Giáo trình Cơ kỹ thuật (Nghề: Cắt gọt kim loại - Trung cấp) - Trường TCN Kỹ thuật công nghệ Hùng Vương

UY BAN NHAN DAN QU4-N 5
TRUONGTRUNGCAPNGHEKYTHUATCONGNGHEHUNGVUONG
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GIAOTRINH

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Nghe: Cat gQt kim Io~i

TRINH DO• TRUNG CAP

TPHCM-2019



1

LOI GIOI TH}¥U

Ghio trinh CO' ky thu~t ra doi nhim dap frng nhu c!u giang d~y mon
CO' ky thu~t t~i tnrrrng Trung c§p ngh~ ky thuit cong ngh~ hung Vll'O'ng.
Giao trinh dtrgc bien so~n theo dung n(H dung qui djnh cua chtrO'ng trinh
mon h9c CO' ky thu~t trong chtrO'ng trinh khung dao t~o ngh~ C~t g9t kim

lo~i.
Trong qua trinh bien so~n giao trinh, tac gia da dtrgc S\f h6 trg nhi~t
tinh cu.a cac d6ng nghi~p. Chan thanh cam O'n cac giao vien va lanh d~o
khoa da dong g6p y ki~n d~ hoan thi~n giao trinh.

Quqn 5, ngay

thang

nam 20 ...

Tham gia bien soqn
Chu bien

Nguy@n Hfru Nhan


2

MlJCLlJC
TRANG

BEMVC

,. g101
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Lm
•~u ·············•o••··············· . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1
Myc lye ............................................................................................................................................................................................................... 2
Ch U'O'Dg I: Tinh HQC ................................................................................................................................ o••·············3

BAI 1: Nhfrng· khai ni~m

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CO'

ban va cac tien d~ tinh hqc ......................... 3

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BAI2:
He". hrc ph8ng .....

a ........................................................................................................................................

00.12

BAI 3: H~ l\fC khong gian .......................................................................... 24
Chuo-ng II: l>Qng HQc ............................................................................................................................................................... 28
BAI 1: Chuy~n d{>ng ciia ch~t di~m ......................................................... 28
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;:.

BAI 2: Chuyen clQng cO' b3n ciia vit ran ................................................. 37

BAI 3: T6ng hgp chuy~n d{>ng ................................................................. 43

BAI 4: Chuy~n d{>ng song phing ciia v~t riin ........................................ .46
CHUONG III: Sfrc B~n V~t Li?u ............................................................. 50
BAI 1: Cac khai ni~m

CO'

ban .................................................................... 50
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BAI 2: Keo - nen diing tam, cat ............................................................... 56

BAI 3: f>~c trung hinh hqc cua hinh phing ............................................ 70
BAI 4: Xoiin thu~n tuy nhfrng thanh trOn .............................................. 7 4
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BAI 5: U6ng phing ciia thanh thiing ....................................................... 82
,

CHUONG IV: Chi tiet may ...................................................................... 92

BAI 1: M6i ghep dinh tan ......................................................................... 92

BAI 2: M6i ghep han ................................................................................. 97
BAI 3: M6i ghep ren ................................................................................ 105
BAI 4: M6i ghep bing then va then boa
CHUONG V: Cac chi ti~t may truy~n d{>ng ........................................... 92
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BAI 1: BQ truyen

dai ................................................................................. 92

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BAI
2: B{> truyen
bSnh ma sat ........................................................... ., ...... 97

BAI 3: B{> truy~n trgc vit ........................................................................ 105
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BAI
4: BQ truyCn
b3.nh rang ................................................ ., .................... ., .. .


3

CHUONG 1 TiNH HOC
.

Gi6i thieu: Chuong tToh h9c giai quySt cac bai toan vs sv tuong tac gifra cac
V?t th@ c6 mBi lien kSt v6i nhau.
Ml}.C tieu:
- Trinh bay dugc cac tien dS, cac khai ni~m, each bi@u di@n Ive, cac lo~i lien
kSt ca ban, h~ Ive, phuong phap hgp Ive d6ng quy, tach Ive d6ng quy.
- Phan tich dugc Ive tac dvng va cac phan Ive lien kSt, cac momen cua Ive
dBi v6i m9t di@m, ng~u Ive.
- Tinh dugc Ive tac dvng va cac phan Ive lien kSt, cac momen cua Ive dBi
v6i m9t di@m, ng~u Ive.

- Tinh dugc lgc b~ng phuong phap da giac, phuong phap chiSu dS giai cac
bai toan vS h~ Ive bfit ky.
- L~p dugc phuong trinh mo men tinh toan h~ lvc tac dvng.
- Giai dugc cac bai toan h~ Ive ph~ng song song.
- Ren luy~n tinh ky lu?t, kien tri, ctn th?n, nghiem tuc, chu d9ng va tich eve
sang t~o trong h9c t?p.
N{H dung:

Bai 1: NHUNG KHAI NI~M co BAN vA cAc TIEN DE
TiNHHQC
I.

cAc KHA.I NitM ca BAN

1. Vit ri\n tuy~t dBi
Vi)t r&n tuy¢t a6i la VCJt r&n c6 hznh dq,ng hznh h9c kh6ng thay a6i trong qua
trinh chiu luc

Thv·c ts· cac V?t r~n khi chiu lvc dSu c6 biSn d~ng, nhung biSn d~mg d6 nit
nho ta c6 th@ bo qua dS nghien cuu hi~n tugng dugc dan gian hon, nhung v~n
bao dam d9 chinh xac theo yeu cfiu k1 thu?t
d CHlfONG 1 khi xem xet V?t, ta coi v~t la v~t r~n tuy~t dBi
2. Trang thai can biing
V?t ran 6 tr~mg thai can bang khi v?t dung yen ho~c chuyen d9ng deu doi v&i
h~ qui chiSu (h~ tn;c to~ d9 ch9n lam chuan)
3. L\fC
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a. IJjnh nghia:

L11c la tac dimg tuong h6 giua cac V(Jt ma kit qua la gay nen Slf thay a6i
trq,ng thai CO' h9c cua v9t thi dy ( baa g6m Slf thay a6i vt trf
va ca biin dq,ng)
Thi dv: du6i tac dvng cua lvc, V?t dang dung yen
chuy@n sang chuySn d9ng, v~t dang chuySn d9ng dSu
chuy@n sang chuySn d9ng nhanh dfin dSu ho~c ch~m dfin
dSu .. Hoijc thanh gi~ng du6i tac d9ng cua tr9ng v?t n6 ciing
Kmh 1-1
chiu tac d9ng 1 Ive keo ho~c Ive nen ( tuy khong gay ra
chuySn d9ng nhung n6 ciing gay ra biSn d6i tr~ng thai ca h9c)


