Question

A square window with 80 meter sides is located on the vertical side of a large rectangular pool. The depth of the pool is 5 meters. The top of the window is located 43 meters below the surface. Assuming the specific weight of the water in the pool is 9.8 kN/m^3, what is the resultant force (in kN) on the window? Round to the nearest kN.

Answer 1

See explanation

Step-by-step explanation:

Solution:-

- We will first set a datum as the free surface of water in the pool.

- There is a square window with side length ( L = 8 m ) on the vertical side of the pool.

- The depth of the pool is given to be 54 m. The top of the window is ( d = 43 m ) from the free surface.

- We are to determine the resultant hydro static force acting on the window.

- Hydro static force ( F ) acting on an object of area ( A ) fully immersed in fluid is given by the following relationship as follows:

F = Pc*A

Where,

Pc: The hydrostaic pressure acting on the centroid of the obect.

- The hydrostatic pressure acting on the centroid of the object immersed in any fluid can be expressed by the following defining relationship:

Pc = γ*yc

Where,

γ: The specific weight of the fluid

yc: The vertical distance from free-surface to the centroid.

- Assuming homogeneous distribution of material used for the window of square shape. The centroid coincides with the geometric center of the window which is as a distance ( yc ) from the free-surface:

`y_c = (d + (L)/(2) ) \\\\y_c = (43 + (8)/(2) ) \\\\y_c = 47 m`

- Now we can calculate the resultant hydro static force ( F ) acting on the window. The specific gravity of fluid ( water ) is γ = 9.8KN/m^3.

`F = ( 9.8 ) * ( 47 )* ( 8 )^2\\\\F = 29479KN`

Note: The values given in the posted question seem unreasonable. I have assumed values in order of convenience. However, the procedure of solving the problem remains exactly the same.