On the afternoon of January 15, 1919, an unusually warm day in Boston, a 26.0 m high, 27.4 m diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded the streets in a 9 meter deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 kg/m3. If the tank was full before the accident, what was the total outward force the molasses exerted on its sides?
Hint: Consider the outward force on a circular ring of the tank wall of height dy and depth y below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside.
F=[tex]\Delta[/tex]p [tex]\times[/tex] A
A = 2[tex]\pi[/tex]r^2 + [tex]\2 pi[/tex]rh
where A is area of the cylinder, h is height, r is radius
The Attempt at a Solution
From the hint my professor gave and from an example he did in class I tried Integrating the force with respect to y, the depth below the the surface, and got (from 0 to h)[tex]\int[/tex]pi gy(2pir^2 + 2 pi r y) dy . I'm pretty sure I did the integration right, but I got an answer 1.42x10^10 and he says the answer should be in Mega-Newtons :S any help??
P.s. sorry if the integral is sloppy, I find it very hard to use the symbols on this text editor :P