Question

A town has a 1.1-million-gallon storage capacity water tower. If the density of water is 62.4 lb/ft³ and local acceleration of gravity is 32.1 ft/s², what is the force, in lbf, the structural base must provide to support the water in the tower?

Answer 1

Answer:

9175863.84 lbf

Explanation:

$\rho$ = Density of water = 62.4 lb/ft³

v = Volume = 1.1 million gallon

g = Acceleration due to gravity = 32.1 ft/s²

Mass is given by

$m=\rho v\\\Rightarrow m=62.4\times 1.1\times 10^6\times 0.133681\\\Rightarrow m=9175863.84\ lb$

The weight of the water will be the force on the structure

$W=mg\\\Rightarrow W=9175863.84\times 32.1\\\Rightarrow W=294545229.264\ lb ft/s^2$

Converting to lbf

$W=\dfrac{294545229.264}{32.1}\\\Rightarrow W=9175863.84\ lbf$

The force on the structural base is 9175863.84 lbf

Answer 2

The force that the structural base must provide to support the water in the tower is 9.17 x 10⁶ lbf.

The given parameters:

• Volume of the water, V = 1.1 x 10⁶ gallons
• Density of water, = 62.4 lb/ft³
• Acceleration due to gravity

The volume of the water in cubic feet is calculated as follows;

$V = \frac{1 \ ft^3}{7.48 \ gal} \times 1.1 \times 10^6 \ gal\\\\ V = 1.47 \times 10^5 \ ft^3$

The mass of the water is calculated as follows;

$m = \rho V\\\\ m = 62.4 \ lb /ft^3 \ \times 1.47 \times 10^5 \ ft^3\\\\ m = 9.17 \times 10^6 \ lb$

The force that the structural base must provide to support the water in the tower is calculated as;

$F = mg\\\\ F = 9.17 \times 10^6 \ lb \ \times 32.1 \ ft/s^2\\\\ F = 2.94 \times 10^8 \ lb.ft/s^2\\\\ W = \frac{ 2.94 \times 10^8 \ lb.ft/s^2}{32.1 \ ft/s^2} \\\\ W = 9.17 \times 10^6 \ lbf$

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