Solutions Manual for Unbalanced Forces and Spring Dynamics

Chapter 1 SOLUTIONS MANUAL unbalanced forces Radiator Shock absorber 111 a x downward deflection of point A xs resulting deformation of spring Potential energy equivalence gives frac12 keq x2 frac12 k xs2 keq k left fracxsx right2 But x 2 sqrta2 left fracb2 right2 2 sqrta2 fracb24 quad a2 left fracb2 xs right2 a2 left fracb2 right2 frac12 x2sqrta2 fracb24 Rightarrow left a2 fracb24 right2 frac b xs 2a2 fracb24 1 Using the relation 1 heta12 approx 1 frac heta2 we obtain x 2lefta2 fracb24 right12 left 1 fracb xs4a2 fracb2412 1 right fracb xs2a2 fracb2412 herefore keq k left fracxsx right2 4 k left fraca2 fracb24b2 right k left frac4 a2 b2b2 right b Here x xs spring deflection herefore keq k 112 Let x vertical displacement of mass M xs resulting deformation of each inclined spring 115 a F Fx0 fracd Fdxx0 x x0 500 x 2x3x10 approx 1100 x 4000 b at x 9 mm Exact Fg 500 x 9 2 93 5958 N Approximate Fg 1100 x 9 4000 5900 N Error 09735 c at x 11 mm Exact F11 500 x 11 2 113 8162 N Approximate F11 1100 x 11 4000 8100 N Error 07596 V A202679 368408 in3 Let h 2 inch π4 D2 2 V D 48429 inch F ax b x3 2 104 x 4 107 x3 Around x Fx Fx dFdx xx x When x 102 m Fx 2104102 4107106 240 N Hence Fx 240 32000x 001 32000 x 80 N Since the linearized spring constant is given by Fx keq x we have keq 32000 Nm For helical spring k G d4 64 n R3 Spring 1 k1 12 x 10624 64 1083 138889 lbin Spring 2 k2 4 x 10614 64 1053 5000 lbin a Spring 2 inside spring 1 parallel keq k1 k2 143889 lbin b Spring 2 on top of spring 1 series 1 keq 1 k1 1 k2 k2 k1 k1 k2 which gives keq 482625 lbin Assume small angles θ1 and θ2 θ2 p1 r2 θ1 x1 horizontal displacement of CG of mass m1 θ1 r1 x2 vertical displacement of CG of mass m2 θ2 r2 p1 θ1 r2 r2 y1 horizontal displacement of springs k1 and k2 θ1 r1 k1 y2 vertical displacement of springs k3 and k4 θ2 r2 p1 k2 θ1 r2 Equivalence of kinetic energies gives 12 Jeqθ12 12 J1θ12 12 J2θ22 m1x12 m2x22 Jeq J1 J2 p1 p22 m1 r12 m2 r22 p1 p22 Equivalence of potential energies gives 12 keq θ12 12 k12 y12 12 k34 y22 12 kt1 θ12 12 kt2 θ22 with k12 k1 k2 k34 k3 k4 keq k1 k2 θ1 θ2 k1 k2 k3 k4 k3 k4 p1 p22 kt1 kt2 1 p2 Let θ1 angular velocity of the motor input Angular velocities of different gear sets are Jmotor J1 J2 J3 J4 J5 θ2N Jload θ1 θ1 n1 n2 θ2 n1 n2 n3 n4 θ3 n1 n2 n3 n4 n2 n4 Equivalence of kinetic energies gives 12 Jeq θ12 12 Jmotor θ12 12 Σ2N κ1 Jk θk2 12 Jload θload2 Jeq Jmotor J1 J2 J3n1 n22 J4 J5n1 n22 J2N 1 Jload n1 n2 n3 n4 n2N 1 n2N c μ 3 π D³ ℓ 1 2 dD 4 d³ Assuming x Dd as the unknown with ℓ 2 in Eq 1 can be written as c μ 3 π ℓ x³ 4 1 2x or 1 4 10⁹3 π 2 4 1 2x This gives x³ 2 x² 5305152 0 Using a trial and error procedure the solution of this cubic equation can be found as x 3692 Using D 3 in we get d 33692 008128 in F a x² b x 02 x² Fx₀ Fx₀ dFdx x₀ x x₀ At x₀ 5 ms Fx₀ 5 5 02 25 30 N dFdx x₀ 5 04 x s 7 