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Another Simple Rope Tensions Question

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A 4.0kg magpie (bird) flies towards a very tight plastic wire on a clothes line. The wire is perfectly horizontal and is stretched between poles 4.0m apart. The magpie lands on the centre of the wire, depressing it by a vertical distance of 4.0cm. What is the magnitude of the tension in the wire?

    2. Relevant equations
    W=mg
    trignometric functions
    W(x) = Wsin(theta)


    3. The attempt at a solution
    This is off a textbook question, and their answer was 1000N. I got 0.8N =='

    I had the following steps:

    - tan(θ) = 4/200
    - θ = 1.15 degrees.
    - W(x) = mgsin(1.15)
    (where W(x) was the required component of weight, m = 4.0kg and g = 9.8ms^-2)
    - W(x) = 4.0kg x 9.8 x sin(1.15) = 0.8N

    What am i doing wrong? :(
    Any help is appreciated.
     
  2. jcsd
  3. Jan 10, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper

    Hi amnestic, welcome to PF.
    You have to resolve the tensions into two components.
    And mg = 2*T*cosθ, where θ = (90 - 1.15) degrees.
     
  4. Jan 10, 2010 #3
    Thanks for the reply rl.blat,
    two tensions?
    I had a triangle, with the opposite as 2m and adjacent as 0.04m

    Then I found the angle, which allowed me to find the hypothenuse (which is the "line" of force that I'm after)

    I'm not sure what you mean by "mg = 2*T*cosθ" isn't mg = 4.0 x 9.8?

    PS: I also changed the angel within the triangle for the working out of this post, as compared to my first post. so its a bit different now.
     
  5. Jan 10, 2010 #4

    rl.bhat

    User Avatar
    Homework Helper

    After depression of the wire due to landing of the bird at the mid point, two segments of the wire are stretched producing tension in the segments. In the equilibrium condition, the weight of bird is balanced by the vertical components of the tensions.
    So "mg = 2*T*cosθ".
     
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