P=qa vga 3解:由∑MB=0得R 4 由∑MA=0得RB pga gd Q1= RA M2=M=R B 4 23=O 2=Ra-ga ga 2 go -ga M3=RA·2a A -q·a 2 R ga B 4 4解：由 MB = 0得 R qa A = 5 4 由 MA = 0得 R qa B = 7 4 Q R qa 1 A 5 4 = = M R a qa 1 A 2 5 4 =  = Q R qa qa 2 A 4 = − = M2 = Q3 = M R a qa a qa 3 A 2 2 3 2 =  −  = Q qa R qa 4 B 3 4 = − = − , M qa 4 2 5 4 =