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By: Anthony Bedford and Wallace Fowler Engineering Mechanics: STATICS Fifth Edition in SI Units Chapter 2: Chapter 2: Vectors Vectors

Chapter 02 - 向量

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Page 1: Chapter 02 - 向量

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By: Anthony Bedford and Wallace Fowler 

Engineering

Mechanics:

STATICS

Fifth Edition

in SI Units

Chapter 2:Chapter 2: VectorsVectors

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Page 2

Chapter 2 Engineering Mechanics: STATICS

Fifth EditionCopyright 2008 Pearson Education South Asia Pte Ltd

 學習目標

如果一個物體受到不同大小與不同方向的外力作用時,如何能決定作用於個物體

 !"力的大小#方向$%  

&'()*+向,的-.,如何/0,123向,,與43向,&567用8的9:;

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Chapter 2 Engineering Mechanics: STATICS

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'<大=

•  >,與向,

•   ?@0,

•   A@0,

•   >,B

•   向,B

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Chapter 2 Engineering Mechanics: STATICS

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•  –>, CDEFGHI的物J,

 – 9如K時L,M,

•  –向, NO大小PQRFGST方向

 – 9如KULV一個WXY於Z一W的[\,作用力

 – E]體^_23K U, V, W, …

 –  向, U 的大小 = |U|

>,#向, 

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Chapter 2 Engineering Mechanics: STATICS

Fifth EditionCopyright 2008 Pearson Education South Asia Pte Ltd

 – 向,的`a23Kbc

•  bc方向 =  向,方向

•  bcde 向,大小

• 9如K

 – r AB =  W  B  XY於W  A 的[\向,

 – r AB  的方向 =  fW  A  到W B 的方向

>,#向, 

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Chapter 2 Engineering Mechanics: STATICS

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>,#向, 

 – |r AB| = gW!L的hi 

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Chapter 2 Engineering Mechanics: STATICS

Fifth EditionCopyright 2008 Pearson Education South Asia Pte Ltd

>,#向, 

j向,kl  "

mn一個物體  受到[o # fUL一[\op到qr[\$

 – [o向,K U

 –  U 的方向 =  [o的方向

 – |U| =  sop的hi

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Chapter 2 Engineering Mechanics: STATICS

Fifth EditionCopyright 2008 Pearson Education South Asia Pte Ltd

 –  t?個[o V

 – suvw[\x一y,z{| U  } V ,~X•€;

 –  U  T V  ‚ƒ於„一[o 

W:  U + V = W

>,#向, 

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Chapter 2 Engineering Mechanics: STATICS

Fifth EditionCopyright 2008 Pearson Education South Asia Pte Ltd

>,#向, 

j向,kl的定… "

 –  向,f†‡ U  到ˆ V

jA‰a定Š 

mk‹與向,LŒ†X的€zŽ; 

j‘’a定Š 

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Chapter 2 Engineering Mechanics: STATICS

Fifth EditionCopyright 2008 Pearson Education South Asia Pte Ltd

>,#向, 

m向,kl“NO”–的K

U + V = V + U #2'&$

m向,“klNO—"–的 "

(U + V) + W = U + (V + W) #2'2$

m如果g個~/8的向,的#‚於˜,™š›一個œ的ž’a; 

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Chapter 2 Engineering Mechanics: STATICS

Fifth EditionCopyright 2008 Pearson Education South Asia Pte Ltd

>,#向, 

 – 9如 "

jf A 到 C 的向,  r AC ‚於 

r AB  與 r BC 的向,#; 

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Chapter 2 Engineering Mechanics: STATICS

Fifth EditionCopyright 2008 Pearson Education South Asia Pte Ltd

•   >,與向,的ŸBK

 – >, # FG $ a  與向, U  的ŸB =  向, aU

 –  大小 = |a||U| ,qV  |a| >, a 的¡Y¢

 –  aU  的方向同於 U ,n  a £

 –  aU  的方向•於 U ,n a R

>,#向, 

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Chapter 2 Engineering Mechanics: STATICS

