Meriam, Kraige & Bolton — Rigid Bodies: Equivalent Systems of Forces: Problem 3_21

⚡ Mecademy AIENG정역학 · ch3  Problem Statement The 180-lb exerciser is beginning to execute some slow, steady bicep curls. As the tension is developed against an exercise machine (not shown), determine the normal reaction forces at the feet and . Friction is sufficient to prevent slipping, and the exerciser maintains the position shown with center of gravity at . Problem 3/21 (a) Normal reaction forces at the feet A and B 1. Formula: To determine the normal reaction forces, we use the equilibrium equations for a rigid body in a 2D plane: We define a coordinate system with the origin at foot . The weight acts vertically downward at . The tension at point has both horizontal () and vertical () components. 2. Substitution: From the diagram, the given values are: Weight of exerciser: Tension in cable: at an angle of Horizontal distance from to : Horizontal distance from to : Horizontal distance from to : Vertical height of point : First, resolve force into components: T=15 lb AB G F = ∑ y 0 M = ∑ A 0 AW GTCT x T y W=180 lb T=15 lb20 ∘ AGd = AG 10 in. ABd = AB 10+10=20 in. ACd = AC 10+10+14=34 in. Ch = C 62 in. T T = x Tcos20= ∘ 15cos20 lb ∘ T = y Tsin20= ∘ 15sin20 lb ∘ Summing moments about point (taking clockwise as positive): 3. Calculation: Calculate components of : Substitute into the moment equation: Apply vertical force equilibrium to find : 4. Result: , ● Final Conclusion: The normal reaction forces at the feet are at the left foot and at the right foot. ✨ Final Answer Summ