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Tension in Cable

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data


    Two Cables tied together at C are loaded as shown. Determine the range of values of Q for which the tension will not exceed 270N in either cable. Solve graphically and Numerically.

    2. Relevant equations
    law of cos, law of sine
    sum of Fx = 0
    sum Fy = 0

    3. The attempt at a solution
    I'm really not sure where to start with this one, probably because I don't fully understand the question. I am assuming that I should set AC = 270N and BC= 270 N. And then find all components such that sum of Fx = 0 and Sum Fy = 0. This will be the Max vaule and the resultant of BC and AC plus P should give me Q? I am getting an answer like this, but it's not making sense.
  2. jcsd
  3. Sep 16, 2007 #2


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    Please check your attachment. I think you're not getting help because your picture won't download into your post and that is essential to working on the problem...
  4. Sep 16, 2007 #3


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    Don't assume anything for the tension in the cables... call them T1 and T2... then write the sum of the forces in the x-direction... y-direction...

    Solve for T1 and T2, in terms of Q... that will give you the limits on Q.
  5. Sep 16, 2007 #4
    I thought that I had to assume the tension in the cables, otherwise how would I solve for the parameters given, ie: "range of values of Q which the tension will not exceed 270 in either cable."
    Do you not solve by setting each tension to max (270 N) and then solving for Q... let's say I get Magnitude Q = xN then I would say 0<= Q <= xN?

    So far I have done this

    sum Fx: 340 cos 0 + 270 cos 90 + 270 cos 180 + Qx cos 240
    I then solved for Qx cos 240 = -70

    sum Fy: 340 sin 0 + 270 sin 90 + 270 sin 180 + Qy sin 240
    I then solved for Qy sin 240 = -270

    I then took the Qx component and Qy component and solved for Q using pythogras to yield 279.92 or 280 N at 60 degrees South of West.
    So my final answer would be 0 N=< Q <= 280 N

    I have no idea if this is correct, so any help with my methodology is welcome. Also if somone is able to confirm that they can't see the diagram I posted? I can view it without issue using several ISP's.
    Last edited: Sep 16, 2007
  6. Sep 16, 2007 #5


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    Doesn't Q have to be downward? Why do you have Qx and Qy?
  7. Sep 16, 2007 #6


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    Could be the browser on my ancient computer at home (I've hit a couple other postings today I couldn't read all of). I'll have to see if I have the same trouble on the computer at work...
  8. Sep 16, 2007 #7
    Does Q have to be downward?
    I don't see why it does, gravity does not come into play. I just rotated that picture above and my FBD 30 degrees clockwise. That made all relevant angles( of vectors) 90 degrees with the exception of Q which was 240 degrees. Just a rotation.
  9. Sep 16, 2007 #8


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    I'm able to see the picture.
  10. Sep 16, 2007 #9


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    I don't understand Qxcos240... Do you mean Qcos240? ie: Qx = Qcos240?

    I don't think this method is correct... It is impossible for both tensions to be 270N at the same time.

    I recommend going with what I said earlier... use T1 and T2 as variables.... solve for them in terms of Q... then you can see what limits are required to make T1 and T2 remain less than 270.
  11. Sep 17, 2007 #10
    Ok, I am doing this completely wrong if it is impossible for both tensions to be 270N at the same time. Can you tell me why? Is it because they have different angles, and therefore regardless of the tension on P or Q it will never be "evenly distributed" over both cables?
  12. Sep 17, 2007 #11


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    You should solve it for each cable separately.
    a) Pick a cable (say, A)
    b) Calculate the tension in A due to the combined force P as drawn and Q downward
    c) Solve for Q, this will give you a maximum value Q(A) on Q
    d) Repeat for B, getting a maximum value Q(B)
    e) Write down the answer: min(Q(A), Q(B))
  13. Sep 17, 2007 #12
    Thanks learningphysics and CompuChip, I think I've done it correctly (I hope).
    Can you let me know if the methodology is correcT?

    I rotated my FBD, the diagram in the text by 30 degrees clockwise. I like it this way because it gives us some special angles to get easy values for (for cancelling unknown TBC and TAC), like angle 0, 90 and 180. I assume to NOT know the values for the tension, and solve for the two unknowns (just like you both indicated I should initially!)

    Now I sum the Fx Forces.
    0= 340 Cos 0 + TAC cos 90 + TCB cos 180 + Q cos 240 0= 340(1) + 0 + TCB(-1) + Q(-1/2) 0= 340 -TCB -Q(1/2) 340 -Q(1/2)= TCB Now we say TCB <= 270N So; 340 -Q(1/2) <= 270N
    -Q(1/2) <= 270N - 340N
    Q(1/2) >= 340N - 270N
    Q(1/2) >= 70N
    Therefore Q >= 140N

    Now I sum the Fy forces in similar fashion
    0= 340 sin 0 + TAC sin 90 + TCB sin 180 + Q sin 240
    Etc... And I end up with;
    Q<= 311.77
    for a final answer of

    140N <= Q <= 312N
  14. Sep 17, 2007 #13


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    Looks good to me.
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