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    Topics || Problems

    A tank open at the top, is made of sheet of iron 1 in thick. The internal dimensions of the tank are 4ft 8 in long; 3ft 6 in wide; 4 ft 4 in deep. Find the weight of the tank when empty, and find the weight when full of saltwater. Salt water weights \(64 \frac{lbs}{ft^3}\) and iron is 7.2 times as heavy as salt water.
    A tank open at the top, is made of sheet of iron 1 in thick

    \(W_{uw} = 64 \frac{lbs}{ft^3}\) is the unit weight of water and \(W_{ui} = 7.2 (640 = 460.8 \frac{lbs}{ft^3}\) is the unit weight of iron.

    Weight of tank when empty, \(W_e = W_{ui}Vol_{iron}\)

    Weight of tank when full, \(W = W_e + W_(uw)Vol_{water}\)

    Calculate the volume of the tank.

    \(Vol_{tank} = Vol_{out} - Vol_{in}\)

    \(Vol_{out} = (3.5+\frac{1}{6})(\frac{14}{3} +\frac{1}{6})(\frac{13}{3} + \frac{1}{12}) \)

    \(Vol_{out} = 78.27\)

    \(Vol_{in} = (\frac{14}{3})(3.5)(\frac{13}{3})\)

    \(Vol_{in} = 70.78\)

    \(Vol_{tank} = 78.27-70.78 = 7.50~ ft^3\)

    The volume of water is same volume as the \(Vol_{in}\)

    Weight of the tank when empty.

    \(W_e = 460.8 (7.50) \)

    \(W_e = 3453.87 ~lbs\)

    Weight of the tank when full with saltwater.

    \(W = 3453.87 + 64 (70.78) = 3453.87 ~lbs\)

    \(W = 7983.64 ~lbs\)