Engineering Mechanics Statics and Dynamics 14th Edition Hibbeler SOLUTIONS MANUAL People also search: e.com/download/engineering-m wnload/engineering-mechanics-statics-and-dynam echanics-statics-and-dynamicsics14th-edition-hibbeler-solutions-manual/ 22 – 1. A spring is stretched 175 mm by an 8-kg block. If the block is displaced 100 mm downward from its equilibr ium position and given a downward downward velocity of 1.50 m s, determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t = 0.22 s.

• Engineering Statics 13th Edition Pdf
• Statics And Dynamics Hibbeler 13th Edition Solutions Manual Pdf
• Statics And Dynamics 13th Edition Solution Manual

Engineering Statics 13th Edition Pdf

SO LUT ION + T Σ F = ma; y mg k ( y y + y ) = - y st $ k y + Hence p = = k $ my m = mg st y = 0 Where k = Bm where ky 8(9.81) = 448.46 N m 0.175 448.46 = 7.487 B 8 $ $ y + (7.487)2 y = 0 6 y + 56.1 y = 0 Ans. The solution of the above differen tial equatio equation n is of the form: y = A sin pt + B cos pt # v = y = Ap cos pt (1) - Bp sin pt (2) At t = 0, y = 0.1 m and v = v0 = 1.50 m s From Eq.

(1) 0.1 = A sin 0 + B cos 0 v0 1.50 = = 0.2003 m p 7.487 From Eq. (2) v0 = Ap cos 0 Hence y = 0.2003 sin 7.487t + 0.1 cos 7.487t At t = 0.22 s, y = 0.2003 sin 7.487(0.22) + 0.1 cos 7.487(0.22) = 0.192 m - 0 B = 0.1 m A = Ans. Ans: $ y + 56.1 y = 0 y 0 t = 0.22 s = 0.192 m 1190 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently publisher. No portion of this material may be reproduced, in any form or by any means, without with out permis permission sion in i n writing writing from the publish 22 – 2. A spring has a stiffness of 800 N m.

If a 2-kg block is br ium attached to the spring, pushed 50 mm above its equili br position, and released from rest, determine the equation that describes the block ’s motion. Assume that positive displacement is downward. SO LUT ION k p = Am = 800 = 20 A 2 x = A sin pt + B cos pt x = - 0.05 - 0.05 m when t = 0, = 0 + B; v = Ap cos pt - Bp B = - 0.05 sin pt v = 0 when t = 0, 0 = A(20) - 0; A = 0 Thus, x = - 0.05 cos (20t ) Ans.

Mechanics dynamics 13th edition by hibbeler, rc hibbeler engineering mechanics dynamics solution manual 13th edition, engineering mechanics statics. Access Engineering Mechanics 13th Edition solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality!

Ans: x = - 0.05 cos (20t ) 1191 © 2016 Pearson Education, Inc., Upper S Saddle addle River, NJ. All rights reserved.

This Thi s material is i s protected under all copyright laws as they currently publisher. No portion of this material may be reproduced, in any form form or by any means, without permission in writing from the publish 22 – 3.

A spring is stretched 200 mm by a 15-kg block. If the block is displaced 100 mm downward from its it s equili br br ium position and given a downward velocity of 0.75 m s, determine the equation which describes the motion.

What is the phase angle? Assume that positive displacement is downward. SOLUTION k = F y = 15(9.81) = 735.75 N m 0.2 k vn = Am = 735.75 = 7.00 A 15 y = A sin v n t + B cos v n t y = 0.1 m when t = 0, 0.1 = 0 + B; v = A v n cos vn t B = 0.1 - Bvn sin vn t v = 0.75 m s when t = 0, 0.75 = A(7.00) A = 0.107 y = 0.107 sin (7.00t) + 0.100 ccos os (7.00 t) f = tan - 1 a B 0.100 b = tan - 1 a b = 43.0° A Ans. 0.107 Ans: y = 0.107 sin (7.00 (7.00 t ) + 0.100 cos (7.00 t ) 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 currently publisher. No portion of this material may be reproduced, in any form or by any means, without with out permis permission sion in i n writing writing from the publish.22 – 4. When a 20-lb weight weight is suspended from a spring, the spring spr ing is stretched a distance of 4 in. Determine the natural frequency and the period of vibrati vibration on for a 10-lb weigh t attached to the same spring.

