On a vessel of 9,000 tons displacement,...

Question 1 of 2707066636eec28f7522a1c51d0a5

On a vessel of 9,000 tons displacement, compute the reduction in metacentric height due to free surface in a hold having free water on the tank tops. The hold is 20 feet long and 30 feet wide. What is the reduction in metacentric height?

Back

Question 1 of 27070

🔍 Key Concepts

• Free surface effect and how it reduces metacentric height (GM) • Formula for free surface moment of inertia: I = (l × b³) / 12 for a rectangular tank (using consistent units) • Relationship between free surface moment, vessel displacement (Δ), and GM reduction: GM reduction = Free surface moment / Δ

💭 Think About

• First, compute the second moment of area (I) of the free surface in the hold using its length and breadth. Which dimension is cubed in the formula? • Next, convert all dimensions and displacement into a consistent unit system (for example, feet and tons-long tons, and remember water density if needed). How will you turn I into a free surface moment for the entire ship? • Finally, divide the appropriate free surface moment by the ship’s displacement to get the reduction in GM. Which option is closest to your computed value?

✅ Before You Answer

• Be sure you use the correct dimension as the breadth (b) in the I = (l × b³) / 12 formula; breadth is normally taken athwartships (side to side). • Check that your units are consistent when you compute GM reduction; the result should come out in feet. • Confirm you are using the displacement of the vessel (9,000 tons) as Δ in the final step, not the volume of the tank.