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Tension on two cables

  1. Apr 8, 2015 #1
    20150408_145501.jpg 1. The problem statement, all variables and given/known data
    There is a girder hanging from a crane by two cables. The girder is 2.6m long and hanging at an angle. The weight of the girder is 0.50 kN. Calculate the tensions of each cable.

    2. Relevant equations


    3. The attempt at a solution
    I would divide the two cables into their x and y components and use trigonometry to find the resultant vector for each...but how does the length of the girder affect this?
     
  2. jcsd
  3. Apr 8, 2015 #2
    Let's see you show some equations on this, and maybe the answer to this question will become clear to you.

    Chet
     
  4. Apr 8, 2015 #3
    Ok, would first divide the total weight of the girder by the number of cables, so each cable would be experiencing an equal force?
     
  5. Apr 8, 2015 #4
    No. Leave T1 and T2 as unknowns. Let's see the force balances in the x and y directions.

    Chet
     
  6. Apr 8, 2015 #5
    Force balances? How do I work them out?
     
  7. Apr 8, 2015 #6
    What are the components of the tensions in the x and y directions?
     
  8. Apr 8, 2015 #7
    For T2 X component it will equal half of t2 correct?
     
  9. Apr 8, 2015 #8
    And T1y component will equal half of T1?
     
  10. Apr 8, 2015 #9
    Yes. Correct so far.
     
  11. Apr 8, 2015 #10
    Yes
     
  12. Apr 8, 2015 #11
    And T1y + T2y must equal 0.50 kN?
     
  13. Apr 8, 2015 #12
    But now I'm not sure where to go with this information. Am I missing something important?
     
  14. Apr 8, 2015 #13
    So i think I have it...

    The x component of T1 is T1cos30
    and the x component of T2 is t2cos60
    And T1x - T2x = 0

    the y component of T1 is t1sin30
    and the y component of t2 is t2sin60

    and t1y+t2y-500=0

    so if T1cos30-T2cos60=0 then T1cos30=T2cos60
    therefore T1 = T2cos60/cos30
    = 0.577T2

    substituting this into the equation for the y components

    0.577T2sin30 + T2sin60 = 500N
    therefore T2(0.577sin30 + sin 60) = 500N
    T2 = 433N

    T1 = 0.577T2
    = 250N

    Is this correct?


    I'm still not sure where the length of the girder comes in?
     
  15. Apr 8, 2015 #14
    manrkin!!!! To be honest, I didn't think you were going to be able to do it. But you did. Congrats.

    I haven't checked your arithmetic, but your methodology is perfect. Nice job.

    Regarding the length of the girder: it might also be possible to solve this problem (maybe easier) using moment balances. That's where the length might come in ( to help you solve it). If you take moments about the connection of cable 2 with the girder, you can get T1 directly, and, if you take moments about the connection of cable 1 with the girder, you can get T2 directly. If this problem is solved using these moment balances, they will give the same results as those which you already have. But, maybe you would like to solve it that way just for practice. Any interest?

    Chet
     
  16. Apr 8, 2015 #15
    Thank you Chet! I really appreciate your help. It took me three cups of tea and some serious head scratching.

    I will come back to you on solving it using the moments, however for now I'm currently half way through a rather large assignment and I'm going to keep on plugging away.

    Kind regards Chet!
     
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