Question

2) A traffic light of weight 100 N is supported by two ropes as shown. Let T1 and T2 are the tensions.

Answer 1

Hi there!

a.

We know that:

$\Sigma F_y = 0 \\\\\Sigma F_x = 0$

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

$T_1 = T_2 = \frac{W}{2sin\theta} = \frac{100}{2sin(37)} = \boxed{83.08 N}$