Forces exerted by a liquid during a tank rupture

[kvan]

Homework Statement

On the afternoon of January 15, 1919, an unusually warm day in Boston, a 28.1 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?

Homework Equations

F= A\deltaP
P= \rho gh
F= \int (\rho g)ydy
F= (\rho g) 1/2y^2

The Attempt at a Solution

I tried using the above equations other than the integral equations to solve. Which led me to having F= (9* \pi27.4)(1600)(9.81)(9)
I was told this was incorrect and should try to integrate but I am not really sure how to use the integration formulas.

 

[CheckMate]

If P is potential energy, then P = density x volume x gravitational acceleration x height

The derivative of potential energy with respect to position is Force, not the other way around. The indefinite integral of Force is potential Energy, all with respect to position.

You would need a negative sign in front of the derivative because as Potential Energy (function of position) increases with distance to arbitrary zero then force decreases. Remember a planet that goes further away from a star would be less and less attracted.

 

[kvan]

But how would that relate to the force of a liquid acting on a container that then breaks?

 

[Skrambles]

Check your units on those equations, and use the height of the tank instead of the height of the stream.

 

[kvan]

If I use the height of the tank instead of the height of the stream I get ~1067MN. That number seems a little high. Please not that I don't actually have the answer. I was just told by my prof that my original method was wrong and recommended I use integration.

 

[Skrambles]

You need to integrate the pressure over the area inside the tank, and the 9 meter figure has nothing to do with that. The third equation above should have dA instead of dy.