Problem Set

NBPhO 2025

6. Birds 4 pts

Mechanics · Statics, Buoyancy, Torque equilibrium

Find the maximum number of identical birds a floating square-cross-section beam can hold above water, given a specific one-bird tipping condition.

High-level summary by Claude.

Ingredients torque equilibriumArchimedes principlecentre of buoyancyfull-submersion buoyancy ceiling
Tags mechanicsstaticsbuoyancyarchimedes-principletorquecentre-of-masscentre-of-buoyancytilted-floatfull-submersion-limit

Difficulty medium

Prerequisites

  • Archimedes' principle and buoyancy on a partially submerged body
  • Static force and torque balance for a rigid body
  • Centre of mass of a uniform rigid body
  • Centroid of a triangular region (centre of buoyancy of a wedge)
  • Small-angle approximation, sin⁡θ≈θ≪1\sin\theta\approx\theta\ll 1 for a long thin object

Learning objectives

  • Pin a tilted-float configuration from two geometric edge-at-water-line conditions, getting tan⁡θ=a/L\tan\theta = a/L
  • Compute submerged volume and centre of buoyancy of a triangular wedge by integration and as the centroid of a right triangle
  • Use force balance and torque balance as a 2×22\times 2 system to extract two unknowns (ρb\rho_b and MM) from one critical configuration
  • Recognise that the full-submersion buoyancy ρwgVbeam\rho_w g V_\mathrm{beam} is an absolute upper bound on the load a body can carry
  • Distinguish active constraints from existence checks: which equation pins the count NN, and which only restricts where the birds may sit

Watch out for

  • Forgetting that the bird's mass MM is not given — it must be deduced from the one-bird configuration via both force and torque balance, not from force balance alone.
  • Treating only one of the two stated edge-at-water-line conditions as active. Both are needed: one fixes the tilt geometrically, the other closes the system of equilibrium equations.
  • Believing the answer depends on where the birds are placed. Force balance alone bounds N≤4N\le 4; torque balance is only an existence check on whether some placement works (it does, e.g. all birds at the midpoint).
  • Overlooking that the buoyancy ceiling is the fully submerged beam's displacement ρwga2L\rho_w g a^2 L. There is no way to extract more upward thrust by tilting or clever placement.
  • Computing the centre of buoyancy of the tilted-beam wedge as the beam's mid-length L/2L/2 rather than the triangle's centroid at 2L/32L/3 from the dry end (equivalently L/3L/3 from the bird's end). The buoyancy acts at the centroid of the submerged volume, not the beam.