Meriam, Kraige & Bolton — Rigid Bodies: Equivalent Systems of Forces: Problem 3_18

⚡ Mecademy AIENG정역학 · ch3  Problem Statement When the 0.05-kg body is in the position shown, the torsional spring at is pretensioned so as to exert a 0.75-N · m clockwise moment on the body. Determine the force required to break contact at . Complete solutions for (a) including the effect of the weight and (b) neglecting the weight. Problem 3/18 (a) Including the effect of the weight 1. Formula: To break contact at , the normal force at the wall must be zero. We apply the moment equilibrium equation about the pivot , taking counter-clockwise as positive: 2. Substitution: Moment of force at : (acting counter-clockwise). Moment of weight : Assuming the mass center of the body lies on its vertical axis of symmetry, the weight acts at a distance of from . (acting counter-clockwise). Moment of the torsional spring: (given as clockwise). 3. Calculation: Calculate the weight moment: . Simplify the equation: Isolate : Final division: O P C C O M = ∑ O M + P M + W M = s 0 PAM = P P⋅(60 mm+60 mm)=P⋅0.12 m W W=mg60 mmO M = W (0.05 kg)(9.81 m/s)(0.06 m) 2 M = s −0.75 N⋅m P(0.12)+(0.05)(9.81)(0.06)−0.75=0 M = W 0.05⋅9.81⋅0.06=0.02943 N⋅ m 0.12P+0.02943=0.75 P0.12P=0.75−0.02943=0.72057 P= = 0.12 0.72057 6.00475 N 4. Result: (to three significant figures) ● Final Conclusion: Including the weight of the 0.05-kg body, the force required to break contact at is . (b) Neglecting the weight 1. Formula: Neglecting the weight means setting in the moment equilibrium equation: 2. Substitution: 3. Ca