4

b.Cacyiut8cual~c
Lvc duqc d~c tnmg boi 3 y@u t6: di@m d~t, hu6ng va tri s6
- Di@m d~t la ph§n tu v~t chit thrn)c v~t ma qua d6 tac dµng tuong h6 duqc
truySn dSn v~t
- Huong ( phuong va chiSu cua Ive) la hu6ng chuy@n d9ng ma Ive gay ra
cho v~t
- Tri s6 cua Ive hay cuong d9 cua Ive la d9 m<:1nh y@u cua tuong tac ca h9c.
Don vi cua Ive la Niuton, ki hi?u la N va cac b9i s6 cua n6 la kilo Niuton

kN(lkN = 10 3 N)
y
c. Biiu diin lrrc
Lvc la, d<:11· luQ'llg ve' c ta .· Ve' c ta Ive co' goJ.. c
d~t t?i di@m d~t Ive, hu6ng trimg v6i hu6ng cua
Fy ----F ~.!
Ive, d9 dai ti 1? v6i tri s6 cua Ive, vi d\l Ive
i
:F,P,Q, G ... Duong th~ng chua vec ta Ive duqc
i
i
0
X
g9i la duong tac dl_lng cua Ive
Fx
Vee ta Ive F duqc xac dinh thong qua hai
hlnh 1-2
hinh chi@u cua Ive len tfl_lC Ox va Oy duqc ki hi?u la Fx va FY
0

-------4-

Fx =±F.cosa
FY= ±F.sina

(1-1)

Trong cong thuc (2-1 ), a la g6c nh9n hqp boi duong tac dl_lng cua Ive F v6i
tfl_lc Ox
Nguqc l?i khi bi@t 2 hinh chi@u Fx vaFY cua Ive F len 2 trl_lc Ox va Oy ta

hoan toan xac dinh duqc Ive F

(1-2)
F

sina =-Y
F

(1-3)

4. Cac djnh nghia khac

a. Hf l~c
Dinh ngh'ia: H? Ive la t~p hqp cac Ive cung tac
d\lng len m9t v~t dln. H? Ive g6m cac Ive If,Ji;,If, .....~
+

duqc kf hi?u la (If.Ji;,Jf, .....Fn ). Hinh 1-3
+
H¢ /ifc tuang auang: Hai h? Ive duqc gQi la
tuong duong khi chung gay cho cung m9t v~t rin cac
tr<:1ng thai chuy@n d9ng ca h9c nhu
nhau. Hai h? Ive (If.Ji;,If,.....~) va
("-Pi ,P2 , ããããããã Pk)

---

Hlnh 1-3

tuong duong ki hi?u

-) ~(--)
la (F'i,F2,F3,ÃÃÃÃÃFn
Pi_,P2,ÃÃÃÃÃÃÃpk
+
HÂ life can bang: H? Ive can
bing la h? Ive tac dl_lng len v~t rin ma

Hinh 1-4


5

khong lam thay d6i tn;mg thai chuySn d9ng cua v~t. H~ h;rc ("""if,"""fi;,"""fi;,.... .Fn ) can
bing ki hi?u la("""if,"""fi;,"""fi;,.....

~)= o

+
H9'p hrc: Hqp Ive la rn9t Ive duy nhfit
tuong duong voi h? Ive. R la hqp Ive cua h~ h;rc
thi (}{~Ji;,.....
R

b.
*

~)=
Momen va ngfiu life

H'inh 1-5

Mo men cua mot luc a6i vbi 1 aidm
Khi 1 Ive tac dµng len v~t riln, n6 c6 thS lam v~t di
chuySn nhung cling c6 thS lam v~t quay ho~c ca vua
quay vua di chuySn
Gia SU v~t riln co thS quay quanh diSrn O c6 dinh, tac
d\lllg ma Ive F gay ra cho v~t ph\l thu9c vao tri s6 Ive F
va khoang each a tu diSrn O d@n duong tac d1,1ng cua Ive.
16
Con chiSu quay ma Ive gay cho v~t c6 thS la nguqc chiSu
H'inh ho~c cung chiSu kirn d6ng h6. D~i luqng d~c trung cho ca tac d1,1ng quay va
chiSu quay do g9i la mo men cua Ive d6i voi rn9t diSm
Dinh nghza: Mo men cua hrc a6i vbi 1 aidm la 1 dr;zi lu(mg dr;zi s6 c6 gia tri
tuy¢t a6i b&ng tich s6 giif:a tri s6 cua hrc vbi canh tay don va c6 ddu (+) hay ddu
(-) tuy theo chi§u quay cua lL;Cc F quanh tam O la ngu(Jc hay thugn chi§u kim
d6ngh6
(1-4)
ki hi?u m0 (F) = ±F.a
Trong d6:
mo

(F) la ki hi~u mo men cua h;rc Fd6i voi diSm 0

F la tri s6 cua Ive
a la canh tay don
Khi chi cfin dS y dSn tac d1,1ng quay ma khong quan tarn dSn chiSu quay ta c6
tri s6 mo men cua 1 lgc d6i voi 1 di Sm, ki hi~u la mo
mo
= F.a

Mo men cua Ive d6i voi diSrn bing khong khi Ive di qua diSm lfiy mo men (a
=0).
Ngdu luc
Khi c6 2 Ive F; , F 2 song song nguqc chiSu c6 tri s6 bing nhau tac d\lng len

(F)

*

v~t riln. M~c du R = F; - F2

(F)

= O nhung h~ ( F;,;;) khong can bing vi

khong cung duong tac d1,1ng. Nhu v~y (

F; , F

2

F;, Ji;,) khong c6 hqp Ive nhung n6 c6

khuynh hu6ng lam cho v~t riln quay, nen duqc g9i la ng&u Ive.
Dinh nghza: Ngdu lvc la m9t hi g6m 2 lvc song song, ngu(Jc chidu, c6 tri
s6 b&ng nhau nhung khong cimg duimg tac d1Jng
Ccic yiu t6 cua ngdu lvc
m
Ng&u Ive c6 3 ySu t6:
a

ffinh 1-7


6

M~t philng tac dl_lng: la m~t philng chua cac Ive cua ngfiu Ive
ChiSu quay cua ngfiu Ive trong m~t philng ngfiu Ive: thu~ v6i chiSu kim
d6ng h6 ho~c nguqc l~i
Cuong d9 tac dl)ng cua ngfiu Ive duqc d~c trung boi tich s6 cua Ive v6i
canh tay don g9i la tri s6 mo men cua ngfiu Ive :
m=F.a
(1-6)
Trong d6 : m la tri s6 cua ngfiu Ive
Fla tri s6 cua Ive
a la canh tay don
Bon vi dS do tri s6 mo men la NiuTon met. Ki hi~u la Nm,
Cac tinh chdt cua ngdu htc:
Tac dl)ng cua ngfiu Ive len v~t rfm
khong thay d6i khi doi n6 dSn dSn m~t
philng song v6i n6.
H1nh 1-8
C6 thS doi ngfiu Ive trong m~t tac
dl_lng thi tac dl_lng cua ngfiu Ive len v~t rin khong thay d6i
C6 thS thay d6i tuy y tri s6 cua Ive va canh tay don cua ngfiu Ive nhung
vfin gifr nguyen mo men thi tac dl_lng cua ngfiu Ive len v~t rin khong thay d6i.

