Assume Fa is 0.4m and Fb is 0.3 m from the axis of rotation 2.
mass of apparatus = 19 kg. At rest,
Fa + mg = Fb
Fb - Fa = 19*9.8 = 186.2 N
0.4Fa = 0.3Fb
Fa = 558.6 N Fb = 744.8
Στ = Iα, I = mr2 = 9.8 kg * 0.15 m = 1.47 kgm, assume α = 30 rad/s2
torque = 1.47*30 = 4140 Nm
This torque generated is big enough to balance out the torque created by the weight of the wheel and make it feel lighter.