bài tập cơ học lý thuyết tập 1-hvktqs

8/8/2019 bi tp c hc l thuyt tp 1-hvktqs

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Hc vin k thut qun s

B mn c hc vt rn khoa c kh

Anh Cng (Ch bin)V Quc Tr L Nho Thit

T Hu Vinh L Hi Chu

Bi tpC hc l thuyt

Tp 1

Tnh hc v ng hc

i tng s dng : i hc v Cao ng

H Ni - 2005


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3

Mc lc

Li ni u 5

Chng 1 : Bi ton phng 7

1.1- C s l thuyt 7

1.2- Hng dn p dng 14

1.3- Bi gii mu 16

1.4- Bi tp 29

Chng 2 : Bi ton khng gian 45

2.1- C s l thuyt 45


2.3- Bi gii mu 51

2.4- Bi tp 61Chng 3 : Bi ton ma st 69

3.1- C s l thuyt 69


3.3- Bi gii mu 75

3.4- Bi tp 84

Chng 4 : Bi ton trng tm 91

4.1- C s l thuyt 91

4.2- Hng dn p dng - bi ton gii mu 94

4.3- Bi tp 100

Chng 5 : Chuyn ng ca im 104

5.1- C s l thuyt 104

5.2- Phng php gii 105

5.3- Bi gii mu 106

5.4- Bi tp 116

Chng 6 : Chuyn ng quay ca vt rn quanh mt trc c nh 1276.1- C s l thuyt 127


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4


6.3- Bi gii mu 131

6.4- Bi tp 137

Chng 7 : Hp chuyn ng ca im 143



7.3- Bi gii mu 146

7.4- Bi tp 160

Chng 8 : Chuyn ng song phng ca vt rn 173



8.3- Bi gii mu 181

8.4- Bi tp 195

Chng 9 : Chuyn ng ca vt rn quanh mt im c nh 208



9.3- Bi gii mu 213

9.4- Bi tp 219

Chng 10 : Hp chuyn ng ca vt rn 224

10.1- C s l thuyt 224


10.3- Bi gii mu 228

10.4- Bi tp 235

Ti liu tham kho 241


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5

Li ni u

C hc l thuyt l khoa hc v cc quy lut cn bng v chuyn ng cacc vt th di tc dng ca lc, l mt trong nhng mn hc trng im cho

sinh vin cc trng i hc k thut.

Vic vn dng cc kin thc hc vo gii cc bi tp c hc l thuyt l

yu cu hng u i vi sinh vin, qua gip h hiu su thm v l thuyt,

ng thi nng cao kh nng tduy v rn luyn k nng trong hc tp.

Gio trnh Bi tp c hc l thuyt c bin son theo chng trnh ging

dy mn C hc l thuyt cho sinh vin ca hu ht cc ngnh uc o to tiHc vin k thut qun s, n cng ph hp vi chng trnh mn hc ca B

Gio dc & o to.

Gio trnh Bi tp c hc l thuyt c phn thnh hai tp. Tp 1 gm 2

phn: Tnh hc (t chng 1 n chng 4) v ng hc (t chng 5 n

chng 10). Trong mi chng u c phn tm tt l thuyt, phn loi bi tp v

phng php gii, cc v d mu v phn bi tp. Cui mi bi tp c tr li hoc

p s sinh vin tham kho v t kim tra li gii ca mnh. Cn l

u trongphn tr li, i vi cc i lng vct (lc, vn tc, gia tc ...), kt qu c

cho dng tr i s.

Gio trnh Bi tp C hc l thuyt Tp 1 do Nhm mn hc C hc l

thuyt thuc B mn C hc vt rn Khoa C kh Hc vin K thut qun s

bin son : V Quc Tr, T Hu Vinh (phn Tnh hc), L Nho Thit, L Hi

Chu (phn ng hc), Anh Cng ch bin.


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6

Trong sch chc chn cn c thiu st, chng ti mong nhn c kin ca

bn c, xin trn trng cm n v tip thu b sung, sa cha cho gio trnh

c tt hn. Cc nhn xt, gp xin gi v : B mn C hc vt rn, Khoa C

kh, Hc vin KTQS.

Cc tc gi


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7

Chng 1. Bi ton phng

1.1. C s l thuyt

1.1.1. Thu gn h lc phng:

H lc phng khi thu gn v mt tm (O), nhn c mt vc t chnh R '

v mt m men chnh OM , vc t m men chnh lun vung gc vi mt phng

tc dng ca lc, xy ra cc trng hp sau:

OR ' 0;M 0 = , H lc tng ng vi mt lc kk

R F= .

OR ' 0;M 0= , H lc tng ng vi mt ngu ( )O O kk

M m F= r

.

OR ' 0;M 0 , H lc tng ng vi mt lc kk

R F= vi tm thu

gn t ti im Occh O mt onOM

dR '

= r .

Ch : Vc t chnh l mt bt bin, n c gi tr v phng chiu khng thay i

v khng ph thuc vo tm thu gn.

1.1.2. iu kin cn bng ca h lc phng:

T cc kt qu thu gn h lc nu trn, ta c nh l v iu kin cn bng

ca h lc phng nh sau:

nh l(v iu kin cn bng ca h lc phng): iu kin cn v h

lc phng cn bng l vect chnh v mmen chnh ca h lc i vi im O no

trit tiu.

( )n n

k O O k

k 1 k 1

R ' F 0; M m F 0= =

= = = = r r r

. (1.1)

T nh l trn c th suy ra rng, h lc phng c ba h phng trnh cn

bng vit theo ba dng sau:

Dng 1: ( )n n n

kx ky O k

k 1 k 1 k 1

F 0; F 0; m F 0;= = =

= = = r Ox Oy ; (1.2)


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Dng 2: ( ) ( )n n n

k A k B k

k 1 k 1 k 1

F 0; m F 0; m F 0;= = =

= = = r r

(1.3)

iu kin: ng niAB khng vung gc vi .

Dng 3: ( ) ( ) ( )n n n

A k B k C k

k 1 k 1 k 1

m F 0; m F 0; m F 0;= = =

= = = r r r

(1.4)

iu kin: Cc imA, B, C khng thng hng.

Trng hp cc h lc c bit:

- H ng quy: c hai phng trnh cn bng:n n

kx ky

k 1 k 1

F 0; F 0;= =

= = (1.5)

i vi h lc ng quy, chng ta cn s dng iu kin cn bng di

dng hnh hc:iu kin cn v h lc ng quy cn bng l a gic lc t

khp kn.

- H lc song song(gi thit vi trc y): c hai h phng trnh cn bng:

( )n n

ky O k

k 1 k 1

F 0; m F 0;= =

= = r

(1.6)

hoc: ( ) ( )n n

A k B k

k 1 k 1

m F 0; m F 0;= =

= = r r

(1.7)

iu kin: ng niAB khng song song vi trcy

Ch : i vi h vt, c hai loi iu kin cn bng:

iu kin cn bng ca tng vt tch ring.

iu kin cn bng ca ton h ho rn (xem ton h nhmt vt rn duy

nht) hay cn gi l iu kin cn bng ca cc ngoi lc (v khi ho rn, h

cc ni lc xem nhcn bng).

Nhng iu kin cn bng ca ton h ho rn l h qu ca cc iu kin

cn bng ca tng vt. Chng ta cng c th xt ring mt phn h, ho rn v lp

cc iu kin cn bng tng ng.


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Nh th i vi h vt, kh nng lp cc iu kin cn bng l rt rng ri,

vn t ra l lp iu kin cn bng thch hp c th gii quyt bi ton t

ra mt cch thun li nht.

H qu:

H ba lc phng cn bng, khng song song l h ng quy phng;

H n lc cn bng, trong c (n-1) lc song song l h lc song song (lc

thn song song vi n-1 lc u);

Hai lc cn bng vi mt ngu lc phi to thnh mt ngu lc ngc chiu

quay v cng tr s mmen.

Kt qu thu gn h lc phng c tng kt trong bng 1.1.

Bng 1.1

Hlc

Kt qu thu gn h lciu kin

cn bng

Phng trnh

cn bng

Btk

( ) ( )1 2 nF , F ,..., F R, M n

j

j 1

R R ' F=

= = r r r

( )n

O O k

k 1

M M m F=

= =

r

k

O O

R F 0

M m 0

= =

= =

O

X 0

Y 0

m 0

=

=

=

ngquy ( ) ( )1 2 nF , F ,..., F R

n

k

k 1

R R ' F=

= = r r r

kR F 0= = X 0

Y 0

=

=

Ngu

( ) ( )1 2 nF , F ,..., F M

( )n

O O k

k 1

M M m F=

= = r

O kM M 0= = kM 0=


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Songsong

( ) ( )1 2 nF , F ,..., F R, M k

O O

R F 0

M m 0

= =

= =

O

Y 0

m 0

=

=

1.1.3. Cc php bin i v xc nh lca. Lc hot ng v phn lc lin kt:

1. Lc hot ng: l lc c quy lut xc nh, hoc tp trung, hoc phn b.

Lc phn b xc nh bi biu v cng phn b v thng c thu gn.

Vi h lc song song cng chiu phn b u hoc theo tam gic, kt qu thu gn

trn hnh 1.1(a), (b). Trng hp phn b theo hnh thang, c th quy v phn b

u v tam gic. Kt qu thu gn h lc song song cng chiu phn b tng qut

trn chiu di ta c lc thu gn song song cng chiu vi h phn b, t ti

trng tm v c cng bng s o din tch S (theo n v thch hp) ca biu

phn b.

(a) (b)Hnh 1.1

2. Phn lc lin kt: Phn lc lin kt t vt gy lin kt (vt c gch cho)

t vo vt kho st (v trng) c biu din di dng lc v ngu lc tp

trung.

Khi gp cc lin kt phc tp, chng ta phi phn tch cu to ca lin kt,

trng thi chu lc ca vt rn, iu kin lm vic ca lin kt, cc di chuyn b

lin kt cn tr quy v cc lin kt n gin gii thiu trn nh quy tc

sau: Tng ng mt di chuyn ( thng, quay) b cn tr, lin kt to c mt

phn lc (lc, ngu lc) ngc chiu di chuyn.

qG

Q=ql

l/2l/2 l/2

q

G

l/2

Q=ql/2


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Bng 1.2 nu c im ca cc lin kt v cc phn lc lin kt thng gp.

Bng 1.2

Lin kt Cu to v cch biu din c im phn lc

1 2 3

Ta trnThng gc vi mt ta,

hng vo vt kho st, k

hiu: N

Dy Nm dc theo dy, hng ra

ngoi vt kho st, k hiu: T

ThanhNm dc theo thanh (ng

ni hai u thanh), k hiu: S

.

Bn l,

gi c

nh

Phn lc R t ti bn l,

c chia thnh hai thnh

phn X, Y theo hai trc x, y.

NgmPhn lc gm: hai thnh

phn lc X, Y v mt ngu

lc M.

N2N1

T2T1T

BA

SA SB

S

x

R

Y

X

y

A X

MA

Y


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1 2 3

tr

ngn

Phn lc vung gc vi trc

(trong mp tc dng ca h

lc).

ci

Phn lc gm 2 thnh phn:

dc trc v vung gc vi

trc (trong mp tc dng ca

h lc).

tr

di

Phn lc bao gm: thnh

phn lc vung gc vi trc

v mt ngu lc nhngm(trong mp tc dng ca h

lc).

Gi di

ng

Phn lc vung gc vi nn,

c mt thnh phn X (trong

mp tc dng ca h lc).

b. Ni lc v ngoi lc:

Khi xt h vt, cn phn bit ngoi v ni lc. Ngoi lc l nhng lc t bn

ngoi tc dng vo h. Ni lc l nhng lc tc dng tng h gia cc vt trong

h. c im ca ni lc l xut hin tng i, cng ng tc dng, ngc chiu

v cng tr s (nhng khng cn bng v t vo hai vt khc nhau). Ch rng

ngoi lc cng nhni lc u c th l lc hot ng hoc lc lin kt.

c. Phng php xc nh lc v vc t chnh:

Xc nh lc: Cho lc F nghing vi trcxmt gc (nhn).

Y

X

o

Y

X

x

X

y

o1

X

o

X

X

o

o1

M M

X


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Hnh chiu ca lc ln cc trc (H.1.2) cho bi

cng thc:

x yF Fcos , F Fsin= = ,

trong : F - tr s ca lc ;

du +(-) khi lc thun (ng

c) vi chiu trc.Xc nh vc t chnh:

Cho h lc kF (k 1,2, ..,n)= . Vct chnh, k hiu R ', l tng hnh hc cc

vect lc:n

k

k 1

R ' F=

= r r

.

xc nh vect chnh, c th p dng hai phng php:

1. Phng php gii tch: Hnh chiu ca vect chnh ln mt trc to

bng tng hnh chiu cc lc ln trc .n n

x kx y ky

k 1 k 1

R ' F ;R ' F ;= =

= =

Tr s ca vect chnh: ( ) ( )2 2

' '

x yR ' R R= +

Cc c sin ch phng ca vect chnh:

( ) ( )''yx

RRcos x,R ' ; cos y,R '

R ' R '= =

r r

2. Phng php hnh hc: Vect chnh'

R l vect khp kn a gic lc, l vect ni im cui ca mt ng gy khc c cc cnh tng ng (song

song, cng chiu, cng di) vi cc vect lc. Trng hp hai im u v

cui ca a gic lc trng nhau, vect chnh R ' 0= , chng ta ni a gic lc t

khp kn.

d. Phng php xc nh mmen lc v mmen chnh:

M men lc: Cho lc F v im O (H.1.2).

