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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  = 12N and   F_2  = 30N are 150° apart. Calculate    the angle between F_2  and the resultant   force.
$$\:\mathrm{Two}\:\mathrm{concurrent}\:\mathrm{forces}\:\boldsymbol{{F}}_{\mathrm{1}} \:=\:\mathrm{12N}\:\mathrm{and} \\ $$$$\:\boldsymbol{{F}}_{\mathrm{2}} \:=\:\mathrm{30N}\:\mathrm{are}\:\mathrm{150}°\:\mathrm{apart}.\:\mathrm{Calculate}\: \\ $$$$\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\boldsymbol{{F}}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{resultant} \\ $$$$\:\mathrm{force}. \\ $$
Answered by Olaf_Thorendsen last updated on 18/Jul/21
F_1 ^→  =  ((F_1 ),(0) )  F_2 ^→  =  (((F_2 cos150°)),((F_2 sin150°)) )  F^→  = F_1 ^→  + F_2 ^→   =  (((F_1 +F_2 cos150°)),((F_2 sin150°)) )  ∣∣F^→ ∣∣^2  = (F_1 +F_2 cos150°)^2 +F_2 ^2 sin^2 150°  ∣∣F^→ ∣∣^2  = (12+30cos150°)^2 +30^2 sin^2 150°  ∣∣F^→ ∣∣^2  ≈ 420,46  ∣∣F^→ ∣∣ ≈ 20,51 N    F_2 ^→ •F^(→)  = F_2 Fcos(F_2 ,F) ≈ 615,15cos(F_2 ,F)  F_2 ^→ •F^(→)  = F_2 cos150°(F_1 +F_2 cos150°)  +F_2 ^2 sin^2 150° ≈ 588,23  cos(F_2 ,F) = ((588,23)/(615,15)) = 0,956  ⇒ arg(F_2 ,F) = 17,01°
$$\overset{\rightarrow} {\mathrm{F}}_{\mathrm{1}} \:=\:\begin{pmatrix}{\mathrm{F}_{\mathrm{1}} }\\{\mathrm{0}}\end{pmatrix} \\ $$$$\overset{\rightarrow} {\mathrm{F}}_{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{F}_{\mathrm{2}} \mathrm{cos150}°}\\{\mathrm{F}_{\mathrm{2}} \mathrm{sin150}°}\end{pmatrix} \\ $$$$\overset{\rightarrow} {\mathrm{F}}\:=\:\overset{\rightarrow} {\mathrm{F}}_{\mathrm{1}} \:+\:\overset{\rightarrow} {\mathrm{F}}_{\mathrm{2}} \:\:=\:\begin{pmatrix}{\mathrm{F}_{\mathrm{1}} +\mathrm{F}_{\mathrm{2}} \mathrm{cos150}°}\\{\mathrm{F}_{\mathrm{2}} \mathrm{sin150}°}\end{pmatrix} \\ $$$$\mid\mid\overset{\rightarrow} {\mathrm{F}}\mid\mid^{\mathrm{2}} \:=\:\left(\mathrm{F}_{\mathrm{1}} +\mathrm{F}_{\mathrm{2}} \mathrm{cos150}°\right)^{\mathrm{2}} +\mathrm{F}_{\mathrm{2}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{150}° \\ $$$$\mid\mid\overset{\rightarrow} {\mathrm{F}}\mid\mid^{\mathrm{2}} \:=\:\left(\mathrm{12}+\mathrm{30cos150}°\right)^{\mathrm{2}} +\mathrm{30}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{150}° \\ $$$$\mid\mid\overset{\rightarrow} {\mathrm{F}}\mid\mid^{\mathrm{2}} \:\approx\:\mathrm{420},\mathrm{46} \\ $$$$\mid\mid\overset{\rightarrow} {\mathrm{F}}\mid\mid\:\approx\:\mathrm{20},\mathrm{51}\:\mathrm{N} \\ $$$$ \\ $$$$\overset{\rightarrow} {\mathrm{F}}_{\mathrm{2}} \bullet\overset{\rightarrow} {\mathrm{F}}\:=\:\mathrm{F}_{\mathrm{2}} \mathrm{Fcos}\left(\mathrm{F}_{\mathrm{2}} ,\mathrm{F}\right)\:\approx\:\mathrm{615},\mathrm{15cos}\left(\mathrm{F}_{\mathrm{2}} ,\mathrm{F}\right) \\ $$$$\overset{\rightarrow} {\mathrm{F}}_{\mathrm{2}} \bullet\overset{\rightarrow} {\mathrm{F}}\:=\:\mathrm{F}_{\mathrm{2}} \mathrm{cos150}°\left(\mathrm{F}_{\mathrm{1}} +\mathrm{F}_{\mathrm{2}} \mathrm{cos150}°\right) \\ $$$$+\mathrm{F}_{\mathrm{2}} ^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{150}°\:\approx\:\mathrm{588},\mathrm{23} \\ $$$$\mathrm{cos}\left(\mathrm{F}_{\mathrm{2}} ,\mathrm{F}\right)\:=\:\frac{\mathrm{588},\mathrm{23}}{\mathrm{615},\mathrm{15}}\:=\:\mathrm{0},\mathrm{956} \\ $$$$\Rightarrow\:\mathrm{arg}\left(\mathrm{F}_{\mathrm{2}} ,\mathrm{F}\right)\:=\:\mathrm{17},\mathrm{01}° \\ $$
Commented by nadovic last updated on 18/Jul/21
Thank you Sir
$${Thank}\:{you}\:{Sir} \\ $$

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