Meriam, Kraige & Bolton — Internal Forces and Moments: Problem 7_61

⚡ Mecademy AIENG정역학 · ch7  Problem Statement In the mechanism shown, the spring of stiffness is uncompressed when . Also, the masses of the parts are small compared with the sum of the masses of the two cylinders. The mechanism is constructed so that the arms can swing past the vertical, as seen in the right-hand side view. Determine the values for equilibrium and investigate the stability of the mechanism in each position. Neglect friction. Problem 7/61 (a) Determination of Total Potential Energy 1. Formula: The total potential energy of the system is the sum of the gravitational potential energy and the elastic potential energy of the spring : 2. Substitution: Let the pivot be the datum for gravitational potential energy (), with the vertical coordinate directed downwards. The vertical position of the total mass (at point ) is . Thus, . The spring is uncompressed when . The vertical position at this angle is . The spring deformation (compression) is . The elastic potential energy is . Substituting these into the total energy expression: 3. Calculation: To find equilibrium, we take the first derivative of with respect to : kθ=60 ∘ m θ V V g V e V=V + g V e OV= g 0 y mBy = B 2acosθV = g −mgy = B −2mgacosθ θ=60 ∘ y = B,0 2acos60= ∘ ax=y − B y = B,0 2acosθ−a=a(2cosθ−1) V = e kx= 2 1 2 ka(2cosθ− 2 1 2 1) 2 V=−2mgacosθ+ ka (2cos θ − 2 1 2 1) 2 Vθ = dθ dV 2mgasinθ+ ka ⋅ 2 1 2 2(2cosθ−1)⋅(−2sinθ) = dθ dV 2mgasinθ−2kasinθ(2cosθ− 2 1) Factoring out : 4. Result: The potential energy