and hence Fx 30 7 x 5 7 x 5 x 5 2 i A e iθ A cos θ i A sin θ A cos θ 5 A sin θ 2 A A cos θ² A sin θ² 5² 2² 53852 θ tan¹25 218014 1 1 2 2 cos 2 ω t 2 ω t 2 φ cos 2 ω t 2 ω t 2 φ 2 xt A cos 50 t α m where A is in mm E1 xt Re 133802 e i3t 168 xt frac12 cos fracpi2 t cos pi t xt x1t x2t 3 sin 30t 3 sin 29t xt 2X sin left omega t fracdelta omega2 right cos left fracdelta omega2 t right xt xt odd function hence a0 an 0 bn 2T 0T xt sin nωt dt 2T A sin nωt dt A sin nωt dt 2AT cos nωtnω 0T 2AT cos nωtnω πT 2Anω 2 cos nπ cos 0 cos 2nπ xt 4Aπ n1 to 12n1 sin 2n1 ωt xt xt even function hence b0 2T 0T xt dt 2T AtT4 AtT34 AtT24 0 an 2T 0T xt cos nωt dt 2Anω2 sin nωt T4 sin nωt 3T24 sin nωt T2 0 4Anω2 sin nπ2 sin 2nπ sin 3nπ2 4Aπ n1 to 1n2n1 cos 2π2n1tT with ω 2πT xt 2A tT 0 t T2 2A tT 2A T2 t T a0 2T 0T xt dt 2T 0T2 2A tT dt T2T 2A tT 2A dt 2AT T24 3AT4 A A an 2T 0T xt cos nωt dt 2T 0T2 2A tT cos nωt dt T2T 2A tT 2A cos nωt dt 2AT tcos nωt nω2 tnT T2T2 2An2 ω2 cos nπ 1 0 bn 2AT 0T xt sin nωt dt 2AT 0T2 t sin nωt T2T t sin nωt dt 0 xt A2 4Aπ² 1n² cos nωt n135 xt 8Aπ 1n1 sin nωtn² n135 t x₁ x₂ x₃ x₄ x₅ x₆ x₇ Speed 100 rpm In a minute a point will be subjected to the maximum pressure A pmax 100 psi 100 x 4 400 times Hence period T 60 400 015 sec pt A 0 t T4 0 T4 t T a0 2 T 0T4 pt dt 2 T A t0 A 2 50 psi am 2 T 0T4 pt cos mωt dt 2 A T sin mωt T4 A m sin mπ2 bm 2 T 0T4 pt sin mωt dt 2 A T cos mωt 0 A cos mπ2 1 Evaluation of am and bm m1 m2 m3 a1 A π sin π2 A 318309 psi a2 A 2π sin π 0 0 psi a3 A 2π sin 3π2 106103 psi b1 A π cos π2 1 318309 psi b2 A 2π cos π 1 318309 psi b3 A 3π cos 3π2 1 106103 psi pt a0 2 Σm1am cos mωt bm sin mωt psi 1 00005 770 7437627 15929212 6668391 3850000 5447310 1 00010 810 7014802 4050000 4049988 10000 8100000 1 00015 850 6010398 6010417 00000 3500000 6104426 2 00020 910 4549798 7880485 4550001 7880000 9100000 5 00025 1010 2614043 9755859 874689 504995 714171 714184 6 00030 1170 00000 1700000 1700000 00000 1170000 3 00035 1370 345547 3423169 14216 6250000 968179 968725 1 00025 1010 2614043 9755859 874689 504995 714171 714184 5 00020 910 4549798 7880485 4550001 7880000 9100000 6 00030 1170 1700000 1700000 00000 1170000 2 00040 710 00000 1700000 1700000 00000 1170000 7 00045 770 7437627 15929212 6668391 3850000 5447310 8 00050 810 7014802 4050000 4049988 10000 8100000 Subsequent rows omitted for brevity TIME AT VARIOUS STATIONS TI 050000000E03 099999973E03 015000000E02 020000001E02 024999999E02 030000000E02 035000000E02 040000007E02 045000172E02 050000002E02 055000010E02 060000110E02 064999983E02 069999993E02 075000003E02 080000013E02 084999986E02 089999962E02 095000006E02 100000002E02 010499999E01 011000000E01 011500001E01 011999998E01 TIME AT VARIOUS STATIONS TI 020000000E01 039999999E01 059999999E01 079999983E01 100000002E00 010000002E01 012000000E01 013999999E00 016000003E00 018000001E00 019999979E00 022000003E00 024000001E00 025999997E00 030000000E00 