Archives: Solution Manual

978-1259690877 Chapter 2 Part 2

978-1259690877 Chapter 2 Part 2

IM – 2 | 14 Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 3. a 4. ▲b 5. a Exercise 2-17 1. ▲ 1. Your distributor is the […]

9 Pages | January 6, 2021
978-1259690877 Chapter 2 Part 1

978-1259690877 Chapter 2 Part 1

IM – 2 | 1 Chapter 2 Two Kinds of Reasoning Chapter Recap The main ideas of the chapter are as follows: • Arguments always have two parts, a premise (or premises) and a conclusion. • The same statement can […]

9 Pages | January 6, 2021
978-1259690877 Chapter 1

978-1259690877 Chapter 1

IM – 1 | 1 Chapter 1 Don’t Believe Everything You Think Chapter Recap People think critically when they evaluate reasoning used in coming to conclusions. Conclusions are beliefs; when they are expressed using true-or-false declarative sentences, they are claims […]

9 Pages | January 6, 2021
978-0133915426 Chapter 22 Part 5

978-0133915426 Chapter 22 Part 5

1261 *22–72. SOLUTION Since the system is underdamped. F rom Eq. 22–32 Appling the initial condition at and . (1) (2) Solving Eqs . (1) and (2) yields: Ans.y = 0.622[e–0.939tsin (11.9t + 1.49)] f=85.5° =1.49 rad D=0.622 ft 11.888 […]

9 Pages | January 6, 2021
978-0133915426 Chapter 22 Part 4

978-0133915426 Chapter 22 Part 4

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

9 Pages | January 6, 2021
978-0133915426 Chapter 22 Part 3

978-0133915426 Chapter 22 Part 3

1230 22–41. If the block-and-spring model is subjected to the periodic force , show that the differential equation of motion is , where xis measured from the equilibrium position of the block. What is the general solution of this equation? […]

14 Pages | January 6, 2021
978-0133915426 Chapter 22 Part 2

978-0133915426 Chapter 22 Part 2

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 22 Part 1

978-0133915426 Chapter 22 Part 1

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 21 Part 5

978-0133915426 Chapter 21 Part 5

1180 21–69. The top has a mass of 90 g, a center of mass at G,and a radius of gyration about its axis of symmetry.About any transverse axis acting through point Othe radius of gyration is .If the top is […]

9 Pages | January 6, 2021
978-0133915426 Chapter 21 Part 4

978-0133915426 Chapter 21 Part 4

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

9 Pages | January 6, 2021
978-0133915426 Chapter 21 Part 3

978-0133915426 Chapter 21 Part 3

1149 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, […]

Pages | January 6, 2021
978-0133915426 Chapter 21 Part 2

978-0133915426 Chapter 21 Part 2

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 21 Part 1

978-0133915426 Chapter 21 Part 1

1109 21–1. SOLUTION However,,where ris the distance from the origin Oto dm. Since is constant, it does not depend on the orientation of the x,y,zaxis. Consequently, is also indepenent of the orientation of the x,y,zaxis. Q.E.D.Ixx +I yy +Izz ƒrƒ […]

14 Pages | January 6, 2021
978-0133915426 Chapter 20 Part 4

978-0133915426 Chapter 20 Part 4

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

9 Pages | January 6, 2021
978-0133915426 Chapter 20 Part 3

978-0133915426 Chapter 20 Part 3

1085 20–35. Solve Prob. 20–28 if the connection at B consists of a pin as shown in the figure below, rather than a ball-and- socket joint. Hint: The constraint allows rotation of the rod both along the bar (j direction) […]

9 Pages | January 6, 2021
978-0133915426 Chapter 20 Part 2

978-0133915426 Chapter 20 Part 2

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 20 Part 1

978-0133915426 Chapter 20 Part 1

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 19 Part 3

978-0133915426 Chapter 19 Part 3

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 19 Part 2

978-0133915426 Chapter 19 Part 2

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 19 Part 1

978-0133915426 Chapter 19 Part 1

985 19–1. SOLUTION Q.E.D.rP>G=k2 G rG>O However, yG=vrG>Oor rG>O=yG v rP>G=k2 G yG>v rG>O(myG)+rP>G(myG)=rG>O(myG)+(mk2 G)v HO=(rG>O+rP>G)myG=rG>O(myG)+IGv, where IG=mk2 G The rigid body (slab) has a mass mand rotates with an angular velocity about an axis passing through the fixed point […]

14 Pages | January 6, 2021
978-0133915426 Chapter 18 Part 4

978-0133915426 Chapter 18 Part 4

969 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, […]

Pages | January 6, 2021
978-0133915426 Chapter 18 Part 3

978-0133915426 Chapter 18 Part 3

952 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, […]

Pages | January 6, 2021
978-0133915426 Chapter 18 Part 2

978-0133915426 Chapter 18 Part 2

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 18 Part 1

978-0133915426 Chapter 18 Part 1

912 18–1. SOLUTION T=1 2my2 G+1 2IGv2where yG=vrG>IC Q.E.D.=1 2IIC v2 =1 2 A mr2 G>IC +IG B v2However mr2 G>IC +IG=IIC =1 2m(vrG>IC)2+1 2IGv2 At a given instant the body of mass mhas an angular velocity and its mass […]

