Meriam, Kraige & Bolton — Rigid Bodies: Equivalent Systems of Forces: Problem 3_101

⚡ Mecademy AIENG정역학 · ch3  Problem Statement The drum and shaft are welded together and have a mass of 50 kg with mass center at G. The shaft is subjected to a torque (couple) of 120 N·m and the drum is prevented from rotating by the cord wrapped securely around it and attached to point C. Calculate the magnitudes of the forces supported by bearings A and B. Problem 3/101 (a) Magnitude of the force supported by bearing A 1. Formula: First, establish the coordinate system with the origin at bearing , the - axis along the shaft towards , the -axis vertical upward, and the -axis horizontal to the right (right-handed system). Use the moment equilibrium about the shaft axis (- axis) to find the cord tension , then use 3D force and moment equilibrium to find reaction components. 2. Substitution: Weight of the assembly: acting at . Applied torque: (clockwise when viewed from to ). Cord geometry: The cord is tangent to the drum of radius . From the drawing, the cord is tangent at the bottom point and goes to point . Cord vector: . Length of cord segment: . Tension vector: . Moment about -axis from tension: . Ax Bzy x T M = ∑ x 0, M = ∑ A 0,F= ∑0 W=mg=50×9.81=490.5 N G(0.4,0,0) M = applied −120i N⋅mA B R = wrap 140 mm=0.14 m D(0.2,0,−0.14) mC(0.44,0.18,−0.14) m = DC(0.44−0.2)i+(0.18−0)j+(−0.14− (−0.14))k=0.24i+0.18j m L = DC = 0.24+0.18 22 0.3 m T=T = 0.3 0.24i+0.18j 0.8Ti+0.6Tj xM = T,x (r T − D,yz r T )= D,zy (0)(0)−(−0.14)(0.6T)=0.084T Solving . Tension components: . 3. Calculatio