A cargo of 30 tons is to be loaded on deck 30 feet from the ship's centerline. The ship's displacement including the 30 tons cargo will be 9,000 tons and the GM will be 5 feet. What would be the list of the vessel after loading this cargo?
• Transverse shift of center of gravity (G) due to an off-center weight • Relationship between tangent of list angle and the horizontal shift of G: ( \tan(\theta) = \frac{GG_1}{GM} ) • How to compute GG₁ from an off-center deck load: ( GG_1 = \frac{w \times d}{\Delta} )
• First, figure out how far the ship’s center of gravity moves sideways (GG₁) when you put 30 tons at 30 feet off the centerline on a 9,000-ton ship. Is that shift going to be big or small? • Use GM = 5 ft and your calculated GG₁ to get ( \tan(\theta) ). Then convert the result into degrees. Which choice matches that angle? • Check if your final angle is reasonable: is a few degrees of list consistent with a relatively small off-center load on a large displacement vessel?
• Be sure to use total displacement including the new weight (9,000 tons) in the ( GG_1 = \frac{w d}{\Delta} ) formula. • Confirm that you use tangent (tan), not sine or cosine, for small-angle list calculations: ( \tan(\theta) = \frac{GG_1}{GM} ). • After finding ( \theta ) in radians, convert correctly to degrees before comparing to the multiple-choice answers.
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