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Moment and forces

  1. Nov 22, 2013 #1

    adjacent

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    1. The problem statement, all variables and given/known data

    Find the other forces present.
    Assume the triangle to be weightless
    attachment.php?attachmentid=64194&stc=1&d=1385125275.gif


    2. Relevant equations
    τ=Force x perpendicular distance


    3. The attempt at a solution
    The force,10N provides a 50Nm Torque. (10 x 5)
    The triangle is in equilibrium so,anticlockwise moment is equal to clockwise moment
    50Nm=2 x F
    F=25N at p

    But within the two meters,not only p touches but other parts of the triangle(in contact with the wall) touches too.
    Should I account for that?

    As the net force should be zero,pivot point exerts a force of 25N to the left.and 10N upwards.
    Is this correct?
     

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    Last edited: Nov 22, 2013
  2. jcsd
  3. Nov 22, 2013 #2

    Doc Al

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    Your attachment seems unviewable. Try again.
     
  4. Nov 22, 2013 #3

    adjacent

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    I think it's okay now
     
  5. Nov 22, 2013 #4

    Doc Al

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    It's not clear exactly how the triangle is attached to the wall or how it makes contact. So I would just consider points O and P as making contact.

    Can we assume the only fixed pivot is at point O and that the triangle just rests against the wall? (And that point P cannot exert an upward force?)

    Looks reasonable to me.
     
  6. Nov 22, 2013 #5

    adjacent

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    Yes.
    And all the parts of line Op touches the wall
     
  7. Nov 22, 2013 #6

    adjacent

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    I want the answer to this question:

     
  8. Nov 22, 2013 #7
    If the triangle presses against the wall,then you also need to consider Normal force from the wall .
     
  9. Nov 22, 2013 #8

    adjacent

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    How?Here only moment equation can be used.
     
  10. Nov 22, 2013 #9
    What do you mean by the above statement? When writing force and torque equations for the triangle ,you need to consider all the forces acting on the triangle .

    The forces on the triangle are

    1)Applied force -10 N downwards
    2)Hinge force
    3)Normal force from the wall
     
  11. Nov 22, 2013 #10

    adjacent

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    I'm worried about the Normal force from the wall.How do I know it?I can calculate Normal force at point p. (50Nm=2*F)
    But there are infinite number of normal forces acting by the wall.(In the distance 2m,it's 25N.In the distance 1m ,its 50.In the distance 0.0000001 meter,It's is ... and so on.Do I need to add all these?
     
  12. Nov 22, 2013 #11
    When two objects are in contact,there exists a contact force between them.The component of this force parallel to the surfaces in contact is called friction .The component of force perpendicular to the surfaces is called Normal force .

    In this problem,there is no friction force,but there exists a normal force.

    Even though there are infinite number of points on the surfaces in contact ,there is only one normal force ,which you may consider to be the net force which the bodies exert on each other ,perpendicular to the surfaces.

    Well..now the question is where is the point of application of this Normal force ?

    @Doc Al :Would you mind chipping in ? I hope I am not talking nonsense and misguiding the OP
     
  13. Nov 22, 2013 #12

    adjacent

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    Yes.That was my question.
     
  14. Nov 22, 2013 #13

    Doc Al

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    I agree with that.

    There's the rub. To find the effective point of application of the normal force.

    No, you're fine. (Sorry, I was AFK for most of the day.)

    I'll have to ponder just how to find that point of application.
     
  15. Nov 23, 2013 #14
    Thanks Doc...

    If the normal force acts at a distance 'x' from O then we have the relation (N)(x) = 50 . But this gives us several possibilities .
     
  16. Nov 23, 2013 #15

    adjacent

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    And ##0<x\leq2##

    Eating my head.Help please
     
  17. Nov 23, 2013 #16

    Doc Al

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    Is this a textbook problem? If so, please state the textbook and problem number. (If I have the book, I can check the context of the question.)
     
  18. Nov 23, 2013 #17

    adjacent

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    This is not a text book question.I made it. :wink:
     
  19. Nov 23, 2013 #18

    PhanthomJay

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    You are getting into all sorts of deformation and indeterminate complexities if you consider any normal forces from the wall acting between O and P. The vertical member is not glued to the wall anyway. In addition to the applied force, the only other forces acting on the triangle occur ay joints O and P. Solve for them. There is no normal force from the wall between those points.
     
  20. Nov 23, 2013 #19

    adjacent

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    You are wrong.p is not a joint,it is just a point in the wall.O is the joint or pivot.Did you read the posts?
     
  21. Nov 23, 2013 #20

    Doc Al

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    Exactly. I don't think there is any simple way to address those forces, which are very dependent on exactly how the surfaces makes contact.
     
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