∑W1=mgp(1-sin60° mgp(1-sin 60)=mv 2/2 n=(2-√3) B mac=0 B|60° F+FB-mg FB(b2)-E4(b2)=0 mg 即可解出F和FB而 b A B 60º FA C FB mg aCn aCτ ∑Wi = mgρ(1－sin 60º) mgρ(1－sin 60º) = mvC 2 / 2 ( ) 2 3 C Cn a a g = = − maCτ = 0 maCn = FA+FB – mg FB (b/2) – FA (b/2) = 0 即可解出FA和FB