Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_21

⚡ Mecademy AIENG정역학 · ch6  Problem Statement The inverted track with freely floating cylinder comprise a system which is designed to hold paper or other thin materials in place. The coefficient of static friction is for all interfaces. What minimum value of ensures that the device will work no matter how heavy the supported material is? Problem 6/21 (a) Minimum value of the coefficient of static friction ● Calculation Process 1. Present Final Formula: To ensure the device works (holds the material) regardless of the weight , the system must be self-locking. For a wedge-like system with a total wedge angle and identical coefficients of friction on all surfaces, the self-locking condition is: where is the angle between the inclined track and the vertical wall. In this problem, . 2. Substitute Values: Substitute the given wedge angle into the self-locking condition: 3. Partial Operations: Perform the trigonometric calculation for the half-angle: 4. Final Calculation: Using the value for : TC Pμ μ P μ W θμ μ≥ tan ( 2 θ ) θ θ=30 ∘ θ=30 ∘ μ = min tan ( 2 30 ∘ ) = 2 θ = 2 30 ∘ 15 ∘ μ = min tan(15) ∘ tan15 ∘ μ ≈ min 0.2679 ● Final Conclusion: The minimum coefficient of static friction required is . At this value, the jamming effect created by the wedge geometry ensures that the material is held by friction forces that increase proportionally with its weight, preventing slippage at any load. ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 6/21 μ= tan15≈ ∘ 0.268 μ = min 0