13880_Lecture Notes 2.5 - 2.6

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Mechanics for Engineers: Statics, 13th SI EditionR. C. Hibbeler and Kai Beng Yap

© Pearson Eduation South !sia Pte "td#$13. !ll rights reser%ed.

CARTESIAN VECTORS ANDTHEIR ADDITION & SUBTRACTION

In&Class !ti%ities:

• Reading 'ui(

• !ppliations ) Rele%ane

• ! *nit +etor

• 3& +etor -ers

• !dding +etors

• Conept 'ui(

• E/aples

• !ttention 'ui(

-oda0s 2beti%es:

Students 4ill be able to5

a6 Represent a 3& %etor in aCartesian oordinate s0ste.

b6 7ind the agnitude andoordinate angles o8 a 3&

%etor6 !dd %etors 98ores6 in 3&

spae

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READING QUIZ

1. +etor algebra, as 4e are going to use it, is based on a ::::::::::: oordinate s0ste.

!6 Eulidean B6 "e8t&handed

C6 ;ree< 6 Right&handed E6 Eg0ptian

#. -he s0bols α, β, and γ designate the :::::::::: o8a 3& Cartesian %etor.

!6 *nit %etors B6 Coordinate diretion angles

C6 ;ree< soieties 6 =, Y and > oponents

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APPLICATIONS

In this ase, the po4erpole has gu0 4ireshelping to

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APPLICATIONS 9ontinued6

In the ase o8 this radio to4er, i8 0ou

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CARTESIAN UNIT VECTORS

-he unit %etors in the Cartesiana/is s0ste are i , j , and k . -he0are unit %etors along the positi%e /,0, and ( a/es respeti%el0.

Charateristis o8 a unit %etor :

a6 Its agnitude is 1.

b6 It is diensionless 9has no units6.6 It points in the sae diretion as the

original %etor 9 A6.

7or a %etor A, 4ith a agnitude

o8 !, an unit %etor is de8ined asu A = A ) ! .

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CARTESIAN VECTOR REPRESENTATION

Consider a bo/ 4ith sides !=, !Y, and !> eters long.

-he %etor A an be de8ined as

A = (!= i + !Y j + !> k )

-he proetion o8 %etor A in the /&0 plane is !A. -he

agnitude o8 !A is 8ound b0 using the sae approah as a #& %etor5 !A 9!=

# !Y#61)# .

-he agnitude o8 the position %etor A an no4 be obtained as

! 99!A6# !>#6 ½ 9!=

# !Y# !>

#6 ½

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*sing trigonoetr0, Ddiretion osines are 8ound using

-hese angles are not independent. -he0 ust satis80 the 8ollo4ing eFuation.

cos ² α + cos ² β + cos ² γ =

-his result an be deri%ed 8ro the de8inition o8 a oordinate diretion anglesand the unit %etor. Reall, the 8orula 8or 8inding the unit %etor o8 an0position %etor5

or 4ritten another 4a0, u A os α i + os β j + os γ k !

DIRECTION O" A CARTESIAN VECTOR

-hese angles are easured bet4een the %etorand the positi%e =, Y and > a/es, respeti%el0.-heir range o8 %alues are 8ro $G to 1$G

-he diretion or orientation o8 %etor A isde8ined b0 the angles ά, , and γ.

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ADDITION O" CARTESIAN VECTORS (S#c$%o '!)

7or e/aple, i8

A != i !Y j !> k and

B B= i BY j B> k , then

A + B = 9!= B=6 i 9!Y BY6 j + 9!> B>6 k

or

A B = 9!= & B=6 i 9!Y & BY6 j + 9!> & B>6 k .

2ne indi%idual %etors are 4ritten in Cartesian 8or, it iseas0 to add or subtrat the. -he proess is essentiall0 thesae as 4hen #& %etors are added.

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I*PORTANT NOTES

Soeties 3& %etor in8oration is gi%en as5

a6 @agnitude and the oordinate diretion angles, or,

b6 @agnitude and proetion angles.

You should be able to use both these sets o8in8oration to hange the representation o8 the%etor into the Cartesian 8or, i.e.,

F J1$ i #$ j 3$ k L M.

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EA*PLE

16 *sing geoetr0 and trigonoetr0, 4rite F 1 and F 2 inCartesian %etor 8or.

