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 Question from tom, a student: A tank, open at the top is made up of thin sheet iron 1in. thick. The internal dimensions of the tank are 4 ft. 8in. in long ;3 ft. 6 in.in wide ;4ft 4 in. in deep Find the weight of the tank when empty, and find the weight when full of salt water. (Salt water weighs 64 lbs. per cu. ft., and iron is 7.2 times as heavy as salt water)

Tom,

I would start by finding the internal volume of the tank, $4.666 \times 3.5 \times 4.333 = 70.762$ cubic feet

Salt water weighs 64 pounds per cubic foot so the weight of the salt water is $70.762 \times 64 = 4528.8$ pounds.

Now find the external volume of the tank by adding 2 inches to the length, 2 inches to the width and 1 inch to the depth. (The tank has no top.) Subtract the internal volume to obtain the volume of iron in the tank. The density of the iron is $7.2 \times 64$ pounds per cubic foot so calculate the weight of the tank.

I hope this helps,
Harley

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