Answer:
the tension in each side of the cable is 3677.57 N
Explanation:
given data
traffic light = 20 kg
cable between two poles = 30 m
sag in the cable = 0.40 m
solution
by the free body diagram
tan θ = [tex]\frac{0.4}{15}[/tex] .............1
θ = 1.527 °
and
tension = mg
The net force is along x - axis is express as
T2 cosθ = T1 cosθ .................2
so T2 - T1 ..............3
and
when we take it along y axis that is express as
( T1 + T2) sinθ = mg ...................4
so by equation 3 we put here
2 × T1 sin(1.527) = 20 × 9.8
T1 = 3677.57 N
The tension in each side of the cable is 3677.57 N.
This is defined as the pulling force transmitted axially by the means of a cable or similar object.
mass of traffic light = 20 kg
distance of cable between two poles = 30 m
sag in the cable = 0.40 m
We can infer that:
tan θ = 0.4/15
θ = 1.527 °
Tension = mg
T2 cosθ = T1 cosθ
( T1 + T2) sinθ = mg
2 × T1 sin(1.527) = 20 × 9.8
T1 = 3677.57 N
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