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Finding tension in a wire in our static equilibrium unit

  1. Dec 6, 2011 #1
    This is the final question on a static equilibrium quiz i had today (physics 12). when comparing answers (as you tend to do after the quiz "did you get x for question y?"), half of my friends (all of equal ability in the course) got a different answer from me, and the other half the same. I am wondering if i have done this correctly.

    1. The problem statement, all variables and given/known data
    a wire is suspended by 2 telephone poles 40m apart. a 0.5kg bird lands directly at the midpoint of the wire, causing it to sag 2m below the original height. what is the tension in the wire.
    bird mass = 0.5kg
    wire length = 40m
    wire displacement at center = 2m

    2. Relevant equations
    [tex]\tau=F(r\sin{\theta})[/tex]
    [tex]\tau=F\times r\bot[/tex]
    3. The attempt at a solution
    I figured I wouldn't need θ, because I already have the leaver arm (r⊥). calculating both of these (x-axis as r⊥ and y-axis as F, and vice-versa respectively) should each yield a value with the units in m∙N (torque). Am I correct in assuming these torques will be equal (or opposite?)

    Right half of wire freebody diagrams (only components pictured, no hypotenuse drawn):
    ___________20m___________ (distance : r⊥)
    |
    |
    | 0.5kg∙9.8m/s² = 4.9N (force : F)
    |
    |

    ____________T____________ (force : F)
    |
    |
    | 2m (distance : r⊥)
    |
    |


    [tex]4.9N \times 20m = T \times 2m[/tex]
    [tex]\frac{4.9N \times 20m}{2m} = T[/tex]
    [tex]49N = T[/tex]


    others in my class insist that i should have divided 49N by 2, but wouldn't that be finding the tension in just half of the wire?
    or have i made assumptions that just don't work (using torque to solve the question to begin with, which is where i suspect i may have went wrong) elsewhere in my work?

    thank you very much for your help.
     
  2. jcsd
  3. Dec 6, 2011 #2
    I was stumped at first, but now realize you assume the wire is horizontal to begin with when its shape is that of a catenary, see,

    http://en.wikipedia.org/wiki/Catenary

    but if we simplify the problem and assume the wire is horizontal then,

    2Tsin(theta) = mg

    where theta = inverse tangent(.1)

    I think your answer is off by a factor of two. Draw a free body diagram of the situation.
     
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