Meriam, Kraige & Bolton — Internal Forces and Moments: Problem 7_57

⚡ Mecademy AIENG정역학 · ch7  Problem Statement Determine the equilibrium values of and the stability of equilibrium at each position for the unbalanced wheel on the incline. Static friction is sufficient to prevent slipping. The mass center is at . Given: , . Problem 7/57 (a) Equilibrium Positions 1. Formula: The system is in equilibrium when the first derivative of the total potential energy with respect to the coordinate is zero (). Let be the incline angle. Taking the potential energy reference at the initial contact point, the height of the center of mass is: The equilibrium condition is: Using the trigonometric identity , we get: 2. Substitution: Substitute the given values , , and : 3. Calculation: First solution: Second solution: θ 10 ∘ Gr=100 mm=rˉ60 mm VθdV/dθ=0α=10 ∘ G V=mg(rθ−sinθ)sinα+(r−cosθ)cosα[rˉrˉ] = dθ dV mgrsinα−cosθsinα+sinθcosα=[rˉrˉ]0 sinθcosα−cosθsinα=sin(θ−α) rsinα+sin(θ−rˉα)=0 r=100 mm=rˉ60 mmα= 10 ∘ 100sin10+ ∘ 60sin(θ−10)= ∘ 0 sin(θ−10)= ∘ − sin10 60 100 ∘ sin(θ−10)= ∘ −1.6667×0.17365=−0.28941 θ−10= ∘ arcsin(−0.28941)=−16.825 ∘ θ = 1 10− ∘ 16.825= ∘ −6.825 ∘ θ−10= ∘ 180− ∘ (−16.825)= ∘ 196.825 ∘ 4. Result: and (or ). ● Final Conclusion: The equilibrium positions are and . (b) Stability Analysis 1. Formula: The stability of each equilibrium position is determined by the sign of the second derivative of the potential energy, . If , the equilibrium is stable. If , the equilibrium is unstable. 2. Substitution and Calculation: For : Since the result is po