# SolSec 2.7 - 2-44 Problems and Solutions Section 2.7(2.63...

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2- 44 Problems and Solutions Section 2.7 (2.63 through 2.79) 2.63 Consider a spring-mass sliding along a surface providing Coulomb friction, with stiffness 1.2 × 10 4 N/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 = 1.2x 10 4 N/m, F o = 50 N, ω =10(2 ! ) = 20 ! rad/s ω = k m = 34.64 rad/s for lubricated steel, μ = 0.07 From Equation (2.109) X = F o k 1 ! 4 μ mg " ( F o ) # \$ % % & ( ( 2 (1 ! r 2 ) X = 50 1.2 ! 10 4 1 " 4(.07)(10)(9.81) # (50) \$ % & ( ) 2 (1 " 20 # 34.64 * + , - . / 2 ) X =1.79 × 10 ! 3 m 2.64 A 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, ω n = k m = 14.14 rad/s, X = 5 cm Equation (2.108) X = F 0 k (1 ! r 2 ) 2 + 4 μ mg " kX # \$ % & ( 2 ) F 0 = Xk (1 ! r 2 ) 2 + 4 μ mg " kX # \$ % & ( 2 F 0 = (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 2.65 A system of mass 10 kg and stiffness 1.5 × 10 4 N/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 = 1.5x 10 4 N/m, F 0 = 90 N, ω = 25(2 ! ) = 50 ! rad/s, ω n = k m = 38.73 rad/s, μ = 0.1 Steady-state Amplitude using Equation (2.109) is X = F 0 k 1 ! 4 μ mg " ( F o ) # \$ % & ( 2 (1 ! r 2 ) = 90 1.5 ) 10 4 1 ! 4(0.1)(10)(9.81) " (90) # \$ % & ( 2 1 ! 50 " 38.73 * + , - . / 2 = 3.85 ) 10 ! 4 m From equation (2.105), the equivalent Viscous Damping Coefficient becomes: c eq = 4 μ mg !" X = 4(0.1)(10)(9.81) ! (50 ! )(3.85 # 10 \$ 4 ) = 206.7 Ns/m