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resultant force

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 Known: $P_1=\\var{P1}$N, $P_2=\\var{P2}$N.

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If the resultant force is directed along $BC$, its component in the direction perpendicular to $BC$ should be zero.

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Let the axis $y$ be directed along the line $BC$ and the axis $x$ be perpendicular to $BC$:

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The $x$ component of the resultant:

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$P_2 \\cos 30^\\circ+P_1 \\cos 55^\\circ-F_{AC}\\cos 10^\\circ = 0$

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Solving this equation for $F_{AC}$:

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$F_{AC}=\\dfrac{P_2 \\cos 30^\\circ+P_1 \\cos 55^\\circ}{\\cos 10^\\circ}=\\dfrac{(\\var{P2}\\text{N})\\cos 30^\\circ+(\\var{P1}\\text{N}) \\cos 55^\\circ}{\\cos 10^\\circ}=\\var{FAC}\\,\\text{N}\\quad\\blacktriangleleft$

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Determine the required tension in cable AC (in N), knowing that the resultant of the three forces exerted at Point C  on the bar BC (i.e. $F_{AC}$, $P_1$ and $P_2$) is directed along BC.

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