Answer to Question #240471 in Mechanics | Relativity for Anas

Question

The distance between two telephonepoles 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. (a) Draw a free-body diagram of the bird. (b) How much tension does the bird produce in the

wire? Ignore the weight of the wire.

Answer 1

We use Newton’s second law with the forces in the x and y directions in equilibrium.

(a) At the point where the bird is perched, the wire’s midpoint, the forces acting on the wire are the tension forces and the force of gravity acting on the bird. These forces are shown below.

(b) The mass of the bird is m=1.00kg, so the force of gravity on the bird, its weight, is mg=(1.00kg)(9.80m/s²)=9.80N. To calculate the angle α in the free-body diagram, we note that the base of the triangle is 25.0m, so that

    

tanα=0.200m/25.0m →α=0.458⁰

Each of the tension forces has x and y components given by

Tx​=Tcosα and Ty​=Tsinα

The x components of the two tension forces cancel out. In the y direction,

∑Fy=2Tsinα−mg=0

which gives

T=mg/2sinα=9.80N/2sin0.458⁰=613N

T=613N