# Finding tension in a wire in our static equilibrium unit

1. Dec 6, 2011

### MeckWarrior

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
$$\tau=F(r\sin{\theta})$$
$$\tau=F\times r\bot$$
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⊥)
|
|

$$4.9N \times 20m = T \times 2m$$
$$\frac{4.9N \times 20m}{2m} = T$$
$$49N = T$$

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. Dec 6, 2011

### Spinnor

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.