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>,#向, 

m向, U ¤/>, a ¥定…如¦K

UU

  

 =

aa

1

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Chapter 2 Engineering Mechanics: STATICS

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>,#向, 

mŸBY>,的ŸlN—"–K

a(bU) = (ab)U #2'3$

mŸBY>,klN0§–K

(a + b)U = aU + bU #2'4$

mŸBY向,klN0§–Ka(U + V) = aU + aV #2'5$

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Chapter 2 Engineering Mechanics: STATICS

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>,#向, 

j向,¨lK

U – V = U + (−1)V #2'$

j„[向,K

mde‚於 1

m©定一個方向 

m如果„[向,  e /T向,  U 的方向X同K U = |U|e

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Chapter 2 Engineering Mechanics: STATICS

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ª9 2.1 

向,的-. 

«3!向,的de“  |U| = 8  與 |V| = 3  ; 向, V  “¬的;®用`¯的方°決定向,  U + 2V 的de;

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Chapter 2 Engineering Mechanics: STATICS

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方± 

²³9´µg個向,}-用向,kl的A‰a定Š,()能,µ向,  U + 2V 的de;

ª9 2.1( ¶ )

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Chapter 2 Engineering Mechanics: STATICS

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¯·

 ²³9´µ U  # 2V ,¸¹)Œ†X;

ª9 2.1( ¶ )

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Chapter 2 Engineering Mechanics: STATICS

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¯·

|U + 2V|  º,的¢“ 13.0 ;

ª9 2.1( ¶ )

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»習¼ 

«3的向,的de“  |U| = 8  與 |V| = 3  ;向, V  “¬ 的; ®/`¯的方°,µ向,  U − 2V 的de; 

·½K |U − 2V| = 5.7

ª9 2.1( ¶ )

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ª9 2.2 

向,Xk 

( ¾¿Àu )

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Chapter 2 Engineering Mechanics: STATICS

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?@0, 

jn向,/XÁ¬的0,123時,向,ÂÃÄÅÆJ

ÞK

 –  ÇÈ向, U K

mÉ\‰Ê標,Ë向,  U 與 x-y Ì

 –  U = 0Í於 x Î# Ï Î的X¬0, U x  # U y 

的#K U = U x + U y

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Chapter 2 Engineering Mechanics: STATICS

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mÐÑ£ x Î方向的„[向,  i #£ y Î方向的„[向

,  j"

U = U  xi + U  y j #2'!$

  qV U  x  與  U  y  “ U 的的>,0,; 

 –  U 的de用ÒÓÔ定JE¹的0,.µ  "

#2'8$

?@0, 

22

 y xU U   +=U

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Chapter 2 Engineering Mechanics: STATICS

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?@0, 

j/0,作向,-.K

 – g個向, U  與 V  的# "

U + V = (U xi + U y j) + (V xi + V y j)

= (U x + V x)i + (U y + V y) j #2'%$

 – 如` "

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Chapter 2 Engineering Mechanics: STATICS

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?@0, 

j/0,作向,-. "

 –  G^ a  與向, U 的ŸB“ "

  aU = a(U xi + U y j) = aU xi + aU y j

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?@0, 

j/向,0,23[\向,  "

mÇÈÊ標  # x A, y A$ !W A /TÊ標 # x B, y B$ !W

B ; 

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Chapter 2 Engineering Mechanics: STATICS

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?@0, 

j/向,0,23[\向,  "

 –  Õ r AB Ö2 B XY於 A 的[\向,K

r AB = ( x B −  x A)i + ( y B −  y A) j #2'&0$

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ª9 2.3 

×.0, 

ØÙfW A 到W B ÚkÛ 900-N 的力於ÜÝÞßà的áâã/向,  ( Ö2;

®ä  ( /«3!Ê

標åæ¦的0,123; 

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ª9 2.3( ¶ )

 

方±

 ()ä用gç方°×.向, F 的0,;t一個方l,() äË用A‰學×. F 與 Y Î!L的è‰1.µ0,;t?

