SIE1007 Dynamics of Machines

Dr Eicher Low

65922052e-: Eicher.Low@SingaporeTech.edu.sg

Introduction

Basic Vector Properties

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.2

Introduction

Dynamics: the study of motion of objects and the cause of the motion

Two main topics under Dynamics:Kinematics: the study of motion without regard to the cause of motion

Kinetics: the study of the relation between forces and motion

Will discuss both in this course!

Scope limited to Newtonian (classical nonrelativistic) mechanics Breaks down at speeds comparable to speed of light and dimensions comparable to the atoms sizeSufficient for practical engineering applications

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.3

Major Topics Covered

Vector kinematics Non-Cartesian coordinate systems

Particle dynamics Newtons laws, work & energy, impulse & momentum

Relative motion Coriolis theorem

Dynamics of system of particles Rigid body kinematics Rigid body dynamics

Many examples related to engineering applications will be given!

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.4

SIE1007 Module Administration

Lecture : Wednesday 09.30 11.30 (Room SR2C) Tutorial : Friday 09.30 11.30 (Room SR2C) Main textbook:

Bedford & Fowler, Engineering MechanicsDynamics (SI edition), Pearson-Prentice Hall (2008)

Other books:Hibbeler and Yap, Engineering MechanicsDynamics, 13th edition (SI units), Pearson-Prentice Hall (2013)Beer, Johnston & Cornwell, Vector Mechanics for EngineersDynamics, 9th edition, McGraw Hill (2010)

CA (50%): Quiz & AssignmentQuiz: closed book with formulas givenAssignment: four to be given during the semester

Final exam (50%): closed book, BUT all the formulas will be given

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.5

To Successfully Complete This Module

Dont memorize, understanding is key! Active learning!

Participate actively in lectures/tutorials

Expand lecture notes with your own notes Dont hesitate to ask questions if necessary

Find other material that can help you understand or enhance your understanding

Consult textbook or other materials Use information super highway

Internet is a wonderful resource, but use it with care!

Remember: it may take time to digest the course material, so dont wait until the last minute to revise

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.6

Particle

Particle: body of negligible dimensions Dimensions of the body irrelevant to the description of its

motion

When can we treat a body as a particle? When its orientation is unimportant or irrelevant!

3 degrees-of-freedom (DOF) in translation only

Examples: Planets can be treated as particles in the context of planetary

motion around the Sun

Aircraft can be treated as particle in the context of aircraft trajectory

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.7

Rigid Body

Rigid body: undeformable body with finite size Deformation of the body is negligible compared to the

overall motion

When can we treat an object as a rigid body? When its orientation is important!

In general, it has 6 DOF, 3 translation and 3 rotation

Examples: Satellite can be treated as rigid body if we concern about its

attitude orientation in orbit

Aircraft can be treated as rigid body if we concern about its attitude motion along its flight path

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.8

Scalars and Vectors

Scalar: entity expressible as a single number Useful to describe the reading of a physical property on a

scale/unit

Examples: mass, temperature, time, length, speed

Vector: entity having both direction and magnitude Exists in a multi-dimensional space

Examples : velocity, force, moment, acceleration

Notation : A or or A

Indicates a vector of magnitude A (a scalar) Unit vector is a vector whose magnitude is 1

A

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.9

Vector Algebra (1)

Multiplication by a scalar Simply multiply the magnitude of the vector without affecting the

direction

Vector addition Tail-to-tip or parallelogram methods

Vector addition is commutative and associative

)0(

AB

)0(

AB

A AB

A A

BBC C

cos2222 ABBAC

ABBA )()( CBACBA

CBA

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.10

What if more than two vectors ?

Consider the three vectors U, V & W. Their vector sum can be described by any of the following combinations:

Is there any other combinations besides these ? The vector sum is independent of the ordering. If the sum of 2 or more vectors = 0, they form a closed polygon.SIE1007 Dynamics of Machines Introduction and Basic Vector Properties

V + U + WU + V + W

U + W + V

Lecture 1.11

Vector Algebra (2)

Vector subtraction

Scalar product (dot product) The result is a scalar

Scalar product is commutative:

Vector product (cross product) The result is a vector

)( BABA B

BBA

A

cosABBAA B

ABBA

sinABC BAC

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.12

Direction is determined by right hand rule

Triple product Result is scalar

Double vector product Result is a vector

C is equal to area of theparallelogram formedby A and B

Vector Algebra (3)

C

ABBA )( CBA

)()()( BACACBCBA

CBABCACBA )()()(

)( CBA

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.13

Components of a Vector

Often desirable to express a vector in terms of its components along a set of perpendicular axes E.g. Cartesian right-handed coordinate system

i, j, k unit vectors along x, y, z axes

Ax, Ay, Az are components of vector A in the xyz reference frame

x

z

B

C

D

E

OAx

Ay

Az

A

y

z

y

x

zyx

zyx

AAA

AAA kji

AAAAjikikjkji

ik

j

xy

z ij

k

222zyx AAAA

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.14

Components in Three Dimensions

Direction Cosines :

