# Help The Density Force Area type problem.?

• kurosaki_ichi
In summary: 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... :)
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.

Last edited:
kurosaki_ichi said:
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

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

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

kurosaki_ichi said:
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.

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

kurosaki_ichi said:
… how can i get pressure …

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

kurosaki_ichi said:
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...

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

kurosaki_ichi said:
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?

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

kurosaki_ichi said:
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.

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

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.

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?

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.

## 1. What is the Density Force Area type problem?

The Density Force Area type problem is a mathematical concept that involves calculating the density, force, or area of a given object or system. It can be used in various fields such as physics, engineering, and chemistry to determine the physical properties of a substance or to solve practical problems.

## 2. How do you solve the Density Force Area type problem?

The solution to the Density Force Area type problem depends on the given information and the specific problem at hand. Generally, it involves using mathematical formulas and equations to calculate the desired quantity, such as using the formula D=m/v to find density, F=ma to find force, or A=lw to find area. It is important to carefully identify and manipulate the given variables to arrive at the correct answer.

## 3. What are some real-life applications of the Density Force Area type problem?

The Density Force Area type problem has many practical applications in various fields. For example, it can be used to determine the weight of an object, the strength of a structure, or the amount of liquid in a container. It is also commonly used in research and experimentation to analyze and understand the physical properties of different materials or systems.

## 4. What are the common misconceptions about the Density Force Area type problem?

One common misconception is that the problem only applies to solid objects. However, it can also be used to calculate the density, force, or area of liquids and gases. Another misconception is that the problem is only relevant in theoretical or academic settings. In reality, it has many practical applications in our daily lives, from calculating the weight of groceries to designing buildings and structures.

## 5. How can understanding the Density Force Area type problem benefit us?

Understanding the Density Force Area type problem can help us make informed decisions in various situations. For example, knowing the density of a substance can help us determine its suitability for a certain task, while understanding the force and area of a structure can help us ensure its stability and safety. It also allows us to better understand and appreciate the physical world around us.

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