Example Given: A = A x i + A y j + A Z k and B = B x i + B y j + B Z k Vector Addition Resultant R =...

ExampleGiven: A = Axi + Ayj + AZk

and B = Bxi + Byj + BZk

Vector AdditionResultant R = A + B

= (Ax + Bx)i + (Ay + By )j + (AZ + BZ) kVector SubstractionResultant R = A - B

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

2.6 Addition and Subtraction of Cartesian

Vectors


Vectors


Vectors


VectorsConcurrent 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 k components of each force in the system


Vectors


Vectors

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, y and z


Vectors


Vectors

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

Force F has a magnitude of FF = Fu = Fcosαi + Fcosβj + Fcosγk


Vectors


VectorsExample 2.8Express the force F as Cartesian

vector


Vectors


VectorsSolutionSince two angles are specified, the third angle is found by

Two possibilities exit, namelyor

605.0cos

5.0707.05.01cos

145cos60coscos

1coscoscos

1

22

222

222

1205.0cos 1


Vectors


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

directionGiven 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


Vectors


Vectors Example 2.9 Determine the magnitude and

coordinate direction angles of resultant force

acting on the ring


Vectors


VectorsSolutionResultant force

FR = ∑F

= 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


Vectors


VectorsSolutionUnit vector acting in the direction of FR

uFR = FR /FR

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

= 0.1617i - 0.2094j + 0.9422kSo 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


Vectors


VectorsExample 2.10Express the force F1 as a Cartesian

vector.


Vectors


VectorsSolutionThe angles of 60° and 45° are not

coordinate direction angles.

By two successive applications ofparallelogram law,


Vectors


VectorsSolutionBy trigonometry,

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

F1y has a direction defined by –j, Therefore

F1 = {35.4i – 35.4j + 86.6k}kN

SolutionChecking:

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


Vectors


Vectors

N

FFFF zyx

1006.864.354.35 222

21

21

211

Solutionα1 = cos-1(0.354) = 69.3°

β1 = cos-1(-0.354) = 111°

γ1 = cos-1(0.866) = 30.0°

Using the same method, F2 = {106i + 184j - 212k}kN


Vectors


Vectors


Vectors


VectorsExample 2.11Two forces act on the hook. Specify the coordinate direction angles of F2, so that the

resultant force FR acts along the positive y axis and has a magnitude of 800N.


Vectors


VectorsSolutionCartesian vector formFR = F1 + F2

F1 = F1cosα1i + F1cosβ1j + F1cosγ1k

= (300cos45°N)i + (300cos60°N)j + (300cos120°N)k

= {212.1i + 150j - 150k}NF2 = F2xi + F2yj + F2zk

View Free Body Diagram


Vectors


VectorsSolutionSince FR has a magnitude of 800N and actsin the +j directionFR = 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


Vectors


VectorsSolutionHence,

0 = 212.1 + F2x F2x = -212.1N 800 = 150 + F2y F2y = 650N 0 = -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 point- Points are measured relative to the origin, O.

2.7 Position Vectors2.7 Position Vectors

x,y,z CoordinatesEg: 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) 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. Eg: If r extends from the origin, O to point P (x, y, z) then, in Cartesian vector form

r = xi + yj + zk


Position VectorNote 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)


2.7 Position Vectors2.7 Position Vectors Position Vector

- Position vector maybe directed from point A to point B - Designated by r or rAB

Vector addition givesrA + 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 r may be formed by taking the coordinates of the tail, A (xA, yA, zA) and subtract them from the 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 axes

Position vector r can be established

Magnitude r represent the length of cable


Angles, α, β and γ represent the direction of the cable

Unit vector, u = r/r


Example 2.12An elastic rubber band is attached to points A and

B. Determine its length and

its direction measured from

A towards B.


SolutionPosition vector

r = [-2m – 1m]i + [2m – 0]j + [3m – (-3m)]k = {-3i + 2j + 6k}m

Magnitude = length of the rubber band

Unit vector in the director of ru = r /r = -3/7i + 2/7j + 6/7k

mr 7623 222



Solutionα = cos-1(-3/7) = 115°

β = cos-1(2/7) = 73.4°

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 2.13The man pulls on the cord with a force of 350N. Represent this force

acting on the support A, as a Cartesian vector and determine its direction.