031999999E00 TIME AT VARIOUS STATIONS TI 024999979E01 050000001E01 074999998E01 010000000E00 012500000E00 014999999E00 017500001E00 019999999E00 022500002E00 025000000E00 026000000E00 028000000E00 029000000E00 032000000E00 034000000E00 036000000E00 037000000E00 038000000E00 039000000E00 040000000E00 022000000E02 000000000E00 VALUES OF I AI AND BI ARE 1 020158B76E02 0235252B4E02 2 033099722E01 012246438E02 3 037719228E01 040640424E00 4 095843577E00 03247425E01 5 011377630E01 018716125E01 6 01167604E01 012500324E01 174 The main program and the output are given below C C MAIN PROGRAM FOR CALLING THE SUBROUTINE FORIER C C FOLLOWING 4 LINES NEED TO BE CHANGED FOR A DIFFERENT PROBLEM DIMENSION X14T14XSIN14XCOS14A10B10 DATA NMTIME 1410035 DATA X 45 8 6 75 7 55 175 25 11 14 105 00 DATA T 025 05 075 1 125 175 2 225 275 3 35 C END OF PROBLEMDEPENDENT DATA CALL FORIER NMTIMEXTAZEROABXSINXCOS PRINT 100 100 FORMAT 46H FOURIER SERIES EXPANSION OF THE FUNCTION XT 200 FORMAT 6H AZERO 37H NUMBER OF DATA POINTS IN ONE CYCLE 2 42H NUMBER OF FOURIER COEFFICIENTS REQUIRED 15 3 14H TIME PERIOD E158 PRINT 300 TII1N 300 FORMAT 33H TIME AT VARIOUS STATIONS TI 4E1581X PRINT 400 XII1N 400 FORMAT 31H KNOWN VALUES OF XI AT TI 4E1581X PRINT 500 FORMAT 29H RESULTS OF FOURIER ANALYSIS PRINT 600 AZERO 600 FORMAT 8H AZERO 2XE15B 31H VALUES OF I AI AND BI ARE 2 DO 700 I 1M 700 PRINT 800 I AI BI 800 FORMAT 152XE1582XE158 FOURIER SERIES EXPANSION OF THE FUNCTION XT DATA NUMBER OF DATA POINTS IN ONE CYCLE 14 NUMBER OF FOURIER COEFFICIENTS REQUIRED 10 TIME PERIOD 035000002E00 TIME AT VARIOUS STATIONS TI 02499999E01 05000001E01 07499998E01 010000002E00 01250000E00 014999998E00 01750001E00 01999999E00 02250000E00 02500000E00 02749999E00 03000001E00 KNOWN VALUES OF XI AT TI 04499999E00 08000000E00 08999999E00 06000002E00 07500000E00 06979999E00 05500001E00 01750000E00 01649995E01 02500000E00 01100004E01 01379999E01 RESULTS OF FOURIER ANALYSIS AZERO 037857169E00 VALUES OF I AI AND BI ARE 1 061951119E00 035046142E00 2 046422496E00 072408919E00 3 041494656E00 017127156E00 4 022273250E01 024691588E00 5 045431590E01 010430527E00 6 020434087E01 056009147E01 7 020436046E01 035784263E05 8 020434732E01 010429442E00 9 022727403E01 024690533E00 Unbalanced force developed P 2 m ω² r cos ω t range of force 0 100 N range of frequency 25 50 Hz 15708 31416 radsec Parameters to be determined m r ω Let r 01 m To generate 100 N force at 25 Hz set Pmax 100 2 m 15708² 01 which gives m 100 00202641 kg 202641 g 2 15708² 01 To generate 100 N force at 50 Hz set Pmax 100 2 m 31416² 01 which yields m 100 000506860 kg 50680 g 2 31416² 01 Goal Weight to be maintained at 10 01 lbmin Parameters to be determined Angular velocity of crank ω lengths of crank and connecting rod dimensions of the wedge dimensions of the orifice in the hopper dimensions of the actuating rod and dimensions of the lever arrangement Procedure