Question

An underwater observation room at an offshore oil rig has a 50-cm diameter window. The center of the window is located at a depth of 6 m below the water surface. The specific gravity of sea water is 1.028. Determine a) hydrostatic force acting on the window and b) the location where the force is acting (center of pressure).

Answer 1

`F = 3.17 * 10^4 N`

Part b)

Force will act at the center of the window

Step-by-step explanation:

Pressure at the centre of the window is given as

`P = P_o + \rho gH`

here we know that

`P_o = 1.01 * 10^5 Pascal`

`\rho = 1.028 * 10^3 kg/m^3`

g = 9.81 m/s/s

H = 6 m

`P = (1.01* 10^5) + (1.028* 10^3)(9.81)(6)`

`P = 1.615* 10^5 Pa`

now the area of the window is given as

`A = \pi r^2`

`A = \pi(0.25)^2 = 0.196 m^2`

Now force on the window is given as

`F = 0.196(1.615* 10^5) = 3.17* 10^4 N`

Part b)

Force will act at the center of the window