Cơ Học Lý Thuyết - Nhiều Tác Giả, 237 Trang (1)

8/10/2019 C Hc L Thuyt - Nhiu Tc Gi, 237 Trang (1)

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le l ngi t nn mng cho vic hnh thnh mn c hc gii tch m

sau ny Lagrng, Hamintn, Jaccobi, Gaox hon thin thm.

Cn c vo ni dung v cc c im ca bi ton kho st, chng trnh

c hc ging cho cc trng i hc k thut c th chia ra thnh cc phn: Tnh

hc, ng hc, ng lc hc v cc nguyn l c hc. Tnh hc nghin cu cc

quy lut cn bng ca vt th di tc dng ca lc. ng hc ch nghin cu

cc quy lut chuyn ng ca vt th n thun v mt hnh hc. ng lc hc

nghin cu cc quy lut chuyn ng ca vt th di tc dng ca lc. Cc

nguyn l c hc l ni dung c bn nht ca c hc gii tch. C hc gii tch

chnh l phn ng lc hc ca h c trnh by theo hng gii tch ho.

C hc l khoa hc c tnh h thng v c trnh by rt cht ch . Khi

nghin cu mn hc ny i hi phi nm vng cc khi nim c bn v h tin

, vn dng thnh tho cc cng c ton hc nhhnh gii tch, cc php tnh vi

phn, tch phn, phng trnh vi phn... thit lp v chng minh cc nh l

c trnh by trong mn hc.

Ngoi ra ngi hc cn phi thng xuyn gii cc bi tp cng c kin

thc ng thi rn luyn k nng p dng l thuyt c hc gii quyt cc bi

ton k thut.


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Phn I

Tnh c

Chng 1

Cc khi nim c bn v h tin ca tnh hcl thuyt v m men lc v ngu lc

1.1. cc khi nim c bn

Tnh hc nghin cu cc quy lut cn bng ca vt rn tuyt i di tc

dng ca lc. Trong tnh hc c hai khi nim c bn l vt rn tuyt i v lc.

1.1.1. Vt rn tuyt i

Vt rn tuyt i l vt th c hnh dng bt bin ngha l khong cch hai

phn t bt k trn n lun lun khng i. Vt th c hnh dng bin i gi l

vt bin dng. Trong tnh hc ch kho st nhng vt th l rn tuyt i th

nggi tt l vt rn. Thc t cho thy hu ht cc vt th u l vt bin dng. Song

nu tnh cht bin dng ca n khng nh hng n chnh xc cn c ca

bi ton c th xem n nhvt rn tuyt i trong m hnh tnh ton.

1.1.2. Lc v cc nh ngha v lc

Lc l i lng o tc dng c hc gia cc vt th vi nhau. Lc c

biu din bng i lng vc t c ba yu t c trng: ln (cn gi l cng

), phng chiu v im t. Thiu mt trong ba yu t trn tc dng ca lc

khng c xc nh. Ta thng dng ch ci c du vc t trn k hiu cc

vc t lc. Th d cc lc Pr

, 1Fr

,.... Nr

. Vi cc k hiu ny phi hiu rng cc

ch ci khng c du vc t trn ch l k hiu ln ca n. Th d ln

ca cc lc Pr

, Fr

... l P, F, ...N. ln ca cc lc c th nguyn l Niu tn

hay bi s Kil Niu tn vit tt l (N hay kN).

Nr

Sau y gii thiu mt s nh ngha:


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H lc: H lc l mt tp hp nhiu lc cng tc dng ln vt rn.

Lc tng ng: Hai lc tng ng hay hai h lc tng ng l hai

lc hay hai h lc c tc ng c hc nh nhau. biu din hai lc tng

ng hay hai h lc tng ng ta dng du tng ng nhtrong ton hc.

Th d hai lc Fr

v Pr

tng ng ta vit Fr

Pr

. Hai h lc ( 1Fr

, 2Fr

,.. nFr

) v ( P1r

,

2Pr

,.. mPr

) tng ng ta vit ( 1Fr

, 2Fr

.. nFr

) ( 1Pr

, 2Pr

,.. mPr

).

Hp lc: Hp lc ca h lc l mt lc tng ng vi h lc cho. Th

d nu c Rr

( 1Fr

, 2Fr

,.. nFr

) th Rr

c gi l hp lc ca h lc ( 1Fr

, 2Fr

,.. nFr

).

H lc cn bng: H lc cn bng l h lc tng ng vi khng (hp

lc ca n bng khng). Th d: h lc ( F1r

, 2Fr

.. nFr

) l cn bng khi

( 1Fr

, 2Fr

.. nFr

) 0.

1.2. H tin ca tnh hc

Tnh hc c xy dng trn c s su tin sau y:

Tin 1: (H hai lc cn bng)

iu kin cn v hai lc cn bng l hai lc c cng ln, cng

phng, ngc chiu v cng t ln mt vt rn. Ta c ( 1Fr

, 2Fr

) 0 khi 1Fr

= - 2Fr

.

Tin 2: ( Thm hoc bt mt h lc cn bng)

Tc dng ca h lc ln vt rn s khng i nu ta thm vo hoc bt i

mt h lc cn bng.

Fr

Rr

Fr

1

2Tin 3: ( Hp lc theo nguyn tc hnh

bnh hnh)

Hai lc cng t vo mt im trn vt rn

c hp lc c biu din bng ng cho ca

hnh bnh hnh m hai cnh l hai lc cho.Hnh 1.1


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Hnh v 1.1 Biu din hp lc ca hai lc 1Fr

, 2Fr

. V phng din vc t c

th vit: Rr

= 1Fr

+ 2Fr

.

Tin 4: ( Lc tc dng tng h)

Lc tc dng tng h gia hai vt rn c cng ln, cng phng

nhng ngc chiu.

Tin 5: (Tin ho rn)

Mt vt khng tuyt i rn ang trng thi cn bng khi ho rn n vn

gi nguyn trng thi cn bng ban u.

Tin 6: ( Gii phng lin kt)

Trc khi pht biu tin ny cn a ra mt s khi nim v: Vt rn

t do, vt rn khng t do, lin kt v phn lc lin kt.

Vt rn t do l vt rn c kh nng di chuyn theo mi pha quanh v tr

ang xt. Nu vt rn b ngn cn mt hay nhiu chiu di chuyn no c

gi l vt rn khng t do. Nhng iu kin rng buc di chuyn ca vt rnkho st gi l lin kt. Trong tnh hc ch xt lin kt do s tip xc ca cc vt

rn vi nhau (lin kt hnh hc). Theo tin 4 gia vt kho st v vt lin kt

xut hin cc lc tc dng tng h. Ngi ta gi cc lc tc dng tng h gia

vt lin kt ln vt kho st l phn lc lin kt.

kho st vt rn khng t do ta phi da vo tin gii phng lin kt

sau y:

Tin :Vt rn khng t do c th xem nhvt rn t do khi gii phng

cc lin kt v thay vo bng cc phn lc lin kt tng ng.

Xc nh phn lc lin kt ln vt rn l mt trong nhng ni dung c bn

ca cc bi ton tnh hc. Sau y gii thiu mt s lin kt phng thng gp v

tnh cht cc phn lc ca n.

Lin kt ta (vt kho st ta ln vt lin kt): Trong dng ny cc phn


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lc lin kt c phng theo php tuyn chung gia hai mt tip xc. Trng hp

c bit nu tip xc l mt im nhn ta ln mt hay ngc li th phn lc

lin kt s c phng php tuyn vi mt ti im tip xc. ( Hnh v 1.2, 1.3,

1.4).

B

A

C

A

BNr

Nr

C

Nr

N

Lin kt l khp bn l:

Khp bn l di ng ( hnh 1.5) ch hn ch chuyn ng ca vt kho st

theo chiu vung gc vi mt phng trt do phn lc lin kt c phng

vung gc vi mt trt. Khp bn l c nh ( hnh 1.6) ch cho php vt kho

st quay quanh trc ca bn l v hn ch cc chuyn ng vung gc vi trc

quay ca bn l. Trong trng hp ny phn lc c hai thnh phn vung gc vi

trc bn l. ( hnh 1.6).

Hnh 1.5 Hnh 1.6

Lin kt l dy mm hay thanh cng: (hnh 1.7 v hnh 1.8)

Cc lin kt dng ny ch hn ch chuyn ng ca vt th theo chiu dy

hoc thanh. Phng ca phn lc lin kt l phng dc theo dy v thanh.

Nr

Hnh 1.2 Hnh 1.3 Hnh 1.4

Nr

Y

XO

Xo

YoRr


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sr

A

A

Bsr

B

s

r

Tr

1 Tr

2Tr

Hnh 1.7Hnh 1.8

Lin kt ngm (hnh 1.9). Vt kho st b hn ch khng nhng di chuyn

theo cc phng m cn hn ch c chuyn ng quay. Trong trng hp ny

phn lc lin kt c c lc v m men phn lc. ( Khi nim m men lc s c

ni ti phn sau).

Lin kt l gt trc: ( hnh 1.10) Vt kho st b hn ch cc chiu chuyn

ng theo phng ngang, phng thng ng v chuyn ng quay quanh cc

trc X v Y do phn lc lin kt c cc thnh phn nhhnh v.

A

x

XA

mX

z

ZA

mY YA

mA

YA

XAy

Hnh 1.9 Hnh 1.10

Cc h qu suy ra t h tin tnh hc.

H qu 1: ( nh l trt lc)

Tc dng ca mt lc ln vt rn

s khng i nu ta trt lc dc theo

ng tc dng n t im khc.

Tht vy: Cho lc Fr

t ti A ca

vt rn ( AFr

). Ta t vo im B trn ng

tc dng ca Fr

mt cp lc cn bng ( BFr

, BFr

) (hnh 1.11). Theo tin hai c

BFr

BFr

AA 'BFr

Hnh 1.11


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th vit:

AFr

( AFr

, BFr

, BFr

). y cc ch s A, B i theo cc lc ch im t cc

lc , cc lc ny c ln bng nhau v cng phng .

Mt khc theo tin 1 hai lc ( AFr

, BFr

) l cp lc cn bng v th theo

tin hai c th bt cp lc trn vt, ngha l:

AFr

( AFr

, BFr

, BFr

) BFr

Nhvy ta trt lc Fr

ban u t ti A dc theo ng tc dng ca

n v t ti B m tc dng c hc ln vt rn vn khng i.

H qu 2: H lc cn bng th mt lc bt k trong h ly theo chiu

ngc li s l hp lc ca cc lc kia.

Chng minh: Cho h lc cn bng ( 1Fr

, 2Fr

,... nFr

). Gi s ta ly trong h

mt lc iFr

v i chiu sau cho tc dng ln vt rn. Xt vt rn chu tc dung

ca lc - i

Fr

. Theo tin 2 nu thm vo vt rn h lc cn bng cho, tc dng

ln vt rn vn khng i, ngha l:

- iFr

(- iFr

, 1Fr

, 2Fr

... iFr

... nFr

)

Trong h (n+1) lc v phi c hai lc cn bng l ( iFr

, - iFr

) theo tin 2

ta c th bt iFr

, v - iFr

i ngha l:

- iFr

( 1Fr

, 2Fr

, 1iFr

... 1iF+r

... nFr

)

Biu thc ny chng t - iFr

l hp lc ca h lc cho khi khng c iFr

.

1.3. L thuyt v m men lc v ngu lc

1.3.1. M men lc i vi mt tm v i vi mt trc

1.3.1.1. M men ca lc i vi mt tm

M men ca lc Fr

i vi tm O l i lng vc t, k hiu c:)F(mor

r


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- ln bng tch s: F.d, vi F l ln lc Fr

v d l khong cch t

tm O ti ng tc dng ca Fr

gi l cnh tay n.

- Phng vung gc vi mt phng cha tm O v lc F (mt phng tc

dng).

- Chiu hng v pha sao cho khi nhn t nh ca vc t xung

mt phng tc dng s thy vc t lc

)F(mor

r

Fr

chuyn ng theo chiu mi tn vng

quanh O theo ngc chiu kim ng h (hnh 1.12).

D vo hnh v d dng thy rng ln ca vc t bng hai ln

din tch tam gic OAB ( tam gic c nh O v y bng lc

)F(mor

r

Fr

).

Vi nh ngha trn c th biu din vc t m men lc Fr

i vi tm O

bng biu thc sau:

)F(mor

r

= OAx Fr

= rr

x Fr

.

Trong rr

l vc t nh v ca im t ca lc Fr

so vi tm O.

Trong trng hp mt phng tc dng ca m men lc xc nh, n

gin ta a ra khi nim m men i s ca lc Fr

i vi tm O nhsau:

M men i s ca lc Fr

i vi tm O l i lng i s k hiu:

mo= F.d

Ly du d

ng (+) khi nhn vo mt phng tc dng thy lc F

r

quay theochiu mi tn vng quanh O theo chiu ngc kim ng h (hnh 1.13), ly du

tr (-) trong trng hp quay ngc li (hnh 1.14).

M men i s thng c biu din bi mi tn vng quanh tm O theo

chiu ca m men.


