Hi there!
a.
We know that:
[tex]\Sigma F_y = 0 \\\\\Sigma F_x = 0[/tex]
Begin by determining the forces in the vertical direction:
W = weight of traffic light
T₁sinθ = vertical component of T₁
T₂sinθ = vertical component of T₂
b.
The ropes provide a horizontal force:
T₁cosθ = Horizontal component of T1
T₂cosθ = Horizontal component of T2
Thus:
0 = T₁cosθ - T₂cosθ
T₁cosθ = T₂cosθ
T₁ = T₂
c.
Since the angles for both ropes are the same, we can say that:
T₁ = T₂
Sum the forces:
ΣFy = T₁sinθ + T₁sinθ - W = 0
2T₁sinθ = W
d.
Now, we can begin by solving for the tensions:
2T₁sinθ = W
[tex]T_1 = T_2 = \frac{W}{2sin\theta} = \frac{100}{2sin(37)} = \boxed{83.08 N}[/tex]
