Centre of Pressure and Centroid Fluid Mechanics

[iq97ca]

Homework Statement

a 2m by 4m window is to be installed under water at a depth of 50 m with seawater on one side and atmospheric pressure on the other. Calculate and plot the distance between the centroid and the centre of pressure starting from when the window is installed at surface such that 1m is above the water until the bottom of the window is flush with the bottom of the enclosure.

Homework Equations

y'cp=density*sin(theta)*I_xx/P_cg*A

I_xx=B*L^3/12

The Attempt at a Solution

so this is a pretty confusingly worded questiona nd I'm not even entirely sure what it's asking for

the centroid would exist at 2m,1m relative to the bottom left corner of the window

 

[Valkarie]

the centre of pressure would be at the same x coordinate and at a depth of 50m. I'm assuming we need to calculate the distance between the two points for different depths of the window since it's asking for a plot? for that we need to use the equation given above, where: y'cp = distance between the centroid and centre of pressuredensity = density of seawater theta = angle of window with respect to horizontal I_xx = second moment of area of the window P_cg = atmospheric pressure A = area of window B = width of window L = length of window so we can calculate the distance between the centroid and centre of pressure by plugging in the values and solving the equation y'cp = density*sin(theta)*I_xx/P_cg*Ay'cp = (density)*sin(theta)*(B*L^3/12)/(P_cg)*(B*L)y'cp = (density)*sin(theta)*(L^2/12)/(P_cg)then we can plot the distance between the centroid and centre of pressure as a function of the depth of the window.