A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to to a support ,as shown . The upper cables make angles 41° and 63° with the horizontal . Calculate the tension in of the three cables

Explanation:

Let T and U represent the tensions in the 41° and 63° cables, respectively. In order for the system to be stationary, the horizontal components of these tensions must balance, and the vertical components of these tensions must total 200 N.

  Tcos(41°) =Ucos(63°) . . . . . balance of horizontal components

  U = Tcos(41°)/cos(63°) . . . . write an expression for U

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The vertical components must total 200 N, so we have

  Tsin(41°) +Usin(63°) = 200

  Tsin(41°) +Tcos(41°)sin(63°)/cos(63°) = 200

  T(sin(41°)cos(63°) +cos(41°)sin(63°))/cos(63°) = 200

  T = 200cos(63°)/sin(41° +63°) ≈ 93.6 . . . newtons

  U = 200cos(41°)/sin(41° +63°) ≈ 155.6 . . . newtons

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The vertical cable must have sufficient tension to balance the weight of the traffic light, so its tension is 200 N.

Then the tensions in the 3 cables are ...

  41°: 93.6 N

  63°: 155.6 N

  90°: 200 N

explanation:

newton's third law, for every action there is an equal and opposite reaction.

volts times amps 6 x 0.5 3w