Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_105

⚡ Mecademy AIENG정역학 · ch6  Problem Statement A 100-lb block is placed on a 30° incline and released from rest. The coefficient of static friction between the block and the incline is 0.30. (a) Determine the maximum and minimum values of the initial tension in the spring for which the block will not slip when released. (b) Calculate the friction force on the block if . Problem 6/105 (a) Maximum and minimum values of the initial tension T 1. Formula: First, determine the normal force from the equilibrium perpendicular to the incline: The maximum available static friction force is: From equilibrium parallel to the incline (taking upward as positive): For no slipping to occur, the required friction must satisfy: 2. Substitution: Substitute given values , , and : The inequality becomes: 3. Calculation: Calculate numerical values: T FT=40 lb N N=Wcosθ F = max μ N s F = ∑ x 0⟹T+F−Wsinθ=0⟹F=Wsinθ−T ∣F∣≤F ⟹ max −F ≤ max Wsinθ−T≤F max W=100 lbθ=30 ∘ μ = s 0.30 N=100cos30 lb ∘ F = max 0.30×(100cos30) lb ∘ Wsinθ=100sin30 lb ∘ −0.30(100cos30)≤ ∘ 100sin30− ∘ T≤0.30(100cos30) ∘ N=100×0.8660=86.60 lb Set up the range for : Min : Max : 4. Result: , ● Final Conclusion: The block will not slip as long as the initial spring tension is between and . (b) Friction force F on the block if T = 40 lb 1. Formula: From the equilibrium equation parallel to the incline: where a positive result indicates the friction force acts up the incline. 2. Substitution: Substitute , , and : 3. Calculation: Since ,