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Equilibrium

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Two cables are tied together at C and are loaded as shown. Knowing that $\\alpha=\\var{a}^\\circ$ and $M=\\var{M}\\text{kg}$, determine the tension (in N)

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Free Body Diagram:

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Force triangle:

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\\[W=Mg=(\\var{M}\\,\\text{kg}) (9.81\\,\\text{m}/\\text{s}^2)=\\var{W}\\,\\text{N}\\]

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Law of sines:

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\\[\\dfrac{T_{AC}}{\\sin (90^\\circ-\\alpha)}=\\dfrac{T_{BC}}{\\sin 50^\\circ}=\\dfrac{W}{\\sin (40^\\circ+\\alpha)}\\]

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(a) \\[T_{AC}=(\\var{W}\\,\\text{N})\\dfrac{\\sin (90^\\circ-\\var{a}^\\circ)}{\\sin (40^\\circ+\\var{a}^\\circ)}=\\var{tAC}\\,\\text{N}\\quad\\blacktriangleleft\\]

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(b) \\[T_{BC}=(\\var{W}\\,\\text{N})\\dfrac{\\sin 50^\\circ}{\\sin (40^\\circ+\\var{a}^\\circ)}=\\var{tBC}\\,\\text{N}\\quad\\blacktriangleleft\\]

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 in cable AC

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 in cable BC.

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