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Re: buoyancy puzzle (long!)



I wouldn't go as far as J.Denker in saying that the problem is
without merit. I certainly tests whether the askee has a grasp of
buoyancy AND is able to sift out extraneous information.

For those who still COVER fluids, I think that the 'Milk Bottle with
Cream in the Neck' is a much more interesting scenario. It was
thoroughly discussed here some years ago but a newbie or two might
enjoy it??



At 12:00 AM -0400 on 5/1/02, Carl Mungan wrote
Three of us have been discussing the following question. We each have
a different point of view. This may challenge you to clarify your
understanding of Archimedes' principle.

Reference: Physics Challenges, Jan. 2002 issue, Dense and Tense Story.

Scenario (simplified slightly from the above reference without
altering the essential physics):

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)?

Notation for ease of discussion:

volume of each cube V=L^3
surface area of joined face A=L^2
mass of cubes m1=d1*V and m2=d2*V
weights of cubes W1=m1*g and W2=m2*g
force that fluid exerts on a horizontal face at depth h is
Ff(h)=d*g*h*A (neglect P_atm for simplicity)


---SOLUTION A (revised version of what appears in the April 2002
issue of TPT)---

Draw a free-body diagram (FBD) for cube 2. We have Ff(H+L) up on the
bottom surface, its weight W2 down, and the glue force F up, so that:

d*g*(H+L)*A - d2*g*V + F = 0 --(1)

A similar reasoning for cube 1 gives:

d*g*(H-L)*A + d1*g*V + F = 0 --(2)

Add (1)+(2) and substitute d=(d1+d2)/2 to get solution A:

H = [(d2-d1)*g*V - 2F]/[(d1+d2)*g*A] --(3)

Comment: Each of the three terms in Eq. (2) is positive. Hence cube 2
is NOT in equilibrium. Thus this solution is plainly wrong.

--- SOLUTION B (presented to me by an unnamed smart fellow)---

Consider cube 2. We have Ff(H+L) up on the bottom surface and its
weight W2 down as in solution A. But we do not have a glue force F
up. Instead we have a contact force F21 (on cube 2 due to cube 1)
down. Hence we rewrite Eq. (1) as:

d*g*(H+L)*A - d2*g*V - F21 = 0 --(1')

and noting that F12 = F21 by Newton's third law (N3), for cube 1 we get:

d*g*(H-L)*A + d1*g*V - F21 = 0 --(2')

Hence there is no dependence on F. The cubes will not fall apart at
any depth, provided fluid does not seep into the glued interface.
Just as in solution A, without fluid between the cubes we cannot
write down a buoyant force on the cubes. In fact, no glue is even
needed, provided you can squeeze every last molecule of water out.

-- MY SOLUTION C---

First suppose block 2 alone were immersed in the fluid and that a
support pedestal were placed under it to keep it from sinking to the
bottom. The pedestal exerts an upward support force Fs. Since fluid
presses on both the top and bottom surfaces, we have:

d*g*(H+L)*A - d*g*H*A - d2*g*V + Fs = 0 --(1")

Next, place block 1 in its final position. Place a support beam above
it (projecting up out of the fluid and attached to a ring stand) to
keep it from floating up. From a FBD we get:

d*g*(H-L)*A - d*g*H*A + d1*g*V + Fs = 0 --(2")

where it is easy to prove that the support beam must exert the same
force Fs downward that the pedestal for cube 2 exerted upward.
(Proof: subtract the two equations and rearrange to get an equality.)

Now imagine replacing all of the fluid in the gap between the two
blocks with liquid glue. (Suppose that it's fast drying so that a
negligible amount escapes from the interface between the two blocks.)
Noting that it has the same density as fluid, Pascal's principle
tells us that Eqs. (1") and (2") will be unchanged. Next the glue
solidifies. The equations will still continue to hold: I've "frozen
in" the buoyant pressure in the glue. Finally, let's remove the
pedestal and support beam. The glue will stretch infinitesmally. This
will give rise to an elastic force whose magnitude must exactly equal
Fs if the blocks are not to break apart. I claim the only reasonable
name for this force is the glue force F.

Hence my solution is:

I agree with B that there is no dependence on H. Simplify Eqs. (1")
and (2"): H in the first term in each equation is precisely canceled
by H in the second term. (Of course. I have balanced the buoyant
force against the weight and glue force, for each cube.)

But I agree with A that the blocks can be torn apart by the fluid. It
is just that the condition for this, as obtained by adding (1")+(2"),
is:

F = (W2-W1)/2 --(3")

instead of (3). I claim that there IS a buoyant force on each cube.
The physical reason we have the force term d*g*H*A in Eq. (2"), say,
is that the fluid presses upward on the bottom of cube 2 and this is
partly transmitted up to block 1. My solution is a combination of
solutions A and B. To see this explicitly, simply let F21 = d*g*H*A -
F.

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/

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