S O LUT IO N k = 20 12 vn = t = 4 = 60 lb ft 60 k = 13.90 rad  s = 10 A 32.2 Am Ans. 2p = 0.452 s vn Ans. Ans: vn = 13.90 rad s t = 0.452 s 1193 22 – 5. When a 3-kg block is suspended from a spring, the spring is of 60 mm. Determine the natural stretched a distance frequency and the period of vibratio n for a 0.2-kg block attached to the same spring. S O LUT IO N k = vn = f = t = 3(9.81) F = = 490.5 N m ¢x 0.060 490.5 k = 49.52 = 49.5 rad s = m A A 0.2 Ans.

Vn 49.52 = 7.88 Hz = 2p 2p 1 f = 1 = 0.127 s 7.88 Ans. Ans: vn = 49.5 rad s t = 0.127 s 1194 22 – 6. An 8-kg block is suspen suspended ded from a spring having h aving a stiffness k = 80 N m.

If the block is given an upward velocity of 0.4 m s when when it is 90 mm above its equilibrium position, motion and the determine the equation which describes the motion maximum upward displacement of the block measured from the equilibriu m position. Assume that positive displacement is measured downward. S O LUT IO N vn = 80 k = A 8 = 3.162 rad  s Am 0.4 m s, x y = - 0.09 m at t = = 0 - Ans: vn = 19.7 rad s C = 1 in. Y = (0.0833 cos 19.7t ) f t.22 – 8. A 6-lb weight weight is suspended from a spring having a stiffness b in. If the weight is given an upward k = 3 l b upward velocity of 20 ft s when it is 2 in.

Above its equilibrium position, determine the equation which describes the motion and the maximum upward displacement of the weight, measured from the equilibrium position. Assume positive displacement is downward.

Statics And Dynamics Hibbeler 13th Edition Solutions Manual Pdf

S O LUT IO N k = 3(12) = 36 lb ft 36 k vn = = 13.90 rad  s = 6 A Am 32.2 t = 0, y = - 20 ft s, y = - 1 ft 6 From Eq. 22 – 3, 3, - 1 = 0 + B 6 B = - 0.167 From Eq. 22 – 4, 4, - 20 = A(13.90) + 0 A = - 1.44 Thus, y = - 1.44 sin (13.9t) - 0.167 cos (13.9t) ft Ans. 22 – 10, 10, C = 2 2A + B2 = 2 2 (1.44) + ( - 0.167)2 = 1.45 ft Ans. Ans: y = - 1.44 sin (13.9t ) C = 1.45 f t 1197 - 0.167 cos (13.9 t ) ) f t currently currently © 2016 © 2016 Pearson Pearson Inc., Inc., Upper Upper S addle Saddle Saddle River, River, NJ.NJ.

AllAll rights rights reserved. ThiThis This s material material i s protected is is protected under under all all copyright copyright laws laws as they as they Education, Education, reproduced, reproduced, publisher. Publish publisher. No No portion portion of this of this material material maymay be be in any in any form form form or by or any by any means, means, without with out without permission permission permis sion in writing i n writing in from from thethe 22 – 9.

A 3-kg block is suspend ed from a spring having a stiffn ess of k = 200 N m. If the block is pushed 50 mm upward from its equilibriu m position posit ion and then released from rest, determine the equation that describes the motion. What are the amplitude and the frequency of the vibration? Assume that positive displacement is downward. SOLUTIO N vn = k 200 = 8.16 rad s A 3 = Am Ans.

X = A sin v n t + B cos vn t x = - 0.05 m when t = 0, - 0.05 = 0 + B; v = Ap cos vn t B = - 0.05 - Bv n sin vn t v = 0 when t = 0, 0 = A(8.165) - 0; A = 0 Hence, x = C = 2 2A + B2 = - 0.05 2 2 (0) cos (8.16 t) Ans. + ( - 0.05) = 0.05 m = 50 mm Ans. Ans: vn = 8.16 rad s x = - 0.05 cos (8.16t ) C = 50 mm 1198 currently currently © 2016 © 2016 Pearson Pearson Inc., Inc., Upper Upper S addle Saddle Saddle River, River, NJ.NJ. AllAll rights rights reserved. ThiThis This s material material i s protected is is protected under under all all copyright copyright laws laws as they as they Education, Education, reproduced, reproduced, publisher.

Publish publisher. No No portion portion of this of this material material maymay be be in any in any form form form or by or any by any means, means, without with out without permission permission permis sion in writing i n writing in from from thethe 22 – 10. The uniform rod of mass m is supported by a pin at A and a spring at B.

If B is given a small sideward displacement and released, determine the natural period of vibr ation. A L Solution Equation of Motio n. The mass moment of inertia of the rod about A is I A = a + Σ M A However; = I a; - mg A a sin u b - mL 2. 3 Referring to the FBD. Of the rod, Fig.

