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EngineeringMechanics Dynamics 14th Edition Hibbeler Solutions Manual 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they -solutions-manual/No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.Full Download:22–1.A spring is stretched 175 mm by an 8-kg block. If the block isdisplaced 100 mm downward from its equilibrium positionand given a downward velocity of 1.50 m s, determine thedifferential equation which describes the motion. Assumethat positive displacement is downward. Also, determine theposition of the block when t 0.22 s.SOLUTION mg - k(y yst) my T ΣFy may;where kyst mgk y y 0mHencep kBmBWhere k 8(9.81)0.175448.46 7.4878 y (7.487)2y 06 448.46 N m y 56.1y 0Ans.The solution of the above differential equation is of the form:y A sin pt B cos pt(1)#v y Ap cos pt - Bp sin pt(2)At t 0, y 0.1 m and v v0 1.50 m sFrom Eq. (1)0.1 A sin 0 B cos 0B 0.1 mv01.50 0.2003 mp7.487From Eq. (2)v0 Ap cos 0 - 0Hencey 0.2003 sin 7.487t 0.1 cos 7.487tAt t 0.22 s,y 0.2003 sin [7.487(0.22)] 0.1 cos [7.487(0.22)]A Ans. 0.192 m1190Ans: y 56.1 y 0y 0 t 0.22 s 0.192 mFull download all chapters instantly please go to Solutions Manual, Test Bank site: testbanklive.com

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–2.A spring has a stiffness of 800 N m. If a 2-kg block isattached to the spring, pushed 50 mm above its equilibriumposition, and released from rest, determine the equationthat describes the block’s motion. Assume that positivedisplacement is downward.SOLUTIONp k800 20AmA 2x A sin pt B cos ptx - 0.05 mwhen t 0,-0.05 0 B;B -0.05v Ap cos pt - Bp sin ptv 0 when t 0,0 A(20) - 0;A 0Thus,Ans.x - 0.05 cos (20t)Ans:x -0.05 cos (20t)1191

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–3.A spring is stretched 200 mm by a 15-kg block. If the blockis displaced 100 mm downward from its equilibriumposition and given a downward velocity of 0.75 m s,determine the equation which describes the motion. What isthe phase angle? Assume that positive displacement isdownward.SOLUTIONk 15(9.81)F 735.75 N m y0.2vn k735.75 7.00AmA 15y A sin vn t B cos vn ty 0.1 m when t 0,0.1 0 B;B 0.1v A vn cos vn t - Bvn sin vn tv 0.75 m s when t 0,0.75 A(7.00)A 0.107Ans.y 0.107 sin (7.00t) 0.100 cos (7.00t)f tan - 1 aB0.100b tan - 1 ab 43.0 A0.107Ans.Ans:y 0.107 sin (7.00t) 0.100 cos (7.00t)f 43.0 1192

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–4.When a 20-lb weight is suspended from a spring, the springis stretched a distance of 4 in. Determine the naturalfrequency and the period of vibration for a 10-lb weightattached to the same spring.SOLUTIONk vn t 20412 60 lb ftk60 13.90 rad s10AmA 32.2Ans.2p 0.452 svnAns.Ans:vn 13.90 rad st 0.452 s1193

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–5.When a 3-kg block is suspended from a spring, the spring isstretched a distance of 60 mm. Determine the naturalfrequency and the period of vibration for a 0.2-kg blockattached to the same spring.SOLUTIONk vn 3(9.81)F 490.5 N m x0.060k490.5 49.52 49.5 rad sAmA 0.2f vn49.52 7.88 Hz2p2pt 11 0.127 sf7.88Ans.Ans.Ans:vn 49.5 rad st 0.127 s1194

