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Calculating torque with buyouant force and hardly definiable arm

  1. Mar 9, 2013 #1
    Hej Guy,

    Here I am again with a question on torque basically.

    Please see the image below.

    I have a half tube underwater. The tube have air inside. Now I would like to calculate the torque for this system. As you see the tube will move to a balanced position, but I don't how much torque it will have on the axis of rotation..

    My question is it legal to approximate the torque as you see in the 3rd image that I divide the tube volume and use those volume segments with the distance to the center.

    I am having this problem because basically inside nothing can act as an arm, but the tube have its volume and this volume have distance from the center of rotation and well the volume creating the buoyant force...

    Or is it just a whole volume as a force upwards and the toque is basically the Torque =buoyant force x center-cylinder-radius ? This seems to be not valid for me.

    Anybody can help me on this?
    Thanks in advance!

    problem.jpg
     
  2. jcsd
  3. Mar 9, 2013 #2

    tiny-tim

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    hej losbellos! :smile:

    (pleeeeeeeeeease don't post such wide pictures :redface:)
    the total buoyant force acts through the centre of buoyancy (the centre of mass of the displaced fluid)

    so the only two net forces on the tube are …

    its weight, acting through the centre of mass

    the buoyant force, acting through the centre of buoyancy​
     
  4. Mar 9, 2013 #3
    This would be valid for calculating the torque due to bouyancy: remember also to calculate the torque in the opposite direction due to gravity.

    Just as the torque due to gravity can be simplified as if the object's whole mass was positioned at its centre of mass, the torque due to bouyancy (in an incompressible fluid) can be simplified as if the object's whole bouyancy was positioned at its centre of bouyancy. Note that "bouyancy" is determined by the mass of fluid displaced, and so the centre of bouyancy is determined in the same way as the centre of mass and for an object of uniform density is at the same point.

    For your object the centre of mass lies some way along the radius of the half-pipe, although exactly where depends on the density of its walls, and the centre of bouyancy lies along the same radius.
     
  5. Mar 9, 2013 #4
    Why is it a post can go unanswered for 7 hours and then two people decide to post an answer at the same time (although I was slower writing mine than tiny-tim)?
     
  6. Mar 9, 2013 #5
    thanks for the answers...
     
    Last edited: Mar 9, 2013
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