Mmen lc F i vi O l lng i s: ( )Om F Fh= Trong : F tr s lc ;

Hnh 1.2

O

y

x

F

H


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h = OH tr s ca cnh tay n.

Ngoi cch p dng trc tip nh ngha, c th dng hai phng php sau:

1. Phng php phn lc: Phn tch lc F ra hai thnh phn 1 xF F= v

2 yF F= song song vi hai trc tng ng, sau ta tnh tng mmen cc thnh

phn:

( ) ( ) ( )O O 1 O 2m F m F m F= +

2. Phng php gii tch: Gix, y l to im t A ca lc, x yF , F l cc

hnh chiu ca lc ln cc trc to tng ng, khi m men lc i vi tm

O l:

( )O y xm F xF yF=

M men chnh: Mmen ca h lc phng i vi tm O (mmen chnh ca hlc) k hiu OM l tng mmen cc lc i vi im O:

( )n

O O k

k 1

M m F=

= r

i vi ngu lc, mmen chnh khng ph thuc tm O nn c gi l

mmen ngu lc v bng: ( )*M m F,F Fd= = . Vi h ngu lc, mmen chnh

bng tng mmen cc ngu lc.

1.2. Hng dn p dng

gii bi ton phng, ta tin hnh theo cc bc sau:

Bc 1:Chn vt kho st: (vt kho st c th l: mt vt; h vt; mt vt

do nhiu vt ghp li; mt phn tch ra t mt vt - phn khng kho st t

vo phn kho st nhng lc gi vai tr ca phn lc lin kt; mt nt, im

tp trung cc dy, cc thanh...).t lc: Trc tin, biu din tt c cc lc

cho, tip theo biu din cc phn lc lin kt (phn lc lin kt ta v lin kt

dy c phng chiu xc nh; phn lc cc lin kt thanh, bn l, ngm c

chiu cha bit trc c v theo gi nh - p s dng, chiu ga nh


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ng; p s m - chiu ngc li). Trong biu din lc, cn ch im t,

phng chiu, s thnh phn v cch phn tch cc phn lc. Cn phn bit

lc hot ng v lc lin kt, lc tc dng vo tng vt v vo ton h ho rn,

ni lc v ngoi lc.

Bc 2:Phn tch c im h lc kho st: (ng quy, song song hay bt

k), t xc nh s phng trnh cn bng c lp c th lp c (2 hoc

3) theo cc iu kin trong mc 1.1.2, sau tin hnh gii cc phng trnh

nhn c.

Ch :

1. Nu h lc kho st l ng quy phng, c th dng iu kin cn bng

dng hnh hc (a gic lc t khp kn). Trong trng hp s lc t, c th dng

phng php bin i lc. bi ton n, vt lt, phng trnh cn bng lp c

ch cha cc lc hot ng

2. i vi bi ton h vt, v c hai loi iu kin cn bng nn tng ng

c hai cch thnh lp cc phng trnh cn bng: phng php tch vt v phng

php ho rn.

Phng php tch vt: ln lt xem xt ring v lp phng trnh cn bng

ca tng vt. Phng php ny c bit thun li nu chn c th t tch

vt sao cho s phng trnh cn bng c lp ca vt c tch bng s lc

l n s.

Phng php ho rn: trc ht lp phng trnh cn bng ca cc ngoi

lc v sau mi tch vt. Phng php ny thng c p dng khi tch

vt khng thun li; khi phng trnh cn bng cc ngoi lc c th cho

ngay mt s ngoi lc n, nh qu trnh tch vt tr nn d dng.

3. Cn ch rng, mi cp ni lc d c v ngc chiu, cng k hiu

nhng vn cng gi tr (cng dng hay m tu theo chiu gi nh l ng hay

sai).


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1.3. Bi gii mu

1.3.1. Bi ton tm iu kin cn bng

Bi ton c t ra nh sau: Cho mt vt rn (hay h vt rn phng) chu

tc dng ca h lc choF1, F2, ..., Fn , tm v tr cn bng ca vt (h vt) hoc

tm iu kin rng buc gia cc lc cho vt (h vt) cn bng mt v trno .

Khi gii cc bi ton dng ny, ta cn ch chn phng trnh cn bng

sao cho khng cha phn lc lin kt, lp c phng trnh cn bng cho php

tm ngay ra iu kin cn bng.

Cc bi ton n v vt lt l cc dng bi ton c bit ca phn ny. V d 1-1:

Thanh ng cht AB nngP, di2l

ct ta hai u A v B vo hai cnh nhn ca

mt gc vung OCD. Trn thanh AB ti E

treo mt vt nng trng lng Q. Cho bit

AE=l/2, cnh OC nghing vi ng ngang

mt gc , tm gc nghing ca thanh AB

vi ng ngang khi thanh AB cn bng?

Bi gii:

Kho st s cn bng ca thanh AB (H. 1-3): thanh chu tc dng ca cc

lcP, Qv cc phn lcNA, NB, ta c h lc cn bng:

( )A BP,Q, N , N 0 .

Ta thy rng, ng tc dng ca cc phn lc lin kt ct nhau ti I, vy

chng khng c mt trong phng trnh cn bng ta lp phng trnh m men i

vi im ny:

( ) ( ) ( )I k I I

m F m P m Q 0= + =

. (a)

Sau khi tnh cc gi tr m men v thay vo (a), ta nhn c phng trnh:

Hnh 1-3

P Q

E

I

N

NA

O

AC

B

D


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( ) ( )l

P 2lcos lcos Q 2lsin sin cos 02

= . (b)

Gii phng trnh (b) theo , ta c kt qu:

( )( )

Q 2 P Q cos 2tg

2 P Q sin2

+ =

+ .

V d 1-2:

Thanh ng cht AB, di2l, trng lng P c u A ta trn tng nhn, u

B dc gi bi dy BC nghing vi tng mt gc 300. Tm gc nghing ca

thanh vi tng thanh nm trng thi cn bng?

Bi gii:

Kho st s cn bng ca thanh AB: gii phng cc lin kt, thanh chu tc

dng ca cc lcP, N, T. Do bi ton ch yu cu tm gc nghing ca thanh,

nn ta s xy dng phng trnh cn bng khng cha cc thnh phn phn lc

cha bit. Ly m men vi giao im ca cc ng tc dng ca cc lc TvN

(im E, H. 1-4), ta c phng trnh sau:

( )E km F P.HE 0= = .

T phng trnh trn ta c:HE = 0, hay:

0 2 3HE l sin BFtg30 l sin 2l cos

3

2 3l sin cos .

3

= =

=

T ta xc nh c:2 3

tg3

= .

Bi ton n:

nh ngha: n l vt rn quay c quanh mt trc c nh O v chu

tc dng ca mt h lc cho nm trong mt mt phng vung gc vi trc

quay ca vt.

Hnh 1-4

P

H

FE

300

C

N

T

B

A


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iu kin cn bng ca n: n cn bng, cc lc ch ng tc dng

ln n phi tho mn phng trnh cn bng cc m men i vi trc quay:

( )O km F 0= .

V d 1-3:

n OA c th quay c quanh trc nm ngang i qua O. im B can c ni vi van ni hi D, chu p sut p, u A ca n phi treo vt c

trng lng Q bng bao nhiu khi p sut hi vt qu mt gi trp no th

van s b m ra? Cho bit OA=a, OB=b, din tch ca van l S. B qua trng

lng ca n v ma st.

Bi gii:

Kho st n OAB (H. 1-5), cc lc ch ng tc dng ln n gm c lc

Q v lc yP ca hi nc.

Vit iu kin cn bng cho m men i vi im O, ta c:

( )O km F Pb Qa 0= = .

T ta c: Pb Qa= , thay gi trP = pS, ta nhn c:b

Q pSa

= .

Bi ton vt lt:

Vt lt l vt rn c th b mt cn bng do b lt quay quanh mt ng

ta no , gi trc quay khi vt lt l O, ta thy c hai nhm lc to nn hai

chiu m men khc nhau. Gi tng m men thun chiu quay ca vt khi b lt l

P1

D

C

Q

P2BA

Hnh 1-6Hnh 1-5

p

Q

P

RO B

A

D


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m men lt ML v tng m men ngc li l m men chng lt MCL, ta c iu

kin vt khng b lt l:

CL LM M , hay: ( ) ( )O kCL O kLm F m F .

T s: CL

L

Mk

M

= , c gi l h s n nh chng lt, khi iu kin

vt khng b lt l k1.

V d 1-4:

Cho cn trc nm trn ng ray. Trng lng ca cn cu lP1 nm gia

khong cch hai bnh. Vt cu v tr xa nht l CD=l, trng lng l P2. Tm

trng lng ca i trng Q cn trc khng b lt? Cho khong cch t i

trng n tm cn trc l2a, khong cch gia hai bnh xe l2b.

Bi gii:Kho st cn trc trng thi lm vic bnh thng, cn trc chu cc lc

hot ng l P1, P2, Q v hai phn lc ti A v B. C th xy ra hai kh nng lt:

lt quanh A hoc quanh B (H. 1-6).

Xt kh nng lt quanh A: trng hp ny s nguy him nht khi P2 = 0,

lc lc Q l lc gy m men lt. Thit lp phng trnh m men quanh trc lt

i qua A, ta c:

( ) ( )A k 1m F Q 2a b P b= .

khng b lt ta phi c iu kin: ( )1P b Q 2a b hay 1b

P Q2a b

.

Xt kh nng lt quanh B, lp phng trnh m men i vi im B, ta c:

( ) ( )B k 1 2m F Q 2a b P b P l= + + . khng b lt ta phi c iu kin:

( )1 2P b Q 2a b P l+ + hay2 1P l P bQ2a b

+.

Nhvy, i trng Q phi c gi tr: 2 11b P l P b

P Q2a b 2a b

+

.


19/240

20

V d 1-5 (Bi ton h vt):

Treo mt a trn ng cht v mt vt nng vo cng mt im O c nh

bng hai si dy mm, mnh, khng dn v c trng lng khng ng k. H

cn bng v tr nh hnh v, hy xc nh gc gia dy treo a vi phng

thng ng. Cho bit: bn knh a r, trng lng aP, chiu di on dy treo

a l l, trng lng vt nng l Q.

Bi gii:

Kho st h cn bng gm a, vt nng v cc dy treo (H. 1-7). Cc lc

tc dng ln h gm: cc trng lc P v Q, phn lc lin kt ti O RO. xc

nh gc lch gia dy treo a vi phng thng ng, ta coi c h l mt vt rn

cn bng di tc dng ca h lc k trn, iu kin cn bng nh sau:

( )OP,Q,R 0 .

Lp phng trnh cn bng i vi mmen ti O, ta c:

( ) ( ) ( )O km F P l r sin Q r l r sin 0= + + = .Gii phng trnh trn, ta nhn c:

Q rsin

P Q r l =

+ +.

Hnh 1-7

B

O

A

QP

C300

Hnh 1-8

F

M

600

D

C

BA

O1O


20/240

21

V d 1-6:

Cho c cu phng chu tc dng ca m men Mv lc y F nh hnh v.

B qua khi lng ca cc thanh v ma st, tm mi lin h gia Mv F c

cu cn bng? Cho bit OA=a, im C nm gia thanh O1B (Hnh 1-8).

Bi gii:

Kho st h bao gm: cc thanh OA, AB, O1B, CD v con chy D cn

bng di tc dng ca cc lc ch ng v cc phn lc lin kt ti O, O1, D.

gii bi ton ta p dng phng php tch vt:

Trc tin ta xt cn bng ca thanh OA: ti A do thanh AB ch b ko hoc

nn do ch c mt thnh phn phn lc S1, ti O c phn lc RO. Vit phng

trnh cn bng m men ti O, ta c:

( )O k 1m F M OA.S 0= + = . (a)

Tip theo, kho st s cn bng ca con chy D cng vi thanh CD: ho

rn h ang xt, ti C do thanh ch chu ko hoc nn nn ch c mt thnh phn

phn lc S2 hng dc theo thanh, lp phng trnh cn bng theo phng ngang,

ta c:

0

2F S cos 30 0 = . (b)

Sau cng, ta xt phn cn li ca h bao gm thanh O1B v thanh AB. Ho

rn h ang xt, ti C c thnh phn phn lc S2, ti A c S1. Vit ph

ng trnhcn bng m men i vi O1, ta nhn c:

RO

S 1

M

A

O

S 2 N

300 F

D

C

RO1

S1

S2

C

600

B

O1


21/240

22

( ) ' 0 'O1 k 1 1 2 1m F S .O B.sin 60 S .O C 0= = . (c)

Kt hp 3 phng trnh trn vi ch rng '2 2S S= , kh cc thnh phn

phn lc, cui cng ta nhn c iu kin:

F.a M 3= .

1.3.2. Bi ton tm phn lc lin kt

Bi ton c t ra nh sau: Cho mt vt rn (hay h vt rn phng) cn

bng di tc dng ca h lc choF1, F2, ..., Fn , hy xc nh cc thnh phn

phn lc lin kt.

