To check stability, a weight of 35 tons is lifted with... - USCG Deck Safety

To check stability, a weight of 35 tons is lifted with the jumbo boom, whose head is 35 feet from the ship's centerline. The clinometer shows a list of 7.0° with the weight suspended. Displacement including the weight is 14,000 tons. What would the length of GM in this condition?

A 0.71 foot

B 0.95 foot

C 1.26 feet

D 2.01 feet

AI Explanation

The correct answer is A) 0.71 foot. The length of GM (the distance between the center of gravity and the metacenter) can be calculated using the formula: GM = (W * h) / (Δ * tan θ), where W is the weight lifted, h is the distance of the weight from the centerline, Δ is the displacement, and θ is the list angle. Plugging in the given values, we get: GM = (35 * 35) / (14,000 * tan 7.0°) = 0.71 foot. The other options are incorrect because they do not correctly apply the formula or use the provided values.

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Question 2467 Deck Safety

To check stability, a weight of 35 tons is lifted with the jumbo boom, whose head is 35 feet from the ship's centerline. The clinometer shows a list of 7.0° with the weight suspended. Displacement including the weight is 14,000 tons. What would the length of GM in this condition?

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