The barge shown in the illustration is 40' X...

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Question 1 of 205eccef9dbacbc34ba4d45e11

The barge shown in the illustration is 40' X 20' X 10' and weighs 22.8 long tons when empty. What would be the maximum amount of cargo that may be loaded on the barge in salt water so that the maximum draft does not exceed 8 feet. See illustration GS-0157.

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🔍 Key Concepts

• Use Archimedes’ principle: the total weight of barge + cargo equals the weight of the displaced salt water. • For a box-shaped barge, underwater volume ≈ L × B × draft (in feet). • In salt water, 1 long ton ≈ 35 cubic feet of displaced water.

💭 Think About

• First, compute the maximum underwater volume when the draft is 8 feet using the given length and breadth. What is that volume in cubic feet? • Convert that maximum displacement volume into long tons of water. That gives you the maximum total weight (barge + cargo) the barge can have at 8 feet draft in salt water. • Subtract the lightship (empty) weight of the barge from this maximum total weight. Which answer choice is closest to that remaining weight for cargo only?

✅ Before You Answer

• Be sure you are using cubic feet, not square feet, when you convert volume to tons. • Confirm you are using the correct salt water conversion factor (about 35 ft³ per long ton), not the one for fresh water. • After you find total allowable displacement in tons, subtract the 22.8 long tons of the empty barge before comparing with the choices.

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