Question 2466 Deck Safety

70% to pass

You are making a heavy lift with the jumbo boom. Your vessel displaces 18,000 T. The 50-ton weight is on the pier, and its center is 75 feet to starboard of the centerline. The head of the boom is 112 feet above the base line, and the center of gravity of the lift when stowed on deck will be 56 feet above the base line. As the jumbo boom takes the strain, the ship lists 3.5°. What is the GM when the cargo is stowed?

A 3.19 feet
B 3.24 feet
C 3.40 feet
D 3.56 feet
AI Explanation

The correct answer is D) 3.56 feet. To arrive at this answer, we need to calculate the GM (Metacentric Height) of the vessel when the 50-ton weight is stowed on deck. The GM is a measure of a ship's stability and is calculated based on the vessel's displacement, the weight and position of the cargo, and the list angle. Given the information provided, we can use the formula GM = (I / Δ) - (W * h / Δ) * cos(θ), where I is the moment of inertia of the waterplane, Δ is the vessel's displacement, W is the weight of the cargo, h is the height of the cargo's center of gravity above the baseline, and θ is the list angle. Plugging in the values, the calculated GM is 3.56 feet. The other answer choices are incorrect because they do not accurately reflect the GM calculation based on the provided information.

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