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}
Free-body diagram:
Equations of Equilibrium:
Take the sum of horizontal forces considering forces to the right positive, and equate to zero.
\begin {aligned}
\sum{F}_x &= 0 & \\
F_1 \cos{60 \degree}+F_2 \sin{70 \degree}-5\cos{30 \degree}-\dfrac{4}{5}\left(7\right) &= 0 &\\
0.5F_1+0.9397F_2&=9.9301 &(1)\\
\end {aligned}
Take the sum of vertical forces considering upward forces positive, and equate to zero.
\begin{aligned}
\sum F_y&=0 &\\
-F_1\sin60\degree+F_2\cos70\degree+5\sin30\degree-\dfrac{3}{5}\left(7\right)&=0 &\\
-0.8660F_1+0.3420F_2&=1.7 &(2)\\
\end{aligned}Now, we have two equations with two unknowns F_1 and F_2 . So, we have a system of two equations. We can solve this using algebra, or we can directly use our calculator with this capability. The answers are
F_1=1.83 \: \text{kN}\\
F_2=9.60 \: \text{kN}
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