Understanding output torque

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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.
 

BvU

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Is torque increased or decreased
It stays the same.

What you should worry about is the force needed to lift the weight over 20 cm in 1 s:
##{1\over 2} ({F\over m} - g) t^2 = 0.2 ## m ## \Rightarrow F = 208 ## Newton at least. With a 10 cm crank you can generate a maximum lifting force of ##6/0.1 = 60 ## Newton.
Not enough by a long shot.

[edit] You might consider using a counterweight
[edit] The calculation is approximate - you'll need more because you also need time for braking
 
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The online Calculator I used only said I needed 2 newton to achieve the acceleration of .2 m/s/s but I’m guessing it didn’t account for a vertical lift. A counter weight could work. Also I still confused as to the output. In the above answer it’s quoted that a 6 newton metre output is equal to 60 newton with a 10 centimetres crank. I thought the 6 newton metre quoted by the manufacturer meant that if a 1 metre lever was attached to the shaft a force of 6 Newton’s was available at the end of the metre long lever. Also the answer states that torque stays the same but it’s increased 10 times. What exactly does 6 newton metre mean.
 
The online Calculator I used only said I needed 2 newton to achieve the acceleration of .2 m/s/s but I’m guessing it didn’t account for a vertical lift. A counter weight could work. Also I still confused as to the output. In the above answer it’s quoted that a 6 newton metre output is equal to 60 newton with a 10 centimetres crank. I thought the 6 newton metre quoted by the manufacturer meant that if a 1 metre lever was attached to the shaft a force of 6 Newton’s was available at the end of the metre long lever. Also the answer states that torque stays the same but it’s increased 10 times. What exactly does 6 newton metre mean.
That is correct. So the lever is 0.1 m and the torque M can be expressed as M = F*r (so F= M/r), where r is the length of the lever. As such, the force at the end of the 10 cm lever is 60 N, as BvU explained.

Let's say you were to fasten a bolt. Then a spanner with a longer lever would make it easier to do so because the lever "amplifies" the force you apply as it results in applied torque to overcome the resistance that keeps the bolt from rotating. Lets say the resistance that needs to be overcome is 6 Nm, then a spanner with a length of 1 meter would require you to apply more than 6 N to the spanner and if it would have a length of 10 meter it would only require more than 0.6 N.

Because in this case the torque originates in the motor, which in the above example would be in the bolt's location, it would work the other way around. So if the motor would generate a torque of 6 Nm it would result in a force of 6 N on the end of a 1 meter lever and in a force of only 0.6 N on the end of a 10 meter lever. In case the lever would be only 0.1 meter, than it would result in 60 N on the end of the lever.
Hope this makes it a bit more clear to you!
 

BvU

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the answer states
So, is this homework ?
A counter weight could work
Or an actual design ?
With a counterweight you would have to accelerate/decelerate twice the mass, but you still have plenty torque to do it. Can you do the calculations with what you now know ?
 
It’s a build. I have a motor and mount driving a saw cutting boards. The saw has to cut through 20 centimetres of board in about 1 second and return to its home position in another 1 second so one revolution per cut with a 10 centimetre crank. I realise the 50 rev per minute motor will not achieve this but it’s close enough. The problem I have is power to drive the saw. I have limited options as it’s a mobile plant so I’m looking at a 5 hp petrol motor with manufacturer spec weight of 16.5 kg plus mount and others such as bearings and worked on a total of 20 kg to move. The other option is a 12 volt electric which is a much lighter option and about 1/5 of the H.P ( the electric is about 1.2 hp ) so speed of cut will be slower. The saw is expected to move up and down within 2 seconds and do this every 5 seconds all day long
 

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