Cơ lí thuyết 1

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C HOC L THUY T 1

TR NG I H C K THU T CNG

NGHI P THI NGUYN

KHOA KHOA H C C B N-B MN C H C
http://en.wikipedia.org/wiki/Image:Rack_and_pinion_animation.gif

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M U

C h c l khoa h c nghin c u chuy n ng c h c c a v t ch t. Trong , chuy n ng c h c l s d i ch c a v t ch t t v tr ny sang v tr khc

trong khng gian, theo th i gian.C h c l thuy t l m t ph n C h c nghin c u cc

quy lu t chung nh t v chuy n ng c h c .

C h c l thuy t l mn h c c s cho hng lo t cc mn k thu t c s v k thu t chuyn ngnh khc.

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Mn c h c l thuy t c chia lm ba ph n: TNH H C V T R N NG H C NG L C H C

C H C L THUY T 1 g m hai ph n TNH H C V T R N v NG H C .

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TI LI U THAM KH O

GS.TSKH Sanh -Ch c ( t p 1) , - NXB Gio d c . GS.TSKH Sanh-Bi t p ch c ( t p 1) , - NXB

Gio d c .

Chu T o oan-C h c l thuy t (t p 1),-NXB Giao thng v n t i. C h c l thuy t GS.TSKH o Huy Bch, Ph m

Huy n NXB i h c Qu c gia H N i, 1999.

Tuy n t p bi t p c l thuy t T p 1 : I.VMestcherski- NXB i h c v THCN,1980.

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CC THNH PH N I M i m gi a k: 20% i m th o lu n: 20% i m cu i k: 60%

HNH TH C THI Thi gi a k: T lu n 1 bi t p

th i gian: 45 Thi cu i k: Tr c nghi m (kho ng 20 cu)

(th i gian: 90)

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Ph n I

TNH H C V T R N

Tnh h c v t r n l ph n nghin c u tr ng thi cn b ng c a v t r n tuy t i d i tc d ng c a cc l c .

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Ph n I

TNH H C V T R N Ch ng 1 : Cac khai nim c ban

va h tin tinh hoc

Ch ng 2 : Cn bng cua h l c khng gian Ch ng 3 : Tr ng h p ring: H l c phng

Ch ng 4 : Ma sat

Ch ng 5 : Trong tm cua vt rn

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Ch ng 1

CC KHI NI M C B N V H TIN TNH H C 1. M u . t bai toan tinh hoc

2. Cc khi ni m c b n v l c

3. H tin tnh h c

4. Lin k t. Ph n l c lin k t.

Tin gi i phng lin k t

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1. M U. T BI TON TNH H C

1.1. i t ng nghin c u 1.2. S cn b ng c a v t r n 1.3. L c

1.4. Bi ton tnh h c

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1.1. i t ng nghin c u

- V t r n tuy t i l cac vt m kho ng cch

gi a cc i m c a n khng thay i khi ch u tcd ng c a v t khc. - V t r n tuy t i l m hnh cua cc v t r n

th c t khi cc bi n d ng c a chng th b qua c do qu b ho c khng ng vai tr quan tr ng trong qu trnh kh o st. V t r n tuy t i c g i t t l v t r n.

- i t ng nghin c u cua tinh hoc la vt rn tuyt i.

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1.2. S cn b ng c a v t r n - Khi ni m chuy n ng hay cn b ng c a v t

r n c tnh t ng i.

- Kh o st s cn b ng m t v t r n lun lun

g n li n v i v t lm m c no . -H quy chi u : V t lm m c dng kh o st

s cn b ng hay chuy n ng c a cc v t c g i l h quy chi u.

Trong cc bi ton k thu t thng th ng h quy chi u c ch n l cc v t t trn m t t.


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V t A: H quy chi u

V t B

N Cn b ng c a v t r n : M t v t r n c g i l cn b ng (ho c ng yn ) i v i

m t v t no n u kho ng cch t m t i m b t k c a v t n i m g c c a h quy chi u lun lun khng i.

1.2. S cn b ng c a v t r n


O

Mco

nst

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1.3. L c

L c l i l ng dng o tc d ng t ng h (t ng tc) gi a cc v t, m k t qu c a n l lm cho cc v t thay i tr ng thi chuy n ng ho c b bi n d ng i.

Cc c tr ng c a l c: i m t c a l c

Ph ng chi u c a l c

C ng c a l c

A

F

ng tc d ng c a l c (gi c a l c).

...,, QRF L c c bi u di n b ng vc t . K hi u


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Biu din l c trong h toa cac

cos .Z

F=

x y zF X e Y e Z e= + +

Trong h to cc vung gc vc t l c

c bi u di n d i d ng:

F

trong :

, ,x y z

e e e l cc vc t n v trn cc tr c to x, y, z.

, ,X Y Z l hnh chi u c a F ln cc tr c t a .

l n cua : F 2 2 2F X Y Z= + +

H ng cua c xac inh b i :F

cos ,X

F= cos ,

Y

F=

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1.4. Bi ton tnh h cBi ton tnh h c t ra l thi t l p cc i u

ki n cn b ng c a v t r n ch u tc d ng c a m t

h l c .

T p h p cac l c tc d ng ln cung m t v t r n g i l h l c.

K hi u h l c l: 1 2( , ,..., )nF F F

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2. CC KHI NI M B SUNG V L C

2.1. C

c nh ngh a v h l c 2.2.Mmen c a l c i v i m t i m.

2.3. Mmen c a l c i v i m t tr c.

2.4. Vct chnh v mmen chnhc a h l c khng gian

2.5. Ng u l c.

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2.1. CC NH NGHA V H L C

Hl c t ng ng: Hai h l c t ng ng l hai h l c c cng tc d ng c h c ln m t v t r n . K hiu:

( ) ( )1 2 1 2

, ,..., , ,...,n m

F F F P P P:

H p l c c a h l c: N u m t h l c t ng ng v i m t v ch m t l c th l c g i l h p l c c a h l c, hay h l c cho c h p l c. K

hi u h p l c c a h l c l : AR

1 2( , ,..., )n AF F F R:

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nh l :i u ki n c n v v t r n cn b ng l h l c tc d ng ln n cn b ng.

H l c cn b ng: H l c cn b ng l h l c khng lm thay i tr ng thi c h c c a v t r n . K hi u: 1 2( , ,..., ) 0nF F F :

2.1. CC NH NGHA V H L C

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2.2.MMEN C A L C I V I M T i MKhi l c tc d ng ln v t, n c th lm cho v t

quay quanh m t i m no . Tc d ng c a l c c c tr ng y b ng mmen c a l c i v i m t i m.

nh ngha : Mmen c a

l c i v i i m O l mt vect , k hi u l xc nh b ng cng th c:

( )Om F

( )Om F r F =

rtrong o l vct nh v cua i m t l c so v i im O.

r OA=

AO

F

( )o

m F

r

B

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Ta xac inh vec t nh sau: ( )om F

)(Fmo

Ph ng :vung gc v i m t ph ng ch a i m O v l c Chi u :C chi u sao cho khinhn t u mt c a n xu ng

g c th y vng quanh O theochi u ng c chi u kim ng h .

l n: dFFmo .)( =

F

F

V i d l kho ng cch vung gc l y t tm l y mmen O n ng tc d ng c a l c.

