Tag Archives: concurrent force system

Statics 3.6 – Equilibrium of Truss Members Connected to a Gusset Plate | Hibbeler 14th Edition

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}

Free-body diagram:

Equations of Equilibrium:

The summation of forces in the x-direction:

\begin{aligned} \sum F_x & = 0 &\\ 6 \sin 70 \degree + F_1 \cos \theta - 5 \cos 30 \degree - \dfrac {4}{5} \left(7 \right) & = 0 & \\ F_1 \cos \theta & = 4.2920 & (1) \end{aligned}

The summation of forces in the y-direction:

\begin{aligned} \sum F_y & =0 & \\ 6 \cos 70 \degree+5 \sin 30 \degree - F_1 \sin \theta - \dfrac{3}{5} \left( 7 \right ) & =0 & \\ F_1 \sin \theta &=0.3521 & (2)\\ \end{aligned}

We came up with 2 equations with unknowns $F_1$ and $\theta$ . To solve the equations simultaneously, we can use the method of substitution.

Using equation 1, solve for $F_1$ in terms of $\theta$ .

\begin{aligned} F_1 \cos \theta & = 4.2920 &\\ F_1 & =\dfrac{4.2920}{ \cos \theta } & (3) \\ \end{aligned}

Now, substitute this equation (3) to equation (2).

\begin{aligned} F_1 \sin \theta & = 0.3521 \\ \left ( \dfrac {4.2920}{\cos \theta} \right) \sin \theta & =0.3521 \\ 4.2920 \cdot \dfrac{\sin \theta}{\cos \theta} & = 0.3521 \\ 4.2920 \tan \theta & = 0.3521 \\ \tan \theta & = \dfrac{0.3521}{4.2920} \\ \theta &= \tan ^{-1} \dfrac{0.3521}{4.2920} \\ \theta & = 4.69 \degree \end{aligned}

Substitute the solved value of $\theta$ to equation (3).

\begin{aligned} F_1 & = \dfrac{4.2920}{\cos \theta} \\ F_1 &= \dfrac{4.2920}{\cos 4.69 \degree} \\ F_1 & = 4.31 \text{kN} \end{aligned}

Therefore, the answers to the questions are:

\begin{aligned} F_1= & \:4.31 \: \text {kN} \\ \theta = & \: 4.69 \degree \end{aligned}