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(C) Let $N$ be the normal force exerted by each blade on the wire and $f = \mu N$ be the limiting friction force.For the wire to be on the verge of sl

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$	an \alpha$

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(C) Let $N$ be the normal force exerted by each blade on the wire and $f = \mu N$ be the limiting friction force.For the wire to be on the verge of sliding away from the hinge,the components of the forces along the axis of the scissors must balance.The normal force $N$ acts perpendicular to the blade,making an angle $\alpha$ with the axis of the scissors.The friction force $f$ acts along the blade,also making an angle $\alpha$ with the axis of the scissors.Resolving forces along the axis of the scissors:$N \sin \alpha = f \cos \alpha$Substituting $f = \mu N$:$N \sin \alpha = \mu N \cos \alpha$$\mu = \frac{\sin \alpha}{\cos \alpha} = 	an \alpha$Thus,the coefficient of friction is $	an \alpha$.

$A$ person is using scissors to cut a wire of circular cross-section and negligible weight. The wire slides in the direction away from the hinge until the angle between the scissors blades becomes $2 \alpha$. The friction coefficient between the blades and the wire is:

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