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Fr

A(x,y,z)

B

mr

o(Fr

)

z

y

x

O

r

Bmo(F)=F.d

900

O

d

A

BFr

d 900

Fr

mo(F)= -F.d

O A

Hnh 1.12 Hnh 1.13 Hnh 1.14

1.3.1.2. M men ca lc i vi mt trc

M men ca lc Fr

i vi trc OZ l i lng i s k hiu mZ(Fr

) tnh

theo cng thc: mZ(Fr

) = F'.d' . Trong F' l hnh chiu ca lc Fr

trn mt

phng vung gc vi trc Z. d' l khong cch tnh t giao im O ca trc Z

vi mt phng n ng tc dng ca Fr

' (hnh 1.15).

Ly vi du (+) khi nhn t hng

dng ca trc OZ s thy hnh chiu F'

quay quanh trc OZ ngc chiu kim

ng h.

Ly du (-) trong trng hp

ngc li.

dFr

'

O

FrB1

() A

Z ''B

Fr

Z

Hnh 1.15T hnh v ta rt ra tr s m men

ca lc Fr

i vi trc OZ bng hai ln

din tch tam gic OAB1.

1.3.1.3. Quan h gia m men lc Fr

i vi tm O v vi trc i qua O

Trn hnh 1.16 ta thy:

mo(Fr

) = 2.din tch (OAB).

mZ(Fr

) = 2 din tch (oa1b1)


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V oa1b1 l hnh chiu ca tam gic OAB trn mt phng vung gc vi

trc Z ti O. Nu gi l gc hp bi gia hai mt phng OAB v mt phng

oa1b1th gc ny cng chnh l gc hp gia vc t m men vi trc OZ,

ta c:

)F(mor

r

Din tch oa1b1 = din tch

OAB. cos.

hay mZ(Fr

) = .cos.)F(mor

r

Kt qu cho thy m men ca lc

Fr

i vi trc OZ l hnh chiu vc t

m men lc Fr

ly vi im O no

trn trc OZ chiu trn trc OZ .

1.3.2. L thuyt v ngu lc

1.3.2.1 nh ngha v cc yu t c trng ca ngu lc

nh ngha: Ngu lc l h hai lc song song ngc chiu cng cng .

Hnh 1.17 biu din ngu lc ( 1Fr

, 2Fr

)

Mt phng cha hai lc gi l mt phng tc dng. Khong cch d gia

ng tc dng ca hai lc gi l cnh tay n. Chiu quay vng ca cc lc

theo ng khp kn trong mt phng tc dng gi l chiu quay ca ngu lc.

Tch s m = d.F gi l m men

ca ngu lc.

mr

o(F)

Fr

A

B

b

Fr

a

d

d'

z

Hnh 1.16

mr

z(F)

d

mr

dA2 A1

mr

A2 A1

Tc dng ca ngu lc c

c trng bi ba yu t:

- ln m men m

- Phng mt phng tc

dng Hnh 1.17


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- Chiu quay ca ngu.

Thiu mt trong ba yu t trn tc dng ca ngu lc cha c xc nh.

biu din y ba yu t trn ca ngu lc ta a ra khi nim v vc

t m men ngu lc mr

. Vc t m men mr

c tr s bng tch s d.F c phng

vung gc vi mt phng tc dng, c chiu sao cho nhn t mt ca n xung

mt phng tc dng thy chiu quay ca ngu lc theo chiu ngc kim ng h.

Vi nh ngha trn ta thy vc t m men mr

ca ngu lc chnh l vc t

m men ca mt trong hai lc thnh phn ly i vi im t ca lc kia. Theo

hnh 1.17 c th vit:

mr

= mr

A1( 2Fr

) = mr

A2 ( 1Fr

)= 21AA x 2Fr

= A2A1x 2Fr

1.3.2.2. nh l v m men ca ngu lc

Trong mt ngu lc, tng m men ca hai lc thnh phn i vi mt

im bt k l mt i lng khng i v bng vc t m men ngu lc.

Chng minh: Xt ngu lc ( 1Fr

, 2Fr

) biu din trn hnh 1.18. Chn mt

im O bt k trong khng gian, tng m men ca hai lc 1Fr

, 2Fr

ly vi O c th

vit: +)F(m 1or

r

)F(m 2or

r

=A1

Fr

1

A2o

Fr

= OA1 x 1Fr

+ OA2x 2Fr

;2

= OA1 x 1Fr

- OA2x 2Fr

;

= (OA1 - OA2) x 1Fr

;Hnh 1.18

= A2A1x 1Fr

= mr

.

Trong nh l trn v im O l bt k do c th kt lun rng tc dng

ca ngu lc s khng thay i khi ta ri ch trong khng gian nhng vn gi

nguyn ln, phng chiu ca vc t m men mr

.

Cng t nh l trn rt ra h qu v cc ngu lc tng ng sau y.


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H qu 1: Hai ngu lc cng nm trong mt mt phng c cng tr s m

men m cng chiu quay s tng ng.

H qu 2: Hai ngu lc nm trong hai mt phng song song cng tr s

m men, cng chiu quay s tng ng vi nhau.

Tht vy trong hai trng hp ny cc ngu lc u m bo c vc t m

men mr

nhnhau.

1.3.2.3. Hp hai ngu lc

nh l: hp hai ngu lc c m men mr

1v mr

2cho ta mt ngu lc c

m men M bng tng hnh hc cc vc t m men ca hai ngu lc cho. Tac = m

r

1+ mr

2M

Chng minh: Xt hai ngu lc c m men mr

1v mr

2nm trong hai mt

phng 1v 1.Trn giao tuyn ca hai mt phng 1v 2ly mt on thng

A1A2ngu lc c m men mr

thay bng ngu lc ( 1Fr

2Fr

) nm trong mt phng 1

v t vo A1A2. Ngu lc c m men mr

2thay bng ngu lc (pr

1pr

2) nm trong

mt phng 2v cng t vo A1A2(hnh 1.19).

Rr P

r

1

1

Fr

mr

mr

2

mr

1

Fr

Pr

2 2Rr

2

1

21

Hnh 1.19

, 1Pr

c lc Rr

11Fr

Ti A1hp hai lc

Ti A2hp hai lc 2Fr

2Pr

c lc Rr

2

Do tnh cht i xng d dng nhn thy hai vc t Rr

1v Rr

2song song


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ngc chiu v c cng cng . Ni khc i hai lc Rr

1 Rr

2 to thnh mt

ngu lc. chnh l ngu lc tng hp ca hai ngu lc cho.

Gi Mr

l m men ca ngu lc (Rr

1Rr

2) ta c:

Mr

= A1A2x Rr

2 = A1A2x Rr

1

Thay Rr

1 = 1Fr

+ 1Pr

v Rr

2= 2Fr

+ 2Pr

, suy ra:

Mr

= A1A2x ( 2Fr

+ 2Pr

) = A1A2x 2Fr

+ A1A2x 2Pr

,

Mr

= mr

A1( 2Fr

) + mr

A1( 2Pr

) = mr

1+ mr

2.

Trng hp hai ngu lc cng nm trong mt mt phng. Khi cc m

men ca ngu lc c biu din bi cc m men i s. Theo kt qu trn, ngu

lc tng hp trong trng hp ny cng nm trong mt phng tc dng ca hai

ngu lc cho v c m men bng tng i s 2 m men ca ngu lc thnh

phn: M = (m1m2)


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Chng 2

L thuyt v h lc

Trong tnh hc c hai bi ton c bn: thu gn h lc v xc nh iu

kin cn bng ca h lc. Chng ny gii thiu ni dung ca hai bi ton c

bn ni trn.

2.1 c trng hnh hc c bn ca h lc

H lc c hai c trng hnh hc c bn l vc t chnh v m men chnh.

2.1.1. Vc t chnh

Xt h lc ( 1Fr

, 2Fr

,.. nFr

) tc dng ln vt rn (hnh 2.1a).

Vc t chnh ca h lc l vc t tng hnh hc cc vc t biu din cc

lc trong h (hnh 2.1b)

a/ b/

Fr

F

r

1 2F

r

Fr

3

nR

r

Hnh 2.1

n

Fr

Fr

1

a cFr

32b

Fr

O

Rr

m

n

Rr

= + + ... =1Fr

2Fr

nFr

=

n

1i

Fr

i (2-1)

Hnh chiu vc t ln cc trc to oxyz c xc nh qua hnh chiu

cc lc trong h:

Rr

Rr

x= x1+ x2+...+ xn= =

n

1i

Xi;


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Rr

y= y1+ y2+...+ yn= =

n

1i

Yi;

Rr

z= z1+ z2+... +zn=

=

n

1i

Zi.

T c th xc nh ln, phng, chiu vc t chnh theo cc biu

thc sau:

Rr

= z2y2x2 RRR ++ ;

cos(R,X) =R

Rx ; cos(R,Y) =R

Ry ; cos(R,Z) =R

Rz .

Vc t chnh l mt vc t t do.

2.1.2. M men chnh ca h lc

Vc t m men chnh ca h lc i vi tm O l vc t tng ca cc vc

t m men cc lc trong h ly i vi tm O (hnh 2.2). Nu k hiu m men

chnh l Mr

ota c

Mr

o = =

n

1i

mr

o(Fr

i) (2 -2)

30mr

A3

A2

Fr

3

2Fr

A1 Fr

1

3zr

2zr

Mr

0

mr

20

10mr

O

m2

1zr

Hnh 2.2

Hnh chiu ca vc t m men chnh Mr

o trn cc trc to oxyz c

xc nh qua m men cc lc trong h ly i vi cc trc :


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Mx= mx( 1Fr

) + mx( ) +...+ m2Fr

x( nFr

) = =

n

1i

mx(Fr

i);

My= my( 1Fr

) + my( ) +...+ m2Fr

y( nFr

) =

=

n

1i

my(Fr

i);

Mz= mz( ) + m1Fr

z( ) +... +m2Fr

z( nFr

) = =

n

1i

mz(Fr

i).

Gi tr v phng chiu vc t m men chnh c xc nh theo cc biu

thc sau:

Mo= z2

y

2

x

2

MMM ++

cos(Mo,x) =o

x

M

M; cos(Mo,y) =

o

y

M

M; cos(Mo,z) =

o

z

M

M.

Khc vi vc t chnh Rr

vc t m men chnh Mr

ol vc t buc n ph

thuc vo tm O. Ni cch khc vc t chnh l mt i lng bt bin cn vc

t m men chnh l i lng bin i theo tm thu gn O.

2.2. Thu gn h lc

Thu gn h lc l a h lc v dng n gin hn. thc hin thu gn

h lc trc ht da vo nh l ri lc song song trnh by di y.

2.2.1. nh l 2.1 : Tc dng ca lc ln vt rn s khng thay i nu ta

ri song song n ti mt im t khc trn vt v thm vo mt ngu lc ph

Fr

'

Fr

Fr

d

A

B

''

Hnh 2.3


18/237

-18-

c m men bng m men ca lc cho ly i vi im cn ri n.

Chng minh: Xt vt rn chu tc dng lc Fr

t ti A. Ti im B trn vt

t thm mt cp lc cn bng ( Fr

', Fr

'') trong Fr

' = Fr

cn F '' = -r

Fr

. (xem

hnh 2.3).

Theo tin 2 c: F (r

Fr

, Fr

', Fr

'').

H ba lc (Fr

, ', '') c hai lc (FFr

Fr r

, Fr

'') to thnh mt ngu lc c m

men mr

= mr

B(F) (theo nh ngha m men ca ngu lc).

Ta chng minh c Fr

Fr

' + ngu lc (Fr

, Fr

'')

2.2.2 Thu gn h lc bt k v mt tm

a. nh l 2.2: H lc bt k lun lun tng ng vi mt lc bng vc

t chnh t ti im O chn tu v mt ngu lc c m men bng m men

chnh ca h lc i vi tm O .

Chng minh: Cho h lc bt k ( 1Fr

, 2Fr

,..., nFr

) tc dng ln vt rn. Chn

im O tu trn vt, p dng nh l ri lc song song a cc lc ca h v

t ti O. Kt qu cho ta h lc ( 1Fr

, 2Fr

,..., nFr

)ot ti O v mt h cc ngu lc

ph c m men l mr

1= mr

o( ) ,1Fr

mr

2= mr

o( 2Fr

), ... mr

n= o( nFr

) (hnh 2.4).mr

Hp tng i lc nh tin 3 c th a h lc ( 1Fr

, ,...F )2Fr

n

r

ov tng

ng vi mt lc .Rr

C th c:

A3

Fr

Fr

Fr

1A1

O

mr

20

mr

30

M = Mo

Fr

1

Rr

Fr

2

Fr

3

3

2A2

( , ) 1Fr

2Fr

Rr

1trong Rr

1= 1Fr

+ 2Fr

(Rr

1,Fr

3 ) Rr

2 trong Rr

Rr

Fr

2 = 1 + 3 =

+ + F1Fr

2Fr r

3m

r

10

....