(=0 khi F = 0 ho c d = 0)

2.2.MMEN C A L C I V I M T i M

AO

d

F

( )o

m F

B

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Nu t tai O h toa Oxyz, va ky hiu:

( )

x y z

o

e e e

m F r F x y z

X Y Z

= =

Trong : , ,x y ze e e l cc vct n v trn cc tr c t a .

{ }ZYXF ,,={ }zyxr ,,=

Hnh chi u c a )(Fmo ln ba tr c t a :

( )oxm F yZ zY =

( )oym F zX xZ =

( )oz

m F xY yX =

thi


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Vi du 1.1

A D

CB

A' D'

C'B'

y

z

x

a

1F

2F

xe

ye

ze

( )2Am F

( )1Am F

Khi hnh lp ph ng canh a, chiu tc dung cua cc l c nh hinh ve. Tm cc vc t mmen c a cc l c i v i nh A.

1 2,F F

( )1 1( )A xm F aF e=

2 2

2

2( )

2

2

2

A x

y

am F F e

aF e

=

ap s:


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M men c a l c i v i m t tr c c tr ng cho

tc d ng c a l c lm v t quay quanh tr c .

2.3. MMEN C A L C I V I M T TR C

O

A

B

A'

B'

F

Fd'

F)(Fm

F

F

inh nghia: Mmen cua l c i v i truc , ky hiu la

, , la s ai s bng tich

hinh chiu cua ln mt phng vung goc v i truc va khoang cach d' t giao im O cua truc v i mt phng n , ly du cng nu quay xung quanh O

theo chiu ng c chiu kim ng h va ly du tr trong tr ng h p ng c lai.

F

F

' '( ) .m F F d = (= 0 khi no?)

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nh l lin h gi a m men c a l c i v i m t i m v m men c a l c i v i m t

tr c. F

( ) ( )Om F hc m F =

Mmen cua l c i v i truc i qua dim O la hinh chiu ln truc cua mmen cua no i v i im O.

O

A

B

A'

B'

F

Fd'

( )Om F

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xyF X

Z

2,F FCho l c tc dung vo khi lp ph ng, canh a, im t tai inh A. Tm m men cua cac l c i v i truc ba truc toa .

Vi du 1.2

( ) 22

xm F aF =

( )2

2ym F aF =

( )2

2z

m F aF =

ap s

( )2 2 sin ,xm F F a =

( )2 2 sin ,ym F F a =

( )2 0zm F =1

sin3

=

O

A

y

z

x a

F2F

B

C

B' A'

C'O'

2.3. MMEN C A L C I V I M T TR C

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2.4.1 Vect chinh c a h l c khng gian nh ngha: Vct chnh c a h l c khng gian, k hi u l t ng hnh h c c a cc vect bi u di n cc l c c a h l c:

R

1 2

1

n

n k

k

R F F F F

=

= + + + =

2.4. VC T CHNH V MMEN CHNHC A H L C KHNG GIAN.

Ph ng phap xac nh vect chinh a. Ph ng phap v (hinh hoc) b. Ph ng phap chi u (giai tich)

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Chu y : Vect chinh la vec t t do.

a. Ph ng phap ve

Vc t chinh c a h l c b ng vect khep kin c a a giac vect l c.

1F

2F

3F

1F

2F

3F

R

O

3F

2F

2.4.1 Vect chinh c a h l c khng gian

2.4. VC T CHNH V MMEN CHNHC A H L C KHNG GIAN.

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b. Ph ng phap chi u

1 21

n

x n kkR X X X X== + + + =

1 2

1

n

y n k

k

R Y Y Y Y=

= + + + =

1 2

1

n

z n k

kR Z Z Z Z

== + + + =

1 2

1

n

n k

kR F F F F

== + + + =

Ky hiu:

( ), ,x y zR R R R=


( )1 1 1 1, ,F X Y Z= ( )2 2 2 2, ,.........................

F X Y Z=

( ), ,n n n nF X Y Z=

Ta c:

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2 2 2

x y zR R R R= + +cos ;x

R

R= cos ;y

R

R= cos .z

R

R=

Vy m un v ph ng chiu cua vc t chnh c xc inh b i:

( ), ,x y zR R R R=

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V d :Xc nh vc t chnh c a h l c g m ba l c sau: ( )1 1, 2, 3F =

( )2 4, 5, 7F =

( )3 2, 8, 1F =Bi gi i:

( )7, 5, 11R =Ta c: 2 2 27 5 11 195R = + + =

( ) ( )

( )

7 5

cos , ; cos , ;195 195

11cos ,

195

R Ox R Oy

R Oz

= =

=


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nh ngha: Mmen chinh c a h l c khng gian i v i tm O, k hi u l m t vect b ng t ng hinh h c cac vect mmen c a cac l c thu c h l c i v i tm O:

OM

1 1

( )

n n

O O k k k

k k

M m F r F= =

= =

2.4.2 Mmen chnh c a h l c khng gian i v i m t tm

Cach xac nh a. Ph ng phap v Mmen chinh c a h l c i v i m t tm b ng

vect khep kin c a a giac vect mmen. b. Ph ng phap chi u

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Cc thnh ph n c a vect m men chnh

theo cc tr c to cc: ( ) ( )Ox Ox k k k k k M m F y Z z Y= = ( ) ( )Oy Oy k k k k k M m F z X x Z= =

( ) ( )Oz Oz k k k k k M m F x Y y X= =


b. Ph ng phap chi u ( ), ,O Ox Oy OzM M M M=

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V d 1: Cho h l c g m ba l c, trong :

( )1 1, 2, 3F =

( )2 4, 5, 7F =

( )3 2, 8, 1F =

Bi gi i:

( )2, 1, 0 ;OA =

Ta c cc vc t nh v c a cc l c so v i i m O:

( )0, 2, 0 ;OB =

t t i A (2,-1,0)

t t i B (0,-2,0)

t t i C (3,1,2) Xc nh mmen chnh c a h l c trn i v i g c to O.

( )3, 1, 2OC=


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( )1 1, 2, 3F = ( )2 4, 5, 7F = ( )3 2, 8, 1F =

p d ng CT:

( )2, 1, 0 ;OA =

V y cc l c v cc vc t nh v t ng ng l:

( )0, 2, 0 ;OB = ( )3, 1, 2OC=

( ) ( )Ox Ox k k k k k M m F y Z z Y= = ( ) ( ) ( )1 1 1 1 2 2 2 2 3 3 3 3OxM y Z z Y y Z z Y y Z z Y= + +

OxM = ( )( 1).3 2.0 ( )( 2).7 ( 5).0+ ( )1.1 2.8+ 32=

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Vi du Khi hinh lp ph ng chiu tac dung cua

cac l c nh hinh ve. Hay tinh vect chinh va mmen chinh cua h l c o i v i tm A.

