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?

To solve this problem, we can divide the cable into two segments: one between the left pole and the traffic light, and the other between the right pole and the traffic light. Since the cable sags in the middle, we can ignore the weight of the traffic light when calculating the tension.

Let's denote the tension in each side of the cable as T. Using the formula for the sag of a cable under tension:

s = (L/2) + (w*L^2) / (8*T)

where:
s = sag in the cable = 0.40 meters
L = distance between the poles = 40 meters
w = weight per unit length of the cable
T = tension in the cable

Since the weight of the traffic light is negligible compared to the weight of the cable, we can ignore it and set w = 0. Then, the formula simplifies to:

0.40 = (40/2) / (2T)
0.40 = 20 / (2T)
0.40 = 10 / T
T = 10 / 0.40
T = 25 N

Therefore, the tension in each side of the cable is 25 N.