One way to describe the direction of a vector is by specifying the angles x, y & z between the vector & the positive coordinate axes:

Ux = |U| cos x, Uy = |U| cos y, Uz = |U| cos SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.15

Components in Three Dimensions

Direction Cosines : Direction cosines: cos x, cos y & cos z Direction cosines satisfy the relation:

cos2 x + cos2 y + cos2 z = 1 Suppose that e is a unit vector with the same direction as U:

U = |U| e

In terms of components :

Uxi + Uyj + Uzk = |U| (exi + eyj + ezk)

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.16

Components in Three Dimensions

Direction Cosines: Thus:

Ux = |U| ex, Uy = |U| ey, Uz = |U| ez By comparing these equations:

cos x = ex, cos y = ey, cos z = ez The direction cosines of a vector U are the components of a

unit vector with the same direction as U

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.17

Vector Operations in Component Form (1)

Summation and multiplication by scalar

Scalar product Note:

BAC

zz

yy

xx

z

y

x

BABABA

CCC

zzyyxx

z

y

x

z

y

x

BABABABBB

AAA

BA

1 kkjjii 0 ikkjji

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.18

Vector Operations in Component Form (2)

Vector productNote:

jikikjkji ,,0kkjjii

zyx

zyx

xyyxzxxzyzzy

BBBAAA

BABABABABABA

kji

kjiBA

)()()(ik

j

)()()(det egdhcfgdibfheiaihgfedcba

ihgfedcba

Recall:

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.19

Cross-Product Matrix

Cross product of two vectors can be expressed as matrix-vector product using cross-product matrix

Cross-product matrix is skew-symmetric

xyyx

zxxz

yzzy

z

y

x

xy

xz

yz

BABABABABABA

BBB

AAAA

AA

0

0

0

BA

xx AA T

Cross-product matrix [Ax]

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.20

Derivative of Vector (1) Derivative of a vector

Derivative of a vector consists of components due to magnitudechange and due to direction change

Component due to magnitude change only (constant direction):

ttttt

dtd

tt

AAAA

00lim

)()(lim

AA tAeA

AA

tt

t dtdA

tA

dtd

eeA

0directionconstant lim

te

nete

ne: unit vector parallel to A

: unit vector perpendicular to A

Example:

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.21

Component due to direction change only (constant magnitude):

General expression for derivative of a vector:

For the example above:

For

Derivative of Vector (2)

0tA

AAA

AdA A

A AAA

te

ne very smallneA //

From trigonometry:

nnn

tA

dtdA

tA

dtd

eeeA

0magnitudeconstant lim

magnitudeconstant directionconstant dtd

dtd

dtd AAA

Example:

nt AdtdA

dtd

eeA

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.22

Vector Differentiation Properties

Some rules of vector differentiation:

dtd

dtd

dtd

dtd

dtd

dtd

dtd

dtd

dtd

dtd

dtd

dtd

BAB

ABA

BAB

ABA

BABA

AAA

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.23

Free-Body Diagrams

Free-Body Diagrams: Serves to focus attention on the object of interest & helps to

identify the external forces acting on it

Used in dynamics to study the motions of objects

Drawing of an isolated or freed object & the external forces acting on it

Drawing a free-body diagram involves 3 steps:1. Identify the object to isolate the choice is often dictated by

particular forces you want to determine/analyse

2. Draw a sketch of the object isolated from its surroundings & show relevant dimensions & angles

3. Draw & label vectors representing all the external forces acting on the isolated object dont forget to include the gravitational force

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.24

Free-Body Diagrams

Equilibrium equation:

F = TABj Wj = (TAB W)j = 0Tension in cable AB is TAB = W

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture1.25

Free-Body Diagrams

A coordinate system is necessary to express the forces on the isolated object in terms of components

E.g. to determine the tensions in the 2 cables:

Isolate lower block & part of cable AB Indicate the external forces: W & TAB Introduce a coordinate system

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.26

Free-Body Diagrams

Isolate upper block

External forces: W, TCD & TAB Equilibrium equation:

F = TCDj TABj Wj= (TCD TAB W)j= 0

Since TAB = W, TCD = 2W

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.27

Free-Body Diagrams

Alternatively, treat the 2 blocks & cable AB as a single object:

Equilibrium equation:

F = TCDj Wj Wj= (TCD 2W)j = 0

Again, TCD = 2W

SIE1007 Dynamics of Machines Introduction and Basic Vector Properties Lecture 1.28