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Homework Help: Vector problem: flagpole with wires at different angle

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Two supporting wires keep a flagpole in its
    vertical position. The horizontal components of the tensions in
    the supporting wires must be equal and opposite,
    otherwise the pole will move laterally. The
    vertical components may be different.
    The longer wire forms a 30 degree angle with the ground, and
    the shorter wire forms a 45 degree angle with the ground.
    a Calculate the ratio of the tension in the
    shorter wire to the tension in the longer wire.
    b If the tension in the shorter wire is 600 N, calculate the magnitude of the tension
    in the longer wire.

    (Refer to picture) in the word document.

    2. Relevant equations

    basic trigonometry:

    sine, cosine, tan,

    possibly cosine/sine rules

    3. The attempt at a solution

    I am not sure, what to do with this problem.
    I have attempted it, but haven't a clue
    how to do it correctly. Please help.
    I've just started this topic, and haven't been
    taught this yet.

    Attached Files:

    • Doc1.doc
      File size:
      23.5 KB
  2. jcsd
  3. Feb 2, 2009 #2
    Re: vectors

    Do you know how to resolve a vector into two component vectors? For example, if you are driving North East at speed V you have a North vector and an equal East vector. These two vectors would each be of size V*cos(45°). If you now turn a bit to the left so you are driving at 30° East of North then the North vector will now be V*cos(30°) and the East vector will be V*cos(60°). Does this help?
  4. Feb 2, 2009 #3
    Re: vectors

    I am having trouble understanding.

    So with my question, V would be the magnitude of the vector, and I would have:

    r.cos(30) , and r.cos(45) ?

    I don't really get it. Because I don't have a magnitude.
  5. Feb 2, 2009 #4
    Re: vectors

    You need tensions; just call them T1 and T2. They won't be equal. The horizontal components are equal and the vertical ones are different.
  6. Feb 2, 2009 #5
    Re: vectors

    Ahhh :)


    So let horizontal components equal x, and vertical components equal T1, T2

    =>cos (theta1) = x/T1
    T1 = x / cos (theta1)

    =>cos(theta2) = x/T2
    T2 = x / cos (theta1)

    Then lets say I choose an arbitrary value...2

    2/ cos(30) = 2.30

    2/ cos(45) = 2.82

    => 2.82 : 2.3 = 1: 1.2(approx) , so that's my ratio?

    Or I could have just done cos(30) : cos(45)
  7. Feb 2, 2009 #6
    Re: vectors

    You answer is incorrect I fear.
    That wasn't quite what I had in mind.

    Longer cable has tension T1.
    The horizontal component will then be T1*cos(60°)
    The vertical component will then be T1*cos(30°)
    Repeat this exercise for the shorter rope with tension T2 and equate the horizontal components.
    From here you can get the ratio of tensions.
  8. Feb 2, 2009 #7
    Re: vectors

    I did that and got:

    cos(30)(T1) = cos(45)(T2)

    and get the same: 1.224:1

    which is the same as what I get doing it the way you feared was incorrect.

    It ends up being the same, because I was equating the horizontal components
  9. Feb 2, 2009 #8
    Re: vectors

    My horizontal component is T1*cos(60°); yours is cos(30)(T1). Not the same!
  10. Feb 3, 2009 #9
    Re: vectors


    I broke the golden rule and didn't draw a diagram.
    Your solution was correct.
  11. Feb 3, 2009 #10
    Re: vectors

    Hey. No problems :). I appreciate your help nonetheless. I needed to resolve to resolve the components and you reminded me too, so Thankyou.
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