Two concurrent forces F_1 = 12N and F_2 = 30N are 150° apart. Calculate the angle between F


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Question Number 147149 by nadovic last updated on 18/Jul/21

$Two concurrent forces F_{1} = 12 N and$ $F_{2} = 30 N are 150 ° apart . Calculate$ $the angle between F_{2} and the resultant$ $force .$
	Answered by Olaf_Thorendsen last updated on 18/Jul/21

	${\vec{F}}_{1} = (\begin{matrix} F_{1} \\ 0 \end{matrix})$ ${\vec{F}}_{2} = (\begin{matrix} F_{2} \cos 150 ° \\ F_{2} \sin 150 ° \end{matrix})$ $\vec{F} = {\vec{F}}_{1} + {\vec{F}}_{2} = (\begin{matrix} F_{1} + F_{2} \cos 150 ° \\ F_{2} \sin 150 ° \end{matrix})$ $∣∣ \vec{F} ∣ ∣^{2} = {(F_{1} + F_{2} \cos 150 °)}^{2} + F_{2}^{2} \sin^{2} 150 °$ $∣∣ \vec{F} ∣ ∣^{2} = {(12 + 30 \cos 150 °)}^{2} + 30^{2} \sin^{2} 150 °$ $∣∣ \vec{F} ∣ ∣^{2} \approx 420, 46$ $∣∣ \vec{F} ∣∣ \approx 20, 51 N$ ${\vec{F}}_{2} ∙ \vec{F} = F_{2} Fcos (F_{2}, F) \approx 615, 15 \cos (F_{2}, F)$ ${\vec{F}}_{2} ∙ \vec{F} = F_{2} \cos 150 ° (F_{1} + F_{2} \cos 150 °)$ $+ F_{2}^{2} \sin^{2} 150 ° \approx 588, 23$ $\cos (F_{2}, F) = \frac{588, 23}{615, 15} = 0, 956$ $\Rightarrow \arg (F_{2}, F) = 17, 01 °$
	Commented by nadovic last updated on 18/Jul/21

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