Friction of a scissor How can the friction by a scissor blade be related to its angle?
To be more specific this is the question that I came across.

Someone is using a pair of scissors to cut a wire of circular cross section and negligible weight. The wire is in equilibrium when the angle between the scissors blades is $2\alpha$. What is the friction coefficient between the wire and the blades of the scissors?


 A: I am not sure what you actually want to ask. So I would recommend that you put more effort into your question.
Assuming that this is the setup that you have, a scissor in gray and the cross section of the rope in orange:

Say the toque applied to the joint of the scissors is $\tau$. What is this force $F$, then? The distance of the rope contact point to the joint shall be $d$. Then we should have $F d = \tau$ and therefore $F = \tau / d$.
The force that actually drives the rope forward is the sum of the two force vectors. Drawing a few helping lines (already did that) one can figure out that $F_\text{forward} = 2 \tau \sin(\alpha) / d$.
The normal force on the rope will be just $F$ assuming a coefficient of friction between metal and rope of $\mu$, we have a friction of $F \mu$ parallel to the scissor blade. The sum of the frictional forces on the upper and lower part will be $2 \mu \tau \cos(\alpha) / d$.
Beyond here, I am not really sure what you are asking, you should expand on your question. If you take a quotient of the two forces, you will obtain some $\tan(\alpha)$ or $1/\tan(\alpha)$ behavior. Does this make sense? Check with the extremes (closed and fully open scissors) to check whether the extreme angle behavior matches your intuition.

Ah, now the question makes sense. The rope is put into the pair scissors and pushed away. At some point, the forward moving force is not sufficient any more to overcome the friction. The angle where this happens is $2 \alpha$.
So we want to find the angle $\alpha$ where
$$ 2 \mu \tau \cos(\alpha) / d = 2 \tau \sin(\alpha) / d \,.$$
This is solved by
$$ \mu = \tan(\alpha) \,. $$
