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Equilibrium

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Knowing that $\\alpha=\\var{a}^\\circ$ and $W=\\var{W}\\,\\text{N}$, determine the tension 

", "advice": "

Free-Body Diagram:

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Force triangle:

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Law of sines:

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$\\dfrac{T_{AC}}{\\sin (90^\\circ+\\alpha)}=\\dfrac{T_{BC}}{\\sin 5^\\circ}=\\dfrac{W}{\\sin (85^\\circ-\\alpha)}$

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$\\dfrac{T_{AC}}{\\sin \\var{a_90}^\\circ}=\\dfrac{T_{BC}}{\\sin 5^\\circ}=\\dfrac{\\var{W}\\,\\text{N}}{\\sin \\var{a_85}^\\circ}$

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(a)

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$T_{AC}=\\dfrac{\\var{W}\\,\\text{N}}{\\sin \\var{a_85}^\\circ}\\sin \\var{a_90}^\\circ=\\var{tAC}\\,\\text{N}\\quad\\blacktriangleleft$

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(b)

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$T_{BC}=\\dfrac{\\var{W}\\,\\text{N}}{\\sin \\var{a_85}^\\circ}\\sin 5^\\circ=\\var{tBC}\\,\\text{N}\\quad\\blacktriangleleft$

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in cable AC (in N),

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in rope BC (in N).

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