Khi gii cc bi ton dng ny, ta cn ch biu din y cc lc hot

ng tc dng ln h v cc phn lc lin kt, t nhn c mt h lc cn

bng. tip theo, lp phng trnh cn bng i vi h lc trn v gii cc phng

trnh .

Ch :

gim bt s lng cc lc phi tm, nu cc n ng quy, ta nn lp

phng trnh m men i vi im ng quy ; nu cc n song song, ta

lp phng trnh hnh chiu trn trc vung gc vi cc n .

i vi bi ton h vt, lc u ta nn xc nh tnh tnh nh ca h, sau

xc nh s lng n v s phng trnh cn bng c lp c th lp

c t tm cc v tr tch h tng ng.V d 1-7:

Dm AB c chiu di l, u A ngm cht vo tng, u B t do. Dm chu

tc dng ca lcF to vi phng ngang mt gc ti B v ngu lc c m men

bngM. Tm phn lc ti A, b qua trng lng ca dm (Hnh 1-9)?

Bi gii:

Xt cn bng ca dm AB. Gii phng lin kt ti ngm A, thay bng cc

phn lc lin kt c chiu gi thit nhhnh v. Ta c h lc cn bng:( )A A AF,M,M ,X ,Y 0 (a)


22/240

23

T (a), ta lp cc phng trnh cn

bng:

AX X Fcos 0= =

,

AY Y Fsin 0= = , (b)

A Am M M Flsin 0= = .Gii h phng trnh trn ta nhn c:

A A AX Fcos ; Y Fsin ; M Fsin M= = =

V d 1-8:

Thanh ng cht AB trng lng P, u A gn bn l v c gi cn bng

v tr nm ngang nh dy buc vt nng D, dy to gc 600 vi phng ngang.

B qua trng lng dy, ma st trc bn l v rng rc. Tm trng lng ca vt

D v phn lc ti A (Hnh 1-10)?

Bi gii:

Xt cn bng ca thanh AB chu tc dng ca h lc ( )AP,R ,T , lp phng

trnh cn bng i vi m men ti A, ta c:

0

A

l

m P Tlsin 60 02= + = , t ta nhn

c:

P

T 3= . (a)

Chiu h lc theo hai phng x v y, ta c:

0

AX AX

TT cos60 R 0, R

2 = = (b)

0

AY AY

T 3Tsin 60 R P 0, R P

2+ = = (c)

Trng lng ca vt D cn bng vi sc cng dy T, vy ta c:P

Q3

= .

Hnh 1-9

l

XA

YA

MA

BA

FM


23/240

24

Ta cng c th gii bi ton trn theo cch khc: thy thanh AB

cn bng di tc dng ca ba lc P, RA v T, trong hai lc P vT c ngtc dng ct nhau ti O, vy thanh cn bng ba lc trn phi ng quy. Biu

din tam gic lc nhhnh trn ta s xc nh c cc gi tr cn tm.

V d 1-9:

Dm ng cht AB c trng lngP v chiu di l, du A ngm cht trong

tng, to vi tng mt gc . Trn dm t khi tr trn ng cht, trng

lng Q. Tm p lc ca khi tr ln dm v tng, phn lc lin kt ti A? Cho

bit 2BD l3

= (Hnh 1 - 11).

Bi gii:

Chn h to nh hnh v, h kho st bao gm hai vt: dm AB v khi

tr, ti A c cc phn lc lin kt: XA, YA, ngu MA; ti v tr khi tr tip xc vi

tng c phn lc N2 vung gc vi tng. Vy h c 4 n trong khi ta ch thit

lp c ti a 3 phng trnh cn bng, do ta phi p dng phng php tch

h gii.

Hnh 1- 10

300Q

RA

T300

CP

T600RA

B DA

O


24/240

25

Tch hai vt v gii phng cc lin kt, ta c: Khi tr chu tc dng ca h

lc ng quy ( )1 2Q,N , N ; dm AB chu tc dng ca h lc ( )'1 A A AP,N ,X ,Y ,M .

Phng trnh cn bng ca khi tr:

1 2

1

X N cos N 0

Y N sin Q 0

= + =

= =

(a)

Phng trnh cn bng ca dm:

'

1 A

'

1 A

'

A A 1

X N cos X 0

Y N sin P Y 0

l lm M P sin N 0

2 3

= + =

= + =

= =

(b)

Gii h cc phng trnh (a) v (b) vi ch '1 1N N= , ta c kt qu sau:

( )21 2 A A A

l 2Q 3PsinQN ; N Qctg ; X Qctg ; Y P Q; M

sin 6sin

+ = = = = + =

.

V d 1-10:

Cho kt cu gm hai thanh ABC v CD cn bng di tc dng ca h lc

nhhnh v. Cho cc gi tr:F1=10kN, F2=12kN, M=25kNm, q=2kN/m, =600.

Cc kch thc trn hnh v o bng m, b qua trng lng hai thanh. Tm

phn lc lin kt ti A, C, D (Hnh 1 - 12)?

Bi gii:

Hnh v d 1-9

D

A

N2

N1

Q

PD

B

XAA

N1

MA

YA

y

x

D

B

A

P

Q


25/240

26

Chn h to nhhnh v, h kho st gm hai thanh ABC v CD, ho rn

h ta c mt vt cn bng di tc dng ca h lc:

( )1 2 A A O OF , F , Q, X , Y , X , Y , M 0 (a)

Trong : Q=4q.

T (a), ta thit lp cc phng trnh cn bng cho h:

A 1 D

A 1 2 D

A 1 1 2 D

X X F cos X Q 0

Y Y F sin F Y 0

m 2Q 4F cos 3F sin M 5F 7Y 0

= + + =

= + =

= + + + =

(b)

Ba phng trnh lp c cha xc nh 4 n. Ta phi s dng thm

phng php tch vt: Gii phng lin kt cho thanh CD, lc ny phn lc ti C

xut hin v c hai thnh phn.

Vit cc phng trnh cn bng cho thanh CD, ta c:

C D

C 2 D

D 2 C C

X X X 0

Y Y F Y 0

m 2F M 4X 4Y 0

= + =

= + =

= =

(c)

T (b) v (c) ta s tm c 6 n nu trn:

3

Hnh 1-12

YD

XD

XC

YCF2

C

M

D

Q

YD

XDXA

YA

B

F2

C

F1

D

A

B

2

3

4

2

F2

C

F1

DAq


26/240

27

XA=-3.39kN; XC=-4.39kN; XD=4.39kN

YA=11.8kN; YC=4.14kN; YD=7.86kN.

Bi ton gin:

Khi nim gin: Gin l cu trc cng c cu to t cc thanh thng lin

kt vi nhau bng cc khp hai u, cc v tr ny c gi l cc nt.

Trong bi ton gin chp nhn cc gi thit sau:

- Tt c cc ti trng ngoi t ln gin ti cc nt.

- B qua ma st ti cc nt v trng lng ca cc thanh (hoc trng lng

ca cc thanh c phn b ra cc nt), v vy cc thanh gin c coi l

ch chu ko hoc nn.

Cu to ca gin: S lng thanh (k) v s nt (n) lin h vi nhau theo h

thc:k=2n-3. Nu s thanh t hn, gin s khng cng; ngc li nu s thanh

nhiu hn, gin s siu tnh.

Bi ton gin: l bi ton xc nh phn lc cc gi ta v ng lc cc

thanh gin.

Bi ton xc nh phn lc: Xem gin nh mt vt rn, vit cc phng

trnh cn bng thng thng v gii.Bi ton xc nh ng lc: C hai phng php:

- Phng php tch nt:nh s nt v thanh (dng s La m nh s

nt), k hiu cc ng lc tng ng vi cc thanh l Sk. Ln lt xt cc iu

kin cn bng ca cc lc ng quy ti tng nt ca gin. Phng php ny

c s dng khi cn xc nh ng lc tt c cc thanh. Khi c ng lc

no bng khng th thanh c gi l thanh khng.

- Phng php mt ct: Ct gin qua mt s thanh, vit phng trnh cn

bng cho tng phn gin. Phng php ny thng c s dng khi cn


27/240

28

xc nh cc ng lc trong cc thanh ring r cn kim tra bn, hoc khi

gp nt c s n ln hn 2.V d 1-11:

Cho gin nh hnh v, gin chu cc lc tc ng sau: F1=F2=F3=F. Xc

nh cc phn lc ti cc gi v ng lc trong cc thanh (Hnh 1-13)?Bi gii:

u tin, ta xt tnh tnh nh ca bi ton: nh s nt v thanh nh trn

hnh v, ta c: k=9 v n=6, nh vy s thanh v s nt tho mn h thc k=2n-3,

vy gin l tnh nh.

Xc nh phn lc: Coi gin nh mt vt rn cn bng, chu tc ng ca h

lc bao gm cc lc hot ngF1, F2, F3 v cc phn lc lin kt ti A v B. Ta

lp cc phng trnh cn bng cho h lc nu trn, nhn c:

A A

3X 3F; Y N F

2= = = .

Xc nh ng lc trong cc thanh: K hiu ng lcSk tng ng vi ng lc

ti thanh thk, chiu ca cc ng lc nh trn hnh v v cc thanh coi nh bko, nu nhkt qu tnh ton l m, khi thanh s b nn.

b

a

Hnh 1-13

F3

y

x

9

8

7

6

5

3

21

4

F2

YA

F1

N

B

A

XAV

IVIII

II

I

VI


28/240

29

Phng php tch nt: Ti mi nt ta nhn c mt h lc ng quy v

tng ng ta thit lp c cc phng trnh cn bng:

kx kyF 0, F 0.= = Bt u t ntI, ti y c hai thanh v tng ng vi hai phng trnh ta

xc nh c: 1 2FS , S F 22

= = . Tip theo, ti ntIIc ba thanh nhng ta

bit gi tr caS1v vy ta cng s xc nh c hai gi tr cn li. Ln lt

nhvy, ta s xc nh c ton b ng lc trong cc thanh.

Phng php mt ct: Gi s i hi xc nh ng lc trong thanh s 6, ta

dng mt ct ab qua cc thanh 2, 3, 5, 6. xc nh S6, ta lp phng trnh cn

bng m men i vi nt VI, ta nhn c:

VI 6 A Am S a Y a X a 0= + = .

Thay gi tr caXA v YA vo phng trnh trn, ta nhn c 63

S F2

= . S6

c gi tr m, nhvy thanh s 6 chu nn.

1.4. Bi tp

1-1. Dng lc ko Q nm ngang ko bnh xe ng cht bn knh R trng

lng P t mt ng A vt ln mt ng B; bc AB=h=R/2. Xc nh phn

lc ti A v B vi tr s Q no bnh xe s vt qua bc?

Tr li: 33

2

3

3QN;QPN BA == ; Xe vt qua bc khi

3

3PQ

Hng dn: Gc AOB=600, xe vt qua bc khi bt u ri mt A (phn

lc NA=0).

BQ

A

O

Hnh bi 1-1 Hnh bi 1-2

F300

P

BA

O


29/240

30

1-2. Trc ca rng rc O c bi hai thanhOA v OB, trng lng khng

ng k, nghing u 600 vi ng nm ngang. Vng qua rng rc l si dy,

u treo vt nng P, u kia nghing 300 vi ng nm ngang v chu lc F(=P)

gi vt cn bng. Xc nh ng lc ca hai thanh?

Tr li:3

32

3

3PS;PS BA ==

1-3. Thanh AB c mc vo tng nh bn l A v c gi nm ngang nh

thanh CD; thanh sau c hai u l bn l ni vo thanh u v trn. Gc nghing

gia hai thanh l 600, trng lng 2 thanh u khng ng k. Cho AC=2m,

CB=1m. Tm phn lc ti bn l A v C khi u B chu lc thng ng P=10kN.

Tr li: kNS;kNY;kNX CAA 3105

2

310 ===

Hng dn: CD l thanh gy lin kt.

1-4. Cu (xem nh ng cht) AB=2a, trng lng P, nm gi c nh A v gi

di ng B. tm cao h c lc gi nm ngang Q. Xc nh phn lc ti A v B

trong hai trng hp sau:

a) Gi di ng B di chuyn trn mt phng ngang.

b) Gi di ng B di chuyn trn mt phng nghing gc 300 vi mt nm

ngang.

Tr li:a)

a

hQ

PN;

a

hQ

PY;QX BAA

2222+===

600

D

CBA

Hnh bi 1-3

a)

h

Q

BA

B

b)

h

Q

A

Hnh bi 1-4


30/240

31

b)3

3

221

6

3

6

3)

a

hQP(N;

a

hQ

PY);

a

h(Q

PX BAA +==+=

1-5. Xc nh phn lc ti ngm ca dm nm ngang, trng lng khng ng k,

chu lc nhhnh v.

Tr li: X0 = 2,8kN, Y0 = 1,7kN, m0 = -5,35kNm.

1-6. Xc nh phn lc ti ngm ca dm nm ngang, trng lng khng ng

k, chu lc nhhnh v.Tr li: X0=12,5kN, Y0=13,7kN, m0=-27kNm.

1-7. Lc ko nm ngang Q t u E ca mc

ABCDE b ngm u A v c dng nh hnh v

(AB=DE v bng bn knh R ca na ng trn

BCD). Tm phn lc ngm A ni lc cc thit

din B, C v D .

