The distance between two telephone poles is 49 m. When a 1.6 kg bird lands on the telephone wire midway between the poles, the wire sags 0.149 m. The acceleration of gravity is 9.8 m/s2 . How much tension in the wire does the bird produce? Ignore the weight of the wire. Answer in units of N
613
To find the tension in the wire produced by the bird, we can use the concept of the sagging of a wire due to the weight applied to it.
Step 1: Determine the amount of sagging caused by the weight of the bird.
Given:
Distance between two poles (L) = 49 m
Sagging due to the bird's weight (s) = 0.149 m
Since the bird lands midway between the poles, the distance from the pole to the bird (x) would be half of the distance between the poles:
x = L/2
x = 49/2
x = 24.5 m
The sagging can be calculated using the equation for the sag of a wire under its own weight:
s = (w * L^2) / (8 * T), where:
s = sag of the wire
w = weight per unit length of the wire
L = distance between the poles
T = tension in the wire
Since we need to find the tension, we can rearrange the equation as:
T = (w * L^2) / (8 * s)
Step 2: Calculate the tension in the wire.
Given:
Mass of the bird (m) = 1.6 kg
Acceleration due to gravity (g) = 9.8 m/s^2
The weight per unit length of the wire (w) can be calculated by multiplying the mass of the bird with the acceleration due to gravity:
w = m * g
w = 1.6 kg * 9.8 m/s^2
w = 15.68 N/m
Now, we can substitute the values into the equation for tension:
T = (w * L^2) / (8 * s)
T = (15.68 N/m * (49 m)^2) / (8 * 0.149 m)
T = (15.68 N/m * 2401 m^2) / (8 * 0.149 m)
T = (37592.48 N m) / (1.192 N m)
T ≈ 31525.69 N
Therefore, the tension in the wire produced by the bird is approximately 31525.69 N.
To find the tension in the wire produced by the bird, we can use the concept of equilibrium.
First, let's consider the forces acting on the bird when it lands on the wire. There are two main forces: the tension force in the wire and the gravitational force acting downwards.
The gravitational force can be calculated using the formula F = mg, where m is the mass of the bird (1.6 kg) and g is the acceleration due to gravity (9.8 m/s^2). Therefore, the gravitational force is:
F_gravity = (1.6 kg)(9.8 m/s^2) = 15.68 N
Since the bird is in equilibrium, the tension force in the wire (T) must be equal and opposite to the gravitational force. Therefore, T = F_gravity.
However, because the wire sags when the bird lands, some of the tension force is also being used to counteract the sagging of the wire. This can be calculated using the concept of a simple harmonic oscillator.
The sagging of the wire is given as 0.149 m, which is the displacement from the equilibrium position. The distance between the poles is 49 m, so the total length of the wire is 49 m + 49 m = 98 m.
The sag of the wire is proportional to the tension force, as given by the formula:
sag = (T/(2L)) * (L^2)/g
Where sag is the sagging of the wire, T is the tension force, L is the distance between the poles, and g is the acceleration due to gravity.
Rearranging this formula to solve for T:
T = (2L^3 * sag) / (L^2 * g)
Plugging in the given values:
T = (2 * (49 m)^3 * 0.149 m) / ((49 m)^2 * 9.8 m/s^2)
T = (2 * 49^3 * 0.149) / (49^2 * 9.8)
T = 0.616 N
Therefore, the tension in the wire produced by the bird is approximately 0.616 N.
weight = 1.6 * 9.8 = 15.68 N
T in each side of wire
do right half, left the same
1/2 weight = 7.84 N
half distance = 24.5
theta angle between vertical and wire
Tan theta = 24.5/.149
theta = 89.65155
so cos theta = .00608152
T cos Theta = half the weight
T = 7.84 N/.00608152 = 1290 N