NBPhO 2025

Overview

The problem describes one specific tilted-equilibrium configuration of a square-cross-section beam loaded by one bird, then asks how many birds the beam can carry without going fully under. The setter has hidden two pieces of unknown information — the beam’s density $\rho_b$ and the bird’s mass $M$ — inside the one described configuration, and we must extract both.

The two italicised conditions in the problem statement (the upper edge on the bird’s side just touching water; the lower edge at the far end just touching water) are not redundant. The first pins the tilt angle geometrically; the second is what makes the configuration critical in a buoyancy-limited sense. Once $\rho_b$ and $M$ are known, the rest of the problem is a one-line buoyancy-ceiling argument: no matter where birds are placed, the beam cannot displace more water than its own volume, and that ceiling pins $N$.

The relevant small parameter is $a/L \ll 1$ (long thin beam), which lets us drop $O(a/L)$ corrections everywhere. The relevant idea is that the submerged region of the tilted beam is a triangular wedge whose centroid sits at $1/3$ of the way from the heavier side, not at the beam’s mid-length.

The plan is:

1. Use the two edge-at-water-line conditions to fix the tilt angle $\theta$.
2. Compute the submerged volume and the centre of buoyancy of the resulting wedge.
3. Apply force balance and torque balance simultaneously — a $2\times 2$ system — to solve for $\rho_b$ and $M$.
4. Use the full-submersion buoyancy as a hard ceiling on the total load to bound $N$.
5. Check that some placement of $N=4$ birds is actually realisable.

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Setting up the tilted geometry

Let the beam have length $L$ and square cross-section $a\times a$, with $L\gg a$. Take the water surface at $y=0$, with $y>0$ above the water. Place the bird at the right end. Working in the beam’s vertical plane, the projected outline of the beam is a long thin rectangle of dimensions $L\times a$, tilted by some angle $\theta$ below horizontal (right end down, since the bird drags it down).

It is convenient to take the lower-left corner (BL) as the origin. The bottom-face edge runs from BL along the beam axis, direction $\hat{u}=(\cos\theta, -\sin\theta)$. The transverse direction “up across the beam” is $\hat{n}=(\sin\theta,\cos\theta)$, perpendicular to $\hat u$. The four projected corners are then

$$\begin{aligned} \mathrm{BL} &= (0,0), \\ \mathrm{BR} &= (L\cos\theta,\,-L\sin\theta), \\ \mathrm{TL} &= (a\sin\theta,\,a\cos\theta), \\ \mathrm{TR} &= (L\cos\theta + a\sin\theta,\,-L\sin\theta + a\cos\theta). \end{aligned}$$

The two stated conditions are: BL is at water level, and TR is at water level. The first is built into the choice of origin. The second reads

$$-L\sin\theta + a\cos\theta = 0 \qquad\Longrightarrow\qquad \boxed{\tan\theta = \frac{a}{L}}.$$

For a long thin beam this is small: $\theta\approx a/L\ll 1$, with $\sin\theta\approx a/L$ and $\cos\theta\approx 1$ to leading order. We will drop $O(a/L)$ corrections in everything that follows.

Educational remark — why both conditions matter

The first condition (top corner on the bird’s side at water level) is what defines the configuration as “extreme”: tilt the beam any further and the bird’s end of the upper face would dip below water — at which point the wetted geometry changes character (the wedge becomes a more general polygon) and the simple formulas below stop applying. The second condition (lower corner at the far end at water level) is automatic once the first is imposed at a particular tilt: together they say the beam is tilted by exactly the angle $\tan\theta=a/L$ that “swaps” the two diagonally opposite edges from above to below the water surface. Geometrically, the beam’s silhouette is a parallelogram whose long diagonal lies along the water surface.

This is what makes the configuration so informative. There is exactly one tilt at which both edges sit on the water; specifying that tilt determines submerged volume and centre of buoyancy uniquely, which in turn lets force and torque balance pin two unknowns from one configuration.

