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. Draw a free-body diagram of the bird. How much tension does the bird produce in the wire? Ignore the weight of the wire.
The correct answer and explanation is:
To draw a free-body diagram of the bird, we should focus on the forces acting on the bird.
Free-Body Diagram of the Bird
In the free-body diagram, the bird experiences two forces:
- Weight (W): This is the force due to gravity acting downward on the bird. It is equal to the bird’s mass (1.00 kg) multiplied by the acceleration due to gravity (9.8 m/s²). So, the weight of the bird is: W=mg=1.00 kg×9.8 m/s2=9.8 NW = mg = 1.00 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N}
- Tension (T): The wire exerts an upward tension force on the bird. Since the wire sags, the tension force in the wire has two components. There is a vertical component that balances the weight of the bird and a horizontal component that balances out the forces along the wire. The tension in the wire at both ends can be considered equal due to symmetry.
Calculation of Tension in the Wire
Given that the wire sags 0.200 m and the distance between the poles is 50.0 m, the bird is placed 25.0 m away from each pole (half of the total distance). The sag creates a triangle, with the vertical distance being 0.200 m, the horizontal distance being 25.0 m, and the wire forming the hypotenuse.
We can use the geometry of the situation to determine the tension in the wire.
Step 1: Find the angle of the wire
The angle θ\theta of the wire with respect to the horizontal can be found using the tangent of the angle: tan(θ)=vertical displacementhorizontal distance=0.200 m25.0 m=0.008\tan(\theta) = \frac{\text{vertical displacement}}{\text{horizontal distance}} = \frac{0.200 \, \text{m}}{25.0 \, \text{m}} = 0.008
So, θ=tan−1(0.008)≈0.46∘\theta = \tan^{-1}(0.008) \approx 0.46^\circ
Step 2: Resolve the Tension
Since the tension in the wire has both vertical and horizontal components, we can resolve the tension TT into its components:
- The vertical component of the tension TvT_v must balance the bird’s weight: Tv=Tsin(θ)T_v = T \sin(\theta) Since the weight of the bird is 9.8 N, we can set up the equation: Tv=9.8 NT_v = 9.8 \, \text{N} Therefore, using Tv=Tsin(θ)T_v = T \sin(\theta): Tsin(0.46∘)=9.8 NT \sin(0.46^\circ) = 9.8 \, \text{N} Solving for TT: T=9.8 Nsin(0.46∘)≈9.80.008=1225 NT = \frac{9.8 \, \text{N}}{\sin(0.46^\circ)} \approx \frac{9.8}{0.008} = 1225 \, \text{N}
Conclusion
The tension in the wire, due to the bird’s weight, is approximately 1225 N. This force is distributed between the two sides of the wire, and the wire forms a slight angle due to the sag caused by the bird’s weight.