Hi everyone,

So I know the diameters of the pulleys (5 cm), and the weight that will be attached on the other end (4 kg). I need to know how much torque I am going to need in kg * cm, so that I am able to lift this object up. I have tried solving it, however I'm stuck on what to do with two pulleys. Any help would be greatly appreciated!

-John B

 

Depends on the diameter of the motor pulley.

 

The pulleys in your diagram only serve to change the direction of force, the diameters aren't relevant to your torque calculation.

What is important is the diameter of the pulley on the motor.

 

Another factor to consider is how fast you want to lift it and, possibly, how fast you want to accelerate it (how long you are willing to let it take to go from rest to full speed).

 

Torque = 4 kg x radius of motor pulley, in theory. In practice, allow for extra torque needed to overcome frictional losses in the pulley system.

 

Torque = 4 kg x radius of motor pulley, in theory. In practice, allow for extra torque needed to overcome frictional losses in the pulley system.

Another factor to consider is how fast you want to lift it and, possibly, how fast you want to accelerate it (how long you are willing to let it take to go from rest to full speed).

The pulleys in your diagram only serve to change the direction of force, the diameters aren't relevant to your torque calculation. What is important is the diameter of the pulley on the motor.

Depends on the diameter of the motor pulley.

Thank you for replying, the diameter of the motor shaft is 0.25 inches or 0.635 cm. However in order to pull the rope down and wind it up, the motor is connected to a cylindrical object with a radius of 0.517 inches or 1.313 cm. so based on this, I have found that I need a motor with more than 5.85 kg*cm of torque. Am I right? or is there some loss of torque because of the change (using the cylinder)?

 

How is the cylinder connected to the motor shaft? Directly? Or is there some kind of gearing or a belt drive?

You will need more torque (which should be in newton·cm) than just the load (which should be in newtons, not kilograms) times the radius of the drive pulley (and then taking any drive ratio into account) because there are frictional losses just about everywhere. If you aren't pulling too fast (since the frictional load will increase with speed), then you will probably be okay by doubling the no-friction torque unless it makes an unacceptable difference in motor options.

 

How is the cylinder connected to the motor shaft? Directly? Or is there some kind of gearing or a belt drive? You will need more torque (which should be in newton·cm) than just the load (which should be in newtons, not kilograms) times the radius of the drive pulley (and then taking any drive ratio into account) because there are frictional losses just about everywhere. If you aren't pulling too fast (since the frictional load will increase with speed), then you will probably be okay by doubling the no-friction torque unless it makes an unacceptable difference in motor options.

The cylinder is connected to the shaft directly, and thank you for the advice!