Question 4817Deck General70% to pass
What is the transverse shift in the center of gravity if 200 short tons are placed 10 feet to port of the centerline on a MODU with TCG 0.7 foot starboard of the centerline, and the displacement is 9,000 short tons?
AI Explanation
The correct answer is C) 0.23 foot.
To calculate the transverse shift in the center of gravity, we can use the formula:
Transverse shift = (Weight x Distance) / Displacement
Given:
- Weight placed 10 feet to port = 200 short tons
- TCG = 0.7 foot starboard of the centerline
- Displacement = 9,000 short tons
Plugging these values into the formula:
Transverse shift = (200 x 10) / 9,000
= 2,000 / 9,000
= 0.23 foot
The other options are incorrect because they do not match the result calculated using the formula. Option A (0.03 foot) and Option B (0.20 foot) are too low, while Option D (0.62 foot) is too high.
Related Questions
#4815 What is the longitudinal shift in the center of gravity if 200 short tons is moved ten feet to port and 30 feet forward on a MODU with a displacement of 8,960 long tons? #4816 A semisubmersible displacing 17,600 long tons has an LCG 3.2 feet forward of amidships. Bulk, weighing 400 long tons, is loaded into P-tanks located 50.8 feet aft of amidships. What is the new LCG? #4818 A semisubmersible, while floating level, displaces 20,000 long tons. LCB is 3.0 feet forward of amidships. Bulk, weighing 300 long tons, is placed in P-tanks located 40 feet aft of amidships. What is the new LCG? #4819 A semisubmersible, while floating level, displaces 18,000 long tons. Bulk, weighing 400 long tons, is placed in P-tanks located 80 feet to starboard of the centerline. What is the new TCG? #4820 A semisubmersible, while floating level, displaces 25,000 long tons. LCG is 2 feet forward of amidships. Bulk, weighing 300 long tons, is placed in P-tanks located 50 feet aft of amidships. What is the new LCG?