Stable Floating Orientation of a Square Timber: 04-BS-7, December 2013, Q11

Question 11: Stable Floating Orientation of a Square Timber (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.

Orientation A (flat side up) vs. B (edge up, diamond); G = centre of gravity, B = centre of buoyancy, waterline shown for SG = 0.50.

With specific gravity SG = 0.50, exactly half the bar's cross-sectional area is submerged in either orientation, since a freely floating body always displaces its own weight. Writing the side length as s, the centre of gravity G sits at the geometric centre of the square in both orientations — but where the centre of buoyancy B sits, and how the submerged cross-section's shape changes as the bar tips, differ completely between the two.

Orientation A (flat side up). The submerged region is a rectangle of draught $0.50s$, so its centroid — the centre of buoyancy — sits at depth $B_A = 0.25s$ below the waterline. Because the waterplane at this orientation is only as wide as the square itself (width s), the metacentric radius $BM=I/V=(s^3/12)/(0.5s^2)=s/6$ is small, while $G$ sits a full $BG=0.25s$ above $B$: the restoring arm $GM=BM-BG=s/6-s/4=-0.083s$ is negative — orientation A is unstable.

Orientation B (edge up, diamond). At SG = 0.50 the waterline sits exactly at the diamond's widest point (the horizontal diagonal, width $s\sqrt2$), and the submerged shape is the lower triangular half of the diamond. A triangle's centroid sits one-third of its own height up from its base (here, the apex-down triangle's base is at the waterline), so the centre of buoyancy is at depth $B_B=\tfrac13\!\left(\tfrac{s\sqrt2}{2}\right)=0.236s$ below the waterline — shallower than orientation A's $0.25s$. But the waterplane here is far wider ($s\sqrt2\approx1.41s$ vs. just $s$), so the metacentric radius is much larger: $BM=I/V=\big((s\sqrt2)^3/12\big)/(0.5s^2)=0.471s$, comfortably exceeding $BG=0.236s$, giving $GM=+0.236s$ — positive and stable.

Conclusion: orientation B (edge up) is the stable one at SG = 0.50; orientation A (flat side up) is unstable and would roll toward B if disturbed.

Quantity	Value
Depth of B, orientation A (flat)	0.250 s
Depth of B, orientation B (edge up)	0.236 s
Metacentric height GM, orientation A	−0.083 s (unstable)
Metacentric height GM, orientation B	+0.236 s (stable)

Topic: Metacentric stability of a floating prismatic body.

Key relations: $GM=BM-BG$; $BM=I_{waterplane}/V_{displaced}$; stable iff $GM>0$ (metacentre above centre of gravity).

Why this works: the depth of B alone is NOT the deciding factor — here B is actually shallower in the stable orientation (B) than in the unstable one (A). What matters is $GM$, which weighs how far B sits below G ($BG$) against how strongly the WATERPLANE's second moment of area resists tipping ($BM=I/V$); a wider waterplane at the same displaced volume gives a much larger righting moment for the same disturbance, and that effect can (and does, here) outweigh a shallower B.

Common pitfall: assuming "the orientation with the deeper B is always more stable" — this problem is a direct counter-example: B is shallower in the STABLE orientation, because the wide diamond waterplane's moment of inertia dominates the comparison.

Source: Crowe, Elger & Roberson, Engineering Fluid Mechanics, Ch. 3 (stability of floating bodies); White, Fluid Mechanics, Ch. 2.

Question 11 of 13: Stable Floating Orientation of a Square Timber

Question 11: Stable Floating Orientation of a Square Timber (5 marks)