Outward force on a tank of water.

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SUMMARY

The discussion centers on calculating the outward force on a cylindrical tank containing molasses, with a height of 30 m and a diameter of 30 m. The pressure at a depth in the tank is given by the equation 101325 + 1000 * 9.8 * d, where d is the depth. Participants highlight the importance of integrating the force rather than the pressure and clarify that the net force on the tank is zero due to cylindrical symmetry. The focus is on determining local forces exerted by the fluid inside the tank.

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  • Familiarity with integration techniques in calculus.
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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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