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#### Solid Geometry Solutions

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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.

$$W_{uw} = 64 \frac{lbs}{ft^3}$$ 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.

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

$$W = 7983.64 ~lbs$$