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Unformatted text preview: 244Problems and Solutions Section 2.7 (2.63 through 2.79) 2.63Consider a springmass sliding along a surface providing Coulomb friction, with stiffness 1.2 ×104N/m and mass 10 kg, driven harmonically by a force of 50 N at 10 Hz. Calculate the approximate amplitude of steadystate motion assuming that both the mass and the surface that it slides on, are made of lubricated steel. Solution: Given: m= 10 kg, k= 1.2x104N/m, Fo= 50 N, ω=10(2!) = 20!rad/s ω= km= 34.64 rad/s for lubricated steel, μ= 0.07 From Equation (2.109) X=Fok1!4μmg"(Fo)#$%%&’((2(1!r2)X=501.2!1041"4(.07)(10)(9.81)#(50)$%&’()2(1"20#34.64*+,./2)X=1.79 ×10!3m 2.64A springmass system with Coulomb damping of 10 kg, stiffness of 2000 N/m, and coefficient of friction of 0.1 is driven harmonically at 10 Hz. The amplitude at steady state is 5 cm. Calculate the magnitude of the driving force. Solution: Given: m= 10 kg, k= 2000 N/m, μ= 0.1, ω=10(2!) = 10(2!) = 20!rad/s, ωn= km= 14.14 rad/s, X= 5 cm Equation (2.108) X=Fk(1!r2)2+4μmg"kX#$%&’(2)F=Xk(1!r2)2+4μmg"kX#$%&’(2F=(0.05)(2000)1!20"14.14#$%&’(2)*++,..2+4(0.1)(10)(9.81)"(2000)(.05))*+,.2=1874 N2452.65A system of mass 10 kg and stiffness 1.5 ×104N/m is subject to Coulomb damping. If the mass is driven harmonically by a 90N force at 25 Hz, determine the equivalent viscous damping coefficient if the coefficient of friction is 0.1. Solution:Given: m= 10 kg, k= 1.5x104N/m, F= 90 N, ω= 25(2!) = 50!rad/s, ωn= km= 38.73 rad/s, μ= 0.1 Steadystate Amplitude using Equation (2.109) is X=Fk1!4μmg"(Fo)#$%&’(2(1!r2)=901.5)1041!4(0.1)(10)(9.81)"(90)#$%&’(21!50"38.73*+,./2=3.85)10!4mFrom equation (2.105), the equivalent Viscous Damping Coefficient becomes: ceq=4μmg!"X=4(0.1)(10)(9.81)!(50!)(3.85#10$4)= 206.7 Ns/m2462.66a. Plot the free response of the system of Problem 2.65 to initial conditions of x(0) = 0 and !x(0) = F/m = 9 m/s using thesolution in Section 1.10.b.Use the equivalent viscous damping coefficient calculated in Problem 2.65 and plot the free response of the “equivalent” viscously damped system to the same initial conditions. Solution: See Problem 2.65 (a) x(0) = 0 and !x(0) = Fom= 9 m/s !=km=1.5x10410=38.73 rad/s From section 1.10: m!!x+kx=μmgfor !x<m!!x+kx=!μmgfor !x>Let Fd=μmg= (0.1)(10)(9.81) = 9.81 N To start, !x(0)=!nB1=9Therefore, A1=Fdkand B1=9!nSo, x(t) = Fdkcos!nt+9!sin!nt"FdkThis will continue until !x= 0, which occurs at timet1: x(t) = A2cos!nt+B2sin!nt+Fdk!x(t) = !"nA2sin"nt+"nB2cos"ntx(t1)=A2cos!nt1+B2sin!nt1+Fdk!x(t1)==!"nA2sin"nt1+"nB2cos"nt1Therefore, A2=x(t1)!Fd/k( )cos"nt1and B2=x(t1)!Fd/k( )sin"nt1So, x(t)=x(t1)!Fd/k( )cos"nt1#$%&cos"nt+x(t1)!Fd/k( )sin"nt1#$%&sin"nt+FdkAgain, when !...
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This note was uploaded on 03/31/2009 for the course MECHANICAL MAE351 taught by Professor J.g.lee during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 J.G.Lee

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