The members of a truss are pin connected at joint O. Determine the magnitude of F1 and its angle θ for equilibrium. Set F2=6 kN.
Solution:
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}