A 25-kg traffic light hangs midway on a cable between two poles 50 meters apart.If the sag in the cable is 0.40 meters,what is the tension in each side of the cable?
A) 120 N
B) 15,310 N
C) 7,660 N
D) 780 N
E) 30,630 N

Question

A 25-kg traffic light hangs midway on a cable between two poles 50 meters apart.If the sag in the cable is 0.40 meters,what is the tension in each side of the cable?
A) 120 N
B) 15,310 N
C) ​7,660 N
D) ​780 N
E) ​30,630 N

Quizplus · Accepted Answer

To solve this problem, we can use the formula for the tension in a cable:
Tension = weight / (2 * sin(theta))
where theta is the angle that the cable makes with the horizontal.

First, we need to find the angle that the cable makes with the horizontal. We can use the right triangle formed by the cable, the pole, and the horizontal distance between the two poles:

sin(theta) = sag / (distance between poles/2)
sin(theta) = 0.4 / 25
theta = 0.958 radians

Now we can use the tension formula to find the tension in each side of the cable:
Tension = (25 kg * 9.81 m/s^2) / (2 * sin(0.958 radians))
Tension = 7,660 N

Therefore, the answer is C, ​7,660 N.

A 25-Kg Traffic Light Hangs Midway on a Cable Between