What is the tension in the 2 cables holding a traffic light? It is given that the light weighs 980N. Cable 1 is inclined downwards by an angle of 30 degrees and cable 2 is inclined downwards by an angle of 40 degrees.

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The weight of the light is 980 N. Cable 1 is at an angle 30 degrees to the horizontal, let the tension be T1. And cable 2 is at an angle 40 degrees to the horizontal, let the tension be T2.

The tension in the cables can be divided into their vertical and horizontal components. The horizontal components of both is equal and the sum of the vertical components is equal to the weight of the light.

T1*cos 30 = T2*cos 40

=> T1 = T2*cos 40/cos 30

T1*sin 30 + T2*sin 40 = 980

=> T2*cos 40*sin 30/cos 30 + T2*sin 40 = 980

=> T2*(cos 40*sin 30/cos 30 + sin 40) = 980

=> T2*1.085 = 980

=> T2 = 903 N

T1 = 798 N

The tension in cable 1 is 798 N and the tension in cable 2 is 903 N.

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