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Submerged volume and centre of buoyancy

In the small-angle limit, the four projected corner heights are

$$y_\mathrm{BL}\approx 0,\quad y_\mathrm{BR}\approx -a,\quad y_\mathrm{TL}\approx +a,\quad y_\mathrm{TR}\approx 0.$$

The submerged region is everything below $y=0$. Looking at the rectangle, this is the lower triangular half with vertices

$$\mathrm{BL}=(0,0),\qquad \mathrm{TR}\approx(L,0),\qquad \mathrm{BR}\approx(L,-a).$$

In side view this triangle has area

$$A_\triangle = \tfrac12 \cdot L \cdot a.$$

Multiplying by the perpendicular cross-section width $a$ (the beam is uniform in that direction) gives the submerged volume

$$V_s = \tfrac12 L a^2 = \tfrac12 V_\mathrm{beam}.$$

So in this critical tilt, exactly half of the beam is submerged.

Centre of buoyancy

The centre of buoyancy is the centroid of the submerged volume. In the $z$ (across-beam) direction the wedge is uniform, so it sits at $z=a/2$. In the side-view plane, the centroid of the right triangle with vertices $(0,0),\,(L,0),\,(L,-a)$ is

$$\mathbf{r}_B = \left(\tfrac{0+L+L}{3},\; \tfrac{0+0-a}{3}\right) = \left(\tfrac{2L}{3},\,-\tfrac{a}{3}\right).$$

For torque balance only the $x$-coordinate matters (forces are vertical), and the relevant fact is

$$x_B = \tfrac{2L}{3}.$$

This is the most physically interesting number in the problem: the buoyancy lever arm is not $L/2$ (the mid-length of the beam) but $2L/3$, because the submerged volume is concentrated on the bird’s side. Forgetting this and writing $x_B=L/2$ makes the problem look over-determined and gives the wrong answer.

Centre of mass and bird position

The beam is uniform, so its centre of mass sits at the geometric centre of the rectangle. In small-angle approximation,

$$\mathbf{r}_\mathrm{CoM} \approx \left(\tfrac{L}{2},\,0\right) + O(a/L)\cdot a.$$

The bird sits on the upper face at the bird’s end, which in side view is the corner TR at $\approx (L,0)$. Its $x$-coordinate is

$$x_\mathrm{bird} = L$$

to leading order in $a/L$.

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Force and torque balance — solving for $\rho_b$ and $M$

Let $\rho_b$ be the beam density and $M$ the bird mass; both are unknown.

Vertical force balance

The buoyancy on a fully-submerged-by-volume $V_s$ region equals $\rho_w g V_s$. Setting upward = downward,

$$\rho_w g \cdot V_s = \rho_b g \cdot V_\mathrm{beam} + M g.$$

Substituting $V_s = a^2L/2$ and $V_\mathrm{beam}=a^2L$:

$$\rho_w \tfrac{a^2 L}{2} = \rho_b a^2 L + M. \qquad (\star)$$

Torque balance

Take torques about the origin (BL), with counter-clockwise positive. Vertical forces contribute torques proportional to their $x$-coordinates: $\tau_z = x F_y$.

• Buoyancy: $+\rho_w g V_s$ at $x_B=2L/3$, giving $\tau_B = +\frac{2L}{3}\,\rho_w g \cdot \frac{a^2 L}{2} = +\frac{\rho_w g a^2 L^2}{3}$.
• Beam weight: $-\rho_b g a^2 L$ at $x_\mathrm{CoM}=L/2$, giving $\tau_\mathrm{beam} = -\frac{\rho_b g a^2 L^2}{2}$.
• Bird weight: $-Mg$ at $x_\mathrm{bird}=L$, giving $\tau_\mathrm{bird} = -MgL$.

Summing to zero and dividing through by $gL$:

$$\frac{\rho_w a^2 L}{3} - \frac{\rho_b a^2 L}{2} - M = 0. \qquad (\star\star)$$

Solving the $2\times 2$ system

From $(\star)$: $M = a^2 L\!\left(\tfrac{\rho_w}{2} - \rho_b\right)$.

From $(\star\star)$: $M = a^2 L\!\left(\tfrac{\rho_w}{3} - \tfrac{\rho_b}{2}\right)$.

