Meriam, Kraige & Bolton — Rigid Bodies: Equivalent Systems of Forces: Problem 3_63

⚡ Mecademy AIENG정역학 · ch3  Problem Statement A uniform steel ring 60 in. in diameter and weighing 600 lb is lifted by the three cables, each 50 in. long, attached at points A, B, and C as shown. Compute the tension in each cable. Problem 3/63 (a) Geometric Configuration and Force Components 1. Formula: Define the coordinate system with the origin at the center of the ring. Let be the radius of the ring and the length of each cable. The height of the common hook point above the center is found via the Pythagorean theorem: The vertical component factor for each tension is . 2. Substitution: 3. Calculation: Define the coordinates of the attachment points: (at from ) (at from ) The vertical component of each tension is . 4. Result: The system geometry is established with a vertical component factor of for all cables. r Lh H r= ,h= 2 D L−r 22 cosφ= L h r= = 2 60 30 in.,L=50 in. h= = 50−30 22 = 2500−900 = 160040 in. cosφ= = 50 40 0.8 A=(30,0,0) in. B=(0,30,0) in.90 ∘ A C=(30cos210,30sin210,0)= ∘∘ (−15 ,−15,0) in. 3120 ∘ B T i T = iz 0.8T i 0.8 ● Final Conclusion: The common hook is located at and the vertical force contributed by each cable is . (b) System of Equilibrium Equations 1. Formula: For equilibrium of the ring, the sum of vertical forces and the moments about the center must be zero: 2. Substitution: 3. Calculation: Substituting the coordinates: Force: Moment : Moment : 4. Result: The relations are and . ● Final Conclusion: The ratios between tensions are determined by m