Determine the magnitude and direction θ of F so that the particle is in equilibrium.
Solution:
Free-body Diagram:
Equilibrium Equation:
Summation of forces in the x-direction:
\begin{aligned}
\xrightarrow{+} \: \sum F_x & = 0 & \\
5 \: \text{kN}+F \sin \theta - 8 \: \text {kN} \cos 30 \degree - 4\: \text{kN} \cos 60 \degree & = 0 & \\
F \sin \theta &= 3.9282 & (1)
\end{aligned}Summation of forces in the y-direction:
\begin{aligned}
+\uparrow \sum F_y & = 0 &\\
8 \sin 30 \degree - 4 \sin 60 \degree - F \cos \theta & =0 &\\
F \cos \theta & = 0.5359 & (2)\\
\end{aligned}We now have two equations. Divide Eq (1) by (2)
\begin{aligned}
\dfrac{F \sin \theta}{F \cos \theta} &= \dfrac{3.9282}{0.5359} \\
\dfrac{ \sin \theta}{ \cos \theta} & = 7.3301 \\
\end{aligned}We know that \tan \theta = \dfrac{\sin \theta}{\cos \theta} :
\begin{aligned}
\tan \theta &=7.3301 \\
\theta & = \tan^{-1}7.3301\\
\textcolor{blue}\theta & \textcolor{blue}{=82.2\degree}\\
\end{aligned}Substituting this result to equation (1), we have
\begin{aligned}
F\sin 82.2 \degree & = 3.9282 \\
\textcolor{blue}F & \textcolor{blue}{=3.96 \ \text{kN}}
\end{aligned}