Meriam, Kraige & Bolton — Rigid Bodies: Equivalent Systems of Forces: Problem 3_17

⚡ Mecademy AIENG정역학 · ch3  Problem Statement When the 0.05-kg body is in the position shown, the linear spring is stretched 10 mm. Determine the force P required to break contact at C. Complete solutions for (a) including the effect of the weight and (b) neglecting the weight. Problem 3/17 (a) Including the effect of the weight 1. Formula: To break contact at , the normal reaction force at must be zero ( ). We apply the equilibrium condition for moments about the pivot point : Summing moments (taking counter-clockwise as positive): 2. Substitution: Mass of the body, Weight, Spring constant, Spring stretch, Spring force, Moment arm for , Moment arm for (assuming uniform body, center of gravity is at the vertical bar's horizontal location), Moment arm for , Substituting into the moment equation: 3. Calculation: Moment of weight: CC N = C 0O M = ∑ O 0 P⋅d + A W⋅d − G F ⋅ s d = B 0 m=0.05 kg W=mg=0.05 kg× 9.81 m/s= 2 0.4905 N k=1750 N/m δ=10 mm=0.010 m F = s kδ=1750 N/m×0.010 m=17.5 N Pd = A 120 mm=0.12 m WG d = G 60 mm=0.06 m F s d = B 40 mm=0.04 m P(0.12)+0.4905(0.06)−17.5(0.04)=0 M = W 0.4905×0.06=0.02943 N⋅m Moment of spring force: Equation: 4. Result: ● Final Conclusion: The force required to break contact at , considering the weight of the body, is 5.59 N. (b) Neglecting the weight 1. Formula: By neglecting the weight, the moment term for is set to zero in the equilibrium equation: 2. Substitution: Using the previously calculated values: 3. Calculation: 4. Result: ● Final