Question 2753Deck Safety70% to pass
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 _______________.
AI Explanation
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°.
Related Questions
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