tana=-=0.1200.a=6052a为与x方向的夹角 F=61N.F.=0.73N.F +F2=6.14N 解法二应用余弦定理、正弦定理解三角形 F mmvf 2mv; v cOS105=6.14×102Ns F△t 6·14N △ sin o sin 1050 sinb=0.7866.0=5152 =F△ .a=51052-4506052 v my (下一页)解法二 应用余弦定理、正弦定理解三角形 （下一页） tan = = 01200, = 6 0 52' 为 I 与 x 方向的夹角 x y I I Fx 6 1N,Fy 0 73N,F Fx Fy 6 14N 2 2 =  =  = + =  I F t m v m v m v v Ns 0 2 1 2 2 2 2 2 2 1 2 2 cos105 6 14 10− =  = + − =     N t I F = 614  =   0 2 sin sin 105 mv Ft =  0 ' sin  = 07866, = 51 52 51 52' 45 6 52' 0 0 0  = − = mv2 mv1 mv1 I = Ft   