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[Statics] Finding the Tension

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    [PLAIN]http://img830.imageshack.us/img830/1091/unled2cjt.jpg [Broken]


    2. Relevant equations
    I used sum of forces in x y direction and sum of moments


    3. The attempt at a solution

    So my problem here is that I solved for Ay (1275.3N) and now I have to solve for T

    The obvious approach would be to use sum of forces in x direction after doing sum of forces in y to find B. Well I did that and found that B = 2550.6 which is double Ay and equal to mg. B is also perpendicular to the bridge thing. Now when I do sum of forces in x direction i do T = Bcos30 but that's wrong apparently.

    Which part am I doing wrong? Is my B wrong?

    I calculated B like so:
    Fy:0 = Ay - mg + Bsin30

    Please help :)
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 10, 2011 #2
    Does this help? %^)
     

    Attached Files:

  4. Oct 11, 2011 #3
    unfortunately it does not. I figured that much. When i break down the N2 into it's components I still get a wrong answer for T
     
  5. Oct 11, 2011 #4
    i think that it is important to know whether there is friction with the rollers at A and B.
     
  6. Oct 11, 2011 #5
    There is not, so you would think that the x component of the normal force at B would be equal to T, but it's not!
     
  7. Oct 11, 2011 #6
    If you explain more explicitly how you got B then I can try to give some help.
     
  8. Oct 11, 2011 #7
    I calculated B like so:
    Fy:0 = Ay - mg + Bsin30

    Direct copy from my first post... i don't think i can be more explicit than this..
     
  9. Oct 11, 2011 #8

    It has to be. What other forces in the x direction are there?
     
  10. Oct 11, 2011 #9
    From what I understand there are 2 forces. B and T both pointing in the opposite direction. Is that correct?
     
  11. Oct 11, 2011 #10
    I think that sin must be replaced by cos.
     
  12. Oct 11, 2011 #11
    One must be careful here. The SIN30 component of B is in the opposite direction of T.
     
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