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Unformatted text preview: ME 530.343: Design and Analysis of Dynamic Systems Spring 2009 Lecture 9 - Coulomb Friction Friday, February 20, 2009 Todays Objectives Dealing with Coulomb friction 2nd order example (if time) Coulomb Friction Equation of Motion W = mg N = W F k = kx F f = N + N if mass moving right- N if mass moving left Case 1 . Half cycle where mass moves left to right x > 0, x > or x < 0, x > m x =- F k- F f =- kx- N m x + kx =- N (inhomogeneous!) Solution: x ( t ) = A 1 cos ( n t ) + A 2 sin ( n t )- N k Note that n = k m Constants A 1 and A 2 to be determined from initial conditions of this half cycle. Case 2. Half cycle where mass moves right to left x > 0, x < or 1 x < 0, x < m x =- kx + N m x + kx = N Solution: x ( t ) = A 3 cos ( n t ) + A 4 sin ( n t ) + N k Still have n = k m Constants A 3 and A 4 to be determined from initial conditions of this half cycle. Complete Solution The complete solution is a combination of separate solutions for each half cycle. 1. assume initial conditions x ( t = 0) = x x ( t = 0) = 0 system starts going right to left (case 2) evaluate constants A 3 = x- N k A 4 = 0 x ( t ) = ( x- N k ) cos ( n t ) + N k only valid for 0 t n (since n = 2 n ) 2. at t = n , mass will be at the extreme left...
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This note was uploaded on 03/02/2010 for the course MECH 530.343 taught by Professor Sun during the Spring '08 term at Johns Hopkins.
- Spring '08