Question 10 of 13: Stable Floating Orientation of a Square Bar
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examinations, December 2014. Three (3) hours, closed book. Section A: Calculative (9 questions, do 7); Section B: Analytical/Graphical (4 questions, do 3). Ten questions constitute a complete paper (50 marks). Every printed question is solved below, including the two "extra" questions in each Section beyond the minimum required.
Fig. Q10 — the two candidate orientations at SG = 0.5 (both float exactly half-submerged by area/volume): flat-face-down (left) vs. corner-down/diamond (right).
Because the bar's density is exactly half that of water, BOTH orientations satisfy floating equilibrium (each is symmetric about its own half-submerged waterline, so the centre of gravity G sits exactly at the waterline in both cases). Equilibrium alone does not tell us which is stable — that requires comparing the metacentric height GM = BM − BG for each, where BM = Iwaterplane/Vsubmerged (I about the longitudinal roll axis) and BG is the vertical distance from the centre of buoyancy B up to G. A floating body is stable only if GM > 0, i.e. the metacentre M sits above G.
Given. Square bar, side length a, SG = 0.5 (working in units of the side length a, so the sign of GM — and hence the comparison — is independent of the bar's actual size).
Find. Which orientation (horizontal or vertical/diamond) is stable, proven via GM.
Horizontal (flat-face-down) orientation. Waterline at mid-height; G is at the centroid of the whole square (exactly at the waterline). The submerged rectangle (width a, depth a/2) has its centroid a/4 below the waterline, so BG = a/4. The waterplane is a strip of width a, giving BM = I/V = (a³/12)/(a²/2) = a/6:
$$GM_{horizontal} = BM-BG = \frac{a}{6}-\frac{a}{4} = \boxed{-\,0.0833\,a \;\;(\text{negative} \Rightarrow \text{UNSTABLE})}$$
Vertical (diamond, corner-down) orientation. Let p = a/√2 (half-diagonal). The diamond is symmetric about its horizontal midline (the widest point), so G again sits exactly at the waterline. The submerged triangle (base 2p, height p) has centroid p/3 below the waterline, so BG = p/3 = 0.2357a. The waterplane at the widest point has width 2p, giving BM = I/V = [(2p)³/12]/p² = 2p/3 = 0.4714a:
$$GM_{vertical} = BM-BG = 0.4714a - 0.2357a = \boxed{+\,0.2357\,a \;\;(\text{positive} \Rightarrow \text{STABLE})}$$
Conclusion. The VERTICAL (diamond, corner-down) orientation is the stable one — despite its centre of buoyancy sitting slightly SHALLOWER (0.2357a < 0.250a) than in the horizontal case, its much wider waterplane at the critical level gives a metacentric radius BM large enough to overcome BG, flipping GM positive.
