The distance between two telephone poles is
46 m. When a 1.4 kg bird lands on the
telephone wire midway between the poles, the
wire sags 0.151 m.
The acceleration of gravity is 9.8 m/s
2
.
How much tension in the wire does the bird
produce? Ignore the weight of the wire.
Answer in units of N
To find the tension in the wire produced by the bird, we can use the concept of equilibrium. When the bird lands on the wire, it creates a downward force due to its weight. This force causes the wire to sag. The tension in the wire must balance out this downward force, resulting in equilibrium.
Since the bird lands midway between the two poles, it creates equal angles on either side. Therefore, we can consider half of the wire's length between the bird and one of the poles.
First, let's calculate the weight of the bird:
Weight = mass x gravity
Weight = 1.4 kg × 9.8 m/s^2 = 13.72 N
Now, let's consider the equilibrium of the forces acting on the segment of the wire between the bird and one of the poles. We need to find the tension in the wire.
Considering each side of the wire segment, there are three key forces acting:
1. The tension in the wire pulling upwards.
2. The weight of the bird pulling downwards.
3. The vertical component of the forces from the two wire segments on either side of the bird supporting the sag.
Since the wire is in equilibrium, the sum of these three forces in the vertical direction should equal zero.
Tension + Tension - Weight = 0
2 * Tension - Weight = 0
Now, we can solve for the tension in the wire:
2 * Tension = Weight
Tension = Weight / 2
Tension = 13.72 N / 2
Tension = 6.86 N
Therefore, the tension in the wire produced by the bird is approximately 6.86 N.