A D

CB

A' D'

C'B'

y

z

x

a

1F2F

xe

ye

ze

3F

4F

ap s

( )1 2 4

2

3 4

2 ;2

2;

2

2;

2

x

y

z

R F F F

R F

R F F

= +

=

= +

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F

Fd

2.5. Ng u l c.

a. nh ngha

Ng u l c l h g m hai l c song song ng c chi u, cng c ng v khng cng ng tc d ng.

b. Cc c tr ng c a ng u l c + M t ph ng tc d ng + Chi u quay+ C ng tc d ng:

m = F.d.(d c g i l cnh tay n c a ng u l c)

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biu din cc c tr ng cua ngu l c ng i ta dng vect mmen ngu l c : m

Chu :Vect mmen c a ng u l c l vect t do v i m

t.

l n : m = F.d

m {Ph ng : vung gc v i m t ph ng tc d ng. Chi u : C chi u sao cho khinhn t u mt c a n xu ng

g c th y ng u l c quay theo chi u ng c chi u kim ng h .

( , )m m F F AB F BA F = = =

B

A

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Nh n xt: Vect m men c a ng u l c b ng t ng m men c a cc l c t o thnh ng u l c i v i i m b t k .

Tc d ng c a ng u l c khng thay i n u ta tu thay i cc l c t o thnh ng u l c mi n sao vect m men c a ng u l c khng i, hay ni khc i, vect m men c a ng u l c hon ton c tr ng cho ng u l c .

( , ) ( ) ( )O Om m F F m F m F = = +

d1

1F

1F

d22F

2F

F

1

.d

1

= F

2

.d

2

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3. H TIN TNH H C . CC H QU 3.1. H tin tnh h c

3.1.1. Tin 1 (Tin v h hai l c cn b ng). i u ki n c n v

h hai l c cn

b ng l hai l c ny c cng ng tc d ng, ng c chi u v cng

c ng .

ABF

'F

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3. H TIN TNH H C . CC H QU

3.1.2 Tin 2 (Tin thm b t hai l c cn b ng).Tc d ng c a m t h l c khng thay i n u

thm ho c b t hai l c cn b ng.

( ) ( )1 2 1 2, ,..., , ,..., , , ; ( , ) 0n nF F F F F F F F F F : :

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3.1.3 Tin 3 (Tin hnh bnh hnh l c). H hai l c cng t t i m t i m t ng

ng v i m t l c t t i i m t chung v c vect l c b ng vect cho hnh bnh hnh m hai

c nh l hai vect bi u di n hai l c thnh ph n. F

1F

2FO

( )1 2,F F F:

1 2F F F= +v


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3.1.4 Tin 4 (Tin tc d ng v ph n tc

d ng). L c tc d ng v l c ph n tc d ng gi a hai v t c cng ng tc d ng, h ng ng c chi u nhau v c cng c ng .

Ch :

L c tc d ng v l c ph n tc d ng khng ph i l hai l c cn b ng v chng tc d ng vo hai v t r n khc nhau.

BA F

F


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3.1.5 Tin 5 (Tin ho r n). M t v t bi n d ng cn b ng d i tc

d ng c a m t h l c th khi ho r n l i n v n

cn b ng.


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3.2. CC H QU

3.2.1. H qu 1 :

Tc d ng c a l c khng thay i khi tr t l c d c theo ng tc d ng c a n.

B BF

A AF

( ) ( ), , ;

A A B B

F F F F :( , ) 0 ;B B B A

F F F F =:

BF ( ) ( ) .A BF F :

L i c: ( , ) 0A BF F :

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3.2.2.K t qu thu g n h l c ng quy.

1 2

1

n

n k

k

R F F F F=

= + + + =

L

H qu 2: H l c ng quy c h p l c

t t i i m ng quy v bi u di n

vect chnh c a h

n u vect chnh khc khng, v cn

b ng n u vect chnh c a h b ng khng.

O

1F 2F

nF

3F

R

nF

1nR

O

3.2. CC H QU

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3.2.3. K t qu thu g n h ng u l c.

T p h p nhi u ng u l c t o thnh h ng u l c.

H qu 3. N u mmen chnh c a h ng u l c khc

khng, h ng u l c t ng ng v i m t ng u l c c m men b ng m men chnh c a h ; cn n u m men chnh c a h b ng khng h ng u l c cn

b ng .

3.2. CC H QU

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4. LIN K T, PH N L C LIN K T.

TIN Gi I PHNG LIN K T.

4.1 V t r n t do v v t r n khng t do. V t r n t do l v t r n c th th c hi n c

m i di chuy n v cng b t v tr ang xt sang v tr ln c n c a n.

Ng c l i, n u m t hay m t s di chuy n c a v t b c n tr b i nh ng v t khc th v t g i l v t r n khng t do.

V t khng t do cn g i l v t ch u lin k t, cn cc v t khc c n tr v t c kh o st g i l v t gy lin k t.

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Nh ng i u ki n c n tr di chuy n c a v t kh o st c g i l lin k t t ln v t y .

Trong tnh h c, ta ch nghin c u lo i lin k t c th c hi n b ng s ti p xc hnh h c gi a v t th c kh o st v i v t th khc, l

nh ng lin k t hnh h c.



4.1 V t r n t do v v t r n khng t do.

4 LIN K T PH N L C LIN K T

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4.2. Ph n l c lin k t

V t gy lin k t ngn c n chuy n ng c a v t kh o st, t c l v m t c h c n tc d ng vo

v t kh o st cc l c .Cc l c do cc v t gy lin k t tc d ng ln

v t kh o st g i l cc ph n l c lin k t .



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4.3. Cc tnh ch t c a ph n l c lin k t.

Tnh ch t th ng. Ph n l c lin k t xu t

hi n khng xc nh tr c m ph thu c vo cc l c cho tr c tc d ng ln v t kh o st v k t c u lin k t (t a, b n l , dy bu c,) c a v t gy lin k t.

C D

B

A

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51#

Ph ng, chi u c a cc ph n l c lin k t. Theo nh ngha,

ph n l c lin k t ph i cchi u ngn c n chuy n ng c a v t nn ng cv i xu h ng chuy n ng

c a v t.

4.3. Cc tnh ch t c a ph n l c lin k t.

C D

B

A

Dy ngn c n chuy n ng c a qu c u d c theo ph ng AB c a dy.

T ng khng cho qu c u di chuy n theo ph ng CD n m ngang.

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52#

Lin k t t a Lin k t t a xu t hi n khi v t r n kh o st t a

ln v t gy lin k t.

N2

N1N

4.4. Cc lin k t th ng g p v cc ph n l c lin k t t ng ng.

AN

BN

CN

N u b qua ma st th ph n l c lin k t t a c ph ng vung gc v i m t t a ho c ng t a v c chi u h ng vo v t kh o st.

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53#

T

1

T

2T

2T 1T

Lin k t dy m m, th ng Ph n l c lin k t n m d c theo dy, i m t

ch bu c dy v h ng ra ngoi v t kh o st. Ph n l c lin k t c a dy cn c g i l s c cng .

7/31/2019 C l thuyt 1

54/155

7/31/2019 C l thuyt 1

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55#

Lin k t g i Lin k t g i dng cc d m

v

khung

G i c nh : c ph n l c lin k t t ng t nh lin k t b n l .

G i di ng : Ph n l c lin k t c a g i di ng vung gc v iph ng di ng c a g i , gi ngnh lin k t t a.

BYAY

AX

R

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56#

Lin k t g i c u Lin k t g i c u c th th c hi n nh qu c u

g n vo v t ch u lin k t v c t trong m t v qu c u g n li n v i v t gy lin k t . Ph n l c g i c u i qua tm O c a c a v c u. Thng th ng ph n l c g i c u c c phn tich thnh 3 thnh ph n vung gc nhau.