A, L 1 B 1 k (kx cos u)( L L) = a mL 2 b a 2 3 x = L sin u. Then - mg L sin u 2 - 1 2 mL a 3 kL 2 sin u cos u = Using the trigonometry identity sin 2u = 2 sin u cos u, 2 - mg L sin u 2 1 KL - 2 sin 2u = 3 2 mL a $ Here since u is small sin u equation becomes u and sin 2u mg L 1 2 $ mL u + a + 3 2 2u. Then the above 2 kL b u = 0 $ L 3mg + 6k u + u = 0 2m L Comparing to that of the Standard form, vn = t = 2p vn = 2 p L 3mg + 6k. Then 2m L A 2m L A 3mg + 6kL Ans. Currently currently © 2016 © 2016 Pearson Pearson Inc., Inc., Upper Upper S addle Saddle Saddle River, River, NJ.NJ.

AllAll rights rights reserved. ThiThis This s material material i s protected is is protected under under all all copyright copyright laws laws as they as they Education, Education, reproduced, reproduced, publisher. Publish publisher. No No portion portion of this of this material material maymay be be in any in any form form form or by or any by any means, means, without with out without permission permission permis sion in writing i n writing in from from thethe Ans: 2m L t = 2 p L A 3mg + 6k 1199 © 2016 Pearson Education, Inc., Upper S Saddle addle River, NJ. All rights reserved. This Thi s material is i s protected under all copyright laws as they currently publisher. No portion of this material may be reproduced, in any form form or by any means, without permission in writing from the publish 22 – 11.

Statics And Dynamics 13th Edition Solution Manual

While standing in an elevator, the man holds a pen pendu dulum which consists of an 18-in. Cord cord and a 0.5-lb bob.

If the elevator is descending with an an acceleration a = 4 f t s2, determine the natural period of vibration for small amplitudes of swing. A 4 f t/s2 S O LUT IO N Since the acceleration of the pendulum is (32.2 - 4) = 28.2 ft s2 Using the result of Example 22 – 1, 1, We have vn = t = g Al 2p vn = = 28.2 = 4.336 rad  s A 18 12 2p = 1.45 s 4.336 Ans. Ans: t = 1.45 s 1200 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

This material is protected under all copyright laws as they currently publisher. No portion of this material may be reproduced, in any form or by any means, without with out permis permission sion in i n writing writing from the publish.22 – 12. Determine the natural period of vibration of the uniform bar of mass m when it is displaced downward slightly and released. O k L — L — 2 2 Solution Equation of Motion.

The mass moment of inertia of the bar about O is I 0 = 1 mL 2. 12 Referring to the FBD of the rod, Fig.

A, a + Σ M 0 = I a; - ky L 1 2 12 cos u a 0 However, y = L 2 b = a sin u. Then L -k mL 2 b a a 2 1 L sin u b cos u a 2 b = 12 2 mL a Using the trigonometry identity sin 2u = 2 sin u cos u, we obtain 1 mL2a + 12 2 L k 8 sin 2u = 0 Here since u is small, sin 2u 12 12 2u. Then the above equ ation becomes $kL 2 mL u + 4 u = 0 $ 3k u + u = 0 m Comparing to that of the Standard form, vn = t = 2p vn = 2 p m A 3k 3k. Ans: t = 2 p m A 3 k 1201 © 2016 Pearson Education, Inc., Upper S Saddle addle River, NJ. All rights reserved. This Thi s material is i s protected under all copyright laws as they currently publisher.

No portion of this material may be reproduced, in any form form or by any means, without permission in writing from the publish 22 – 13. The body of arbitrary shape has a mass m, mass center at G, and a radius of gyration about G of k G. If it is displaced a slight amount u from its equilibriu m position and released, determine the natural period of vibration. O d u G SOLUTION a + ©M O = IO a; - mgd $ u + $ 2 2 + md Du sin u = C mk G gd 2 2 k G sin u = 0 + d However, for small rotation sin u L u. Hence $ gd u + 2 u = 0 2 k G + d gd From the above differential equation, vn = t = 2p vn = 2p gd B k 2G + d 2 = 2 p. 2 k G + d 2 C gd Ans. A k G2 + d 2 Ans: t = 2 p C k 2G + d 2 gd 1202 22 – 14.

The 20-lb rectangul ar plate has a natural period of vibration t = 0.3 s, as it oscillates around the axis of rod AB. Ad, Determine the torsional stiffness k, measured in lb ft r ad, of the rod. Neglect the mass of the rod. A k B Solution T = k u 1 Σ M z = I za; - k u = 20 2 4 ft $ a b (2) u 12 32.2 $ u + k (4.83)u (4.83)u = 0 2 ft t = 2p 2k (4.83 ) = 0.3 k = 90.8 lb # ft  rad Ans.

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