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A fixed rigid wire forms a V-shape with vertex at A and angle BAC = 60 deg. The wire is rough. Two identical rings, each of mass m, are threaded onto the two arms of the wire and connected by a light elastic string of natural length 2a and spring constant mg/a. The rings are in equilibrium at positions where PA = AQ = 3a (P on arm AB, Q on arm AC). If mu is the minimum coefficient of friction between a ring and the wire, find the value of mu + sqrt(3).
- 2
- 3
- 4
- 7
Correct answer: 2
Solution
With the bisector vertical, each arm makes 30 deg with the vertical. The horizontal spring force (mg) and the vertical weight (mg) give net along-wire force (outward) = mg*(sqrt(3)-1)/2 and normal force = mg*(1+sqrt(3))/2. Then mu = (sqrt(3)-1)/(sqrt(3)+1) = 2 - sqrt(3), so mu + sqrt(3) = 2.
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