X X AYZ

z

y

x

AX

AY

RA

Z

Spherical joint

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57#

Lin k t c i Lin k t c i cho php v t r n quay quanh tr c

Oz. Ph n l c lin k t c i c c phn thnh 3 thnh ph n vung gc nhau.

x

y

z

OX

OZ

OY

z

y

xO

X

OY

OZ R

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58#

Lin k t ngmHai v t c lin k t ngm khi chung c

gn c ng v i nhau.

Ngm ph ng:

AY

Am

AX

AX

AY

Am

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59#

Ngm khng gian:

O

z

y

x AX

AY

AZ

xm

ym

zm

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60#

Lin k t thanhLin k t thanh c hnh thnh nh th a mn

cc i u ki n sau: Ch c l c tc d ng hai u Tr ng l ng thanh khng ng k Nh ng lin k t hai u thanh c th c hi n nh b n l , g i c u.

Ph n l c lin k t thanh n m d c theo ng th ng n i hai u thanh, h ng vo

thanh khi thanh ch u ko vh ng ra kh i thanh khi thanh ch u nn. ( ng l c )

O1 O2

A B

AS

BS

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61#

4.5. Tin gi i phng lin k t. V t r n khng t do ( t c v t ch u lin k t)

cn b ng c th c xem l v t r n t do cn b ng n u gi i phng cc lin k t, thay th tc d ng c a cc lin k t c gi i phng b ng cc ph n l c lin k t t ng ng.

A

BC

O D

E

E

C

O DA

CB

1P

2P

AY

CN

1P

AX

CN D

NEN

2P

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62#

+ Thu g n h l c khng gian.

+ Tm i u ki n cn b ng c a h l c khng gian.

Ch ng 2

CN BNG CUA H L C KHNG GIAN

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63#

1. THU G N H L C KHNG GIAN V M T TM

1.1. Thu g n h l c khng gian v m t tm

1.2. Bi n i tm thu g n.

1.3. Cc kt qua thu gon ti gian

1.4. nh l Varinhng

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64#

1. THU G N H L C KHNG GIAN V M T TM

1.1. Thu g n h l c khng gian v m t tm

1.1.1 nh l d i l c song song.

AF BF

( )B AF F=L c t t i i m A t ng ng v i l c

t t i i m B b t k v ng u l c c m men b ng m men c a i v i i m B. AF

B AF F=( ), ;A BF F m:

( )B Am m F=

v i:

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65#

1.1.1 nh l d i l c song song.

AF BF

( )B AF F=

AF

L c t t i i m A t ng ng v i l c

t t i i m B b t k v ng u l c c m men b ng m men c a i v i i m B.

BB

F AFAF

( ); ( , )B A BF F F

:

( ), 0;B B B AF F F F =:

( ), ,A B BF F F:

Ch ng minh:

BF

T i B t:

( ), ;A BF F m: ( )B Am m F=

m

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66#

B

BF

m

NH N XT:

N u ta c: B mF th: ( ), AB mF F:

v i: ;A BF F=BAF

B

md

F=

A c v tr sao chong c v i chi u c a

( )A Bm Fm

1 1 2 Thu g n h l c

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67#

1.1.2 Thu g n h l c . ),...,,( 21 nFFFXt h l c khng gian:

p d ng nh l d i l c song song ta d i t ng l c v i m O.

( ) ( )' '1 1 1 1 1 1 1, ; , OF F m F F m m F= =:

( ) ( )' ', ; ,n n n n n n O nF F m F F m m F= =:.........................................

( ) ( ) ( )( )1 2 1 2 1 2, ,..., , ,..., ; ( ), ( ),..., ( )n n O O O nF F F F F F m F m F m F :

Hay

( ) ( )1 2, ,..., ,n O OF F F R M:

1 1 2 Th h l

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68#

( ) ( )1 2, ,..., ,n O OF F F R M:

V y h l c khng gian b t k t ng ng

v i m t l c t t i O v m t mmen ng u l c

. L c b ng vct chnh c a h , cn

b ng mmen chnh c a h i v i i m O.

OR

OM OR OM

1

n

k

kR F

==

1( )

n

O O k

kM m F

==

1.1.2 Thu g n h l c .

1 2 Bi n i tm thu g n

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69

#

1.2. Bi n i tm thu g n. 1.2.1. Bi n i tm thu g n.

),...,,( 21 nFFFXt h l c khng gian:

( ) ( )( )1 2 ' '( , ,..., ) , , ;n O O O O O OF F F R M R m R M: :Ta thu g n h l c ny v O v O':

M t khc: ( ) ( )1 2 ' ' '

1'( , ,..., ) , ;

n

O kO OOk

nF F M m FF R M ==

:

trong : ' .O OR R R= =( )( )' ', O OO OR m R M+:

( ) .O O O OM M m R = +Suy ra:

V y khi thay i tm thu g n ta c m t l c

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70

#

V y khi thay i tm thu g n ta c m t l c t tm m i, c gi tr khng i (b ng vct chnh), cn ng u l c m i c lin h v i ng u l c thu g n ban u theo bi u th c:

( )' ' .O O O OM M m R= +1.2.2. Cc b t bi n c a h l c khng gian.

Vct chnh l m t i l ng b t bi n.

Tch v h ng c a vct chnh v mmen chnh l m t i l ng b t bi n (ng khi vc t chnh khc khng).

. .( ( ) ) .O O O OR M R m R M R M = + =

1 3 Cc kt qua thu gon ti gian

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71

#

0, 0OR M= =

0, . 0OR M R =

0, 0OR M=

1.3. Cc kt qua thu gon ti gian

H l c cn bng.

H l c t ng ng v i mt ngu l c.

H l c co h p l c.

OOOR

OMOR

'O

R1

OR

O'0OM =0OM

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72

#

0, . 0OR M R H t ng ng v i h inh c ng l c.

O

OM OR

OM

OM

'OM

'O

ROR

OR

O

O'

//O OR MT c la

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73

#

1.4. nh l Varinhng

Trong tr ng h p h l c khng gian c h p l c th mmen c a h p l c i v i m t tm b t k

b ng t ng mmen c a cc l c thnh ph n i v i

tm y.

( ) ( )1

n

O O k O

k

m R m F M =

= =

2 I U KI N CN B NG C A H L C

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74

#

2.1. inh ly

2. I U KI N CN B NG C A H L C KHNG GIAN

2.2. Cc ph ng trnh cn b ng c a

h l c khng gian.

2.3. Ph ng trinh cn bng cua mt vai h l c c bit.

2 I U KI N CN B NG C A H L C

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75

#

2.1. inh ly iu kin cn va u h l c khng gian cn bng la vect chinh va mmen chinh cua h l c i v i mt im bt ky ng th i bng khng.

