CELE Hydraulics & Fluid Mechanics — Buoyancy and FlotationExam Answer Templates

Marks

1

Topic

Archimedes' Principle

Difficulty

easy

Template Id

T1

Examiner Tip

The single non-negotiable phrase is 'weight of the fluid displaced.' Missing this key phrase and writing only 'volume displaced' earns zero for a 1-mark definition item.

Model Answer

Archimedes' Principle states that any body fully or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces: F_B = γ_fluid × V_displaced.

Question Type

very_short_answer

Answer Structure

• One complete sentence: state the principle with the formula F_B = γ_fluid × V_disp [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'volume of fluid displaced' instead of 'weight of fluid displaced' — zero marks.
• Omitting the direction (upward) of the buoyant force.
• Writing the formula with body volume instead of displaced volume for a floating body.

Key Phrases To Include

• upward buoyant force
• weight of fluid displaced
• F_B = γ_fluid × V_displaced

Marks

1

Topic

Flotation — Draft and Fraction Submerged

Difficulty

easy

Template Id

T2

Examiner Tip

This is a recall-plus-application 1-mark item. Write the rule first, then the answer. It shows the examiner you know the principle, not just the number.

Model Answer

For a floating homogeneous body, the fraction submerged equals its specific gravity relative to the fluid. Since SG_wood = 0.75 and SG_water = 1.0, the fraction submerged = 0.75/1.00 = 0.75 (i.e., 75% of the block is below the waterline).

Question Type

very_short_answer

Answer Structure

• State the flotation rule: fraction submerged = SG_body / SG_fluid [0.5 mark]
• Substitute and state the answer: 0.75/1.00 = 0.75 or 75% [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Answering 25% submerged (confusing fraction exposed with fraction submerged).
• No reasoning given — just writing '0.75' without justification may receive partial credit only.

Key Phrases To Include

• fraction submerged = SG_body / SG_fluid
• 0.75
• 75%

Marks

2

Topic

Stability — Metacenter and Metacentric Height

Difficulty

easy

Template Id

T3

Examiner Tip

Many students lose the definition mark by vaguely saying 'M is a point on the ship.' The examiner expects the geometric construction: intersection of the shifted buoyancy line of action with the body's original centerline.

Model Answer

The metacenter M is the point where the line of action of the buoyant force (after a small angular tilt) intersects the original vertical axis through the center of buoyancy B. The metacentric height is GM = BM − BG, where BM = I / V_disp (I = second moment of area of the waterplane about the tilting axis). A floating body is stable when M lies above G (GM > 0); unstable when M lies below G (GM < 0); and in neutral equilibrium when M coincides with G (GM = 0).

Question Type

short_answer

Answer Structure

• Sentence 1: Define the metacenter M geometrically — intersection of tilted buoyancy line with original vertical axis [1 mark]
• Sentence 2: State stability condition — GM > 0 for stable, GM < 0 for unstable, GM = 0 for neutral [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining M as the center of buoyancy — confusing B and M.
• Stating stability condition as 'B above G' instead of 'M above G'.
• Omitting the formula GM = BM − BG when asked for a mathematical definition.

Key Phrases To Include

• metacenter
• line of action of buoyant force after tilt
• original vertical axis
• GM = BM − BG
• stable when GM > 0
• M above G

Marks

2

Topic

Flotation — Draft Computation

Difficulty

easy

Template Id

T4

Examiner Tip

The shortcut d = SG × h works only for a homogeneous block floating in water (SG_fluid = 1.0). Show this shortcut as a check after the full solution to demonstrate understanding.

Model Answer

Given: cube side h = 0.5 m, SG = 0.60, γ_w = 9.81 kN/m³. (a) Weight of cube: W = SG × γ_w × V_body = 0.60 × 9.81 × (0.5)³ = 0.60 × 9.81 × 0.125 = 0.7358 kN. For a floating body in equilibrium, F_B = W = 0.7358 kN. (b) Draft d: V_disp = W / γ_w = 0.7358 / 9.81 = 0.07500 m³. Plan area A = 0.5 × 0.5 = 0.25 m². d = V_disp / A = 0.0750 / 0.25 = 0.300 m. Check: d = SG × h = 0.60 × 0.5 = 0.30 m ✓ Answers: F_B = 0.736 kN; Draft d = 0.30 m.

