as the motor assembly is 240 mm off centre to hinge
both pulleys are of the same diameters.
the angle of wrap = 180 deg
taking moment at the hinge point
case 1
if the motor is rotating in the clockwise direction
hence the upper part will be the Tight side and the lower part will be the slack side
so the tension will be generated in the tight one
total tension on the belt (T1) * eccentricity = mass of motor *g*eccentricity
where g is the acceleration due to gravity = 9.81m/s^2
T1 *(300*10^-3)= 20.0*9.81*(240*10^-3)
T1 = 156N
Torque which can be transmitted = T1*r = 156*0.1 = 15.6Nm
case 1
if the motor is rotating in the counterclockwise direction
hence the upper part will be the slack side and the lower part will be the tight side
total tension on the belt (T2) * eccentricity = mass of motor *g*eccentricity
T2 *(100*10^-3)= 20.0*9.81*(240*10^-3)
T2 = 470.4 N
Torque which can be transmitted = T2*r = 470*0.1 = 47.6Nm
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