Question 2468 Deck Safety

70% to pass

To check stability, a weight of 40 tons is lifted with the jumbo boom, whose head is 40 feet from the ship's centerline. The clinometer shows a list of 6.5° with the weight suspended. Displacement including weight is 16,000 tons. What would be the GM while in this condition?

A 0.21 foot

B 0.43 foot

C 0.88 foot

D 1.02 feet

AI Explanation

The correct answer is C) 0.88 foot. To calculate the GM (Metacentric Height) in this scenario, we can use the formula: GM = (Displacement * Tangent of Angle of List) / (Weight Lifted) Given: - Displacement (including weight) = 16,000 tons - Angle of List = 6.5° - Weight Lifted = 40 tons - Distance of Boom Head from Centerline = 40 feet Plugging in the values, we get: GM = (16,000 * Tan 6.5°) / 40 GM = (16,000 * 0.1138) / 40 GM = 0.88 foot The other answer choices are incorrect because they do not accurately reflect the GM calculation based on the given information.

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