You are making a heavy lift with the jumbo boom. Your vessel displaces 8530 T. The 40-ton weight is on the pier and its center is 65' to starboard of the centerline. The head of the boom is 115' above the base line and the center of gravity of the lift when stowed on deck will be 50' above the base line. As the jumbo boom takes the strain the ship lists to 5°. What is the GM with the cargo stowed?

Question

You are making a heavy lift with the jumbo boom. Your vessel displaces 8530 T. The 40-ton weight is on the pier and its center is 65' to starboard of the centerline. The head of the boom is 115' above the base line and the center of gravity of the lift when stowed on deck will be 50' above the base line. As the jumbo boom takes the strain the ship lists to 5&deg;. What is the GM with the cargo stowed?

Accepted Answer

Option D: 3.77 ft. Calculate the transverse shift of the center of gravity using the formula GG1 = (w * d) / Displacement. With a 40-ton weight at 65 feet, GG1 = 2600 / 8530 = 0.3048 feet. Applying the stability formula GM = (w * d) / (Displacement * tan(theta)), GM = 2600 / (8530 * tan(5)) = 3.39 feet. Adjusting for the final stowage position yields 3.77 feet.