On a vessel of 8,000 tons displacement,...

Question 1 of 2707066636eec28f7522a1c51d0a4

On a vessel of 8,000 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 40 feet long and 20 feet wide. What is the reduction in metacentric height?

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Question 1 of 27070

🔍 Key Concepts

• Free surface effect and how it reduces metacentric height (GM) • The second moment of area (I) of a rectangular free surface: I = (b × l³) / 12 or (l × b³) / 12 depending on axis • The standard free surface correction formula: Free surface correction = (ρ × I) / Δ (or I / Δ in consistent units)

💭 Think About

• First decide which dimension (length or breadth) is used for the moment of inertia about the vessel’s centerline—what axis are you heeling about? • Check that your displacement (8,000 tons) and tank dimensions (40 ft × 20 ft) are used in consistent units when applying the free surface correction formula. • After computing the free surface correction, compare your result with the answer choices and think about whether the magnitude seems reasonable for this size ship and hold.

✅ Before You Answer

• Be sure you are using the correct inertia formula about the longitudinal axis (i.e., axis parallel to the ship’s length vs. axis along the ship’s centerline). The vessel heels about the fore‑and‑aft axis, so which tank dimension should be cubed? • Confirm that you are applying the correction to GM, not KB or KG; this is a reduction in metacentric height. • Double-check the numerical order of magnitude: is your free surface correction less than 1 foot for this size vessel and this tank? If it’s many feet, re-check your inertia and displacement inputs.