# 6161103 11.7 stability of equilibrium

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<ul><li> 1. 11.7 Stability of Equilibrium Once the equilibrium configuration for a body or system of connected bodies are defined, it is sometimes important to investigate the type of equilibrium or the stability of the configurationExample Consider a ball resting on each of the three paths Each situation represent an equilibrium state for the ball </li> <li> 2. 11.7 Stability of EquilibriumWhen the ball is at A, it is at stableequilibriumGiven a small displacement up the hill, it willalways return to its original, lowest, positionAt A, total potential energy is a minimumWhen the ball is at B, it is in neutralequilibriumA small displacement to either the left orright of B will not alter this conditionThe balls remains in equilibrium in thedisplaced position and therefore, potentialenergy is constant </li> <li> 3. 11.7 Stability of EquilibriumWhen the ball is at C, it is in unstableequilibriumA small displacement will cause the ballspotential energy to be decreased, and so itwill roll farther away from its original, highestpositionAt C, potential energy of the ball is maximum </li> <li> 4. 11.7 Stability of EquilibriumTypes of Equilibrium1. Stable equilibrium occurs when a small displacement of the system causes the system to return to its original position. Original potential energy is a minimum2. Neutral equilibrium occurs when a small displacement of the system causes the system to remain in its displaced state. Potential energy remains constant3. Unstable equilibrium occurs when a small displacement of the system causes the system to move farther from its original position. Original potential energy is a maximum </li> <li> 5. 11.7 Stability of EquilibriumSystem having One Degree of Freedom For equilibrium, dV/dq = 0 For potential function V = V(q), first derivative (equilibrium position) is represented as the slope dV/dq which is zero when the function is maximum, minimum, or an inflexion point Determine second derivative and evaluate at q = qeq for stability of the system </li> <li> 6. 11.7 Stability of EquilibriumSystem having One Degree of Freedom If V = V(q) is a minimum, dV/dq = 0 d2V/dq2 > 0 stable equilibrium If V = V(q) is a maximum dV/dq = 0 d2V/dq2 < 0 unstable equilibrium </li> <li> 7. 11.7 Stability of EquilibriumSystem having One Degree of Freedom If d2V/dq2 = 0, necessary to investigate higher- order derivatives to determine stability Stable equilibrium occur if the order of the lowest remaining non-zero derivative is even and the is positive when evaluated at q = qeq Otherwise, it is unstable For system in neutral equilibrium, dV/dq = d2V/dq2 = d3V/dq3 = 0 </li> <li> 8. 11.7 Stability of EquilibriumSystem having Two Degree of Freedom A criterion for investigating the stability becomes increasingly complex as the number for degrees of freedom for the system increases For a system having 2 degrees of freedom, equilibrium and stability occur at a point (q1eq, q2eq) when V/q1 = V/q2 = 0 [(2V/q1q2)2 (2V/q12)(2V/q22)] < 0 2V/q12 > 0 or 2V/q22 >0 </li> <li> 9. 11.7 Stability of EquilibriumSystem having Two Degree of Freedom Both equilibrium and stability occur when V/q1 = V/q2 = 0 [(2V/q1q2)2 (2V/q12)(2V/q22)] < 0 2V/q12 > 0 or 2V/q22 >0 </li> <li> 10. 11.7 Stability of EquilibriumProcedure for AnalysisPotential Function Sketch the system so that it is located at some arbitrary position specified by the independent coordinate q Establish a horizontal datum through a fixed point and express the gravitational potential energy Vg in terms of the weight W of each member and its vertical distance y from the datum, Vg = Wy Express the elastic energy Ve of the system in terms of the sketch or compression, s, of any connecting spring and the springs stiffness, Ve = ks2 </li> <li> 11. 11.7 Stability of EquilibriumProcedure for AnalysisPotential Function Formulate the potential function V = Vg + Ve and express the position coordinates y and s in terms of the independent coordinate qEquilibrium Position The equilibrium position is determined by taking first derivative of V and setting it to zero, V = 0 </li> <li> 12. 