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.
<h3>left cable (a)</h3>
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
<h3>right cable (b)</h3>
Tb = (400 lb)sin(72°)/sin(90°) = 380.423 lb
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<em>Additional comment</em>
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)
was moved 4 units right and 1 unit down