Tr li:

XA= -Q; YA= 0; mA= QR; YC = 0; XC = Q

MC = 2QR; XD = Q; YD = 0; mD = QR

1-8. Nhn A lng ngoi ng trn A (nm trong

mt thng ng). Treo vo nhn vt nng P. Mt

khc buc vo nhn si dy ABC, u C (sau khi

vt qua rng rc B t v tr cao nht ca ng

trn) treo vt nng Q. Xc nh gc tng ng v

tr cn bng?

Hnh bi 1-5

q=1,5kN/m

4504kN

AO

2kNm

2m3m

q=2kN/m

3m

Hnh bi 1-6

q=4kN/m

300

5kN

A O

4kNm

4,5m

Hnh bi 1-7

R

QE

D

C

BA

Hnh bi 1-8

B

A

C


31/240

32

Tr li:P

Qarccos

22=

1-9. Thanh ng cht AB trng lng P=20N

c trc quay nm ngang A, u B treo vt nng

P=200N. Cng ti u B, buc si dy, u dy

kia treo vt nng Q = 200N sau khi vt qua rng

rc nh D. Bit AD nm ngang, AD=AB. Tm

gc nghing ca thanh khi cn bng.

Tr li: 508002

,sin

1-10. n ABC trng lng 80N gm hai

tay n AB = 0,4m v BC = 1m vung gc

vi nhau ti trc nm ngang B ca n. Tihai u A v C buc hai si dy, u treo

hai vt nng P1= 310N, P2= 100N, sau khi

vng qua hai rng rc nh E v F. bit v

tr cn bng gc EAB = 1350, trng tm G ca n cch ng thng ng BD

mt on 0,212m.

Xc nh gc = BCF.

Tr li: =1

450; =2

1350.

1-11. Khi ln trn, mt ng nghing gc vi mt nm ngang (tm ln

pha thp). Bit khong cch hai bnh xe AB = 2a, trng lng xe l P (trng

lng tm C nm trn trung trc ca AB v cch mt ng mt on h).

Xc nh vn tc ti a v xe khng b lt bit

rng lc qun tnh li tm t ti C, nm ngang v c tr

s:

R

v

.g

P

Q

2

=

(g gia tc trng trng, R bn knh ln).

Hnh 1-11

Hnh bi 1-9

D

B

A

Q

Hnh bi 1-10

E

G

D

B

A

1350

21

Hnh bi 1-11


32/240

33

Hnh bi 1-13

BD

A

600

C

Tr li:+

atghhtga

.gRv2

1-12. Lc nm ngang Q t vo u A ca cn OA, cn ny quay c quanh

trc O v p B vo khi tr c nm trong gc vung gia nn ngang v tng

ng. B qua trng lng ca cn, bit trng

lng khi tr ng cht l P v OB = BA, =600;

A nm trn ng thng ng qua C.

Tm cc phn lc trc O, phn lc ca

nn v tng, lc p ti B.

Tr li:2

3

23 QY;

QX;QN ooB ===

2

3

2

3 QN;QPN ED =+=

1-13. Thanh AB ng cht trng lng P gn vi nn bng bn l A v ta ln

qu cu C, qu cu ny ng cht, trng lng Q, ta ln nn v c gia bi

dy AC (ni tm C vi cht bn l A). Bit thanh

nghing 600 vi nn. Tm cc phn lc ti A, D v

sc cng ca dy .

Tr li:

4

3

4

3

22

PY;PX;

PT;Q

PN

AAD

===+=

1-14. Mt ng dc nghing gc 300 gm hai on AB = 60m v BC = 30m

ni vi nhau bng bn l B v c gi bi gi c nh A (bn l) v hai ct CC

v . B qua trng lng ca dm v cc ct.

Trn ng c on tu AE, trng lng (thng

ng) mi mt di l 20kN. Tm phn lc ti A,

ng lc cc ct v lc tc dng tng h ti b.

Hnh bi 1-14

Hnh bi 1-12

O

D

E

A Q

600

C

B


33/240

34

Cho AD = 40m, AE = 70m, chiu cao ca on tu khng ng k.

Tr li: SC = 50KN; YA = 225KN;

YE = 150KN; SD = 1125KN.

1-15. Trn ng nm ngang c xe AB trng lng Q mang cn BC trng lng

P, quay c quanh trc B v c gi bi dy ED,

vng qua u mt C l dy mang vt nng P c u

dy kia buc vo A. Cho AE=EB=BD=DC v cn BC

nghing 600 vi mt ng. Tm phn lc hai bnh

xe A, B sc cng ca dy ED v lc tc dng tng

h ti bn l B.

Tr li:

A B

Q P 1 Q 3P 5N p; N p;

2 4 2 4

+= = +

( ) ( )B B3 3

T P; X P p ; X P p .2 2

= = + = +

1-16. Trn nn nm ngang t thang hai chn ni vi nhau nh bn l C v dy

EF. Trng lng mi chn thang (ng cht) l 120N. Ti D c ngi nng 720N,

kch thc ghi trn hnh v. Tm phn lc ti A, B v sc cng ca dy.

Hnh bi 1-16

Tr li: NA=408N; NB=552N; T=522N.

1-17. Lc thng ng Pt vo n OBA truyn tc dng qua thanh BC xung

cn CK, cn ny trt theo mng thng ng DE p vt K. Cho OB=BA,

BCOA, CD=DE=EK=a; gc nghing ca OBA vi ng nm ngang OC l .Tm lc nn ti K v phn lc ca mng DE (qui v hai lin kt ta D v E)

Hnh 1-15


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35

Tr li: NK=2Pcos2; ND=4Psincos; NE=2Psincos.

P

K

D

E

CO

B

A

a

a

a

1-18. Lc thng ng Pt vo n by truyn tc dng xung cn BD thng

qua con trt B. Cn BD trt theo mng trt C thng ng. Bit n OA

nghing 300

vi

ng nm ngang, OB=BA, CD=CB=a. Tm lc nn ti D vphn lc mng trt (xem nh loi ngm).

Tr li:2

3

2

3

2

3aPm;.PN;PN CCD === .

1-19. Vt nng P c treo vo nt (1, 2) ca dn

gm 5 thanh (1, 2, 3, 4, 5) b tr nh hnh v v c

gi c nh nh 3 thanh 6, 7, 8.

Tm ng lc cc thanh (b qua trng lng xem

nh cc thanh ni vi nhau v vi nn v tng bngbn l).

Tr li: 1 2 43 2 3

S P; S S P;2 3

= = =

5 6 7 8

3S 0; S P S ; S P.

2= = = =

Hnh bi 1-18Hnh bi 1-17

300

P

8

7

6

5

4

3 2

1

600600

Hnh bi 1-19


35/240

36

1-20. Ba dm ng cht AB, BC, CD trng lng 2P, P, P ni vi nhau bng bn

l B, C v c nm ngang nh gi c nh A v cc gi di ng E, G,

D. Knh thc cho trn hnh v.

Tm phn lc cc gi v lc tc dng

tng h ti B v C.

Tr li:

PN;P

N;P

N GCD 222

===

PN;PN;

PN AEB

6

5

3

2

2

===

1-21. Cho c cu nh hnh v, ngu lc m t vo

tay quay OA, thng qua con chy A, cn BC, con

chy C v cn ED, truyn tc dng n vt G. Bit

OB thng ng OA=OB=r: OBC=300; BA=AC, cnED nm ngang v on CC thng ng. Tm lc t

vo vt G khi cn bng, Tm phn lc trc O, B v

con chy A, C.

Tr li: GCoA Nrm

N;r

mY;

r

mX;

r

mN =====

3

2

3

3

3

320

r

mY;

r

mX BB

3

2

3==

1-22. Hai khi tr ng cht C1, C2, trng lng P1=10N, P2=30N, nm ta ln

nhau trong mt gc vung to nn bi hai mt OA v OB nghing 300 v 600 vi

mt nm ngang. Tm gc nghing ca on ni hai tm C1, C2 khi cn bng,

tm lc p tng h gia hai qu cu.

Tr li: =0; N=17,3N

Hnh bi 1-21

Hnh bi 1-20


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37


1-23. Thanh ng cht OA, trng lng P quay c quanh trc O v ta ti

im B ca n ln qu cu ng cht C. Qu cu ny c trng l

ng Q, bn knhR, c treo vo trc nh dy OD=R. Bit OD nghing 300 vi OA, tm gc

nghing (vi ng thng ng) ca dy OD khi cn bng.

Tr li:PQ

Ptg

34

3

+=

1-24. Ngu m t vo tay quay OA quay c quanh trc

O. Nh con chy A c th trt dc CB m lc truyn tc

dng sang cn CB. Bit OC thng ng, OA=R nm ngang0

30=C , CB=3R. Tm lc Q nm ngang cn t vo B

h cn bng.

Tr li:R

mQ

9

38=

Hnh bi 1-24


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38

1-25. Cam A l mt khi lng tr thit din tam gic

vung trt c theo mt nhn nm ngang di tc dng

ca lc dy nm ngang P, cam c mt nghing, gc

nghing nh y c cn trt BC, trt theo mng

trt thng ng. Xc nh lc thng ng Q phi t vo

mng trt c cn bng.

Tr li: Q=Pctg.

1-26. Qu cu ng cht c trng lng P c treo

trn tng nhn thng ng nh si dyAC. Dy lp vi tng gc . Xc nh

lc cng T v p lc ca qu cu ln tng.

Tr li:P

T ; N P.tgcos

= =

1-27. Gi ABC nng, h ti trng P=20kN qua rng rc A v ti D. Ti D

c gn vo tng c gc DCA=300. Gc gia cc thanh ca cn trc l

ABC=60o, ACB=300. Hy xc nh lc dc trong cc thanh AB v AC.

Tr li: NAB= 0; NAC= +34,6 kN.

1-28. Bn trong mt bn cu lm nhn cht im M trng lng P gi cn bng nh

dy vt qua rng rc A, u cui dy treo vt nng c trng l ng Q nh hnh v. Xc

nh p lc N ca cht im M ln mt cu v gc lp bi bn knh OM vi phng

ngang. Cho bit OA=R

Hnh bi 1-25

P

B

A

C

D

Hnh bi 1-27

C

A

O

Hnh bi 1-26


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Tr li:Qsin

2Ncos

=

;2 2Q Q 8P

sin2 4P

+ +=

1-29. Dm AB chiu di l, u B treo vt M trng lng P, u A ta vo tng

nhn thng ng, ti C ta vo g nh hnh v. Xc nh phn lc ta ti A v C,

khong cch AC khi h cn bng. B qua ma st v trng lng ca dm. Cho

bit dm lp vi phng ngang gc .

Tr li: N=P.tg; cP

N ;cos

=

AC=lcos2

1-30. Tm vn AB di 2l trng lng P1 treo trn hai dy AC v BC c di

bng nhau, dy lp vi tm vn gc . Mt ngi trng lng P2 ng ti im D

vi AD=a. Xc nh gc nghing ca tm vn vi phng nm ngang ti v tr

cn bng v lc cng TA, TB ca hai dy.

Tr li: 2

1 2

(l a )Ptg Ctg ;

l(P P )

= +

A 1 2cos( )

T (P P )sin2

= +

;

B 1 2

cos( )T (P P )

sin2

+ = +

.

M

A

Q

O

R

Hnh bi 1- 28 Hnh bi 1-29

x

C

B

QA

P

Hnh bi 1-31a

BA

ba

Hnh bi 1-31b

P

BA

ba

D

C

a

A

B

Hnh bi 1-30


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1-31. Dm AB cn bng di tc dng ca lc P nh hnh v. Cho bit cc kch

thc a, b, v cc gc , . Xc nh phn lc lin kt ti A, B trong hai trng

hp. B qua trng lng ca dm.

Tr li: a, XA=P.cos; A BP.bsin P.a sin

Y ; Na b a b

= =

+ +.

b,

A A B

P.a sin .tg Pbsin Pa sinX Pcos ; Y ; N

a b a b (a b)cos

= + = =

+ + + .

1-32. Dm cng xn cn bng di tc dng ca h lc nh hnh v. Xc nhphn lc ti ngm A. B qua trng lng ca dm.

Tr li: XA=Pcos; YA=Psin+q.a;2

A

5qaM P.a sin M

2= + .

1-33. Cho kt cu cn bng di tc dng ca h lc nhhnh v (kch thc cho

bng mt). Xc nh phn lc ti A, C v B. Bit P1=10kN, P2=12kN, M=25kNm,

q=2kN/m, =600.

Tr li: XA=-7,39kN; YA=12,8kN; XB=4,39kN;

YB=7,86kN; XC=-4,39kN; YC=4,14kN.

PqM

BA

aaa


2233

B

P2MC

4

A

P1

q


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1-34. Hai dm AB, BC lin kt khp ti B cn bng di tc dng ca cc lc

nh hnh v. Bit cc kch thc: a, b, d, v cc gi tr F1, F2, M1, M2, q, , . B

qua trng lng ca cc dm. Tm cc phn lc lin kt ti A, D, B, C.

1-35. Dm lin hp gm hai thanh AB, BC cn bng di tc dng ca cc lc

nhhnh v. Xc nh phn lc ti A, B, C.

1-36. Khung gm hai thanh AB, BC ni bn l ti B cn bng di tc dng ca

cc lc nhhnh v. Tm phn lc ti A, B v C.

1-37. Dm lin hp gm ba thanh AB, BC, CD cn bng di tc dng ca cc

lc nh hnh v. B qua trng lng ca cc dm. Tm cc thnh phn phn lc

ti A, B, E, D, C.