1

1 2

1

0

( , ,..., ) 0

( ) 0

n

k

k

n n

O O kk

R F

F F F

M m F

=

=

= =

= =

2. I U KI N CN B NG C A H L C KHNG GIAN

2 2 Cc ph ng trnh cn b ng c a

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76

#

2.2. Cc ph ng trnh cn b ng c a

h l c khng gian.

giai cac bai toan, ta th ng s dung cac ph ng trinh hinh chiu cua h ph ng trinh vect trn trong h truc toa cac:

1

0n

x k

k

R X=

= =

1

0n

y k

k

R Y=

= =

1

0n

z k

k

R Z=

= =

1

( ) 0n

x x k

k

M m F=

= =

1

( ) 0n

y y k

k

M m F=

= =

1

( ) 0n

z z k

k

M m F=

= =

2 3 Ph ng trinh cn bng cua mt vai

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77

#

2.3. Ph ng trinh cn bng cua mt vai h l c c bit.

00

0

x

y

z

RR

R

= = =

H ngu l c:

0

00

x

y

z

M

MM

=

= =

H l c ng quy:

H l c song song

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78

#

H l c song song Chon h truc toa sao cho truc Oz song

song v i ph ng cua cac l c. Ta co ba ph ng trinh cn bng:

1

1

1

0

( ) 0

( ) 0

n

z k

k

n

x x k

k

n

y y k

k

R Z

M m F

M m F

=

=

=

= =

= =

= =

x

y

z 1F

2F

nF

3 CAC BAI TOAN VA VI DU

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79

#

3. CAC BAI TOAN VA VI DU 3.1. Cac b c giai bai toan cn bng.

Cc bi ton tnh h c c th c chia thnh hai lo i sau:

Hy tm m i quan h gi a cc l c ho t ng

cho v t cn b ng, ho c n u bi t cc l c ho t ng hy tm cc v tr cn b ng c a v t.

V t cn b ng d i tc d ng c a cc l c ho t ng cho tr c, hy tm m t ph n ho c ton b cc ph n l c lin k t tc d ng ln cc v t.

CC B C GI I BI TON CN B NG

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80

#

CC B C GI I BI TON CN B NG B c 1: Ch n v t kh o st cn b ng.

B c 2: Gi i phng lin k t cho v t kh o st.

V t c ch n xt cn b ng l v t ch u tc d ng c a cc l c c n tm:

- m t v t r n.

- m t v t do nhi u v t ghp l i. - m t ph n t ng t ng tch ra t m t v t. - m t nt, i m t p trung cc dy, cc thanh.

V ring v t kh o st, thay cc lin k t b ng cc ph n l c lin k t t ng ng.


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81

#

B c 1: Ch n v t kh o st cn b ng. B c 2: Gi i phng lin k t cho v t kh o st.

B c 3: Thnh l p cc ph ng trnh cn b ng .

B c 4: Gi i h ph ng trnh cn b ng va nhn xet kt qua.


3.2. CC V D .

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82

#

Vi du 3.1 3.2. CC V D .

T m hnh ch nh t c tr ng l ng P = 1kN, cgi cn b ng v tr n m ngang nh hai b n l A,B

v dy treo IK t o gc =300 v i m t ph ng c a t m nh hnh v . Cc kch th c o b ng mt. Tm

cc ph n l c t i A, B v s c cng c a dy.

p s

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83

#

1 ;T kN=

p s

7 3 1, .

12 12

B BY kN Z kN = =

3 7, ;

12 12A AY Z kN = =

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Vi du 3 2

7/31/2019 C l thuyt 1

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85

#

C

B D

A

O

X

Y

Z

45o

30o

BI GI I:

AS

CS

DS

P

OAS

H

K

Kh o st nt O

Phn tch l c L p h PT cn b ng

Gi i h PT

Vi du 3.2

Vi du 3 3

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86

#

Vi du 3.3

Q

( ), ,4 6

b ax y

=

Tm ph n l c t i cc chn bn. Cc

kch th c cho trnhnh v .

M t chi c bn ba chn, c t trn m t

ph ng ngang. Tr ng l c c a bn t t i giao i m c a hai ng cho c a m t bn. T i i m K trn

m t bn, c t a ch u tc d ng c a

l c th ng ng .

Vi du 3.3

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87

#

Vi du 3.3

2

, , .3 4 6 4 6 4A B C

Q P Q P Q P

N N N= + = + = +p s :

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88

#

Bai tp :3-1 3-12; 3-16 3-18.trang 72 79, sach Bai tp c hoc (tp 1),

Sanh

Ch ng 3

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89

#

Ch ng 3TR NG H P RING: H L C PHNG

1. KHAI NIM MMEN AI S

2. H P L C CUA H L C PHN B PHNG

3. CN BNG CUA H L C PHNG 4. IU KIN CN BNG CUA H VT RN

1. KHAI NIM MMEN AI S

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90#

i v i h l c phng, ta a ra khai nim mmen ai s cua l c i v i mt im:

Mmen ai s cua l c i v i im O, ky hiu , la mt s ai s:

( )Om FF

( ) .Om F F d=

trong F l tri s cua l c, d l khoang cch t O

n ng tc dung cua l c, ly du "+" khi l cquay quanh O theo chi u ng c chi u kim ng h , v l y d u "-" trong tr ng h p ng c l i.

F

O

B

Ad

F

hi

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91#

( ) ( ) ( ) ( )1 21

...N

O O O O N O k

k

M m F m F m F m F

=

= + + + =

OM

M men chnh c a h l c ph ng i v i i m O l m t s i s , k hi u , b ng t ng m men i s c a cc l c c a h l c i v i

i m O:

Khi o


7/31/2019 C l thuyt 1

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92#


Xet oan dm AB dai l, chiu tac dung cua h

l c phn b song song cng chi u v i c ng phn b q(x):

q(x)

B

d l

Q

Sau y ta s xc nh h p l c :

H l c song song ny c h p l c, k hi u l . Q

Q

A


7/31/2019 C l thuyt 1

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93#

0

( ) ;l

AR q x dx= 0

( )l

AM q x x dx=

Thu gon h l c nay v im A:

( )A Am Q M=

Gi s h p l c t t i C cch A m t o n AC = d.

Theo nh l Varinhng:

q(x)

Bd

l

Q

A C

( ) . .A A

m Q Q d R d= =

.A A

M R d=


7/31/2019 C l thuyt 1

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94#

0

( ) ;

l

A

R q x dx=

0( ) ;

l

AM q x x dx=

V y:

q(x)

Bd

l

Q

A C

.A A

M R d=

0

0

( )

( )

l

l

q x xdx

dq x dx

=

0

( ) ;l

Q q x dx=

K t lu n:

7/31/2019 C l thuyt 1

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95#

K t lu n:

Q

c ng tc d ng i qua tr ng tm c a hnh phn b l c C chi u cng chi u v i cc l c thnh ph n c a h l c phn b .

C l n b ng di n tch c a hnh phn b l c.

