- #1
Ripcrow
- 56
- 5
- TL;DR Summary
- I have an electric motor that produces 6 nm (4.4ftlb) of torque and rotates at 50 rpm through gearing. I have a weight I need to move. It’s a 20 kg weight that needs move up and down 20 cm with one second to move down and one second to go back up.
Summary: I have an electric motor that produces 6 nm (4.4ftlb) of torque and rotates at 50 rpm through gearing. I have a weight I need to move. It’s a 20 kg weight that needs move up and down 20 cm with one second to move down and one second to go back up.
The electric motor rotates at 50 rpm and has a torque output of 6nm. I need to attach a crank to the output shaft to create 20 cm of movement. My problem is will this motor be enough to produce the required power at the end of the crank. Online calculators say I need 2 nm to move this 20 kg weight with an acceleration of .2 m/s/s. I understand how applying torque to a shaft makes its rotate but I’m a bit confused as to how torque acts when it is the shaft applying torque to a lever. Is torque increased or decreased. If I attach a 10 centimetre crank will the available torque be increased or decreased.
The electric motor rotates at 50 rpm and has a torque output of 6nm. I need to attach a crank to the output shaft to create 20 cm of movement. My problem is will this motor be enough to produce the required power at the end of the crank. Online calculators say I need 2 nm to move this 20 kg weight with an acceleration of .2 m/s/s. I understand how applying torque to a shaft makes its rotate but I’m a bit confused as to how torque acts when it is the shaft applying torque to a lever. Is torque increased or decreased. If I attach a 10 centimetre crank will the available torque be increased or decreased.