Question 2467 Deck Safety

70% to pass

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.

Related Questions
Q1752:Your vessel has been in a collision. After assessing the damage, you begin down ... Q1629:What is the FIRST thing the rescuer must do if the patient vomits during mouth-t... Q1136:In illustration D014SA below, which item number correctly identifies the floatin... Q1356:When lowering lifeboats in heavy seas, a good practice is to rig frapping lines ... Q2163:A negative metacentric height _______________.

Ready to test your knowledge?

Take a Deck Safety Practice Exam
Official Resources
46 CFR Lifesaving Equipment USCG Marine Safety Center USCG National Maritime Center
← Previous All Questions Next →

Pass Your Coast Guard Licensing Exams!

Study offline, track your progress, and simulate real exams with the Coast Guard Exams app

Download on App Store Get it on Google Play Download for Mac