Select ω based on available motor Determine the dimensions of the orifice in the hopper which delivers approximately 10 lbmin assuming continuous flow of material For trial dimensions of the wedge determine the increasedecrease in the size diameter of the orifice Choose the final dimensions of the wedge such that the material flow rate delivered by the orifice lies within the specified range Force to be applied 200 lb frequency 50 Hz 31416 radsec Procedure 1 Select a motor that provides either directly or through a gear system the desired frequency Assume that it is connected to the cam 2 Determine the sizes and dimensions of the plate cam and the roller 3 Choose the weight as 200 lb 4 Select the weight as 200 lb From the geometry determine the range of displacement vertical motion of the weight 5 Determine the force exerted due to the falling weight Considerations to be taken in the design of vibratory bowl feeders 1 Suitable design of the electromagnet and its coil 2 Radius of the bowl and the pitch of the spiral helical delivery track 3 Tooling to be fixed along the spiral track to reject the defective or outoftolerance or incorrectly oriented parts 4 Design of elastic supports 5 Size and location of the outlet Chapter 2 Free Vibration of Single Degree of Freedom Systems 21 st 5 x 103 m omegan left fracgst right12 left frac9815 x 103 right12 442945 radsec 70497 Hz 22 tn 021 sec 2 pi sqrtfracmk sqrtm 021 sqrtfrack2 pi i tnnew frac2 pi sqrtmsqrtknew frac2 pi sqrtm15 k 2 pi left frac021 sqrtk2 pi right1 01715 sec ii tnnew frac2 pi sqrtmsqrt05 k 2 pi left frac021 sqrtk2 pi right105 02970 sec 23 omegan 62832 radsec sqrtm sqrtfrackm sqrtm sqrt62832 when spring constant is reduced omegan decreases omegannew 055 omegan 345576 radsec sqrtfrack800k fracxk800k 055 frack 800k 24 k 100 left frac1001000 right 10000 Nm omegan sqrtfrackeqm sqrtfrac4 k10 632456 radsec 25 m 2000 3864 Let omegan 75 radsec omegan sqrtfrackeqm keq m omegan2 frac20003864 752 2911491 lbin 4 k where k is the stiffness of the air spring Thus k 2911491 4 727873 lbin 26 x A cos omegan t phi dot x omegan A sin omegan t phi ddot x omegan2 A cos omegan t phi a omegan A 01 gn 003183 m d x0 xt0 A cosphi 002 m 06283 b dot x0 dot xt0 omegan A sinphi 01 sin510724 007779 msec c ddot xmax omegan2 A 314162 003183 0314151 msec2 27 frac1keq frac1k12eq frac1k3 keq frack12eqk12eq k3 k1fracl2l32 k2fracl1l32k3 Let x₁ x₂ displacements of pulleys 1 2 x 2 x₁ 2 x₂ E₁ Let P tension in rope For equilibrium of pulley 1 2 P k₁ x₁ E₂ For equilibrium of pulley 2 2 P k₂ x₂ E₃ where 1k₁ 14k 14k k₁ 2k and k₂ k₁ k₂ 2k Combining Eqns E₁ to E₃ x 2 x₁ 2 x₂ 2 2 Pk₁ 2 2 Pk₂ 4 P12k 12k 4 Pk Let kₑ₉ equivalent spring constant of the system kₑ₉ Px kx Equation of motion of mass m m x kₑ₉ x 0 ωₙ kₑ₉m kx4m For a displacement of x of mass m pulleys 1 2 and 3 undergo displacements of 2x 4x and 8x respectively The equation of motion of