Outward force on a tank of water.

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Homework Help Overview

The problem involves a cylindrical tank storing water, specifically focusing on calculating the outward force exerted on the tank's sides due to the pressure of the fluid inside. The tank has specific dimensions and the density of the fluid is provided, leading to questions about pressure calculations and integration methods.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the integration of pressure versus force, questioning the original poster's method and the interpretation of the problem statement. There are inquiries about whether the total outward force includes contributions from both the fluid and external air pressure.

Discussion Status

The discussion is ongoing with participants providing feedback on the original poster's calculations and questioning the clarity of the problem statement. Some guidance has been offered regarding the integration approach, but no consensus has been reached on the correct interpretation of the problem.

Contextual Notes

There is ambiguity in the problem regarding whether to consider the net force from both the fluid and air or just the fluid's contribution. Additionally, the symmetry of the pressure loading is noted, suggesting that the overall net force may be zero.

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Homework Statement



There's a 30 m high, 30 m diameter cylindrical metal tank used for storing water. The molasses has a density of 1000 kg/^3. If the pressure at the surface was equal to the air pressure outside the tank, find the force pushing outward on the sides of the tank.

Homework Equations



Surface area of a cylinder's sides: 2*Pi*r*h

The Attempt at a Solution



The pressure at a depth d in the water should be equal to 101325+1000*9.8*d (1 atmosphere plus the weight of the molasses above it). I tried integrating this from 0 to 30 and then multiplying it by the surface area but that didn't work. What should I have done?
 
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Hi PrestonBlake! :smile:

(molasses? :confused:)
PrestonBlake said:
… find the force pushing outward on the sides of the tank.

The pressure at a depth d in the water should be equal to 101325+1000*9.8*d (1 atmosphere plus the weight of the molasses above it). I tried integrating this from 0 to 30 and then multiplying it by the surface area …

(is it asking for the net force, or just the force from inside?)

should work …

show us your full calculations, and then we'll see what went wrong, and we'll know how to help! :smile:
 
It says the total outward force, which I take it means just the force from inside.

I integrate 101325+1000*9.8*d from 0 to 30 and get 7.44975*10^6, then I multiply by 2*Pi*15*30 and get 2.10637*10^10 which is wrong.
 
why are you integrating the pressure? you should be integrating the force

(i think you've counted the height twice :redface:)
 
tiny-tim said:
why are you integrating the pressure? you should be integrating the force

(i think you've counted the height twice :redface:)

I'm not that good at integrals, what should I be integrating?
 
PrestonBlake said:

Homework Statement



There's a 30 m high, 30 m diameter cylindrical metal tank used for storing water. The molasses has a density of 1000 kg/^3. If the pressure at the surface was equal to the air pressure outside the tank, find the force pushing outward on the sides of the tank.

Homework Equations



Surface area of a cylinder's sides: 2*Pi*r*h

The Attempt at a Solution



The pressure at a depth d in the water should be equal to 101325+1000*9.8*d (1 atmosphere plus the weight of the molasses above it). I tried integrating this from 0 to 30 and then multiplying it by the surface area but that didn't work. What should I have done?

This is a very poorly worded ambiguous question. If isn't clear whether it asks you to find the net force outward exerted by the combination of the fluid on the inside and the air on the outside, or just the force exerted by the fluid on the inside. Secondly, since the pressure loading by the fluid on the tank is cylindrically symmetric, the overall net force on the tank is zero. If they are not looking for the overall forces, but instead, the local forces, this is just the force per unit area exerted by the local pressure (either including just the fluid inside, or both the fluid inside and the air outside).
 
Chestermiller said:
This is a very poorly worded ambiguous question. If isn't clear whether it asks you to find the net force outward exerted by the combination of the fluid on the inside and the air on the outside, or just the force exerted by the fluid on the inside. Secondly, since the pressure loading by the fluid on the tank is cylindrically symmetric, the overall net force on the tank is zero. If they are not looking for the overall forces, but instead, the local forces, this is just the force per unit area exerted by the local pressure (either including just the fluid inside, or both the fluid inside and the air outside).

From what I understand, they mean the local forces.
 

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