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Finding tensions in cables

  1. Aug 13, 2013 #1
    1. The problem statement, all variables and given/known data


    A 5m long uniform beam AC with mass of 600kg is being lifted from the ring at B by two chains (AB, 3m long and CB, 4m long). Determine the tension in chain AB and CB when the beam is clear of the ground.


    2. Relevant equations



    3. The attempt at a solution

    I tried dividing it into 2 right angle triangles but doesnt the non-equal length sides mean it's not a right angle triangle? I'm struggling to find the angles because of this.

    Picture_3.jpg
     
  2. jcsd
  3. Aug 13, 2013 #2
    If there's tension in the two chains, then you are going to have a 3-4-5 triangle. This is a right triangle.

    Do you think that the beam is going to remain horizontal after both ends are off the ground? If not, which end do you think will be higher? How many unknowns do you assess are required to solve this problem?

    Chet
     
  4. Aug 13, 2013 #3
    Use the Law of cosines to find the angles. Then do a free body diagram of point B (keeping in mind that the horizontal components of the tension in each wire are equal but opposite, and that the sum of the vertical components equal the tension in the upper wire above B) and solve for the vertical components using vectors and the angles.
     
  5. Aug 13, 2013 #4
    The 3 side will be higher, no? But how do we find the missing side of the right angle triangle? c^2 = a^2 + b^2 would give the missing height, but a different answer depending on if the 3m side is used or the 4m side is used.

    To solve this, all i would need is the angles of the right angle triangles right? I can solve the tensions from there?
     
  6. Aug 13, 2013 #5
    As I already said, if there is tension in the chains, then the triangle is going to be a right triangle, with sides 3, 4, and 5. Let θ be the angle that the beam makes with the horizontal after it loses contact with the ground, and let T1 and T2 be the tensions in the two wires. Note that θ also represents the angle through which the entire 3-4-5 triangle rotates. Thus, specifying θ completely specifies where all points on the triangle will move to.

    Chet
     
  7. Aug 13, 2013 #6
    This is what I think about how much the load swings...
     

    Attached Files:

  8. Aug 13, 2013 #7

    gneill

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    Staff: Mentor

    Surely the center of mass of the beam should end up directly below B? It would seem to represent the lowest energy state (gravitational PE).

    attachment.php?attachmentid=60896&stc=1&d=1376445974.gif
     

    Attached Files:

  9. Aug 13, 2013 #8
    Good point. My mistake. It seems so obvious: I don't know what I was thinking!
     
    Last edited: Aug 13, 2013
  10. Aug 13, 2013 #9

    gneill

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    Staff: Mentor

    Been there, done that! :smile:
     
  11. Feb 14, 2017 #10
    Then what's next?
     
  12. Feb 14, 2017 #11

    gneill

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    Staff: Mentor

    The last post when the thread was active was from August 2013. The OP (Original Poster) was last active in January 2014. Somehow I doubt that he or she will interested in continuing the problem here.

    If you are working on the same problem and need help you'll need to show your own attempt at solution just as though you were posting the problem yourself. No help can be given until effort is shown (it's in the forum rules!).
     
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