mass m can be written as m x F₀ 0 1 where F₀ 2 F₁ 4 F₂ 8 F₃ as shown in figure Since F₃ 8x k Eq 1 can be rewritten as m x 8 F₃ 8 8k 0 from which we can find ωₕ 64 km 8 km a ωn 4kM b ωh 4kM m Initial conditions velocity of falling mass m v 2gℓ x₀ xt0 weightkeq mg4k Conservation of momentum M mx₀ mv or x₀ xt0 m2gℓM m ωₙ kₑ₉m k₁ k₂ l₂² k₂ k₃ l₂²m k₁ l₁² k₂ l₂² k₃ l₃² k₁ k₂ l₁² k₂ k₃ l₂²m m 2000 kg δₛₜ 002 m ωₙ g δₛₜ12 98100212 221472 radsec xt A₀ sinωh t φ₀ where A₀ x₀² x₀ωh² m²g²16k² m²v²2kM m and φ₀ tan¹x₀ωhx₀ tan¹mg4k4k Let x be measured from the position of mass at which the springs are unstretched Equation of motion is m x k₁ x δₛₜ k₂ x δₛₜ W sin θ E₁ where δₛₜ k₁ k₂ W sin θ Thus Eq E₁ becomes m x k₁ k₂ x 0 ωₙ k₁ k₂m k 62832 m1 125664 m m1 2 m m 43 kg κ 125664² m 526381 Nm Thus keq can be expressed as keq k1 k2 tan² θ Equation of motion m x keq x 0 Natural frequency ωh keqm k1 k2 gW tan θ 223 a Neglect masses of rigid links Let x displacement of W Springs are in series keq k2 Equation of motion m x keq x 0 Natual frequency ωh keqm k2m b Under a displacement of x of mass each spring will be compressed by an amount xs x 2b l² b²4 Equivalent spring constant 12 keq x² 2 12 k x² or keq 2 k xsx² 2 k 4b²l² b²4 Equation of motion m x keq x 0 Natural frequency ωh keqm 8 kb² m l² b²4 F1 F3 k1 x cos 45 F2 F4 k2 x cos 135 F force along x F1 cos 45 F2 cos 135 F3 cos 45 F4 cos 135 2 x k1 cos² 45 k2 cos² 135 keq Fx 2k12 k22 k1 k2 Equation of motion m x k1 k2 x 0 225 Let αi denote the angle made by ith spring with respect to xaxis Let x displacement of mass along the direction defined by θ If keq equivalent spring constant the equivalence of potential energy gives 12 keq x² 12 Σkix cosθ αi² Σ ki cos θ cos αi sin θ sin αi² keq Σ ki cos² αi cos² θ sin² αi sin² θ 2 Σ ki cos αi sin αi cos θ sin θ Natural frequency ωh keqm For ωh to be independent of θ and Σ ki cos αi sin αi 0 In the present example E3 and E4 become k1 cos 60 k2 cos 240 k3 cos 2α3 k1 cos 420 k3 cos 360 2α3 0 k1 sin 60 k2 sin 240 k3 sin 2α3 k1 sin 420 k3 sin 360 2α3 0 ie k1 k2 2 k3 sin 2α3 0 Squaring E5 and E6 and adding 4 k3² k2 k1² 1 3 k3 12 k2 k1 k3 k2 k1 Dividing E6 by E5 tan 2α3 3 α3 12 tan¹3 30 226 T1 xa T T2 xb T a m x T1 T2 0 m x Ta Tb 0 Equation of motion m x 1a 1b 0 Natural frequency ωh Tm a b Tmab a b 227 m 160 lbsec²inch k 10 lbinch Velocity of jumper as he falls through 200 ft m g h 12 m v² or v 2 g h 3864 200 12 13618811 insec About static equilibrium position x0 xt0 0 x₀ ẋt0 13618811 insec Response of jumper xt A0 sinωh t φ0 The natural frequency of a vibrating rope is given by see Problem 226 ωh Tabmab where T tension in rope m mass and a and b are lengths of the rope on both sides of the mass For the given data 10 T 80160120 386412 T 0060375 which yields T 1000060375 16563147 lb when ω0 0 total vertical height 2l h when ω 0 total vertical height 2l cos θ k spring force k2l k 2l cos θ k 2k1 