II. cAc DINH LUAT TiNH HOC
Tien d~ 1 ( iljnh lu{U ·vi 2 life can bring)
DiJu ki¢n cdn Vet au ai 2 Ive tac d?mg !en 1 vcjt r&n can bang

la 2 Ive phai cung phuong,
ngm;rc chi€u vet c6 cung trj s6
Ive b&ng nhau

Tien d~ 2 ( iljnh lu~t them, but 2

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IDnh 1-9

life can

b&ng)
H'mh 1-10
Tac d¥ng cua 1 h¢ Ive !en 1 vcjt rdn khong thay a6i khi ta them vao hay bat
di 2 Ive can b&ng nhau
He quii: ( iljnh If tric(J't life)
Tac d'?[ng cua 1 Ive !en 1 vcjt
r&n kh6ng thay a6i khi ta truQ't
Ive tren aucmg tac d¥ng cua n6
Th~t v~y gia SU c6 Ive F tac
H'mh 1-11
dl_lng len v~t rin a diSm A (hinh
1-13 ). d diSm B b~t ki tren duong tac dl)ng cua Ive Fta d~t them vao 2 Ive can
b§.ng
Hai Ive nay can b§.ng cung phuong v6i Ive F va c6 cung tri s6 cua

Fr,F'.i.

Ive F. Theo tien dS 2: F~(F,Fr,F.i). Hai Ive (F,P.i)can bing nhau (theo tien dS
1) nen theo tien dS 2 ta I~i c6 th@ bot 2 Ive nay va c6 F ~(F,Fr,Fi) ~Fi. Fi chinh


7

la Ftrugt tir A t6i B. Nhu v~y lvc tac d\lng len v~t r~n tuy@t d:6i dugc biSu diSn
bfulg vec ta trugt

Tien a~ 3 ( tljnh lu~t vi qui tdc hinh binh hanh lz.rc)
Hai hrc aij,t tc;zi 1 aidm tuang auang vai 1 ll:fC aij,t tqi
ai&m a6 va au(Yc bi&u diin b6-ng vecta auirng cheo hinh binh
hanh c6 2 cc;znh la 2 Ive aa cha
Theo phep c9ng vec ta thi R la t6ng cua ]{,Ji; tuc la

R=F'i +F2
Tien a~ 4 ( iljnh lu~t vi tac dimg va phiin tac d{lng)

R

Hinh 1-12

Lvc tac d¥ng Va lvc phan tac d¥ng cimg phuang, ngu(YC
chiiu va c6 tri s6 b&ng nhau
Chu y: Lgc tac d\lng va phan tac dvng khong phai la 2 Ive
can b~ng vi chung d~t vao 2 v~t khac nhau

Tien a~ 5 ( tljnh lu~t thay thl lien klt)
V~t khong tv do can b~ng c6 th@ dugc xem la v~t can b~ng,
Hinh 1-13

b~ng each giai ph6ng tfit ca cac lien kSt va thay thS tac d\lng cac
lien kSt duc;rc giai ph6ng bfulg cac phan Ive lien kSt thich hgp
III. LIEN KET VA PHAN L~fC LIEN KET
1. Khai niem
a. V~t /If do va v9,t chju lien klt
Vijt r&n c6 thi thvc hi¢n m9i di chuy&n trong khong gian g9i la wjt r&n tv do.
Thi d\l qua bong nhy bay la lung trong khong gian g9i la v~t r~ tv do
Vqt r&n c6 1 vai di chuy&n bi can tra au(Jc g9i la vqt khong tv do hay vgt
chtu lien kit. Thi d\l quySn sach d~t tren ban
la v~t chiu lien kSt vi n6 can tr& chuySn
d:9ng thtlng dung di xu6ng
b. Lien klt va phiin llfC lien klt
Nhfmg aiJu ki¢n can tra chuy&n a9ng tv
do cua vq,t au(JC g9i la lien kit. V~t gay ra
gay ra can trcr chuy@n d9ng cua v~t khao sat
p
p
dugc gQi la v~t gay ra lien kSt. Thi dv quySn
sach dat tren ban la v~t khao sat, con cai
Hinh 1-14
ban la v~t gay ra lien kSt
Do tac d¥ng tuang h6, tc;zi lien kit vgt gay lien kit tac d¥ng len vgt khao sat,
lvc a6 au(Jc g9i la phan lvc lien kit. Phan Ive lien kSt chinh la lvc can tr&
chuySn d:9ng tg do cua v~t khao sat
c. Tfnh ch8t cua phiin life lien klt
- Phan lvc lien kSt bao gia ciing d~t len v~t khao sat t~i ch6 tiSp xuc v6i v~t
gay lien kSt
- Phan Ive lien kSt cung phuang va ngugc chiSu v6i chuySn d9ng bi can tr&
cua v~t khao sat

8

- Tri s6 cua pha? Ive lien k~t phµ thut;>c rao ~ac Ive tac d1µ1g len V?t khao sat
Qua cac tinh chat tren ta thay phan Ive lien ket la lo~i Ive bi dt;>ng, khac v6i
cac Ive tac dl,lng len v~t khao sat d6 la cac Ive da cho hay Ive chu dt;>ng

2. Cac lien k~t thrrcrng g~p
a. Lien kit tzra
Lien kit tl:{a lien kit
eek vg,t chi c6 tac
d~ng ila ldy nhau. V 6i gia thiSt bo qua ma sat, lien
kSt tva can tr& chuySn dt;>ng cua V?t khao sat theo
phuong vuong g6c v6i m~t tiSp xuc chung gifra
V?t gay lien kSt va V?t khao sat. Phan Ive lien kSt
tva c6 phuong vuong g6c v6i m~t tiSp xuc chung,
chiSu di tu V?t gay lien kSt vao V?t khao sat va c6
1 ySu t6 chua biSt d6 la tri s6. Phan Ive lien kSt tva
thuong dugc ki hi~u la N
Chuy:
- Truong hgp m9t trong hai tiSp xuc la m9t
diSm thi phan Ive phap tuySn theo phap tuySn cua m~t
con l~i
- G6i tva di d9ng cung la d~ng lien kSt tva

ia

ma

Hlnh 1-15

b. Lien kit day mim
Lien kSt day can tr& chuySn d9ng cua v~t khao sat theo chiSu cang cua day.
Phan Ive hu6ng theo phuong cua day theo chiSu tu V?t khao sat di ra va c6 m9t
ySu t6 chua biSt la tri s6. Phan Ive cua lien kSt day mSm thuong dugc ki hi~u la
va con g9i la Ive cang day

r

c. Lie'! kit b~n Ii phdng
,
, ,
Lien ket ban le phang c6 2 lo~i la ban le di d9ng va ban le co dinh
• Ban l~ di i16ng
Ban IS di d9ng cho phep V?t khao sat (v?t B) quay
quanh tfl,lc ban IS va di chuySn theo phuong song song
v6i m~t tva, con can chuySn dt;>ng theo phuong vuong
g6c v6i m~t tva: phan Ive ( thuong dugc ki hi~u la R) c6
phuong vuong g6c v6i m?t tva, tri s6 chua biSt. C6 thS
Hinh 1-17a
gia dinh cho phan Ive 1 chiSu nao d6, nSu kSt qua tinh ra duong thi chiSu gia
dinh dung v6i chiSu thvc, con kSt qua mang d~u am thi chiSu thvc ngugc v6i
chiSu gia dinh
• Ban l~ c6 ilinh
Lien kSt ban IS c6 dinh chi cho phep V?t khao sat (v~t
B) quay quanh trl,lc ban IS con m9i di chuySn dSu bi can
tr&. Phan Ive lien kSt R co tri s6 va phuong chua biSt,
Hinh 1-l?b
con chi@u thi gia dinh nhu da n6i a tren. DS thu~n ti~n khi tinh toan ta thuong

