## The floating beam

I have NOT invented this problem, but saw it many years ago.
Can't recollect where. Possibly the German edition of Scientific American,
possibly another journal of a similar kind.
###
Bonus track session in CCMB 217, Wed Jan 31 and Thu Feb 01, 2001, 7-9pm

How does a beam with quadratic cross section float in water (if it floats)?
With one side being horizontal or rather with a corner pointing downwards?
We will assume that this is a 2-dimensional problem, so we only consider
the quadratic cross section of the beam.
The answer depends on the density of the beam. Gravity acts on the center
of mass of the beam, whereas the buoyance force acts on the center of mass
of the displaced water.

Depending on whether the center of mass of the displaced water lies left to
the center of mass of the beam (i.e., square) or right, the beam tends to
be turned clockwise or counterclockwise. So we want the two centers of mass
(the one of the beam and the one of the water) to be on top of each other.
Then there will be no torque, and the beam can swim in any configuration
satisfying this condition. -- Really?

Hey, wait a moment: there is not only a question of equilibrium, but also of
stability. If we turn the beam a bit away from its position (that happens
all the time, say by bird's droppings on the beam, or by a slight breeze
that agitates the water a bit), there will be a torque caused by the fact that
the centers of mass are no longer on top of each other.
Will this torque turn the beam further away or will it turn it back?
Actual beams can only float in such a way that the torque turns the beam back.

Which coordinates do we want to choose for our calculations? Neither choice
will avoid tedious calculations altogether. And, no, I will not show the
calculations on the www page. I don't have the time to code all this in html,
and you will want to do it yourself. But here is the result:

Normalize the density of water to 1, and consider beams of density at
most 1/2. (The other case can be handeled very similarly, or, with
sophisticated reasoning, reduced to the considered case.)
If the density of the beam is
at most (3-sqrt(3))/6, approximately 0.21133, then the beam will float
with a side of the square horizontal. This is what you will find intuitive
if you think of an actual beam made of styrofoam floating.
However, if the density is larger than 9/32, namely 0.28125, then the beam
floats with a corner pointing downward.
But in between, the beam will do neither. It will rather tilt to one side
increasingly. At density 1/4, the water level will actually be a line
connecting one corner of the square with the midpoint of one square.

Last modified: Jan 29, 2001