ENGI 1313 Mechanics I

Shawn Kenny, Ph.D., P.Eng.Assistant ProfessorFaculty of Engineering and Applied ScienceMemorial University of [email protected]

ENGI 1313 Mechanics I

Lecture 05: Cartesian Vectors

2 ENGI 1313 Statics I – Lecture 05© 2007 S. Kenny, Ph.D., P.Eng.

Chapter 2 Objectives

to review concepts from linear algebra to sum forces, determine force resultants

and resolve force components for 2D vectors using Parallelogram Law

to express force and position in Cartesian vector form

to introduce the concept of dot product


Lecture 05 Objectives

to further examine Cartesian vector notation and extend to representation of a 3D vector

to sum 3D concurrent force systems


Recall 2D Cartesian Vector

Coplanar Force Vector Summation Unit vector dimensionless

Force VectorsComponent

VectorsResultant Force

Vector

321R FFFF

jFFFiFFFF y3y2y1x3x2x1R

Unit Vector; i = FX

FX

^

Unit Vector; j = Fy

Fy

^


Extend to 3D

Why Use Vectors Simplifies

mathematical operations

Rectangular Coordinate System Right-hand rule


Cartesian Vector in 3D

Three Unit Vectors Component magnitude

• Ax, Ay, Az scalar

Component sense• +, -

• Cartesian quadrant Component direction

• i, j, k unit vectors


Summation Cartesian Vectors in 3D

Use 2D Principles Vector A

Vector B

kBAjBAiBABA zzyyxx

kAjAiAA zyx

kBjBiBB zyx

kBAjBAiBABA zzyyxx


Cartesian Vector Magnitude

Successive Application of 2D Principle Pythagorean theorem Find components Ax and Ay

Combine with z component

• Magnitude A of vector A

2y

2x AAA

2z

2y

2x

2z

2 AAAAAA


Cartesian Vector Direction

Coordinate Direction Angle Vector tail with

coordinate axis, , and

• Range from 0 to 180 Visualization aid using

right rectangular prism


Cartesian Vector Direction (cont.)

Coordinate Direction Angle, Measured +x-axis

to tail of vector A

A

Acos x



Coordinate Direction Angle, Measured +y-axis

to tail of vector A

A

Acos x

A

Acos y



Coordinate Direction Angle, Measured +z-axis

to tail of vector A

A

Acos x

A

Acos y

A

Acos z



Direction Cosines Coordinate direction angles

(, , & ) determined by cos-1

A

Acos x

A

Acos y

A

Acos z

A

Acos x1

A

Acos y1

A

Acos z1



Express as Cartesian Vector, A

Recall Unit Vector

Form Unit Vector, uA

uA

kAjAiAA zyx

A

^Unit Vector; uA = A

^

A

AuA

2z

2y

2x AAAAA

kA

Aj

A

Ai

A

A zyx



Unit Vector, uA

Relate to Direction Cosines

Therefore

uA

^

kA

Aj

A

Ai

A

A

A

Au zyx

A

A

Acos x

A

Acos y

A

Acos z

kcosjcosicosA

AuA



Find Unit Vector Amplitude, | uA | Recall general case for vector and magnitude

Where the unit vector and magnitude is

Therefore

^

kcosjcosicosuA

2z

2y

2x AAAAA

kAjAiAA zyx

1coscoscosu 222A

1coscoscos 222


Cartesian Vector Orientation (cont.)

Typical Cartesian Vector Problems Magnitude and

coordinate angles• Example 2.8

Magnitude and projection angles• Example 2.10

Only need to know 2 angles


Comprehension Quiz 5-01

If you only know uA (unit vector) you can determine the ________ of A uniquely.

A) magnitude (A) B) angles (, , and ) C) components (Ax, Ay, & Az) D) All of the above

Answer B Unit vector (uA) defines direction

|A| defines magnitude


Example Problem 5-01

Forces F and G are applied to a hook. Force F makes 60° angle with the X-Y plane. Force G has a magnitude of 80 lb with = 111° and = 69.3°.

Find the resultant force in Cartesian vector form

G= 80lb = 111 = 69.3


G= 80lb = 111 = 69.3

Example Problem 5-01 (cont.)

Resolve Force F components

Cartesian Vector Notation

lb60.8660sinlb100Fz

Fz

F

lb00.5060coslb100F

lb36.3545coslb50Fx

lb36.3545sinlb50Fy Fy

Fx

lbk60.86j36.35i36.35F


G= 80lb = 111 = 69.3


Determine for Vector G1coscoscos 222

1cos3.69cos111cos 222

8641.03535.03584.01cos 22

78.149or22.30



Coordinate Direction Angles =111

= 111

y

z

x

-xG = 80lb

cosGGx

lb111cos80Gx

lb111180cos80Gx

G

Gcos x

Unit circle

lb67.28Gx



Coordinate Direction Angles and

y

z

x

-xG = 80lb

= 69.3 = 30.2



Cartesian Vector G

= 111

y

z

x

-x

= 69.3 = 30.2

lbk22.30cos80j3.69cos80i111cos80G

lbk13.69j28.28i67.28G

G = 80lb


G= 80lb = 111 = 69.3


Combine Force Vectors

Resultant Vector

lbk13.69j28.28i67.28G

lbk60.86j36.35i36.35F

GFR

lbk13.6960.86j28.2836.35i67.2836.35R

lbk156j08.7i69.6R


Group Problem 5-01

Problem 2-57 (Hibbeler, 2007) Determine the magnitude and coordinate

direction angles of F11 and F22. Sketch each force on an x, y, z reference.

Nk40j50i60F1

Nk30j85i40F2


Group Problem 5-01 (cont.)

Force F11

Nk40j50i60F1

N7.87N40N50N60F 2221

9.46N75.87

N60

F

Fcos 1

1

x11

9.62N75.87

N40

F

Fcos 1

1

z11

125N75.87

N50

F

Fcos 1

1

y11


Group Problem 5-01 (cont.)

Force F22

Nk30j85i40F2

N6.98N30N85N40F 2222

114N62.98

N40

F

Fcos 2

2

x22

3.72N62.98

N30

F

Fcos 2

2

z22

150N62.98

N85

F

Fcos 2

2

y22


Group Problem 5-02

Problem 2-59 (Hibbeler, 2007) Determine the

magnitude and coordinate angles of F acting on the stake.


Group Problem 5-02

Determine F and components

N504

5N40F

N7.1370cosN40Fx

N6.3770sinN40Fy

N304

3N40Fz

Fz

Fy

Fx

F


Group Problem 5-02

Determine Coordinate Direction Angles

1.74N50

N68.13

F

Fcos x

1.53N50

N30

F

Fcos z

3.41N50

N58.37

F

Fcos y


Classification of Textbook Problems

Hibbeler (2007)Problem

SetConcept

Degree of Difficulty

Estimated Time

2-57 Cartesian: Magnitude & direction Medium 10-15min

2-58 to 2-60 Cartesian: Forces Easy 5-10min

2-61 to 2-69 Cartesian; Force components & resultant Easy 5-10min

2-70 to 2-71 Cartesian; Force components & resultant Hard 20-25min

2-72 to 2-78 Cartesian; Force components & resultant Medium 15-20min


References

Hibbeler (2007) http://wps.prenhall.com/

esm_hibbeler_engmech_1 en.wikipedia.org

ENGI 1313 Mechanics I

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