Torque required to rotate sand mixer [closed]

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...

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 FYI this site supports Mathjax for rendering math. I've done it for your equations here. Have a look at the FAQ if you want to know how to do it yourself. – Michael Brown Jan 7 at 6:00 Thanks Michael... – Sushil Shinge Jan 7 at 6:16 Your expression for the force is wrong, as it has units of $kg$. You probably saw the conversion to Newtons for the gravitational force. Why is this wrong here? – Bernhard Jan 7 at 6:32 I really can't tell without a sketch or a free body diagram. – ja72 Jan 7 at 7:25 Hi Sushil, and welcome to Physics Stack Exchange! Generally we discourage questions that just ask for someone to check your work. Once you have identified the specific concept that you're not sure about, that's the point at which it's appropriate to ask a question here. – David Zaslavsky♦ Jan 7 at 7:37
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closed as too localized by David Zaslavsky♦Jan 7 at 7:37

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