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Resultant force

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Two forces are applied as shown to a hook support. Knowing that the magnitude of $\\mathbf{P}$ is {P}N, 

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Using the triangle rule and law of sines:

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(a) Solve

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$\\dfrac{\\sin \\alpha}{50\\,\\text{N}}=\\dfrac{\\sin 25^\\circ}{P}$

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$\\sin \\alpha=\\sin 25^\\circ \\dfrac{50\\,\\text{N}}{\\var{P}}=\\var{sina}$

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$\\alpha=\\arcsin(\\var{sina})=\\var{a}^\\circ$

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$\\alpha=\\var{a}^\\circ\\,\\,$  \u25c0

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(b) 

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$\\alpha+\\beta+25^\\circ=180^\\circ$

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$\\beta=180^\\circ-25^\\circ-\\var{a}^\\circ=\\var{b}^\\circ$

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$\\dfrac{R}{\\sin \\var{b}^\\circ}=\\dfrac{\\var{P}\\,\\text{N}}{\\sin 25^\\circ}$

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$R=\\var{R}\\,\\text{N}$   \u25c0

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Determine the required angle $\\alpha$ (in degrees) if the resultant $\\mathbf{R}$ of the two forces applied to the support is to be horizontal.

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Determine the corresponding magnitude of $\\mathbf{R}$ (in N)

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