For that 15.0 kg traffic light, the wire at 30 degrees carries tension T1 and the wire at 45 degrees carries tension T2. What are their values?

1. Balancing the horizontal components gives the first tension times cos 30 equal to the second times cos 45. 2. That relation makes the second tension 1.225 times the first. 3. Balancing the vertical components gives the first tension times sin 30 plus the second times sin 45 equal to the weight. 4. Substitute the relation: first tension x 0.500 plus 1.225 x first tension x 0.707 equals the weight. 5. That collects to 1.366 times the first tension equals (15.0 kg)(9.80 m/s^2) = 147 N. 6. Divide: the first tension is 108 N. 7. Multiply by 1.225: the second tension is 132 N. 8. Splitting the 147 N weight equally between the wires ignores that only vertical components share the load. _Source: OpenStax College Physics (CC BY 4.0), Ch 4 "Dynamics: Force and Newton's Laws of Motion", section 4.7 Further Applications of Newton's Laws of Motion_