Question 2753 Deck Safety
70% to passA 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 _______________.
The correct answer is C) 5.94°. To determine the list of the ship with the cargo suspended from the boom, we can use the formula for calculating the list angle: Tan(list angle) = (cargo weight x boom length) / (ship's displacement x GM). Given: - Cargo weight: 75 tons - Boom length: 50 feet - Ship's displacement: 6,000 tons - GM: 6 feet Substituting the values into the formula: Tan(list angle) = (75 tons x 50 feet) / (6,000 tons x 6 feet) Tan(list angle) = 3,750 / 36,000 Tan(list angle) = 0.1042 List angle = arctan(0.1042) = 5.94° The other options are incorrect because they do not match the calculated list angle of 5.94°.
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