Question

Motor 200 mm 200 mm 5. A 20.0-kg motor provides power to a machine using a belt around a 200-mm pulley as shown. The motor is mounted as shown to produce tension in the belt. The coefficient of friction between the belt and the pulley is 0.400. a) Calculate the maximum torque that can be transmitted with the motor rotating clockwise. b) Calculate the maximum torque that can be transmitted with the motor rotating counter clockwise. --- 240 mm

Answer 1

Motor 1200 mm 200 mm TUITA 240 mm mg

as the motor assembly is 240 mm off centre to hinge

both pulleys are of the same diameters.

the angle of wrap $\theta$ = 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