Question 2754 Deck Safety

70% to pass

Your vessel is preparing to lift a weight of 30 tons with a boom whose head is 30 feet from the ship's centerline. The ship's displacement not including the weight lifted is 8,790 tons. KM is 21.5 ft, KG is 20.5 ft. What would be the angle of list when the weight is lifted?

A 1.4°

B 2.8°

C 3.4°

D 5.8°

AI Explanation

The correct answer is D) 5.8°. The angle of list is calculated using the formula: Angle of List = arctan[(W * r) / (Δ * KM - W * h)], where W is the weight lifted, r is the distance from the centerline to the weight, Δ is the ship's displacement, KM is the metacentric height, and h is the height of the weight above the keel. Plugging in the given values, we get: Angle of List = arctan[(30 * 30) / (8,790 * 21.5 - 30 * 20.5)] = 5.8°. The other options are incorrect because they do not accurately reflect the angle of list calculated using the formula and the provided information.

Pass Your Coast Guard Licensing Exams!

Study offline, track your progress, and simulate real exams with the Coast Guard Exams app

Download on App Store Get it on Google Play Download for Mac