Question

The distance between two telephone poles is 40.0 m. When a 1.00-kg bird lands on the telephone wire midway between the poles, the wire sags 0.220 m. How much tension does the bird produce in the wire?

Answer 1

Answer:

$$F = \frac {m * g}{\sin (\theta)}$$ $$t g (\theta) = \frac {0 . 2 2 m}{2 0 m} = 0. 0 1 1$$ $$\theta = 0. 6 3 0 2$$ $$F = \frac {1 k g * 9 . 8 \frac {m}{S ^ {2}}}{\sin (0 . 6 3 0 2)} = 1 6. 6 3 N$$