$$\begin{gathered} \therefore =(-405 \mathrm{mm}){\mathrm{F}}_2=\mathrm{P}(400 \mathrm{mm}) \\ {\mathrm{F}}_2=-(0.98765)\mathrm{P} \\ =-(0.98765)(80 \mathrm{kN}) \\ =-79.012 \mathrm{kN} \\ { \\ }_{ \\ } \end{gathered}$$

Because we initially assumed ${\mathrm{F}}_2$ was in tension, the negative sign in the answer indicates that ${\mathrm{F}}_2$ is actually in compression.

Backsubstituting ${\mathrm{F}}_2$ into the equation $$\begin{gathered} {\mathrm{F}}_1=-0.14{\mathrm{F}}_2{ \\ }_{ \\ } \end{gathered}$$ gives:

$$\begin{gathered} {\mathrm{F}}_1=-(0.14){\mathrm{F}}_2 \\ =-(0.14)(-79.012 \mathrm{kN}) \\ =11.077 \mathrm{kN}{ \\ }_{ \\ } \end{gathered}$$