11.7 Stability of EquilibriumProcedure for AnalysisStability Stability at the equilibrium position is determined by evaluating the second or higher-order derivatives of V If the second derivative is greater than zero, the body is stable If the second derivative is less than zero, the body is unstable If the second derivative is equal to zero, the body is neutral </li> <li> 13. 11.7 Stability of EquilibriumExample 11.5The uniform link has a mass of10kg. The spring is un-stretchedwhen = 0. Determine theangle for equilibrium andinvestigate the stability at theequilibrium position. </li> <li> 14. 11.7 Stability of EquilibriumSolutionPotential Function Datum established at the top of the link when the spring is un-stretched When the link is located at arbitrary position , the spring increases its potential energy by stretching and the weight decreases its potential energy </li> <li> 15. 11.7 Stability of EquilibriumSolution V = Ve + Vg 1 2 l = ks W s + cos 2 2 Since l = s + l cos or s = l (l cos ), 1 2 Wl V = kl (1 cos ) (2 cos ) 2 2 2 </li> <li> 16. 11.7 Stability of EquilibriumSolutionEquilibrium Position For first derivative of V, dV Wl = kl 2 (1 cos )sin sin = 0 d 2 or W l kl (1 cos ) sin = 0 2 Equation is satisfied provided sin = 0, = 0o W 10(9.81) = cos 1 1 = cos 1 1 = 53.8o 2(200)(0.6) 2kl </li> <li> 17. 11.7 Stability of EquilibriumSolutionStability For second derivative of V, d 2V Wl = kl 2 (1 cos )cos + kl 2 sin sin cos d 2 2 Wl = kl 2 (cos cos 2 ) cos 2 Substituting values for constants d 2V d 2 = 0o = ( ) 200(0.6) 2 cos 0o cos 0o 10(9.81)(0.6) 2 cos 0o = 29.4 < 0 unstable equilibrium d 2V d 2 =53.8o = ( 200(0.6) 2 cos 53.8o cos 53.8o ) 10(9.81)(0.6) 2 cos 53.8o = 46.9 > 0 stable equilibriu m </li> <li> 18. 11.7 Stability of EquilibriumExample 11.6Determine the mass m of theblock required for equilibriumof the uniform 10kg rod when = 20. Investigate the stabilityat the equilibrium position. </li> <li> 19. 11.7 Stability of EquilibriumSolution Datum established through point A When = 0, assume block to be suspended (yw)1 below the datum Hence in position , V = Ve + Vg = 9.81(1.5sin/2) m(9.81)(y) </li> <li> 20. 11.7 Stability of EquilibriumSolution Distance y = (yw)2 - (yw)1 may be related to the independent coordinate by measuring the difference in cord lengths BC and BC B C = (1.5)2 + (1.2)2 = 1.92 BC = (1.5 cos )2 + (1.2 sin )2 = 3.69 3.60 sin y = B C BC = 1.92 3.69 3.60 sin 1.5 sin V = 9.81 ( m(9.81) 1.92 3.69 3.60 sin ) 2 </li> <li> 21. 11.7 Stability of EquilibriumSolutionEquilibrium Position For first derivative of V, 3.60 cos = 73.6 cos dV m(9.81) =0 d 2 3.69 3.60 sin dV = 69.14 10.58m = 0 d = 20 o For mass, 69.14 m= = 6.53kg 10.58 </li> <li> 22. 11.7 Stability of EquilibriumSolutionStability For second derivative of V, d 2V m(9.81) 1 (3.60 cos ) 2 d 2 = 73.6 sin 2 (3.69 3.60 sin )3 / 2 2 m(9.81) 3.60 sin 2 3.69 3.60 sin For equilibrium position, d 2V = 47.6 < 0 unstable equilibrium d 2 </li> <li> 23. 11.7 Stability of EquilibriumExample 11.7The homogenous block having a mass mrest on the top surface of the cylinder. Showthat this is a condition of unstable equilibriumif h > 2R. </li> <li> 24. 11.7 Stability of EquilibriumSolution Datum established at the base of the cylinder If the block is displaced by an amount from the equilibrium position, for potential function, V = Ve + Vg = 0 + mgy y = (R + h/2)cos + Rsin Thus, V = mg[(R + h/2)cos + Rsin] </li> <li> 25. 11.7 Stability of EquilibriumSolutionEquilibrium Position dV/d = mg[-(R + h/2)sin + Rsin +Rcos]=0 =mg[-(h/2)sin + Rsin] = 0 Obviously, = 0 is the equilibrium position that satisfies this equation </li> <li> 26. 11.7 Stability of EquilibriumSolutionStability For second derivative of V, d2V/d2 = mg[-(h/2)cos + Rcos - Rsin] At = 0, d2V/d2 = - mg[(h/2) R] Since all the constants are positive, the block is in unstable equilibrium If h > 2R, then d2V/d2 < 0 </li> </ul>