Question

A hemispherical tank of radius 2 feet is positioned so that its base is circular. How much work is required to fill the tank with water through a hole in the base when the water source is at the base? (The weight-density of water is 62.4 pounds per cubic foot.)

Answer 1

To fill the hemispherical tank with water through a hole in the base, the work required is 418.88 foot-pounds.

Step-by-step explanation:

To calculate the work required to fill the hemispherical tank with water through a hole in the base, we can use the concept of work done against gravity.

The volume of the tank can be calculated using the formula for the volume of a hemisphere, which is (2/3)πr^3. In this case, the radius is given as 2 feet.

The weight of the water can be found by multiplying the volume by the weight-density of water, which is 62.4 pounds per cubic foot.

The work done is then the weight of the water multiplied by the height it is lifted, which is equal to the radius of the hemisphere.

So, the work required to fill the tank with water is (2/3)π(2^3)(62.4)(2) = 418.88 foot-pounds.

Answer 2

784 ft·lb

Step-by-step explanation:

The amount of work required is equivalent to the work required to raise the total weight of water to the height of the centroid of the hemisphere.

The weight of the water is ...

Volume × density = (2/3)πr³ × 62.4 lb/ft³

= (2/3)π(2 ft)³(62.4 lb/ft³) ≈ 1045.5 lb

A reference page tells us that the centroid of a hemisphere is 3/8r above the circular base, so we're raising this weight of water by (3/8)(2 ft) = 3/4 ft.

The work done is ...

(3/4 ft)(1045.5 lb) ≈ 784 ft·lb