個方l,()äË用éê的ØÙ AB 的ëìãË用XíA

‰a1×. F 的0,;

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¯· 

t一個方l 

×. F 與 y Ê標Î的è‰K

°=   

  = 62680

40arcta   .α

ª9 2.3( ¶ )

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¯·

î用A‰ï何學×.µ F ãð¹的0,23°K

( )

( ) N805402

 N6.26c!"9006.26"#900c!""#

!!

 ji

 ji

 jFiFF

−=−=

−=   α α 

ª9 2.3( ¶ )

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¯·

t?個方l

®Ë用«ñ定的òó,×.f A

到 B 的hiK

( ) ( ) $489$80$4022

.=+

ª9 2.3( ¶ )

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¯·

î用XíA‰a×. F 的0,K

$89.4

$80 a% 

$89.4

$40

== FF y x

  F  F 

( ) ( )

( ) N805402

 N9004.89

40 N900

4.89

40

 ji

 jiF

−=

−=

ª9 2.3( ¶ )

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»習¼ 

ØÙfW A 到W B ÚkÛ 900-N 的力於ÜÝÞßà的àáã/向,  F 1Ö2;ô定õöpW B 的[\Ëà F 的 y

0,的大小“ x 0,÷的 3 ø;®ä  F /¹的0,123

;W B ùú& x Î8iûWžü的ý方% 

·½K F = 285i - 854 j (N) ;W B ùú& x Î8iûW 26.7

$ !Æ; 

ª9 2.3( ¶ )

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ª9 2.4

f‰e×.0, 

( ¾¿Àu )

ª9 2.5×.þê向,的de 

( ¾¿Àu )

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A@0, 

&   *+´µA@V的物體K

(a)  ÝÙ¬於ÿ方體一Ì時«見到的ÿ方體;

(')  ÿ方體的ë側Ý;

(c)  一個‰Ê標åæ與ÿ方體的’疊";

(%) Ê標åæ的A@23;

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j一個右手Ê標åæK

m/0, 

U x) U y * U z  # 

0Í於  x) y  與  z   Î $ 23 向, U  K

U = U x + U y + U z    #2'&&$

A@0,

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A@0,

mË用©向 x , y # 方向的„[向,  i  、  j  # k   ,

()可/ä U î用>,0,23如¦ 

 –  U = U  xi + U  y j + U  z k    #2'&2$

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A@0,

j/向,0,23向,的deK

jÇÈ一個向,  U /T¹的0,K

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A@0,

j/向,0,23向,的deK

mE  U y  , U z /Tg÷的‹#  U y + U z «構›的‰A

‰aK

|U y + U z |2 = |U y|2 + |U z |2   #2'&3$

m向,  U  ‚於向,  U x 與  U y + U z !#; 3 個向,

a›一個‰A‰a K

|U|2 = |U x|2 + |U y + U z |2

@0,

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A@0,

j/向,0,23向,的deK

mä° (2.13) Ö入K

|U|2 = |U x|2 + |U y| + |U z |2 = U  x2 + U  y2 + U  z 2

m因此,向, U 的大小K

#2'&4$222

 z  y x   U U U    ++=U

@0,

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A@0,

j方向餘ÔK

 –   HI一個向,方向的方l!一,“標3µ向,與Ê標ΣÎ向!L的è‰ θ  x)、θ  y  與 θ  z ;

U  x = |U| c!" θ  x, 

U  y = |U| c!" θ  y, 

U  z  = |U| c!" θ  z   #2'&5$

A@0,

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A@0,

&   方向餘ÔK

A@0,

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A@0,

&   方向餘ÔK

 –   方向餘Ô  K c!" θ  x) c!" θ  y * c!" θ  z 

 –   方向餘Ô滿足¦IŽ係°K

c!"2 θ  x + c!"2 θ  y + c!"2 θ  z  = 1 #2'&$

 –   ô定 e “一個與 U 同方向的„[向,,因此

 

