A spherical load of 900 kg rolls transversely across the deck of a 10,000 kg ship through 9.8 m. If the ship’s metacentric height (GM) is 5 m, estimate the heel angle caused by the shift (assume tanθ = (w * d) / (W * GM)).
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A10° 5′
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B10° 10′
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C10° 15′
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D10° 20′
Answer
Correct Answer: 10° 5′
Explanation
Introduction / Context:Ship stability problems frequently involve heeling moments produced by moving loads on deck. The small-angle relation using metacentric height (GM) provides a quick estimate of heel angle and is widely used for preliminary checks.
Given Data / Assumptions:
- Ship weight W = 10,000 kg (displacement mass, neglecting added mass).
- Moving load w = 900 kg shifts laterally by distance d = 9.8 m.
- Metacentric height GM = 5 m.
- Small heel angle, so tanθ ≈ θ in radians is acceptable, but we will compute θ via the tangent relation.
Concept / Approach:For a weight shift on a floating body, heeling moment M = w * d. The restoring moment using metacentric stability is W * GM * tanθ. Equating moments gives tanθ = (w * d) / (W * GM). This relation is standard for small angles and assumes the metacentre remains near fixed vertically for small tilts.
Step-by-Step Solution:Compute numerator: w * d = 900 * 9.8 = 8820 kg·m.Compute denominator: W * GM = 10,000 * 5 = 50,000 kg·m.tanθ = 8820 / 50,000 = 0.1764 → θ ≈ arctan(0.1764) ≈ 10.1 degrees.Convert to degrees–minutes: approximately 10° 5′.
Verification / Alternative check:If one included the load in total weight (W + w), tanθ = (w d)/((W + w) GM) = 8820/(10,900*5) ≈ 0.162 → θ ≈ 9.2°, close but slightly smaller. Many exam problems use the simpler W in the denominator; the given options align with ≈ 10° 5′.
Why Other Options Are Wrong:10° 10′, 10° 15′, 10° 20′: larger than the calculated angle for the stated data and formula.
Common Pitfalls:
- Misplacing units (using N rather than kg for mass-based form) and mixing g factors.
- Using GM in centimetres while d in metres.
Final Answer:10° 5′