6161103 ch02b

Example

Given: A = Axi + Ayj + AZk

and B = Bxi + Byj + BZk

Vector Addition

2.6 Addition and Subtraction of Cartesian Vectors


Vector Addition

Resultant R = A + B

= (Ax + Bx)i + (Ay + By )j + (AZ + BZ) k

Vector Substraction

Resultant R = A - B

= (Ax - Bx)i + (Ay - By )j + (AZ - BZ) k



� Concurrent Force Systems- Force resultant is the vector sum of all the forces in the system

FR = ∑F = ∑Fxi + ∑Fyj + ∑Fzk

where ∑Fx , ∑Fy and ∑Fz represent the algebraic sums of the x, y and z or i, j or kcomponents of each force in the system



� Force, F that the tie down rope exerts on the ground support at O is directed along the rope

� Angles α, β and γ can be solved with axes x, yand z



� Cosines of their values forms a unit vector u that acts in the direction of the rope

� Force F has a magnitude of F

F = Fu = Fcosαi + Fcosβj + Fcosγk



Example 2.8

Express the force F as Cartesian vector



Solution

Since two angles are specified, the third

angle is found by

Two possibilities exit, namely

or

( ) ( )

( ) o

oo

605.0cos

5.0707.05.01cos

145cos60coscos

1coscoscos

1

22

222

222

==

±=−−=

=++

=++

−α

α

α

γβα

( ) o1205.0cos 1 =−= −α



Solution

By inspection, α = 60° since Fx is in the +x

direction

Given F = 200N

F = Fcosαi + Fcosβj + Fcosγk

= (200cos60°N)i + (200cos60°N)j

+ (200cos45°N)k

= {100.0i + 100.0j + 141.4k}N

Checking:

( ) ( ) ( ) N

FFFF zyx

2004.1410.1000.100 222

222

=++=

++=



Example 2.9

Determine the magnitude and coordinate

direction angles of resultant force acting on the ringthe ring



Solution

Resultant force

FR = ∑F

= F1 + F2= F1 + F2

= {60j + 80k}kN

+ {50i - 100j + 100k}kN

= {50j -40k + 180k}kN

Magnitude of FR is found by

( ) ( ) ( )kN

FR

1910.191

1804050 222

==

+−+=



Solution

Unit vector acting in the direction of FR

uFR = FR /FR

= (50/191.0)i + (40/191.0)j + (180/191.0)k(180/191.0)k

= 0.1617i - 0.2094j + 0.9422k

So that

cosα = 0.2617 α = 74.8°cos β = -0.2094 β = 102°cosγ = 0.9422 γ = 19.6°

*Note β > 90° since j component of uFR is negative



Example 2.10

Express the force F1 as a Cartesian vector.



Solution

The angles of 60° and 45° are not coordinate

direction angles.

By two successive applications of

parallelogram law,



SolutionBy trigonometry,

F1z = 100sin60 °kN = 86.6kNF’ = 100cos60 °kN = 50kNF = 50cos45 kN = 35.4kNF1x = 50cos45 °kN = 35.4kNF1y = 50sin45 °kN = 35.4kN

F1y has a direction defined by –j, Therefore

F1 = {35.4i – 35.4j + 86.6k}kN

Solution

Checking:



( ) ( ) ( )

FFFF zyx

222

21

21

211

=+−+=

++=

Unit vector acting in the direction of F1

u1 = F1 /F1

= (35.4/100)i - (35.4/100)j + (86.6/100)k

= 0.354i - 0.354j + 0.866k

( ) ( ) ( ) N1006.864.354.35 222 =+−+=

Solution

α1 = cos-1(0.354) = 69.3°

β1 = cos-1(-0.354) = 111°

γ1 = cos-1(0.866) = 30.0°



γ1 = cos-1(0.866) = 30.0°

Using the same method,

F2 = {106i + 184j - 212k}kN



Example 2.11

Two forces act on the hook. Specify the

coordinate direction angles of F2, so that the

resultant force F acts along the positive y axis resultant force FR acts along the positive y axis

and has a magnitude of 800N.