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–6.An 8-kg block is suspended from a spring having a stiffnessk 80 N m. If the block is given an upward velocity of0.4 m s when it is 90 mm above its equilibrium position,determine the equation which describes the motion and themaximum upward displacement of the block measuredfrom the equilibrium position. Assume that positivedisplacement is measured downward.SOLUTIONvn k80 3.162 rad sAmA8y - 0.4 m s,x - 0.09 m at t 0x A sin vn t B cos vn t- 0.09 0 BB -0.09y Avn cos vn t - Bvn sin vn t-0.4 A(3.162) - 0A -0.126Thus,x - 0.126 sin (3.16t) - 0.09 cos (3.16t) mAns.C 2A2 B2 2( -0.126)2 ( - 0.09) 0.155 mAns.Ans:x 5 - 0.126 sin (3.16t) - 0.09 cos (3.16t) 6 mC 0.155 m1195

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–7.A 2-lb weight is suspended from a spring having a stiffnessk 2 lb in. If the weight is pushed 1 in. upward from itsequilibrium position and then released from rest, determinethe equation which describes the motion. What is theamplitude and the natural frequency of the vibration?SOLUTIONk 2(12) 24 lb ftvn k24 19.66 19.7 rad s2AmA 32.2y -1,12Ans.y 0 at t 0From Eqs. 22–3 and 22–4,-1 0 B12B -0.08330 Avn 0A 0C 2A2 B2 0.0833 ft 1 in.Ans.y (0.0833 cos 19.7t) ftAns.Position equation,Ans:vn 19.7 rad sC 1 in.y (0.0833 cos 19.7t) ft1196

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–8.A 6-lb weight is suspended from a spring having a stiffnessk 3 lb in. If the weight is given an upward velocity of20 ft s when it is 2 in. above its equilibrium position,determine the equation which describes the motion and themaximum upward displacement of the weight, measuredfrom the equilibrium position. Assume positive displacementis downward.SOLUTIONk 3(12) 36 lb ftvn k36 13.90 rad sAmA 632.2t 0,y - 20 ft s,1y - ft6From Eq. 22–3,-1 0 B6B -0.167From Eq. 22–4,-20 A(13.90) 0A - 1.44Thus,Ans.y [-1.44 sin (13.9t) - 0.167 cos (13.9t)] ftFrom Eq. 22–10,C 2A2 B2 2(1.44)2 (- 0.167)2 1.45 ftAns.Ans:y [-1.44 sin (13.9t) - 0.167 cos (13.9t)] ftC 1.45 ft1197

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–9.A 3-kg block is suspended from a spring having a stiffnessof k 200 N m. If the block is pushed 50 mm upwardfrom its equilibrium position and then released from rest,determine the equation that describes the motion. What arethe amplitude and the frequency of the vibration? Assumethat positive displacement is downward.SOLUTIONvn k200 8.16 rad sAmA 3Ans.x A sin vn t B cos vn tx -0.05 m when t 0,- 0.05 0 B;B -0.05v Ap cos vn t - Bvn sin vn tv 0 when t 0,0 A(8.165) - 0;A 0Hence,x - 0.05 cos (8.16t)Ans.C 2A2 B2 2(0)2 ( - 0.05) 0.05 m 50 mmAns.Ans:vn 8.16 rad sx - 0.05 cos (8.16t)C 50 mm1198

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–10.The uniform rod of mass m is supported by a pin at A and aspring at B. If B is given a small sideward displacement andreleased, determine the natural period of vibration.ALSolution1Equation of Motion. The mass moment of inertia of the rod about A is IA mL2.3Referring to the FBD. of the rod, Fig. a,L1a ΣMA IAa ;-mg a sin u b - (kx cos u)(L) a mL2 ba23BkHowever; x L sin u. Then- mgL1sin u - kL2 sin u cos u mL2a23Using the trigonometry identity sin 2u 2 sin u cos u,- mgLKL21sin u sin 2u mL2a223 Here since u is small sin u u and sin 2u 2u. Also a u . Then the aboveequation becomes mgL1mL2 u a kL2 bu 032 3mg 6kLu u 02mLComparing to that of the Standard form, vn t A3mg 6kL. Then2mL2p2mL 2p vnA 3mg 6kLAns.Ans:t 2p11992mLA 3mg 6kL

