On a vessel of 15,000 tons displacement,...

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Question 1 of 2066636eec28f7522a1c51d099

On a vessel of 15,000 tons displacement, compute the reduction in metacentric height due to free surface in a hold having free water in the tank tops. The hold is 50 feet long and 60 feet wide. What is the reduction in metacentric height?

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🔍 Key Concepts

• Free surface effect and how it reduces metacentric height (GM) • Formula for free surface moment of a rectangular tank: I = (b × l³) / 12 or I = (l × b³) / 12 (which dimension goes with the cube?) • Relationship between free surface moment, displacement (Δ), and reduction in GM: ΔGM = (ρ × I) / Δ

💭 Think About

• For a rectangular hold 50 ft by 60 ft, which dimension should you cube to get the second moment of area for free surface? Think about which axis the ship is rolling about. • Once you get I, how do you combine it with the vessel’s displacement (15,000 tons) to find the reduction in GM? Pay attention to unit consistency (feet vs tons). • Compare your computed ΔGM to the answer choices – is your value closer to 1.5 ft or 1.7 ft? That can help catch arithmetic mistakes.

✅ Before You Answer

• Be sure you used the correct second moment of area formula for a free surface about the ship’s centerline (roll axis). • Confirm that you used the displacement in the denominator, not the deadweight or some other weight figure. • Double‑check that your units are consistent: feet for dimensions and tons for displacement, and that your ΔGM result is in feet.

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