"

U = |U| e

 –   /0,23K

U  xi + U  y j + U  z k = |U| (e xi + e y j + e z k )

A@0,

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A@0,

&   方向餘ÔK

 – 因此K

U  x = |U| e x,  U  y = |U| e y,  U  z  = |U| e z 

 –   ä些°:與° (2.15) 作³較K

c!" θ  x = e x, 

c!" θ  y = e y, 

c!" θ  z  = e z 

 –  U 的方向餘Ô‚於與q同方向!„[向,的0,; 

A@0,

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A@0,

&   /0,23[\向,K

 –   ÇÈÊ標  # x A) y A, z  A$ 的W A

/TÊ標  # x B) y B, z  B$ 的W

B ; 

A@0,

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A@0,

&   /0,23[\向,K

 –   f A 到 B 的[\向,  r AB K

r AB = ( x B −  x A)i + ( y B −  y A) j + ( z  B −  z  A)k    #2'&!$

A@0,

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A@0,

&   於ñ定Ù!向,的0

,K –  &A@7用V,常/與向

, U !Ù8的g個W  A  與  B 的Ê標©定,

1定…向, U 的方向

 –   î用° (2.17) 決定[\向,

r AB

A@0,

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Chapter 2 Engineering Mechanics: STATICS

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A@0,

j於ñ定Ù!向,的0,K

mä r AB ¤/¹的deÃ到一個fW A ©向W B 的„

[向,  e AB;

 –  e AB 的方向與  U X同 

m用 U 的de與

 

e AB XŸ的ŸB K U = |U| e AB

ª9 2 6

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ª9 2.6 

方向餘Ô 

椼架的 C W的Ê標“  xC  = 4 $) yC  = 0) z C  = 0 , W的Ê標

“  x D = 2 $) y D = 3 $) z  D = 1 $'

f C W到 W的[\向,  rCD 的方向餘Ô“

什麼% 

ª9 2 6 ( ¶ )

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方±

  ê道 C W# 的Ê標,()能ðµ  rCD 的0,;然w

()能×.  rCD 的de  ( f C 到 的hi ) ã且Ë用

° (2.15) .µ的方向餘Ô; 

ª9 2.6 ( ¶ )

ª9 2 6 ( ¶ )

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¯·

E向,0,1.µ[\向,  rCD ;

rCD = ( x D −  xC )i + ( y D 

−  yC ) j + ( z  D −  z C )k 

= (2 − 4)i + (3 − 

0) j +

(1 − 0) k  ($)= −2i + 3 j + k  ($)

ª9 2.6 ( ¶ )

ª9 2 6 ( ¶ )

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¯·

 ×. rCD 的de;

( ) ( ) ( )

$3.74

$1$3$2222

222

=++−=

++= z CD yCD xCDCD rrrr

ª9 2.6 ( ¶ )

ª9 2 6 ( ¶ )

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¯·

×.方向餘Ô;

267.0$3.74

$1 c!"

,802.0$3.74

$3 c!"

,535.0

$3.74

$2 c!"

===

===

−=−==

CD

 z CD z 

CD

 yCD

 y

CD

 xCD x

rr

r

rr

r

θ 

θ 

θ 

ª9 2.6 ( ¶ )

ª9 2 6 ( ¶ )

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»習¼ 

椼架的W B 的Ê標“  x B = 2.4 $) y B = 0) z  B = 3 $ ;®×.f

W B ©向 W!„[向,  * BD 的0,;

 ·½K * BD = − 0.110i + 0.827 j − 0.551k 

ª9 2.6 ( ¶ )

ª9 2 7 2 ! " 2 #

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ª9 

2.7 2.! " 2.#

×.A@向,0, 

( ¾¿Àu )

 

ª9 2.$%×.力的0, 

( ¾¿Àu )

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Chapter 2 Engineering Mechanics: STATICS

®¾¿Àut 2 'ö詳細的內Ä