Question Type

numerical

Answer Structure

• State given data clearly [no mark, but good practice]
• Part (a): W = SG × γ_w × V — compute weight = buoyant force [1 mark]
• Part (b): d = SG × h or d = V_disp / A — correct draft computation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the full cube volume as V_disp in a floating (not submerged) problem — overestimates F_B.
• Forgetting to check units: mixing kN and N in the same equation.
• Not showing the equilibrium statement F_B = W before computing draft.

Key Phrases To Include

• F_B = W (equilibrium condition)
• W = SG × γ_w × V_body
• d = V_disp / A
• d = SG × h

Marks

3

Topic

Archimedes' Principle — Fully Submerged Body and Apparent Weight

Difficulty

medium

Template Id

T5

Examiner Tip

State 'fully submerged' explicitly to justify using V_disp = V_body. This one phrase protects you from losing the first method mark even if your arithmetic has a minor error.

Model Answer

Given: D = 0.10 m, L = 0.20 m, SG_steel = 7.85, γ_w = 9.81 kN/m³ = 9810 N/m³. Step 1 — Volume of cylinder: V = π/4 × D² × L = π/4 × (0.10)² × 0.20 = 1.5708 × 10⁻³ m³. Step 2 — Buoyant force (Archimedes' Principle, fully submerged, V_disp = V_body): F_B = γ_w × V = 9810 × 1.5708 × 10⁻³ = 15.41 N. Step 3 — Weight of the cylinder: W = SG × γ_w × V = 7.85 × 9810 × 1.5708 × 10⁻³ = 120.95 N. Step 4 — Tension T (vertical equilibrium: T + F_B = W): T = W − F_B = 120.95 − 15.41 = 105.54 N ≈ 105.5 N. Answers: F_B = 15.4 N; T = 105.5 N.

Question Type

numerical

Answer Structure

• Step 1: Compute cylinder volume V = πD²L/4 [0.5 mark]
• Step 2: F_B = γ_w × V (full volume since fully submerged) [1 mark]
• Step 3: W = SG × γ_w × V [0.5 mark]
• Step 4: Equilibrium T = W − F_B, correct answer with unit [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using V_disp ≠ V_body — applying floating formula to a submerged body.
• Equilibrium error: writing T = W + F_B (adding instead of subtracting F_B).
• Using diameter instead of radius in the area formula without squaring correctly.
• Not stating 'fully submerged' as the basis for V_disp = V_body.

Key Phrases To Include

• fully submerged — V_disp = V_body
• F_B = γ_w × V
• W = SG × γ_w × V
• T + F_B = W (vertical equilibrium)
• apparent weight = W − F_B

Marks

3

Topic

Stability — Center of Buoyancy and Center of Gravity

Difficulty

medium

Template Id

T6

Examiner Tip

For a 3-mark conceptual question, write exactly 3 focused points — one per mark. Avoid writing a paragraph that buries the key phrases. Examiners scan for specific terms.

Model Answer

Center of Buoyancy B: B is the centroid of the displaced fluid volume (the submerged portion of the floating body). The buoyant force F_B acts vertically upward through B. As the body tilts, B shifts because the shape of the displaced volume changes. Center of Gravity G: G is the centroid of the entire body's mass distribution. The weight W acts vertically downward through G. For a rigid body with fixed mass distribution, G does not move when the body tilts. Role in Stability: When a floating body tilts by a small angle θ, B shifts to B', creating a righting couple (W × GZ) if M is above G. Stability is governed by the metacentric height GM = BM − BG. If GM > 0 (M above G), the righting moment W·GM·sinθ restores the body to upright — stable. If GM < 0 (M below G), the moment overturns the body — unstable.

Question Type

short_answer

Answer Structure

• Define B — centroid of displaced volume, buoyancy acts upward through B, shifts when tilted [1 mark]
• Define G — centroid of body mass, weight acts downward through G, fixed for rigid body [1 mark]
• Stability role — GM = BM − BG; stable if M above G (GM > 0); righting moment W·GM·sinθ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing B and G definitions — stating G is the centroid of displaced volume.
• Not mentioning that B shifts when the body tilts (key to understanding why metacenter matters).
• Omitting the stability conclusion or formula GM = BM − BG.

