# Tough problem on tension

1. Jul 11, 2006

### webren

"The distance between two telephone poles is 50.0 m. When a 1.00-kg bird lands on the telephone wire midway between the poles, the wire sags 0.200 m. How much tension does the bird produce in the wire? Ignore the weight of the wire."

Below are my steps in solving the problem. My problem is my answer does not match the book's and the book's answer doesn't quite make sense. Here is my solution:

In solving this problem, I drew a free body diagram and realized there were two tensions: left and right of the bird. I assumed both tensions are equal, because the problem states the bid sat "midway" on the wire.

I made a right triangle, with my hypotenuse being the tension, my opposite side (or y component) being .2 m, and my adjacent side (or x component) 25 m. I used the inverse tangent to find what the angle would be, which came up to be 0.46 degrees. Intuitively, that makes sense, because the weight of the bird can only be so much, and the weight wouldn't cause a large angle.

Next, I stated that the sum of forces equals T1 + T2 - W = 0 (because of no accleration - the bird isn't moving).

The X force component is T2Cos 0.46 degrees.
The y force component is T1Sin 0.46 degrees.

My final equations comes to be T2(1) + T2(0.008) = W
I simplified it to be T2(1.008) = W. Because W = mg, I made the equation T2(1.008) = (1.0)(9.80). In solving for the tension, I get 9.72 N. Because there are two tensions and because the tensions are equal, I multiplied 9.72 by 2 and got 19.44 N as my final answer.

The book's answer is 613 N. That seems extremely heavy for a bird. Is the book correct, am I, or neither?

Thanks for your time.

2. Jul 11, 2006

### Cyrus

First, nice post. You wrote down your thought process, and it is correct. Good job.

However, I am afraid your book is correct.

The problem is that you should have said:

$$\sum F_y: 2T_2sin( \theta) = W$$

The equation above should clear it up for you, it not ask. Look over your Free body a second time and you will see why.

EDIT: Sorry that should be a sine not cosine.

Last edited: Jul 11, 2006
3. Jul 11, 2006

### webren

Thanks for the quick response.

Was my error in defining a y component for the force? Oh, and in my y component, I made it T2Sin(0.46)degrees.

4. Jul 11, 2006

### Cyrus

Well, I do not know how you got this:

T2(1) + T2(0.008) = W

T2(1) - I do not see where the (1) came from.

No. It is exactly the two components of tension in the y-direction, (one for each segment of wire) that counters the weight of the bird in the downward direction.

You should have said:

$$\sum F_x: T_1 cos(\theta) - T_2 cos(\theta) = 0$$

From there, $$T_1 = T_2$$

Now, you can solve:

$$\sum F_y: T_1 sin( \theta) + T_2 sin( \theta) - mg = 0$$

Last edited: Jul 11, 2006
5. Jul 11, 2006

### webren

I got the 1 from cos(0.46) being 0.999.

6. Jul 11, 2006

### Cyrus

Ah, I see.

No, that is correct cos(0.46)~=1; but, that is not necessary for the problem.

Does my post above your last make sense to you?

7. Jul 11, 2006

### webren

Yes, your post cleared everything up. I've realized and corrected my mistakes. Thank you.

8. Jul 12, 2006

### Cyrus

No problem.

9. Jul 12, 2006

### BobG

The book's answer is correct, but the bird still only weighs 9.8 N.

This is an old trick for pulling your Jeep out of the mud. Even with a mechanical winch, you can wrap one end of the cable around a rock or tree and hand crank the winch until the cable is taut. Pushing or pulling the cable perpendicular to the length of the cable allows you to apply a huge amount of force to pulling your Jeep out of the mud.

Of course you're not actually moving it very far each time and have to keep recranking the winch, but ....