Meriam, Kraige & Bolton — Internal Forces and Moments: Problem 7_29

⚡ Mecademy AIENG정역학 · ch7  Problem Statement The potential energy of a mechanical system is given by , where is the position coordinate associated with its single degree of freedom. Determine the position or positions of equilibrium of the system and the stability condition of the system at each equilibrium position. Problem 7/29 (a) Determination of Equilibrium Positions 1. Formula: Equilibrium occurs at positions where the first derivative of the potential energy function with respect to the system coordinate is zero: 2. Substitution: Substitute the given potential energy function : 3. Calculation: Differentiate the expression and solve for : First derivative: Factoring out : Setting each factor to zero: 4. Result: The equilibrium positions are , , and . ● Final Conclusion: The system has three equilibrium positions located at and . (b) Stability Analysis at Equilibrium Positions V=6x− 4 3x+ 2 5x = dx dV 0 V(x)=6x− 4 3x+ 2 5 (6 x − dx d 4 3x+ 2 5)=0 x 24x− 3 6x=0 6x6x(4x− 2 1)=0 6x=0⟹x = 1 0 4x− 2 1=0⟹x= 2 ⟹ 4 1 x = 2 0.5, and x = 3 −0.5 x=0x=0.5x=−0.5 x=0 x=±0.5 1. Formula: The stability of an equilibrium position is determined by the sign of the second derivative of the potential energy: Stable Equilibrium: (Potential energy is at a local minimum) Unstable Equilibrium: (Potential energy is at a local maximum) 2. Substitution: First, find the general second derivative expression: 3. Calculation: Evaluate the second derivative at each equilibrium position: At : Since , the