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Minumum rope length

  1. Apr 10, 2008 #1
    The rope connected to the triangle is pulling away at 3000N
    i need to determine the minimum length of ac so that the tension in either ab or ac does not exceed 4000N. AC and BC are the same length.
  2. jcsd
  3. Apr 10, 2008 #2
    can someone please just tell me the equation that relates rope length to tension.
  4. Apr 10, 2008 #3


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    Hi Ry122! :smile:

    I don't understand … your diagram says BC = 5m, and angles BCA and CBA are 30º.

    So AC is fixed, isn't it?

    What am I missing? :confused:
  5. Apr 10, 2008 #4
    oops, I made a mistake. angle CBA is unknown and length BA is different from AC
  6. Apr 10, 2008 #5
    Is the tension in the rope pulling away (3000N) equal to the tension in BC? if so ill know how to answer the question
  7. Apr 10, 2008 #6


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    No, certainly not.

    You have to use Newton's second law: sum of the forces in any particular direction is zero.

    But I still don't understand what stops the triangle from rotating. :confused:
  8. Apr 10, 2008 #7
    a large mass is attached to the triangle at BC.
  9. Apr 10, 2008 #8


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    Sorry … this is no good.

    Will you please type out the whole original question?
  10. Apr 10, 2008 #9
    The tension in the tow rope pulling the car in Newtons is 3000N. Determine the minumum length of the rope l, between A and B, so that the tension in either AB or AC does not exceed 4000N
  11. Apr 10, 2008 #10


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    So where does the 30º come into it? Or is that gone too?

    Is angle CAB unknown as well as angle CBA?
  12. Apr 10, 2008 #11
    angle CBA = theta
    angle BCA = 30 degrees
    BA = l
    BC = 1.2m
  13. Sep 15, 2008 #12
    can anyone help me with this?
  14. Sep 15, 2008 #13

    Attached Files:

    Last edited: Sep 15, 2008
  15. Sep 15, 2008 #14


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    You need to show your attempt first.

    HINT: Start by breaking the forces down into x and y components, then solve the equations for the desired unknown. Initially, the desired unknown will be theta. Once you have theta, you can use a little trig to find the length, L, which is what you truly desire.

  16. Sep 15, 2008 #15
    I get:
    Sum of Fx
    0 = Fab cos (theta) + Fac cos (30)

    Sum of Fy
    0 = Fab sin (theta) + Fac cos (30) - 3000N

    Would Fab and Fac equal max. tension (4000N)? otherwise, I'm completely stuck
  17. Sep 16, 2008 #16


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    Check your FBD again, your component equations do not look correct. Once you solve the x-component equation you will be able to substitute that result into the y-component equation, which will leave only one unknown, theta (along with the assumption that one of the lines is at the max allowable tension of 4000N).

  18. Sep 16, 2008 #17
    I've been staring at those equations for quite a while now...
    all i can see is that:

    In the x-axis:
    Fab cos(theta) = -Fac cos(30)

    and in the y-axis:
    3000N = Fab sin(theta) + Fac sin(30)

    I know through the sin rule that:

    sin (theta)/ Fac = sin (30)/Fab

    but I don't know how the unknown length, l, can fit in
  19. Sep 16, 2008 #18


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    I’ll get you started on your x-components:

    [tex] F_{ac}cos(30) - F_{ab}cos(\theta) = 0 [/tex]

    Do you see why this is so? If you do, then try the y-components and lets see what you get.

  20. Sep 16, 2008 #19
    ah true
    that's because they have opposite directions right?

    for the y-components:
    [tex] F_{ac}sin(30) + F_{ab}sin(\theta) + 3000 = 0 [/tex]

    is that correct?
  21. Sep 16, 2008 #20


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    Just remember that it is the x-components that are pointing in different directions.

    Not quite. The same applies for the vertical axis. In what direction is T shown in the problem? Now what about the y-component of the force (tension) Fab and Fac?

    HINT: Draw a right triangle and resolve the resultant (Fab and Fac) into components to find the direction (remember vectors are drawn tip to tail).

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