How do you calculate the amount of torque you need to pull an object?

  • Thread starter qpham26
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  • #1
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HI guys, I am trying to build a pulley with a gear box.

So I want my pulley, which is operate by a 3V DC motor with max speed of 8300 RPM.
I think the size of the output shaft is 3mm

Ok, so Can someone help me with how to calculate the original amount of torque I can get out of this motor?

and let say the I can connect the shaft to a gearbox with ratio of 300 to 1.

Will I be able to pull about 10 lbs with this gear ratio?
Assuming I am pulling it over flat surface?

Thanks alot.
 

Answers and Replies

  • #2
Simon Bridge
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You'd want the internal dimensions of the motor - the intrinsic torque depends on the EM force applied to the coils and how wide the coils are.

It's probably faster, and more accurate, to measure it.
Put a wheel on the axle to it will wind a string - use the string to pull a spring.
When the motor stops - measure the extension of the spring.

Probably less drastic - just get it to lift a known weight a known height and time it.
 
  • #3
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I only know that the motor is the regular small size 3v DC one, 3mm shaft i believe. Other than the speed of 8300 RPM, the cover of the motor doesn't say anything.

And I googled for the equation for calculating the power or torque of the motor. However, all of them involve either knowin the power and get torque or the opposite. That is why I dont know where to start, so that I can choose the best gear ratio and output shaft for the lifting purpose.
 
  • #4
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So, let say I know the radius of the pulley which I will use to pull the object. Lets say we are doing it vertically.
then the torque that I need to overcome the weight of the object will have to be
T = r x mg ???

and with this, what are the other information that I need to obtain in order to find out the right gear ratio for the job?
So far I have the RPM of the motor, 3V, 0.98A
I know nothing about electricity so I really dont know what would be the output torque for this tiny motor.

sry, I have to order the gearbox online and cant really return it, so I really want to do this right =.=
 
  • #5
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I actually have a physics question on this:


What limits the motion of a gear of a motor?

Assuming we have no air resistance, no friction.

and assuming that your motor has a constant EM force, shouldn't it be the case that no matter how large the inertia you are trying to rotate, the angular velocity is always increasing? just slower if the inertia is large?


thank you
 
  • #6
Simon Bridge
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I actually have a physics question on this:
...
assuming that your motor has a constant EM force, shouldn't it be the case that no matter how large the inertia you are trying to rotate, the angular velocity is always increasing? just slower if the inertia is large?

thank you
This is one of those "yes and no" answers.
The angular velocity will continue to increase (under idealized conditions: no friction, perfectly matching gears, rigid components etc) only if there is a net unbalanced torque - and only in classical physics - so instantaneous speeds will need to be much slower than light.

qpham26 said:
the torque that I need to overcome the weight of the object will have to be
T = r x mg ???
If you used a light pulley of radius R, to lift mass M through a distance D, in time T, then, ignoring friction for back-of-envelope calculations, you can deduce the torque from Newton's laws and kinematics. You don't have to bother finding the weight that stops it.

If you needed to be exact then you'd have to include the moment of inertia (second fbd for the pulley/bobbin) and find some way to account for friction.

What you do exactly depends on how accurate you need to be.
 
  • #7
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This is one of those "yes and no" answers.
The angular velocity will continue to increase (under idealized conditions: no friction, perfectly matching gears, rigid components etc) only if there is a net unbalanced torque - and only in classical physics - so instantaneous speeds will need to be much slower than light.


If you used a light pulley of radius R, to lift mass M through a distance D, in time T, then, ignoring friction for back-of-envelope calculations, you can deduce the torque from Newton's laws and kinematics. You don't have to bother finding the weight that stops it.

If you needed to be exact then you'd have to include the moment of inertia (second fbd for the pulley/bobbin) and find some way to account for friction.

What you do exactly depends on how accurate you need to be.



But isn't classical mechanics a "good enough" model in this case?
 
  • #8
Simon Bridge
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But isn't classical mechanics a "good enough" model in this case?
I believe I was quite clear about the circumstances where the situation in your question (post #5) would hold.

Since no part of the system described will be relativistic - the classical regime would be "good enough".

But you asked a physics question for a hypothetical situation where there was no friction - and other implied idealizations - which does not apply to this case. This case will have friction and other losses which may be important, and there is an indication that a gearbox may be involved which won't be lossless or have perfectly matching teeth. Real life electric motors do not have a uniform emf or move smoothly. To avoid possible misunderstandings, I chose to be specific.

You'll see that I have described the measurement suggestions as suitable for back-of-envelope" calculations. Real life is messy.
 
  • #9
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I believe I was quite clear about the circumstances where the situation in your question (post #5) would hold.

Since no part of the system described will be relativistic - the classical regime would be "good enough".

But you asked a physics question for a hypothetical situation where there was no friction - and other implied idealizations - which does not apply to this case. This case will have friction and other losses which may be important, and there is an indication that a gearbox may be involved which won't be lossless or have perfectly matching teeth. Real life electric motors do not have a uniform emf or move smoothly. To avoid possible misunderstandings, I chose to be specific.

You'll see that I have described the measurement suggestions as suitable for back-of-envelope" calculations. Real life is messy.


Understood, sorry for not reading your post throughoutly before asking another question.

I can certainly see how loses from damping that is acted on the inerita(the object we are rotating) can be handled , simply by adding a damper parallel to the mass in the linear graph.

however, as you have brought up, the gears and not perfect , and I am assuming in real life there are going to be frictions in in the gear box as well, so can i just add a rotational damper/friction effect parallel to the transformer port in my linear graph?

so net torque to transformer/gear = Torque supplied by motor - counter damping torque.



thank you
 
  • #10
Simon Bridge
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qpham26 is the OP in this thread - I am reluctant to discuss someone elses question without OPs express permission as it would constitute a hijack.

Basically you can fiddle with your classical model to bring it closer to the reality - but you need a reality to compare it with or it's meaningless. It is usually better to come up with a model for the loss-mechanisms. You do know that, with no load, the motor will accelerate quickly to it's top speed - it does not keep accelerating.
 
  • #11
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qpham26 is the OP in this thread - I am reluctant to discuss someone elses question without OPs express permission as it would constitute a hijack.

Basically you can fiddle with your classical model to bring it closer to the reality - but you need a reality to compare it with or it's meaningless. It is usually better to come up with a model for the loss-mechanisms. You do know that, with no load, the motor will accelerate quickly to it's top speed - it does not keep accelerating.

Without a load, it will be accelerating to its top speed and it will not accelerate anymore - > which is due to the lost in reality am I correct? thanks!
 

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