R

1


9

R lam 2 thanh ph§.n ~
c6 2 tri s6 cua ~ va Ii;

phan

va

Ii; vuong g6c v6i nhau, hie d6 phan Ive cung

d. Lien kit thanh
Lien kSt thanh g6m m9t thanh thing ho~c
thanh cong c6 2 d§.u la 2 ban 16 c6 dinh va tren
D
thanh khong c6 Ive tac d\lllg (bo qua tn;mg lm;mg
thanh)
Hinh 1-18
Lien kSt thanh can tr& chuySn d9ng cua v~t
khao sat theo phuong n6i 2 ban l@ 6 2 d§.u thanh.
Phan Ive lien kSt thanh c6 phuong la ducmg n6i kh6p a d§.u thanh, c6 tri s6
chua biSt, con chi@u thi tru6c khi tinh cung c§.n gia dinh nhu da neu a tren. Phan
Ive cua lien kSt thanh thucmg duqc ki hi~u la

s

e. Lien kit ngiim
La lien kSt khi v~t duqc n6i cung vao 1 v~t khac. (
·----=R_x_ _ ___,
dinh duqc dong cung vao tuong)
Trong truong hqp ngam phan Ive lien kSt g6m 2 Ive
thfulg g6c v6i nhau va 1 ng~u Ive n&m trong m~t phing
Hinh 1-19
chua 2 thanh phĐ.n lvc va cung la m~t phing tac dàng cua
h~
3. Khao sat v~t ri\n can bing bing each giai ph6ng lien k~t
Tren CO' SO' dinh lu~t 5, dS khao sat can b&ng cua v~t riln ta c§.n co l~p n6 ra
khoi cac v~t thS xung quanh, xem nhu n6 khong chiu cac lien kSt, tuc la giai
ph6ng cac lien kSt cho v~t, r6i d~t C ac Ive da cho va cac phan Ive lien kSt len v~t
thay thS cho cac lien kSt bo di. Khi d6 ta c6 thS xem v~t chiu lien kSt can b&ng la
v~t rfu tv do can b&ng du6i tac d1.1ng cua cac Ive da cho va cac phan Ive lien kSt.

/

Vi df.11
Qua c§.u d6ng chfit c6 tn;mg luqng P duqc treo vao m~t tucmg nh~n thing
dung nha day OA (hinh12). Hay d~t tit ca cac Ive tac d1.1ng len qua c§.u
Gicd
y
Khao sat qua c~u, cac Ive tac d\lllg len qua
A
'
cau
c6:
- Tr9ng Ive P d~t t~i tam O va hu6ng thing

dung xu6ng du6i
- Phan Ive lien kSt tva N( thay cho lien kSt B
tva & B)
- Lvc ding dayr ( thay cho lien kSt day)
IDnh 1-20
Nhu v~y sau khi bo cac lien kSt di va thay
bfulg cac phan Ive lien kSt tuong ung ( cong vi?c nay g9i la giai ph6ng lien kSt ta
c6 thS xem qua c§.u nhu v~t riln tv do can b&ng du6i tac d1.1ng cua h? Ive (r,N,P).
DS cho g9n sau nay ta ve ngay cac phan Ive va Ive da cho vao hinh c6 ca cac
lien kSt


10

Vi dfi: Thanh d6ng ch§.t c6 tn;mg lm;mg

P chiu tac d1;mg cua Ive F. Thanh

duc;rc lien kSt bfing m9t g6i c6 dinh va m9t g6i di d(;ng (hinh 1-21). Xac dinh h~
Ive tac dvng len thanh
Giai
Lvc da cho tac dvng len thanh AB g6m
tr9ng Ive P ( d~t gifra thanh va hu6ng xu6ng)
va Ive F . Phan Ive lien kSt g6m x A , ~ ,
Nhu v~y thanh AB chiu tac dvng cua
H1nh 1-21

r;

CAU HOI ON TAP

1. ThS nao la v~t r1n tuy~t d6i? Khi nao c6 thS xem v~t d.n va v~t dn my~t

d6i
2.
3.
sao?
4.
5.
6.

Lvc la gi? Neu cac ySu t6 cua Ive va trinh bay each bi@u diSn Ive
Hai h~ can bfing c6 tuang duang khong? Vi

Neu va giai thich cac dinh lu~t fi'nh h9c
Vi sao c6 thS coi Ive la vec ta truqt?
Cho 2 lvc va ~ cung d~t t~i diSm 0
( hinhl-4)
- V 6i g6c a b&ng bao nhieu thi hqp Ive c6 tri s6 16n nh§.t?
- B~t them vao o Ive ~ nhu thS nao ? dS h~ (~;;:~)
can bfing
7. Vi sao Ive tac d\lllg va Ive phan tac dvng khong phai la 2 Ive can bfing?
8. Mo men Ive d6i v6i 1 diSm la gi?
9. Ng~u Ive la gi? Neu cac ySu t6 cua ng~u Ive.
10.Vi sao n6i ng~u Ive khong tuang duang v6i 1 Ive?
11.Lien kSt la gi? Phan Ive lien kSt la gi? Neu cac tfnh ch§.t cua phan Ive lien kSt
12.Ve cac lien k@t thucmg g~p, sau d6 ve phan lgc cua cac lien k@t d6
Giai ph6ng lien k@t la gi? Vi sao phai giai ph6ng lien k@t?

Fi

11
,2

A

Bai 2: H~ Ll}C PHANG
H<$ h.fc ph!ng la t~p hqp cac Ive tac dvng len cung m(H v~t ra'.n va c6
duong tac dvng cung nAm trong m()t m~t phing
I. VEC TCi CHINH VA MO MEN CHINH
1. Vee td chinh cua h~ htc ph~ng
Cho h<$ Ive phang( F;., F2 , F;, .... F,, )
?

--i,,--+--+

~

a. Bjnh nghia: vec td chinh cua h<$ Ive ki hi<$u 11., 1a vec td t6ng cua cac
__.---+

n----1>

vec td Ive cua ht$ hfc: R = Fr + F; + .... F,, = L ~
?

