Answer
The force required to open the bottle is between 20N and 50N
Work Step by Step
We sum the torques about the left end of the bottle opener;
$(+ \circlearrowleft) \sum \tau =0$
$F_C*9mm-79mm*F=0$
$F=(9mm*F_C)/(79mm)$
We then substitute in 200N and 400N for F;
$F=(9mm*200N)/(79mm)$
$F\approx20N$
$F=(9mm*400N)/(79mm)$
$F\approx50N$
