Meriam, Kraige & Bolton — Analysis of Structures: Problem 6_74

⚡ Mecademy AIENG정역학 · ch6  Problem Statement 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 6/74 (a) Derivation of the expression for tension 1. Formula: To raise the cylinder at a steady speed, the pulley unit is in equilibrium. The sum of moments about the center of the pulley must be zero. Considering vertical force equilibrium for the pulley unit, the reaction force from the bearing must balance all downward forces: The friction moment resists the rotation. For journal friction, it is defined by the radius of the friction circle . Given the small-angle approximation : 2. Substitution: Based on the diagram, to raise mass , the pulley must rotate clockwise. The tension creates a clockwise moment, while the weight and the friction moment create counter-clockwise moments. Substitute and : 3. Calculation: Expand and group terms involving : 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 M = ∑ O 0 R R=T+mg+m g 0 M f r = f r sinφ 0 sinφ≈μ M = f R⋅r = f R⋅μr 0 m Tmg M f T⋅r−mg⋅(kr)−M = f 0 RM f T⋅r−mg⋅kr−(T+mg+m g)⋅ 0 μr = 0 0 T Solve for : \[T = \frac{g [