04-BS-7 · May 2018

Question 10 of 13: Stability of a Floating Square Bar

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

04-BS-7 Mechanics of Fluids — National Examination, 2018-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Graphical & Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.

Reference texts: White, F.M., Fluid Mechanics (8th ed.) — fluid statics and manometry (Ch. 2), Bernoulli and the energy equation (Ch. 3), viscous flow in ducts and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), potential flow and the Magnus effect (Ch. 8), open-channel flow and the hydraulic jump (Ch. 10), turbomachinery and jet propulsion (Ch. 11).

Check — assumptions used across this paper:

Q1's manometer chain is read off the extraction as a two-stage water–mercury–glycerine–(air)–glycerine–mercury system. The enclosed air pocket between the two glycerine columns is treated as weightless (uniform pressure), so only the one described open end is needed to close the hydrostatic chain back to pipe P; the second "opening" is not load-bearing for this calculation.
Q5's wave/hydraulic-jump analysis takes the depth "in front of the wave" (0.15 m, undisturbed, at rest) as the upstream state and "behind the wave" (0.75 m) as the downstream state, per the question's own prose (the raw figure-label ordering in the extraction is a reconstruction and is not used to override the stated text). The classic hydraulic-jump head-loss formula is applied to the given depths, and the swept flow rate uses the measured wave celerity directly — a standard engineering estimate, not a fully momentum-self-consistent bore solution.
Q6's air properties are taken at 20°C (domestic ambient, ρ=1.19 kg/m³) since no duct-air temperature is stated.
Q7(c)'s "discharged at right angles to the initial direction (20° becomes 0°)" is read as: the reverser's exhaust jet is normally angled 20° forward of the fully-radial (right-angle) direction; part (c) removes that forward lean entirely, leaving a purely radial (90° to the engine axis) discharge with zero axial velocity component.
Q8's terminal velocity and Q9's cable drag coefficient are obtained from the Reynolds-number relations the attached charts themselves plot (Morrison's sphere-drag correlation for Q8; the flat subcritical Cd≈1.2 plateau of the smooth-cylinder curve for Q9, since Re≈3×10⁴ falls solidly within it).

Question 10: Stability of a Floating Square Bar (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.

Both orientations float with the water level bisecting the section, since SG=0.5 ⇒ 50% submergence regardless of orientation.

Given. Square bar, side $a$, density = half that of water (SG=0.5), floating half-submerged in either the flat (horizontal) or edge-up/diamond (vertical) orientation.

Find. Which orientation is stable, with proof via the metacentric height $GM=BM-BG$.

Approach. For each orientation, locate the centre of buoyancy $B$ (centroid of the submerged area) and the fixed centre of gravity $G$ (geometric centre, since the bar is homogeneous), compute the metacentric radius $BM=I_{waterplane}/V_{submerged}$, and check the sign of $GM=BM-BG$.

Horizontal (flat) orientation. Waterline bisects the square at mid-height; draft $=a/2$, submerged rectangle width $a$. $$BG=\frac{a}{4}\qquad BM=\frac{I}{V}=\frac{a^3/12}{a\cdot(a/2)}=\frac{a}{6}\qquad GM=\frac{a}{6}-\frac{a}{4}=\boxed{-\frac{a}{12}\ (\text{unstable})}$$
Vertical (edge-up, diamond) orientation. The waterline again bisects the shape (a diamond's horizontal mid-diagonal always splits it into equal areas), so the submerged region is the lower triangular half, base $=a\sqrt2$ at the waterline: $$BG=\frac{a\sqrt2}{6}=0.236a\qquad BM=\frac{I}{V}=\frac{(a\sqrt2)^3/12}{a^2/2}=\frac{\sqrt2}{3}a=0.471a$$ $$GM=0.471a-0.236a=\boxed{+0.236a\ (\text{stable})}$$
Conclusion. The VERTICAL (edge-up, diamond) orientation is the stable one: its much wider waterplane at the waterline (diagonal $a\sqrt2$ versus the flat orientation's side $a$) gives a metacentric radius $BM$ nearly three times larger, decisively outweighing its slightly smaller $BG$ — the opposite of the intuitive "wider base is more stable" reasoning one might apply to a rigid block on land.

Orientation	BG	BM	GM	Stable?
Horizontal (flat)	0.250a	0.167a	−0.083a	No
Vertical (diamond)	0.236a	0.471a	+0.236a	Yes

Concept & Theory

Topic. Floating-body stability, metacentric height.

Key relations. $GM=BM-BG$, $BM=I_{waterplane}/V_{submerged}$; positive GM = stable (righting moment on tilt), negative GM = unstable (capsizing moment).

Why this works. Stability of a floating body depends on how the centre of buoyancy shifts relative to G when the body tilts slightly; that shift is governed almost entirely by the waterplane's second moment of area (via $I\propto\text{width}^3$), so a wider waterline strip — even one that also raises B closer to G — usually wins because the cube in the width dominates the linear terms in BG.

Common pitfall. Assuming the orientation with the shallower (higher, closer-to-G) centre of buoyancy is automatically more stable — here the diamond orientation is stable primarily because of its much larger $BM$, not because of a large gap between B and G.

Source. White, Fluid Mechanics, Ch. 2 (Buoyancy and stability).

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