Gioi thi~u
- Bai h9c nay cung cdp cha h9c sinh vi each phan tich h? hrc va ai€u ki?n
can bling cita h? Ive
Ml}.C tieu thl}.'C hi~n:
- Trinh bay ilu(YC hai aij,c trung CO' ban cita h? Ive.
- Tinh toan au(Yc CClC ag,i lu(YYlg vie tO' theo CClC cong thuc giid tfch eta h9c.
- Viit au(YC CClC phucmg trinh can bling cita h? Ive khong gian va h? Ive ph¢ng aon gian.
N{,i dung chinh:
5.1. Hai d~c tnmg CO' ban cua h~ h;rc
5.2. Thu g9n h~ h;rc
5.3. f>iSu ki~n can bfulg va h~ phuong trinh can bfulg cua h~ Ive khong g1an
5.4. f>iSu ki?n can brtng va h? phuong trinh can brtng cua h? Ive phfulg
Cac hinh thfrc hqc t,p:
- Di€u ki~n can bfulg cua h~ lµc phfulg, h~ lµc khong gian nqi dung: 5.1. Hai dJc trung CO' ban cua h~ I-.,c 5.1. Hai dJc trung CO' ban cua h~ I-.,c
Hai d~c tnmg CO' ban cua h~ li;rc la vecta chinh va momen chinh.
5.1.1. Vee ta chinh cua h? hrc:
Cho h~ lµc phfulg ( F1 , F2 , •• ••• Fn) (ffinh 11.5 .1)
Hinh 11.5.2
V ecta chinh cua h~ lµc ki hi~u R , la vecta t6ng cua cac vecta lµc cua h~ lµc:
N _
R' = F; + F2 + ... + FN = LFK
K=l (5-1)
Xac dµih vecta chinh: c6 the sir d1,1ng phuang phap ve da giac lµc
Trong truong hqp nay da giac lµc la da giac phfulg (Hinh 11.5.2). Cung c6 the xac dµih vecta chinh qua cac hinh chi~u cua n6 tren hai trl)c to..i d9 vuong g6c :
N
R'x = f1x + f2x + ... +fNx = L F kx k=I
N
R'y= F1y + F2y + ... +FNy= Iry
k=l
(5-2)
Tri s6, phuang va chi€u cua vecta chinh duqc xac dµih theo cong thuc:
- R' - R'
R'= .J R'2 + R'2
· cos(Ox R' )= _..::.. · cos(Oy R' )= -2:..
X y 1 J R',, ' R' (5-3)
5.1.2. M6men chinh cua h? lvc:
V ecta mo men chinh cua h~ Ive d6i v6i tam O la vec to t6ng cua cac vec ta momen cac lµc trong h~ liy d6i v6-i tam O (Hinh 11.5.3). N~u kf hi~u momen chinh la.Mo ta c6:
n