Lecture 4- 1page

8/14/2019 Lecture 4- 1page

http://slidepdf.com/reader/full/lecture-4-1page 1/20

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics S ev en t h

E d i t i on

2 - 1

Rectangular Components in Space (3-D Vectors)

• The vector is

contained in the

plane OBAC .

F r

• Resolve into

horizontal and

vertical components.

F r

y y F F θ cos=

yh F F θ sin=

• Resolve into

rectangular componentshF

φ θ

φ

φ θ

φ

sinsin

sin

cossin

cos

y

h y

y

h x

F

F F

F

F F

=

=

==





E d i t i on

2 - 2

Rectangular Components in Space

• With the angles between and the axes,F r

( )

k ji

F

k jiF

k F jF iF F

F F F F F F

z y x

z y x

z y x

z z y y x x

rrrr

r

rrr

rrrr

θ θ θ λ

λ

θ θ θ

θ θ θ

coscoscos

coscoscos

coscoscos

++==

++=

++=

===

• is a unit vector along the line of action of

and are the directioncosines for

F r

F r

λ r

z y x θ θ θ cosand ,cos,cos





E d i t i on

2 - 3

A UNIT VECTOR

Characteristics of a unit vector :

a) Its magnitude is 1.

b) It is dimensionless.

c) It points in the same direction as the

original vector ( A).

The unit vectors in the Cartesian axissystem are i, j, and k. They are unit

vectors along the positive x, y, and z

axes respectively.

For a vector A with a magnitude ofA, an unit vector is defined as U A =

A / A .


http://slidepdf.com/reader/full/lecture-4-1page 4/20© 2003 The McGraw-Hill Companies, Inc. All rights reserved.


E d i t i on

2 - 4

3-D CARTESIAN VECTOR TERMINOLOGY

Consider a box with sides AX,

AY, and AZ meters long.

The vector A can be defined as

A = (AX i + AY j + AZ k) m

The projection of the vector A in the x-y plane is A´. The

magnitude of this projection, A´, is found by using the same

approach as a 2-D vector: A´ = (AX2 + AY

2)1/2 .

The magnitude of the position vector A can now be obtained as

A = ((A´)2

+ AZ2

) ½

= (AX2

+ AY2

+ AZ2

) ½




E d i t i on

2 - 5

The direction or orientation of vector A isdefined by the angles α, β, and γ.

These angles are measured between the vector

and the positive X, Y and Z axes,

respectively. Their range of values are from

0° to 180°Using trigonometry, “direction cosines” are found using the formulas

These angles are not independent. They must satisfy the following equation.

cos ² α + cos ² β + cos ² γ = 1

This result can be derived from the definition of a coordinate direction anglesand the unit vector. Recall, the formula for finding the unit vector of any

position vector:

or written another way, u A = cos α i + cos β j + cos γ k .

TERMS (continued)




E d i t i on

2 - 6

ADDITION/SUBTRACTION OF VECTORS (Section 2.6)

Once individual vectors are written in Cartesian form, it is easy

to add or subtract them. The process is essentially the same as

when 2-D vectors are added.

For example, if

A = AX i + AY j + AZ k and

B = BX i + BY j + BZ k , then

A + B = (AX + BX) i + (AY + BY) j + (AZ + BZ) k

or

A – B = (AX - BX) i + (AY - BY) j + (AZ - BZ) k .





E d i t i on

2 - 7

IMPORTANT NOTES

Sometimes 3-D vector information is given as:

a) Magnitude and the coordinate direction angles, or

b) Magnitude and projection angles.

You should be able to use both these types of

information to change the representation of

the vector into the Cartesian form, i.e.,

F = {10 i – 20 j + 30 k} N .





E d i t i on

2 - 8

EXAMPLE

Given:Two forces F and G are applied

to a hook. Force F is shown in

the figure and it makes 60°

angle with the X-Y plane. ForceG is pointing up and has a

magnitude of 80 lb with α =

111° and β = 69.3°.

Find: The resultant force in the

Cartesian vector form.

Plan:

1) Using geometry and trigonometry, write F and G in the


2) Then add the two forces.

G

S E




Vector Mechanics for Engineers: StaticsS ev en t h

E d i t i on

2 - 9

Solution : First, resolve force F.Fz = 100 sin 60° = 86.60 lb

F' = 100 cos 60° = 50.00 lb

Fx = 50 cos 45° = 35.36 lb

Fy = 50 sin 45° = 35.36 lb

Now, you can write:

F = {35.36 i – 35.36 j + 86.60 k} lb

S E





E d i t i on

2 - 10

Now resolve force G.

We are given only α and β. Hence, first we need to find the value

of γ.

Recall the formula cos ² (α) + cos ² (β) + cos ² (γ) = 1.

Now substitute what we know. We have

cos ² (111°) + cos ² (69.3°) + cos ² (γ) = 1.

Solving, we get γ = 30.22° or 120.2°. Since the vector is pointing

up, γ = 30.22°

Now using the coordinate direction angles, we can get U G, and

determine G = 80 U G lb.

