Outward force of molasses on a cylinder

[lluhi]

Homework Statement

On the afternoon of January 15, 1919, an unusually warm day in Boston, a 27.4-m-high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 9-m-deep stream, killing pedestrians and horses, and knocking down buildings. The molasses had a density of 1600 kg/ m^3

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 width dy and at a 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 the tank.)

Homework Equations

F=PA p=p0+roh*g*h

The Attempt at a Solution

if F=PA then dF=dPdA and dA for a cylinder is pi*d (integration of pi*r^2) and for d=27.4, dA=86.1

i have F=86.1*int[101325pascals + (1600kg/m^3)(9.8m/s^2)hdh] from 0 to 27.4

integrating gives

86.1 (101325h+7840h^2) and plugging in 27.4 for h gives 7.46e^8 which is wrong... help?

 

[ehild]

The atmospheric pressure acts both inward and outward. I do not think you should count with it. And make your calculation a bit more clear. dA is the area of a strip on the wall of the cylinder: dA=2pi R dy.

ehild

 

[RTW69]

dF=p*dA; p=rho*g*y;dA=pi*dia*dy

F=rho*g*dia*pi*integrate[y*dy] from 0 to h

 

[lluhi]

got it, thanks