Question

What is the total force of a cylindrical water tower 35 meters high and 8 meters in radius?

Answer 1

Question #37072

What is the total force of a cylindrical water tower 35 meters high and 8 meters in radius?

Answer

The total force is equal to the weight

$$F = P$$

The weight is

$P = mg$ where $m$ is the mass of the tower $g$ is the acceleration due the gravity

$m = V\rho$ where $V$ is the volume of the tower $\rho$ is the density of water $(1000\mathrm{kg} / \mathrm{m}^3)$

For cylindrical tower

$V = \pi R^2 H$ where $H$ is the height $R$ is the radius of the tower

The final equation is

$$F = \pi R ^ {2} H \rho g$$ $$F = 3.14 * 8^{2} * 35 * 1000 * 9.88 = 69,000,000 N = 6.9 * 10^{7} N$$

Answer $6.9 * 10^{7} N$.