Next, consider the $\mathrm{FBD}$ for the small gear.

The torque acting on the small gear is related to the torque on the large gear by the ratio of the gear radii.

$$\begin{gathered} \sum _{}{\mathrm{M}}_{\mathrm{B}}={\mathrm{T}}_{\mathrm{small}}-\mathrm{F}\times {\mathrm{R}}_{\mathrm{small}}=0 \\ {\mathrm{T}}_{\mathrm{small}}=\mathrm{F}\times {\mathrm{R}}_{\mathrm{small}} \\ =\frac{{\mathrm{T}}_{\mathrm{large}}}{{\mathrm{R}}_{\mathrm{large}}}\times {\mathrm{R}}_{\mathrm{smal}} \\ \therefore {\mathrm{T}}_{\mathrm{small}}=\left(\frac{{\mathrm{R}}_{\mathrm{small}}}{{\mathrm{R}}_{\mathrm{large}}}\right){\mathrm{T}}_{\mathrm{large}} \end{gathered}$$