need a help

Should be around - 0.175m

how to got this answer? please give me the detail? thanks

I found this picture (see the first attachment). Without it, the problem could not be solved.
From the picture we see that the cup 1 is placed at the left edge of the board.

Because the system is in balance, the total torque should be 0.

I will upload a picture were I draw all the forces at the system (see second attachment)
W is the weight of the board. It is placed at the center of gravity. Since the board is uniform, the distance of the center of gravity from either edge of the board is 2/2=1m. That means that W is applied 0,05m to the left of the balance point. Lets name this distance d.


As we said, the total torque should be 0. So ΣT=0.
We have: ΣT= W1*d1+W*d -W2*d2 <=> 0= 0,2*10*1,05+2*10*0,05- 0,4*10*d2<=> 0= 2,1 +1 -4,5125 -4*d2 <=> -3,1=-4*d2 <=> d2=0,775

That means that the distance of cup 2 from the balance point is 0,775 to the right,