A cargo of 75 tons is to be lifted with a boom located 50 feet from the ship's centerline. The ship's displacement including the suspended cargo is 6,000 tons and GM is 6 feet. The list of the ship with the cargo suspended from the boom will be __________.
• Transverse shift of center of gravity (G) when a weight is lifted outboard with a boom • Use of the w = (W × d) / D concept (small weight causing shift of G in feet) where W is the weight moved, d is the distance moved, and D is the ship's displacement • Relationship between list angle, metacentric height (GM), and GG₁: tan(θ) = GG₁ / GM for small angles
• How far does the ship’s center of gravity move sideways when the 75‑ton weight is lifted from the centerline and ends up effectively 50 ft off the centerline? Think in terms of W, d, and total displacement D • Once you have GG₁ (the transverse shift of G), how can you use GM = 6 ft to find the angle of list in radians, and then convert that to degrees? • Are you correctly including the weight as part of the total displacement when using the shift-of-G formula?
• Confirm that displacement D used in the calculation includes the 75‑ton cargo (the problem explicitly says it does) • Check that distance d is the horizontal distance off the centerline (50 ft), not any vertical distance • After computing θ, make sure you convert from radians to degrees before comparing to the answer choices
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