You are making a heavy lift with the jumbo boom. Your vessel displaces 8390 T. The 40 ton weight is on the pier and its center is 55' to starboard of the centerline. The head of the boom is 110' above the base line and the center of gravity of the lift when stowed on deck will be 45' above the base line. As the jumbo boom takes the strain the ship lists to 3.5°. What is the GM with the cargo stowed?
• transverse metacentric height (GM) and its relation to list angle for a suspended weight • how a suspended load effectively acts at the head of the boom until landed and stowed • using moment = force × distance to find the heeling moment and relate it to GM
• First, think about how to calculate the heeling moment caused by a 40‑ton weight acting 55 feet off the centerline. Where is that weight’s effective center of gravity while it is hanging? • Next, recall the relationship between the heeling moment, the vessel’s displacement, GM, and the angle of list (in radians). How can you rearrange that relationship to solve for GM at the time the ship is listed 3.5°? • Finally, after the load is stowed at 45 feet above the baseline on deck (no longer suspended), how does that change the ship’s vertical center of gravity, and how does that new KG relate back to the GM you just found?
• Be sure you convert 3.5° to radians before using it in a trigonometric or stability formula. • Confirm that you are using the ship’s displacement in long tons consistently with the 40‑ton load so your units match. • Double‑check that you distinguish between the temporary rise in virtual KG with a suspended load and the final KG when the cargo is stowed on deck, because the question asks for GM with the cargo stowed.
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