cos θ For vertical equilibrium of mass m mg T2 cos θ T1 cos θ E1 For horizontal equilibrium Fc T1T2 sin θ T2 Fc T1 sin θsin θ E2 From E2 E1 can be expressed as mg Fc T1 sin θ cos θ T1 cos θ ie T1 mg Fc cot θ mg m ω² l cos θ2 cos θ 2 T2 Fc T1 sin θsin θ m x ω² l mg tan θ2 sin θ mg2 cos θ Spring force 2k l 1 cos θ 2 T2 cos θ m l ω² cos θ mg 50 a2 00 k 0S h² a²2 when each wire stretches by xₑ let the resulting vertical displacement of the platform be x 0S xₑ hxₑ² a²2 xₑ h a²2 h xₑ² a²2 h a²2 1 h a²2 1 2hx x²h² a²2 1 For small x x² is negligible compared to 2hx and 1 θ 1 θ2 and hence xₑ h a²21 hh² a²2 1 hh² a²2x Potential energy equivalence gives 12 kₑq x² 4 k k²h² a²2 Equation of motion of M M x kₑq x 0 υn 2πωn kₑqM 2πkₑqM π2k² a²2k12 Assume same area of cross section for all segments of the cable Speed of blades 300 rpm 5 Hz 31416 radsec ω² keq keq mω² 250 62832² 988965 10⁴ Nm 1 AD 05² 05² 07071 m OD 2² 07071² 21213 m Stiffness of cable segments kpo A E ρpo A 207 10⁹ 1 207 10⁹ A Nm kOD A E ρOD A 207 10⁹ 21213 975817 10⁹ A Nm The total stiffness of the four inclined cables kic is given by kic 4 kOD cos² θ 4 975817 10⁹ A cos² 1947100 3469581 10⁹ A Nm Equivalent stiffness of vertical and inclined cables is given by 1 keq 1 kpo 1 kic ie keq kpo kic 207 10⁹ A 3469581 10⁹ A 1296494 10⁹ A Nm 2 Equating keq given by Eqs 1 and 2 we obtain the area of cross section of cables as A 988965 10⁴ 1296494 10⁹ 76126 10⁶ m² 1 2 π k1 1 2 5 k1 4 π² 2² 9869651 1 2 π k1 m 1 2 40825 k1 4 π² 168688 6579822 Using k1 A E ℓ1 we obtain k1 m A E ℓ1 m A 207 10⁹ 2 m ie A 95359 10⁹ m Also k1 m 5000 A E ℓ1 m 5000 ie A 63573 10⁹ m Using Eqs 1 and 2 we obtain A 95359 10⁹ m 63573 10⁹ m 317865 10⁶ ie 31786 10⁹ m 317865 10⁶ ie m 100001573 kg A 95359 10⁹ m 95359 10⁹ 100001573 09536 10⁴ m² transverse vibration spring constant of a fixedfixed beam with offcenter load k 3EI a³ b³ 3EI a³ la³ ωn km 3EI w a² la 12 with I πd⁴ 64 moment of inertia torsional vibration If flywheel is given an angular deflection θ resisting torques offered by lengths a and b are GJθ a and GJθ b Total resisting torque Mt GJ 1a 1b kt Mt θ GJ 1a 1b where J πd⁴ 32 polar moment of inertia where Jo mass polar moment of inertia of the flywheel meq equivalent mass of a uniform beam at the free end see Problem 238 33 140 m 33140 1 1 150 x 12 0283 3864 03107 Stiffness of tower beam at free end kb 3 E I L³ 3 30 x 10⁶ 112 1 1³ 0001286 lbin Length of each cable OA 2 14142 ft OB 2 15 212132 ft AB OB OA 197990 ft TB TA² AB² 100² 197990² 1019412 ft tan θ AT AB 100 197990 50508 θ 788008 Axial stiffness of each cable k A E ℓ 05 30 x 10⁶ 1019412 x 12 12261971 lbin Axial extension of each cable yc due to a horizontal displacement of x of tower l12 l2 x2 2 l x cos 180circ heta l2 x2 2 l x cos heta or l1 l sqrt1 fracx2l2 2 fracxl cos heta Weight of sign 0283frac1874512 2664 extlb kt 533 fraca b3l3 Gleft1 063 fracbarightleft1 fracb412 a4right