Step 1: Geometry and Forces

Let the angle between the two blades be $\theta$. The angle of each blade with the central axis (bisector) is $\alpha = \theta/2$.

Consider the forces exerted by the top blade on the wire:

• Normal Force ($N$): Acts perpendicular to the blade edge, pushing into the wire. This force has a vertical component (squeezing) and a horizontal component (pushing the wire away/outward).
• Friction Force ($f = \mu N$): Acts along the blade surface. Since the wire tends to slide away from the hinge, friction acts towards the hinge (inward) to oppose this motion.

Step 2: Equilibrium Condition

For the wire to be on the verge of sliding away, the outward component of the Normal force must be balanced by the inward component of Friction.

Resolving forces along the horizontal axis (central bisector):

• Outward component of $N$: $N \sin(\alpha)$
• Inward component of $f$: $f \cos(\alpha)$

At the limit of stability:

$$N \sin(\alpha) = f \cos(\alpha)$$ $$N \sin(\alpha) = \mu N \cos(\alpha)$$ $$\mu = \frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha)$$

Step 3: Conclusion

Since $\alpha = 0.5\theta$:

$$\mu = \tan(0.5\theta)$$