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

 

[rwisz]

Your attempt at a solution is on the right track. However, there are a few things that need to be corrected in your equations and calculations.

Firstly, the pressure at the surface of the molasses should not be assumed to be equal to the air pressure outside the tank. This is because the molasses is exerting a downward force on the air above it, creating a pressure gradient. Therefore, the pressure at the surface of the molasses should be calculated using the equation p=p0+roh*g*h, where p0 is the air pressure outside the tank.

Secondly, when calculating the total outward force, you need to consider the pressure on the entire surface area of the tank, not just a single circular ring. This means that the integration should be done over the entire surface area of the tank, which is 4*pi*r^2. In your calculation, you are only considering a single ring of width dy, which is not enough to find the total outward force.

Finally, when integrating, you need to take into account the units of each term in the equation. The units for pressure (P) are N/m^2, and the units for area (A) are m^2. Therefore, the units for the total outward force should be in Newtons (N).

To correct your calculation, you can use the following equation:

F = int[ (p0 + roh*g*h)*4*pi*r^2 ] dh

where p0 is the air pressure outside the tank, roh is the density of molasses, g is the acceleration due to gravity, h is the depth below the surface, and r is the radius of the tank.

This should give you the total outward force exerted by the molasses on the tank.