To check stability, a weight of 35 tons is lifted with the jumbo boom, whose head is 35 feet from the ship's centerline. The clinometer shows a list of 7.0° with the weight suspended. Displacement including the weight is 14,000 tons. What would the length of GM in this condition?
• Use of the heeling moment formula when a weight is moved off the centerline • Relationship between tangent of the heel angle, displacement, GM, and heeling arm • Correct unit consistency for tons, feet, and angles in degrees vs radians
• How can you express the sideways (heeling) moment caused by a 35‑ton weight hanging 35 feet off the centerline? • What is the relationship between that heeling moment and the ship’s righting moment, which depends on displacement, GM, and the sine (or tangent) of the heel angle? • Once you set up the equality of heeling and righting moments, how can you rearrange the equation to solve for GM?
• Convert the heel angle to the correct trigonometric function (decide whether to use sin or tan of the angle, and stay consistent with the standard small‑angle stability formula you learned). • Be sure that the ship’s displacement includes the lifted weight, as stated in the problem, when you plug into your formula. • After computing GM, check that your result is a reasonable size (on the order of a few feet, not inches or tens of feet) and matches one of the options.
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