Solution

Cartesian vector form

FR = F1 + F2

F = F cosα i + F cosβ j + F cosγ kF1 = F1cosα1i + F1cosβ1j + F1cosγ1k

= (300cos45°N)i + (300cos60°N)j

+ (300cos120°N)k

= {212.1i + 150j - 150k}N

F2 = F2xi + F2yj + F2zk



Solution

Since FR has a magnitude of 800N and acts

in the +j direction

FR = F1 + F2FR = F1 + F2

800j = 212.1i + 150j - 150k + F2xi + F2yj + F2zk

800j = (212.1 + F2x)i + (150 + F2y)j + (- 50 + F2z)k

To satisfy the equation, the corresponding

components on left and right sides must be equal



SolutionHence,

0 = 212.1 + F2x F2x = -212.1N800 = 150 + F2y F2y = 650N0 = -150 + F2z F2z = 150N0 = -150 + F2z F2z = 150N

Since magnitude of F2 and its components are known, α1 = cos-1(-212.1/700) = 108°β1 = cos-1(650/700) = 21.8°γ1 = cos-1(150/700) = 77.6°

� x,y,z Coordinates

- Right-handed coordinate system

- Positive z axis points upwards, measuring the height of an object or the altitude of a

2.7 Position Vectors2.7 Position Vectors

the height of an object or the altitude of a point

- Points are measured relative to the origin, O.

� x,y,z Coordinates

Eg: For Point A, xA = +4m along the x axis, yA = -6m along the y axis and zA = -6m

along the z axis. Thus, A (4, 2, -6)


along the z axis. Thus, A (4, 2, -6)

Similarly, B (0, 2, 0) and C (6, -1, 4)

� Position Vector

- Position vector r is defined as a fixed vector which locates a point in space relative to another point.


point.

Eg: If r extends from the

origin, O to point P (x, y, z)

then, in Cartesian vector

form

r = xi + yj + zk

� Position Vector

Note the head to tail vector addition of the

three components


Start at origin O, one travels x in the +i direction,

y in the +j direction and z in the +k direction,

arriving at point P (x, y, z)


� Position Vector

- Position vector maybe directed from point A to point B

- Designated by r or rABAB

Vector addition gives

rA + r = rB

Solving

r = rB – rA = (xB – xA)i + (yB – yA)j + (zB –zA)k

or r = (xB – xA)i + (yB – yA)j + (zB –zA)k

� Position Vector

- The i, j, k components of the positive vector rmay be formed by taking the coordinates of the tail, A (xA, yA, zA) and subtract them from the head B (xB, yB, zB)


head B (xB, yB, zB)

Note the head to tail vector addition of the

three components


� Length and direction of cable AB can be found by measuring A and B using the x, y, z axesthe x, y, z axes

� Position vector r can be established

� Magnitude r represent the length of cable


� Angles, α, β and γrepresent the direction of the cableof the cable

� Unit vector, u = r/r


Example 2.12

An elastic rubber band is

attached to points A and B. attached to points A and B.

Determine its length and its

direction measured from A

towards B.


Solution

Position vector

r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k

= {-3i + 2j + 6k}m= {-3i + 2j + 6k}m

Magnitude = length of the rubber band

Unit vector in the director of r

u = r /r

= -3/7i + 2/7j + 6/7k

( ) ( ) ( ) mr 7623 222 =++−=


Solution

α = cos-1(-3/7) = 115°

β = cos-1(2/7) = 73.4°

γ = cos-1(6/7) = 31.0°γ = cos-1(6/7) = 31.0°

2.8 Force Vector Directed along a Line


� In 3D problems, direction of F is specified by 2 points, through which its line of action lies

� F can be formulated as a Cartesian vector

F = F u = F (r/r)

Note that F has units of

forces (N) unlike r, with

units of length (m)



� Force F acting along the chain can be presented as a Cartesian vector by

- Establish x, y, z axes

- Form a position vector r along length of chain



� Unit vector, u = r/r that defines the direction of both the chain and the force

� We get F = Fu



Example 2.13

The man pulls on the cord

with a force of 350N. with a force of 350N.

Represent this force acting

on the support A, as a

Cartesian vector and

determine its direction.



Solution

End points of the cord are A (0m, 0m, 7.5m)

and B (3m, -2m, 1.5m)

r = (3m – 0m)i + (-2m – 0m)j + (1.5m –r = (3m – 0m)i + (-2m – 0m)j + (1.5m –7.5m)k

= {3i – 2j – 6k}m

Magnitude = length of cord AB

Unit vector, u = r /r

= 3/7i - 2/7j - 6/7k

( ) ( ) ( ) mmmmr 7623 222 =−+−+=



Solution

Force F has a magnitude of 350N, direction

specified by u

F = Fu

= 350N(3/7i - 2/7j - 6/7k)

= {150i - 100j - 300k} N

α = cos-1(3/7) = 64.6°β = cos-1(-2/7) = 107°γ = cos-1(-6/7) = 149°



Example 2.14

The circular plate is

partially supported by

the cable AB. If the the cable AB. If the

force of the cable on the

hook at A is F = 500N,

express F as a

Cartesian vector.



