When the liquid surface remains horizontal, the resultant force $\overrightarrow{{\mathrm{F}}_{\mathrm{R}}}$  on the molecule must be vertically downward as shown in fig-(1) and the $\overrightarrow{{\mathrm{F}}_{\mathrm{R}}}$ must always remain perpendicular to the free surface of liquid. In this case  ${\mathrm{F}}_{\mathrm{A}}={\mathrm{F}}_{\mathrm{C}}\sin {45}^0=\frac{{\mathrm{F}}_{\mathrm{c}}}{\sqrt{2}}$
                                  If  ${\mathrm{F}}_{\mathrm{A}}>\frac{{\mathrm{F}}_{\mathrm{c}}}{\sqrt{2}}$ , the resultant force $\overrightarrow{{\mathrm{F}}_{\mathrm{R}}}$  on the molecule will be as shown in fig-(2). Since $\overrightarrow{{\mathrm{F}}_{\mathrm{R}}}$  must always be perpendicular to the free surface of liquid, the free surface will be concave in this case.