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Sum of forces

  1. Feb 20, 2015 #1
    1. The problem statement, all variables and given/known data

    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 [Broken]

    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

    2. Relevant equations


    3. 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 [Broken]

    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 [Broken]


    I thought it is the gas but the pipe can be in that position:

    http://imageshack.com/a/img540/2986/WqMDvD.png [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Feb 20, 2015 #2

    Svein

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    You have forgotten the force from P2 on the pipe outside the box.
     
  4. Feb 20, 2015 #3
    From P2 or P1 ? if it's outside it's P1, no ? The exercice guess P1=0.
     
  5. Feb 20, 2015 #4

    Svein

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    You figure says "full pipe". I expect that means that the pipe is full of gas at pressure P2.
     
  6. Feb 20, 2015 #5
    No, sorry, it's a pipe full of material, I would like to say cylinder
     
  7. Feb 20, 2015 #6

    Svein

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    The the pressure is still P2 on the inside and P1 on the outside (including the pressure on the end of the pipe).
     
  8. Feb 20, 2015 #7
    Yes, and what I can understand ? One end of the cylinder is inside and the other end is outside. What force is false ?
     
  9. Feb 21, 2015 #8

    Svein

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    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?
     
  10. Feb 21, 2015 #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 ?
     
    Last edited: Feb 21, 2015
  11. Feb 21, 2015 #10

    Svein

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    1. From where do you get the forces F1/F5?
    2. Torque is not a force.
     
  12. Feb 21, 2015 #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
     
    Last edited: Feb 21, 2015
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