91Ó°ÊÓ

The net force

${\stackrel{→}{F}}_{net}={\stackrel{→}{F}}_1+{\stackrel{→}{F}}_2+{\stackrel{→}{F}}_3+{\stackrel{→}{F}}_4=4\widehat{i}$

If a vector F is inclined by an angle $\mathrm{θ}$with respect to the x- axis, and the x-component is $\text{F}\cos \mathrm{θ}$and its y - component is $\text{F}\sin \mathrm{θ}$.

Resolving$F_2$ into two components, along x-axis and along the y-axis as shown in the following figure:

$\begin{array}{l}{\stackrel{→}{F}}_{net}={\stackrel{→}{F}}_1+{\stackrel{→}{F}}_2+{\stackrel{→}{F}}_3+{\stackrel{→}{F}}_{4}4\widehat{i}+0\widehat{j}\\ {\stackrel{→}{F}}_{net}=\left[5\right(−\widehat{j}\left)\right]+\left[6\cos {20}^{°}\widehat{i}+6\sin {20}^{°}\widehat{j}\right]+F_3\widehat{j}+\text{ }\left[F_4(−\widehat{i})\right]4\widehat{i}+0\widehat{j}\\ 4\widehat{i}=\left(6\cos {20}^{°}−F_4\right)\widehat{i}+\left(6\sin {20}^{°}+F_3−5\right)\widehat{j}\text{ â¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠâ¶ÄŠ}\mathrm{.....}\text{ }\left(1\right)\end{array}$

Comparing both y-component of expression (1)

$\begin{array}{l}0\widehat{j}=\left(6\sin {20}^{°}+F_3−5\right)\widehat{j}\\ ⇒{\stackrel{→}{F}}_3=\left(2.9\text{ N}\right)\widehat{j}\end{array}$

Comparing x-component of expression (1)

$\begin{array}{l}4\widehat{i}=\left(6\cos {20}^{°}−F_4\right)\widehat{i}\\ {\stackrel{→}{F}}_4=(6\cos {20}^{°}−4)\widehat{i}\\ {\stackrel{→}{F}}_4=\left(1.64\text{ N}\right)\widehat{i}\end{array}$