Floating Square Bar — Stable Orientation: 04-BS-7, May 2017, Q11

Question 11: Floating Square Bar — Stable Orientation (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.

Given. Long prismatic bar, square cross-section, side $w$; $SG=0.5$ (so exactly half the cross-sectional area is submerged in either orientation, by Archimedes' principle); orientation A: flat side up/down (upright square); orientation B: rotated 45°, corner up/down (diamond).

Both orientations submerge exactly half the cross-sectional area (SG=0.5); orientation B has a much wider waterline, giving it a larger metacentric radius.

Find. (i) The stable orientation. (ii) Depth of centre of buoyancy below the waterline, in terms of $w$, for each orientation. (iii) Explanation linking buoyancy-centre depth to stability.

Approach. For a long prismatic bar, roll stability is governed by the 2-D metacentric height $GM=BM-BG$, with $BM=I_{wl}/A_{sub}$ (waterline second moment per unit length over submerged area per unit length) computed for each orientation, then compared to decide which is positive (stable).

Orientation A (flat side up). Submerged depth $=0.5w$ (SG=0.5, uniform section), so $A_{sub}=0.5w^2$; waterline width $b_{wl}=w$ (full side, since the waterline cuts the vertical sides). Centre of buoyancy depth below the waterline: $\tfrac12(0.5w)=\boxed{0.25w}$. The centre of gravity G is at mid-height, which is exactly on the waterline, so G lies $0.25w$ above B and $BG=0.25w$. $$BM_A=\frac{I_{wl}}{A_{sub}}=\frac{w^3/12}{0.5w^2}=0.1667w \quad\Rightarrow\quad GM_A=BM_A-BG_A=0.1667w-0.25w=\boxed{-0.083w}$$
Orientation B (corner/diamond up). The waterline passes exactly through the geometric centre (by symmetry, since $SG=0.5$ submerges exactly half the diamond's area, which is exactly the bottom triangular half). Waterline width $b_{wl}=w\sqrt2$ (the full horizontal diagonal); submerged area $A_{sub}=0.5w^2$ (same as A). The submerged part is a triangle of height $w/\sqrt2=0.707w$ with its apex at the bottom, and its centroid lies one third of that height below the waterline: $$\text{depth of B below waterline}=\frac13\cdot\frac{w}{\sqrt2}=\boxed{\frac{w}{3\sqrt2}=0.236w}$$ (equivalently, apex-to-centroid $=\tfrac23(w/\sqrt2)=0.471w$, so the depth below the waterline is $0.707w-0.471w=0.236w$).
Metacentric height for B. $BM_B=I_{wl}/A_{sub}=(w\sqrt2)^3/12\big/0.5w^2=0.471w$; since G sits exactly at the waterline for this symmetric case, $BG_B=0.707w-0.471w=0.236w$: $$GM_B=BM_B-BG_B=0.471w-0.236w=\boxed{+0.236w}$$
Compare. $GM_A=-0.083w<0$ (unstable — any small tilt grows) while $GM_B=+0.236w>0$ (stable — any small tilt is restored). $\boxed{\text{Orientation B (corner/edge up) is the stable float.}}$

Orientation	Depth of B below waterline	Metacentric height GM	Stable?
A (flat side up)	0.25 w	−0.083 w	No
B (corner up)	0.236 w	+0.236 w	Yes

How the depth of the centre of buoyancy decides the orientation. Because $SG=0.5$, every line through the centroid cuts the square into two equal halves. So the waterline passes through G at every angle of roll, and the bar neither rises nor sinks as it turns. The only thing that changes with orientation is where the displaced water came from. That half-section of water, with its centroid at the depth of B, has effectively been lifted to the free surface, so the potential energy of the system is $PE(\theta)=mg\,d_B(\theta)+\text{constant}$. The bar settles where $d_B$ is smallest, i.e. with the centre of buoyancy as high as possible. Flat side up, $d_B=0.25w$ is the largest value over all roll angles, an energy maximum, so any small roll releases energy and the bar rolls over: unstable. Edge up, $d_B=w/(3\sqrt2)=0.236w$ is the smallest value, an energy minimum: stable. This matches the metacentric result, $GM_A<0$ and $GM_B>0$. The shallower centre of buoyancy (orientation B) marks the stable float. The difference in depth is small (0.25w versus 0.236w), but its sign is what matters, and the metacentric heights show how decisive it is.

Concept & Theory

Topic. Stability of floating bodies; the metacentric-height criterion for a long prismatic (2-D) section.

Key relations. $GM=BM-BG$; $BM=I_{wl}/A_{sub}$ (per unit length for a prism); stable iff $GM>0$ (metacentre above centre of gravity).

Why this works. When a floating body tilts slightly, the centre of buoyancy shifts sideways as the submerged wedge shape changes; $BM$ measures how strongly that shift restores an upright position, while $BG$ measures how much the body's own weight, acting through a fixed $G$, works against that restoration. A body needs a wide waterplane (large $I_{wl}$) relative to its submerged bulk to be stable, which is exactly why ships are wide and flat rather than tall and narrow.

Common pitfall. Assuming a deeper centre of buoyancy means a more stable float. When G stays at the same height relative to the waterline in every orientation, as it does here, the opposite is true: the stable orientation is the one with the shallowest centre of buoyancy (minimum potential energy). Orientation B has the shallower B and is the stable one, confirmed independently by $GM_B>0$.

Source. White, Fluid Mechanics, Ch. 2 (Stability of Immersed and Floating Bodies).

Question 11 of 13: Floating Square Bar — Stable Orientation

Question 11: Floating Square Bar — Stable Orientation (5 marks)