解:重物A mya m,g-FTI 鼓轮B: moa= f T2 F m18 m2p2c= Fror+FR TI B 运动学关系: ao= re aA=(R+r)e mA R • r O FN F m2g FT2 ε FT1 m1g aA 解: 重物A: m1aA= m1g – FT1 鼓轮B: m2aO = FT2 – F m2ρ 2 ε = FT2r + FR 运动学关系: aO = Rε FT1 = FT2 aA=(R+ r)ε aO