Solve the problem.
-One end of a pool is a vertical wall $16 \mathrm { ft }$ wide. What is the force exerted on this wall by the water if it is $7 \mathrm { ft }$ deep? The density of water is $62.4 \mathrm { lb } / \mathrm { ft } ^ { 3 }$.
A) 3490 lb
B) 48,900 lb
C) 24,500 lb
D) 12,200 lb

Question

Quizplus · Accepted Answer

The force exerted by the water on the wall is calculated using the formula for hydrostatic force: $ F = \rho \cdot g \cdot A \cdot h_c $, where $\rho$ is the density of water, $g$ is the acceleration due to gravity (which is included in the density here), $A$ is the area of the wall, and $h_c$ is the depth to the centroid of the area. Here, $A = 16 \times 7 = 112 \, \text{ft}^2$ and $h_c = 7/2 = 3.5 \, \text{ft}$. Thus, $ F = 62.4 \times 112 \times 3.5 = 24,500 \, \text{lb}$.

Solve the Problem 16ft16 \mathrm { ft }16ft Wide What Is the Force Exerted on This Wall by Vertical

Solve the Problem $16 \mathrm { ft }$ Wide What Is the Force Exerted on This Wall by Vertical