density,
> d=(d1+d2)/2, is equal to that of an incompressible fluid filling a
> beaker. The two cubes are glued face-to-face with the lighter cube 1
> positioned directly above cube 2. (For simplicity, suppose that the
> glue has a density equal to that of the fluid, both when it is liquid
> and when it is solid.) The glued combination is thus neutrally
> buoyant in the fluid. It is immersed so that the (solid) glue plane
> is at depth H > L (but H+L is less than the depth of the beaker, so
> that the pair of cubes is "hovering" between the top and bottom
> surfaces of the fluid). The maximum tension force that the glue can
> withstand before tearing apart is F.
>
> 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)?

....

> Can anybody propose another argument, or even better an experimental
> test, to get the three of us out of our current deadlock? Carl
> --
> Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)
>        U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
> mungan@usna.edu    http://physics.usna.edu/physics/faculty/mungan/

I spent a minute or so reading this puzzle in part at least.
It appears to be indeterminate - in that the means of placing the conjoint
blocks at depth is unspecified: is there a down push, a down pull, a
combination?  The submersion method affects the stress at the glue joint.


Brian Whatcott
  Altus OK                      Eureka!