Solution

End points of the cable are (0m, 0m, 2m) and B

(1.707m, 0.707m, 0m)

r = (1.707m – 0m)i + (0.707m – 0m)jr = (1.707m – 0m)i + (0.707m – 0m)j

+ (0m – 2m)k

= {1.707i + 0.707j - 2k}m

Magnitude = length of cable AB

( ) ( ) ( ) mmmmr 723.22707.0707.1 222 =−++=

Solution

Unit vector,

u = r /r

= (1.707/2.723)i + (0.707/2.723)j – (2/2.723)k



= (1.707/2.723)i + (0.707/2.723)j – (2/2.723)k

= 0.6269i + 0.2597j – 0.7345k

For force F,

F = Fu

= 500N(0.6269i + 0.2597j – 0.7345k)

= {313i - 130j - 367k} N

Solution

Checking



( ) ( ) ( )N

F

500

367130313 222

=

−++=

Show that γ = 137° and

indicate this angle on the

diagram

N500=



Example 2.15

The roof is supported by

cables. If the cables exert

F = 100N and F = 120NFAB = 100N and FAC = 120N

on the wall hook at A,

determine the magnitude of

the resultant force acting at

A.



Solution

rAB = (4m – 0m)i + (0m – 0m)j + (0m – 4m)k

= {4i – 4k}m

FAB = 100N (rAB/r AB)

= 100N {(4/5.66)i - (4/5.66)k}

= {70.7i - 70.7k} N

( ) ( ) mmmrAB 66.544 22 =−+=



Solution

rAC = (4m – 0m)i + (2m – 0m)j + (0m – 4m)k

= {4i + 2j – 4k}m

FAC = 120N (rAB/r AB)

= 120N {(4/6)i + (2/6)j - (4/6)k}

= {80i + 40j – 80k} N

( ) ( ) ( ) mmmmrAC 6424 222 =−++=



Solution

FR = FAB + FAC

= {70.7i - 70.7k} N + {80i + 40j – 80k} N

= {150.7i + 40j – 150.7k} N= {150.7i + 40j – 150.7k} N

Magnitude of FR

( ) ( ) ( )N

FR

217

7.150407.150 222

=

−++=

2.9 Dot Product2.9 Dot Product

� Dot product of vectors A and B is written as A·B (Read A dot B)

� Define the magnitudes of A and B and the angle between their tailsangle between their tails

A·B = AB cosθ where 0°≤ θ ≤180°

� Referred to as scalar

product of vectors as

result is a scalar


� Laws of Operation

1. Commutative law

A·B = B·AA·B = B·A

2. Multiplication by a scalar

a(A·B) = (aA)·B = A·(aB) = (A·B)a

3. Distribution law

A·(B + D) = (A·B) + (A·D)


� Cartesian Vector Formulation

- Dot product of Cartesian unit vectors

Eg: i·i = (1)(1)cos0° = 1 andEg: i·i = (1)(1)cos0° = 1 and

i·j = (1)(1)cos90° = 0

- Similarly

i·i = 1 j·j = 1 k·k = 1

i·j = 0 i·k = 1 j·k = 1


� Cartesian Vector Formulation- Dot product of 2 vectors A and B

A·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk)

= A B (i·i) + A B (i·j) + A B (i·k)= AxBx(i·i) + AxBy(i·j) + AxBz(i·k)

+ AyBx(j·i) + AyBy(j·j) + AyBz(j·k)

+ AzBx(k·i) + AzBy(k·j) + AzBz(k·k)

= AxBx + AyBy + AzBz

Note: since result is a scalar, be careful of including any unit vectors in the result


� Applications

- The angle formed between two vectors or intersecting linesintersecting lines

θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180°

Note: if A·B = 0, cos-10= 90°, A is

perpendicular to B


� Applications- The components of a vector parallel and perpendicular to a line

- Component of A parallel or collinear with line aa’ is - Component of A parallel or collinear with line aa’ is defined by A║ (projection of A onto the line)

A║ = A cos θ

- If direction of line is specified by unit vector u (u = 1),

A║ = A cos θ = A·u


�Applications- If A║ is positive, A║ has a directional sense same as u

- If A is negative, A has a directional - If A║ is negative, A║ has a directional sense opposite to u

- A║ expressed as a vector

A║ = A cos θ u

= (A·u)u

� ApplicationsFor component of A perpendicular to line aa’

1. Since A = A║ + A┴,

then A = A - A


then A┴ = A - A║2. θ = cos-1 [(A·u)/(A)]

then A┴ = Asinθ

3. If A║ is known, by Pythagorean Theorem

2||

2 AAA +=⊥


� For angle θ between the rope and the beam A,

- Unit vectors along the beams, uA = rA/rAbeams, uA = rA/rA

- Unit vectors along the ropes, ur=rr/rr

- Angle θ = cos-1

(rA.rr/rArr)

= cos-1 (uA· ur)


� For projection of the force along the beam A

- Define direction of the beam

u = r /ruA = rA/rA

- Force as a Cartesian vector

F = F(rr/rr) = Fur

- Dot product

F║ = F║·uA


Example 2.16

The frame is subjected to a horizontal force

F = {300j} N. Determine the components of

this force parallel and perpendicular to the this force parallel and perpendicular to the

member AB.


