Question

A 400 pound metal star is hanging on two cables which are attached to the ceiling. The left hand cable makes a 18° angle with the ceiling while the right hand cable makes a 72° angle with the ceiling. What is the tension on each of the cables? (Round your answers to three decimal places.)

Answer 1

Answer:

• left: 123.607 lb
• right: 380.423 lb

Step-by-step explanation:

By balancing horizontal and vertical forces, we find the cable tensions to be ...

  Ta = W·sin(b)/sin(a+b) . . . . . where W is the weight being held

  Tb = W·sin(a)/sin(a+b)

Where Ta is the tension in the cable that makes an angle of 'a' with respect to the vertical, and Tb is the tension in the cable that makes an angle of 'b' with respect to the vertical.

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The given angles are with respect to the ceiling, so the angles with respect to the vertical will be their compmements.

left cable (a)

  angle 'a' is 90° -18° = 72°

  angle 'b' is 90° -72° = 18°

  a+b = 72° +18° = 90°

  Ta = (400 lb)sin(18°)/sin(90°) = 123.607 lb

right cable (b)

  Tb = (400 lb)sin(72°)/sin(90°) = 380.423 lb

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Additional comment

The nice expressions for cable tension come from the balance of forces.

  vertical: Ta·cos(a) +Tb·cos(b) = W

  horizontal: Ta·sin(a) = Tb·sin(b)

Solving the horizontal equation for Ta, we get ...

  Ta = Tb·sin(b)/sin(a)

Substituting into the vertical equaiton gives ...

  Tb·sin(b)cos(a)/sin(a) +Tb·cos(b) = W

Multiplying by sin(a) gives ...

  Tb(sin(b)cos(a) +sin(a)cos(b)) = W·sin(a)

Using the trig identity for the sine of the sum of angles, we can rewrite this in the form shown above:

  Tb = W·sin(a)/sin(a+b)

The problem is symmetrical with respect to 'a' and 'b', so the other tension is found by interchanging 'a' and 'b' in the equation:

  Ta = W·sin(b)/sin(a+b)