1-38. C h gm ba thanh AB, BC v CD cn bng di tc dng ca cc lc

nh hnh v. Lc F2 song song vi mt phng nghing. B qua trng lng ca

cc thanh. Xc nh phn lc ti A, B, E, C, D.


ba

d

cCF2

BM

AF1

d

baa

M2B

M1

D

q

F2

C

F1

F1F2

ccbba

M3M2M1

DCEBA

Hnh bi 1-37

F2

C

B

A

M2

F1

M1

b

a

a

2a

Hnh bi 1-36

q

F2

Hnh bi 1-38

a

a

b

ccb

M3

q

F1

M2M1

D

CEBA

D

Hnh bi 1-39

BA

aaaaaaa

FEGP

C


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42

1-39. Khung 3 nhp c cc kch thc nhhnh v. Xc nh phn lc ti cc gi

A, B, C, D nu lc tc dng vo khung theo phng nm ngang. B qua trng

lng ca cc thanh trong khung.

Tr li: A B C DP 2 P 2

R ; R P; R P; R2 2

= = = = .

1-40. Xc nh phn lc lin kt cc gi ta v lc dc trong cc thanh ca dn

phng gm 7 thanh gn vi nhau, chu tc dng cc lc P; v Q ti cc nt C v

E nhhnh v.

B qua trng lng cc thanh. Cho bit gc 0ACF DEF , CAF 90= = = .

Tr li: T1=-(P+Q); T2=T5=0.

1-41. Xc nh phn lc ti cc gi A, B, v lc dc trong tt c cc thanh ca dn

phng, chu tc dng ca h lc nh hnh v.

Tr li: XA=-1kN; YA=3kN; NB=1kN

Ch s thanh 1 2 3 4 5 6 7 8 9

ng lc (kN) -2 -2 -1 1,41 2 4,24 -4 1,41 -1

P

C

F

Q

E

DB

A

Hnh bi 1-40

Q

P

a

aaa

Hnh bi 1-41

aa


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1-42. Ti hai bc B treo vt D trng lng Q v ni vi trc C ca con ln A

trng lng P nh hnh v. Nu h cn bng, tm p lc do A tc dng vo tng

v gc to bi nhnh dy ni gia hai vt A, B vi phng nm ngang. B qua

ma st ti O v I.

Tr li:Q.r.cos PR

N ; sinR Qr

= =

1-43. Cho c cu tay quay cult nh hnh v. Xc nh m men M1 ca ngu lc

t vo AB cn bng vi mmen ca ngu lc M t trn CD v phn lc ti

A, C khi ACAB. B qua ma st v trng lng cc thanh.

Tr li: 2A C 1Msin

R R ; M Msina

= = =

1-44. Cho mt s tm g c cng trng lng v cng chiu di 2l. Cc tm g

c xp chng ln nhau trn mt phng ngang nh hnh v. Hy xc nh di

gii hn ca cc phn nh ra, cc tm c th trng thi cn bng.

Tr li: l; l/2; l/3; l/4; ...

Ch : Khi gii bi ton, cng lin tip trng lng ca cc tm bt u t tm

trn cng.

DB

rR

A

I

C


C

D

BA

a

M1

M

Hnh bi 1-44 B

C

D

FE

A

Hnh bi 1-45


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1-45. Thang xp ch A t trn sn nhn nm ngang gm hai phn AC v BC,

mi phn di 3m nng 120N, ni vi nhau bng khp bn l C v dy EF.

Khong cch BF=AF=1m, trng tm mi phn AC v BC im gia. Mt ngi

nng 720N ng ti im D vi khong cch CD=0,6m. Xc nh phn lc ti A

v B v lc cng ca dy, nu gc =450.

Tr li: RA=408N; RB=552N; T=522N.


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45

Chng 2. Bi ton khng gian

Bi ton khng gian l bi ton a ti vic kho st h lc khng gian tc

dng ln vt rn hay h vt rn. Trong qu trnh kho st, bi ton ny a n

hai vn chnh sau: thu gn h lc khng gian v dng ti gin; tm iu kin

cn bng ca h lc khng gian.

Phng php v trnh t gii quyt bi ton khng gian cng tng t nh

bi ton phng.

2.1. C s l thuyt

2.1.1. Thu gn h lc khng gian

Khi thu gn mt h lc khng gian v mt tm O bt k, ta nhn c mt

lc bng vc t chnh ca h lc v mt ngu lc bng vc t m men chnh ca

h lc, xy ra cc trng hp sau:

0M;0'R o == , h lc tng ng khng hay h lc khng gian cn bng.

0M;0'R o = , h lc tng ng vi mt lc =k

kFR .

0M;0'R o = , h lc tng ng vi mt ngu ( )==k

kOO FmMM

.

0M;0'R o , trong trng hp ny, tu theo quan h gia lc thu gn v

ngu lc thu gn ta c cc kt qu sau:

- Khi: ( ) 0Mx'Ro= , lc thu gn nm trong mt phng ca ngu, h lc s

tng ng vi mt lc =k

kFR vi tm thu gn t ti im Occh

O mt on'R

Md

o

= .

- Khi: ( ) 0Mx'Ro , lc thu gn khng nm trong mt phng ca ngu, h

lc khng gian tng ng vi mt h xon.


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2.1.2. iu kin v cc phng trnh cn bng ca h lc khng gian

1. iu kin cn bng ca h lc khng gian:

nh l: iu kin cn v h lc khng gian ( ))..1 nkFk =

cn bng l vc t chnh ( 'R ) v vc t m men chnh ( oM ) i vi mt tm bt

k ca h lc u trit tiu.

.0;0' == OMR (2.1)

2. Phng trnh cn bng ca h lc khng gian bt k: T nh l nu

trn, ta c h 6 phng trnh sau:

( )

( )

( )

==

==

==

=

==

==

==

=

=

=

=

=

=

=

n

k

kzOz

n

k

kyOy

n

k

kxOx

O

n

k

kzz

n

k

kyy

n

k

kxx

FmM

FmM

FmM

M

FR

FR

FR

R

1

1

1

1

1

1

;0

;0

;0

0;

;0'

;0'

;0'

0'

(2.2)

3. Phng trnh cn bng ca cc h lc c bit:

H lc ng quy: ly im ng quy ca h lc lm tm m men, iu

kin 0MO = s t ng c p ng, ta cn li ba phng trnh:

.0;0;0111

===

===n

k

kz

n

k

ky

n

k

kx FFF (2.3)

H lc song song (gi thit song song vi trc z): cng cn li baphng trnh:

( ) ( ) .0;0;0111

===

===n

k

ky

n

k

kx

n

k

kz FmFmF

(2.4)

Trng hp h ngu lc: iu kin 0'R = t ng c p ng, ta

cn li ba phng trnh tng hnh chiu ca cc ngu lc ln ba trc to

ln lt trit tiu.

( ) ( ) ( ) === ======n

k

kzz

n

k

kyy

n

k

kxx FmMFmMFmM111

0;0;0

. (2.5)


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4. iu kin cn bng ca vt rn khng t do:

i vi vt rn khng t do (vt rn chu lin kt), p dng cc phng

trnh nu trn, ta cn gii phng cc lin kt, thay vo cc phn lc lin kt

tng ng sau tin hnh thit lp cc phng trnh cn bng nu trn.

Cn ch rng, trong nhiu bi ton, c bit l cc bi ton tm iu kincn bng ca vt rn hay h vt rn, nu ta ch ti cc lin kt v kh nng

chuyn dch ca vt rn th c th gii bi ton mt cch nhanh gn hn. Di

y l mt s dng thng gp:

iu kin cn bng ca vt rn quay c quanh mt trc c nh:

( )=

==n

k

kzz FmM1

0

. (2.6)

iu kin cn bng ca vt rn quay c quanh mt tm c nh:

( ) == ,0kOO FmM

Hay: ( ) ( ) ( )===

======n

k

kzz

n

k

kyy

n

k

kxx FmMFmMFmM111

0;0;0

(2.7)

2.1.3. Cc php bin i v xc nh lc:

1. Cc lin kt khng gian: Cc lin kt ta, dy v thanh tng t nh

trong bi ton phng, cc lin kt khc c thm phng z. Bng di y ghi c

im mt s lin kt .

Bng 3-1

Lin

ktCu to v biu din

c im

phn lc

1 2 3

Bn

ltr

(ng

n)

Vc t R vung gc vi

trc quay, phn thnh hai

thnh phn X v Y.

Y

X


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Gicu

(vci)

Ba thnh phn phn lc

X, Y v Z.

Ngm

Ba thnh phn phn lc X,

Y, Z v ba ngu lc MX,

MY, MZ.

ngtrdi

Hai thnh phn phn lc

X, Y v hai ngu lc MX,

MY, MZ.

2. Phng php xc nh lc v vect chnh:

Trong bi ton khng gian ch yu s dng phng php chiu xc nhcc lc.

Hnh chiu ca lc F ln cc trc to c xc nh bi cc cng

thc sau:

,cosFF;cosFF;cosFF zyx ===

Trong : - , , l cc c sin ch phng ca lc F vi cc trc to .

- Du + (-) tng ng khi lc thun (ngc) chiu trc.

i vi h lc, vct chnh ca h, k hiu R, l tng hnh hc cc vectlc c xc nh theo cng thc:

1 2 3

MY

Y

MX

X

Z

X

Y

Z

X

MY

MZ

MX

Y

X

Z


48/240

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=

=n

k

kFR1

'

T ta tnh c tr s v cosin ch phng ca vect chnh:

===

===n

k

kzz

n

k

kyy

n

k

kxx FRFRFR111

';';'

( ) ( ) ( ) .'

'',cos;

'

'',cos;

'

'',cos,''''

222

R

RRz

R

RRy

R

RRxRRRR z

yxzyx ===++=

2.1.4. Phng php xc nh mmen lc v mmen chnh:

1. M men lc: Cho lc F v im O, gi r l vc t nh v im t lc

tnh t tm O.

Mmen lc Fi vi O l mt vc t, k hiu ( )FmO

c xc nh theo

cng thc:

( ) FrFmO =

Trong :

- Gi tr m men: ( ) FhFmo =

F - tr s ca lc;

h - tr s ca cnh tay n.

- Chiu ca vc t m men hng v pha t ta thy lc F quay

quanh O ngc chiu kim ng h.

xc nh m men ca lc i vi cc trc qua im O, tO, ta k cc

trc to Ox, Oy, Oz, ta c m men ca lc F vi cc trc nh sau:

( ) ( ) ( ) ( ) ( ) ( )zxOyyzOxxyOz FmFm,FmFm,FmFm ===

Trong zxyzxy F,F,F l hnh chiu ca vc t lc F xung cc mt phng

xOy, yOz, zOx tng ng.

Cc biu thc trn cho php ta a bi ton xc nh m men ca mt lc

i vi mt tm tr v xc nh m men ca mt lc vi cc trc to .


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xc nh m men ca mt lc vi mt trc, ngoi cch p dng trc tip

nh ngha, c th dng phng php phn lc v phng php gii tch nh trong

phn h lc phng:

( ) ( ) ( ) xyzzxyyzx yFxFFm;xFzFFm;zFyFFm ===

Trong : x, y, z to im t lc,Fx , Fy , Fz hnh chiu ca lc ln cc trc to .

2. M men chnh ca h lc: Vc t m men chnh ca h lc khng gian

i vi tm O (mmen chnh ca h lc) k hiuoM l tng mmen cc lc i

vi im O:

( )=

=n

1kkoo FmM

Hnh chiu ca vc t m men chnh ca h lc trn cc trc to (vi O

l gc to ) bng tng m men ca cc lc i vi cc trc:

( ) ( ) ( )===

===n

1kkzz

n

1kkyy

n

1kkxx FmM;FmM;FmM

Vi h ngu lc, vc t mmen chnh khng ph thuc tm O v bng tng

hnh hc cc vc t mmen ngu lc.

2.2. Hng dn p dng

gii bi ton khng gian, ta tin hnh cc bc tng t nh trong biton phng nh sau:

Bc 1:Chn vt kho st; t lc tc dng ln vt kho st (nu vt rn

chu lin kt, trc tin, ta gii phng cc lin kt v t thay vo cc

phn lc lin kt tng ng).

Bc 2: Phn tch c im h lc kho st: (ng quy, song song hay bt

k), t xc nh s phng trnh cn bng c lp c th lp c (3

hoc 6) theo cc iu kin trong mc 2.1.2, sau tin hnh gii ccphng trnh nhn c.


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2.3. Bi gii mu

Trong phn ny chng ta ch gii hn xem xt dng cc bi ton xc nh

phn lc lin kt cho bi ton mt vt khng gian v h vt khng gian.

2.3.1. Bi ton mt vt

V d 2-1:

Treo mt vt nng c trng lng P vo u O ca thanh OA, thanh c

gn vo tng bng bn l ti A v c gi cn bng nh hai dy nm ngang OB

v OC. Cho bit cc gc ti nh A, B, C bng nhau v bng 45 0 (xem hnh v).

B qua ma st v trng lng ring ca thanh, xc nh ng lc trong thanh v

sc cng ca cc dy?

Bi gii:

Chn h trc to nh hnh v, xc nh

ng lc trong thanh OA v sc cng cc dy, ta

kho st iu kin cn bng ca nt O. Nt chu

tc dng ca cc lc: trng lc P; ng lc S; cc

lc cng dy T1, T2.

