You have 200 tons of below deck tonnage. There is no liquid mud aboard. If you have 140 tons of cargo above deck with a VCG above the deck of 4.2 feet, what is the maximum allowed VCG of the remainder of the deck cargo that is permitted? See illustration D036DG below.
• Use the loading diagram D036DG to find the maximum total deck cargo (above deck tons) allowed when you already have 200 tons below deck and no liquid mud. • The loading instructions on the diagram limit maximum deck cargo VCG to 3.0 ft above deck – use this as the combined VCG limit for all deck cargo together. • Use a combined center of gravity (VCG) formula for two weights: existing 140 tons at 4.2 ft and the unknown remaining deck cargo at some VCG, such that the overall deck-cargo VCG = 3.0 ft.
• From the graph, for 200 tons below deck, what is the maximum permitted tons above deck in the safe loading zone? Once you know that, how many tons of deck cargo are still available beyond the 140 tons already loaded? • If the average VCG of all deck cargo must be 3.0 ft, and 140 tons are already higher than that (4.2 ft), does the remaining deck cargo need a VCG higher or lower than 3.0 ft to bring the average down? • Can you set up the equation for the combined VCG of two weights, ( (w_1 h_1 + w_2 h_2)/(w_1 + w_2) ), and solve it for the unknown height of the remaining deck cargo, using 3.0 ft as the target combined VCG?
• From D036DG, read the correct maximum above-deck tons at 200 tons below deck (be careful which curve applies: no liquid mud). • Use 3.0 ft as the limiting VCG for total deck cargo, exactly as stated in the loading instructions on the diagram. • Before choosing an answer, re‑check your algebra when solving for the unknown VCG and verify that the resulting number matches one of the multiple‑choice options.
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