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Torque Equilibrium to find center of mass.

  1. Apr 14, 2014 #1
    1. The problem statement, all variables and given/known data
    1. A non-uniform beam with a mass of 85kg has a length of 6.2m and is attached to a wall and supported at an angle of 28 degrees from the wall by a horizontal rope which is attached to the wall above the beam . The tension in the rope is found to be 310N. How far from the pivot point is the center of mass of the beam?

    2. Relevant equations
    T= d x F
    Trigonometry (sin=o/h, cos=a/h, tan=o/a)
    Fg= m x g

    3. The attempt at a solution
    Well I am pretty lost in the question but I gave it a shot anyways.
    Fg= 85 x 9.8= 833 N
    Fg(perpendicular)= 833sin28= 391.1 N
    Ft= 310 N
    Ft(perpendicular)= 310cos28= 273.7 N
    ∑= 391.1-273.7= 117.4 N
    T= 117.4 x 6.2= 727.88

    This is where I think I messed up but don't know where else to go from here.
     
  2. jcsd
  3. Apr 15, 2014 #2

    Simon Bridge

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    Welcome to PF;
    The trick with this sort of question is to sketch it out and draw in the forces, and label points.

    So call the pivot O, the other end A and the point the rope joins the wall is B ... pick an arbitrary point off-center on the beam to represent the center of mass and call it C.

    OC=x (which you want to find).
    I don't see any reference to an unknown distance in your analysis.
    Sketch in the forces at O, A, B, and C, don't forget the directions.

    Now you should be in a position to write the expression for the total torque about O, which will include the unknown x. Solve for x.

    Note - torque is usually given as a cross product as in T = d x F, but the weight is just a scalar product Fg=mg.
    Formally: ##\vec \tau = \vec r \times \vec F## and ##\vec F\!_g=m\vec g##
     
  4. Apr 15, 2014 #3
    Maybe a free body diagram would help us follow what you are trying to do.

    But you got the right equations, just need to execute them properly.
     
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