You have 520 tons of below deck tonnage....

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Question 1 of 2066636eec28f7522a1c51befa

You have 520 tons of below deck tonnage. There is no liquid mud. If you have 160 tons of cargo above deck with a VCG above the deck of 3.2, what is the maximum allowed VCG of the remainder of the deck cargo that is permitted? Illustration D036DG

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🔍 Key Concepts

• Use the D036DG loading diagram to find the maximum total ABOVE DECK tonnage allowed when you have 520 tons BELOW DECK and no liquid mud. • Remember the note on the diagram: maximum deck cargo VCG is 3.0 ft above deck; relate this to the combined VCG of all deck cargo. • Use the combined VCG formula: overall VCG = (sum of each weight × its VCG) ÷ (total weight).

💭 Think About

• From the graph, what is the maximum permitted total deck cargo (in tons) for 520 tons below deck? How much additional deck cargo ("remainder") does that leave after the first 160 tons? • If the average VCG of all deck cargo must be 3.0 ft or less, how can you set up an equation using the 160 tons at 3.2 ft and the unknown VCG of the remaining deck cargo? • Is the VCG of the remaining deck cargo expected to be higher or lower than 3.0 ft in order to bring the total average down to 3.0 ft, given that part of the cargo is already at 3.2 ft?

✅ Before You Answer

• From the D036DG graph, correctly read the allowed ABOVE DECK load corresponding to 520 tons BELOW DECK (don’t estimate too roughly—this value drives the whole problem). • Compute the remainder deck cargo weight as: (max allowed ABOVE DECK tons from the graph) minus 160 tons already on deck. • In your combined VCG equation, double‑check that you multiply each weight by its own VCG, add the moments, then divide by the total deck weight, and set this equal to 3.0 ft before solving for the unknown VCG.

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