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Torque/moment of beam in equilibrium

  1. May 25, 2012 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/5plXa.png



    2. Relevant equations
    ƩFx = 0, ƩFy = 0, ƩM = 0



    3. The attempt at a solution
    I realize that this is an equilibrium problem and that the sum of the forces in the x and y directions is 0, and the sum of the moments is 0. I draw two free body diagrams, one at point A (the contact point between the left wall and beam), and one at point B (the edge). Each FBD has the forces exerted by the bar on the wall - I call it Fa and Fb respectively. Also, both points have normal forces which are perpendicular to the surface. FNa is perpendicular to the wall while FNb is perpendicular to the rod/edge. Also incorporating the weights (Wa and Wb), I get four equations.

    What I can't figure out is how to formulate an equation for the sum of the moments, and subsequently solving the problem for theta. Any help would be tremendously appreciated.
     
  2. jcsd
  3. May 25, 2012 #2

    tiny-tim

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    Hi Hyperfluxe! :smile:
    (you don't need to separate the weight, one W will do :wink:)

    You should be able to find two equations for FNa and θ.

    What is your moments equation (about B)?
     
  4. May 28, 2012 #3
    Ok, what I did instead is making a sum of the moments equation about point A. So, from my FBD, I get 3 equations. F1 is the force between the left contact point and the wall, and F2 is the force from the edge contact point. W is the weight (assuming it acts down at the center of the bar), I get:

    1 - F1 = F2sin(theta) ----> I didn't need this equation

    2 - F2cos(theta) = W

    3 - (F2cos(theta))(0.325)-(W)(x) + (F2sin(theta))(y)
    where x=0.5cos(theta) and y=0.325tan(theta) from geometry.

    I equate W, cancel out F2, and solve for theta using a some algebra and a trig identity to get theta = 29.98degrees.

    Is my method and answer correct? Thanks!
     
  5. May 28, 2012 #4

    tiny-tim

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    Hi Hyperfluxe! :smile:
    Yes, your method is fine. :smile:

    (though you could just have said F20.5secθ, instead of (cosθ + sinθtanθ) :wink:)

    (i haven't chekced your numerical result)
     
  6. May 28, 2012 #5
    Thank you very much! Turns out I didn't have enough knowledge on moments on Friday, so I wasted 2 hours trying to solve that question. It feels really satisfying right now though!
     
  7. May 28, 2012 #6

    tiny-tim

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    (You obviously needed practice on moments, so I didn't suggest this before …)

    There are only three external forces, and you know the directions of all of them

    so can you see a way to find θ just by drawing some lines on the diagram? :wink:
     
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