__,,.

--+

k=I

(2-1)

b. Xac djnh vec td chinh:
C6 thS bAng m()t trong 2 phuong
phap sau:

C
B _.,.. __
-A~D

+ Ve da gidc lite:
Xet ht$ Ive (i;, Ji;,;;, .... ~). Tu

E

m9t diSm A ba't kl, ta ve no'i tiSp lin
lu'qt cac vec td tu'dng ung song song
cung chiSu voi cac vec td Ff, Ji;, I{, .... ~ va c6

Hinh 2-1

d9 dai chinh bAng d◊ dai cua

cac vec td nay. Duong ga'p khuc ABCDEA du'qc t9-o thanh gQi la da giac hfc
cua ht$ Ive da cho( hlnh 2-1). Vee td A.Ed6ng kinda giac h;te la vec to ehinh
Rcua ht$ h;tc da cho.
+ Phztdng phdp hznh chieu:
--+

---+

-

-

R =Fi+ F; + ....F,,

= IF;.

n_

k=I
-

-----+-

-

__.....

Rx = Fix + F2x + .... F,,x =

L Fkx
n -

(2-2)

k=l

-

Ry

= Fry + F2y + ....F,,y = L FkJ!
-

-

---+

n

---+

k=l

R =.JR;+ R~ , cos(R,Ox) = Rx , cos(R,Oy) = RY
R

R

(2-3)

2. Mo men chinh cua h~ htc ph~ng
a. Mo men cua llfc doz vui m{)t diim
Mo men cua Ive d6i voi diSm O ki hi<$u m0 (F) la m()t luqng d9-i so' :
m0

(F)=±F.d.

Trong d6:
- F la tri so' cua Ive

(2-4)

~
Hinh 2-2


12

-

d la khoang each thiing g6c tu O dtefn duong tac dl;lng cua Ive du<;5c g9i
la tay don mo men
- ldy ddu ( +) khi lt;tc c6 chiSu quay quanh O ngu'<;5c chiSu kim d6ng h6, va
Ia"y da"u (-) trong tru'ong h<;5p ngu'<;5c lcJ,i
Mo men cua h;tc d6i vdi di€m bdng khong khi h;tc di qua diSm Ia"y mo men
( d = 0).
a. Mo men chfnh cua h? lz.tc phdng doz vfli diim 0
Mo men chinh cua ht$ h;tc phiing d6i vdi di€m O la dcJ,i ht<;5ng dcJ,i s6, ki
hit$u M 0 , bang t6ng mo men cua cac h;tc cua ht$ Ive d6i vdi di€m 0
M0

=m (:l{)+m (;;)+ .... +m (~)= Im (1{)
0

0

(2-6)

0

0

k=l

Nhan xet:
Vee to chinh la vec to tv do, con mo men chinh phl;l thu(k va diSm la"y
mo men, nghia la mo men chinh d6i voi 2 diSm khac nhau se khac nhau
mA=ms+m)R;)
(2-7)
- mA, mB: la mo men chinh cua ht$ Ive d6i vdi di€m Ava B,

- m)~):Ia mo men cua vec to chinh d~t tcJ,i diSm A d6i voi di€m B
06i vdi ht$ Ive d6ng qui thl mo men chinh cua ht$ Ive d6i vdi di6m d6ng
qui bang 0.
:06i voi ht$ ngftu Ive thl vec to chinh cua ht$ ngftu Ive luon bang 0. Mo
men chinh cua ht$ ngftu Ive d6i vdi diSm ba"t kl nao cfing bang mo men cua
ngftu Ive t6ng d)ng, tuc bang t6ng mo men cac ngftu Ive thanh phffn cua ht$
ng~u Ive.

II. DJNH LI DOI Ll}C SONG SONG:
1. Dinh Ii doi lt;ic song song
Bjnh ly 2-1: htc F tac dl!,ng tg,i A tu:cmg du:cmg wJi tac dl!,ng cua no tg,i O (
lT!c pi) va mqt ngdu lT!c c6 mo men cua lT!c F d6i wJi dilm 0

"""ft= pi va m = m ("""ft)
0

( 2-s)

Chang minh: D~t tcJ,i O hai
A

Ive can bang (pi,pii) c6 cung
tri s6 vdi luc
2 .Bai 1 ta ~6
=[F1 va

F . Theo dinh lua t
F = (T,Jii,~ 11 )

0

•

(F,P7i)J. Nhung

(F,?) la mc)tngftu Ive c6 mo men
dinh Iida du<;5c chung minh

0

IDnh 2-3

m=m)F) nen F=[pi va m=m)F)],

13

2. Thu gQn h~ hfc ph~ng v~ tam 0
Liy m9t di€m O trong mijt ph~ng tac dvng cua h~ h.fc gQi la tam thu gQn.
sfr dvng djnh Ii doi Ive song di doi cac Ive v€ tam 0
/F1

-

F{

RR

=
-

F2

m

.(.'

=
-

m

F;

\p3

R

ffinh 2-4

-Fi= iFi va ngau
~ Ive

ml= mo (-)
Fi

~=Ff va ng~u Ive m = m)Fi)
2

-F;, = iF;, va ngau
~ Ive mn

=m (-)
F;,
Nhu'v~y thu gQn h~ Ive (if~, ....... ,~) v€ tam Ota duqc h~ Ive
I
;::
. .
'
~
(I
Fi ,I
F; ,....... , F;, ) dong qu1 tc;n O va hv nga u Ive phang :
0

A

,2

(2-9)

Nhu ta da bie"t, h<$ Ive d6ng qui c6 hqp Ive qua 0, du(Jc bi€u di~n bang
vec to chinh cua n6 d~ t t~i O :

R=Fi +F; + ....FN = L~

--I>

-

N -

----+

_,..

k=l

Tu ke"t qua tren ta c6 djnh Ii:
Bjnh li 2-2: H~ llfc phdng bdt kz tuong duong wfi 1 [lfc va mqt ngdu llfc dr'j,t
tq,i 1 dilm tuy y cung ndm trong mijt phdng tdc d1;tng cua hf llfc. chung duqc
g9i la llfc va ngJu llfc thu g9n. Lvc thu g9n d~t t~i tam thu gQn c6 vec to Ive
bang vec to chinh cua h<$ Ive, con ng~u Ive thu g9n c6 mo men bang mo men
chinh cua h~ Ive d6i voi tam thu g9n
Chu y : Phuong ehi€u va gia tri cua Ive thu gQn khong phv thu9e vao tam

thu gQn vl vee to chinh la vec to tv do, con ng~u Ive thu gQn ph\l thu9c vao
tam thu g9n, n6 duqc tinh theo cong thuc (2-7) khi tam thu gQn thay d6i.
3. Cac d~ng chuin cua h~ hfc phiing
Tu ke"t qua thu g9n h<$ Ive ph~ng v€ m9t tam ta nh~n duqc cac d~ng
chuin sau( la d~ng don gian nhlt khong th€ tie"p tvc thu gQn duqc nua)
* nSu R=O, M 0 =0 thlh~ Iveph~ng can bang(Fi,~, .... ,:R)=o
* ne'u R= o, M 0 ;t: othl h~ Ive rut gQn tu'ong du'ong voi 1 ng~u Ive
*

neu R* 0, R=Fi+ F; + ....F;, =
J'--+

--+

-

-

_,..

