Help The Density Force Area type problem.?

[kurosaki_ichi]

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 -9m-deep stream, killing pedestrians and horses, and knocking down buildings. The molasses had a density of 1000kg/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 and at a depth 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.)

The Attempt at a Solution

I really don't know how to start, if their is anyone who is willing to help me with guidelines and how to begin this question much appreciated.

 

[tiny-tim]

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 -9m-deep stream, killing pedestrians and horses, and knocking down buildings. The molasses had a density of 1000kg/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 and at a depth 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.)
…
I really don't know how to start, if their is anyone who is willing to help me with guidelines and how to begin this question much appreciated.

Hi kurosaki_ichi

Hint: follow the hint in the question …

what is the pressure at height h below the surface?

and then use force = pressure times area, and integrate the force

 

[kurosaki_ichi]

Still not quite understand , how does that work integrating force?

do I integrate dF=P * A ? am so lost help...:(

 

[berkeman]

Still not quite understand , how does that work integrating force?

do I integrate dF=P * A ? am so lost help...:(

The pressure right at the surface of the liquid against the wall is pretty much zero, right? And the pressure against the wall at the bottom of the tank is maximum, right? So it varies from zero to max, as you descend down the tank from the surface to the bottom of the wall.

You need to figure out what the pressure is at a depth D (or height H), and do an integration to add up all the force on the wall that results from this varying pressure.

 

[kurosaki_ichi]

I have to cacuulter the pressure first then intergrate the sum of the all forces acting on the tank? how can i get pressure, i wish if its possible for steps for me to follow, and this is supposed to be a simple problem! headache

 

[tiny-tim]

… how can i get pressure …

same as for water … see your other thread

 

[kurosaki_ichi]

i found the pressure i don't know what to do, man am i really this lost i feel so confused

 

[berkeman]

i found the pressure i don't know what to do, man am i really this lost i feel so confused

Show us your diagrams and calculations so far, so we can help. You have to actually do the work here... we can only provide hints or look for math errors...

 

[kurosaki_ichi]

Caculated pressure

p=pa +density(g)(h)
p=1.00 atm + (1000)(9.8)(27.4)
1.00atm=101325 Pa

p=369845 Pa---> 369.845 --370 kPa

according to what i wwas told am supposed to integrate?

so force=pressure times area

dF=pdA dF=(370kPa) dA

my limits of integration integrating the forces are 9m and 0

 

[berkeman]

Caculated pressure

p=pa +density(g)(h)
p=1.00 atm + (1000)(9.8)(27.4)
1.00atm=101325 Pa

p=369845 Pa---> 369.845 --370 kPa

according to what i wwas told am supposed to integrate?

so force=pressure times area

dF=pdA dF=(370kPa) dA

my limits of integration integrating the forces are 9m and 0

How does the shape of the cylindrical metal tank come into play in the integration?

 

[kurosaki_ichi]

INtegrating a cylinder would thatmean i need to take a cross section of thinkness delta x and add the volumes of cylinders togethor

 

[berkeman]

INtegrating a cylinder would thatmean i need to take a cross section of thinkness delta x and add the volumes of cylinders togethor

Not volumes, areas. Divide the cylinder up into many thin bands of height dh. For each band, the pressure is the same all the way around the band, because the whole band is at a height h.

So you need to figure out the pressure as a function of height, P(h), and then for a band of height dh, what that translates into in terms of force. Then integrate accordingly from height 0m to 9m to get the total force. Be sure to keep track of units, to be sure you are calculating the correct quantities.

 

[kurosaki_ichi]

I finaly got it! it the force answe i think i got is
5.07E8

 

[nvn]

kurosaki_ichi: Your current answer is incorrect. You should not integrate to 9 m; go back to the problem statement and rethink to what depth you should integrate. Do not use absolute pressure; use gauge pressure (i.e., do not add 1 atm).

Your differential force, dF, in post 9 is wrong. The pressure varies as a function of fluid depth, y, as berkeman explained. Your integral needs to show the change in pressure as a function of fluid depth, p(y). Therefore, shouldn't your integral contain the expression p(y)*dA? What is dA? It is the area of one differential horizontal ring of cylinder wall surface area. The differential ring has a height of dy. Therefore, what is dA? Figure out dA. And figure out p(y). Post your current integral, and show all your work.

 

[kurosaki_ichi]

dF = (p0+rohgy) (pid)dy

F = Integral[dy (p0+rohgy) (pid)]

= 2.38×10E8 N+5.07×108 N
= 7.45×10E8 N

Threfoe the total force exterted on the tank wall is 7.45×10E8 N, their was also an inward force so what i did is found that force as well which i caculated to be pipdp0h=2.38×10E8N, this is from the air outside... so the net force on the wall i caculuted to be:

=5.07×10E8N

I think that is what the question is looking for?

 

[nvn]

kurosaki_ichi: Your current answer is incorrect; 2.3898e8 N is correct due to the atmospheric pressure, but 5.07e8 N is incorrect. Your integral before integration looks fine. Therefore, for your F equation, add another line showing the symbolic solution to your integral, and show the limits of integration. Then on the next line for your equation, show all numeric values you are substituting into your solution, before you multiply them together, to obtain 5.07e8 N. In other words, show all your work, if you want help.