the given data is-
Mixer having one impeller(rotor) rotating in horizontal plane about vertical axis.
capacity of mixer to design $C = 250kg$
rotor dia. $D = 45" = 1143mm$
friction coefficient (assumed) $\mu = 0.4$
Rotor mass $M_R = 30kg$
My solution -
Force required $$\begin{array}{lcl} F & = & \mu \times \text{Normal reaction} \\ & = & \mu \left(C+M_R\right) \\ & = & 0.4 \times 280 kg \\ & = & 112 kgf\\ & = & 1120N \end{array}$$
Torque required $$\tau = F D = 1120 N \times 1143 mm = 1280160 N\cdot mm =1280.16 Nm$$
I want to confirm this solution. Is it OK ??? If not then please give me hint...