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Equilibrium

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

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Free-Body Diagram:

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Force triangle:

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Law of sines:

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$\\dfrac{T_{AC}}{\\sin (90^\\circ-\\beta)}=\\dfrac{T_{BC}}{\\sin (90^\\circ-\\alpha)}=\\dfrac{P}{\\sin (\\alpha+\\beta)}$

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$\\dfrac{T_{AC}}{\\sin \\var{right_beta}^\\circ}=\\dfrac{T_{BC}}{\\sin \\var{right_alpha}^\\circ}=\\dfrac{\\var{P}\\,\\text{N}}{\\sin \\var{alpha_beta}^\\circ}$

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(a) 

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$T_{AC}=\\dfrac{\\var{P}\\,\\text{N}}{\\sin\\var{alpha_beta}^\\circ}\\sin\\var{right_beta}^\\circ=\\var{tAC}\\,\\text{N}\\quad\\blacktriangleleft$

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(b)

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$T_{BC}=\\dfrac{\\var{P}\\,\\text{N}}{\\sin\\var{alpha_beta}^\\circ}\\sin\\var{right_alpha}^\\circ=\\var{tBC}\\,\\text{N}\\quad\\blacktriangleleft$

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 in cable AC,

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 in cable BC.

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