A 20kg traffic light hangs midway on a cable between two poles 40meters apart. If the sag in the cable is .4 meters, what is the tension in each side of the cable?
Consider the tension in the sag part as T, then the horizontal component is T*sinTheta, and the vertical component is T cos Theta, where theta is the angle between the vertical and the wire at the center.
Now, because there are two sides, each holding half the weight, then
T cos Theta=10kg*9.8N/kg*cosTheta
but looking at the dimensions, cosTheta is .4/20
So, you can solve for tension T.
4900N
To find the tension in each side of the cable, we need to use the concept of balancing vertical forces.
Given:
Mass of the traffic light (m) = 20 kg
Distance between two poles (L) = 40 meters
Sag in the cable (s) = 0.4 meters
First, let's calculate the weight of the traffic light.
Weight (W) = mass (m) × gravitational acceleration (g)
Taking gravitational acceleration as 9.8 m/s^2:
W = 20 kg × 9.8 m/s^2
W = 196 N
Now, let's draw a free body diagram of the traffic light hanging in the center of the cable.
```
|
|
|
Tension 1 === Tension 2
(left side) (right side)
|
|Traffic Light (W)
|
```
As the traffic light is in equilibrium, the vertical forces on both sides of the cable must balance each other.
Considering the left side of the cable:
Tension 1 (left side) + Weight (W) = 0 (net vertical force is zero)
Similarly, considering the right side of the cable:
Tension 2 (right side) + Weight (W) = 0 (net vertical force is zero)
Since the weight (W) is acting downwards, the tension forces are acting upwards.
Now, let's calculate the tension in each side of the cable.
Tension 1 (left side) = Weight (W) = 196 N
Tension 2 (right side) = Weight (W) = 196 N
Therefore, the tension in each side of the cable is 196 N.
To find the tension in each side of the cable, you can use the formula for the tension in a hanging cable:
T = w/2 + √((L^2/4) - (s^2))
Where:
T = tension in each side of the cable
w = weight of the traffic light
L = distance between the poles
s = sag in the cable
Let's substitute the given values into the formula:
w = 20 kg (weight of the traffic light)
L = 40 meters (distance between the poles)
s = 0.4 meters (sag in the cable)
Now we can calculate the tension in each side of the cable:
T = (20 kg) / 2 + √((40^2/4) - (0.4^2))
T = 10 kg + √((1600/4) - (0.16))
T = 10 kg + √(400 - 0.16)
T ≈ 10 kg + √399.84
T ≈ 10 kg + 19.996
T ≈ 29.996 kg
So, the tension in each side of the cable is approximately 29.996 kg (or approximately 300 N).