解:①FA=G=98N;SA=a2=(0.1m)2=0.01m²
PA=FA/SA=98N/0.01m² =9800Pa
②由于AB受到的重力均为98牛,因此AB的质量之比为1:1,即mA:mB=1:1;
VA=a3=(0.1m)3=0.001m³;
VB=b3=(0.2m)3=0.008m;
VA/VB =0.001m³/0.008m³=1/8;
ρA/ρB=(mA/VA)/(mB/VB )=1/(1/8)=8;
③设切去厚度为h时PA′=PB′;
即:ρAg(0.1-h)=ρBg(0.2-h);
解得h=3/35m≈0.086m;
当h<0.086m时,PA′>PB′;
当h>0.086m时,PA′<PB′;
当h=0.086m时,PA′=PB′;