Fluid force problem (hydrostatic force). Boston Aquarium?

The viewing portion of the rectangular glass window in a typical fish tank at the New England Aquarium inBoston is 63 in wide and runs from 0.5 inches below the water's surface to 33.5 inches below the surface. The weight-densit of seawater is p= 64lb/ft^3. Find the fluid force against the viewing portionof the window. (Be careful of your units).Fluid force problem (hydrostatic force). Boston Aquarium?
The force on the window is proportional to the depth of the water. The force is the (density of the water) * (depth^2/2) * (width of the window) all in consistent units, and this acts at the centroid of the water pressure diagram. The centroid of the triangular pressure diagram is 1/3 up from the base or (33.5-.5)/3 = 11 inches from the base.



The depth of the water is 33.5-.5 = 33 inches = 2.75 ft. The width of the panel = 63" = 5.25 feet.



The resulting force on the glass panel = 64*2.75^2/2*(5.25)= 1271 lbsFluid force problem (hydrostatic force). Boston Aquarium?
Ok, here's how you do it:



First you draw a diagram like in any problem you attempt to make.



From that you can get the next integral:



F=integral(p*h*w*derivative(h))



where p is the weight density, h is the depth and w is the width



integrating you get



F=p*w(hf^2-ho^2)/2



and substituting



F=(64lb/ft^3)(63in)(33.5in^2-0.5in^2)/鈥?br>
F=1309lb



as you may see, the formula for the force is like getting the area of a triangle and multiplying it by the depth, where one side of the triangle is the density and the height is the depth.



hope this helps.