Cac tr ng h p c bit:

7/31/2019 C l thuyt 1

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96#

H l c phn b c c ng phn b l c tuy n tnh 1

2

Q ql=

ld3

2=

H l c phn b c c ng phn b l c u

q(x) = q = const.Q ql=

l

q

l/2

Q

d = l/2

Cac tr ng h p c bit:

Q

A B

3. CN BNG CUA H L C PHNG

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97#

3. CN BNG CUA H L C PHNG

3.1. Cac ph ng trinh cn bng cua h l c phng

T iu kin cn bng cua h l c khng gian ta co cac ph ng trinh cn bng cua h l c phng sau y:

1 1 1

0 0 ( ) 0n n n

k k O k

k k k

X Y m F

= = =

= = =

trong o: x y, O la im bt ky .

x

y

z

1F

2FnF

O

Dang 1:

Dang 2:

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98#

1 1 1( ) 0 ( ) 0 0

n n n

A k B k kk k km F m F X = = == = =

Dang 2:

1 1 1

( ) 0 ( ) 0 ( ) 0n n n

A k B k C k

k k k

m F m F m F

= = =

= = =

Dang 3:

trong o: tr c x khng AB.

trong o: Ba i m A, B, C khng th ng hng .

3.2. Cc v d

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99#

Cho d m AB, c u A ngm vo t ng, cn b ng d i tc d ng c a cc l c v ng u l c nh hnh v .

Bi t: 0200 ; 180 ; 30 / ; 60 ; 1 .F N M Nm q N m a m= = = = =B qua tr ng l ng c a d m. Tm ph n l c lin k t

t i u A.

V d 3.1

3.2. Cc v d

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100#

( )

( )

( )

100 ,

100 3 60 ,

400 3 .

A

A

A

X N

Y N

M Nm

=

=

=

0

200 ; 180 ; 30 / ; 60 ; 1 .F N M Nm q N m a m= = = = =

4. IU KIN CN BNG CUA H VT RN

7/31/2019 C l thuyt 1

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101#

Ni l c : la l c t ng tac gi a cac vt trong h

.

Ch :

Vect chnh v mmen chnh c a h n i l c b ng khng.

0 ( ) 0i i i i

k O O k

k k

R F M m F= = = =

Ngoai L c: La cac l c do cac vt ngoai h tac dung ln cac vt thuc h vt ang khao sat.

i

kFKy hiu:

e

kFKy hiu:

CC I U KI N CN B NG C A H V T R N

7/31/2019 C l thuyt 1

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102#

i u ki n cn b ng c a t ng v t tch ring.

i u ki n cn b ng c a ton h ho r n ( xemton h nh m t v t r n duy nh t ), hay cn g il i u ki n cn b ng c a cc ngo i l c ( v khi

ho r n l i h n i l c tri t tiu ).

Vy ta co hai ph ng phap giai bai toan h vt:

Ph ng phap tach vt Ph ng phap hoa rn

Vi du 3.2:

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103#

Cho c h nh hnh v : = 30

o

, AB = 60m,BC = 20m. Trn o n AE ch u tc d ng c a h l c phn b u v i c ng q=20kN/m, AD = 40m, AE = 70m. B qua tr ng l ng c a d m v cc c t ch ng. Xc nh ph n l c lin k t t i A, l c t ng h t i B v ng l c trong cc thanh ch ng CC, DD.

q

A

B

C

C

D

D

E

Ho r n :

7/31/2019 C l thuyt 1

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104#

A

D

B

q

CS

BY

BX

DS

AY

AX

C

q

B

E

Nguyn Thi Kim Thoa

Tch v t BC: q

A

B

C

C

D

D

E

Tch v t AB:

7/31/2019 C l thuyt 1

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105#

BY

A

D

q

AX

AY

DS

Nguyn Thi Kim Thoa

'

BXBCS

BY

BXC

q

B

E

Vi du 3.3 :

7/31/2019 C l thuyt 1

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106#

Cho c h nh hnh v . Thanh b qua tr ng

l ng, hai qu c u c cng tr ng l ng P. Kchth c ghi trn hnh. Xc nh ph n l c lin k t t i A, D, C v l c t ng h t i B.

2m 2m

1m 1m 1m

A B C D

DYCNAY

AX

P P

DX

Tch v t

7/31/2019 C l thuyt 1

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107#

B

2m 2m

A

P

AX

AYBN

BN

1m 1m 1m

B C D

P

CNDY

DX

Nguyn Thi Kim Thoa

V d 3.4Gi h 5 th h h l h

7/31/2019 C l thuyt 1

108/155

108#

Gin ph ng g m 5 thanh ch u l c nh hnh v v c b ng g i di ng A, g i c

nh B. Tm ph n l c t i cc g i v ng l c trong cc thanh. Bi t:1 2 31 , 2 , 3 .F kN F kN F kN= = =

C

A BM

1m1m

1

F

2F 3F

060 030

Nguyn Thi Kim Thoa

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1311 ; 0, .4 4A B B

N X Y= = =Ho r n h ta tm c:

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110#

4 4

A

AMS

060

ACS

AN

2F

0

0

2

cos 60 0sin 60 0

k AM AC

k A AC

X S SY F N S

= + = = + + =

M'AMS

060

CMS

BMS

' 0

0cos 60 0

sin 60 0

k AM BM CM

k CM

X S S SY S

= + == =

B

BY

BX

3F

030

'

BMS

BCS

' 0cos 30 0k B BM BC X X S S= =

Tch nt:

Nguyn Thi Kim Thoa

ACS2FCM

S 3FS

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111#

A

AMS

060

ANM

'

AMS

060

CM

BMS B

BY

BX

030

'BMS

BCS

p s :

3 3, , 0,

2 4AC AM CM

S kN S kN S= = =

3 1, .

4 2BM BC

S kN S kN = =

1311 ; 0, .4 4A B B

N X Y= = =

KL: T d u c a cc nghi m, ta c th k t lu n cc thanh AM, BM ch u ko, cn cc thanh BC, AC ch u nn.

Nguyn Thi Kim Thoa

V d 3.5 :

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112#

Cho c h nh hnh v . Cho P = 50kN,

q = 5kN/m, M=200kNm, KB = KD. Tr ng l ng c a thanh

ACB l P = 100kN, tr ng l ng c a thanh BD l P = 60kN.

Tm ph n l c lin k t t i ngm A, b n l B v g i di ng D. 1m 2m

1m

3m

q

A

B

C

D

K

PM

V d 3.6 :

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113#

V t n ng P c treo vo nt (1,2) c a gin g m

5 thanh (1,2,3,4,5) b tr nh hnh v v c gi c nh nh ba thanh 6, 7, 8. Thanh b qua tr ng l ng v

c xem nh n i v i nhau v n i v i t ng b ng b n

l . Tm ng l c c a cc thanh. 1

2

3

4

5

6

7

8P

600

600

300

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114#

Bai tp :

1-1 1-20; trang 24 32;

2-1 2-25 trang 48 58;

sach Bai tp c hoc (tp 1), Sanh

BI T P N P (l p 44M)

7/31/2019 C l thuyt 1

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115#

Nhm 1: A & B & C: 1-9, 2-6, 2-8, 3-4, 4-14.

Nhm 2: D & : 1-5, 2-2, 2-12, 3-5, 4-5.

Nhm 3: H: 1-10, 2-8b, 1-13, 3-9, 4-4.

Nhm 4: L & N: 1-13, 2-9, 2-13, 3-8, 4-3.

Nhm 5: Q & S: 1-14, 2-15, 2-23, 3-12, 4-1.

Nhm 6: TH & TR: 1-17, 2-14, 2-16, 3-14, 4-2.Nhm 7: TU & V: 1-18, 2-21, 2-24, 3-23, 4-13.