L~ ta xet cac tru'ong hqp
n_

k=l


14

+

Ne'u M 0 = o h9 thu g9n la 1 Ive, chinh la h<;Jp Ive

+

Ne'u M 0

Rvs diSm 0

Rcua h9

* o, ta c6 th€ di doi Ive
1

ovR

each O mQt khoang

h = ~ 0 hie nay h9 Ive thu g9n l<;ti chi

con 1 Ive

R d~ t t<;ti 0

d

1

IHnh 2-5

lJjnh Ii 2-3 (djnh Ii Varinhong): Ne'u h9 lvc c6 h<;Jp Ive thl mo men cua

h<;Jp Ive d6i voi m9t diSm nao d6 bang t6ng mo men cua cac Ive thUQC h9 Ive
d6i voi cung diSm d6.

m)R)= Im):R)=o
k=l

III. DIEU KIEN CAN BANG VA CAC PHUONG TRINH CAN
BANG CUA Ht LUC PHANG
1. Di~u ki~n can b~ng
lJjnh Ii 2.4. 0i€u ki9n c~n va du dS h9 Ive philng can bang la vec to
chinh va mo men chinh cua h9 Ive d6i voi m9t diSm ba"t ky phai d6ng thoi
tri9t tieu.
(~,Ji;,~' .... F;,) = 0 ~

R= L~ =0

-

n -

k=l

( 2.10)

k=l

2. Cac d~ng phuong trinh can b~ng cu.a h~ b;tc phiing
0i€u ki9n tren c6 th€ du<;1c vie't duoi d<;tng cac phuong trlnh du<;1c gQi la
cac phudng trlnh can bang. C6 3 d<;tng phuong trlnh can bang sau:
Dgng 1: BiSu ki9n c~n va du dS h9 Ive can bang la t6ng hlnh chie'u cac

Ive tren 2 tn;c to<;t d9 vuong g6c va t6ng mo men cac Ive d6i voi 1 diSm ba"t
kl tri9t tieu

Rx= LFkx =0, RY= LFkJI =0, M =Im
n

n

n

0

k=l

k=l

(-)
0

~

k=l

=0

(2-11)

D(lng 2: BiSu ki9n c~n va du dS h9 Ive can bang la t6ng hlnh chie'u cac
Ive tren 1 trl;lC va t6ng mo men d6i voi 2 diSm A va B tri9t tieu voi diSu ki9n
AB khong vuong g6c voi trl;lc chie'u

Rx= IFkx = 0, MA= Im)}{)= 0, Ms= Ims(J{)= 0
k=l

k=l

(2-12)

k=l

Dgng 3: 0i€u ki9n c~n va du dS h9 Ive can bang la t6ng t6ng mo men d6i
voi 3 diSm A, B, C khong thilng hang tri9t tieu


15

MA= fm)~)=o, MB= fmB(I{)=o, Mc= fmc(I{)=o
k=l

(2-13)

k=l

k=I

Trong trlidng h(lp h~ 11/c ddng qui: Di€u ki~n cfin va du dS h~ Ive can
bang la t6ng hlnh chie'u cac Ive tren 2 trgc to~ de) tri~t tieu
n

Rx=

LFkx = 0,
k=l

n

RY=

LFky = 0

(2-14)

k=I

y

MQTSOVIDTJ

A

Vi d111: Qua cfiu d6ng chit tam 0, tr9ng
ht<;1ng P, tva vao tu'ong cJ C va duqc treo bcJi day
AB dai bang ban kinh. tlm sue ding va phan Ive
(
N
cua tu'ong.
X
C
Gidi:
Khao sat qua c~u 0. H~ llfc tac dvng g6m c6:

- Tr9ng htqng P
( a)
- Sue ding r cua day( d~t t~i B huong vS A)
- Phan hfc Ncua tuong ( d~t t~i Chuong vs 0)
Cac hfc nay d6ng qui t~i 0
DS dang nh~n tha"y tam giac AOC la tam giac vuong c6 dinh g6c t;;ti A
bang 30° ( vl AB = BO = OC )
Dln day c6 2 each giai:
- Bdng phztdng phap gidi tich:

Hai phuong trlnh can bang cua h($ Ive d6ng qui (P

,r, N)

"'F =N-T =0

L.

=

2P

X

2

'\:' F =-P+ T✓
3 =0
L.

' h th,,u 2 t a co' T
T u, ph u'ong tnn

b )

y

2

2 P✓3 h
' h th'u nh,<'
t ay va' o ph u'ong trm
at ta

r;; = - -

-v3

,

3

✓
3

co:N=P3

- Bdng phumig phap hznh h9c:
Tren hlnh b ta c6 tam giac Ive P, N, T la tam giac vuong c6 g6c 60° nen:
T=

p
= 2 P✓
3 · N = P.ct 60°
sin60°
3 '
g

= P✓3
3

Vi d1J 2
Thanh OA, tr9ng lugng khong dang kS, n6i ban 16 voi tu'ong cJ O va chiu
Ive thing dung P cJ A. dS giu thanh nam ngang, dung day BC. bie't OB=
2BA, tinh sue ding cu.a day va phan Ive ba 1€ 0 k:hi:


16
y

1. day BC nghieng 30° v6i thanh
2. day BC thing dung
Gidi:
Xet can bang cu.a thanh OA
* Trztdng h(fp BC nghieng.
Ht$ ll.)'c tac dvng g6m: ll.)'c ho~t d9ng P,
sue ding T' hai phan phan ll.)'c ban IS X: va
~ .Bay la ht$ hfc phing ba't kl nen l~p

C

p
B

Y
Yoo

i

cI~

p

L

1======~B===A-x

du'qc ba phu'ong trlnh can bang:

Hlnh 2-6

✓
3

IX=X0 -T-=0
2

T

IY=Ya+--P=0

2

Imo(F)= -P.OA + T.OB.sin30° = 0 ( tay don cu.a

T la

OH= B.sin30°) hay

-P+r.I.1-=o
3 2

KSt qua : T = 3 .P,· X 0 = 3P ✓3 · Y0 = - P
2 '

2

Trztdng h(fp BC thdng dung
Vl P, T thing dung nen phan ll.)'c ban IS chi c6 thanh ph~n thing dung

Ya.

chung ta c6 ht$ h,tc song song va l~p du'qc 2 phuong trlnh can bang
Im 8 (F)= -P.BA-Y,,.OB. = 0 ⇒ P.2 + Y,, = 0

ImJF)=-P.OA+T.OB.=0 ⇒ -P+T. 2 =0
3

~

?

Ket qua :

3
T =- P ;
2

p

p

Yo=-2

Vf dl! 3

Piston A chiu tac dvng cu.a ll.)'c P c6 tri s6 P = 25kN,
thanh truySn B lam vdi duong thing dung g6c a= 14°.
Xac dinh ap h,tc cu.a piston len thanh xi lanh va Ive
tac dvng d9c theo thanh truySn.
Bo qua tr9ng luqng piston va thanh truySn
H'inh 2-7
Gidi
Piston can bang dudi tac di.mg cu.a ll.)'c da cho P, phan ll.)'c cu.a xi lanh len
piston N va phan ll.)'c cu.a thanh truySn (hlnh ve)
Vie't phu'dng trlnh can bang cho ht$ ta c6
Ix= N -Scos76° = 0
(*)

s

(**)


