The distance between two telephone poles is 50.0 m. When a 0.500-kg bird lands on the telephone wire midway between the poles, the wire sags 0.14 m. Ignore the weight of the wire.
1)How much tension does the bird produce in the wire? (Express your answer to two significant figures.)
To find the tension produced by the bird in the wire, 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 is balanced by the tension in the wire, which pulls in the opposite direction to keep the wire from sagging too much.
First, let's find the total weight of the bird. The weight is given by the formula:
Weight = mass × acceleration due to gravity
Given that the mass of the bird is 0.500 kg and the acceleration due to gravity is approximately 9.8 m/s², we can calculate the weight:
Weight = 0.500 kg × 9.8 m/s² = 4.90 N
Since the bird is located midway between the poles, the sag in the wire forms a right triangle with half the distance between the poles (25.0 m) as the base and the sag distance (0.14 m) as the height. We can use the Pythagorean theorem to calculate the length of the wire:
Length of the wire = √(base² + height²)
= √(25.0 m)² + (0.14 m)²)
≈ √(625 m² + 0.0196 m²)
≈ √(625.0196 m²)
≈ 25.0 m
Now that we have the length of the wire, we can proceed to find the tension produced by the bird. The tension in the wire is the same throughout, so we can use the equation for the tension in a hanging object:
Tension = Weight / Length of the wire
Tension = 4.90 N / 25.0 m
Carrying out the division, we find:
Tension ≈ 0.196 N
Therefore, the tension produced by the bird in the wire is approximately 0.196 N.