Yx a₁ a₂x a₃x² a₄x³ mu coefficient of friction x displacement of CG of block Since the joint between column and floor does not permit rotation each column will bend with inflection point at middle 242 From problem 241 Restoring force without springs μ F2 F1 μ W x cμa spring restoring force 2 k x Total restoring force μ W x cμa 2 k x Equation of motion W g x μ W c cμa 2 k x 0 ω n ω s μ W 2 k cμa g 12 cμa W Solution of this equation gives μ ag 2 W c 2 k q c W g W ω s 2 a 2 k g a 243 a Natural frequency of vibration of electromagnet without the automobile ω a kM 100000 386430000 358887 radsec b When the automobile is dropped the electromagnet moves up by a distance x0 from its static equilibrium position x0 static deflection elongation of cable under the weight of automobile W auto 2000 02 in k 10000 x0 0 Resultant motion of electromagnet x upwards xt A0 sin ω n t φ0 where A0 x 20 x0ω n 2 12 x0 02 and φ0 tan1 x0 ω nx0 tan1 90 Hence xt 02 sin 358887 t 90 02 cos 358887 t c Maximum xt xt max A0 02 in Maximum tension in cable during motion k xt max 10000 02 3000 5000 lb 244 a Newtons second law of motion Ft k1 x k2 x m x or m x k1 k2 x 0 b DAlemberts principle Ft m x 0 or k1 x k2 x m x 0 Thus m x k1 k2 x 0 c Principle of virtual work When mass m is given a virtual displacement δx Virtual work done by the spring forces k1 k2 x δx Virtual work done by the inertia force m x δx According to the principle of virtual work the total virtual work done by all forces must be equal to zero m x δx k1 k2 x δx 0 or m x k1 k2 x 0 d Principle of conservation of energy T kinetic energy 12 m x2 U strain energy potential energy 12 k1 x2 12 k2 x2 T U 12 m x2 12 k1 k2 x2 c constant ddt T U 0 or m x k1 k2 x 0 245 Equation of motion Mass m m g T m x Pulley J0 J0 θ T k4 r θ φ0 4 r where θ0 angular deflection of the pulley under the weight mg given by mg r k 4 r θ0 4 r or θ0 mg16 r k Substituting Eqs 1 and 3 into 2 we obtain J0 θ m g m x r k 16 r2 θ mg16 r k Using x r θ and x r θ Eq 4 becomes J0 m r2 θ 16 r2 k θ 0 246 Consider the springs connected to the pulleys by rope to be in series Then 1keq 1k 15k or keq 56 k Let the displacement of mass m be x Then the extension of the rope springs connected to the pulleys 2 x From the free body diagram the equation of motion of mass m m x 2 k x keq 2 x 0 or m x 113 k x 0 247 T kinetic energy Tmass Tpulley 12 m x2 12 J0 θ2 12 m r2 J0 θ2 U potential energy 12 k x2 12 k 4 r θ2 12 k 16 r2 θ2 Using ddt T U 0 gives m r2 J0 θ 16 r2 k θ 0 248 T kinetic energy 12 m x2 12 J0 θ2 U potential energy 12 k x2 where θ x r x g extension of spring 4 r θ 4 x Hence T 1 2 m J 0 r 2 x 2 U 1 2 18 k x 2 Using tt yx P a 4 E I ℓ x 2 ℓ x 2 2 ℓ 3 a 1x ℓ 3 ℓ x ℓ a Using a 02 ℓ 08 x a ℓ 10 and I m mass of a panel 5 x 12 3 x 12 1 0283 3864 15820 15820 12 12 36 2 1709878 I 0 polar moment of i I₀ polar moment of inertia of cross section of shaft AB π d⁴ 32 π 32 1⁴ 0098175 in⁴