Equilibrium of a Solid Body To find the torques around B, see the figure below, I took the cross product as followed:

and we know that $W=160 N$ but when I insert that value into the the equation in the red box, I get $F_3=46.2 N$ but they get 138.6 N, how do they get that? the only possible way that can happen is if they multiply $F_3$ by 3.


The book is wrong and you are correct.
$$\begin{pmatrix} -\frac{L}{2} \cos 60° \\ \frac{L}{2} \sin 60° \\ 0 \end{pmatrix} \times \begin{pmatrix} 0 \\ -W \\ 0 \end{pmatrix} + \begin{pmatrix} -L \cos 60° \\ L \sin 60° \\ 0 \end{pmatrix} \times \begin{pmatrix} F_3 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$
$$\rightarrow \begin{pmatrix} 0 \\ 0 \\ W \frac{L}{2} \cos 60° - F_3 L \sin 60° =0 \end{pmatrix}$$
$$ F_3 = \frac{W}{2} \frac{\cos 60°}{\sin 60°} =  \frac{80}{\sqrt{3}} = 46.1880 $$