Key Phrases To Include

• centroid of displaced volume
• centroid of body mass
• buoyancy acts upward through B
• weight acts downward through G
• B shifts when tilted
• GM = BM − BG
• stable if M above G

Marks

2

Topic

Flotation — Displaced Volume in Seawater

Difficulty

medium

Template Id

T7

Examiner Tip

Always write out the seawater conversion γ_sw = 1.025 × 9.81 explicitly. Board exam problems often include seawater to test whether you blindly use 9.81.

Model Answer

Given: displacement W = 8,000 tonnes = 8,000 × 9.81 kN = 78,480 kN (or W = 8,000 × 1000 × 9.81 N = 78.48 × 10⁶ N). γ_sw = SG × γ_w = 1.025 × 9.81 = 10.055 kN/m³. By flotation equilibrium, F_B = W: V_disp = W / γ_sw = 78,480 kN / 10.055 kN/m³ V_disp = 7,805 m³. Alternatively: mass = 8,000 × 10³ kg; ρ_sw = 1.025 × 1000 = 1025 kg/m³; V_disp = 8 × 10⁶ / 1025 = 7,805 m³. Answer: V_disp ≈ 7,805 m³.

Question Type

numerical

Answer Structure

• Convert displacement to weight or mass and compute γ_sw [0.5 mark]
• Apply V_disp = W / γ_sw (flotation equilibrium F_B = W) [1 mark]
• State correct answer with unit m³ [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using γ_w = 9.81 kN/m³ (fresh water) instead of γ_sw = 10.055 kN/m³ for seawater.
• Converting tonnes to kN incorrectly: 1 tonne-force = 9.81 kN (not 10 kN).
• Answering in m³ but using fresh water — answer is 5% too high.

Key Phrases To Include

• γ_sw = 1.025 × 9.81 = 10.055 kN/m³
• F_B = W (flotation equilibrium)
• V_disp = W / γ_sw
• 7,805 m³

Marks

5

Topic

Stability — Metacentric Height of a Rectangular Pontoon

Difficulty

hard

Template Id

T8

Examiner Tip

The most penalized mistake in board exams is using I = BL³/12 instead of I = LB³/12 for rolling. Remember: for rolling, the tilting axis is parallel to L (the long axis), so B (the width) goes into the cube: I = LB³/12.

Model Answer

Given: B_width = 5 m, L = 12 m, draft d = 1.5 m, KG = 1.2 m, γ_w = 9.81 kN/m³. Step 1 — Displaced volume: V_disp = L × B × d = 12 × 5 × 1.5 = 90 m³. Step 2 — Second moment of area of waterplane (rolling about the long axis, axis parallel to L): I = L × B³ / 12 = 12 × (5)³ / 12 = 12 × 125 / 12 = 125 m⁴. Step 3 — Metacentric radius: BM = I / V_disp = 125 / 90 = 1.389 m. Step 4 — Elevation of center of buoyancy B above keel (KB): KB = d / 2 = 1.5 / 2 = 0.75 m. Step 5 — BG (distance from B to G): KG = 1.2 m (given), KB = 0.75 m. BG = KG − KB = 1.2 − 0.75 = 0.45 m (G is above B). Step 6 — Metacentric height: GM = BM − BG = 1.389 − 0.45 = 0.939 m. Step 7 — Stability assessment: GM = +0.939 m > 0 → M is above G → the pontoon is STABLE against rolling. Summary: BM = 1.389 m; KB = 0.75 m; BG = 0.45 m; GM = +0.939 m — STABLE.

Question Type

numerical

Answer Structure

• Step 1: V_disp = L × B × d = 90 m³ [0.5 mark]
• Step 2: I = L × B³/12 = 125 m⁴ (waterplane second moment about rolling axis) [1 mark]
• Step 3: BM = I / V_disp = 1.389 m [1 mark]
• Step 4: KB = d/2 = 0.75 m; BG = KG − KB = 0.45 m [1 mark]
• Step 5: GM = BM − BG = +0.939 m; explicit stability conclusion — STABLE [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using I = BL³/12 instead of I = LB³/12 for rolling (swapping B and L in the formula) — most common error, loses 1 mark.
• Taking V_disp as full body volume instead of L × B × d (draft-based).
• Computing BG as KB − KG (getting negative BG) without recognizing G is above B.
• Omitting the stability conclusion — losing the last 1 mark.
• Not identifying the tilting axis (rolling = about the long axis; pitching = about the short axis).