14 Pages | January 6, 2021
978-0133915426 Chapter 17 Part 7

978-0133915426 Chapter 17 Part 7

902 17–111. 4 (0.4) m 3p SOLUTION For roll A. c(1) +©MA=IAa;T(0.09) =1 2(8)(0.09)2aA For roll B a(2) (3) Kinematics: (4) also, (5) Solving Eqs. (1)–(5) yields: Ans. Ans. Ans. aB=7.85 m>s2aO=3.92 m>s2 T=15.7 N aB=43.6 rad>s2 aA=43.6 rad>s2 A […]

9 Pages | January 6, 2021
978-0133915426 Chapter 17 Part 6

978-0133915426 Chapter 17 Part 6

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

9 Pages | January 6, 2021
978-0133915426 Chapter 17 Part 5

978-0133915426 Chapter 17 Part 5

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 17 Part 4

978-0133915426 Chapter 17 Part 4

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 17 Part 3

978-0133915426 Chapter 17 Part 3

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 17 Part 2

978-0133915426 Chapter 17 Part 2

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 17 Part 1

978-0133915426 Chapter 17 Part 1

791 17–1. SOLUTION Thus, Ans.Iy=1 3ml 2 m=rAl =1 3rAl 3 =Ll 0 x2(rAdx) Iy=L M x2dm Determine the moment of inertia for the slender rod. The rod’s density and cross-sectional area Aare constant. Express the result in terms of […]

14 Pages | January 6, 2021
978-0133915426 Chapter 16 Part 9

978-0133915426 Chapter 16 Part 9

781 16–143. SOLUTION Gear Motion:The IC of the gear is located at the point where the gear and the gear rack mesh, Fig. a.Thus, v=vO rO>IC =3 0.15 =20 rad>s Then, Since the gear rolls on the gear rack, .By […]

9 Pages | January 6, 2021
978-0133915426 Chapter 16 Part 8

978-0133915426 Chapter 16 Part 8

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

9 Pages | January 6, 2021
978-0133915426 Chapter 16 Part 7

978-0133915426 Chapter 16 Part 7

750 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, […]

Pages | January 6, 2021
978-0133915426 Chapter 16 Part 6

978-0133915426 Chapter 16 Part 6

730 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, […]

Pages | January 6, 2021
978-0133915426 Chapter 16 Part 5

978-0133915426 Chapter 16 Part 5

710 16–78. If the ring gear A rotates clockwise with an angular velocity of vA =30 rad>s , while link BC rotates clockwise with an angular velocity of vBC =15 rad>s , determine the angular velocity of gear D. SOLUTION […]

14 Pages | January 6, 2021
978-0133915426 Chapter 16 Part 4

978-0133915426 Chapter 16 Part 4

690 *16–60. The slider block C moves at 8 m > s down the inclined groove. Determine the angular velocities of links AB and BC, at the instant shown. SOLUTION Rotation About Fixed Axis. For link AB, refer to Fig. […]

14 Pages | January 6, 2021
978-0133915426 Chapter 16 Part 3

978-0133915426 Chapter 16 Part 3

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 16 Part 2

978-0133915426 Chapter 16 Part 2

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 16 Part 1

978-0133915426 Chapter 16 Part 1

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 15 Part 8

978-0133915426 Chapter 15 Part 8

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

12 Pages | January 6, 2021
978-0133915426 Chapter 15 Part 7

978-0133915426 Chapter 15 Part 7

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

12 Pages | January 6, 2021
978-0133915426 Chapter 15 Part 6

978-0133915426 Chapter 15 Part 6

574 15–95. Determine the angular momentum H p of the 6-lb particle about point P. SOLUTION Position and Velocity Vector. The coordinates of points A, B and P are A(–8, 8, 12) ft, B(0, 18, 0) ft and P(–8, 0, […]

14 Pages | January 6, 2021
978-0133915426 Chapter 15 Part 5

978-0133915426 Chapter 15 Part 5

554 15–78. Using a slingshot, the boy fires the 0.2-lb marble at the concrete wall, striking it at B. If the coefficient of restitution between the marble and the wall is , determine the speed of the marble after it […]

14 Pages | January 6, 2021
978-0133915426 Chapter 15 Part 4

978-0133915426 Chapter 15 Part 4

534 15–59. SOLUTION Conservation of Linear Momentum: The linear momentum of the system is conserved along the xaxis (line of impact). The initial speeds of the truck and car are (vt)1=c30 A 103 B m hda 1h 3600 s b=8.333 […]

14 Pages | January 6, 2021
978-0133915426 Chapter 15 Part 3

978-0133915426 Chapter 15 Part 3

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 15 Part 2

978-0133915426 Chapter 15 Part 2

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 15 Part 1

978-0133915426 Chapter 15 Part 1

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]

14 Pages | January 6, 2021
978-0133915426 Chapter 14 Part 5

978-0133915426 Chapter 14 Part 5

456 © 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, […]

12 Pages | January 6, 2021
978-0133915426 Chapter 14 Part 4

978-0133915426 Chapter 14 Part 4

437 14–61. SOLUTION Equations of Motion: By referring to the free-body diagram of the dragster shown in Fig. a, +©Fx=max; 20(103)=1000(a)a=20 m>s2 Kinematics: The velocity of the dragster can be determined from Power: Ans.= C 400(103)t D W P=F#v=20(103)(20 t) […]

13 Pages | January 6, 2021