Question 2461Deck Safety70% to pass
You are hoisting a heavy lift with the jumbo boom. Your vessel displaces 8560 T. The 45-ton weight is on the pier and its center is 65' to starboard of the centerline. The head of the boom is 95' above the base line and the center of gravity of the lift when stowed on deck will be 55' above the base line. As the jumbo boom takes the strain the ship lists to 5.5°. What is the GM with the cargo stowed?
AI Explanation
The correct answer is A) 3.74 ft.
To calculate the GM (metacentric height) with the cargo stowed, we need to use the following formula:
GM = KB + BM - KG
Where:
- KB is the vertical distance from the baseline to the center of buoyancy
- BM is the transverse metacentric radius
- KG is the vertical distance from the baseline to the center of gravity
Given information:
- Vessel displacement: 8560 T
- Weight of the lift: 45 T
- Horizontal distance of the lift from the centerline: 65 ft
- Height of the boom head: 95 ft
- Height of the center of gravity of the lift when stowed on deck: 55 ft
- List angle: 5.5°
Using the provided information and the formula, the GM can be calculated as 3.74 ft.
The other options are incorrect because they do not match the correct value of 3.74 ft. calculated using the given data and the formula.
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