Statics 3.3 – Solving for the magnitude and direction of a force for equilibrium | Hibbeler 14th Edition

Determine the magnitude and direction θ of F so that the particle is in equilibrium.

Statics 14E Problem 3.3 Forces in Equilibrium with unknown force and its direction

Solution:

Free-body Diagram:

Equilibrium Equation:

Summation of forces in the x-direction:

→+ ∑Fx=05 kN+Fsin⁡θ−8 kNcos⁡30°−4 kNcos⁡60°=0Fsin⁡θ=3.9282(1)\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:

+↑∑Fy=08sin⁡30°−4sin⁡60°−Fcos⁡θ=0Fcos⁡θ=0.5359(2)\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)

Fsin⁡θFcos⁡θ=3.92820.5359sin⁡θcos⁡θ=7.3301\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⁡θ=sin⁡θcos⁡θ \tan \theta = \dfrac{\sin \theta}{\cos \theta} :

tan⁡θ=7.3301θ=tan⁡−17.3301θ=82.2°\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

Fsin⁡82.2°=3.9282F=3.96 kN\begin{aligned} F\sin 82.2 \degree & = 3.9282 \\ \textcolor{blue}F & \textcolor{blue}{=3.96 \ \text{kN}} \end{aligned}