Kvant Physics Problem 32

Verified: no
Verdicts: SKIP + SKIP
Solve time: 7m12s
Source on kvant.digital

Problem

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

Setup and Assumptions

A prismatic wooden block of constant square cross-section of side $a$ and length $L$ floats on the surface of water of density $\rho_w$. The density of the block is $\rho = \rho_w/2$. The block is uniform, so its center of gravity $G$ coincides with its geometric center.

Two equilibrium orientations are considered. In the first, a face of the square cross-section is horizontal. In the second, the square cross-section is rotated by $45^\circ$ so that a diagonal is vertical and the body floats in a “diamond” orientation.

The fluid is incompressible and at rest. Surface tension and viscosity are neglected. Stability is determined using hydrostatic stability theory for small angular perturbations about equilibrium, based on the metacentric criterion.

Physical Principles

Archimedes’ principle gives the buoyant force as $F_b = \rho_w g V$, where $V$ is the submerged volume.

Equilibrium requires equality of weight and buoyant force, $mg = \rho_w g V$.

For small rotations about equilibrium, stability is determined by the metacentric height $GM$, defined through

$GM = BM - BG,$

where $B$ is the center of buoyancy in equilibrium and $M$ is the metacenter.

The metacentric radius satisfies

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

where $I$ is the second moment of area of the waterline cross-section about the tilt axis, and $V$ is the displaced volume.

Stability requires $GM > 0$, instability corresponds to $GM < 0$.

Derivation

The weight of the block is $mg = \rho g a^2 L$. Buoyancy equilibrium yields

$\rho g a^2 L = \rho_w g V,$

so the submerged volume is

$V = \frac{\rho}{\rho_w} a^2 L = \frac{1}{2} a^2 L.$

First equilibrium: face horizontal

The submerged depth is $a/2$, so the center of buoyancy is at the centroid of a submerged rectangular prism of height $a/2$. The center of gravity lies at height $a/2$ from the base of the cube, while the center of buoyancy lies at height $a/4$. Hence

$BG = \frac{a}{2} - \frac{a}{4} = \frac{a}{4}.$

The waterplane is a rectangle of area $a \times L$. Its second moment of area about the longitudinal tilt axis is

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

The metacentric radius is

$BM = \frac{I}{V} = \frac{\frac{L a^3}{12}}{\frac{1}{2} a^2 L} = \frac{a}{6}.$

Thus

$GM = BM - BG = \frac{a}{6} - \frac{a}{4} = -\frac{a}{12}.$

The negative value indicates instability of this equilibrium.

Second equilibrium: square rotated by $45^\circ$

In this orientation the cross-section becomes a diamond. The waterline passes through the center and divides the square into two congruent triangles. The submerged volume corresponds to one triangular half of the cross-section.

The height of the diamond is $a/\sqrt{2}$, so the submerged triangular height is

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

The centroid of a triangle lies at a distance $h/3$ from its base, so the center of buoyancy lies below the center by

$BG = \frac{h}{3} = \frac{a}{3\sqrt{2}}.$

The waterplane intersection is a rectangle of length $L$ and width equal to the horizontal diagonal of the square,

$b = a\sqrt{2}.$

Hence the second moment of area is

$I = \frac{L b^3}{12} = \frac{L (a\sqrt{2})^3}{12} = \frac{L \cdot 2\sqrt{2} a^3}{12} = \frac{L \sqrt{2} a^3}{6}.$

The displaced volume is still

$V = \frac{1}{2} a^2 L.$

Thus

$BM = \frac{I}{V} = \frac{\frac{L \sqrt{2} a^3}{6}}{\frac{1}{2} a^2 L} = \frac{\sqrt{2} a}{3}.$

The metacentric height becomes

$GM = BM - BG = \frac{\sqrt{2} a}{3} - \frac{a}{3\sqrt{2}} = \frac{a}{3\sqrt{2}}.$

Result

For the face-horizontal equilibrium,

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

so this position is unstable.

For the $45^\circ$ rotated equilibrium,

$GM = \frac{a}{3\sqrt{2}}.$

Substituting gives a positive value, so this position is stable.

$\boxed{\text{Stable equilibrium: rotated } 45^\circ \text{ (diamond position)}}$

Sanity Checks

The metacentric height has dimensions of length in both cases, consistent with $GM = BM - BG$.

In the face-horizontal case, the center of buoyancy lies below the center of mass, but the metacentric radius is too small to compensate, producing a negative $GM$ and instability, consistent with the physical expectation that a wide flat float can overturn easily if its center of buoyancy shift is insufficient.

In the rotated case, both contributions scale linearly with $a$, and the positive coefficient $\frac{1}{3\sqrt{2}}$ ensures stability for any nonzero size. The result is consistent with the general principle that a configuration with a higher effective waterline moment of inertia is more stable.