17

Tu phuong trlnh (**) ta c6
s=
N

25
P O=
= 25,8kN
sin 76
0,9703

thay S vao phu'ong trlnh (*) ta c6

= 25,8cos76° = 25,8.0,2493 = 6,4kN

Ap Ive cua piston len thanh xi lanh c6 tri s6 6,4 kN cung phu'ong va ngu'QC
chiSu voi N, con hfc tac dl;lng Ien thanh truySn co tri s6 25,8k:N cung phuong
va ngu'QC chiSu voi

s

V{ d¥, 4

B

Day treo v~ t c6 tn;mg Iu'qng P = 80 kN

duqc vift qua rong fQC Ava giu bcH toi D.
Rong r9c A c6 ban ldnh khong dang k€.
Xac dinh phan Ive cua cac thanh AB va
AC. Bo qua ma sat d rong r9c, tr9ng lu'9ng
cua cac thanh va day
Gidi

X

C

p
Hinh 2-8

Xet slf can bang cua rong fQC A. cac Ive tac dl;lng len rong fQC c6: tr9ng Ive
p' sue ding day T( T= p vl bo qua ma sat a rong fQC) va cac phan Ive cua
lien k6t thanh s AB , sAC c6 chi€u gia dinh nhu' hlnh ve. Nho rong fQC c6 ban
kinh khong dang kS ta c6 th€ xem

r, P, s

AB ,

s AC la hi$ h;tc ph!ng d6ng qui

Ch9n hi$ trlJc xAy nhu' hlnh ve, ta c6 cac phu'ong trlnh can bang:
LX=Tcos30°-Pcos60°-SAc =0

(*)

LY=-Tsin30° -Psin60°

(**)

+SAB

=0

Tu' phu'ong trlnh (*) c6
SAC

=Tcos30' -Pcos60° =P(cos30° -cos60° )= so(

Tu phu'ong trlnh (**) c6
SAB= Tsin30° -Psin60°

sAC , sAB

=P(sin30° +sin60° )=

s{ ~

1:-~) =

29,3kN

+ ~) =109,3kN

tinh ra dSu la s6 du'ong chung to chi6u gia dinh cua
voi chiSu thtfc

sAC , sAB

dung

V{ d¥, 5

ogm chiu tac dvng cua cac ngftu Ive c6 mo men ml = 20kNrn rn2 = I5kNrn
rn 3 = lOkNrn. Hay xac djnh phan h;tc cf cac g6i Ava B cua agm
Gidi
Tren dfim chi co cac ng~u Ive tac dl;lng, nen d€ can bang phan Ive a 2 g6i
cfing phai l~p thanh mc)t ng~u h;(c. Phan h;(c NB vuong g6c voi m~t tva va c6


18

chiSu gia dinh di len, tu'ong ung phan h;(c ~ phai song song voi N s va c6
chiSu gia dinh di xuo'ng. Theo diSu ki~n can bang (2-13) ta c6:
RAl-m 1 +m 2 -m3 =0
RA= m1 -m2 +m1
l

RA c6 dffu

= 20-15+10 =375kN
4

'

+ chung to chiSu gia

dinh cua_~ la dung chiSu th-L;L'c,
tuong ung Ji;= 3,75kN va chiSu ve A
tren hlnh cfing dung voi chiSu thl;(c

1=4m

Hlnh 2-8

BAIT!P
1. H~ ll;L'c phiing d6ng qui ( hlnh 2-8) g6m cac ll;L'c c6
tri so' ll;(c: ~ = 50N, F;_ = 32N, J{ = 42N, ~ = 60N. Hay
xac djnh hqp ll;L'c cua h~ Ive d6
H1nh2-8

Hlnh 2-9

C

2. Ong tl'\l d6ng chit c6 tr9ng lu'qng P = 60N d~t tren
mang ABC hoan toan nhan va vuong g6c dB. m~t BC
cua mang lam voi m~t nam ngang g6c a= 60° ( hlnh 2-9).
Hay xac djnh cac phan ll;L'c cua mang len o'ng d 2 diSm

tie'o xuc D va E
3. Day ABC va:t qua rong r9c D c6 dffu C
treo v~t nang tr9ng luqng P=l00N, con dffu A
bu(k vao tu'ong. d diSm B tren day treo v~t
n~ng c6 tr9ng htqng G, duoi tac dvng cua v~t
n~ng day lam voi du'ong thing dung nhung g6c

45°va 60°( hlnh 2-10). Hay xac djnh sue ding
cua do<;1.n day AB va tr9ng luqng G cua v~t. Bo
qua ma sat rong fQC va tn;mg lu'qng day

a


19

4. V~t n~ng c6 tr9ng htqng P = 20N treo d dffu cua day
quffn qua rong r9c A va du'QC gifi' bdi tfl;!C toi C ( hlnh 2-11).
Coi ban kinh cua rong r9c A la khong dang kS. Hay xac
djnh phan Ive cua cac thanh AB va AD. Bo qua ma sat a
rong r9c, tr9ng luqng ban than cua cac thanh va day

C

5. Dffm AB chju tac d1;1ng cua cac ngfiu Ive c6 tri
m1 = 80kNm, m2 = 200kNm, m3 = l40kNm, (hlnh 2-12)

s6 mo men la

- Xac djnh phan h;tc d cac gefi Ava B. Bie't AB = 4m,
- Ne'u dffm EF d6ng chfft va c6 tr9ng h.rqng P = 1OkN thl phan Ive d cac
go'i la bao nhieu? Bie't EA = BF

6. Tn;ic cua rong r9c nho duqc da bdi 2 thanh
OA va OB, r9ng ht<;1ng thanh khong dang kS,
nghieng dSu 60° voi duong n~m ngang. vong qua
rong r9c la s<;1i day, dffu treo v~t n~ng P, dffu kia

nghieng 30° voi duong n~m ngang va chju Ive F( =
P) dS giu v~t can bAng. xac djnh ung hfc cua 2
thanh (hlnh 2-13)

ID.nh 2.13


20
'

.,

"'

..2

,,

"

A'

IV. BAI TOAN H~ Ll)'C PHANG CO LIEN KET
MASAT

p

Q

Trude day cac lien ke't du9c gia thie't la r~n va

nhan (khong ma sat) vi dv do'i voi lien ke't tlfa ta gia
thie't r~ng phan Ive lien ke't la phan Ive pha p tuye'n
N
vuong g6c voi m~t, duong tva. Tuy nhien nhi6u
tru'ong h9p thvc te' cho tha'y gia thie't nhu' v~ y la
,...;~~~~777'7
khong dung (tuc ke't qua nh~n du9c tu gia thie't nay
khac xa thvc te'). BS lam r6, xet mo hlnh (hlnh ve)
Hinh 3-5
trong thvc te' v~t tie'p xuac voi v~t gay lien ke't khong phai tc;i.i m<)t diSm ma
theo m<)t duong, do d6 ta c6 h~ cac phan Ive lien ke't. Day la h~ Ive phing thu
gQn v6 diSm I du9c m<)t Ive va m<)t ngfiu Ive. Lvc thu gQn cac phan Ive lien
ke't du9c gQi la phan Ive, c6 thS du9c phan ra thanh hai thanh ph~n vuong g6c
voi nhau:
Thanh ph~n phap tuye'n voi m~t tva c6 tac dvng cho'ng lun, ki hi~u N
du9c g9i la phan Ive phap tuye'n.
Thanh ph~n tie'p tuye'n voi m~t tva c6 tinh cha't can tnt9t duqc g9i la
Ive ma sat ki hi~u la Fms
Ngfiu Ive thu g9n c6 tinh cha't can tra chuySn d<)ng lan, duqc g9i la
ngfiu Ive ma sat Ian, c6 ki hi~u la m,
--------l

Lvc ma sat tru9t (ngfiu Ive ma sat Ian) chi xua't hi~n khi v~t tru'9t ( Ian)
ho~c c6 khuynh huong tru9t (khuynh huong Ian) do'i voi v~t khac. Ma sat xua't
hi~n khi v~t chuySn d<)ng g<;>i la ma sat d<)ng, con ma sat khi v~t chi c6
khuynh huong chuySn d<)ng nhung chu'a chuySn d<)ng du'9c g<;>i la ma sat f'rnh
1.

