The situation is shown in the diagram



When the floor is removed, the forces on the person are

(1) weight mg downward

(2) normal force $\mathrm{N}$ due to the wall, towards the centre

(3) frictional force ${\mathrm{f}}_{\mathrm{s}}$, parallel to the wall, upward.

The person is moving in a circle with a uniform speed, so its acceleration is $\raisebox{1ex}{${\mathrm{v}}^2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{r}$}\right.$ towards the centre.

Newton's law for the horizontal direction (2nd law) and for the vertical direction (1st law) give

                      $\mathrm{N} = {\mathrm{mv}}^2/\mathrm{r}$                        ...(i)

and                ${\mathrm{f}}_{\mathrm{s}} = \mathrm{mg}$.                            ...(ii)

​For the minimum speed when the floor may be removed, the friction is limiting one and so equals μ_sN. This gives

                   ${\mathrm{\mu }}_{\mathrm{s}}\mathrm{N} = \mathrm{mg}$

$\text{or,}\frac{{\text{μ}}_{\text{s}}{\text{mv}}^2}{\text{r}}=\text{mg}\left[\text{using (i)}\right]$

$\text{or,}\text{v}=\sqrt{\frac{\text{rg}}{{\text{μ}}_{\text{s}}}}=\sqrt{\frac{\text{2 m}\times {\text{10 m/s}}^2}{\text{0.2}}}={\text{10 m s}}^{-1}$