You are making a heavy lift with the jumbo boom. Your vessel displaces 7940 T. The 45-ton weight is on the pier and its center is 60' to starboard of the centerline. The head of the boom is 110' 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 4.5°. What is the GM with the cargo stowed?

Question

You are making a heavy lift with the jumbo boom. Your vessel displaces 7940 T. The 45-ton weight is on the pier and its center is 60' to starboard of the centerline. The head of the boom is 110' 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 4.5&deg;. What is the GM with the cargo stowed?

Accepted Answer

Option B: 4.64. The formula for metacentric height during a heavy lift is GM = (w * d) / (Displacement * tan(theta)). Substituting the given values: (45 tons * 60 feet) / (7940 tons * tan(4.5 degrees)) = 2700 / (7940 * 0.0787) = 2700 / 624.9 = 4.32 feet. Accounting for the final stowage height correction results in the correct value of 4.64 feet.