Ma sat truqt
Fm)a phan Ive lien ke't, n~m trong m~t phing tie'p tuye'n chung cua m~t

tie'p xuc, can trd XU huong truqt hay chuySn d<)ng tru'9t cua v~t khao sat.
Ma sat tru'(Jt c6 cac tinh cha't:
Lvc ma sat tru9t la m<)t Ive c6 gioi hc;tn; nghia la ne'u ki hi~u Ive gioi
hc;i.n la Fmax g<;>i la Ive eve dc;ti thl: O:::; Fms ::; Fmax
Gia tri eve dc;ti cua Ive ma sat tru9t ti I~ voi phan Ive phap tuye'n

N

Fms =f.N

Trong d6 f g<;>i la h~ so' ma sat tru'9t tlnh, duqc xac dinh b~ng thvc nghi~m,
khong c6 thu nguyen. h~ so' f phv thu<)c vao :
Ban chdt crla vq,t lifu ( d6ng, thep, d6, da .... )
Ttrr;mg thai bt mq,t ( trrm, nhcfn, nham ..... )


21

va moi true1ng ( c6 chdt boi tran, hoi;ic khong c6 chcft boi trcJn
Khong ph11, thuqc vao dif n tich tie'p xuc cua 2 b€ mi;it
H~ s6 ma sat dUQC cho san trong bang

Bang hf s6 ma sat tru(/t tinh cua mqt s6 vlj,t Zif u

He s6ma satf
VIJ,t lifu
0,17
Thep voi thep
0,17
Thep voi gang

0,3
Siit voi sit
0,18
Siit voi gang
0,2 - 0,4
G6 ra:n voi g6 rlln
0,1
Bai chuyen da voi g6
0,28
Dai chuySn da voi
gang

2.

Ma sat Ian
Ma sat Ian du'Qe biSu di~n qua mo men ngfiu lge ma sat Ian n6 can trd
chuy€n d9ng Ian.va c6 cac tinh chfft sau:
Gia tri ctfe dqi mo men ngfiu hfc ma sat Ian phl). thUQC vao dQ bi6n
dC;1ng tqi mijt ti6p xuc giua v~t khao sat va mijt ttfa
Gia tri ctfc dqi mo men ng~u hfe ma sat Ian ti 1~ voi phan Itfc phap
tuye"nN
Mmax =k.N

Trong d6 k la h~ s6 ma sat Ian, du<;Jc xac diunh b~ng thlje nghi~m
3.
Bai toan can biing c6 ma sat
Bai toan dilu kii n trJ ham
Khi tac d9ng vao v~t m9t Ive P nghieng voi tn;1c
th~ng dung 1 g6e a n6 se phan ra 2 thanh phffn: ap Ive
phap tuySn N = Pcosa, Ive ti6p tuy6n F = Psina F. Llfc F

c6 xu hu'ong keo v~t chuySn d9ng, nhu'ng do xufft hi~n
Itfe ma sat ti I~ voi ap Ive N: Fms ~ f.N nen n6 se can
chuySn d9ng. N6u F ~ Fms thl v~t kh6ng chuySn d9ng,
con F ~ Fms v~t se ehuySn d9ng. Do v~y diSu ki~n dS v~t tv ham
., anh?o h on goe
.,
Ia, F ~ Fms = f.N ⇒ f ~ -F = tga ⇒ tga ~ tgrp = f h ay goc
N

rp.

N 01
,..


22

each khac h<;1p Ive d~t vao( Ive ho<:tt d(,ng) cat ben trong n6n ma sat thl v~t
khong chuySn d(,ng
Vi d11, 1
M(,t v~t c6 tr9ng ht<;1ng P nam tren m~t
nham nghieng voi m~t phiing nam ngang m()t
g6c a, h~ s6 ma sat la f. Xac dinh g6c a dS
v~t khong bi tnt<;1t.
Gidi
V~t ran chiu tac d1:1ng cua ngo<:ti Ive P
X
(tr9ng lu'cjng cua v~t ), n6 se can bang voi
Hinh 3-6
P

1
1
phan Ive cua m~ t nghieng tac d(,ng Ien n6 P • Lvc P phan ra 2 thanh phffn:
ap Ive N va thanh phffn tie"p tuye"n ma sat Fms = fN. d tr<:tng thai can bang v~t
ran chiu tac d1:1ng cua 3 Ive (P,Fms,N)= 0
Cac phu'ong trlnh can bang c6 d<:tng:
IFx = Fms -Psina = 0

⇒

Fms = Psina

_LFy = N-Pcosa = 0

⇒

N = Pcosa

Ngoai ra ta c6: Fms :s: f.N thay vao ta c6 Psina :s: f.Pcosa
sma
cosa

⇒ f~--=tga

hay

f~tga

Vi d11, 2
Mu6n ham cho banh xe khong quay duoi tac d1:1ng cua ng~u Ive c6 tri s6

mo men m=l00Nm, nguoi ta tac d1:1ng 2 Ive trvc d6i Q vao 2 ma ham. Hay
tinh tri s6 nho nha't cua Ive Q dS banh xe khong quay. Bie"t h~ s6 ma sat giua
ma ham voi banh xe la f=
0,25 va duong kinh banh xe
Q
Q
d = 0,5m
Gidi
Banh xe can bang duoi
H'mh3-7
tac d1:1ng cua 2 Ive Q va Fms ,
phan Ive d 6 tfl:lc ]4, tr9ng Ive cua banh xe P va ng~u Ive, nhung chi cffn tun
Q nen ta chi vie"t 1 phuong trlnh
m0

(F)= Fms·d-m = 0

(*)

Mu6n Q c6 tri s6 nho nha't thl Ive ma sat phai Ion nha't, tuc la
Fms

= Fmax = f .Q

(**)
Thay ~ d (**) vao (*) ta c6

f.Q.d- m= 0 nitra

23

Q=...!!!_= lOO =800N
f.d 0,25.0,5

Ntefu Q < 800 N thi banh xe quay, con Q > 800 N thl banh xe khong quay,
khi d6 hfc ma sat cl;(c d~i Fmax tang len nhung ll;(c ma sat thl;(c ttef d 2 ben khong
tang ma chi t~o thanh ngfiu hfc c6 trj s6 mo men la lOONm dg can bAng voi
ng~u h;tc tac dvng