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Static Equilibrium questions

  1. Mar 27, 2009 #1
    1. The problem statement, all variables and given/known data
    http://www.physics.upenn.edu/~jrk/p101/Homework/hw7_sp09.pdf

    Here is the link. I am having trouble with questions 8 and 9. I posted the whole link because these questions have a diagram that go with it.


    2. Relevant equations
    Sum of F = 0
    Sum of torque=0



    3. The attempt at a solution

    Well, I just considered rod B, because that's what they asked about.

    So the Forces there are:

    The normal force from the wall
    An upward force to keep it from falling
    And the tension force acting downward on rod b.

    But the problem is, then how will the horizontal forces balance? I think I'm definitely missing something.

    When I did the problem going straight to the torques with these forces,

    I chose an axis of rotation at a and plugged in the corresponding torques for the forces listed above, and I got answer c.

    But I know I'm doing something wrong in the first part because my teacher said to always start with the forces and then pick the appropriate axis of rotation, but my forces don't make sense. Please help.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Mar 27, 2009 #2

    LowlyPion

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    Perhaps the best approach is to examine the Torque about point B?

    You have 2 forces to consider, since any forces along BC may be ignored because they have no moment arm through the pivot.

    Then you should be able to write an equation for the projection of those forces to the x and y.

    mg projects at BC*Cosθ

    And the Tension in AC projects as T*BC*sinθ ... This means then that

    T*BC*sinθ = mg*BC*cosθ

    Since the Tension in AC at the wall is +x, then there must be a corresponding -x force at the wall at the pivot at point B.
     
    Last edited by a moderator: Apr 24, 2017
  4. Mar 27, 2009 #3
    Why would I make the axis of rotation B, if I'm trying to solve for the normal force exerted on B? And which forces would I use in each direction in writing the sum of forces. I don't think I'm really understanding...are you considering the whole system or just rod B?
     
  5. Mar 27, 2009 #4

    djeitnstine

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    The simplest solution would be to consider rods AC and BC along with the hanging mass alone. That way the forces exerted on the wall become apparent. Then sum the torques about A. This equation gives an immediate answer. Also, the answer C is correct. This is because reaction forces are always opposite the member force.

    I can show you the working if you need.
     
  6. Mar 27, 2009 #5
    So does this mean that in the x-direction, there is the normal force from the wall on A and a normal force on the wall on B, and in the y-direction, there is the rope tension and the vertical forces from rod A and B? So then if you consider A the axis of rotation, then there would be no need to consider torque from the normal force on A or the vertical force of A? So then the torques would only be from the normal force of B, the Tension in the rope, and the vertical force of B?
     
  7. Mar 27, 2009 #6

    djeitnstine

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    surely you meant to say horizontal but you are correct. Also you can do the same procedure at B to find the answer to #9

    Edit: I misread there but since the vertical force at B is acting at the point you are summing the torques by, its distance is 0 and this does not show up
     
  8. Mar 27, 2009 #7
    Why isn't the vertical force of B acting at point B...how is it acting at Point A?
     
    Last edited: Mar 27, 2009
  9. Mar 27, 2009 #8

    djeitnstine

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    You should review torques. When summing the torques about A, what is the perpendicular distance acting between the point A and the line of action of the vertical component of force B?
     
  10. Mar 27, 2009 #9
    Oh, I am pretty sure I understand now. Because torque can only act perpendicular by the definition of the cross product, right?

    Would you mind working the problem out, so that I can make sure my work is consistant? Thanks! I just want to make sure everything is correct, so that when I review, I can remember exactly what to do.
     
    Last edited: Mar 27, 2009
  11. Mar 27, 2009 #10

    djeitnstine

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    So I will sum the torques about point A and will use the scalar definition of torques which is the perpendicular distance from the axis of rotation. For simplicity I will denote [tex]B_x[/tex] as the horizontal force at B.

    [tex]\sum \tau =0[/tex]
    [tex]-lmg+dB_x=0[/tex]
    [tex]B_x=\frac{l}{d}mg[/tex]

    (use the right hand rule to find whether the moment is positive or negative)
     
  12. Mar 27, 2009 #11
    So, you always look at the perpendicular distance from the axis of rotation?

    Thank you so much for all of your help! I really appreciate it.
     
  13. Mar 27, 2009 #12

    djeitnstine

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    Yes look for the forces acting perpendicular to your displacement (vector) or distance from axis of rotation (or whatever you want to call it) to make your life easier. If its not so simple, use the vector method with the aforementioned displacement vector crossed with the force
     
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