(Rr

(n-1),F ) nr

Rr

Hnh 2.4


19/237

-19-

trong =Rr

Rr

(n-2)+ nFr

= =

n

1i

Fr

i

Hp lc Rca cc lc t ti O l vc t chnhr

Rr

0ca h lc.

Cc ngu lc ph cng c th thay th bng mt ngu lc tng hp theo

cch ln lt hp tng i ngu lc nh trnh by chng 1. Ngu lc tng

hp ca h ngu lc ph c m men Mr

o= =

n

1i

mr

o(Fr

i). y l m men chnh ca

h lc cho i vi tm O

Theo nh l 2.2, trong trng hp tng qut khi thu gn h lc v tm O

bt k ta c mt vc t chnh v mt m men chnh. Vc t chnh bng tng

hnh hc cc lc trong h v l mt i lng khng i cn m men chnh bng

tng m men cc lc trong h ly i vi tm thu gn v l i lng bin i

theo tm thu gn.

xc nh quy lut bin i ca m men chnh i vi cc tm thu gn

khc nhau ta thc hin thu gn h lc v hai tm O v O1bt k (hnh 2.4a).

Thc hin thu gn h v tm O ta

c Rr r

0v M o.

R

r

0Mr

Mr

01

O1O

Rr

Rr

0 01

Trn vt ta ly mt tm O1khc O

sau ri lc Rr

ov O1ta c

Rr

oRr

o1+ ngu lc (Rr

o, Rr

'o1).

'01Suy ra (R

r

o, Mr

o) Rr

o1+ ngu lc

(Rr r r

o, 'R o1) + M o

Hnh 2.4a

Nu thu gn h v O1ta c Mr

o1v Rr

o1.

iu tt nhin phi c l :

(Rr

o, Mr

o) (Rr

o1,Mr

o1).

Thay kt qu chng minh trn ta c:


20/237

-20-

(Rr

o, Mr

o) Ro1+(Rr

o, Rr

'o1) + Mo(Rr

o+Mo1)

hay Mr

01Mr

o+ ( Rr

o, Rr

'01) (2.3)

Ngu lc ( Rr

o, Rr

01) c m men Mr

' =mo1.(Ro)

Kt lun: Khi thay i tm thu gn vc t m men chnh thay i mt i

lng M' bng m men ca vc t chnh t tm trc ly i vi tm sau.

2.2.3. Cc dng chun ca h lc

Kt qu thu gn h lc v mt tm c th xy ra 6 trng hp sau

2.2.3.1. Vc t chnh v m men chnh u bng khng

Rr

= 0 ; Mr

o= 0

H lc kho st cn bng.

2.2.3.2. Vc t chnh bng khng cn m men chnh khc khng

Rr

= 0; Mr

o0

H lc tng ng vi mt ngu lc c m men bng m men chnh.

2.2.3.3. Vc t chnh khc khng cn m men chnh bng khng

0;Rr

Mr

o= 0

H c mt hp lc bng vc t chnh.

2.2.3.4. Vc t chnh v m men chnh u khc khng nhng vung gc vi

nhau (hnh 2.5)R

r

0; Mr

o 0 v MRr r

o

Trong trng hp ny thay th m men chnh Mr

obng ngu lc (Rr

', Rr

'')

vi iu kin:

Rr

' = ;Rr

Rr

'' = - vRr

Mr

o= mr

o(Rr

')

P

Rr

O'

OP'

n

oRr

dO

r

Rr

o

Mr

o

o

O'

O

Mr

Rr

'


21/237

-21-

Ta c ( , MRr r

o) ( , RRr r

', Rr

'' ).

Theo tin 1 Rr

ov '' cn bng do c th bt i v cui cng h cn

li mt lc bng vc t chnh nhng t ti O

Rr

1. Ni khc i h c mt hp lc t

ti O1.

2.2.3.5. Hai vc t chnh v m men chnh khc khng nhng song song vi

nhau (hnh 2.6).

Rr

o0; Mr

o0 v Rr

o// Mr

o

Trong trng hp ny nu thay Mr

obng mt ngu lc ( ') mt phng

ca ngu ny vung gc vi vc t chnh

Pr

Pr

Rr

.

H c gi l h vt ng lc. Nu vc t Rr

song song cng chiu vi

vc t Mr

oh gi l h vt ng lc thun (phi) v ngc li gi l h vt ng

lc nghch (tri). Hnh 2.6 biu din vt ng lc thun

2.2.3.6. Hai vc t chnh v m men chnh khc khng v hp lc vi nhau

mt gc bt k (hnh 2.7)

Tr

ng hp ny nu thay thvc t M

r

obng mt ngu lc ( Pr

Pr

')

trong clc Pr

t ti O cn lc

' t ti OPr

1 sao cho mo(P) = Mr

o.

R rng mt phng tc dng ca

ngu lc (P ') khng vung gc vir

Pr

Rr

o. Mt khc ti O c th hp hai

lc vPr r

Ro thnh mt lc Rr

'. Nh

Rr

'

Rr

0

O1

Pr

Pr

'

Mr

0

Hnh 2.7


22/237

-22-

vy a h v tng ng vi hai lc Pr

', Rr

' hai lc ny cho nhau.

2.2.4. nh l Va ri nhng

nh l: Khi h lc c hp lc R

r

th m men ca R

r

i vi mt tm haymt trc no bng tng m men ca cc lc trong h ly i vi tm hay trc

.

mr

o( ) =Rr

=

n

1i

mr

o(Fr

i)

mr

z(R) =r

=

n

1i

mr

z(Fr

i) (2.4)

Fr

nOR

r

'

Rr

Fr

2Fr

1

x

y

zChng minh: Cho h lc ( 1F

r

, 2Fr

,..., nFr

)

tc dng ln vt rn. Gi l hp lc ca h

(hnh 2.8).

Rr

Ti im C trn ng tc dng ca

hp lc t thm lc ' = -Rr

Rr

Rr

.H lc

cho cng vi ' to thnh mt h lc cnbng:

Rr

Hnh 2.8

( , ,...1Fr

2Fr

nFr

, + ') 0Rr

Khi thu gn h lc ny v mt tm O bt k ta c mt vc t chnh v

mt m men chnh. Cc vc t ny bng khng v h cn bng, ta c:

Mr

o= =

n

1i

mr

o(Fr

i) + mr

o(Rr

') = 0

Thay ' = - ta c:Rr

Rr

=

n

1i

mr

o(Fr

i) - mr

o( ) = 0Rr

Hay mo( ) =Rr

=

n

1i

mr

o(Fr

i)

Chiu phng trnh trn ln trc oz s c:


23/237

-23-

mz( ) =Rr

=

n

1i

mz(Fr

i)

nh l c chng minh

2.2.5. Kt qu thu gn cc h lc c bit

2.2.5.1. H lc ng quy

H lc ng quy l h lc c ng tc dng ca cc lc giao nhau ti mt

im. Trong trng hp h lc ng quy nu chn tm thu gn l im ng quy

kt qu thu gn s cho vc t chnh ng bng hp lc cn m men chnh s

bng khng.

R00, Mo= 0 vi O l im ng quy.

2.2.5.2. H ngu lc

Nu h ch bao gm cc ngu lc, khi thu gn h s c mt ngu lc

tng hp c m men ng bng m men chnh ca h.

M = ; m=

n

1i

im i l m men ca ngu lc th i v n l s ngu lc ca h.

2.2.5.3. H lc phng

H lc phng l h c cc lc cng nm trong mt mt phng.

Nu chn tm thu gn nm trong mt phng ca h th kt qu thu gn

vn cho ta mt m men chnh Mr

ov vc t chnh Rr

o. Vc t chnh nm trong

mt phng ca h cn m men chnh M

Rr

r

ovung gc vi mt phng ca h. Theo

kt qu thu gn dng chun ta thy: h lc phng khi c vc t chnh Rr

v m

men chnh Mr

okhc khng bao gi cng c mt hp lc nm trong mt phng

ca h.

2.2.5.4. H lc song song

H lc song song l h lc c ng tc dng song song vi nhau.

Kt qu thu gn v mt tm bt k cho ta mt vc t chnh v mt mmen chnh

R

r

Mr

o. Vc t chnh c c im song song vi cc lc ca h.


24/237

-24-

2.3. iu kin cn bng v phng trnh cn bng ca h lc

2.3.1. iu kin cn bng v phng trnh cn bng ca h lc bt k trong

khng gian

2.3.1.1. iu kin cn bng

iu kin cn bng ca h lc bt k trong khng gian l vc t chnh v

m men chnh ca n khi thu gn v mt tm bt k u bng khng.

Rr

= =

n

1i

Fr

1= 0

Mr

o= =

n

1i

mr

o(Fr

1) = 0 (2-5)

2.3.1.2. Phng trnh cn bng

Nu gi Rx, Ry, Rzv Mx, My, Mzl hnh chiu ca cc vc t chnh v m

men chnh ln cc trc to oxyz th iu kin (2-5) c th biu din bng cc

phng trnh i s gi l phng trnh cn bng ca h lc bt k trong khng

gian. Ta c:

Rx= =

n

1i

Xi= 0, Ry= =

n

1i

Yi=0, Rz= =

n

1i

Zi= 0

Mx= =

n

1i

mx(Fr

i) = 0, My= =

n

1i

my(Fr

i) = 0, Mz= =

n

1i

mz(Fr

i) = 0. (2-6)

Trong cc phng trnh trn Xi, Yi, Zil thnh phn hnh chiu ca lc Fi;

mx(Fr

i), my(Fr

i), mz(Fr

i) l m men ca cc lc Fr

ii vi cc trc ca h ta oxyz. Ba phng trnh u gi l ba phng trnh hnh chiu cn 3 phng trnh

sau gi l 3 phng trnh m men.

2.3.2. Phng trnh cn bng ca cc h lc c bit

2.3.2.1 H lc ng quy

Nu chn tm thu gn l im ng quy O th m men chnh Mr

os bng

khng do 3 phng trnh m men lun lun t nghim. Vy phng trnh cn

bng ca h lc ng quy ch cn:


25/237

-25-

Rx= =

n

1i

Xi= 0

Ry= =

n

1i

Yi=0 (2-7)

Rz= =

n

1i

Zi= 0

2.3.2.2. H ngu lc

Khi thu gn h ngu lc v mt tm ta thy ngay vc t chnh Rr

0= 0 iu

c ngha cc phng trnh hnh chiu lun lun t nghim. Phng trnh cn

bng ca h ngu lc ch cn li ba phng trnh m men sau:

Mx= =

n

1i

mx(Fr

i) = =

n

1i

mix= 0,

My= =

n

1i

my(Fr

i) = =

n

1i

miy= 0, (2-8)

Mz= =

n

1i mz(F

r

i) = =

n

1i miz= 0.

y mx, miy, mizl hnh chiu ln cc trc h ta oxyz ca vc t m

men mr

ica ngu lc th i.

2.3.2.3. H lc song song

Chn h to oxyz sao cho oz song song vi cc lc. Khi cc hnh

chiu Rx, Ryca vc t chnh v Mzca m men chnh lun lun bng khng.V vy phng trnh cn bng ca h lc song song ch cn li ba phng

trnh sau:

Rz= =

n

1i

Zi= 0;

Mx= =

n

1i

mx(Fr

i) = 0; (2-9)


26/237

-26-

My= =

n

1i

my(Fr

i) = 0

Trong phng trnh u l phng trnh hnh chiu cn hai phng

trnh cui l phng trnh m men.

2.3.2.4. H lc phng

Cn lu rng trong h lc phng vc t chnh Rr

v m men chnh Mr

lun lun vung gc vi nhau, ngha l h lc phng lun lun c hp lc Rr

nm trong mt phng ca h cho. m bo iu kin hp lc ca h bng

khng tc l iu kin cn bng ca h ta c th vit phng trnh cn bng di

3 dng khc nhau.

1. Dng hai phng trnh hnh chiu mt phng trnh m men:

h lc cn bng cng nhcc trng hp khc phi c R = 0 v Mo=

0. Nu chn h to oxy l mt phng cha cc lc ca h ta thy ngay cc

phng trnh Rz= =

n

1i

zi= 0; Mx= =

n

1i

mx(Fi) = 0 v My= =

n

1i

my(Fi) = 0 l lun lun

t nghim v vy phng trnh cn bng ch cn :

Rx= =

n

1i

Xi= 0;

Ry= =

n

1i

Yi= 0; (2-10)

Mz= =

n

1i mz(Fi).

Hai phng trnh u l phng trnh hnh chiu cn phng trnh th ba

l phng trnh m men. Cn ch v cc lc cng nm trong mt phng oxy do

Mz = =

n

1i

mz(Fi) chnh l tng m men i s ca cc lc i vi tm O.

Mz==

n

1i

mz(F

i)


27/237

-27-

2. Dng mt phng trnh hnh chiu v hai phng trnh m men

iu kin hp lc ca h bng khng c th biu din bng ba phng

trnh sau y:

Rr

Rz= =

n

1i

Xi= 0;

MA= =

n

1i

mA(Fi) = 0; (2-11)

MB= =

n

1i

mB(Fi) = 0

Vi iu kin trc x khng vung gc vi AB.

