Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_18

⚡ Mecademy AIENG정역학 · ch6  Problem Statement The uniform slender bar has an ideal roller at its upper end A. Determine the minimum value of the angle for which equilibrium is possible for and for . Problem 6/18 (a) Minimum angle for ● Calculation Process 1. Present Final Formula: First, we derive the general equilibrium condition for the bar of length and weight . Let the bottom end be the origin . The top end is at . The wall is inclined at from the floor, so the normal force at the roller is perpendicular to this surface, making an angle of with the horizontal. Summing moments about : From force equilibrium: At impending slip, . Substituting and solving for : 2. Substitute Values: Substitute and into the derived formula: θμ = s 0.25μ = s 0.50 μ = s 0.25 LWB(0,0) A(−Lcosθ,Lsinθ)105 ∘ N A 105− ∘ 90= ∘ 15 ∘ B M = ∑ B 0⇒W cosθ − ( 2 L )(N cos15)(Lsinθ)− A ∘ (N sin15)(L A ∘ cos θ = 2 W N (sinθcos15+ A ∘ cosθsin15)= ∘ N sin(θ+ A 15) ∘ N = A 2sin(θ+15) ∘ Wcosθ F = ∑ x 0⇒F = B N cos15 A ∘ F = ∑ y 0⇒N = B W−N sin15 A ∘ F = B μ N sB N A tanθ tanθ= −tan15 2 1 ( μ s 1 ∘ ) μ = s 0.25tan15≈ ∘ 0.26795 tanθ= −tan15 2 1 ( 0.25 1 ∘ ) 3. Partial Operations: 4. Final Calculation: ● Final Conclusion: The minimum angle for equilibrium when is . (b) Minimum angle for ● Calculation Process 1. Present Final Formula: We use the general formula derived in part (a): 2. Substitute Values: Substitute and : 3. Partial Operations: 4. Final Calculation: ● Final Conclusion: The minimum angle for equilibriu