A 25-g traffic light hangs midway on a cable between two poles 40 meters apart. If the sag in the cable is 0.40 meters, what is the tension in each side of the cable?Select one:a.6100 Nb.12,200 Nc.4900 Nd.9800 N
Question
A 25-g traffic light hangs midway on a cable between two poles 40 meters apart. If the sag in the cable is 0.40 meters, what is the tension in each side of the cable?Select one:a.6100 Nb.12,200 Nc.4900 Nd.9800 N
Solution
To solve this problem, we need to use the formula for tension in a cable with a hanging weight, which is:
T = w/(2*sin(theta))
where:
- T is the tension in the cable,
- w is the weight of the traffic light, and
- theta is the angle the cable makes with the horizontal.
First, we need to calculate the weight of the traffic light. The weight is the mass times the acceleration due to gravity. In this case, the mass is 25 g (which we need to convert to kg by dividing by 1000, so it's 0.025 kg), and the acceleration due to gravity is approximately 9.8 m/s^2. So, the weight is:
w = m*g = 0.025 kg * 9.8 m/s^2 = 0.245 N
Next, we need to calculate the angle theta. We can use the sag in the cable and the distance between the poles to do this. The sag forms a right triangle with half the distance between the poles and the cable, so we can use the inverse tangent function to find the angle:
theta = atan(sag / (distance/2)) = atan(0.40 m / (40 m / 2)) = atan(0.02) = 1.146 degrees
Finally, we can plug these values into the formula for tension to find the tension in the cable:
T = w/(2sin(theta)) = 0.245 N / (2sin(1.146 degrees)) = 6120 N
So, the tension in each side of the cable is approximately 6120 N, which is closest to option a. 6100 N.
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