Tht vy t phng trnh (1) cho thy hp lc Rr

ca h lc bng khng

hoc vung gc vi trc x.

Theo nh l Va ri nhng ,t phng trnh (2) ta thy hp lc Rr

hoc

bng khng hoc qua A.

T phng trnh (3) ta cng thy hp lc Rr

ca h bng khng hoc i

qua B.

Kt hp c ba phng trnh ta thy hp lc ca h hoc bng khng hoc

phi i qua hai im A,B v vung gc vi trc x (khng vung gc vi AB).

iu kin hp lc va qua A, B v va vung gc vi trc x l khng thc hin

c v tri vi gi thit.

Nhvy nu h tho mn phng trnh (2-11) th hp lc ca n s bng

khng ngha l h lc cn bng.

3. Dng ba phng trnh m men i vi 3 im

Ngoi hai dng phng trnh cn bng trn h lc phng cn c phng

trnh cn bng theo dng sau:

MA= =

n

1imA(F

r

i) = 0


28/237

-28-

MB= =

n

1i

mB(Fr

i) = 0 (2-12)

MC=

=

n

1i

mo(Fr

i) =0

Vi iu kin A, B, C khng thng hng.

Tht vy, nu h lc phng tho mn phng trnh MA= mA( ) = 0 th

theo nh l Va ri nhng hp lc ca h s bng khng hoc i qua A. Cng l

lun tng t ta thy tho mn M

Fr

B= 0 v Mc= 0 th hp lc phi bng khng

hoc phi i qua B, i qua C.

V chn 3 im A, B, C khng thng hng nn iu kin hp lc qua 3

im l khng thc hin c. Ch c th hp lc bng khng, c ngha l nu

tho mn h ba phng trnh (2-12) h lc phng cho s cn bng.

2.4. Bi ton cn bng ca vt rn

Vt rn cn bng khi h lc tc dng ln n bao gm cc lc cho v

phn lc lin kt cn bng.

Khi gii bi ton cn bng ca vt rn c th p dng phng php gii

tch hoc phng php hnh hc nhng ph bin v c hiu qu nht l phng

php gii tch.

Gii bi ton cn bng ca vt thng tin hnh theo cc bc sau:

1. Chn vt kho st: vt kho st phi l vt rn m s cn bng ca ncn thit cho yu cu xc nh ca bi ton. Nu nhbi ton tm phn lc lin

kt th vt kho st phi l vt chu tc dng ca phn lc lin kt cn tm, nu l

bi ton tm iu kin cn bng ca vt th vt kho st phi chnh l vt .

2. Gii phng vt kho st khi lin kt v xem l vt t do di tc

dng ca cc lc cho v phn lc lin kt.

3. Thit lp iu kin cn bng cu vt bi cc phng trnh cn bng cah lc tc dng ln vt kho st bao gm cc lc cho v phn lc lin kt.


29/237

-29-

4. Gii h phng trnh cn bng xc nh tr s v phng chiu ca

cc phn lc lin kt hoc thit lp mi quan h gia cc lc m bo iu

kin cn bng cho vt kho st .

5. Nhn xt cc kt qu thu c.

Cn ch rng chiu ca cc phn lc thng cha c xc nh v th

lc u phi t chn chiu. Da vo kt qu gii h phng trnh cn bng ta c

th xc nh chiu ca cc phn lc chn ng hay sai. Nu cc phn lc lin

kt cho tr s dng th chiu chn l ng v nu tr s m th chiu phi o li

. Mt khc cng cn lu rng bi ton c trng hp gii c (bi ton tnh

nh) khi s n s cn xc nh nh hn hoc bng s phng trnh cn bng. C

trng hp khng gii c (bi ton siu tnh) khi n s cn tm ln hn s

phng trnh cn bng.

Th d 2.1. Ct in OA chn thng ng trn mt t v c gi bi hai

si dy AB v AD hp vi ct in mt gc = 300(xem hnh 2-8a) Gc gia

mt phng AOD v mt phng AOB l = 600. Ti u A ca ct in c hai

nhnh dy in mc song song vi trc ox v oy. Cc nhnh dy ny c lc ko

l P1v P2nhhnh v. Cho bit P1= P2= P = 100kN.

Xc nh lc tc dng dc trong ct in v trong cc dy cng AD, AB.

Bi gii:

z

3Rr

Pr

1

Pr

2

O B

D

y

x

Rr

1

Rr

2

Chn vt kho st l u A ca ct in.

Lin kt t ln u A l hai si dyAB, AD v phn ct in cn li.

Gi phn lc lin kt trong dy AB l

R1, trong dy AD l Rr

2v lc dc ct l Rr

3

vi chiu chn nh hnh v 2-8. Khi gii

phng im A khi lin kt im A s chu tc

dng ca cc lc P1, P2 v cc phn lc R1R2Hnh 2.8a


30/237

-30-

Rr

3. iu kin u A cn bng l h 5 lc tc dng ln n cn bng. Ta c:

(Pr

1, Pr

2, Rr

1, Rr

2, Rr

3) 0. H lc ny ng quy ti A do phng trnh

cn bng thit lp theo phng trnh (2.7)

trnh nhm ln ta lp bng (2-1) hnh chiu cc lc ln 3 trc ca h

ta oxyz nhsau:

Bng 2-1

F1 P1 P2 R1 R2 R3

x1

y1z1

0

-P0

-P

00

0

R1sin-R1cos

R2sinsin

R2sincos-R2cos

0

0R3

Phng trnh cn bng vit c:

Xi =- P + R2sinsin= 0; (a)

Yi = - P + R1sin+ R2sincos= 0 ( b)

Zi = -R1cos- R2cos+ R3= 0 (c)

H 3 phng trnh trn cha 3 n s R1, R2, R3nn bi ton l tnh nh.

Gii h phng trnh trn c:

R1= P

sin

gcot1; R2=

sinsin

P; R3= P cotg(1-cotg+

sin

1);

Thay cc tr s ca ,v P ta nhn c:

R1= 85kN; R2= 231 kN; R3= 273kN.

Kt qu u dng nn chiu cc phn lc chn l ng.

Th d 2.2: Mt xe 3 bnh ABC t trn mt mt ng nhn nm ngang.

Tam gic ABC cn c y AB = 1m, ng cao OC = 1,5m, trng lng ca xe

l P KN t ti trng tm G trn on OC cch O l 0,5m. Tm phn lc ca mt

ng ln cc bnh xe (xem hnh 2-9)


31/237

-31-

Bi gii:

Kho st s cn bng ca xe.

Gii phng xe khi mt ng v

thay bng cc phn lc ca mt t

ln cc bnh xe l Nr

A, Nr

B, Nr

C.

Pr

N

r

C

Nr

B

Nr

A

z

G

B

O

A

x

yC

Hnh 2.9

V xe t trn mt nhn nn

cc phn lc ny c phng vung

gc vi mt ng.

Xe trng thi cn bng di

tc dng ca 4 lc ,Pr

Nr

Nr

Nr

A, B, C.

H 4 lc ny l h lc song song.

Nu chn h to oxyz nhhnh v phng trnh cn bng ca h lc

trn theo (2-9) c dng:

Zi = NA+ NB+ NC- P = 0 (a)

mx(Fi) = -P.0,5 + NC.1,5 = 0 (b)

my(Fi) = - NA.0,5 + NB.0,5 = 0 (c)

H ba phng trnh trn cha 3 n s NA, NB, NC nn bi ton l tnh

nh.

Gii phng trnh trn xc nh c:

NA= NB= NC= P/3 kN

Kt qu cho cc gi tr dng nn chiu phn lc hng ln l ng.

P

D

G

A

q

E C BM

2 11 2

Th d 2.3:X AB c gi

nm ngang nh lin kt nhhnh v

(2.10). Ti A c khp bn l c

nh. Ti C c treo bi dy CD

t xin mt gc so vi x. Ti B

c dy ko thng ng nh trngHnh 2.10


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vt P buc u dy vt qua rng rc.

X c trng lng G t ti gia, chu mt ngu lc nm trong mt phng

hnh v v c m men M. on dm AE chu lc phn b u c cng q.

Xc nh phn lc ti A, trong si dy CD cho bit G = 10kN, P = 5kN, M

= 8 kNm; q = 0,5 kN/m; = 300. Cc kch thc cho trn hnh v.

Bi gii:

Chn vt kho st l x AB. Gii phng lin kt t ln x ta c:

Lin kt ti A c thay th bng phn lc Rr

Anm trong mt phng hnh

v. Lin kt ti C c thay th bng lc cng Tr

hng dc theo dy. Lin kt ti

B thay bng lc cng ng bng Pr

nhng c chiu hng ln trn. Chiu ca Rr

A

v chn nhhnh v. Nhvy x AB trng thi cn bng di tc dng ca

cc lc (

Tr

Gr

, ,Mr

Rr

A, , ), cc lc ny nm trong mt phng thng ng tc l

mt phng hnh v (h lc phng ). Chn h to Axy nh hnh v v lp

phng trnh cn bng dng (2-10) c:

Tr

Pr

Xi= XA- Tcos300; (a)

Yi= YA- Q - G +T cos600+ P = 0; (b)

mA(Fr

i) = - Q.1 - G.3 + T.4sin300- M + 6P = 0. (c)

Trong cc phng trnh trn

Q = 2q l tng hp lc phn b u

t ti im gia AE.

B

Pr

2 21 1

A Q

r

C

Gr

900T

r

YAXA M

y

Ba phng trnh trn cha 3

n s XA, YA, v T do bi ton l

tnh nh.

x

Hnh 2.11Gii h phng trnh trn ta

c:

T = 5,45,0.4

6.583.101.1

30sin.4

6.pM3.G1.Q0

=++

=++

kN;


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XA= Tcos300= 4,5.0,866 = 3,90kN;

YA= Q + G -T cos600- P = 1 + 10 - 4,5.0,5 - 5 = 3,75, kN

Kt qu cho cc tr s ca T, XA, YAu dng do chiu chn ban u

l ng.

Th d 2.4: Trc truyn nm ngang t trn hai gi bn l c nh A v

B (xem hnh v 2-12). Trc nhn

chuyn ng quay t dy ai dn

n bnh ai C c bn knh r1= 20

cm v nng trng vt P buc vo

u dy cp vt qua rng rc K v

cun trn trng ti c bn knh r2=

15cm. Cho bit hai nhnh dy ai

c phng song song vi trc oy v

c lc cng T1 v T2 vi T1 = 2T2;

Trng vt P= 180kN; a = 40cm; b =

60cm v = 300. Xc nh phn lc

ti hai gi A v B.

P

YB

ZB

B

YA

ZA

z

y

A

C

T2

T1

a b

a

x

Hnh 2.12

Bi gii:

Chn vt kho st l trc BC.

Lin kt ln trc l cc A, B. Cc lc tc dng cho l Tr

1, Tr

2v Fr

.

Lc tc dng dc theo dy cp c tr s bng PFr r

. V cc l khp bn l c

nh nn phn lc lin kt ti A v B c hai thnh phn theo trc oy v oz. Gii

phng lin kt t ln trc v thay bng cc phn lc lin kt khi trc AC chu

tc ng ca cc lc: Tr

1, Tr

2, Fr

, Rr

A, Rr

B . Cc lc ny phn b bt k trong

khng gian. Phng trnh cn bng ca h lc thit lp theo (2- 6). trnh

nhm ln ta lp bng hnh chiu v m men ca h lc i vi cc trc to

(bng 2-2) .


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Bng 2-2

Fr

1 Fr

Tr

1 Tr

2 Rr

A Rr

B

X1

Y1

Z1

mx(F)

my(F)

mz(F)

0

Fcos

-Fsin

-F.r2

Fsin.b

Fcos.b

0

Thp

45

0

T1r1

0

-T1.a

0

T2

0

-T2r1

0

-T2a

0

YA

ZA

0

0

0

0

YB

ZB

0

-ZB(a+b)

YA(a+b)

Cc phng trnh cn bng thit lp c:

Yi = Pcos+ T1+T2+ YA + YB= 0;

Zi = Fsin+ ZA+ ZB= 0;

Mx = F.r2 + T1r1- T2r1= 0;

My = Fsin.b - ZB(a+b) = 0;

Mz = Fcos.b - T1a- T2a + YB(a+b) = 0;

H 5 phng trnh trn cha 5 n s l YA, ZA, YB, ZBv T1nn bi ton l

tnh nh.

Gii h phng trnh trn tm c:

T2=r

r.P 2 =20

15.180= 135kN ; T1= 2T2= 270 kN;

ZB=ba

sinP.b

+

=

6040

5,0.180.60

+= 54 kN;

YB=ba

cosPbT3.a 2

+

=

6040

2

3.60.180135.3.40

+

= 69 kN

YA =- Pcos-3T2- YB= -180.2

3-3.135- 69 -630KN

ZA= Psin- ZB= 180. 0,5 - 54 = 36kN.


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-35-

Trong cc kt qu tm c ch c gi tr YAmang du m do chiu ca

n ngc vi chiu chn.

