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Solution Manual For Kinematics and Dynamics of Mechanical Systems Implementation in MATLAB® and Simscape Multibody 3rd Edition Kevin Russell, John Q. Shen, Raj Sodhi

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This document provides a complete and well-structured Solution Manual for Kinematics and Dynamics of Mechanical Systems: Implementation in MATLAB® and Simscape Multibody, 3rd Edition by Kevin Russell, John Q. Shen, and Raj Sodhi. It includes accurate, step-by-step solutions designed to help students understand key concepts such as motion analysis, kinematics, dynamics, system modeling, and simulation using MATLAB and Simscape Multibody tools. The content is organized chapter-by-chapter, making it easy to follow theoretical concepts and apply them to real-world engineering problems. This resource is ideal for assignments, exam preparation, and in-depth revision. Perfect for students in mechanical engineering and related fields seeking reliable academic support aligned with the latest edition.

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CONTENTS

Preface …………………………………………...……………………………………….. 1

Chapter 2 Mathematical Concepts in Kinematics ……………………………………….. 2

Chapter 3 Fundamental Concepts in Kinematics ……………………………………….. 8

Chapter 4 Kinematic Analysis of Planar Mechanisms …………………………………… 25

Chapter 5 Dimensional Synthesis ………………………………………………………... 86

Chapter 6 Static Force Analysis of Planar Mechanisms …………………………………. 195

Chapter 7 Dynamic Force Analysis of Planar Mechanisms ……………………………… 252

Chapter 8 Design & Kinematic Analysis of Gears ……………………………………….. 330

Chapter 9 Design & Kinematic Analysis of Disk Cams …………………………………. 370

Chapter 10 Kinematic Analysis of Spatial Mechanisms ………………………………….. 409

Chapter 11 Introduction to Robotic Manipulators ………………………………………… 455

, PREFACE

Because all the computed solutions in this manual were produced using the 2021 version of

MATLAB™ (specifically version R2021b, win64). any version of MATLAB™ post 2020 is also

suitable to run the MATLAB™ and Simscape Multibody™ files associated with this textbook

(provided all the required toolkits listed in Appendix A.1 are installed).

As presented in Chapter 5 (in the textbook), there is an infinite number of solutions for a

given dimensional synthesis problem. Because the solution values for a dimensional synthesis

problem depend on the dyad angles specified, the solutions given here for Chapter 5 (which were

calculated using arbitrary dyad displacement angles) are intended to serve as a guide to the

solution calculation process and not as a benchmark for evaluating student solutions.

We encourage and look forward to any feedback you may have. For e-mail correspondence,

we can be reached at . Thank you.



K. Russell

Q. Shen

R.S. Sodhi




1

, CHAPTER 2

Problem 2.1 Statement:

Formulate an equation for the vector loop illustrated in Figure P.2.1. Consider that vector V j

always lies along the real axis.




Figure P.2.1 Vector loop (3 vectors where V j changes length) in 2-D complex space

Problem 2.1 Solution:

Taking the clockwise sum of the vector loop in Figure P.2.1 produces the equation

V1ei( 1 +1 ) − V2 ei( 2 +2 ) + V j = 0 .

When expanded and separated into real and imaginary terms, the vector loop equation becomes

V1 cos ( 1 + 1 ) − V2 cos ( 2 +  2 ) + V j = 0
.
V1 sin ( 1 + 1 ) − V2 sin ( 2 +  2 ) = 0

Problem 2.2 Statement:

Formulate an equation for the vector loop illustrated in Figure P.2.2. Consider that vector V j

always lies along the real axis and vector V3 is always perpendicular to the real axis.




2

, Figure P.2.2 Vector loop (4 vectors where V j changes length) in 2-D complex space

Problem 2.2 Solution:

Taking the clockwise sum of the vector loop in Figure P.2.2 produces the equation

−V1ei( 1 +1 ) + V2 ei( 2 +2 ) − V3 − V j = 0 .

When expanded and separated into real and imaginary terms, the vector loop equation becomes

−V1 cos ( 1 + 1 ) + V2 cos ( 2 +  2 ) − V j = 0
.
−V1 sin ( 1 + 1 ) + V2 sin ( 2 +  2 ) − V3 = 0

Problem 2.3 Statement:

Calculate the first derivative of the vector loop equation solution from Problem 2.2. Consider

angles 1 ,  2 and vector V j from Problem 2 to be time-dependent.

Problem 2.3 Solution:

Differentiating the vector loop equation solution from Problem 2.2 produces the equation

−i1V1ei( 1 +1 ) + i 2V2ei( 2 +2 ) − V j = 0 .

When expanded and separated into real and imaginary terms, the vector loop equation becomes

1V1 sin ( 1 + 1 ) −  2V2 sin ( 2 +  2 ) − V j = 0
.
−1V1 cos ( 1 + 1 ) +  2V2 cos ( 2 +  2 ) = 0


3

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