$\sum _{}{\mathrm{M}}_{\mathrm{C}}={\mathrm{F}}_1(750 \mathrm{mm})-{\mathrm{F}}_2(300 \mathrm{mm})-\mathrm{P}(400 \mathrm{mm})=0$

To solve for the internal forces ${\mathrm{F}}_1$ and ${\mathrm{F}}_2$, a second equation involving these two unknowns must be developed. To do this, the rigid bar geometry and the deformation of the axial members will be considered.