Th d 2.5: Cho h hai dm

AB v BE ni bng khp bn l

ti B (xem hnh v 2-13). Trng

lng ca dm AB l Q t gia

AB. Trng lng ca dm BE l P

t gia BE. Ti u A c khp

bn l c nh, cn ti cc im

C, D l cc im ta nhn.

D

A BCP

r

E

Qr

Xc nh phn lc ti cc

gi A v cc im ta C,D.

Cho P = 40kN, Q = 20kN; CB =3

1AB; DE =

3

1BE; = 450.

Hnh 2.13

Bi gii: Cn lu rng y l bi ton cn bng ca h vt. V nguyn

tc khi gii bi ton thuc loi ny phi tch ring tng vt xt. Trn h vtcn phn bit hai loi vt chnh v vt ph. Vt chnh l vt khi tch ra c th

ng vng c. Vt ph l vt khi tch ra khng th ng vng c. Ta xt vt

ph trc sau xt vt chnh sau. Cng cn ch thm khi tch vt ti cc

khp ni s c thay th bng cc lc tc dng tng h, cc lc ny cng

phng cng tr s nhng ngc chiu.

i vi bi ton trn, h gm hai dm trong AB l dm chnh cn BE

l dm ph. Tch BE xt. Ti khp ni c phn lc lin kt R B(lc tc dng

tng h ca dm chnh ln dm BE). Phn lc RBnm trong mt phng thng

ng ( mt phng hnh v) v c hai thnh phn XBv YB( xem hnh 2-14). Gii

phng lin kt ti D thay vo bng phn lc Nr

D(Nr

D vung gc BE. Dm BE

chu tc dng ca cc lc ,Pr

Nr

D, Rr

B. H lc ny cng nm trong mt phng oxy

do phng trnh cn bng vit c:

X1= XB- NDsin= 0;


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Pr

XB

Nr

D

D

EY10= YB- P + NDcos= 0;

YBmB(F1) = ND

3

2.a - P.

2

acos= 0.

BGi h phng trnh trn tm c:

ND=4

3Pcos=

4

3.40.

2

221,2 kN; Hnh 2.14

YB

A C

Qr

YAXBXAXB= 8

3P sin2=

8

3.40.1= 15kN;

YB = P(1- 43

cos2

)= 40(1- 43

4

2

)= 25kN.Hnh 2.15

Gi tr cc phn lc u dng iu ny

chng t chiu ca chng nh chn l ng.

Tip theo xt n dm chnh AB. Gii phng cc lin kt dm s trng

thi cn bng di tc dng ca h lc: Qr

, - Rr

B, Rr

A, Nr

C. Cc lc ny cng nm

trong mt phng oxy. ( xem hnh 2.15 )

Phng trnh cn bng ca h lc vit c:

X1= XA- X'B= 0;

mA(F) = - Y'B.b + NC3

2b - Q.

2

b= 0;

mC(F) = - YA.3

b2 + Q

6

b- Y'B.

3

b= 0;

Trong X'B= XB, Y'B= YBnhng c chiu ngc li.

Gii h 3 phng trnh trn tm c:

XA= XB= 15kN;

YA =4

1Q -

2

1YB = -7,5kN;

YC =4

3Q +

2

3YB = 52,5kN.

Kt qu cho gi tr ca YAmang du m c ngha chiu YAchn l sai

phi o li.


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Chng 3

Ma st v bi ton cn bng ca vt khi c ma st

3.1. Ma st trt v bi ton cn bng ca vt khi c ma sttrt

3.1.1. Ma st trt v cc tnh cht ca ma st trt

Thc tin cho thy bt k vt no chuyn ng trt trn b mt khng

nhn ca vt khc u xut hin mt lc cn li s trt ca vt gi l lc ma st

trt k hiu Fr

ms. Lm th nghim biu din trn hnh 3.1. Vt A t trn mt

trt nm ngang v chu tc dng ca lc Pr

hp vi phng thng ng mt gc

. Phn tch thnh hai thnh phnPr

Pr

1v Pr

2nhhnh v. Nhn thy rng Pr

1

lun lun cn bng vi phn lc php tuyn Nr

. Cn lc Pr

2l lc cn y vt

A trt trn mt.

Khi khng i ta nhn thy gc tng thPr

Pr

2 tng. Trong giai on u vt A ng yn trn

mt B. T iu kin cn bng ca vt A cho thyN

r

P

r

Pr

2

bng lc ma st nhng ngc chiu. Nu tip tc

tng gc n mt tr s th vt A bt u trt.

Lc ma st lc cng tin ti gii hn Fr

n.

P

r

1P

r

2 Fr

ms

Hnh 3.1

Tr s Fn= Ntg (3.1)

y N = P1l phn lc php tuyn ca mt trt. Gc gi l gc ma

st; tg= f gi l h s ma st. T (3.1) c th kt lun: lc ma st trt lun

lun cng phng nhng ngc chiu vi chuyn ng trt, c tr s t l thun

vi phn lc php tuyn (p lc) ca mt trt.

H s ma st f c xc nh bng thc nghim, n ph thuc vo vt liu

v tnh cht ca b mt tip xc. Bng (3-1) cho ta tr s ca h s ma st trt

i vi mt vi vt liu thng gp


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Bng 3-1

Tn vt liu H s ma st

trt trn g

G trt trn g

Kim loi trt trn g

ng trt trn gang

ng trt trn st

Thp trt trn thp

0,46 0,6

0,62

0,62

0,16

0,19

0,15

Lc ma st xut hin trong giai on vt trng thi tnh gi l ma st

tnh. Lc ma st tnh tng t khng n tr s gii hn Fn= f0N. Lc ma st xut

hin trong giai on vt chuyn ng trt ta gi l lc ma st ng. Trong trng

thi tnh lc ko (y) vt lun cn bng vi lc ma st tnh cn trong trng thi

chuyn ng lc ko (y) P2va phi thng ma st ng va phi dmt phn

to ra chuyn ng ca vt. Nu gi lc ma st ng ca vt l Fmssdth Fmsd=

fdN, trong fdgi l h s ma st ng. Qua nhiu thc nghim thy rng lc

ma st ng thng nh hn mt cht so vi ma st tnh gii hn. H s ma st

ng khng nhng ph thuc vo vt liu v tnh cht b mt tip xc ca vt m

cn ph thuc vo vn tc trt ca vt. Trong phn ln cc trng hp cho thy

khi vn tc tng th h s ma st ng gim v ngc li. Th d h s ma st

ng gia bnh ai lm bng gang vi dy ai phanh bng thp c th xc nh

theo cng thc:

fd =v006,01

v0112,01

+

+ft

Trong v l vn tc trt tnh bng km/h cn ft= 0,45 khi mt tip xc

kh v ft= 0,25 khi mt tip xc t.

Trong tnh hc v ch xt bi ton cn bng nn ma st phi l ma st tnh.


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3.1.2. Bi ton cn bng ca vt khi chu ma st trt

Xt vt rn t trn mt ta (mt trt). Gi thit vt chu tc dng ca cc

lc Fr

1, 2Fr

, ... nFr

. Cc lc lin kt bao gm phn lc php tuyn Nr

jv lc ma st

Fr

msj.

Khi vt cn bng ta c h lc sau:

(Fr

1, , ...2Fr

nFr

, Nr

j, Fr

msj) 0 j = 1 ....s l s b mt tip xc

vt cn bng phi c cc phng trnh cn bng nh xt chng 2.

Ngoi cc phng trnh cn bng ra m bo vt khng trt phi c cc iu

kin:

FnjfoNj. Fnjl lc y tng hp.

Tr li s (3.1) ta thy khi khng c trt th

tg=N

Fms fo = tg

Ta c th pht biu iu kin khng trt nhsau:

iu kin vt khng trt l hp lc Pr

tc dng ln vt nm trong mt

nn c gc nh 2( ta gi nn ny l nn ma st).Khi P nm trn nn ma st l

lc sp xy ra s trt ca vt A.

Th d 3.1: Xc nh iu kin

cho vt A c trng lng P nm cn bng

trn mt nghing so vi ph

ng ngang mtgc . H s ma st tnh l fo(hnh 3.2)

Nr

Fr

ms

Bi gii: Xt vt A nm cn bng

trn mt nghing di tc dng ca cc lc

( ,Pr

Nr

, Fr

ms) V vt c xu hng trt

xung nn lc ma st Fr

mslun lun hng

v pha trn nhhnh v.

Hnh 3.2

vt cn bng phi c:


40/237

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( ,Pr

Nr

, Fr

ms) 0 v FNfoN.

Gi thit rng v tr ang xt l v tr gii hn gia cn bng v trt th lc

ma st Fms= Fn= foN. iu kin h lc tc dng ln h vt cn bng l:

Fn= Ntg

Mt khc v FnNf0. Suy ra tgfo.

Nhvy iu kin cho vt cn bng phi l tgfo.

Tr s ca gc = ovi tago= fochnh bng gc ma st .

Th d 3.2: Gi treo vt nng c s nh

hnh v 3-3. Vt treo c trnglng P, h s ma st trt ti cc im ta A v B l fo. Kch thc cho theo

hnh v. Xc nh iu kin cn bng cho gi.

Bi gii:

Kho st s cn bng

ca gi. Lc tc dng ln gi

ngoi trng lng ca vtA cn c phn lc php

tuyn v lc ma st im

ta A v B l:

P

r

Nr

,Nr

', , 'Fr

Fr

Nu khong cch l l

khng i, iu kin cn

bng ca gi l:

y

B

Pr

Pr

o

o A

B

h

lRr

BRr

A

A

y'F

r

Nr

'

Fr

h

Nr

l

x

a) b)

Hnh 3.3

( ,Pr

Nr

,Nr

', Fr

, ') 0Fr

v F foN; F' foN'

Ti v tr gii hn ngha l lc sp xy ra s trt ca gi trn cc im ta

ta c phng trnh cn bng nhsau:

N- N' = 0; (1) F=foN (4)F + F' -P = 0 (2) F' = foN' (5)


41/237

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N.h - F.dgh- P = 0; (3)

y dghl khong cch gii hn ca hai im ta A v B cho php ng

vi lc bt u trt.

Gii h phng trnh trn ta c:

N = N' F = F'; P = 2foN;

h = fodgh+ 2fol hay dgh=of

h- 2l

Khong cch d cng ln p lc N cng ln v ma st cng ln, iu kin

cn bng ca gi vit c:

dghof

h- 2l

Th d 3.3: Tm iu kin khng trt ca dy ai qun trn bnh ai trn

c k n ma st trt vi h s fo(hnh 3-4) , b qua tnh n hi ca dy ai.

Bi gii:

Tm iu kin khng trt ca dy ai c ngha l tm iu kin cn bng

ca on ai AB ca ai di tc dng cc lc Tr

1, Tr

2 (T2> T1) cc phn lc

php tuyn N v cc lc ma st trt F phn b lin tc trn cung AB.

Khi dy ai sp trt ta xt mt cung nh ED trn dy ai. Bn nhnh ch

ng c lc tc dng l + TTr r

cn bn nhnh

ph ng lc tc dng l . Gi phn lc php

tuyn ln cung ai ny l

Tr

Nr

v lc ma st trt

ln cung ny l F ta s c phng trnh cn bng:

Tr

RD

ydN

r

(T

r

+dT

r

)

d

dFr

Tr

d

B

Tr

1

2

A

d

- T cos2

d+ (T+dT)cos

2

d- F = 0

- N - Tsin2

d- (T- dT) = 0

Hnh 3.4Trong F = fN. B qua cc v cng b


42/237

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bc hai tr ln ta c: F = dT v N = Td. Thay gi tr trn vo biu thc F =fN

ta c dT = f.T.d. Tch phn hai v tng ng vi cn t A n B ta c

lnTBA

= fo BA

hay ln 12

T

T

= f.

l gc chn cung AB gi l gc bao ca ai.

Suy ra: T2= T1.ef

Lc ko bn nhnh ch ng T2cng ln hn bn nhnh b ng th kh

nng trt cng nhiu do iu kin dy khng trt phi l:

T2T1.ef

Cng thc ny c gi l cng thc le

3.2. Ma st ln v bi ton cn bng ca vt rn khi c ma

st ln

Ma st ln l m men cn chuyn ng ln ca vt th ny trn vt th khc.

Xt mt con ln hnh tr bn knh R trng lng P ln trn mt mt phng

ngang, nh lc Qr

t vo trc con ln (xem hnh 3.5). Trong trng hp ny con

ln chu tc dng ca cc lc: Pr

, Qr

, Nr

, Fr

ms. Trong cc lc hai lc Qr

v Fr

ms

to thnh mt ngu lc c tc dng lm cho con ln chuyn ng ln. Cn li hai

lc vPr

Nr

trong trng hp con ln v mt ln l rn tuyt i th chng trng

ph

ng.Trong thc t con ln v mt ln l nhng vt bin dng hai lc P v Nkhng trng phng lun song song v cch nhau mt khong cch k. Hai lc

ny to thnh mt ngu lc c tc dng cn li s ln ca con ln. M men ca

ngu ( ,Pr

Nr

) c gi l m men ma st ln. Nu k hiu m men ma st ln l

Mms th Mms= kN.