Ph n I

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116#

TNH H C V T R N

Ch ng 1: Cac khai nim c ban va h tin

tinh hoc

Ch ng 2: Cn bng cua h l c khng gian

Ch ng 3: Tr ng h p ring: H l c phng

Ch ng 4: Ma sat

Ch ng 5: Trong tm cua vt rn

Ch ng 4 MA ST

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117#

Ch ng 4 MA ST

1. PHAN L C LIN KT TRN CAC MT T A. KHAI NIM V MA SAT VA S PHN LOAI.

2. INH LUT MA SAT COULOMB

3. CN BNG CUA CAC VT RN CHIU CAC 3. CN BNG CUA CAC VT RN CHIU CAC LIN KT CO MA SAT LIN KT CO MA SAT

1.PHAN L C LIN KT TRN CAC MT T A.

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118#

KHAI NIM V MA SAT VA S PHN LOAI.

1.1. M hnh ph n l c lin k t trn cc m t t a Trong th c t , cc v t r n khi ti p xc v i

nhau lun lun x y ra trn m t mi n nh no .

Do , khi hai v t ti p xc v i nhau s xu t hi n m t h cc ph n l c lin k t . Ccl c ny ngn c n cc chuy n ng ho c xu h ng chuy n ng c a v t nay i v i v t kia.

1.2. Khi ni m v l c ma st

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119#

Thu g n h ph n l c tai min tip xuc v

m t i m ti p xc no , ta c l c v ng u l c. Ta phn tch l c v ng u l c thnh cc thnh ph n php tuy n v ti p tuy n:

( , )msR N F:

( , )

l x

ms msM M M:

V y h ph n l c lin k t t ng ng v i 4 th h h h l

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120#

thnh ph n ph n l c:

Thnh ph n ph n l c ti p tuy n k hi u l ngn c n chuy n ng tr t ho c xu

h ng tr t c a v t trn b m t lin k t; g i l l c ma st tr t.

msF

Thnh ph n ph n l c php tuy n nhth ng th y, ngn c n chuy n ng theo ph ng php tuy n c a b m t v t;

N

Thnh ph n ng u l c ngn c n s lM

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121#

Thnh ph n ng u l c ngn c n s ln c a v t trn b m t lin k t; g i l ng u

l c ma st ln .

msM

Thnh ph n ng u l c

ngn c n s xoay c a v t xung quanh php tuy nc a m t lin k t, g i l

ng u l c ma st xoay .

x

msM

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1.3. Phn loai ma sat

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123#

D a vao trang thai c hoc cua vt ta phn loai ma sat thanh: ma sat tinh va ma sat ng.

Ma sat tinh: la ma sat xut hin khi cac vt trang thai ng yn hay khi co cac xu h ng chuyn ng t ng i gi a vt nay va vt kia.

Ma sat ng: la ma sat xut hin khi cac vt chuyn ng t ng i v i nhau.

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124#

D a vao tinh cht cua b mt tip xuc ta co:

ma sat kh va ma sat nh t

Ma sat kh : la ma sat xut hin khi cac b

mt cua cac vt tip xuc tr c tip (khng co cac l p bi tr n nh du, m ).

Ma sat nh t : Khi trn b mt cac vt tip

xuc co cac l p bi tr n ta co ma sat nh t.

2. INH LUT MA SAT COULOMB Cac inh lut ma sat c xy d ng t th c nghim vt

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125#

2.1. inh lut ma sat tr t. L c ma sat tr t tnh xut hin ngn can s tr t hoc xu h ng tr t t ng i cua hai vt tip xuc v th a mn b t ng th c:

trong o, f l h s ma st tr t tnh - ai l ng

khng th nguyn - c tr ng cho ban cht vt ly cua cac mt tip xuc; N l ph n l c php tuy n.

Cac inh lut ma sat c xy d ng t th c nghim vt ly

.ms

F f N

2.2. inh lut ma sat ln. 2.2. inh lut ma sat ln.

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126#

Ngu l c ma sat ln xut hin ngn can s

lnt ng i gi a cac vt tip xuc v th a mn b t ng th c:

trong o, k la h s ma sat ln th nguyn la chiu dai c tr ng cho ban cht vt ly cua cac vt tip xuc.

.lms

M k N

inh lut ma sat xoay cung c phat biu t ng inh lut ma sat xoay cung c phat biu t ng t .t .


7/31/2019 C l thuyt 1

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127#

LIN KT CO MA SAT LIN KT CO MA SAT

3.1. Cac b c giai bai toan cn bng cua vt chiu lin kt co ma sat.

B c 1 : Chon vt khao sat va giai phong lin kt

cho vt nh bai toan khi ch a xet n ma sat.

B c 2 : t thm cac l c, ngu l c ma sat.

Cn xet xu h ng chuyn ng cua vt xac inh ung chiu cua l c, ng u l c ma sat.


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128#

LIN KT CO MA SAT LIN KT CO MA SAT

3.1. Cac b c giai bai toan cn bng cua vt

chiu lin kt co ma sat. B c 3 : Vit ph ng trinh cn bng cho h l c tac

dung ln vt (gm ca cac l c ma sat).

H n n a cac l c ma sat phai thoa man cac BT ma sat.

B c 4 : Giai h gm cac ph ng trinh va cac BPT.

Chu y: Chu y: Nghim cua h gm cac ph ng trinh v Nghim cua h gm cac ph ng trinh va cac bt ph ng trinh la mt min nghim (th hin cac bt ph ng trinh la mt min nghim (th hin d i dan bt n th c . d i dang bt ng th c).

7/31/2019 C l thuyt 1

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Vi du 4.2:

7/31/2019 C l thuyt 1

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130#

Thanh ng cht AB, co

trong l ng P, u B t a vao t ng khng nhn, u A t a vao san nhn nm ngang chiu tac dung

cua l c Q nh hinh ve. H s ma sat gi a thanh va t ng la f. Tim gia tri cua l c Q thanh cn bng

vi tri nghing goc v iph ng nm ngang.

a

A

B

C

P

Q

7/31/2019 C l thuyt 1

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V d 4.4:

7/31/2019 C l thuyt 1

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132#

O

Q

Trn m t n m ngang c bnh xe ng

ch t tm O, bn knh R, tr ng l ng P, ch u l c Q nh hnh v . Xc nh tr s Q bnh xe cn b ng . Bi t h s ma st tr t f, h s ma

st ln k.

7/31/2019 C l thuyt 1

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Ph n I

C

7/31/2019 C l thuyt 1

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134#

TNH H C V T R N Ch ng 1: Cac khai nim c ban va h tin

tinh hoc

Ch ng 2: Cn bng cua h l c khng gian Ch ng 3: Tr ng h p ring: H l c phng

Ch ng 4: Ma sat

Ch ng 5: Trong tm cua vt rn

7/31/2019 C l thuyt 1

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135#

Ch ng 5

TRONG TM CUA VT RN

1.TM CUA H L C SONG SONG.

i h hi

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136#

1.1. inh nghia

i m hnh h c C g i l tm c a h l c song

song c xc nh b i cng th c:

1

1

n

k k

k

C n

k

k

F r

r

F

=

=

=

Cho h l c song song bt ky v i (h co h p l c), co cac im t t ng ng la 0kF

1 2, ,... nF F F1 2

, ,... nM M M

kF kFtrong , l thnh ph n hnh chi u c a l c trntr c song song v i cc l c.

k kr OM=Ky hiu . Ta co inh nghia:

1.2. Tnh ch t

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137#

H p l c c a h l c song song i qua i m C

v n u quay cc thnh ph n quanh cc i m t c a chng m t gc trong i u ki n gi nguyn i m t v gi tr c a cc l c thnh ph n th h pl c c a chng cng quay quanh tm C m t gc .

