Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_70

⚡ Mecademy AIENG정역학 · ch6  Problem Statement The 20-kg steel ring A with inside and outside radii of 50 mm and 60 mm, respectively, rests on a fixed horizontal shaft of 40-mm radius. If a downward force applied to the periphery of the ring is just sufficient to cause the ring to slip, calculate the coefficient of friction and the angle . Problem 6/70 (a) Calculation of the angle 1. Formula: From the equilibrium of the ring, we analyze the forces and moments. Let the center of the ring be and the contact point with the shaft be . The total vertical force is the sum of the applied force and the weight . Since there are no other horizontal forces, the total reaction from the shaft at must be vertical. The friction force is the component of tangent to the surface at . By summing moments about the center , we find . Then, using the vertical force equilibrium, we find . The angle is related to these by: 2. Substitution: Substitute the given values , , , , and : Weight Friction 3. Calculation: Total vertical load P=150 N μθ θ GC PW=mg RC fRC Gf Rθ f=P r i r o f=(P+mg)sinθ P=150 Nm=20 kgg= 9.81 m/s 2 r = o 60 mmr = i 50 mm W=20×9.81=196.2 N f=150× ( 50 60 ) f=(150+196.2)sinθ f=150×1.2=180 N R=150+196.2=346.2 N 180=346.2sinθ⟹sinθ= ≈ 346.2 180 0.51993 θ=sin(0.51993)≈ −1 31.328 ∘ 4. Result: ● Final Conclusion: The angle at which the ring comes to rest on the verge of slipping is . (b) Calculation of the coefficient of friction 1. Formula: Since the ring is on the verge of slipping, the