### Traffic Light Hanging from Horizontal Pole

A traffic light is hanging from a horizontal pole as shown in the figure. The two metallic cables holding the light can be considered massless. The total mass of the traffic light feature is 100 Kg. What is the tension in each of the two cables?
The tension on the cable at 600 with the horizontal
 a) 5432 N b) 1516 N N c) 760 N d) 9800 N e) None of the above.

and the tension on the cable at 400 with the horizontal

 a) 5432 N N b) 494 N c) 2333 N d) 9800 N e) None of the above.

Solution:

This problem involves only one object, the traffic light. However, there are two unknown in this problem, the tension in each of the cables. To solve for the two unknown, two equations are needed it. This two equations will be derived by using Newton's second law by components.

Over this object, there are three forces acting, the weight of the traffic light, mg, the tension of the cable at 400 with the horizontal, T1, and the tension of the cable at 600 with the horizontal, T2, . Notice that the weight of the traffic light is along the vertical axis. On the other side, following step 4 for solving Newton's applications, the tensions must be broken down into their corresponding components along the horizontal and vertical axis. The second drawing of the sequence shows the vertical and horizontal components of the tension T1. These two components are

The third drawing of the sequence adds the breaking down into its components for the tension T2, resulting in the relations

The fourth drawing of the sequence shows the FBD for the traffic light with the vectors decomposed into their components along the x-axis and the y-axis.

 Horizontal Component (x-axis) From the FBD, the two forces acting in the horizontal direction are the horizontal component of the tension T1 pulling the traffic light to the right and the horizontal component of the tension T2 pulling the traffic light to the left. The resulting net force is   The traffic light is supposed to be static, no acceleration in either direction. Thus, Newton's second law applied to the horizontal component is Putting together these two results, the following equation is derived:       Equation 1 Vertical Component (y-axis) From the FBD, the three forces acting in the vertical direction are the vertical components of the tensions T1 and T2 pulling the traffic light up; and, the weight of the traffic light pulling the light down. The resulting net force in this direction is Again, the traffic light is not accelerating in this direction which implies that Newton's second law applied to this component is Combining these two results,      Equation 2 Equations 1 and 2 can be used to solved for the two unknown. In this case, a direct method for solving these equations is to solve for T1 in equation 1, and substitute it in equation 2. After T2 = 760 N has been found, this value can be substituted back in order to solve for T1 = 494 N. The applet on the left shows the different algebraic steps necessary for reaching to the solutions.

 by Luis F. Sáez, Ph. D. Comments and Suggestions: LSaez@dallaswinwin.com