Stability of floating bodies: If I is the second moment of area of the waterline plane about the longitudinal axis and V is the displaced volume, the height of the metacentre above the centre of buoyancy (BM) equals:
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ABM = V / I
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BBM = I / V
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CBM = I * V
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DBM = sqrt(I / V)
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EBM = I / (rho * g)
Answer
Correct Answer: BM = I / V
Explanation
Introduction / Context:
Initial stability of floating vessels is assessed using the metacentric height GM. A key geometric quantity is BM, the distance between the metacentre M and the centre of buoyancy B, determined solely by the waterplane geometry and displaced volume.
Given Data / Assumptions:
- Small heel angles (initial stability).
- Homogeneous fluid with unit weight rho * g.
- Longitudinal axis consideration (roll stability).
Concept / Approach:
For small angles, the shift of the line of action of buoyancy produces a righting moment proportional to the waterplane second moment and inversely proportional to the displaced volume. The canonical relation is BM = I / V. Then overall metacentric height GM = KB + BM − KG, where KB and KG are geometric positions of B and the centre of gravity, respectively.
Step-by-Step Solution:
Write the initial stability formula: BM = I / V.Interpretation: Larger waterplane I increases stability; larger displacement V reduces BM.Use GM = KB + BM − KG to assess overall stability (GM > 0 for stable equilibrium).Verification / Alternative check:
Dimensional check: I has units of m^4, V has m^3, so BM has m—consistent for a length measure.
Why Other Options Are Wrong:
- V/I inverts the relationship.
- IV or I/(rhog) have incorrect dimensions.
- sqrt(I/V) has wrong functional form and dimensions.
Common Pitfalls:
- Confusing waterplane second moment of area I (m^4) with the hull mass moment of inertia (kg·m^2).
Final Answer:
BM = I / V