Question 5.4: A Traffic Light at Rest A traffic light weighing 122 N hangs...

Conceptualize Inspect the drawing in Figure 5.11a. Let us assume the cables do not break and nothing is moving.

Categorize If nothing is moving, no part of the system is accelerating. We can now model the light as a particle in equilibrium on which the net force is zero. Similarly, the net force on the knot (Fig. 5.11c) is zero, so it is also modeled as a particle in equilibrium.

Analyze We construct a diagram of the forces acting on the traffic light, shown in Figure 5.11b, and a free-body diagram for the knot that holds the three cables together, shown in Figure 5.11c. This knot is a convenient object to choose because all the forces of interest act along lines passing through the knot.

From the particle in equilibrium model, apply Equation 5.8 for the traffic light in the y direction:

\sum \overrightarrow{F}=0     (5.8)

Choose the coordinate axes as shown in Figure 5.11c and resolve the forces acting on the knot into their components:

Apply the particle in equilibrium model to the knot:

(1) \sum F_x=-T_1 \cos \theta_1+T_2 \cos \theta_2=0

(2) \sum F_y=T_1 \sin \theta_1+T_2 \sin \theta_2+\left(-F_g\right)=0

Equation (1) shows that the horizontal components of \overrightarrow{T}_1 and \overrightarrow{T}_2 must be equal in magnitude, and Equation (2) shows that the sum of the vertical components of \overrightarrow{T}_1 and \overrightarrow{T}_2 must balance the downward force \overrightarrow{T}_3, which is equal in magnitude to the weight of the light.

Solve Equation (1) for T_2 in terms of T_1:

(3)  T_2=T_1\left(\frac{\cos \theta_1}{\cos \theta_2}\right)

Substitute this value for T_2 into Equation (2):

Solve for T_1:

Substitute numerical values:

Using Equation (3), evaluate T_2:

Both values are less than 100 N (just barely for T_2), so the cables will not break.

Finalize Let us finalize this problem by imagining a change in the system, as in the following What If?