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–11.While standing in an elevator, the man holds a pendulumwhich consists of an 18-in. cord and a 0.5-lb bob. If the elevatoris descending with an acceleration a 4 ft s2, determine thenatural period of vibration for small amplitudes of swing.a4 ft/s2SOLUTIONSince the acceleration of the pendulum is (32.2 - 4) 28.2 ft s2Using the result of Example 22–1,We havevn t g28.2 4.336 rad sAlA 18 122p2p 1.45 s vn4.336Ans.Ans:t 1.45 s1200

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–12.Determine the natural period of vibration of the uniformbar of mass m when it is displaced downward slightly andreleased.OkL—2L—2SolutionEquation of Motion. The mass moment of inertia of the bar about O is I0 1mL2.12Referring to the FBD of the rod, Fig. a,a ΣM0 I0a ;However, y L1- ky cos ua b a mL2 ba212Lsin u. Then2-ka1LLsin u b cos u a b mL2a2212Using the trigonometry identity sin 2u 2 sin u cos u, we obtain1kL2mL2a sin 2u 0128 Here since u is small, sin 2u 2u. Also, a u . Then the above equation becomes 1kL2mL2 u u 0124 3ku 0u mComparing to that of the Standard form, vn t 3k. ThenAmm2p 2p vnA 3kAns.Ans:t 2p1201mA 3k

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–13.The body of arbitrary shape has a mass m, mass center at G,and a radius of gyration about G of kG. If it is displaced aslight amount u from its equilibrium position and released,determine the natural period of vibration.OduGSOLUTIONa MO IO a; -mgd sin u C mk2G md2 D u u gdk2G d2sin u 0However, for small rotation sin u Lu. Hence u gd2kG d2u 0From the above differential equation, vn t 2p vn2pgdA k2G d2gdB k2G 2p d2.k2G d2C gdAns.Ans:t 2p1202k 2G d 2C gd

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–14.The 20-lb rectangular plate has a natural period of vibrationt 0.3 s, as it oscillates around the axis of rod AB.Determine the torsional stiffness k, measured in lb # ft rad,of the rod. Neglect the mass of the rod.AkBSolutionT kuΣMz Iza ;- ku u k(4.83)u 0t 2p2k(4.83)4 ft 1 20ab(2)2 u12 32.22 ft 0.3k 90.8 lb # ft rad Ans.Ans:k 90.8 lb # ft rad1203

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.22–15.A platform, having an unknown mass, is supported by foursprings, each having the same stiffness k. When nothing ison the platform, the period of vertical vibration is measuredas 2.35 s; whereas if a 3-kg block is supported on theplatform, the period of vertical vibration is 5.23 s. Determinethe mass of a block placed on the (empty) platform whichcauses the platform to vibrate vertically with a period of5.62 s. What is the stiffness k of each of the springs?kkSolution T ΣFy may; mtg - 4k(y yts) mty Where 4k yts mtg4k y y 0mtHence P t 4kA mt2pmt 2pPA 4kFor empty platform mt mP, where mP is the mass of the platform.2.35 2pmP A 4k(1)When 3-kg block is on the platform mt mP 3.5.23 2pmP 3 A 4k(2)When an unknown mass is on the platform mt mP mB.5.62 2pmP mB A4k(3)Solving Eqs. (1) to (3) yields :k 1.36 N m mB 3.58 kg Ans.mP 0.7589 kgAns:k 1.36 N mmB 3.58 kg1204

2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.*22–16.A block of mass m is suspended from two springs having astiffness of k1 and k2, arranged a) parallel to each other, andb) a

Solution Equation of Motion. The mass moment of inertia of the rod about A is I A 1 3 mL2. Referring to the FBD. of the rod, Fig. a, a ΣM A I Aa ; -mga L 2 sin ub - (kx cos u)(L) a 1 3 mL2ba However; x L sin u. Then-mgL 2 sin 2u - kL sin u cos u 1 3 mL2a Using the trigonometry identity sin 2u 2 sin u cos u,-mg L 2