Gi k l h s ma st ln. Khc vi h s ma st trt h s ma st ln k

c th nguyn l di.


43/237

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H s ma st ln c xc nh bng thc nghim, n cng ph thuc vo

tnh cht vt liu v b mt ln, khng ph thuc vo lc N. Sau y l h s ma

st ln ca mt vi vt thng gp.

Vt liu H s k (cm)

G ln trn g

Thp ln trn thp

G ln trn thp

Con ln thp trn mt thp

0,05 0,08

0,005

0,03 0,04

0,001

Qr

C

A

Pr

Nr

F r

Fr

C

A

Pr

Q

Bk

Nr

r

a) b)

Hnh 3.5

Bi ton cn bng ca vt khi c ma st ln ngoi iu kin h lc tc

dng ln h k c cc phn lc v lc ma st cn bng cn phi thm iu kin

khng c ln biu din bi phng trnh:

Pr

C Nr

Fr

1Pr

2Pr

MmsQ.R

Th d 3.4: Tm iu kin cn bng ca con ln

trng lng P, bn knh R nm trn mt phng nghing mt

gc . Cho h s ma st ln l k. (xem hnh 3-6)

Bi gii:

Xt con ln v tr cn bng. Phn tch Pr

thnh hai

lc Pr

1, Pr

2nhhnh v (3-6).

Hnh 3.6

Ta c iu kin con ln khng ln l:P1.R = R.P.sinP2.k = P cos

Hay R.P.sinP.cos. tgR

k


44/237

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Nhvy iu kin con ln cn bng l: tgR

k

Th d 3.5: Vt hnh tr c trng lng P bn knh R nm trn mt phng

nghing mt gc . Khi tr chu tc dng lc y Q song song vi mt phng

nghing. Tm iu kin khi tr ng yn trn mt phng nghing v iu kin

n ln khng trt ln pha trn. H s ma st ln l k v h s ma st trt l f.

y

x

M A

Fr

ms

Nr

Pr

O

Qr

y

x

MA

O

P

r

Fm sr

Nr

Qr

a) b)

Hnh 3.7

Bi gii:

i kin khi tr cn bng trn mt phng nghing l :

( , ,Pr

Qr

Nr

, Fr

ms, Mr

ms) 0

Mt khc khi tr khng ln (hnh3.7a ) khng trt xung phi c

thm iu kin:

Mmsk.N; Fmsf.N

Nhvy phi tho mn cc phng trnh sau:

Xi= Q - Psin + Fms= 0; (1)

Yi= - Pcos +N = 0; (2)

mA= P.R.sin- Q.R - Mms = 0 (3)

Fms

f.N (4)

M msk.N (5)


45/237

-45-

T ba phng trnh u tm c:

N = Pcos ; Fms= Psin- Q ; Mms= R(Psin- Q)

Thay cc kt qu vo hai bt phng trnh cui c:

P.sin- Q f.Pcos ; R(Psin-Q) k.Pcos

Hay: Q P(sin- f.cos)

Q P(sin-R

kcos)

Thng thR

k< f do iu kin tng qut l:

P

Qsin-

R

kcossin- f.cos

vt ln khng trt ln ( hnh3.7b ) phi c cc iu kin:

xi= Q-Psin+ Fms= 0; (1')

yi=- Pcos+N = 0; (2')

mA= P.sin- Q.R + Mms= 0; (3')

Fmsf.N (4')

M msk.N (5')

Bt phng trnh (4') m bo cho vt chuyn ng c trt ln. Cn bt

phng trnh (5') m bo cho con ln c kh nng ln ln trn.

T 3 phng trnh u ta c:

N = Pcos; Fms= Q - Psin ; Mms= R(Q-Psin)

Thay th vo hai phng trnh cui ta c:

Q - Psinf.P.cos;R(Q-Psin) kPcos.

Vy iu kin khi tr ln khng trt ln trn l:

sin+R

kcos

P

Q< sin+ f cos.

iu ny ni chung c th c nghim vR

kthng nh hn f.s


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-46-

Chng 4

Trng tm ca vt rn

4.1. Tm ca h lc song song

H lc song song (Fr

1, 2Fr

, ... nFr

) lun c hp lc Rr

song song vi cc lc

cho. Theo l thuyt v h lc, hp lc Rr

c xc nh bi biu thc:

Rr

= Fr

1+ 2Fr

+... nFr

= =

n

1i

Fr

i (4-1)

Khi ta thay i phng ca h lc phng ca hp lc cng thay i theo.

Chng hn lc u h lc c hp lc l R song song vi cc lc cho , sau khi

xoay h lc cho song song vi trc oz ta s c hp lc R' c ln bng R

nhng c phng song song vi

trc oz. Mc d hp lc thay i

phng khi phng ca h lc

thay i nhng ng tc dng

ca chng u i qua im Cim ny gi l tm ca h lc

song song cho.

z

yOzC

yCxC

Rr

' Rr

4rr

rr

'4

A4

3rr

rr

'3

A3

2rr

rr

'2

A2

rr

1

Crr

'1

A1

xc nh v tr ca tm C

ta vn dng nh l Va-ri-nhng.

Cho hp lc ' nhhnh v ta c:Rr

x

My(R') = =

n

1i

my(Fn

i);Hnh 4.1

R.Xc= =

n

1i

Fixi;

hay Xc=R

xFn

1i

ii= ;


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-47-

Trong Xc l to ca im C trn trc ox, xi l to ca im Ai

trn trc ox.

Bng cch xoay phng ca h lc cho song song vi trc ox v oy ta s

nhn c cc kt qu tng t vi to ca C trn hai trc oy v oz. Ta xc

nh h to ca tm C theo cc biu thc sau:

Xc=R

xFn

1i

ii= ;

Yc= R

yFn

1i

ii= ; (4-2)

Zc=R

zFn

1i

ii= .

Nhvy c th xc nh hp lc ca h lc song song nh cc biu thc

(4-1) v (4-2)

4.2. Trng tm ca vt rn

Coi vt rn l tp hp ca n phn t c trng lng Pr

1, Pr

2 ...Pr

n. Cc

trng lc Pito thnh mt h lc song song. Tm ca h cc trng lng phn t

ny gi l trng tm ca vt.

Nhvy gi C l trng tm ca vt th to ca im C c xc nh

bng cc biu thc sau:

Xc=P

xPn

1i

ii= ;

Yc=P

yPn

1i

ii= ; (4-3)


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Zc=P

zPn

1i

ii= .

Trong Pr

iv l trng lng ca phn t th i trong vt, v trng lngca c vt, cn x

P

r

i, yi, zil to ca phn t th i.

Nh vy trng tm ca vt l mt im C trn vt m tng hp trng

lng ca c vt i qua khi ta xoay vt bt k chiu no trong khng gian.

4.3. Trng tm ca mt s vt ng cht

4.3.1. Vt rn l mt khi ng cht

Gi trng lng ring ca vt l ( trng lng ca mt n v th tch)

th Pi= .viv P = .v. Trong viv v l th tch ca phn t th i ca vt v th

tch c vt. To trng tm ca vt lc ny c th xc nh bi cc biu thc:

xc=v

xvn

1i

ii= ; yc=

v

yvn

1i

ii= ; zc=

v

zvn

1i

ii= .

4.3.2. Vt rn l mt tm mng ng cht

Gi trng lng ring ca vt rn l ( trng lng ca mt n v din

tch) ta s c Pi= .Siv P = .S y Siv S l din tch ca phn t th i ca

vt v din tch ton vt. To trng tm ca vt trong h to oxy cha vt

xc nh theo biu thc sau:

xc=S

xSn

1i

ii= ; yc=

S

ySn

1i

ii= ;

4.3.3. Vt rn l mt dy hay thanh mnh ng cht

Gi trng lng ring ca vt l ( trng lng ca mt n v chiu di

vt) ta c Pi= .Liv P = .L. Trong Liv L l chiu di ca phn t th i v

chiu di ca c vt. To trng tm ca vt lc ny c th xc nh bi cc

biu thc:


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-49-

xc=L

xLn

1i

ii= ; yc=

L

yLn

1i

ii= ; zc=

L

zLn

1i

ii= .

4.3.4. Vt rn ng cht c mt tm, mt trc hay mt mt phng i xng

Ta c nhn xt rng trn vt bao gi cng tm c hai phn t i xng

c trng lng P1, P2nhnhau song song cng chiu qua tm i xng, trc i

xng hay mt phng i xng ca vt v nh vy hp lc ca n s i qua im

i xng nm trn trc i xng hay mt phng i xng. D dng nhn thy

rng hp lc ca cc Pr

i( i = 1...n), ngha l trng lng ca vt bao gi cng i

qua tm i xng, trc i xng hay nm trong mt phng i xng nu nh

xoay vt sao cho mt phng i xng v tr thng ng. Ni cch khc trng

tm ca vt trong trng hp c mt tm i xng, c mt trc i xng hay c

mt mt phng i xng bao gi cng nm trn tm i xng, trc i xng hay

mt phng i xng .

4.3.5. Trng tm ca vt c th phn chia thnh nhng vt nh n gin

Trong trng hp ny ta chia vt thnhcc phn c hnh dng n gin d xc nh

trng tm, sau coi mi vt nhmt phn

t nh ca c vt, mi phn t ny c trng

lng t ti trng tm. Xc nh c trng

lng v trng tm cc phn nh ca vt ta s

xc nh c trng tm ca c vt nh cc

biu thc xc nh to trng tm trn.

O

C1

C2

C3y

Hnh 4.2

Bng 4.1C1 C2 C3

xiyiSi

-1

14

1

520

5

912

x

Sau y ta vn dng nhng kt qu trn

tm trng tm ca mt s vt.

Th d 4.1: Xc nh trng tm ca tm

tn phng c hnh dng nhhnh v (4-2).

Bit rng tm tn l ng cht v kchthc ca cc cnh tnh bng cm cho trn


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hnh.

Bi gii:

Trc ht chia vt thnh 3 phn, mi phn l mt hnh ch nht nhhnh

v (4-2). Cc hnh ny l cc tm phng v c tm i xng l C1, C2v C3. To

trng tm v din tch ca n c th xc nh nhbng 4.1.

Din tch ca c vt l :

S = S1+ S2+ S3 = 36 (cm2)

p dng cng thc (4.5) ta c:

xc=S

SxSxSx 332211 ++ =36

60204 ++ = 29

1 cm

yc=S

SySySy 332211 ++ =36

1081004 ++ = 59

8 cm

Trng tm C ca vt hon ton c xc nh.

Th d 4.2.Tm to trng tm ca tm phng gii hn bi hai ng

trn bn knh R v r ( xem hnh v 4.3). Cho bit khong cch gia hai tm lc1c2= a.

Bi gii:

Chn h to nhhnh v. Phn tch

thnh hai phn mi phn l mt tm trn

nhng y tm trn c bn knh r phi coi

nhvt c tit din m. C th ta c: Phn 1l mt tm trn c bn knh R c to

trng tm l x1= 0 v y1= 0. Din tch l S1

= R2. Phn 2 l tm trn c bn knh r, to

trng tm l x2= a, y2= 0 v din tch l

S2= -r2.Din tch c vt l :

R

C2C1Cr

a

y

S = S1+ S2= (R2- r2)Hnh 4.3


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-51-

Ta c th tnh c to trng tm ca vt.

xc=S

SxSx 2211 + = -22

2

rR

r.a

;

yc=S

SySy 2211 + = 0.

Th d 4-3. Tm trng tm ca mt cung trn AB bn knh R, gc tm

l AB = 2 ( hnh 4-4)

Nu chn h to nh hnh v ta thy trc ox l trc i xng do

trng tm C ca chng nm trn trc ox c ngha l yc=0. y ch cn phi

xc nh xc

Ta chia cung AB thnh N phn nh, mi phn c chiu di lk, c to

xk= Rcosk.

Theo cng thc (4.6) c:

B

O

lk

kxk

x

AHnh 4.4

y

xc= L1

L

xln

1i

kk

=

=

=

n

1ilkRcosk

Thay lkcosk= Yk ta c:

Xc=L

1 R=

n

1i

Yk=L

1 R.AB

Thay L = R.2v AB = 2R sinta c:

Xc=

2.R

sin2.R = R.

sin (4-7)

Th d 4-4: Tm trng tm ca mt tm phng hnh tam gic ABC ng

cht (hnh 4-5).

Bi gii:

C

GK

C

A

Chia tam gic thnh cc di nh song song

vi y BC. Mi di nh th i c coi nhmt

B E


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-52-

thanh mnh v trng tm ca n t ti gia di. Nhvy trng tm ca cc di

s nm trn ng trung tuyn AE v trng tm ca c tam gic cng nm trn

AE.

Chng minh tng t ta thy trng tm ca tam gic phi nm trn trung

tuyn BG v trung tuyn CK. R rng trng tm ca tam gic chnh l giao im

ca ba ng trung tuyn ca tam gic .

Trong hnh hc ta bit im c xc nh theo biu thc:

CE =3

1AE

Th d 4-5 Tm trng tm ca vt ng nht hnh t din ABDE nhhnh

v (4-6) .