SolutionEnd 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 –

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



SolutionForce 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 2.14The circular plate is partially supported by the cable AB. If the force of the cable on

the hook at A is F = 500N, express F as a Cartesian vector.



SolutionEnd points of the cable are (0m, 0m, 2m)

and B (1.707m, 0.707m, 0m)

r = (1.707m – 0m)i + (0.707m – 0m)j + (0m – 2m)k

= {1.707i + 0.707j - 2k}mMagnitude = length of cable AB

mmmmr 723.22707.0707.1 222

SolutionUnit vector,

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

= 0.6269i + 0.2597j – 0.7345kFor force F,

F = Fu = 500N(0.6269i + 0.2597j – 0.7345k) = {313i - 130j - 367k} N



SolutionChecking

Show that γ = 137° and indicate this angle on the diagram



N

F

500

367130313 222



Example 2.15The roof is supported by cables. If the cables exert FAB = 100N and FAC = 120N

on the wall hook at A, determine the magnitude of the resultant force acting at A.



SolutionrAB = (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




SolutionrAC = (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



SolutionFR = FAB + FAC

= {70.7i - 70.7k} N + {80i + 40j – 80k} N = {150.7i + 40j – 150.7k} N

Magnitude of FR

N

FR

217

7.150407.150 222

2.9 Dot Product2.9 Dot ProductDot 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 tails A·B = AB cosθ where 0°≤ θ

≤180°Referred to as scalar

product of vectors as result is a scalar

2.9 Dot Product2.9 Dot Product

Laws of Operation1. Commutative law

A·B = B·A2. Multiplication by a scalar

a(A·B) = (aA)·B = A·(aB) = (A·B)a

3. Distribution lawA·(B + D) = (A·B) + (A·D)


Cartesian Vector Formulation- Dot product of Cartesian unit vectorsEg: i·i = (1)(1)cos0° = 1 andi·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 BA·B = (Axi + Ayj + Azk)· (Bxi + Byj + Bzk)

= 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 lines

θ = cos-1 [(A·B)/(AB)] 0°≤ θ ≤180°

Note: if A·B = 0, cos-10= 90°, A is perpendicular to B

2.9 Dot Product2.9 Dot ProductApplications

- The components of a vector parallel and perpendicular to a line- 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 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║

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

3. If A║ is known, by Pythagorean Theorem


2||

2 AAA

2.9 Dot Product2.9 Dot Product For angle θ between the

rope and the beam A, - Unit vectors along the beams, 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

uA = rA/rA

- Force as a Cartesian vector

F = F(rr/rr) = Fur

- Dot product F║ = F║·uA


Example 2.16The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB.


SolutionSince

Then

N

kjijuF

FF

kji

kjirr

u

B

AB

B

BB

1.257

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

429.0857.0286.0300.

cos

429.0857.0286.0

362

362222

2.9 Dot Product2.9 Dot ProductSolutionSince result is a positive scalar,FAB has the same sense of

direction as uB. Express in Cartesian form

Perpendicular component

NkjikjijFFF

Nkji

kjiN

uFF

AB

ABABAB

}110805.73{)1102205.73(300

}1102205.73{

429.0857.0286.01.257

2.9 Dot Product2.9 Dot ProductSolutionMagnitude can be determined From F┴ or from Pythagorean

Theorem

N

NN

FFF AB

155

1.257300 22

22


Example 2.17The 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 parallel and perpendicular to BA.


SolutionFor angle θrBA = {-2i - 2j + 1k}m

rBC = {- 3j + 1k}m

Thus,

5.42

7379.0

103

113202cos

BCBA

BCBA

rr

rr



SolutionComponents of F

N

kjikj

uFF

kji

kjirr

u

BAB

AB

ABAB

590

3.840.5060

31

32

32

0.2539.758

.

31

32

32

3)122(


SolutionChecking from trigonometry,

Magnitude can be determined From F┴

N

N

FFAB

540

5.42cos800

cos

NFF 5405.42sin800sin


SolutionMagnitude can be determined from F┴ or

from Pythagorean Theorem

N

FFF AB

540

590800 22

22

Chapter SummaryChapter Summary

Parallelogram LawAddition of two vectorsComponents 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

formF = Fxi + Fyj + Fzk

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 vector

Force expressed as a Cartesian vectorF = 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 + 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

Example Given: A = A x i + A y j + A Z k and B = B x i + B y j + B Z k Vector Addition Resultant R =...

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