Ta c iu kin cn bng cho h lc trn nh

sau:

( ) 0T,T,S,P 21 (a)V h lc ng quy ti O, nn t iu kin cn bng (a), ta lp c ba

phng trnh sau:

==

==

==

0P45cosS,0F

045cosS45cosT45cosT,0F

045cosT45cosT,0F

0kz

002

01ky

02

01kx

(b)

Gii h phng trnh trn, ta nhn c:

2

2PTT;2PS 21 ===

C T2

T1

S

O y

B

A

P

z

x

Hnh 2-1


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T kt qu nhn c, ta thy S nhn gi tr m vy chiu gi nh trn

hnh v l khng ng, trn thc t thanh OA chu nn.

V d 2-2:

Tm hnh vung ABCD trng lng P c gi cn bng trng thi nm

ngang nh lin kt bn l ti A, khp cu ti B v dy DE. Dy to vi mt phngtm mt gc 300 v BE vung gc vi AB. Xc nh cc phn lc ti A, B v sc

cng dy?

Bi gii:

xc nh cc phn lc, ta kho st iu kin cn bng ca tm ABCD

chu tc dng ca trng lc P ti im O, chn h trc to nh hnh v. S

dng tin gii phng lin kt, thay th tc dng ca cc lin kt bng cc phn

lc lin kt: ti D c sc cng dy T, ti A c cc phn lc R Ax, RAz, ti B c RBx,

RBy, RBz vi chiu ca cc lc c gi thit nh trn hnh v.

Nh vy, tm ABCD cn bng di tc dng ca h lc khng gian bt k,

ta c iu kin cn bng sau:

( ) 0R,R,R,R,R,T,P BzByBxAzAx (a)

T iu kin cn bng trn ta c th lp c 6 phng trnh vi 6 n phi

tm, vy bi ton l tnh nh. n gin, ta phn lc T thnh hai thnh phn: Tz

v Txy. Ta c cc phng trnh sau:

D

E

P

O

C

B

A

x

RBxy

RByRBz

RAz

RAx T

TxyTz

zE

P

O

D C

BA

Hnh 2-2.


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53

,0PTRRF

,045sinTRF

,045cosTRRF

zBzAzkz

0xyByky

0xyBxAxkx

=++=

=+=

=+=

(b)

( )

( )

( ) ,0aR2

2aTFm

,0aT2

aPFm

,0aR2

a

PFm

Bxxykz

zky

Bzkx

==

==

=+=

(c)

Gii cc phng trnh trn vi ch rng:

0z

0xy 30sinTT;30cosTT ==

Ta xc nh c cc gi tr nh sau:

.2/PR;P61.0R;P61.0R;0R;0R;PT BzByBxAzAx ======

V d 2.3

Cnh ca hnh ch nht trng lng P=0.1kN, nm trng thi cn bng

di tc dng ca cc lc: lc ko T=0.04kN t ti im D v to vi phng

ngang mt gc QDN bng 450; mmen cn m. Ca c gi bi cc lin kt: lin

kt ci ti A; lin kt bn l ti B.

Xc nh cc phn lc lin kt v gi tr ca mmen m? Cho bit KD//AE,

gc KDN bng 450, ED=0.8m, AB=2m, AE=1m.Bi gii:

Kho st s cn bng ca ca di tc dng ca cc lc hot ng: trng

lng P; lc ko T; mmen cn m.

Ca c gn bi hai lin kt: lin kt ti A tng ng nh lin kt gi

cu; lin kt ti B l lin kt bn l. Gii phng cc lin kt v thay chng bng

cc phn lc lin kt tng ng (hnh v), phn tch lc T thnh cc lc thnh

phn theo cc phng ca h trc to . Nhvy ca cn bng vi iu kin sau:

( ) 0m,R,R,R,R,R,T,T,T,P ByBxAzAyAxzyx (a)


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54

T iu kin cn bng (a), ta lp c 6 phng trnh cn bng, vi 6 n cntm, vy bi ton l xc nh tnh, ta c:

,0

,0

,0

=+=

=+=

=++=

zAzkz

yByAyky

xBxAxkx

TPRF

TRRF

TRRF

(b)

( )( )

( ) .0mKDTFm

,0ABRDETFm

,0ABRAMPAETAKTFm

xkz

Bxxky

Byzykx

=+=

=+=

=+=

(c)

Thay cc gi tr v gii h trn ta nhn c:

.kNm02.0m,kN0029.0R,kN008.0R

,kN0718.0R,kN0229.0R,kN12.0R

ByBx

AzAyAx

===

===

Cc gi trRAx, RBx, RBy nhn gi tr m, vy chiu ca cc lc ny ngc

vi chiu quy c trn hnh v.

Ch : Trong bi ton trn, nu gi thit m men cn m bit v lin kt ti B

tng ng vi lin kt khp cu, nh vy cnh ca c m hnh ho nh mtvt rn b gi ti hai im A v B. Khi s lng n vn l 6 v thay cho gi tr

M

A

B

D

Q

T

B

A

x

RBx

y

RBy

RAy

RAz

RAx

TyTz

Tx

z

m

E

N

F

D

C

xy

z

m

E

P

Hnh 2-3.


54/240

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m bit ta c thm thnh phn phn lc RBz, hng dc theo trc z. Mc d s

phng trnh cn bng lp c v s n bng nhau nhng bi ton vn l bi

ton cha xc nh v cc thnh phn theo trc z ch xut hin trong phng trnh:

.0TPRRF zBzAzkz =++=

T phng trnh ny ta ch xc nh c tng cc gi tr phn lc theo trc

z: 0TPRR zBzAz ==+ , ch khng xc nh c ring cc gi tr trn.

V d 2.4:

Rto ca my in l mt khi tr trn ng cht quay quanh trc i xng

nm ngang AB. Rto c trng lng P chu tc dng ca mt ngu lc M nm

trong mt phng vung gc vi trc quay. Lp ng trc vi rto mt puli trng

lng Q v c bn knh r. Hai nhnh ai truyn mc ln puli song song vi nhau,

nghing vi phng ngang mt gc , chu hai lc ko l T1 v T2, c cng

T1 = 2T2. Cho OA=10cm, AB=2AC=50cm. Tm m men M ca ngu rto cn

bng v tm cc phn lc ca hai A v B. B qua ma st.

Bi gii:

Chn h trc to nh trn hnh v, h kho st bao gm c rto v puli

nm cn bng. Cc lc hot ng tc dng ln h gm: cc trng lc P v Q, cc

lc ko T1 v T2 v ngu M. Cc phn lc lin kt gm: RA (XA, YA) v RB (XB,

YB).

Ta c iu kin cn bng ca h nh sau:

y

x

T2

T1

T2T1

QXA

ZA

PM

XB

ZB y

x

z

Hnh 2-4


55/240

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( ) .0R,R,T,T,M,Q,P BA21 (a)

T iu kin cn bng (a), ta lp c h phng trnh sau:

( )

=++=

===+++=

=++=

=+++=

.0cosT10cosT10X50F

,0MrTTF,0sinT10sinT10Z50P25Q10m

,0ZZQPsinTsinTF

,0XXcosTcosTF

21Bx

21x

21Bx

BA21y

BA21x

(b)

Gii h phng trnh trn, vi ch rng:

,sin;cos;sin;cos 22221111 TTTTTTTT zxzx ====

ta c kt qu:

( ) ( ) ( )

( ) ( ) .sinTT2.0P5.0Q2.0Z;sinTT8.0P8.0Q5.1Z

;cosTT2.0X;cosTT2.1X;TT10M

21B21A

21B21A21

++=+=

+=+==

2.3.1. Bi ton h vt

V d 2.5

Cho hai bnh rng I v II c bn knh r1 v r2 gn ng tm v thng gc

vi hai trc quay AB v CD, cc bnh rng n khp vi nhau v gc n khp l .

Trn trc ca bnh rng I c gn mt tang ti c bn knh R, trn tang qun

dy mm v nh treo vt G trng lng P. Trc CD chu tc dng ca ngu lc

M gi cho h cn bng.. B qua ma st.1. Tm iu kin cn bng ca h?

2. Cho CD=a, b qua trng lng bn thn ca bnh rng II v trc CD, tm

phn lc ca D.

Bi gii

Kho st h gm hai vt rn: trc AB mang tang ti v bnh rng I; trc CD

mang bnh rng II. Cc lc ch ng tc dng ln h l trng lc P, ngu M. Cc

phn lc lin kt tc dng ln h: RA, RB, RD v ngu lc MD. Cc lc lin kt

trong l cc lc n khp S, S ti tip im K.


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1. iu kin cn bng ca h: Hai vt kho st u quay c quanh nhng trc

c nh AB v CD, chn h to nh hnh v, iu kin cn bng ca tng vtc th vit nh sau:

Vt AB: 0sin1 = SrPR (a)

Vt CD: 0sin2' = rSM (b)

Do S v S l cc lc n khp c gi tr bng nhau nn ta c iu kin cn

bng ca h l:

PRr

rM

1

2=

2. Tm phn lc D: Do trc CD c mt , v vy ta coi D ng vai tr nh

mt lin kt ngm c bit. H phn lc ca n tc dng ln trc CD bao gm

phn lc RD v ngu MD, c hai lc ny u vung gc vi trc CD. Ta phn tch

chng thnh cc thnh phn nh sau (hnh v): RD (XD, ZD); MD (MZ, MX).

Kho st s cn bng ca trc CD mang bnh rng II, ta c iu kin i vi

h lc tc dng:

( ) 0,,,' DD MRMS (c)

AS

K

II

I

P

D

B

A

RA

G

MRD

RB

MDS

C

Hnh 2.5

y1

y2


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T iu kin (c), ta lp c h phng trnh sau:

==

=+=

==

==

0sin

0cos

0cos

0sin

'

'

'

'

aSMm

aSMm

SZF

SXF

ZZ

XX

DZ

DX

(d)

T h phng trnh (d), ta nhn c:

.sin;cos

;cos;sin

''

''

aSMaSM

SZSX

ZX

DD

==

==

Lc S c tnh t phng trnh (a) v (b):

sin'

1r

PRSS == .

V d 2.6

Cho c h nh hnh v (hnh v d 2-6), bnh ai 1 c ng knh l d1, vt3 c trng lng G, bnh rng 4 (bnh vt) c ng knh d4 n khp vi trc vt

5, ng knh d5. Coi dy ai, bnh ai, bnh rng, trc vt c trng lng khng

ng k. Cc kch thc c cho trn hnh v: alallall 7;4;3 54321 ===== .

Hy xc nh gi tr m men cn t vo trc vt 5 h cn bng, ng

thi xc nh phn lc ti cc B, D, A, C.

Ch : Lc tc dng tng h gia trc vt v bnh vt nn phn thnh ba

thnh phn vung gc vi nhau, bao gm:

PtgQPtgTd

MP === ;;

2

4

,

S

K

II

D

M

XD

MZ

C

y2

MX

z

x

ZD

a


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Trong :

- M: Mmen quay tc ng ln bnh vt;

- P: Thnh phn lc theo phng trc ca trc vt;

- T: Thnh phn lc hng tm ca bnh vt;

- Q: Thnh phn lc theo phng trc ca bnh vt;

- : gc n khp gia bnh vt v trc vt,

- : gc nghing ca trc vt.

Bi gii

Trc tin ta nhn thy rng, do dy ai khng gin v khng c ma st

nn nhnh dy bn bnh ai 1 c lc cng ng bng trng lng ca vt nng 3v nghing vi phng ngang mt gc . Kho st h cn li, gm trc mang

bnh vt 4, bnh ai 1 v trc vt 5. Cc gi tr cn tm c xc nh t iu kin

cn bng ca bnh vt v trc vt. gii bi ton, ta gii phng h khi cc lin

kt v tch h thnh hai vt rn nh trn hnh 2-6 (b), sau khi t vo h cc phn

lc lin kt v cc lc tng h, ta thy rng c h c 12 gi tr cha bit:

ZBBAADDCCC MYXZYXYXZYXP ,,,,,,,,,,, .

L5

L4

L3

L2L1 A

D

C

3

5

4

1

B2

Hnh 2-6 (a).


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60

Xt ring trc mang bnh vt, h lc tc dng bao gm: Cc lc tng h

gia bnh vt v trc vt P, T, Q, cc phn lc lin kt XC, YC, ZC, XD, YD, lc

cng ai G. iu kin cn bng ca h ny nh sau:

( ) 0,,,,,,,, GYXZYXQTP DDCCC (a)

T iu kin (a), ta thit lp c 6 phng trnh cn bng sau:

( ) == 022

14 dGd

PFM KZ

(1)

( ) ( ) ( ) =+++++= 0cos 543433 lllGllXPlFM DKY (2)

( ) ( ) ( ) =+++++= 0sin2

54334

43 lllGTld

QllYFM DKX

(3)

== 0QZF CkZ (4)

=++= 0sinGTYYF DCkY (5)

=++= 0cosGPXXF DCkX (6)

MZ

Hnh 2-6 (b).

