How Does Non-Linear Friction Affect Torque in Drilling?

AI Thread Summary
Non-linear friction's impact on torque in drilling is explored through the relationship between frictional torque and normal force. The discussion highlights the need for a formula to determine the torque required to overcome friction when a shaft or drill tip contacts a surface. The equation τ_friction = β F_normal is proposed, but its validity and derivation are questioned, particularly for different shapes. The conversation reveals that frictional torque is not widely covered in literature and depends on the dimensions of the contact area. The participants seek further guidance on deriving expressions for various shapes to better understand this phenomenon.
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Hi everyone.

This is not a homework problem. I am very familiar with the case of linear(direction) friction.
F_{friction}=\mu F_{normal} ...(1)
However my question is closer to something like this:
\tau_{friction}=\beta F_{normal} ...(2)
Where \tau_{friction} is a torque

Imagine if a shaft were to be making contact with the floor at its stub. Assuming the normal force that the shaft exerts on the ground is known. Then how much torque would need to be applied around the shafts axis to make it overcome friction and start turning.

Maybe another analogy would be a drill with a flat tip, how much torque would the flat tip need to overcome friction ?

Is equation 2 of the right form ? what name does this effect go by because so far I could only find references to linear friction.
 
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Frictional torque on flat tip of round solid rod, T = mu*(d/3)*N, where d = rod diameter, and N = normal force rod exerts on floor. This assumes the rod is normal to the floor surface.
 


Hi nvn, thanks that helps a lot. I see "frictional torque" doesn't appear to be a very well covered topic. I was surprised to see that it is a function of the dimensions of the contacting area.
Could you please guide me to a derivation of frictional torque ? I would like to find an expression for other shapes as well.
 
Pressure[/color] p = N/A; rod diameter D = 2*r2.

\begin{equation*}\begin{split}T<br /> &amp;= \int\int r{\cdot}(\mu\,p)\,dA\\[0.5mm]<br /> &amp;=\ \mu\,p\int\int r\,dr\,(r\,d\theta)\\[2mm]<br /> &amp;=\ \mu\,p\int_{0}^{2\,\pi}\int_{0}^{r_2}r^2\,dr\:d\theta\\[3mm]<br /> &amp;=\ (D/3)\,\mu\,N\end{split}\end{equation*}
 


OK Thanks! that helps a lot :)
 
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