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Equilibrium of rigid boy

  1. Jun 25, 2011 #1
    1. The problem statement, all variables and given/known data
    [PLAIN]http://img24.imageshack.us/img24/8382/fm1u.png [Broken]

    2. Relevant equations
    µ ≥ tan Θ

    3. The attempt at a solution
    I'm new here, if I post something wrong here. please correct me!! :)
    I don't understand the question.
    Where is the normal reaction between the rod and the floor? from figure 1.
    Is it in the same direction as the rod acting on the floor? or perpendicular to the floor?
    If the normal reaction is same as the direction as the rod acting on the floor then:
    Resolve vertically - R cos Θ = W
    Resolve horizontally - R sin Θ + F = ?
    F ≤ µR
    F ≤ µW / cos Θ ? is it correct?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 25, 2011 #2

    ehild

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    The normal force is always normal to the surface. Draw the free body diagram with all forces acting on the rod. The normal force from the floor acts upward.

    ehild
     
  4. Jun 25, 2011 #3
    Thanks ehild... :)
    so resolve vertically R = W :)
    I will try to work out this question. I will be back~!!
     
  5. Jun 25, 2011 #4

    ehild

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    It is not that simple. Draw all forces. There is a normal force also at the peg.The resultant torque and the resultant force both have to be zero at equilibrium.

    ehild
     
    Last edited: Jun 25, 2011
  6. Jun 26, 2011 #5
    I forgot there is a normal force at the peg >.<!!

    so R1 + R2 = W
    and how to resolve horizontally?
     
  7. Jun 26, 2011 #6

    ehild

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    The reaction force from the floor is vertical, but that of the peg is not. The forces are vectors, you have to add both the horizontal and vertical components.
    What forces act horizontally?

    ehild
     
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