... ()nnnAkBkCkk1k1k1mF0;mF0;mF0======∑∑∑rrr (3.27) Chứng minh: Ta có: () ()nnABAkBkk=1k =1 M=mF=0;M=mF=0∑∑rrtheo (3.9) ta có: 14 () ()()() ()()() ()()nnnOxOxkxkkkzkkyk1k1k1nnnOyOykykkkxkkzk1k1k1nnnOzOzkzkkkykkxk1k1k1MmFmFyFzFMmFmFzFxFMmFmFxFyF============−===−===−∑∑∑∑∑∑∑∑∑rrrrrr ... ()()nnOOkkkk1k1MmFrF′′==′==∧∑∑uuruurrrr ()()nnOOkkkk1k1MmFrF====∧∑∑uuruurrrr ⇒ OM′uur - OMuur=()nkkk1rF=′∧∑urr - ()nkkk1rF=∧∑urr=()nkkkk1rrF=′−∧∑rrr ... năng lật quanh A. LaätM=P2x( 2-0 ,5); ChoángM= P 1 x0,5 + Qx(a+0,5) LaätM ≤ ChoángM⇔ 1, 5P2≤0,5P 1 +Q.(a+0,5) ⇔ a≥ 5,0 10 605,0 10 5, 1 −×−×= -2 m (< 0). Vậy cần trục không...