Key Phrases To Include

• I = LB³/12 (waterplane about rolling axis)
• V_disp = L × B × d
• BM = I / V_disp
• KB = d/2
• BG = KG − KB
• GM = BM − BG
• GM > 0 → stable

Marks

1

Topic

Flotation — Draft

Difficulty

easy

Template Id

T9

Examiner Tip

For 1-mark definition items, always pair the verbal definition with the formula to secure full marks with minimal risk.

Model Answer

Draft d is the vertical depth of a floating body measured from the waterline to the lowest point of the body (keel). For a prismatic floating body: d = V_disp / A, where V_disp = W/γ_fluid and A is the plan area at the waterline.

Question Type

very_short_answer

Answer Structure

• One sentence: definition of draft as vertical depth to keel [0.5 mark]
• Formula: d = V_disp / A [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Defining draft as the 'total height of the body' — incorrect, it is only the submerged depth.
• Omitting the formula when the question implies a mathematical definition.

Key Phrases To Include

• depth from waterline to keel
• d = V_disp / A
• V_disp = W / γ_fluid

Marks

2

Topic

Archimedes' Principle — Floating Body

Difficulty

medium

Template Id

T10

Examiner Tip

For floating bodies, the fastest and most reliable path to F_B is: F_B = W = SG × γ_w × V_body. There is no need to compute the submerged volume separately.

Model Answer

Step 1 — Floating equilibrium: F_B = W. W = SG × γ_w × V_log. V_log = π/4 × D² × L = π/4 × (0.4)² × 3 = π/4 × 0.16 × 3 = 0.3770 m³. W = 0.55 × 9.81 × 0.3770 = 2.033 kN. Since the log floats, F_B = W = 2.033 kN. Note: Only 55% of the log's volume is submerged (SG = 0.55), confirming it floats. Answer: F_B = 2.03 kN.

Question Type

numerical

Answer Structure

• Compute log volume V = πD²L/4 = 0.377 m³ [0.5 mark]
• Use F_B = W = SG × γ_w × V = 2.03 kN for floating body [1 mark]
• State answer with correct unit [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Computing F_B = γ_w × V_body (ignoring SG) — this gives the buoyant force if fully submerged, not the weight.
• Mixing up radius and diameter: using D = 0.4 m in πr² instead of π(D/2)².
• Not recognizing that for a floating body F_B = W — trying to compute F_B separately without equilibrium.

Key Phrases To Include

• floating equilibrium: F_B = W
• V = πD²L/4
• W = SG × γ_w × V
• 2.03 kN

Marks

3

Topic

Stability — Righting Moment

Difficulty

medium

Template Id

T11

Examiner Tip

Board exam derivation questions reward clear step-by-step logical flow over length. Three focused steps — geometry, formula, interpretation — earn all 3 marks more reliably than a long vague paragraph.

Model Answer

When a floating body heels through a small angle θ, the center of buoyancy shifts from B to B', and the buoyant force F_B = W now acts vertically upward through the new position B'. This line of action intersects the original vertical centerline at the metacenter M. The horizontal distance between the line of action of W (through G) and F_B (through B') is the righting arm GZ: GZ = GM × sin θ. The righting moment (restoring couple) is: M_R = W × GZ = W × GM × sin θ Where: - W = total weight of the floating body (kN) - GM = metacentric height (m) = BM − BG = I/V_disp − BG - θ = heel angle (degrees or radians), assumed small (sin θ ≈ θ for θ < 10°) - A positive GM gives a restoring (righting) moment → stable equilibrium. - A negative GM gives an overturning moment → unstable equilibrium.

Question Type

short_answer

Answer Structure

• Describe the geometry of tilt — B shifts to B', buoyancy line intersects centerline at M [1 mark]
• State righting arm GZ = GM sin θ and righting moment M_R = W × GM × sin θ [1 mark]
• Explain each term and stability implication of sign of GM [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not mentioning the shift of B to B' — missing the geometric basis of the derivation.
• Writing M_R = F_B × GM (incorrect — it is W, not F_B separately, since F_B = W in equilibrium).
• Not explaining the sign convention of GM.

Key Phrases To Include

• center of buoyancy shifts from B to B'
• metacenter M
• righting arm GZ = GM sin θ
• righting moment = W × GM × sin θ
• positive GM → restoring moment
• negative GM → overturning

Marks

5

Topic

Stability — Upright Floating Cylinder

Difficulty

hard

Template Id

T12

Examiner Tip

For circular cross-sections, I = πD⁴/64. The polar moment J = πD⁴/32 is for torsion, not buoyancy. Writing down the formula before substituting protects you from this classic confusion.

