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This time its a statics problem

  1. Nov 14, 2004 #1
    A nonuniform bar is suspended at rest in a horizontal position by two massless cords as shown. One cord makes the angle theta1 = 36.9 with the vertical; the other makes the angle theta2 = 53.1 with the vertical. If the length L of the bar is 6.10m, compute the distance x from the left-hand end of the bar to its center of mass.

    I keep getting stuck on this one. Basically all I know is that it's a statics problem, since it's at rest. So, I know that

    T1 sin(theta1) + T2 sin(theta2) = mg
    T1 cos(theta1) - T2 cos(theta2) = 0
    where T1 and T2 are the tensions on the cords supporting the rod.

    There are so many unknowns that I'm not sure where to go from here, but I've been playing around with:

    T1sin(theta1)*X - T2sin(theta2)*(L-X) = 0
    (Force1 * Distance1) == (Force2 * Distance2)

    I would appreciate some suggestions and hints! (i'm not asking for anyone to solve it because the problem may be on my test tommarow). Thanks a bunch!
  2. jcsd
  3. Nov 14, 2004 #2
    ...sry. didnt mean to post this twice
  4. Nov 14, 2004 #3
    Your first 2 equations is wrong already. sin and cos should exchange. Hint: Take the summation of moment at one of the end.
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