Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_69

⚡ Mecademy AIENG정역학 · ch6  Problem Statement Calculate the tension required to lower the 800-kg load described in Prob. 6/68. Also find . [Data from Prob. 6/68: The 800-kg load is supported by a pulley with a 300-mm diameter and a 30-mm diameter bearing. The coefficient of static friction for the bearing is 0.25. The mass of the cable and pulley is small and may be neglected.] Problem 6/69 (a) Tension required to lower the load 1. Formula: The tension required to lower the load at a constant speed is determined by the equilibrium of moments about the center of the pulley and vertical force balance. Weight of the load: Vertical force balance: Moment equilibrium about pulley center: , where is the friction moment and is the pulley radius. Friction circle radius: , where is the bearing radius and . 2. Substitution: Pulley radius Bearing radius Friction angle Friction circle radius 3. Calculation: Friction moment: Moment balance equation: Using , substitute : TT 0 T W=m⋅g T+T = 0 W T⋅R−T ⋅ 0 R+M = f 0 M = f W⋅r f R r = f rsinφrφ= tan(μ) −1 W=800 kg× 9.81 m/s= 2 7848 N R= = 2 300 mm 150 mm=0.15 m r= = 2 30 mm 15 mm=0.015 m φ=tan(0.25)≈ −1 14.036 ∘ r = f 0.015sin(14.036)≈ ∘ 0.003638 m M = f 7848 N×0.003638 m=28.55 N⋅m (T − 0 T)⋅0.15=28.55⟹T − 0 T= = 0.15 28.55 190.33 N T+T = 0 7848 NT = 0 T+190.33 N 4. Result: ● Final Conclusion: The tension required to lower the load is (or ). (b) Tension in the stationary section of the cable 1. Formula: From the force balance established in par