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Tension Force Question

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    See Attachment

    2. Relevant equations



    3. The attempt at a solution

    I'm unsure how to begin the problem because of the fifth tension force. How does one draw this force on a free body diagram? Does the force of tension on the ball by tension 5 have a direction of towards the left or towards the right?

    I was able to include the rest of the forces in my free body diagram but am unsure how to include the force of tension for the 5th tension. I have yet to proceed further in solving the problem because I'm unsure about the free body diagram.

    Thanks for any help.
     

    Attached Files:

  2. jcsd
  3. Sep 25, 2011 #2
    Good evening!

    Think of the "tension" force as a rope. If one end pulls, what would happen to the other side?

    After that, just sum the forces:
    [itex] \sum F_x [/itex]
    [itex]\sum F_y[/itex]
     
  4. Sep 25, 2011 #3
    I'm unsure how to draw the fifth tension on a free body diagram though but am completely fine with the other forces. My gut feeling is that the fifth string pulls the ball up at the same angle as (7pi)/18 and would be align with the third and fourth tensions but am not completely sure
     
  5. Sep 26, 2011 #4
    So my answers don't match with the answers on the answer sheet
    I got
    T1 = 346 N
    T2 = 346 N
    T3 = 506 N
    T4 = 506 N
    T5 = 302 N

    my work is below i scanned it and uploaded it

    http://imageshack.us/photo/my-images/850/unledvbq.png/

    sorry for some reason I couldn't post it as a image I think because it was to large of a file
     
    Last edited: Sep 26, 2011
  6. Sep 26, 2011 #5
    Anyone?
     
  7. Sep 27, 2011 #6
    B.U.M.P
    it went onto the second page...
     
  8. Sep 27, 2011 #7

    gneill

    User Avatar

    Staff: Mentor

    Regarding the rope carrying T5, you can draw a FBD for the point where tensions meet. So for example the joint where T1, T4 and T5 meet is a natural FBD! Each of the tensions represents an outward directed force, and they all originate at a point.

    You can use symmetry to deduce a few things before getting down to calculations. First, the equal 30° angles that T1 and T2 make with the vertical imply that the triangle they make up along with T5 is an equilateral triangle (with 60° at each vertex). It also implies that T5 will be horizontal. Next, the equal angles of T3 and T4 to the vertical, and the fact that they attach to the system below at the same height implies that T3 and T4 are equal.

    You've got correct values for T1 and T2. How would you go about solving for T3 or T4? A hint would be that T3 and T4 together must support the vertical weight of the ball. And we've already decided that T3 = T4.
     
  9. Sep 27, 2011 #8
  10. Sep 27, 2011 #9

    gneill

    User Avatar

    Staff: Mentor

    I'm having a little trouble sorting out your workings from that image. So why don't we try to do some of it here?

    Since T3 and T4 together must be supporting the weight of the ball, what must the sum of their vertical components be?
     
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