If one person is applying an upward force equal to 60 N at one end, at what location on the board is the other person applying an upward force

When one person applies force f1 = 60 N at one end, the other person applies f2 = 160 - 60 = 100 N. The torques about the center must balance. Thus the distances from the center d1 and d2 must be such that f1*d1 = f2*d2, or 60*1.5 = 100*d2, so d2=0.9. Thus f2 must be applied at a point 1.5 + 0.9 = 2.4 m from the end where f1 is applied.

at right angles to each other