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Minimizing tension in the cables

  1. Jul 24, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A heavy box weighing 25kN is to be lifted by 2 overhead cranes with cables inclined at 40° and 30° with the vertical respectively.
    i) Find tension in each cable during lifting.
    ii) Due to safety concerns, the tension T2 in second cable is to be minimized. Find the angle of inclination of this cable for this purpose.


    2. Relevant equations

    Lami's Theorem

    3. The attempt at a solution

    I'm mainly concerned with the second part of the question. Applying Lami's Theorem, I can easily find the answer to the first part. I'm sure there exists a geometrical method for the second part but I'm not sure how to start. What about triangle inequalities? Will it be of any help?
     
  2. jcsd
  3. Jul 24, 2014 #2

    Simon Bridge

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    For (i) presumably lifting happens at a constant speed?
    For (ii) start by drawing a free-body diagram and applying Newton's laws... you need a relation for the tension in the second cable as a function of angle. Do the whole thing using variables - do not use the values given.

    You should also read Lami's theorem carefully.

    [edit] part (ii) appears under-specified too - are you expected to keep the first cable at 40deg or does this cable change angle as the other one does?
     
  4. Jul 24, 2014 #3

    utkarshakash

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    I think it's not supposed to change.
     
  5. Jul 24, 2014 #4

    Simon Bridge

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    Good - then the angle between the weight and the first cable's tension won't change either.
    What is Lami's theorem?

    Have you drawn the free body diagram?
     
  6. Jul 24, 2014 #5

    utkarshakash

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    It's difficult for me to do that right now as I don't have the required softwares(MS-Paint sucks, btw!!). I'll try to explain the diagram through words. Let T2 be inclined at an angle θ° with the vertical to the left side and T1 make an angle 30° with vertical on the right side. The gravitational force will be directed vertically downwards. Now, applying Lami's Theorem,

    [itex]\dfrac{T_2}{150°} = \dfrac{T_1}{180°-θ°} = \dfrac{25*10^3}{30+\theta} [/itex]
     
  7. Jul 24, 2014 #6

    utkarshakash

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    Sorry for my previous reply. The first cable can also change its direction simultaneously.
     
  8. Jul 25, 2014 #7

    Simon Bridge

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    That makes a difference yeah.
    When you evaluate the forces, do not put any numbers in. Use symbols only.
    So the tensions are T1 and T2, and the weight is W
    If T1 has angle A1 to the vertical, and T2 has angle A2 to the vertical, then you can write the equations that balance the forces.

    You also need to write out what Lami's theorem is telling you.

    Note: - put A1=0 and A2=90deg - what are the tensions?
     
  9. Jul 25, 2014 #8

    utkarshakash

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    t2=2t1
     
  10. Jul 25, 2014 #9
    Make minimal it itself!
     
  11. Jul 26, 2014 #10

    utkarshakash

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    How do I verify that minimum of T2 equals 2T1?
     
  12. Jul 26, 2014 #11
    If both the cables can change their angles, the obvious approach, as Simon Bridge indicated would be to have one of the cable vertical (T1) and the other horizontal (T2) assuming it can support the load, and due to lack of information, you can probably assume does.

    How do you get t2 = 2*t1? If the t2 cable is inclined 90 degrees to the vertical(or in other words, it is horizontal), will it balance any vertical force components? What will its value be then by balancing the horizontal components while cable 1 is vertical?
     
  13. Jul 26, 2014 #12

    utkarshakash

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    Ah! I didn't notice that T2 would be completely horizontal. Rather I blindly plugged in the value in my equation(how stupid :tongue2:!) The tension T2 would become zero.

    What would have been the answer if the direction of first cable wasn't allowed to change?
     
  14. Jul 26, 2014 #13
    Then the first cable would have a tension component in the horizontal direction, which should be balanced by the other cable. Any vertical tension would only lead to an increase in the total tension of cable 2.

    Another interesting case is for the first cable to have a limiting tension, which would imply that you cannot have a tension more than a certain value that the cable can handle, and related to that, you could find a specific angle so that T2 is minimized.
     
  15. Jul 27, 2014 #14

    Simon Bridge

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    How??!
    Didn't I just describe rope 1 going straight up and rope 2 horizontal?
    How does a horizontal rope get tension from the weight?
     
  16. Jul 27, 2014 #15

    ehild

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    In that case T2 is zero when minimum and the first cable is vertical. Solve the problem without changing the 30°angle of the first cable.

    ehild
     
  17. Jul 27, 2014 #16

    Simon Bridge

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    Which was, of course, what I was getting at :(
     
  18. Jul 27, 2014 #17

    ehild

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    You omitted the sines in front of the angles. Compare the horizontal and vertical force components, eliminate T1 and find T2 in terms of theta, find the place of minimum.

    ehild
     

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  19. Jul 27, 2014 #18

    utkarshakash

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    [itex]T_2 = \dfrac{W \sin 30}{\sin (30+\theta)}[/itex]

    If I differentiate this wrt θ, I get θ=60°. Is this answer correct?
     
  20. Jul 28, 2014 #19

    ehild

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    I got the same. You do not need to differentiate. T2 is minimum when the denominator is maximum, (as the numerator is constant) The maximum of the sine function is at angle 90°, so 30+θ=90.:smile:

    ehild
     
  21. Jul 28, 2014 #20

    utkarshakash

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    Thanks for your help!
     
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