机械原理作 业题 解第二章机构的结构分析F 3n 2pl ph33241 0F = 3n(2pl + ph )= 34(2 5+1) =134578291-1F = 3n(2pl + ph p)F= 38 (2 10 + 2 0)1= 14(2-3)16F = 3n(2pl + ph )= 33(2 4+ 0) =1F = 3n(2pl + ph p)F= 34 (2 5 +10)0= 1F = 3n(2pl + ph p)F= 37 (2 8+ 20)2= 1p = 2pl + ph 3n = 2 3 + 0 3 2 = 0= 311(2 17 +0 2)0= 1F = 3n(2pl + ph p)F= 2 10 + 0 3 6 = 2p = 2pl + ph 3n(1) 未刹车时n =6 ,pl =8 ,ph =0 ,F=2(2) 刹紧一边时n =5 ,pl =7 ,ph =0 ,F=1(3) 刹紧两边时n =4 ,pl =6 ,ph =0 ,F=0机械原理作 业题 解第三章平面机构的运动分析P 34题3-1 试求图示各机构在图示位置时全部瞬心。a )AB312P 12P 23P