Kvant Physics Problem 32

Let the side length of the square cross section be $a$.

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Problem

A wooden block with a square cross section floats on the surface of water. Which of the two equilibrium positions shown in Figure 1 is stable? The density of the material from which the block is made is equal to one half of the density of water.

Solution

Let the side length of the square cross section be $a$. Since the density of the wood is one half the density of water, exactly one half of the block's volume is submerged in equilibrium. Thus, in every equilibrium position, the waterline divides the square cross section into two regions of equal area.

We compare the two positions shown in the figure.

In position (a), the square has a horizontal side. The waterline passes through the center of the square and cuts the cross section into a submerged rectangle of height $a/2$ and an emerged rectangle of height $a/2$.

The center of gravity $G$ of the block is at the center of the square. The center of buoyancy $B$ is the centroid of the submerged half square. Hence

$$BG=\frac a4.$$

To determine stability, consider a small rotation through an angle $\theta$. The metacentric height is

$$GM=BM-BG,$$

where

$$BM=\frac{I}{V}.$$

Here $I$ is the second moment of area of the waterline about the longitudinal axis, and $V$ is the submerged cross sectional area.

For position (a), the waterline is a segment of length $a$, so

$$I=\frac{a^3}{12}.$$

The submerged area is

$$V=\frac{a^2}{2}.$$

Hence

$$BM=\frac{I}{V} =\frac{a^3/12}{a^2/2} =\frac a6.$$

Therefore

$$GM=\frac a6-\frac a4 =-\frac a{12}<0.$$

A negative metacentric height means that a small rotation produces an overturning moment. Position (a) is unstable.

Now consider position (b), where the square is turned through $45^\circ$. The vertical diagonal is perpendicular to the water surface. Since exactly half the area must be submerged, the waterline again passes through the center of the square.

The submerged part is the lower half of a diamond. Its centroid lies on the vertical diagonal. The distance from the center of the square to a vertex is

$$\frac{a\sqrt2}{2}.$$

The lower half is an isosceles triangle whose height is

$$h=\frac{a\sqrt2}{2}.$$

The centroid of a triangle lies at one third of its height above the base, so the center of buoyancy is at a distance

$$\frac h3=\frac{a\sqrt2}{6}$$

above the lower vertex. Since the center of the square is at height $h$ above that vertex,

$$BG=h-\frac h3 =\frac{2h}{3} =\frac{a\sqrt2}{3}.$$

The waterline is the horizontal diagonal of the square, whose length is

$$a\sqrt2.$$

Therefore

$$I=\frac{(a\sqrt2)^3}{12} =\frac{a^3\sqrt2}{6}.$$

The submerged area is still

$$V=\frac{a^2}{2}.$$

Thus

$$BM=\frac{I}{V} =\frac{a^3\sqrt2/6}{a^2/2} =\frac{a\sqrt2}{3}.$$

Hence

$$GM=BM-BG =\frac{a\sqrt2}{3}-\frac{a\sqrt2}{3} =0.$$

The vanishing of the metacentric height means that the usual first order test is inconclusive. A higher order analysis is required.

Let the block be rotated by a small angle $\theta$ away from position (b). The waterline must still divide the square into two equal areas, so it continues to pass through the center. The submerged region remains one half of the square bounded by a line through the center. Among all such divisions, the centroid of the submerged half is farthest below the center when the dividing line coincides with a diagonal, which is exactly position (b). Any small rotation moves the centroid of buoyancy upward toward the center, reducing the distance between $B$ and $G$.

Since the buoyant force acts through $B$, raising $B$ raises the potential energy of the floating system. Thus every sufficiently small rotation from position (b) increases the potential energy. The restoring moment is directed back toward the diagonal orientation.

Accordingly, position (b) is stable, although it is a limiting case with zero metacentric height and requires the higher order argument above.

∎

Notes

The horizontal sided position has negative metacentric height,

$$GM=-\frac a{12},$$

so it is unstable.

The diagonal position has

$$GM=0,$$

but examination of the geometry beyond first order shows that it corresponds to a minimum of the potential energy. Hence the block ultimately settles in the orientation with a vertical diagonal.

The stable equilibrium is position (b), and position (a) is unstable.