- Ht phuung trinh ciin bdng:
Bai 06: BAI TOAN 11¥ V!T
Gi6i thi~u
Giup h9c sinh biit GU(IC each giai cac bai toan phuc tl;lp, ung d1;mg
thvc t~ Mvc tieu thvc hi~n:
Gicli GU(IC m(jf s6 bai toan vi hi v(it. Ung d1,tng ai giai m9t bai toan thlfC ti.
Nqi dung chinh:
6.1. Khai ni~m cac bai toan v~ h~ v~t
6.1.1. Bai toan aJn
6. 1. 2. Bai toan hi v(it
6.2. Phuong phap giai cac bai toan vS h~ v~t
6.2.1. Phuong phap hoa
,
ran
6.2.2. Phuong phap tach v(it Cac hinh thuc h9c t, p:
Hoc tren lop vS cac khai ni~m va cac phuong phap giai - hoc theo nh6m.
Nqi dung:
6.1. Khai ni~m cac bai toan vi h~ vit
6.1.1. Bai toan aJn
DSn la m9t v~t rful c6 th~ quay quanh m9t tf\lc c6 djnh O du6i tac d\lllg cu.a m9t h~ lµc nfun trong m(lt phfulg vuong g6c v6i m(lt d6.
Sir c6 m9t dSn quay dugc quanh m9t t11JC O chju b&i h~ lµc phfulg (F1 - ,F 2 , ••••• F_i) (Hinh 11.6.1). ~goai c~c lµc t!c c!ltng tren_t~i_!.rvc quay O xu~t hi~n m9t phan lµc R0 • Vi den can bang nen h~ lµc: (F1, F2 , .. .. .FN ,Ro)~ 0
DiSu ki~n can bfulg cu.a h~ lµc la:
LX=Rox + LFkx: =0 LY=Roy + L~-v =0 LY=Roy + L~-v =0 Lmo (.Fk) = 0 (6-1) (6-2) (6-3) Hinh 11.6.1 6.1.2. Bai toan hf wjt
Djnh nghia: h~ v~t la m<)t h~ g6m dS 4, n<)i lµc c6 tung doi m9t trµc
nhiSu v~t lien ket vai nhau. Cac lµc tac d\lllg lfui cac v~t thu9c h~ g6m hai lo~i lµc la ngo~i lµc va n9i lµc.
Ngo(l,i Ive: La cac lµc tir bfui ngoai h~ tac d\lllg lfui cac v~t thu9c h~.
Npi Ive: La nhfrng lµc do cac v~t thu9c h~ tac dvng lfui nhau. Do v~y theo tien dS 4, n9i lµc c6 tung doi m9t trµc d6i nhau.
a Q
Hinh 11.6.2
Vid1p
M9t du hinh c~u hinh vSm c6 ba kh6p a A, B, C. Tr9ng luQ'llg m6i phful la -
P1, P2.
tr§.n phful AC c6 dijt v~t n~g la Q. (Hinh 11.6.2)
Day la bai toan h? v~t g6m c6: phful AC va BC cua c~u. Ngo~i lgc c6 P1 , P2
, Q va cac phan Tue RA, RB. N9i lgc chi c6 h;rc lien k€t t~C la Reva R-; vai -
Rc=-Rc'
Chu y: tr9ng luQ'llg cua v~t bao giCY ding Ia ngo~i h;rc.
6.2. Phll'ong phap giai cac bai toan v~ h~ v~t
6.2.1. Phuongphap tach vgt
La xet tung v~t rieng re can b&ng du&i tac dl,lng cac lgc tn;rc tiSp dijt Ifill v~t d6. Cu m6i v~t ta l~p duqc ba phuong trinh can b&ng va giai phuong trinh ta duqc kSt qua.
Vi dl,l: Cho m9t h? g6m hai thanh AB va BE nhu hinh 6.3. Thanh AB c6 tr9ng luQ'llg P1 = 200N, thanh be c6 tr9ng luQ'llg P2 = 160N v&i AB = a, BD = .!.a ,
3
BC=b,
EC= .!.b. Tim phan lgc t~i A, D, E. BiSt a= 30°.
3 A D B A D B - ll Hinh 11.6.3 Bai giai: C
Day la h? g6m hai v~t: thanh AB va BC. ap dl,lng phuong phap tach v~t, ta khao sat tung v~t m9t. Xet thanh AB can b&ng. H? h;rc tac dl,lng Ifill thanh g6m c6 Pi phan h;rc t~ A la X A, YA , phan lgc t~i fi la ND (g5i di d9ng) va
n9i lgc ti;ti ban IS B la~, Y-;. H? lgc nay can b&ng: (P1 ,x;, Y;, x-;, ~, fl;
LX=XA+XB=O (1)
LY=YA+YB+Nn-Pi=O (2)
""' - a 2
£...,mA(F)=--Pi +-a.ND +a.YB =0 (3)
2 3
phuong trinh cua h~ h,rc la:
Xet thanh BE can bfulg v&i cac Ive tac dl)ng la: (P2, N2, X B ,YB) ~ 0 Phuong trinh can bfulg la:
Giai phuong trinh (6) v&i chu y la: _ X~=-Xi; Y~=- YB Ta tim duqc kSt qua N = ~ P . Cosa = 60 N
4
LX =X~+NE_sina=O
Ly =-YB +Pi +NE.cosa=O
'°' - 2 b
.i..JmB(F)=--b.NE ---Pi.cosa=O
3 3
Ti'r (4) ta c6: XB= xB· =NE.sina = 51,96N
Tuong tv ta tim duqc:
YB= YB'= -P2+ NE.cosa = -130N
3 3
Nn=--YB+-Pi
2 4
XA =-XB =-5l,96N
6.2.2. Phuong phap hoa rdn
(4)
(5)
(6)
La xem ca h~ nhu m9t v~t dn can bfulg du&i tac dl)ng cua cac ngo<;1i Ive d~t lfui h~ (n9i Ive tri~t tieu 1§.n nhau tung doi m9t) ta chi l~p duqc ba phuong trinh can bfulg. Sau d6 ta c6 thS tach thfim m9t s6 v~t dS khao sat va l~p thfim nhilng phuong
trinh can bfulg m&i cful thiSt dS giai
Vi d\l: Cho m9t h~ g&m hai thanh ab va be can bfulg, chiu cac Ive tac d\lllg J\ ,P2
va kich thu&c nhu hinh ve. Hay xac dinh phan Ive t<;li A, E.
Pi =lOON Pi =200N
D
Giai:
Day la bai toan h~ v~t, g6m hai v~t la AC va BC. D~ giai bai toan nay thu~ ti~n hem, dung phucmg phap h6a dn. NA j
Pi =120N
Pi =200N N-
E
Ngo~i h;rc tac d\,lng lful h~ v~t g6m 0=====
0~=~):
8==:t==:::f c
hai tr9ng luQTig P1 , A_ con c6 phan lµc t~i A va E la: NA, NE va thanh AC
can bfulg nen h~ lµc can bfulg ( P1 ,P2,
NA, NE)~ 0.
L2a----+11+-a --a
Phuong trinh can bfulg : (1)
NE= Pi.2.a+P2.5a = 2.200+5.120 = 250N
4a 4
Tu phucmg trinh (2) ta c6:
NA= 250 -200 -120=30 N