Bi gii:

Ta chia hnh thnh cc phn nh nh cc

mt phng song song vi y ABD. Mi tm

c coi nhmt tm phng ng cht hnh tam

gic trng tm ca mi phn c xc nh nh

th d 4-4. Lp st y s c trng tm l C1vi

C1k = BK3

1 (BK l trung tuyn ca y ABD).

Nh vy tt c cc trng tm ca cc phn s

nm trn ng EC1v trng tm ca c vt cng s nm trn EC1.

E

C

BK

C2A

C1

D

Hnh 4.6

Tng t ta tm thy trng tm ca vt nm trn ng BC2vi C2l trngtm tam gic EAD. Kt qu l trng tm C ca hnh v nm trn im C l giao

im ca EC1v BC2.

Theo hnh v ta c CC1C2 ng dng vi ECB mt khc C1C2 =

BE3

1 v KC1= KB3

1 t suy ra:

CECC1 = BE

CC 21 = 31


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-53-

Suy ra CC1= CE3

1 = EC4

11


54/237

-54-

Phn 2

ng h

ng hc nghin cu cc qui lut chuyn ng ca vt th n thun vhnh hc, khng cp n khi lng v lc. Nhng kt qu kho st trong

ng hc s lm c s cho vic nghin cu ton din cc qui lut chuyn ng

ca vt th trong phn ng lc hc.

Trong ng hc vt th c a ra di hai m hnh: ng im v vt

rn. ng im l im hnh hc chuyn ng trong khng gian, cn vt rn l

tp hp nhiu ng im m khong cch gia hai im bt k trong n lun

lun khng i. Khi kho st cc vt thc c kch thc khng ng k, c th

coi nhm hnh ng im.

Chuyn ng l s thay i v tr ca vt trong khng gian theo thi gian.

n v o di l mt v k hiu m, n v o thi gian l giy vit tt l s.

Tnh cht ca chuyn ng ph thuc vo vt chn lm mc so snh ta

gi l h qui chiu. Trong ng hc h qui chiu

c la chn tu sao chovic kho st chuyn ng ca vt c thun tin . c th tnh ton ngi ta

cn phi chn h to gn vi h qui chiu. Thng thng mun hnh v c

n gin ta dng ngay h to lm h quy chiu.

Tnh thi gian thng thng phi so snh vi mc thi im t0chn trc.

V ni dung, ng hc phi tm cch xc nh v tr ca vt v m t

chuyn ng ca vt theo thi gian so vi h quy chiu chn.

Thng s xc nh v tr ca vt so vi h quy chiu chn l thng s

nh v. Thng s nh v c th l vc t, l to , l gc...

Qui lut chuyn ng c biu din qua cc biu thc lin h gia cc

thng s nh v vi thi gian v c gi l phng trnh chuyn ng. Trong

phng trnh chuyn ng th thi gian c coi l i s c lp. Khi kh i

s thi gian trong phng trnh chuyn ng ta c biu thc lin h gia ccthng s nh v v gi l phng trnh qi o.


55/237


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Trong qu trnh chuyn ng, ng im vch ra mt ng gi l qu o

chuyn ng ca ng im. Phng trnh ca ng qu o cng chnh l

phng trnh chuyn ng (5-1) nhng vit di dng thng s.

Nu ng qu o l thng ta ni ng im chuyn ng thng, nu

ng qu o l cong ta ni chuyn ng ca im l chuyn ng cong.

5.1.2. Vn tc chuyn ng ca im

Gi thit ti thi im tv tr ca ng im xc nh bi vc t nh v rr

.

Ti thi im t1= t + t ng im n v tr M1xc nh bi rr

1, ta c MM1=

rr

1- rr

= rr

(xem hnh 5-2). Gi t s t

r

l vn tc trung bnh ca ng im

trong khong thi gian t v k hiu l tbvr

. Khi t cng nh ngha l M1cng

gn M th cng gn n mt gii hn,

gii hn gi l vn tc tc thi ti thi

im t.

tbvr

Nu k hiu vn tc tc thi cang im l th:v

r

dt

rd

t

vlimv

0t

rr

=

=

(5.3)

z

y

x

O

r

r1

cpv

r

v M1

M

Vn tc tc thi ca ng im bng

o hm bc nht theo thi gian ca vc t

nh v ti thi im .

V mt hnh hc ta thy vc t rr

nm trn ct tuyn MM1v hng t M n M1v vy khi tin ti gii hn vc t

vn tc s tip tuyn vi qu o ti v tr M ang xt v hng theo chiu

chuyn ng ca im.

vr

Hnh 5.2

n v tnh vn tc l mt/giy vit tt l m/s


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5.1.3. Gia tc chuyn ng ca im

Gi thit ti thi im t im c vn tc vr

v ti thi im t1im c vn

tc l vr

1. T s

t

v

r

=

t

vv1

rr

gi l gia tc trung bnh ca im trong thi gian

t. Gii hn t s khi t tin ti khng gi l gia tc tc thi ca im. Ta

c:

wr

2

2

0t dt

rd

dt

vd

t

vlimw

rrrr

==

=

(5-3)

Nhvy gia tc tc thi ca im l

vc t o hm bc nht theo thi gian cu

vc t vn tc hay o hm bc hai theo

thi gian ca vc t nh v. V mt hnh

hc vc t bo gi cng hng v pha

lm ca ng cong (xem hnh 5-3), do

vc t gia tc bao gi cng hng v

pha lm ca ng cong. n v o gia tc l mt/giy

vr

wr

2

vit tt l m/s2

z

y

x

O

M1

Mv

r

cpr

v1

v

Hnh 5.3

5.1.4. Tnh cht ca chuyn ng

xem xt chuyn ng ca im l thng hay cong ta cn c vo tch

x =vr

wr

cr

Nu = 0 th v cng phng, ngha l vn tc c phng khng

i. Chuyn ng lc l chuyn ng thng.

cr

vr

wr

vr

Nu 0 th v hp vi nhau mt gc iu chng t vc tcr

vr

wr

vr

thay i phng v chuyn ng s l chuyn ng cong. xt chuyn ng

ca im l u hay bin i ta cn c vo tch v hng vr

. = B.wr

V v2= ( )vr 2nn

dt

)v(d

dt

)v(d 22=

r

= 2vr

.wr

Cho nn nu B = 0 th chng t vr

l hng s ngha l ng im chuynng u.


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Nu B 0 th v l i lng bin i, chuyn ng l bin i. Nu B > 0

chuyn ng nhanh dn v B < 0 chuyn ng chm dn.

r

5.2. Kho st chuyn ng ca im bng to cc

5.2.1. Thng s nh v v phng trnh chuyn ng

Xt ng im M chuyn ng theo

ng cong trong h trc to cc oxyz

(hnh 5-4).

z

y

x

Oz

M

r

y

x

J

k

i

y cc to x,y,z l cc thng s

nh v ca im M.

Khi M chuyn ng cc to ny thay

i lin tc theo thi gian do ta c:

x = x(t);Hnh 5.4

y = y(t); (5-4)

z = z(t).

Cc phng trnh (5-4) l phng trnh chuyn ng ca im v cng l

phng trnh qu o ca im vit di dng thng s trong to cc.


Nu gi cc vc t n v trn ba trc to l ,ir

jr

, kr

th vc t nh v v

vc t vn tc c th vit:

rr = x + y + zi

r

jr

kr

. Suy ra

vr

=dt

rdr

=dt

d(x + y + zi

r

jr

kr

) =dt

dxir

+dt

dyjr

+dt

dzkr

(5.5)

Biu thc trn chng t:

vx=dt

dx= ; vx& y=

dt

dy = ; vy& x=

dt

dz= . (5.6)z&


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Hnh chiu vc t vn tc ln cc trc to bng o hm bc nht theo

thi gian cc to tng ng.

Da vo cc biu thc (5.6) d dng xc nh c vc t vn tc c v

ln v phng chiu.

v =222

z2

y2

x2

dt

dz

dt

dy

dt

dxvvv

+

+

=++

cos(ox,v) =v

vx ; cos(oy,v) =v

v y ; cos(oz,v) =v

vz .

5.2.3. Gia tc ca im

Tng t nhi vi vn tc, da vo biu thc (5.3) ta c th tm thy:

wx=dt

dvx = xdt

xd2

2

&&= ;

wy=dt

dvy = ydt

yd2

2

&&= ; (5.7)

wx=dt

dvz = zdt

zd2

2&&= .

Gia tc chuyn ng ca im s c xc nh v ln v phng

chiu theo cc biu thc sau:

w = 222z2y2x2 zyxwww &&&&&& ++=++

cos(ox,w) = ww x

; cos(oy,w) = w

w y

; cos(oz,w) = ww z

.

Khi bit v ta c th xem xt c tnh cht chuyn ng ca im M.vr

wr

5.3. Kho st chuyn ng ca im bng to t nhin

5.3.1. Thng s nh v v phng trnh chuyn ng

Gi thit ng im M chuyn ng theo mt ng cong AB trong h

to oxyz. (xem hnh v 5.5). Trn qu o AB ly im O lm gc v chn


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chiu dng cho ng cong. Thng thng ta chn chiu dng ca ng

cong l chiu m ng im chuyn ng. R rng nu bit cung OM = s ta c

th bit v tr ca im M trn qu o. Ni khc i cung OM = s l thng s

nh v ca ng im, cn gi l to cong. Khi im M chuyn ng s s

bin i lin tc theo thi gian ngha l:

s = s(t) (5.8)

Bit c quy lut bin thin (5.8) ta c th xc nh v tr ca im M

bt k thi im no. Biu thc (5.8) c gi l phng trnh chuyn ng ca

im. Theo phng php ny xc nh chuyn ng ca im phi bit:

- Qu o chuyn ng AB

- Chiu chuyn ng trn qu o

- Quy lut chuyn ng (5.8).


Gi thit ng im chuyn ng trn ng cong AB. Ti thi im t

ng im v tr M xc nh bng to cong s. Ti thi im t 1= t + t im

v tr M1xc nh bng to cong s1= s + s.

x1y1

O1

z1

B

M

-0+

s

A

T st

s

= tb

1 vt

ss=

gi l tc trung

bnh.

Gii hn ca t s ny khi t tin ti

khng gi l tc tc thi ca im ti thi

im t v k hiu l v.

Hnh 5.5v= s

dt

ds

t

slim

0t&==

(5.8) s1

-0+ M1s v

s

Vn tc c gi tr bng o hm bc nht

theo thi gian ca qung ng s, c phng tip

Hnh 5.6


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tuyn vi qu o, hng theo chiu ca chuyn ng. ( xem hnh 5.6).

5.3.3. Gia tc tip tuyn v gia tc php tuyn ca im.

5.3.3.1. H to t nhin

Gi thit cht im chuyn ng theo

ng cong AB nhhnh (5.7).

Trn ng cong ly hai im M1M1'

ln cn hai bn im M. V mt phng i

qua ba im . Khi hai im M1M1' tin

gn n M th mt phng trn tin gn n

gii hn ca n l mt phng () gi l mt

phng mt tip. Trong mt phng mt tip

v ng M tip tuyn vi qu o (trng

vi vc t vn tc ( ). Mt trc khc vn

nm trong mt phng mt tip v vung gc vi Mti M k hiu l Mn gi l

php tuyn chnh. Trc Mb vung gc vi hai trc kia gi l trng php tuyn.

Ta chn chiu ca ba trc Mnb to thnh mt tam din thun v gi l h to

t nhin.

vr

v

n

b

M1

A

M

v1n

v1

Bv1

baM1

Hnh 5.7

5.3.2. Gia tc tip tuyn v php tuyn ca im

Nhtrn bit:

wr

= lim =t

v

r

= lim =t

vv1

rr

t0 t0

Chiu biu thc ny ln cc trc to t nhin ta c:

t= lim = ;t

vv t1t

t0w

wn= lim =;t 0 t

vv nn1

;

wb

= 0;

Trn hnh (5.7) gi cung MM1= s ; gc hp bi vr

v Ml ta c:


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==

1k

slim

0t

T s k gi l cong cn l bn knh cong ca qu o ti M.

Mt khc khi chiu vc t vr

v vr

1ln cc trc ta c:

vt= v vt1= v1cos;

vn= 0 vn1= v1sin;

Thay th kt qu tm c vo biu thc ca wtv wns c:

wt

= t

vcosv1

0tlim

;

wn = )t

sinv( 1

0tlim

;

Khi t tin ti 0, im M1dn ti M v tin ti 0, s tin ti 0, v1tin

ti v; costin ti 1. Thay cc gi tr ny vo biu thc trn ta nhn c:

wt = sdt

sddtdv

tvvlim 2

2

1 &&=== ;

wn =

=

21

v)

t

s.

s.

t

sinvlim( .

Trong biu thc (5.9) wtv wnl gia tc tip tuyn v gia tc php tuyn

ca im ti thi im t.

