# Tension on a clothes line

1. Jan 2, 2010

This question came up in my physics holiday homework and I can't seem to get the correct answer :(

1. The problem statement, all variables and given/known data
A 4.0 kg magpie lands in the middle of a perfectly horizontal plastic wire on a clothes line stretched between two poles 4.0 m apart. The magpie lands in the centre of the wire depressing it by a distance of 4.0 cm. What is the magnitude on the tension in the wire?

2. Relevant equations
N/A

3. The attempt at a solution
Created a triangle with angle theta at the centre of the clothes line with a hypotenuse of 2 m (as its half the clothes line) and the side opposite the angle being 0.04 m (the 4cm depression).
Solving for theta yields 1.146 degrees
Now I have a new triangle (same angle to the horizon though) with the opposite side equal to 40 to represent the upwards tension supporting the bird ( we use g=10...) and solve for the hypotenuse: 10/sin(1.146) = 2.0 x 10^3 N

However the answer in the book is 1000.
Thanks for any help!

2. Jan 2, 2010

### sArGe99

Let x denote the extension of the wire downwards and L the natural length of the rope. If theta is the angle b/w x and length of wire in deformed state,
you should get cos theta = x/(L/2) after a bit of approximation.ie. taking (L/2) outside the root from denominator.

2T cos(theta) = mg

T= mg*L/(4*x)

3. Jan 2, 2010

Okay using you notation I do get to cos theta = x/(L/2) = 2x/L
But I can't figure out where the 2 came from in 2T cos(theta) = mg, are we using cos(theta) = adjacent side/ hypotenuse i.e cos(theta) = mg/ tension?
I understand how you got from there to T=mg*L/(4*x) however.

4. Jan 2, 2010

### PhanthomJay

If you draw a free body diagram of the bird, you should note that there are 3 forces acting on it: Its weight acting down, the tension in the left side of the cable acting away from the bird up and to the left, and the tension in the right side of the cable acting away from the bird up and to the right. From the symmetry of the problem, you should note that the tension forces are equal. By summing forces in the y direction, 1/2 the weight must be carried by the vertical component of the left cable tension, and 1/2 the weight must be carried by the vertical component of the right cable tension. Note that you said
when you should have said '20' instead of '40'.

5. Jan 2, 2010