A traffic light of mass $m$ is suspended by two chains of equal length that make an angle $\theta$ with the horizontal. If the system is in equilibrium, what is the angle $\theta$ of the chains with respect to the horizontal if the tension in each chain is $T$?

• A. $\theta ={\sin }^{-1;}\frac{2T}{m}$
• B. $\theta ={\sin }^{-1;}\frac{T}{2m}$
• C. $\theta ={\sin }^{-1;}\frac{m}{2}∕T$
• D. $\theta ={\sin }^{-1;}\frac{m}{2T}$

Answer: (D)

Verified step by step guidance

1. Draw a free-body diagram of the traffic light suspended by two chains. Label the tension in each chain as \(T\), the mass of the traffic light as \(m\), and the angle each chain makes with the horizontal as \(\theta\).
2. Identify the forces acting on the traffic light: the gravitational force downward, \(mg\), where \(g\) is the acceleration due to gravity, and the tension forces \(T\) in the two chains acting along the chains at angle \(\theta\).
3. Since the system is in equilibrium, the net force in both the horizontal and vertical directions must be zero. Write the equilibrium conditions for the vertical and horizontal components of the forces.
4. For the vertical direction, the sum of the vertical components of the tensions must balance the weight: \$2T \sin(\theta) = mg\(. For the horizontal direction, the horizontal components of the tensions cancel out: \)T \cos(\theta)\( from one chain is balanced by \)T \cos(\theta)$ from the other.
5. Solve the vertical equilibrium equation for \(\theta\): rearrange to get \(\sin(\theta) = \frac{mg}{2T}\), then take the inverse sine to find \(\theta = \sin^{-1}\left(\frac{mg}{2T}\right)\).