Outward force on a tank of water.

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The discussion revolves around calculating the outward force on the sides of a cylindrical tank filled with molasses. The pressure at a depth in the tank is derived from the atmospheric pressure plus the weight of the molasses, leading to confusion about whether to integrate pressure or force. Participants highlight the ambiguity in the problem regarding whether it asks for net force or just the force from the fluid inside. It is noted that the pressure loading is cylindrically symmetric, suggesting the overall net force on the tank is zero. Clarification is sought on whether the question pertains to local forces or the net force exerted by both the fluid and the air.
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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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