"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.