L5

L4

L3

x

D

C

4

1

y

G

KP

T

z

QYD

XDZC

XC

YC

L2L1

P

T

Q

x

y

ZA

XB

YBXA

YA

z

K


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Tng t i vi trc vt, h lc tc dng bao gm: M men MZ, cc lc

tng h gia bnh vt v trc vt P, T, Q, cc phn lc lin kt XA, YA, ZA, XB,

YB. Ta c iu kin cn bng i vi h lc ny l:

( ) 0,,,,,,',',' ZBBAAA MYXZYXQTP (b)

T iu kin (b), ta lp c 6 phng trnh cn bng sau:

( ) == 02

5dQMFM ZKZ

(7)

( ) ( ) =+= 0221 QlllXFM AKY (8)

( ) ( ) =+= 02

5221

dPTlllYFM AKX

(9)

=+= 0PZF AkZ (10)

=++= 0TYYF BAkY (11)

=+= 0QXXF BAkX (12)Nh vy t iu kin cn bng ca cc vt, ta lp c 12 phng trnh

cn bng c lp. Gii h 12 phng trnh trn vi 12 n tng ng, ta s xc

nh c cc gi tr cn tm, nh sau:

;8

sin152

4

;8

cos154;

4

4

1

Gtga

dtgP

YGP

Xd

dGP DD

=

==

;cos;sin; DCDCC XGPXYGPtgYPtgZ ===

;122

;2

;2

55

+===

a

dtgPY

PtgX

dPtgM AAZ

.2

;212

; 5 Ptg

Xtg

a

dPYPZ BBA =

==

2-4. Bi tp

2-1. Vt nng P c treo vo u O ca gi treo to bi 3 thanh trng lng

khng ng k, gn vi nhau v vi tng bng cc bn l nh hnh v. Tm ng

lc ca cc thanh.


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Tr li: 22 PSS;PS CBA ===


2-2. Vt nng P=100N c treo vo u O ca gi treo to bi 3 thanh, trng

lng khng ng k, gn vi nhau v vi tng bng cc bn l nh hnh v.Tm ng lc ca cc thanh.

Tr li: NSS;NS DCA3

250

3

3200 ===


2-3. Bn ba chn A, B, C to thanh tam gic u cnh a. Trng lng bn l P

v trng tm nn trn ng thng ng zOO1 qua tm O1 ca tam gic ABC.

Trn mt bn c vt M trng lng p t im c to x, y; trc ng song vi

AB. Xc nh phn lc ti cc chn.

300

450

Ax y

A

OC

D

Bz


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Tr li:a

Pyx

pPN A

+

+=

3

3

3;

a

pxy

pPN B

++

+=

3

3

3;

pa

x.

pPN C

3

32

3

+=

2-4. Cho cn cu nh hnh v (H. 2-4). Bit AD = DB=1m; CD = 1,5m; CM =

1m; KL=4m; trng lng cn cu (k c i trng) l P =100 kN, trng tm G

ca c cu nm trong mt LMFN v cch MN mt on 0,5m trng lng vt

c cu l Q=30kN. Tm phn lc ti 3 bnh xe khi mt thng ng LMFN song

song vi AB.

Tr li: NA=8,3kN; NB78,3kN; NC=43,4kN.

2-5. Tm trn, trng lng khng ng k, nm ngang trn mi nhn tm O

ca tm. Trn chu vi tm trn t ba im nng trng lng P1=1,5N, P2= 1N v

P3=2N sao cho tm trn vn cn bng. Xc nh cc gc =AOB, = AOC.

Tr li: = 750 30 ,= 1510.

2-6. Tm g ng cht hnh vung trng lng P c v tr nm ngang

nh 6 thanh (khng trng lng) nh hnh v. Ton hnh c dng khi lp

phng. Tm ng lc cc thanh.

Tr li: S1 = S3 = S4 = S5 = 0; S2 = S3 =2

P


2-7. Tm g khng trng lng chu lc P v c v tr nm ngang nh 6

thanh (khng trng lng) nh hnh v. Ton hnh c dng khi lp phng, tm

ng lc cc thanh.


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Tr li: S1=P1; .PS;PS;PS 222 542 === S6 = -P.

2-8. Xt tm g nh trong hnh v, b qua trng lng v gi thit chu tc

dng ca ngu lc m nm trong mt phng tc dng ca tm g. Tm ng lc cc

thanh, cho bit tm g c cnh a.

Tr li: ;;2;0 536241 amSS

amSSSS ======

Hnh bi 2-7 Hnh bi 2-82-9. Cnh ca ng cht hnh ch nht, trng lng 200 N lp vo tng nh

gi cu A v bn l B v c gi cn bng v tr nm ngang nh dy cung CE

nghing 600 vi ng thng ng AE. Bit ng cho AC nghing 300vi cnh

AB, tm phn lc A, B v sc cng ca dy.

Tr li: T= 200N; XA = 86,6N; YA= 150; ZA=100N; XB= ZB = 0

2-10. Cnh ca ch nht ABCD ng cht trng lng 120N gn vi nn nh hai

bn l A, B v c o cn bng v tr nghing 600 vi mt nm ngang nh thanh

(khng trng lng) ED(=DA) nm trong mt thng ng qua AD. Tm cc phn

lc bn l v ng lc ca thanh.

Tr li: S = 34,5N; XA=17,3N; ZA=30N; XB=0; XB=60N.


64/240

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2-11. Cnh ca ng cht hnh ch nht ABCD, trng lng P, chiu di

3aAB = , chiu rng AD=a, c trc quay thng ng AB vi hai gi cu A v

bn l B. Ca c m rng gc 1200 vi khun ca, u D chu lc Q nm song

song vi cnh di EA ca khun, u C c gi bi dy CE. Tm sc cng ca

dy v phn lc cc .

Tr li: )(QP

X;QT A 222

3

46 +== ;

3212

3QPZ;

QPY AA +==

12

312

2

3

4

PQY;)(

QPX DB ==

2-12. Trc AB nm ngang trn hai bn l A, B, mang theo bnh xe C v

thanh DE u c trng lng khng ng k. Trc cn bng di tc dng ca 2

vt nng Q= 250 N treo u dy cun quanh vnh bnh xe v P= 1kN gn vo

u E. Bit DE nghing 30

0

vi

ng thng ng, bn knh xe l 20 m, cc kchthc khc ghi trn hnh v. Tm chiu di DE = l v phn lc cc .



65/240

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Tr li: l = 10cm; ZA = 300N; ZB = 950N; XA = XB = 0.

2-13. Hai A, B (bn l) trc nm ngang AB mang theo bnh xe C v khi

tr AB; bn knh bnh xe cng cun dy u t do treo vt P = 60N, sau khi vng

qua rng rc nh D. Kch thc cho trn hnh v, nhnh dy gia bnh xe v rng

rc nm trong mt phng ca bnh xe v nghing vi ng knh nm ngang ca

bnh xe mt gc = 300. Tm Q v phn lc khi cn bng.

Tr li: Q = 3600, ZA = 160N; ZB = 230N; XA =-69,3N; XB =17,3N.

2-14. Trc thng AB c bi bn l A v ci B, mang theo bnh xe C v

rto D. Tng trng lng bnh xe v rto D l Q = 200N. Bnh xe C c bn knh

10 cm vi 2 nhnh ai truyn song song sc cng T= 100 N v T2 = 50 N. Tm

m men ngu lc )P,P(

*

cn tc dng vo rto c v tr cn bng, cc phn lc . Kch thc khc cho trn hnh v.

Tr li: m= 5Nm; ZB = 200N; XA = XB = 0; YB= 10N; YA= -180N.

2-15. Dm nm ngang OC ng cht, trng lng P = 1000N di 2m, chu tc

dng ca ngu lc )Q,Q( ' trong mt nm ngang tr s Q = 100N tay n

EF=20cm. Dm lin kt vi tng nh gi cu O v hai dy AB v CD b tr nh

hnh v. Cho OB = 0,5 m. Tm phn lc ti O v sc cng cc dy.

Tr li: T1 = 1000N; T2 = 80 N; X0= 40N; Y0 = 540 3 N; Z0 = 500N.

X

Z

T1

T210cm

50cm

Y

P

D

A

C



66/240

67

Q

Q'

F

E

C

30o

D

y

z

30o

BA

x

O

Hnh bi 2-15

B B1

A1

A

O

Hnh bi 2-16

2-16. Thanh ng cht AA1 = 2r, trng lng P b treo bi hai dy thng ng

AB = A1B1 = l. Tc dng vo thanh mt mmen ngu lc M trong mt nm ngang

lm thanh b quay mt gc quanh trc trung trc thng ng ca n (thanh

c nng cao nhng vn nm ngang ). Xc nh mmen M theo gc v tm

sc cng ca dy.

Tr li:

24

222

2

=

sinrl

sinPrm ;

242

222

=

sinrl

lPT

2-17. Ti bn nh A, H, B v D ca hnh lp phng t 4 lc bng nhau

P1=P2=P3=P4=P c chiu nh hnh v. Hy thu gn h lc v dng n gin

nht, trong P1 hng theo AC, P2 hng theo HF, P3 hng theo BE, P4 hng

theo DG.

Tr li: Hp lc bng 2P hng theo ng cho DG.2-18. Tm hnh ch nht ng cht ABCD trng lng Q=10N, c gi cn

bng v tr nm ngang nh gi bn l cu ti B, bn l tr ti C v dy EF nh

hnh v. Trn tm tc dng lc P=5N theo phng thng ng v ngu lcc m

men M=20Nm. Xc nh lc cng T ca dy v phn lc ti B v C. Cho bit

DE=BC=0,5m, BC=2m, gc FEH=300, 060= .

Tr li: T=40N; XB=-5,68N; YB=17,3N

ZB=7,5N; XC=35,68N; ZC=-10N.


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2-19. Tm hnh ch nht ABCD ng cht, trng lng P, c gi v tr nm

ngang nh sau thanh nh hnh v, b qua trng lng cc thanh. Tm chu tc

dng ca lc Q hng dc theo CD v ca ngu lc (F, F). Cho bit AD=a,

AB=b, chiu cao DG=c, gc BCA=. Lp cc phng trnh xc nh ng lc

trong 6 thanh.

2-20. H hai trc song song nm ngang AB v CD mang hai bnh rng O v O

n khp vi nhau ti I. Cho AB=1m, BO=CO=OD=0.5m, bn knh hai bnh

rng bng 0.4m, php tuyn chung In ca hai mt rng ti I nghing gc vi

OO. Tng trong ca bnh rng O c bn knh bng 0.2m c qun dy c mt

u treo vt nng P=1kN. B qua trng lng cc bnh rng v r to, tm ngu lc

cn thit M t vo r to trn trc AB gi cho h cn bng, xc nh phn lc

cc (bn l) A, B, C, D v lc tc dng tng h ti I.

Tr li: ;sin2

N;4

3;cotg

4I

PPZZ

PYY DCDC =====

.75.0Z;cotg75.0;4

Z;cotg4

;2.0 BA PPYPP

YPM BA =====

Hnh bi 2-17 Hnh bi 2-18 Hnh bi 2-19

O

A

B

C

D

P

O

M

n

I

Hnh bi 2-20


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Chng 3. Bi ton ma st

Trc y, khi xt lin kt, chng ta b qua ma st. hon chnh by

gi chng ta ch n ma st trong cc bi ton cn bng.

Trong chng ny chng ta tip cn dng bi ton tnh m lin kt t ln

vt l lin kt c ma st. y l dng bi ton kh, phc tp hn nhiu so vi bi

ton khi b qua ma st, trong gii hn c th chng ta ch tip cn bi ton ny

vi cc kt qu thc nghim v ma st v cc nh lut ma st ca Cu lng.

Trong qu trnh xem xt cc bi ton c lc ma st, chng ta s lm r mt

phn ngha v vai tr ca lc ma st trong cc bi ton cn bng trong phn tnh

hc v n s gip ta hiu thu o hn trong cc bi ton ca phn ng lc hc

sau ny.

3.1. C s l thuyt3.1.1. M hnh phn lc lin kt

Cho vt rn A ta ln vt rn B, gi s mt ta kh v khng nhn. Lc ,

ngoi thnh phn phn lc lin kt Nc phng php tuyn nh bit, vt A

cn chu lc ma st trtFms v ngu lc ma st lnMms khi vt A chuyn ng

hoc c khuynh hng chuyn ng i vi vt B (hnh 3.1).

Hnh 3.1 Hnh 3.2

Phn tch cc lc ma st: Thc t, khi hai vt tip xc, b mt tip xc

khng phi l mt im m l mt din tch no , cc lc lin kt xut hin v

to thnh mt h lc {Ni} (hnh 3.2), khi thu gn h lc ny v mt tm ta nhn

c mt thnh phn lc R bng vc t chnh ca h lc trn v mt m men M

bng m men chnh ca h lc trn. Vc t R khng vung gc vi tip tuyn ca

Fms

N

Mms

B

AN2

NnN1

ng tip xc


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mt tip xc v c phn thnh hai thnh phn N v F. Nh vy h lc nu trn

tng ng vi 2 lc N, F v ngu M, lc N chnh l phn lc lin kt khi khng

c ma st, lc F l thnh phn lc sinh ra khi tnh n ma st v ngu M chnh l

ngu lc ma st ln.