C

M1

M2 M3

M4

1F

2F

3F

nF

R1F

2F

3F

nF

R

2. INH NGHIA TRONG TM CUA VT RN.

Kh o st v t r n n m g n tri t V t ch u

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138#

Kh o st v t r n n m g n tri t. V t ch u tc d ng c a l c h p d n c a tri t, g i l tr ng

l c P c a v t .

k k k k

C

k

P r P rrPP

= =

Tm C c a h tr ng l c c xc nh b i cng th c:

CM1M2

Mk

kP

1P

2PP

im C (co vi tri c inh i v i vt) goi la trong tm cua vt rn.

Cng th c xc nh cc t a tr ng tm c a v t r n:

7/31/2019 C l thuyt 1

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139#

; ; .k k k k k k

C C CP x P y P zx y z

P P P = = =

( ) ( ) ( )

1 1 1

; ; .C C CV V V

x xdP y ydP z zdPP P P= = =

k k

C

P r

r P

=

( )

1

.CV

r rdPP =

Dang hinh chiu trong h toa Descarte:

7/31/2019 C l thuyt 1

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3.3. nh l 3 ( nh l Guynanh 1 )

7/31/2019 C l thuyt 1

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141#

( y )Di n tch S c a m t

trn xoay sinh ra do m t ng cong ph ng AB khi quay quanh tr c ng ph ng , nh ng khng c t n, c

c nh b i cng th c:

trong , L l di c a

ng cong AB, cn d lkho ng cch t tr ng tm C c a ng cong n tr c .

2S Ld=x

xc

C

ds

dl

B

A

3.4. nh l 4 ( nh l Guynanh 2 )

7/31/2019 C l thuyt 1

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142#

Th tch V c a m t v t trn xoay sinh ra

b i m t t m ph ng khi quay quanh tr c vkhng c t n, c xc nh b i cng th c:

2V Sd

=trong , S l di n tch t m ph ng; d l kho ng cch t tr ng tm c a t m n tr c .

3.5. Cc ph ng php tm tr ng tm c a v t r n.

3 5 1 Ph ng php i x ng

7/31/2019 C l thuyt 1

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143#

3.5.1. Ph ng php i x ng.

p d ng nh l 1. V d

Thanh th ng, vnh trn, m t trn, m t hnh ch nh t, hnh h p ch nh t, hnh c u ng ch t u c tr ng tm t i tm i x ng c a v t .

3.5.2. Ph ng php phn chia

Chia vt thanh cac phn a bit trong tm ri ap

7/31/2019 C l thuyt 1

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144#

Chia vt thanh cac phn a bit trong tm, ri ap

d ng CT: k kC P rrP=

V d: Tm tr ng tmc a m t t m ph ng ng ch t, hnh ch L, v i cc kch th c nh hnh v .

a

b

d

d

O1

O2

A B

C D

EGH

y

X

V i la vec t inh vi trong tm cua phn th k. kr


P r

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145#

k k

C

P rr

P

=

1 1

2 2

P S

P S

==

1 1 2 2 1 1 2 2

1 2 1 2

c

Px P x S x S xx

P P S S

+ += =

+ +

V d:

a

b

d

d

O1O2

A B

C D

EGH

y

X

1P

2P

1 1 2 2

1 2

C

S x S xx

S S+=+


V d: S x S x+

7/31/2019 C l thuyt 1

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146#

V d:

a

b

d

d

O1O2

A B

C D

EGH

y

X

1P

2P

1 1 2 2

1 2

C

S x S xx

S S

+=

+( )1 2

1 2

. ; . ;

; .2 2 2

S b d S a d d

d a d a d

x x d

= =

+

= = + =

3.5.3. Ph ng php kh i l ng m

(ph ng php b).

7/31/2019 C l thuyt 1

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147#

(p g p p )Khi v t b khot nhi u l c hnh th khc nhau m

tr ng tm c a cc l khot c th tm c, th ta c th p d ng ph ng php phn chia trn, v i i u ki n l cc l khot i c kh i l ng mang d u m.

V d : Tm tr ng tmc a m t t m trn ng ch t, c bn knh R, bntrong t m b c t i m t

mi ng hnh ch nh t c hai c nh a, b v tr nh hnh v .

O

Rb/2

b/2

a

x

y

3.5.4. Ph ng php tch phn.

N u v t l m t kh i ng ch t c th tch V :

7/31/2019 C l thuyt 1

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148#

N u v t l m t kh i ng ch t c th tch V :

( )

1 ;CV

r rdV V

=

N u v t l m t m t ng ch t c di n tch

S : 1;C

S

r rdSS

=

N u v t l m t thanh ng ch t, c chi u di L :

( )

1;C

L

r rdLL

=

( )

1 ;CV

r rdPP

= .P V=uuuuuuuu

V dTm tr ng tm c a n a a trn ng

7/31/2019 C l thuyt 1

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149#

g gch t, c bn knh R

O x

y

RO x

y

R

d

3.5.4. Ph ng php p d ng cc nh l Guynanh.

7/31/2019 C l thuyt 1

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150#

h/2

h/2

R

C

xc

O

B

A

x

y

V d

Tm tr ng tm c a cung trn ng ch t bn knh R, v i gc tm l 2

V dTm tr ng tm c a n a a trn ng

7/31/2019 C l thuyt 1

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151#

Tm tr ng tm c a n a a trn ng

ch t, c bn knh R

O x

y

R

4.TRONG TM CUA MT S VT RN NG CHT Tr ng tm c a m t thanh ng ch t l i m

7/31/2019 C l thuyt 1

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152#

Tr ng tm c a m t thanh ng ch t l i m gi a c a thanh.

C BAa a

Tr ng tm c a cc hnh bnh hnh, hnh ch nh t, hnh vung, ng trn, m t trn, kh i h p ch nh t, kh i l p ph ng ng ch t l tm c a chng.

C rC CC C r

Tr ng tm c a tam gic ng ch t l giao c a cc ng trung tuy n

7/31/2019 C l thuyt 1

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153#

C

Tr ng tm c a cung trn ng ch t AB c bn knh R v gc t i tm: 2AOB =

sinCx R

=

A

R

C

xc

O

B

x

y

Tr ng tm c a qu t trn ng ch t AOB c bn knh R v gc t i tm 2AOB =

7/31/2019 C l thuyt 1

154/155

154#

c bn knh R v gc t i tm 2AOB =

2 sin

3

C

Rx

= RC

xc

O

B

A

x

y

Tr ng tm c a kh i hnh chp, kh i hnh nn ng ch t

7/31/2019 C l thuyt 1

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nn ng ch t

Tr ng tm c a kh i hnh chp, kh i hnh nn u n m trn o n th ng n i t nh S n tr ng tm O c a y, v chia o n theo t l :

C

S

C

O

S

1

4CO SO=

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