Two-concurrent-forces-F-1-12N-and-F-2-30N-are-150-apart-Calculate-the-angle-between-F-2-and-the-resultant-force-

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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