3.1.2. nh lut ma st tr

t1. nh lut ma st trt tnh

nh ngha ma st trt: L sc cn khi c hin tng trt ca mt vt ny trn

b mt ca mt vt khc. K hiuFms.

nh lut 1: Khi dch chuyn mt vt trn b mt mt vt khc s xut hin lc

ma st mt phng tip xc gia cc vt, lc ny c th c tr s t "0" n Fgh

(lc ma st gii hn). Lc ma st c xu hng ngc vi chiu chuyn ng ca

vt m lc ch ng tc dng c xu hng gy ra.

ghms FF (3.1)

nh lut 2: Tr s ca lc ma st gii hn bng tch h s ma st tnh vi p lc

php tuyn (hay phn lc php tuyn).

NfFms 0 (3.2)

Trong f0- h s ma st trt tnh (c xc nh bng thc nghim), tr

s ca h s ma st trt tnh khng ph thuc vo din tch b mt tip xc m

ph thuc vo vt liu ca cc mt tip xc.

2. nh lut ma st trt ng

nh lut: Khi chuyn ng, lc ma st trt hng ngc vi chiu ca vn tc

tng i ca vt so vi vt gy lin kt v c tr s:

fNFms (3.3)

Trong f- h s ma st trt ng (c xc nh bng thc nghim), tr

s ca h s ma st trt ng ph thuc vo b mt tip xc, trng thi mi

trng, vt liu ca cc mt tip xc v vn tc chuyn ng.

Trn hnh 3.3 ch ra s bin i ca ma st trt, khi vt ng yn (V=0), hs ma st c gi tr bng gi tr h s ma st trt tnh f0, khi vt bt u chuyn


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ng h s ma st thay i dn theo s bin i ca vn tc chuyn ng, lc ny

n c gi tr ca h s ma st trt ng f.

Hnh 3.3

3. Gc ma st v nn ma st

Gc nghing ln nht ca phn lc ton phn R khi lc ma st bng lc ma

st ti hn c gi l gc ma st (hnh 3.4a).

a) b)

Hnh 3.4

Trong bi ton khng gian, vt A c th chuyn ng theo mi hng trn

b mt vt B, v vy phn lc lin kt c th nm tu trong mt nn gii hn bi

gc v c gi l nn ma st (hnh 3.4b).p dng gc ma st v nn ma st, ta c iu kin cn bng:

iu kin cn bng khi c ma st trt: iu kin vt cn bng l phn lc

ton phn ca cc lin kt ta c ma st trt phi nm trong nn ma st.

4. Bi ton cn bng khi c ma st trt

Trong bi ton xc nh iu kin cn bng khi c ma st, ta thng gp hai

dng bi ton sau:

Dng 1: Kho st trng thi cn bng ti hn v lc ma st c gi tr bng lc

ma st ti hn, NfFFghms 0== .

0

0

V

A

B

R

Fgh

NR

A

B


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Dng 2: Kho st tt c cc v tr cn bng c th c ca vt, lc ny gi tr

ca lc ma st c xc nh theo cng thc: NfFF ghms 0= .

3.1.3. nh lut ma st ln

1. nh ngha ma st ln: l sc cn xut hin khi mt vt ny ln trn b mt

mt vt khc. K hiuMl.

a) b)

Hnh 3.5

Xt con ln bn knh R, trng lng P nm trn mt phng ngang nhm. Tc

dng vo trc con ln lc Q Fgh. Ti A xut hin lc Fms = Q, ngn khng cho

con ln trt trn mt phng, gi thit phn lc php N cng t ti A, ta c N = P

(hnh 3.5a). Nh vy Q v Fms to thnh mt ngu v cho d Q c nh n my

th con ln vn c th ln trn mt phng. Thc t, ch khi lc Q t n mt gi

tr nht nh no (Qgh) th con ln mi bt u ln. iu ny c th gii thch

nh sau: hai vt tip xc vi nhau trn mt din tch AB, khi tc ng lc Q, vt

s b bin dng p lc ti A gim, ti B tng ln v din tch tip xc cng tng

dn ln, phn lc N lc ny t ti im B, ngu (N, P) cn bng vi ngu (Q,Fms) cn li xu hng ln (hnh 3.5b). Khi lc Q t gi tr gii hn, khong cch

AB = k, khi ta c:

NkRQgh = (3.4)

hay: NR

kQgh = . (3.5)

i lng k trong cng thc trn c gi l h s ma st ln, c th

nguyn di (trong a s trng hp, t sk/R nh hn nhiu so vi h s ma

st tnh f0).

C

N

AQ

P

Fms

B

C

N

AQ

P

Fmsk


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2. nh lut ma st ln: Ngu lc ma st ln xut hin khi c xu hng ln tng

i ca mt vt trn b mt mt vt khc, c chiu ngc vi chiu ca xu hng

ln v c gi tr: kNMl .

3. M hnh phn lc khi c ma st ln: Khi gii cc bi ton c ma st ln, ta p

dng hai dng m hnh sau y:

Dng 1: Ngoi cc lc v phn lc thng thng, ta thm vo i lng ma st

ln (hnh 3.6a). H lc tc dng ln vt lc ny l (P, Q, N, F ms, Ml) v coi

kNMl .

a) b)

Hnh 3.6

Dng 2: Coi phn lc N b di mt on h v pha c xu hng ln ca vt (hnh

3.6b), h lc tc dng lc ny l ((P, Q, N, Fms) v hk.

3.2. Hng dn p dng

3.2.1. Phn loi cc bi ton ma st

Cc bi ton c ma st c phn loi nh sau:

1. Bi ton mt vt v h vt.

2. Bi ton ma st trt v ma st ln.

3. Bi ton thun: xc nh phn lc h cn bng.

4. Bi ton ngc: tm iu kin h cn bng.

Trong bi ton ny ta chia ra cc dng sau:

Bi ton tm v tr cn bng.

Bi ton tm iu kin lc hot ng phi tho mn c cn bng.

Bi ton t hm: tm iu kin (cc thng s hnh hc) c cn bng, d lchot ng rt ln.

B

C

N

AQ

P

Fmsh

Ml

A

C

N Q

P

Fms


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3.2.2. Trnh t v phng php gii

Trnh t gii: Bi ton ma st c gii theo hai bc nhng c nhiu c im

cn ch .

Bc 1: t lc: ngoi lc hot ng v cc phn lc thng thng ta phi

thm lc ma st trt v ngu lc ma st ln. v ng chiu ca lc ma

st, phi xc nh xu hng trt v ln ca cc vt.

Bc 2: Lp cc iu kin cn bng: lp cc iu kin cn bng i vi h

lc nh bit; a vo iu kin i vi cc lc ma st.

Nh vy, cc iu kin cn bng bi ton c ma st tng ln, do vic

gii cng s phc tp hn.

Phng php gii

1. Phng php gii tch: V lc v ngu lc ma st, lp cc phng trnh cn

bng v vit cc iu kin ma st, c th theo mt trong ba dng sau: Dng 1: Gi nguyn bt ng thc ma st v gii h thng hn hp gm cc

phng trnh cn bng v cc bt phng trnh ma st. Phng php ny rt

thun li khi ch c mt bt ng thc ma st.

Dng 2: Xt h trng thi ti hn (h sp trt hoc ln). Khi thay cho

bt ng thc ma st, chng ta c ng thc NfFFghms 0== v kNMl = , h

hn hp ni trn tr thnh h thun phng trnh v cho php tm n l hm

ca cc h s ma st. Chiu ca bt ng thc

c xc nh nh kinh nghimhoc da vo chiu bin thin ca cc hm nhn c.

Dng 3: Xt h trng thi cha ti hn nhng vit iu kin ma st di

dng ng thc vi cc h s ma st (f' hay h) nh hn h s ma st thc.

Chng ta cng c h thun phng trnh v cho n l hm ca f' (h). Thay f'

(h) bng f0 (k) v vit li iu kin cn bng di dng bt ng.

2. Phng php hnh hc: p dng cho bi ton c ma st trt, theo phng

php ny chng ta v nn ma st (khng v phn lc php v lc ma st), xy ra

hai trng hp sau:


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Trng hp c mt nn ma st: Khi cn bng phn lc ton phn cn bng

vi lc hot ng. Nhng phn lc ton phn nm trong nn ma st nn lc

hot ng cng phi c tnh cht . Vy iu kin cn bng l lc hot ng

nm trong nn ma st.

Trng hp c hai nn ma st: Xc nh phn giao ca hai nn ma st. iu

kin cn bng l lc hot ng phi ct phn giao nu trn (v phi ng quy

v giao im ca hai phn lc ton phn l im chc chn nm trong phn

giao ca hai nn ma st).

3.3. Bi gii mu

V d 3-1

Hy xc nh gc nghing ca mt phng nghing vt nm trn n cn

bng, nu h s ma st ca vt vi mt phng l f0?

Bi gii

Ta p dng gii bi ton ny bng phng php gii tch, theo u bi yu

cu phi xc nh tt c cc v tr cn bng ca vt. V vy, trc tin ta xc nh

v tr cn bng ti hn ca vt, lc gc = gh .

H lc tc dng ln vt (P, N, Fgh) cn bng

(hnh 3-7), v vy ta c quan h:

ghghtgNF = (a)

Mt khc theo nh lut ca ma st trt, ti

thi im ti hn gi tr lc ma st bng:

NfFgh 0= (b)

Kt hp (a) v (b) ta c:0

ftggh = (c)

Xt biu thc (c), nu gimf0 th ghcng gim theo, v vy ta kt lun rng

s cn bng cng c th xy ra khi < gh. Vy tt c cc gi tr ca gc m

vt s cn bng c xc nh theo bt ng thc sau:

0ftg (d)

Fgh

N

P

Hnh 3-7


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V d 3-2

Thang AB ta trn tng v sn nh, cho bit h s ma st ca thang vi

tng v sn l f0 . Hy xc nh gc nghing ca thang vi tng c th tro

ti nh thang. B qua trng lng ca thang.

Bi gii

Ta p dng phng php hnh hc gii bi ton ny. Gi s thang nm ti

v tr cn bng ti hn (hnh 3-8), thang chu tc dng ca cc phn lc R A v RB,

cc lc ny nm lch vi cc php tuyn mt gc bng gc ma st .

ng tc dng ca cc phn lc ct nhau ti K.

Do , khi cn bng th lc th ba P tc dng ln

thang cng phi i qua im ny. Bi vy, theo hnh

v ngi ch ln c cao nht n im C. Mun tro

ln n im B th bt buc im giao ca hai phnlc phi nm trn ng BO hoc bn tri ng ny,

iu ny ch xy ra khi ng tc dng ca phn lc

RA ct im B hoc nm di im .

Vy, ln ht thang, gc nghing ca thang phi tho mn iu kin:

V d 3-3

Thanh AD trng lng khng ng k, nm ngang trn hai gi B v C,h s ma st trt gia thanh v cc gi l f. Ti D thanh chu lc ko Q nghing

vi thanh mt gc . Cho bit BC=2CD=2a. Xc nh gc xut hin hin

tng t hm cho d lc Q rt ln.

Bi gii

Ta ln lt p dng c hai phng php gii.

Phng php gii tch: H lc tc dng ln thanh bao gm: (Q, N1, N2, F1, F2),

thanh trng thi cn bng, ta c cc phng trnh sau (hnh 3-9a):

== ;0cos 21 FFQFX (a)

O

RA

C

B

A

RB

P

K

Hnh 3-8


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=+= ;0sin 21 NNQFY (b)

== .02cos)( 1aNaQFm kC (c)

Cc iu kin p t ln cc lc ma st c s dng dng sau:

,;222111

NfFNfF ==

trong : .; 21 ffff

Thay gi tr ca F1 v F2 vo cc phng trnh (a, b, c), ta nhn c:

213

2

fftg

+= . (d)

Hnh 3-9a

Khi tng cc gi tr f1, f2 ln tr s f, v phi ca (d) s gim

ffftg

2

1

3

2=

+ hay

farctg

2

1 . (e)

Trong biu thc (e), iu kin cn bng khng ph thuc lc Q, v vy khi

gc tho mn iu kin trn th lun xy ra hin tng t hm.

Phng php hnh hc: Ti hai im B v C ta v hai nn ma st, hai nn c

phn giao l phn gch cho, nh E. h cn bng, lc Q phi c ng tc

dng ct phn giao ny (hnh 3-9b).

Hnh 3-9b

N2

N1

Q

DCBA F2

F1

H

E

Q

DCBA


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Gc nghing ca ng ED vi phng ngang chnh l gc ti hn ca lc

Q, c xc nh theo cng thc:

fa

actg

HD

EHtg

2

1

2===

(f)

Vy, h lun cn bng khi gc nghing ca lc Q lun ln hn gi tr trong

biu thc (f). T hm lun xy ra v d tng tr s Q, ng tc dng ca lc Q

vn ct phn giao.

V d 3-4 (Ma st ca dy i vi mt tr)

Cho lc P tc dng vo u mt si dy vt qua mt tr trn bn knh R.

Hy xc nh lc Q nh nht cn phi tc dng vo u dy kia dy cn bng,

bit gc m ca dy quanh tr l (hnh 3-10).

Bi gii

Xt dy trng thi cn bng ti hn, trc

tin, ta xt cn bng ca on dy DE c chiu

di dl=Rd, vi R l bn knh ca tr. Hiu sc

cng ca dy ti cc im D v E l dT c cn

bng bi lc ma st dF=f0N. Vy ta c:

DT= f0N

xc nh dN, ta chiu cc lc ln trc y:

TddTddTTdTdN +=++= 22

2sin)(

2sin (a)

Thay gi tr ny vo biu thc ca dT, ta c:

Tdf

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