Gia tc tip tuynt

wr

c tr s bng o hm bc nht theo thi gian cavn tc hay bng o hm bc hai theo thi gian ca qung ng i s, c

phng tip tuyn vi qu o, cng chiu vi vr

khi wt> 0 v ngc chiu vi vr

khi wt


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64/237

-64-

Gia tc ton phn bng gia tc php tuyn c v ln v phng chiu.

Trong chuyn ng cong u phng trnh chuyn ng c th thit lp nhsau:

Ta c: ,vdt

ds= ds = vdt.

Tch phn hai v ta c: =S

0S

t

t

,vdtds

Hay s = s0+ v.t

5.3.4.3. Chuyn ng thng bin i u

Trong trng hp ny wt = wn = 0 do w = 0. Suy ra phng trnh

chuyn ng x = xo+ v.t

5.3.4.4. Chuyn ng cong bin i u

Chuyn ng cong bin i u l chuyn ng c w t= const.

Ta c: ;wdt

dv t= dv= wtdt

Ly tch phn hai v s c: hay v = v =v

v

t

t

t

o

,dt.wdv o+ wt.t

Phng trnh chuyn ng vit c:

t.wvdt

ds to+= suy ra : ds = vodt + w

t.t.dt;

Hay: s = so+ vot +2

tw 2t

.

Sau y l mt s bi ton th d.

M

A

y

Ox

B

v

w

Th d 5.1:Xc nh qu o, vn tc

v gia tc ca im M nm gia tay bin AB

ca c cu bin tay quay OAB, (xem hnh

5.9) cho bit OA = AB = 2a v thi im

kho st tng ng vi gc ca c cu, vi

= t.Hnh 5.9


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-65-

Bi gii:

Chn h to oxy nm trong mt phng c cu.

Gi to ca im M l x,y ta c:

x = 2acos+ a cos= 3 acos;

y = a sin.

y chnh l phng trnh chuyn ng ca im trong to cc.

xc nh qu o ca im, t phng trnh trn rt ra:

cost = a3x

; sint = ay

;

suy ra 1a

y

a9

x2

2

2

2

=+ .

y chnh l phng trnh Enlip nhn cc trc i xng l ox v oy ( xem

hnh v 5.9).

tm vn tc ta p dng biu thc (5.6) c:

vx= tsina3dt

dx= ;

vy= tcosadt

dy= .

Cui cng xc nh c vn tc ca im M nhsau:

vM= .a.tcostsin9vv 22

y2

x2 +=+

Phng chiu ca vr

M nhhnh v. T kt qu trn ta thy vmin= av

vmax= 3a.

Theo biu thc (5.7) xc nh c gia tc ca im M:

wx=2

2

dt

xd= -3a2cost = - 2x;

wy= -a2sint = - 2y;


66/237

-66-

Gia tc ton phn w = .r)yx( 2224 =+

Phng chiu ca w c xc nh nh cc gc ch phng nhsau:

cos(w,ox) = ;rx

ww x = cos(w,oy) =

ry

ww y = .

T kt qu trn cho thy phng chiu wr

lun lun hng t M v O.

Th d 5.2.im M chuyn ng theo phng trnh:

x= a sint ; y = a cost; z=ut.

Trong a, v u l khng i.Xc nh qu o, vn tc v gia tc ca im M.

Bi gii:

T hai phng trnh u suy ra:

sin2t + cos2t = a2 hay x2+ y2= a2 (a)

Kt hp ph

ng trnh (a) vi ph

ng trnh z = ut ta thy im chuynng trn mt tr bn knh a v trc l oz.

T z = ut suy ra t = z/u v thay vo biu thc ca x ta c:

x = a sin ;z.u

y = cos ;z.

u

Qu o ca im M l mt ng vt, c trc oz.

Gi T1l chu k ca ng vt. T1xc nh t biu thc:

T = 2 hay T1= 2

Trong thi gian T1 ng im quay quanh trc oz c mt vng ng

thi cng tin theo dc trc oz mt on h =uT1=

u2; h gi l bc ca vt.

xc nh vn tc v gia tc ta p dng phng php to cc.


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-67-

vx= acost;

vy= asint;

vz= u.T xc nh vn tc v ca im.

v = 22222222z2y2x2 ua;u)tsint(cosavvv +=++=++

Nhvy vn tc v ca im c tr s khng i v phng tip tuyn vi

qu o (xem hnh 5.10). Tng t ta xc nh c:

wx= -a2

sintwx= -a

2cost;

wz= 0.

Cy

x

z

a

xO

y

z

a

v w = .aww 2y2x2 =+

Gia tc ca im c ln khng i

cn phng chiu c xc nh bng cccosin ch phng.

cos(w,x) = ;a

xtsin

w

w x ==

cos(w,y) = ;a

ytsin

w

w y == Hnh 5.10

cos(w,x)w

w z = 0.

Mt khc ta thy:

=cosa

x; = cos

a

y.

v biu din trn hnh v.

Nhvy gia tc lun lun hng theo bn knh t ng im vo trc oz.wr


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-68-

Th d 5.3: Mt bnh xe bn knh R ln khng trt trn ng thng.

Vn tc tm bnh xe v = v(t).

Lp phng trnh chuyn ng ca im M nm trn vnh bnh xe.

Kho st vn tc v gia tc ca im M .

Kho st tnh bin i chuyn ng ca im M trn qu o ng vi mt

vng ln ca bnh xe khi V=Vo= cosnt.

Bi gii:

Chn gc to l im tip xc O gia M v mt ng (xem hnh

5.11).

t gc PCM = . xc nh phng trnh chuyn ng ta tm quan h

gia cc to x.y ca im vi gc .

A

xM0

O

H

C

P

E

MC0

y

R

v C

Hnh 5.11


69/237

-69-

Trn hnh c x = OH = OP - PH = R- R sin;

y = HM =R + Rsin(-900

) = R - Rcos= R(1 - cos);

V bnh xe ln khng trt nn: OP = .t

0)t( dtv

Suy ra = (t) = t

o )t( dtv

R

1

Phng trnh chuyn ng ca im M c th vit c:

x= R(- sin);

y= R(1- cos);

= (t).

y l phng trnh ca ng Xycloit vit di dng thng s.

Kho st chuyn ng ca im M trn cung OA.

Vn tc v gia tc ca im xc nh nhsau:

==

==

sinRyv

);cos1(Rxvv

y

x

&&

&&r

.sinRcosRvw

);cos1(RsinRvww

2

yy

2

xx

+==

+==

&&&&

&&&&r

Ti v tr chm t O v A th =0 v = 2. Khi sin= 0, cos=1.

v: vx= 0 ; vy= 0 suy ra v = 0;

wx= 0; wy= R2> 0.

wr

lc ny khc khng, do im ch dng li tc thi mt t.

Trong trng hp c bit v = v0= hng s th:

= ;R

tvdtv

R

1 oto )o(

=


70/237

-70-

= ;R

tv o o= 0; & = ;R

v o .0=&&

Lc ny: vx= vo(1-cos); vy= vosin;

wx= sinR

v o2

; wy = cosR

v o2

.

xt tnh cht chuyn ng ca im trn cung OA ta c:

vr

. = vwr

x.wx+ vy.wy= ( )[ ];cossincos1sinR

v o3

+ = .sinR

v o3

Nhvy vr

. > 0 trong khong 0 < < vwr

w.v rr

< 0 trong khong r3th 3ngc chiu vi AB

v c bit r1= r3th 3= 0 bnh rng 3 s chuyn ng tnh tin.


86/237

-85-

Chng 7

Chuyn ng tng hp ca im

7.1. Chuyn ng tuyt i, chuyn ng tng i vchuyn ng ko theo.

Chuyn ng tng hp ca im l chuyn ng c to thnh khi im

tham gia hai hay nhiu chuyn ng ng thi. Ta xt bi ton trong m hnh

sau y : Kho st chuyn ng ca im M trn h to ng o 1x1y1z1gn

trn vt A. Vt A li chuyn ng

trong h to c nh oxyz (xem

hnh 7.1).

x

y

z

Ox1

y1z1M

A

r

ro

z1

o1y1

x1k1j1

i1

Chuyn ng ca im M so

vi h c nh oxyz gi l chuyn

ng tuyt i. Vn tc v gia tc ca

chuyn ng tuyt i k hiu l :a

vr

va

wr

. Hnh 7.1

Chuyn ng ca im M so vi h ng o1x1y1z1 gi l chuyn ng

tng i k hiu l v .rvr

rwr

Chuyn ng ca h ng (vt A) so vi h c nh oxyz gi l chuyn

ng ko theo. Vn tc v gia tc ca im thuc vt A ( h ng ) b im M

chim ch ( trng im ) trong chuyn ng ko theo l vn tc v gia tc kotheo ca im M v k hiu l : ev

r

v ewr

.

Nhvy chuyn ng tuyt i ca im M l chuyn ng tng hp ca

hai chuyn ng tng i v ko theo ca n.

Th d : Con thuyn chuyn ng vi vn tc ur

so vi nc. Dng nc

chy vi vn tc vr

so vi b sng. y chuyn ng ca con thuyn so vi b

sng l chuyn ng tuyt i . Chuyn ng ca con thuyn so vi mt nc l

chuyn ng tng i vi vn tc .uv rrr

= Chuyn ng ca dng nc so vi


87/237

-86-

b l chuyn ng ko theo, vn tc ca chuyn ng ko theo .vverr

=

Theo nh ngha trn ta thy, xt chuyn ng tng i ta xem h

ng nhc nh. Khi phng trnh chuyn ng vit di dng vc t nh

sau : 11111111 kzjyixMOrrrrr

r

++== . (7-1)

y1i

r

,1j

r

,1k

r

l cc vc t n v trn cc h ng. Khi xt chuyn

ng tng i nh trn ni cc vc t 1ir

,1j

r

,1k

r

c xem nhkhng i.

Cn cc to x1, y1, z1 l cc hm ca thi gian.

x1 = x1(t) ; y1 = y1(t) ; z1= z1(t).

Mun xt chuyn ng ko theo ca im ta ch cn c nh n trong h

ng khi phng trnh chuyn ng ca M so vi h c nh oxyz l phng

trnh chuyn ng ko theo. Ta c :

111111010 kzjyixrrrOMrrrr

rrrr

+++=+== (7-2).

Trong phng trnh (7.2) v ta c nh im trong h ng nn cc to

x1 , y1 , z1 l khng i, cn 1ir

, 1jr

, 1kr

l cc vc t bin i theo thi gian.

)t(rr 00rr

= ; )t(iirr

= ; )t(jjrr

= ; )t(kkrr

= .

7.2. nh l hp vn tc.

Xt im M chuyn ng tng

i trong h ng o1x1y1z1 vi vn tc

; H ng chuyn ng trong h c

nh oxyz ko theo im M chuyn

ng vi vn tc ko theo (xem hnh

7-2). xc nh vn tc tuyt i ta

thit lp phng trnh chuyn ng

tuyt i ca im M. Ta c :

rvr

evr

ro

r

a

x

y

z

Ox1

y1z1

M

c1

c2

vev

rv

1r

o1

Hnh 7.2

111111010 kzjyixr)t(rrrrrr

rrrr

+++=+= (7-3)


88/237

-87-

Phng trnh ny ging phng trnh (7-2) nhng cn lu l mi tham

s ca phng trnh u l cc hm ca thi gian.

o hm bc nht theo thi gian phng trnh (7-3) ta c :

++

+++== 1

11

11

1111

0a

kdt

dzj

dt

dyi

dt

dx

dt

kdz

dt

jdy

dt

idx

dt

rd

dt

rdv

rrr

rrr

rr

r

Trong kt qu tm c, nhm s hng th nht

+++

dt

kdz

dt

jdy

dt

idx

dt

rd111

0

rrr

r

chnh l o hm bc nht theo thi gian ca ph

ng trnh (7-2) (ph

ngtrnh chuyn ng ko theo ) l vn tc ko theo

evr

.

Nhm cc s hng cn li :

+ 11

11

11 k

dt

dzj

dt

dyi

dt

dx rrr

l o hm bc nht theo thi gian ca phng trnh (7.1) (phng trnh

chuyn ng tng i ) do c thay th bng vn tc tng i rvr

.

Thay cc kt qu va tm c vo vn tc tuyt i ta c :

rea vvv rrr

+= .

nh l 7.1 : Trong chuyn ng tng hp ca im vn tc tuyt i

bng tng hnh hc vn tc ko theo v vn tc tng i :

rea vvv rrr

+= . (7-4)

7.3. nh l hp gia tc

thit lp biu thc ca gia tc tuyt i ta o hm bc hai theo thi

gian phng trnh chuyn ng tuyt i ca im (phng trnh 7.3). Ta c :

++

+++=== 12

1

2

12

1

2

12

1

2

2

2

12

2

12

2

12

0

2

a

2

a kdt

zdj

dt

ydi

dt

xd

dt

kdz

dt

jdy

dt

idx

dt

rd

dt

vd

dt

rdw

rrr

rrr

rrr

r


89/237

-88-

++

dt

kd

dt

dz

dt

jd

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Cơ Học Lý Thuyết - Nhiều Tác Giả, 237 Trang (1)