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Force question

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data

    A sailor tests the tension in a wire rope holding up a mast by pushing against the rope with his hand at a distance L from the lower end of the rope. When he exerts a 'transverse' push N, the rope suffers a transverse displacement s.

    Show that for s << L, the tension in the wire rope is given approximately by the formula:

    T = N L/s




    2. Relevant equations



    3. The attempt at a solution

    I looked up the word transverse and it meant to be perpendicular. So, then we have that the tension vector, and the displacement vector and another side make a right angle triangle. Then, T is the hypotenuse of the triangle displacement^2 + another side^2.. I know that if I can get a proper sine or cosine equation, I can use the opp/hyp identities to come to that solution, but I can't see it.. thanks
     
  2. jcsd
  3. Feb 8, 2010 #2

    rl.bhat

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    Since s << L, tanθ~ sinθ ~ θ = s/L.
    So F/T = sinθ = s/L
     
  4. Feb 8, 2010 #3
    hi, thank you very much, but how is sin theta = F/T? the displacement s would form a right angle triangle with L and where the rope used to be before it was displaced. the applied force could be of varying magnitudes so we can't say that it's the same as S?

    thanks
     
  5. Feb 9, 2010 #4
    or maybe when theta is really small, both triangles formed by N and T and L and S are approximately the same? thanks
     
  6. Feb 9, 2010 #5

    rl.bhat

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    Yes. In a triangle formed by the forces, sides of the triangles are proportional to the corresponding forces.
     
  7. Feb 9, 2010 #6
    thanks! that;s what I thought intuitively.. but how can I show this?

    thanks again
     
  8. Feb 9, 2010 #7

    rl.bhat

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    It is a theorem in the vectors. No need to prove it here.
     
  9. Feb 9, 2010 #8
    thanks, do you know where I can learn about this theorem? I had no idea about it.. and is it a theorem involving vectors in PHYSICS or vectors in general? it seems to rely on the physics behind this question. (that the force vectors will be proportional to the sides L and S) thanks again
     
  10. Feb 10, 2010 #9
    could anybody tell me about this please?
    thanks
     
  11. Feb 10, 2010 #10

    rl.bhat

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    Original position of the rope, transverse force and the tension in the rope firm a right angled triangle. In that
    F/T = sinθ.
    When s <<< L, sinθ = θ. And θ = s/L.
    So F/T = s/L.
     
  12. Feb 10, 2010 #11
    thanks, but how do you know that F/T = sin theta?? what if the applied force vector is longer than the tension, then it would have to be scaled..
     
  13. Feb 10, 2010 #12

    rl.bhat

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    When you apply transverse force in the horizontal direction, the vertical rope is shifted sideways having a tension T. Draw the diagram and find sinθ.
     
  14. Feb 10, 2010 #13
    yes, I understand, but how do we know that these vectors (the applied force and the tension) will make a right angle triangle with the rope position before the force was applied?

    For example, the tension vector might be "too long", and only part of this vector will be included in the triangle.. I hope you understand what I'm asking.. thanks
     
  15. Feb 10, 2010 #14

    rl.bhat

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    Can you draw a figure as described in the problem?
    I presume the rope is straight before applying the transverse force. Even if the tension vector is "too long", the tension is same through out the rope.
     
  16. Feb 10, 2010 #15
    hi,
    72zj1u.jpg
    I also drew in what I understand from this problem.. there's the triangle, and the lengths in red should form a right angle triangle.
    the broken line is where it is before being pushed. thanks

    I was asking about how we knew that N, T and the broken red line part formed a "proper" right angle triangle? what if a right angle triangle is only formed with N, the red broken line part and a part of T (3/4? 1/2?. et c), then sin theta wouldn't be N/T
     
  17. Feb 10, 2010 #16

    rl.bhat

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    If N is the force of push, the vector N should be from broken line to full line. The broken line, N and T form a right angled triangle. Angle between the N and the broken line is 90 degrees.
     
  18. Feb 10, 2010 #17
    hmm, that's what I thought when I first saw it.. but how do I know the length of N is s and the length of T is L? Isn't it possible that you would need more force to move displacement s than a magnitude of s? same with tension..

    thanks
     
  19. Feb 10, 2010 #18

    rl.bhat

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    Since two ends of the ropes are fixed, the change in the length of the rope is very minimum. When you push more, θ increases and hence s increases. T also increases, but L does not change. There must be a braking point for the rope. So you have restrict your N.
     
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