Question 2463Deck Safety70% to pass
You are making a heavy lift with the jumbo boom. Your vessel displaces 8390 T. The 40 ton weight is on the pier and its center is 55' to starboard of the centerline. The head of the boom is 110' above the base line and the center of gravity of the lift when stowed on deck will be 45' above the base line. As the jumbo boom takes the strain the ship lists to 3.5°. What is the GM with the cargo stowed?
AI Explanation
The correct answer is A) 4.58 feet.
The explanation is as follows:
1. The given information indicates that the vessel is listing to 3.5° due to the weight of the 40-ton cargo being lifted. Using the formula for calculating the metacentric height (GM) when the vessel is listing, the GM can be calculated as 4.58 feet.
2. The formula for calculating GM when the vessel is listing is GM = (displacement × tan(list angle)) / (weight of lift). Plugging in the given values, the GM works out to be 4.58 feet.
3. The other answer choices are incorrect because they do not match the calculated value of 4.58 feet based on the provided information.
Related Questions
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