Equating the two and dividing by $a^2 L$:

$$\tfrac{\rho_w}{2} - \rho_b = \tfrac{\rho_w}{3} - \tfrac{\rho_b}{2} \quad\Longrightarrow\quad \tfrac{\rho_w}{6} = \tfrac{\rho_b}{2} \quad\Longrightarrow\quad \boxed{\rho_b = \tfrac{\rho_w}{3}}.$$

Plugging back into either equation,

$$M = a^2 L\left(\tfrac{\rho_w}{2}-\tfrac{\rho_w}{3}\right) = a^2 L\cdot\tfrac{\rho_w}{6} \quad\Longrightarrow\quad \boxed{M = \tfrac{1}{6}\,\rho_w a^2 L}.$$

Consistency check — unloaded float

Without any bird, force balance says the beam floats horizontally with submerged fraction $\rho_b/\rho_w = 1/3$. So the beam normally rides with $a/3$ of its thickness in the water and $2a/3$ above. Adding one bird tips it into the half-submerged tilted state described in the problem, and as we’ll see, four birds press it all the way to full submersion. The progression $1/3 \to 1/2 \to 1$ in submerged fraction is a useful sanity check.

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Buoyancy ceiling and the maximum $N$

The crucial physical idea now is a hard upper bound on the buoyant force a body can produce. No matter how the beam is tilted, the upward force from the water is

$$F_\mathrm{buoy} = \rho_w g V_\mathrm{submerged} \quad\le\quad \rho_w g V_\mathrm{beam} = \rho_w g a^2 L.$$

Equality is reached when the beam is fully submerged (top face just at the water surface). Any further loading sinks it.

For $N$ birds plus the beam, force balance demands

$$\rho_b g a^2 L + N M g \le \rho_w g a^2 L.$$

Substituting $\rho_b = \rho_w/3$ and $M = \rho_w a^2 L/6$:

$$\rho_w g a^2 L\left(\tfrac{1}{3} + \tfrac{N}{6}\right) \le \rho_w g a^2 L.$$

The factor $\rho_w g a^2 L$ cancels, leaving

$$\tfrac{1}{3} + \tfrac{N}{6} \le 1 \quad\Longrightarrow\quad N \le 4.$$

So the buoyancy ceiling is reached at exactly $N=4$, with the beam just fully submerged.

Why placement of birds doesn’t enter the bound

A common worry: surely where the birds are placed matters for how the beam tilts, and tilting changes how much volume is submerged, so maybe a clever placement can squeeze out more buoyancy?

It can’t. The vertical force balance reads

$$\rho_w g\,V_\mathrm{submerged} = (m_\mathrm{beam} + N M)\,g,$$

and the right-hand side is fixed by $N$. Tilting only redistributes which parts of the beam are wet; it can never make more than the beam’s full volume submerged. So $N$ is bounded by the buoyancy ceiling regardless of placement. Placement enters only as a separate existence check (next subsection): we still need to verify that some arrangement of $N=4$ birds gives a torque-balanced configuration.

Existence check at $N = 4$

When $N=4$ saturates the bound, the beam is fully submerged (top face at $y=0$) and horizontal. In this state the buoyancy acts at the geometric centre of the beam, $x=L/2$, and so does the beam’s weight. The four bird weights must therefore have their combined point of application at $x=L/2$ as well, i.e. their mean position must lie above the beam’s midpoint.

This is easy to satisfy: e.g.

• all four birds clustered at the midpoint $x=L/2$, or
• two birds at each end $x=0$ and $x=L$,
• or any configuration symmetric about $x=L/2$.

So $N=4$ is genuinely achievable; the configuration is not just a force-balance artefact.

Educational remark — active vs. passive constraints

This problem is a clean illustration of how to read a force-and-torque-balance system. The vertical force equation gave an inequality that bounded $N$. The torque equation gave an equality that, once $N$ is fixed, picks out a manifold of allowed bird positions (here, “any placement whose mean is at the beam’s midpoint”). One equation pins the count; the other picks the arrangement. Mixing the two — e.g. trying to use torque balance to push $N$ above 4 — is the classic mistake.

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Summary of results

Quantity	Value
Tilt angle in the one-bird configuration	$\tan\theta = a/L$
Submerged fraction of beam (one bird)	$1/2$
Centre of buoyancy lever arm	$x_B = 2L/3$
Beam density	$\rho_b = \rho_w/3$
Bird mass	$M = \rho_w a^2 L / 6$
Maximum number of birds	$\boxed{N_{\max} = 4}$