Physics...anyone good???

Question

1. suppose the motor has a bearing shaft diameter of 3.2 cm and the coefficient of friction of each bearing is 0.008. find the torque of friction of bearings.

a. 0.0320 N-m

b. 0.314 N-m

c. 0.411 N-m

d. 0.481 N-m

2. if the motor is running at 3000 rpm and is capable of developing a torque of 120 N-m and is used to drive 3-to-1 reduction gears, what torque is the output shaft from the reduction gears capable of producing?neglect friction losses.

a. 13.3 N-m

b. 40 N-m

c. 120 N-m

d. 360 N-m

e. 1080 N-m

Thanks! 10 points for the best answer...

Answer 1

(Torque required to overcome bearing friction) =

(Radial or axial load) * (coefficient of friction) * (bearing bore diameter / 2)

(Torque) = (F) * (0.008) * (0.032 / 2)

Without the load value (F), the torque cannot be calculated.

(Output torque) = (Input torque) * (gear ratio)

(Output torque) = (120) * (3 / 1) = 360 N.m

For a reduction gear, the output speed is slower than input speed, therefore gear ratio is always greater than 1.