I Transmission of torque using friction

[erobz]

@Lnewqban This is my approach to the conical section. It's more of a combination of intuition and a leap of faith so let's see if it makes sense.

I first simplified the conical section to two flat angled surfaces.
View attachment 338068

Because it's static we can say:
$$\sum F_x = 0 \rightarrow F+2R\cos (\alpha)=0 \rightarrow R=\frac{-F}{2\cos(\alpha)}$$
We can see from that how the angle causes the reaction to be bigger in magnitude just as expected.

Isn't there a frictional force component of $2 \mu R \cos \theta$ missing from this force balance?

 

[Juanda]

Isn't there a frictional force component of $2 \mu R \cos \theta$ missing from this force balance?

I think you're right. I should have added the frictional force tangent to the surfaces as well.
In that case, the expression I get is:
$$\sum F_x=0\rightarrow F+2R\cos(\alpha)+2\mu R\cos(\alpha+\frac{\pi}{2})=0\rightarrow F+2R(\cos(\alpha)-\mu\sin(\alpha))=0$$
$$R=\frac{-F}{2(\cos(\alpha)-\mu\sin(\alpha))}$$
Then, the normal stress to the surface would be again:
$$\sigma =\frac{R}{A}$$
If I tried to slide the block in the $z$ direction (in or out of the screen) while applying $F$, then the opposing force would be $\mu R$ I believe.

 

[khamlichi]

Please note that there are two regimes of friction: uniform pressure and uniform wear. At the beginning the pressure is uniform and wear is uneven. But, after the running-in phase, the wear becomes uniform and pressure adjustes unevenly. This is important in practice since there exists noticeable difference between the two situations.