Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_75

⚡ Mecademy AIENG정역학 · ch6  Problem Statement Repeat Prob. 6/74 for the case of lowering the cylinder of mass at a steady speed. (Prob. 6/74: The two pulleys are fastened together and are used to hoist the cylinder of mass . The fraction can vary from near zero to one. Derive an expression for the tension required to raise the cylinder at a steady speed if the coefficient of friction for the bearing of radius is , a small enough value to permit the substitution of for , where is the friction angle. The mass of the pulley unit is . Evaluate your expression for if , , , , , and .) Problem 6/75 (a) Derive an expression for the tension required to lower the cylinder 1. Formula: Consider the equilibrium of the pulley unit. Let be the center of the journal bearing. For steady-speed lowering, the pulley rotates clockwise. The vertical forces acting on the pulley are the downward tension , the downward load , the downward weight of the pulley unit , and the upward reaction force at the bearing. Sum of vertical forces: Sum of moments about point : Friction moment: , where 2. Substitution: Substitute the expressions for and into the moment equation: 3. Calculation: Expand and solve for : m m kT r 0 μ μsinφφ m 0 Tm=50 kgm = 0 30 kgr=0.3 mk=1/2r = 0 25 mmμ=0.15 T O Tmg m g 0 R F =∑ y 0⟹R=T+mg+m g 0 OM =∑ O 0⟹mg(kr)−T(r)− M = f 0 M = f R⋅r f r = f r sinφ≈ 0 μr 0 Rr f mg(kr)−T(r)−(T+mg+m g)(μr )= 00 0 T mg⋅kr−T⋅r−Tμr − 0 mgμr − 0 m gμr = 00 0 mg(kr−μr )− 0 m gμr = 00 T(r+μr ) 0 T= r+μr 0 m