Solution

Since

( ) ( ) ( )kji

r

ru

B

BB

362

362222 ++

++==

rrr

r

rr

Then

( ) ( ) ( )

( ) ( )

N

kjijuF

FF

kji

r

B

AB

B

1.257

)429.0)(0()857.0)(300()286.0)(0(

429.0857.0286.0300.

cos

429.0857.0286.0

362 222

=

++=

++⋅==

=

++=

++

rrrrrr

rr

rrr

θ


Solution

Since result is a positive scalar,

FAB has the same sense of

direction as uB. Express in B

Cartesian form

Perpendicular component

( )( )

NkjikjijFFF

Nkji

kjiN

uFF

AB

ABABAB

}110805.73{)1102205.73(300

}1102205.73{

429.0857.0286.01.257

rrrrrrrrrr

rrr

rrr

rrr

−+−=++−=−=

++=

++=

=

⊥


Solution

Magnitude can be determined

From F┴ or from Pythagorean TheoremTheorem

( ) ( )N

NN

FFF AB

155

1.257300 22

22

=

−=

−=⊥

rrr


Example 2.17

The pipe is subjected to F = 800N. Determine the

angle θ between F and pipe segment BA, and the

magnitudes of the components of F, which are magnitudes of the components of F, which are

parallel and perpendicular to BA.


Solution

For angle θ

rBA = {-2i - 2j + 1k}m

rBC = {- 3j + 1k}mrBC = {- 3j + 1k}m

Thus,

( )( ) ( )( ) ( )( )

o

rr

rr

5.42

7379.0

103

113202cos

=

=

+−−+−=

⋅=

θ

θBCBA

BCBA

rr

rr


Solution

Components of Fkji

r

ru

AB

ABAB 3

)122( +−−==

rrr

r

rr

( )

N

kjikj

uFF

kji

BAB

AB

590

3.840.5060

31

32

32

0.2539.758

.

31

32

32

=

++=

+

−+

−⋅+−=

=

+

−+

−=

rrrrr

rrr

rrr


Solution

Checking from trigonometry,

FFAB cos=rr

θ

Magnitude can be determined

From F┴

N

N

540

5.42cos800

=

= o

NFF 5405.42sin800sin ===⊥o

rrθ


Solution

Magnitude can be determined from F┴ or from Pythagorean Theorem

( ) ( )N

FFF AB

540

590800 22

22

=

−=

−=⊥rrr

Chapter SummaryChapter Summary

Parallelogram Law� Addition of two vectors

� Components form the side and resultant � Components form the side and resultant form the diagonal of the parallelogram

� To obtain resultant, use tip to tail addition by triangle rule

� To obtain magnitudes and directions, use Law of Cosines and Law of Sines


Cartesian Vectors� Vector F resolved into Cartesian vector form

F = Fxi + Fyj + Fzk

� Magnitude of F� Magnitude of F

� Coordinate direction angles α, β and γ are determined by the formulation of the unit vector in the direction of F

u = (Fx/F)i + (Fy/F)j + (Fz/F)k

222zyx FFFF ++=


Cartesian Vectors� Components of u represent cosα, cosβ and cosγ

� These angles are related by

cos2α + cos2β + cos2γ = 1

Force and Position Vectors� Position Vector is directed between 2 points

� Formulated by distance and direction moved along the x, y and z axes from tail to tip


Force and Position Vectors� For line of action through the two points, it

acts in the same direction of u as the position vectorposition vector

� Force expressed as a Cartesian vector

F = Fu = F(r/r)

Dot Product� Dot product between two vectors A and B

A·B = AB cosθ


Dot Product� Dot product between two vectors A and B

(vectors expressed as Cartesian form)

A·B = AxBx + AyBy + AzBzA·B = AxBx + AyBy + AzBz

� For angle between the tails of two vectors

θ = cos-1 [(A·B)/(AB)]

� For projected component of A onto an axis defined by its unit vector u

A = A cos θ = A·u

Chapter ReviewChapter Review

6161103 ch02b

Education

tail vector

cartesian

unit vector

cartesian

position vector

coordinatedirection

2m 0m

7 position