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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).

Respuesta :

Answer:

[tex]F = 3.17 \times 10^4 N[/tex]

Part b)

Force will act at the center of the window

Explanation:

Pressure at the centre of the window is given as

[tex]P = P_o + \rho gH[/tex]

here we know that

[tex]P_o = 1.01 \times 10^5 Pascal[/tex]

[tex]\rho = 1.028 \times 10^3 kg/m^3[/tex]

g = 9.81 m/s/s

H = 6 m

[tex]P = (1.01\times 10^5) + (1.028\times 10^3)(9.81)(6)[/tex]

[tex]P = 1.615\times 10^5 Pa[/tex]

now the area of the window is given as

[tex]A = \pi r^2[/tex]

[tex]A = \pi(0.25)^2 = 0.196 m^2[/tex]

Now force on the window is given as

[tex]F = 0.196(1.615\times 10^5) = 3.17\times 10^4 N[/tex]

Part b)

Force will act at the center of the window

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