2.用能量指示图求△Wmx 由能量指示图 △wmax=△web +125 =△Wcb+△Wd+△we 550+(-100)+125 a b f g h =575(Nm) △Wmax 3. JF T8]Om ↑-100+12 △Wm 4 6]-)2 900△Wmax [6]π (IN= lkm/s2) =60.159453≈60.5 0.06×兀2×1202a b -50 c 2. 用能量指示图求DWmax +550 -100 d +125 e -500 f +25 g h DWmax 由能量指示图: DWmax = DWeb = DWcb + DWdc + DWed = 550 + (-100) + 125 = 575 (Nm) 3. JF = ——— DWmax [d] wm 2 = DWmax [d]( —— 2pnm 60 = 900 DWmax [d]p 2 nm 2 = 900×575 0.06×p 2×1202 = 60.159453 ≈60.5 (kgm2 ) ( 1N = 1kgm/s 2 ) M (Nm) f Med Mer a b c d e f g h -50 -100 +125 -500 +25 -50 +550 ● 11 -50 ) 2