Stability of a Floating Square Bar — Horizontal vs Vertical: 04-BS-7, December 2016, Q11

Question 11: Stability of a Floating Square Bar — Horizontal vs Vertical (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Fig. Q11 — centre of buoyancy B and centre of gravity G in the horizontal orientation (both terms comparable, O(side length a)) versus the vertical orientation (BG grows with the full bar length L, BM shrinks — grossly unstable).

Stability of a floating body is governed by the metacentric height GM = BM − BG, where BM = I_waterplane/V_submerged is the "restoring" term (bigger waterplane moment of inertia relative to submerged volume resists tipping) and BG is the fixed vertical offset between the centre of gravity and the centre of buoyancy (a destabilizing term whenever G sits above B, as it does here for a homogeneous half-submerged solid). Both orientations are analysed with the same formula, differing only in which cross-sectional dimension acts as the "beam" for the tipping mode being checked.

Horizontal orientation (long axis horizontal, square cross-section a×a, draft a/2). Rolling about the longitudinal axis, the waterplane is a long rectangle of beam a: $$BM_h = \frac{a^2}{12(a/2)} = \frac{a}{6}, \qquad BG_h = \frac{a}{2}-\frac{a}{4} = \frac{a}{4}$$ $$GM_h = \frac{a}{6}-\frac{a}{4} = \boxed{-\frac{a}{12}}$$ (a small deficit, of order the cross-section size a).
Vertical orientation (long axis vertical, length L, draft L/2). Tipping about a horizontal axis through the (now square, a×a) waterplane: $$BM_v = \frac{a^2}{12(L/2)} = \frac{a^2}{6L}, \qquad BG_v = \frac{L}{2}-\frac{L}{4} = \frac{L}{4}$$ $$GM_v = \frac{a^2}{6L}-\frac{L}{4}$$ For a "long" bar (L≫a), BM_v→0 while BG_v grows with L, so GM_v is deeply negative — e.g. at L=10a, $GM_v = \boxed{-2.48\,a}$, roughly 30× more negative than GM_h.
Conclusion. Both calculated GM values are technically negative for a mathematically perfect square/half-submerged case, but the horizontal deficit is of order the small cross-section dimension a, while the vertical deficit grows without bound as L increases. In practice, and for the two orientations offered, horizontal is overwhelmingly the stable (or least-unstable) orientation: a small disturbance from horizontal is resisted by a restoring term of the same order as the destabilizing one, whereas a bar standing on end is essentially unrestrained and will immediately capsize toward horizontal.

Quantity	Result
GM (horizontal)	−a/12 (small)
GM (vertical, L=10a)	−2.48a (large, grows with L)
Stable orientation	Horizontal

Topic: Metacentric stability of floating prismatic bodies.

Key relations: GM=BM−BG; BM=I_wp/V_sub; for a rectangular waterplane of beam b, I=Lb³/12.

Why this works. BM measures how strongly the waterplane's shape resists tipping (a wide, flat waterplane gives a large restoring moment for a given tip angle); BG measures how far the destabilizing weight sits above the buoyant support. A long bar standing on end presents a small (a×a) waterplane relative to its long submerged length, so BM collapses toward zero exactly as BG grows with the bar's length — the worst possible combination for stability.

Common pitfall. Assuming a floating body is stable simply because it floats (equilibrium exists) — equilibrium and STABILITY are different questions; GM<0 means the equilibrium is real but any small disturbance grows rather than decays.

Source: White, Fluid Mechanics, Ch. 2 (stability of floating bodies, metacentric height).

Question 11 of 13: Stability of a Floating Square Bar — Horizontal vs Vertical

Question 11: Stability of a Floating Square Bar — Horizontal vs Vertical (5 marks)