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2-44 Problems and Solutions Section 2.7 (2.63 through 2.79) 2.63Consider a spring-mass 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 steady-state 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 spring-mass 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=F0k(1!r2)2+4μmg"kX#$%&’(2)F0=Xk(1!r2)2+4μmg"kX#$%&’(2F0=(0.05)(2000)1!20"14.14#$%&’(2)*++,-..2+4(0.1)(10)(9.81)"(2000)(.05))*+,-.2=1874 N
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2-45 2.65A system of mass 10 kg and stiffness 1.5 ×104N/m is subject to Coulomb damping. If the mass is driven harmonically by a 90-N force at 25 Hz, determine the equivalent viscous damping coefficient if the coefficient of friction is 0.1. Solution:Given: m= 10 kg, k= 1.5x104N/m, F0= 90 N, ω= 25(2!) = 50!rad/s, ωn= km= 38.73 rad/s, μ= 0.1 Steady-state Amplitude using Equation (2.109) is X=F0k1!4μmg"(Fo)#$%&’(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/m