Solving a Container Pressure Problem

In summary: Yes, centrifugal forces act to dissipate energy and they can cancel each other out, so the sum of forces on the cylinder is 0.
  • #1
V711
61
0

Homework Statement



A closed container has a pressure P2 inside it. Outside there is the pressure P1 with P2 > P1. A pipe is enrolled on the grey disk. The pipe has a mass. The grey disk can rotate around itself.

http://imageshack.com/a/img905/3235/Egzdk8.png

1/ Draw all forces with P1=0
2/ Is there a torque on the grey disk/pipe ?
3/ Give the sum of forces on the container and on the pipe

Homework Equations

The Attempt at a Solution



1/ The pipe receives F1 and F2. The container receives F2, F4, F5 and F6.

http://imageshack.com/a/img908/8966/4ZFv5a.png

2/ Yes, the disk/pipe receives the torque F1 and F2

3/ The sum of forces on the pipe is 0. The sum of forces of the container is 0 if I don't take in account the centrifugal forces of the pipe because F2+F4+F5+F6 = 0 (in vector). But with centrifugal forces I don't find 0, where is the force that compensate centrifugal forces ?

http://imageshack.com/a/img908/592/R8gXrs.png I thought it is the gas but the pipe can be in that position:

http://imageshack.com/a/img540/2986/WqMDvD.png
 
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  • #2
You have forgotten the force from P2 on the pipe outside the box.
 
  • #3
From P2 or P1 ? if it's outside it's P1, no ? The exercice guess P1=0.
 
  • #4
You figure says "full pipe". I expect that means that the pipe is full of gas at pressure P2.
 
  • #5
No, sorry, it's a pipe full of material, I would like to say cylinder
 
  • #6
V711 said:
No, sorry, it's a pipe full of material, I would like to say cylinder
The the pressure is still P2 on the inside and P1 on the outside (including the pressure on the end of the pipe).
 
  • #7
Svein said:
The the pressure is still P2 on the inside and P1 on the outside (including the pressure on the end of the pipe).
Yes, and what I can understand ? One end of the cylinder is inside and the other end is outside. What force is false ?
 
  • #8
V711 said:
Yes, and what I can understand ? One end of the cylinder is inside and the other end is outside. What force is false ?
An analogy: Think of a plate of spaghetti in normal atmospheric pressure (P2). Put the end of one strand into your mouth. Suck (i.e. create a lower pressure P1 inside your mouth). What happens?
 
  • #9
The spaghetti will move inside the mouth. Here the cylinder move outside the container (P inside > P outside), correct ? There is a torque on the disque/cylinder with forces F1/F3, that torque moves the cylinder outside, correct ?

My problem, is the sum of forces on the container, for me there are forces F2, F4, F5, F6 from pressure, the sum is 0, but when the disk rotates with the cylinder, there are centrifugal forces, and what forces cancel these centrifugal forces for have the sum at 0 ?
 
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  • #10
  1. From where do you get the forces F1/F5?
  2. Torque is not a force.
 
  • #11
1/ Note the cylinder drives the disk like a toothed belt for example. Like there is an axis of rotation all concentric forces can't rotate the cylinder. The only force F1 can rotate the cylinder. If there is F1, this force want to push the cylinder at right, but the axis prevent this, so the axis of rotation receives F2 and gives F3 to the cylinder. Concentric forces push more at left than at right because the surface at right is bigger than at left, so there are F5 and F6 forces (F5 and F6 are the sum of all concentric forces around the disk/cylinder). Like the cylinder pass through the container at left there is the force F4 to the container. The sum of all these forces on the container is 0. And the cylinder receives a torque so the disk rotates and the cylinder moves outside the container.

2/ I'm agree
 
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1. How do I determine the pressure inside a container?

The pressure inside a container can be determined by using the ideal gas law, which states that pressure (P) is equal to the number of moles of gas (n) multiplied by the universal gas constant (R) and the temperature (T), divided by the volume (V). This can be represented as P = (nRT)/V.

2. What factors affect the pressure inside a container?

The pressure inside a container can be affected by several factors, including the amount of gas present, the temperature of the gas, and the volume of the container. Additionally, the type of gas and any external forces acting on the container can also impact the pressure.

3. How can I solve for the pressure inside a container if I know the volume and temperature?

If you know the volume and temperature of a container, you can use the ideal gas law (P = (nRT)/V) to solve for pressure. First, you will need to determine the number of moles of gas present (n) by dividing the mass of the gas by its molar mass. Then, you can plug in the values for n, R, V, and T to solve for pressure.

4. Can the pressure inside a container be negative?

No, the pressure inside a container cannot be negative. Pressure is defined as the force exerted by a gas on the walls of a container, so it is always a positive value. If the pressure inside a container becomes too low, the gas may condense or become a liquid, but it will still exert a positive pressure on the walls of the container.

5. How can I decrease the pressure inside a container?

The pressure inside a container can be decreased by either reducing the number of gas molecules inside the container, decreasing the temperature of the gas, or increasing the volume of the container. Additionally, external forces such as gravity or the weight of the gas itself can also impact the pressure inside a container.

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