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Two cables support a 800​-lb ​weight, as shown. Find the tension in each cable.

Two vertical line segments are connected by a line segment that falls from left to right and a line segment that falls from right to left. At the intersection of these two line segments is a rectangle labeled 800 pounds. The angle formed at the intersection of the left vertical line segment and vertical line segment falling from left to right has a measure of 75 degrees. The angle formed at the intersection of the right vertical line segment and vertical line segment rising from left to right has a measure of 45 degrees.

75 degrees

45 degrees


PICTURE below

Two cables support a 800lb weight as shown Find the tension in each cableTwo vertical line segments are connected by a line segment that falls from left to righ class=

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sqdancefan

Answer:

  • 892 lb (right)
  • 653 lb (left)

Step-by-step explanation:

The weight is in equilibrium, so the net force on it is zero. If R and L represent the tensions in the Right and Left cables, respectively ...

  Rcos(45°) +Lcos(75°) = 800

  Rsin(45°) -Lsin(75°) = 0

Solving these equations by Cramer's Rule, we get ...

  R = 800sin(75°)/(cos(75°)sin(45°) +cos(45°)sin(75°))

     = 800sin(75°)/sin(120°) ≈ 892 . . . pounds

  L = 800sin(45°)/sin(120°) ≈ 653 . . . pounds

The tension in the right cable is about 892 pounds; about 653 pounds in the left cable.

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This suggests a really simple generic solution. For angle α on the right and β on the left and weight w, the tensions (right, left) are ...

  (right, left) = w/sin(α+β)×(sin(β), sin(α))

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