Model Answer

Given: D = 0.8 m, H = 1.6 m, SG = 0.80, γ_w = 9.81 kN/m³. Step 1 — Draft d (depth submerged for upright floating): d = SG × H = 0.80 × 1.6 = 1.28 m. Step 2 — Displaced volume: V_disp = π/4 × D² × d = π/4 × (0.8)² × 1.28 = 0.6434 m³. Step 3 — Second moment of area of waterplane (circular cross-section): I = π D⁴/64 = π × (0.8)⁴ / 64 = π × 0.4096 / 64 = 0.02011 m⁴. Step 4 — Metacentric radius: BM = I / V_disp = 0.02011 / 0.6434 = 0.03126 m ≈ 0.031 m. Step 5 — Elevation of B above keel: KB = d/2 = 1.28/2 = 0.640 m. Step 6 — Elevation of G above keel (homogeneous cylinder): KG = H/2 = 1.6/2 = 0.800 m. Step 7 — BG: BG = KG − KB = 0.800 − 0.640 = 0.160 m (G is above B). Step 8 — Metacentric height: GM = BM − BG = 0.031 − 0.160 = −0.129 m. Step 9 — Stability assessment: GM = −0.129 m < 0 → M is BELOW G → the cylinder is UNSTABLE when floating upright. Conclusion: The cylinder will tend to capsize (fall on its side). This is typical for tall slender cylinders with SG < 1.

Question Type

numerical

Answer Structure

• Step 1: d = SG × H = 1.28 m [0.5 mark]
• Step 2: V_disp = πD²d/4 = 0.643 m³ [0.5 mark]
• Step 3: I = πD⁴/64 = 0.02011 m⁴ (circular waterplane) [1 mark]
• Step 4: BM = I/V_disp = 0.031 m [0.5 mark]
• Step 5-7: KB = 0.64 m; KG = 0.80 m; BG = 0.16 m [1 mark]
• Step 8-9: GM = BM − BG = −0.129 m < 0 → UNSTABLE [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Using I = πD⁴/32 (polar moment of inertia, for torsion) instead of I = πD⁴/64 (area moment) — most common error.
• Setting KG = d/2 (using draft instead of total height H for G of the whole cylinder).
• Computing GM = BG − BM (reversing the formula sign).
• Not stating the explicit instability conclusion.

Key Phrases To Include

• d = SG × H (homogeneous, floating in water)
• I = πD⁴/64 (circular waterplane)
• BM = I / V_disp
• KB = d/2; KG = H/2 (homogeneous)
• GM = BM − BG
• GM < 0 → unstable

Marks

2

Topic

Archimedes' Principle — Apparent Weight

Difficulty

medium

Template Id

T13

Examiner Tip

Memorize the shortcut: W_apparent = W_air(1 − 1/SG) for a fully submerged body in water. It is fast, accurate, and shows formula mastery — which examiners reward.

Model Answer

Given: W_air = 3.0 kN, SG_concrete = 2.4, γ_w = 9.81 kN/m³. Step 1 — Volume of concrete block: V = W_air / (SG × γ_w) = 3.0 / (2.4 × 9.81) = 3.0 / 23.544 = 0.12742 m³. Step 2 — Buoyant force (fully submerged): F_B = γ_w × V = 9.81 × 0.12742 = 1.250 kN. Step 3 — Apparent weight: W_apparent = W_air − F_B = 3.0 − 1.250 = 1.750 kN. Alternative (direct formula): W_apparent = W_air × (1 − 1/SG) = 3.0 × (1 − 1/2.4) = 3.0 × (0.5833) = 1.750 kN. Answer: Apparent weight in water = 1.75 kN.

Question Type

numerical

Answer Structure

• Find volume V from W_air and SG, then compute F_B = γ_w × V [1 mark]
• W_apparent = W_air − F_B = 1.75 kN (or use direct formula) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing W_apparent = W_air + F_B (adding buoyancy instead of subtracting — conceptual error).
• Using the floating formula V_disp ≠ V_body (concrete sinks, it is fully submerged).
• Final answer in N instead of kN causing a factor-of-1000 error.