G = {80 ( cos (111°) i + cos (69.3°) j + cos (30.22°) k )} lb

G = {- 28.67 i + 28.28 j + 69.13 k } lb

Now, R = F + G or

R = {6.69 i – 7.08 j + 156 k} lb

V M h i f E i S i S E





E d i t i on

2 - 11

POSITION VECTOR

A position vector is defined asa fixed vector that locates a

point in space relative to

another point.

Consider two points, A & B, in

3-D space. Let their coordinates

be (XA, YA, ZA) and ( XB,

YB, ZB ), respectively.

The position vector directed from A to B, r AB , is defined as

r AB

= {( XB

– XA

) i + ( YB

– YA

) j + ( ZB

– ZA

) k }m

Please note that B is the ending point and A is the starting point.

So ALWAYS subtract the “tail” coordinates from the “tip”

coordinates!

V t M h i f E i St ti S E





E d i t i on

2 - 12

FORCE VECTOR DIRECTED

ALONG A LINE (Section 2.8)

If a force is directed along a line,

then we can represent the forcevector in Cartesian Coordinates by

using a unit vector and the force

magnitude. So we need to:

a) Find the position vector, r AB , along two points on

that line.

b) Find the unit vector describing the line’s direction,u AB = ( r AB/r AB).

c) Multiply the unit vector by the magnitude of the

force, F = F u AB .

V t M h i f E i St ti S E





E d i t i on

2 - 13

Rectangular Components in Space

Direction of the force is defined by

the location of two points,

( ) ( )222111 ,,and ,, z y x N z y x M

( )

d Fd F

d Fd F

d Fd F

k d jd id d

F F

z zd y yd x xd

k d jd id

N M d

z z

y y

x x

z y x

z y x

z y x

===

++=

=

−=−=−=

++==

rrrr

rr

rrr

r

1

and joiningvector

121212

λ

λ

V t M h i f E i St ti S

E




Vector Mechanics for Engineers: Staticsev en t h

di t i on

2 - 14

APPLICATIONS

Many problems in real-lifeinvolve 3-Dimensional Space.

How will you represent

each of the cable forces in

Cartesian vector form?

V t M h i f E i St ti S e

E d





di t i on

2 - 15

APPLICATIONS (continued)

Given the forces in the cables, how will you determine

the resultant force acting at D, the top of the tower?

Vector Mechanics for Engineers: Statics S e

E d





di t i on

2 - 16

Sample Problem 2.7

SOLUTION:• Determine the unit vector pointing from A

towards B.

( ) ( ) ( )

( ) ( ) ( )

m3.94

m30m80m40

m30m80m40

222

=

++−=

++−=

AB

k ji ABrrr

• Determine the components of the force.

( )( )

( ) ( ) ( )k ji

k ji

F F

rrr

rrr

rr

N795 N2120 N1060

318.0848.0424.0 N2500

++−=

++−=

= λ

k ji

k ji

rrr

rrrr

318.0848.0424.0

3.9430

3.9480

3.9440

++−=

⎟ ⎠ ⎞⎜⎝ ⎛ +⎟ ⎠ ⎞⎜⎝ ⎛ +⎟ ⎠ ⎞⎜⎝ ⎛ −=λ


E d





di t i on

2 - 17

Sample Problem 2.7

The tension in the guy wire is 2500 N.

Determine:

a) components F x , F y , F z of the force

acting on the bolt at A,

b) the angles θ x , θ y , θ z defining the

direction of the force

SOLUTION:

• Based on the relative locations of the

points A and B, determine the unit

vector pointing from A towards B.

• Apply the unit vector to determine the

components of the force acting on A.

• Noting that the components of the unitvector are the direction cosines for the

vector, calculate the corresponding

angles.


E d





di t i on

2 - 18

Sample Problem 2.7

• Noting that the components of the unit vectorare the direction cosines for the vector, calculate

the corresponding angles.

k ji

k ji z y xrrr

rrrr

318.0848.0424.0

coscoscos

++−=++= θ θ θ λ

o

o

o

5.71

0.321.115

=

==

z

y

x

θ

θ θ

Vector Mechanics for Engineers: Statics S ev

E d




Vector Mechanics for Engineers: Staticsv en t h

i t i on

2 - 19

EXAMPLE

Given: 400 lb force along the

cable DA.

Find: The force F DA in the


Plan:

1. Find the position vector r DA and the unit vector u DA.

2. Obtain the force vector as F DA = 400 lb u DA .

Vector Mechanics for Engineers: Statics S ev

E d i




Vector Mechanics for Engineers: Staticsv en t h

i t i on

2 - 20

EXAMPLE (continued)

The figure shows that when relating D

to A, we will have to go -2 ft in the x-

direction, -6 ft in the y-direction, and

+14 ft in the z-direction. Hence, r DA = {-2 i – 6 j + 14 k} ft.

We can also find r DA by subtracting

the coordinates of D from the

coordinates of A.

r DA = (22

+ 62

+ 142

)0.5

= 15.36 ftu DA = r DA/r DA and F DA = 400 u DA lb

F DA = 400{(-2 i – 6 j + 14 k)/15.36} lb

= {-52.1 i – 156 j + 365 k} lb

Documents

Lecture 4- 1page