(2)F=-ma"=-mrwi,F=-ma:=-mra M。=-Ja=-mwa (各2分) 九、解: (1)动能定理 T=0 1 T-7J0i+ 1G g J0+ -(2G+G)w (2分) 2 W=Ghsina-Gh (2分) (2G+G)=(Gsina-G,)h 28 a=(Gsina-G) (2分) (2G+G) (2)∑M(F)=0: Ja+Gar-Gr+Tr-0 (2分) T=G,+(2G+G)sinaG (2分) 2(2G+G) ∑F(F)=0:X。-Tcosa=0X。=Tcosa (2分) >F(F)=0:Y+Ga-(G+G+Tsina)=0 (2分) Y=(G+G,+Tsina)-Ga (1)a=sina-Gig 答 2G+G1 (2)8e 3G+(2G +G)sin ag 2(2G+G1) (3)Nx= Gcosa3+(2+)sin@] 2(2G+G) G N,=2(20+G) (4G+6G+[5G +(2G +)sin a]sin a) 共4页第4页(2) 2 F ma mrw n in = − c = − , F ma mr it = − c = − 2 2 3 M J mr iO = − o = − (各 2 分) 九、解: (1)动能定理 (2 分) (2 分) g G G G G a (2 ) ( sin ) 1 1 + − = (2 分) (2) M (F) = 0 : O 1 0 1 + ar − G r + Tr = g G J o (2 分) G G G G G G T 2(2 ) 3 (2 )sin 1 1 1 + + + = (2 分) F x (F) = 0: X o −T cos = 0 X0 = T cos (2 分) F y (F) = 0: ( 1 sin ) 0 1 + a − G + G +T = g G YO (2 分) a g G Y G G T 1 0 1 = ( + + sin) − 共 4 页第 4 页 W = Ghsin −G1h G G h g G G v ( sin ) 2 (2 ) 1 2 1 = − + 2 1 2 1 2 2 2 2 2 1 1 (2 ) 2 1 2 1 2 1 2 1 2 1 0 G G v g v g G v J g G T J T = A + o + o + c = + = O J0 G YO XO T G1 a g G1