You are hoisting a heavy lift with the jumbo boom. Your vessel displaces 5230 T. The 35-ton weight is on the pier and its center is 60' to starboard of the centerline. The head of the boom is 105' above the base line and the center of gravity of the lift when stowed on deck will be 42' above the base line. As the jumbo boom takes the strain the ship lists to 5°. What is the GM with the cargo stowed?
• GG1 = w × d / Displacement to find the shift in the ship’s center of gravity as the weight is taken by the boom • How virtual rise of G with a suspended weight differs from the final G when the weight is landed and stowed • Using the list formula: w × d = Displacement × GM × tan(list) for the suspended condition
• First, think about the condition at the exact moment the boom takes the full weight and the ship lists 5°. What does this tell you about GM at that instant? • How do you relate the horizontal shift of the 35‑ton weight (from the pier to its final stowed position) to the change in the ship’s center of gravity? • After the weight is stowed at 42’ above the base line, how has the vertical position of G changed compared to the original condition with the weight ashore?
• Be sure you’re using consistent units (tons with tons, feet with feet, inches with inches – do not mix without converting). • Double‑check that you use tan(5°) (not sin or cos) in the list equation. • Verify that you calculate GM for the suspended condition first, then adjust to find GM with cargo stowed on deck as the question asks.
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