(1) The stress diagram of the depressurized counterweight object when the air pressure in the tank is 2 standard atmospheres is as follows:

(2) From p=FS, it can be obtained that the maximum upward pressure of the gas in the tank on the decompression counterweight object is:

F steam =p steam S = 2×/kloc-0 /×105pa× 6×10-6m2 =1.2n;

(3) When the steam has just reached 2 standard atmospheres, the pressure limiting valve is in equilibrium.

Then f steam =F atmospheric pressure +G, that is, p steam S=p atmospheric pressure S+mg,

The mass of the counterweight object:

M=p steam S+p atmosphere Sg

= 2× 1× 105 pa×6× 10？ 6m2？ 1× 105Pa×6× 10？ 6m2 10N/kg

=0.06kg=60g。

Answer: (1) The stress diagram of the decompression counterweight object when the air pressure in the pot is 2 standard atmospheres is as shown in the above figure;

(2) The maximum upward pressure of the gas in the tank on the depressurized counterweight object is1.2n; ;

(3) The mass of the pressure reducing counterweight above the pressure limiting valve shall not exceed 60g. ..