#6 -hen add the t4o 8ores 9b0 adding / and 0&oponents6.

G

G%,#: -4o 8ores F 1 and F 2 are

applied to a hoo

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7( N$$ 93)N6 3$$ M

7/ $ $ M

70 N$$ 9O)N6 O$$ M

Mo4, 4rite F 1 in Cartesian %etor 8or

9dont 8orget the units6.

F 1 J$ i O$$ j 3$$ k L M

Solution5

7irst, resol%e 8ore F 1.

EA*PLE 9ontinued6

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Mo4, resol%e 8ore F 2.

7#( &$$ sin ONG − NQN. M

7# $$ os ONG NQN. M

7# an be 8urther resol%ed as,

7#/ NQN. os 3$G O. M

7#0 NQN. sin 3$G ##. M

-hus, 4e an 4rite5

F 2 JO. i ##. j − NQN. k L M

F2’

F2z

EA*PLE 9ontinued6

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So F R F 1 F 2 and

F 1 J$ i O$$ j 3$$ k L M

F 2 JO. i ##. j − NQN. k L M

F R J O$ i Q3 j − #QQ k L M

EA*PLE 9ontinued6

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CONCEPT QUESTIONS

1. I8 0ou 6 6 !ll o8 the abo%e.

#. 7or a 8ore %etor, the 8ollo4ing paraeters are randol0generated. -he agnitude is $. M, α 3$U , $U , V 1$$U. What is 4rong 4ith this 3& %etor ?

!6 @agnitude is too sall.

B6 !ngles are too large.

C6 !ll three angles are arbitraril0 pi

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GROUP PROBLE* SOLVING

16 *sing the geoetr0 and trigonoetr0, resol%e, and 4rite

F 1 and F 2 in the Cartesian %etor 8or.#6 !dd F 1 and F 2 to get F R.

36 eterine the agnitude and angles α, β, γ .

G%,#: -he sre4 e0e is subeted

to t4o 8ores, F 1 and F 2.

"%-: -he agnitude and theoordinate diretion angleso8 the resultant 8ore.

P./:

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GROUP PROBLE* SOLVING 9ontinued6

7A an be 8urther resol%ed as,

71/ #$O. sin #NG Q.Q M

710 #$O. os #NG 1N.Q M

7irst resol%e the 8ore F 1.

71( & #N$ sin 3NG & 1O3.O M

7A #N$ os 3NG #$O. M

Mo4 4e an 4rite5F 1 JQ.Q i 1N.Q j − 1O3.O k L M

F´

F1z

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F 2 J $$ i ##. j #$$ k L M

-he 8ore F 2 an be represented in the Cartesian %etor 8or as5

F 2 O$$J os 1#$G i os ONG j os Q$G k L M

J $$ i ##. j #$$ k L M

Mo4, resol%e 8ore F 2.

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Mo4 8ind the agnitude and diretion angles 8or the %etor.

7R J9&113.O6# OQ.O# NQ.Q#L1)# ON.# ON M

α os&1 97R/

) 7R

6 os&1 9&113.O ) ON.#6 1$3.NG

β os&1 97R0 ) 7R6 os&1 9OQ.O ) ON.#6 1N.1G

γ os&1 97R( ) 7R6 os&1 9NQ.Q ) ON.#6 3.3G

So F R F 1 F 2 and

F 1 J Q.Q i 1N.Q j − 1O3.O k L M

F 2 J $$ i ##. j #$$ k L M

F R J &113.O i OQ.O j NQ.Q k L M

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ATTENTION QUIZ

1. What is not true about an unit %etor, e.g., u A0

!6 It is diensionless. B6 Its agnitude is one.

C6 It al4a0s points in the diretion o8 positi%e =& a/is.

6 It al4a0s points in the diretion o8 %etor A.

#. I8 F J1$ i 1$ j 1$ k L M and

G J#$ i #$ j #$ k L M, then F G J :::: L N

!6 1$ i 1$ j 1$ k

B6 3$ i #$ j 3$ k

C6 1$ i 1$ j 1$ k

6 3$ i 3$ j 3$ k

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Mechanics for Engineers: Statics, 13th SI EditionR C Hibbeler and Kai Beng Yap

© Pearson Eduation South !sia Pte "td#$13 !ll rights reser%ed

End of the Lecture

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13880_Lecture Notes 2.5 - 2.6