Key Phrases To Include

• V = W_air / (SG × γ_w)
• F_B = γ_w × V
• W_apparent = W_air − F_B
• apparent weight = W_air(1 − 1/SG)

Marks

2

Topic

Stability — Engineering Applications

Difficulty

easy

Template Id

T14

Examiner Tip

Board exam application questions expect you to name a specific structure AND explain the engineering purpose of GM. Generic answers like 'for safety' earn no marks.

Model Answer

1. Design of barges, pontoons, and floating platforms: The metacentric height GM is computed during the design phase to ensure GM > 0 with adequate margin (typically GM ≥ 0.3 m to 1.0 m for different vessel types). This prevents capsizing due to wave action, cargo shifting, or wind loads. 2. Design of caissons and floating bridge components: During towing and launching of large caissons (e.g., for bridge foundations and port works), engineers compute GM to ensure the structure remains stable in its temporary floating condition before it is sunk and founded. A low or negative GM triggers ballasting or geometric modifications.

Question Type

short_answer

Answer Structure

• Application 1: Barge/pontoon design — GM determines safe operating condition [1 mark]
• Application 2: Caisson/floating structure during construction — temporary floating stability check [1 mark]

Scoring Breakdown

Common Mark Deductions

• Vague answers like 'used in ships' without explaining GM's specific role.
• Giving two versions of the same application (e.g., 'ships' and 'boats') — counted as one.

Key Phrases To Include

• metacentric height GM > 0
• prevent capsizing
• barge/pontoon/floating platform
• caisson stability during towing
• ballasting

Marks

5

Topic

Stability — Complete Analysis with Righting Moment

Difficulty

hard

Template Id

T15

Examiner Tip

When GM comes out negative, the examiner expects you to explicitly state 'this is an OVERTURNING moment, not a righting moment.' Many students lose marks by computing |M_R| without explaining the sign.

Model Answer

Given: B = 3 m, L = 8 m, d = 1.0 m, KG = 1.4 m, W = 235.44 kN, θ = 5°, γ_w = 9.81 kN/m³. Step 1 — Displaced volume: V_disp = L × B × d = 8 × 3 × 1.0 = 24 m³. Step 2 — Waterplane second moment of area (rolling about long axis — axis parallel to L = 8 m): I = L × B³ / 12 = 8 × (3)³ / 12 = 8 × 27 / 12 = 18 m⁴. Step 3 — Metacentric radius: BM = I / V_disp = 18 / 24 = 0.750 m. Step 4 — Center of buoyancy elevation: KB = d/2 = 1.0/2 = 0.500 m. Step 5 — BG: BG = KG − KB = 1.4 − 0.5 = 0.900 m (G is above B). Step 6 — Metacentric height: GM = BM − BG = 0.750 − 0.900 = −0.150 m. Step 7 — Stability assessment: GM = −0.150 m < 0 → M is BELOW G → the barge is UNSTABLE. The barge will capsize at this loading condition (G is too high; draft is insufficient for stability). Step 8 — Righting (or overturning) moment at θ = 5°: M_R = W × GM × sin θ = 235.44 × (−0.150) × sin 5° = 235.44 × (−0.150) × 0.08716 = −3.079 kN·m. The negative sign confirms an overturning (capsizing) moment, not a righting moment. Conclusion: GM = −0.15 m (unstable); overturning moment = 3.08 kN·m at 5° heel.

Question Type

numerical

Answer Structure

• Step 1: V_disp = L × B × d = 24 m³ [0.5 mark]
• Step 2: I = LB³/12 = 18 m⁴ (correct axis) [1 mark]
• Step 3-4: BM = 0.750 m; KB = 0.50 m [0.5 mark]
• Step 5-6: BG = 0.90 m; GM = −0.150 m [1 mark]
• Step 7: Stability conclusion — UNSTABLE with explanation [0.5 mark]
• Step 8: M_R = W × GM × sin θ = −3.08 kN·m; negative = overturning [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• I = LB³/12 computed with L and B swapped (I = BL³/12) — wrong rolling axis.
• BG = KB − KG (wrong sign; getting BG = −0.9 m without recognizing G above B).
• Not interpreting the negative righting moment as an overturning moment.
• Using θ = 5 radians instead of sin(5°) = 0.0872.

Key Phrases To Include

• I = LB³/12 (rolling axis)
• BM = I / V_disp
• KB = d/2
• BG = KG − KB
• GM = BM − BG = −0.15 m
• GM < 0 → unstable
• M_R = W × GM × sin θ
• negative moment → overturning