Does Newton's Third law apply to torque/rotation?

[alkaspeltzar]

That's a different drawing and description from what you provided before. So just remove my edit: you need to tell us what gear B is connected to. A shaft? A wheel touching the ground? What? In order for the torques to sum to zero, there needs to be a second torque on gear B, opposing the torque applied by gear A.

Russ, i have edited my picture to show A is connected to a Motor, on the ground. And B is a wheel touching the ground.

 

[russ_watters]

Russ, i have edited my picture to show A is connected to a Motor, on the ground. And B is a wheel touching the ground.
View attachment 276893

So, the ground a applies a torque around B, equal and opposite of what gear A applies.

 

[A.T.]

Russ, i have edited my picture to show A is connected to a Motor, on the ground. And B is a wheel touching the ground.

If there are no horizontal forces between B and ground, it's still the same:

When taken around the same point:

torque_by_MOTOR_on_A = -torque_by_A_on_MOTOR
torque_by_B_on_A = -torque_by_A_on_B

Otherwise there is also:

torque_by_GROUND_on_B = -torque_by_B_on_GROUND

 

[hutchphd]

torque_by_B_on_A = -torque_by_A_on_B

NO

The vector torque is an axial vector. To uniquely define a torque on an extended body one must specify the axis (or equivalently the point of application) of the torque. Within this definition a strong form of Newton's 3rd Law Rotational applies: there is an equal and opposite torque at the same point of application (or along the exact same axis equivalently) from the torquee to the torquer.
The equivalent Newtons Law for forces also tacitly assumes the line of action of the forces is coincident.

The case listed above is not correctly analyzed by the Newton 3rd rotational law because the axes are not coincident.

 

[alkaspeltzar]

NO

The vector torque is an axial vector. To uniquely define a torque on an extended body one must specify the axis (or equivalently the point of application) of the torque. Within this definition a strong form of Newton's 3rd Law Rotational applies: there is an equal and opposite torque at the same point of application (or along the exact same axis equivalently) from the torquee to the torquer.
The equivalent Newtons Law for forces also tacitly assumes the line of action of the forces is coincident.

The case listed above is not correctly analyzed by the Newton 3rd rotational law because the axes are not coincident.

So I asked that at the beginning of my post. N3l for torques only applies if the forces are inline and act at the same point, THerfore same radius.

So with these gears, yes Ang. Momentum must still be conserved, but I cannot say there is a n3l for torques

 

[A.T.]

... because the axes are not coincident.

I have explicitly specified this in all previous posts :

..., when taken around the same point

except in the last update one you quoted. Will add it now.

 

[russ_watters]

So I asked that at the beginning of my post. N3l for torques only applies if the forces are inline and act at the same point, THerfore same radius.

So with these gears, yes Ang. Momentum must still be conserved, but I cannot say there is a n3l for torques

 

[hutchphd]

So I asked that at the beginning of my post. N3l for torques only applies if the forces are inline and act at the same point, THerfore same radius.

"Therefore same radius" is not true. You seem to be applying Newtons 3rd law to any torques that are equal and I have no idea what your question is at this point.
Would you care to try again?

 

[A.T.]

...but I cannot say there is a n3l for torques

How would a N3L for torques look like to you?

 

[alkaspeltzar]

How would a N3L for torques look like to you?

I have been asking where does the 50 inch lbs torque about B get balanced out.

 

[alkaspeltzar]

I have been asking where does the 50 inch lbs torque about B get balanced out.

I agree with this, but what happens to the torque about B. Do we ignore it since it's not relative to axis A?
torque_by_MOTOR_on_A = -torque_by_A_on_MOTOR
torque_by_B_on_A = -torque_by_A_on_B

Otherwise there is also:

torque_by_GROUND_on_B = -torque_by_B_on_GROUND

 

[alkaspeltzar]

I agree with this, but what happens to the torque about B. Do we ignore it since it's not relative to axis A?
torque_by_MOTOR_on_A = -torque_by_A_on_MOTOR
torque_by_B_on_A = -torque_by_A_on_B

Otherwise there is also:

torque_by_GROUND_on_B = -torque_by_B_on_GROUND

I think I'm understand. I cannot compare torque at B since we have to maintain being consistent with A right? That's what I am looking to confirm.

So we look at torques about A and only A, everything balances correct?

Perok that's what I'm trying to get. I am really trying. Not trying to be pig headed

 

[A.T.]

I have been asking where does the 50 inch lbs torque about B get balanced out.

What does "balancing out" have to do with N3L? It seems like you confusing N2L and N3L. Make sure you understand the difference between them for forces, before going into torques.

 

[alkaspeltzar]

What does "balancing out" have to do with N3L? It seems like you confusing N2L and N3L. Make sure you understand the difference between them for forces, before going into torques.

What I am asking is where does the the 50inch lbs torque have its equal pair?

Below what you stated makes perfect snese. Not arguing this at all. But shouldn't there be an equal torque somewhere for the 50 inch lbs, and if no, why?If there are no horizontal forces between B and ground, it's still the same:

When taken around the same point:

torque_by_MOTOR_on_A = -torque_by_A_on_MOTOR
torque_by_B_on_A = -torque_by_A_on_B

Otherwise there is also:

torque_by_GROUND_on_B = -torque_by_B_on_GROUND

 

[russ_watters]

What I am asking is where does the the 50inch lbs torque have its equal pair?

Here:
torque_by_GROUND_on_B = -torque_by_A_on_B = 50

 

[alkaspeltzar]

Here:
torque_by_GROUND_on_B = -torque_by_A_on_B = 50

So the ground applies a horizontal force of 10lbs at the 5 inches about B. This is opposite and equal to the torque by A on B, about B. ANd if you consider gear B, these are the opposite and equal torques for it. Its like Gear A is now acting similar to the ground, and therefore the N3L balance.

Right?

 

[russ_watters]

So the ground applies a horizontal force of 10lbs at the 5 inches about B. This is opposite and equal to the torque by A on B, about B. ANd if you consider gear B, these are the opposite and equal torques for it. Its like Gear A is now acting similar to the ground, and therefore the N3L balance.

Right?

Sounds about right, yes.

 

[alkaspeltzar]

Sounds about right, yes.

THank you

I think i see what is going on here. In the end, torque is dependent on the point of rotation and everything about it, like summing the moments has to be done around the same axis. However, because of conservation of ang momentum, and N3L, it all has to balance one way or another.

This example probably was bad as i didn't start with good unknowns, but in the end i see it all. Makes me understand why in many cases we don't have to care about all the reactions because we identify our system to make life easier.

Thanks again

 

[A.T.]

So the ground applies a horizontal force of 10lbs at the 5 inches about B. This is opposite and equal to the torque by A on B, about B. ANd if you consider gear B, these are the opposite and equal torques for it. Its like Gear A is now acting similar to the ground, and therefore the N3L balance.

Right?

Wrong. The torque_by_A_on_B and the torque_by_GROUND_on_B are not a N3L pair. This is not N3L, but a N2L balance. You are still confusing the laws you are applying.

 

[alkaspeltzar]

Wrong. The torque_by_A_on_B and the torque_by_GROUND_on_B are not a N3L pair. This is not N3L, but a N2L balance. You are still confusing the laws you are applying.

Probably be clearer if you drew it out as I asked nicely but since I have tried and still getting it wrong I guess I just won't care. Your replys are short but leave a lot to be explained to someone trying to learn

 

[A.T.]

Probably be clearer if you drew it out as I asked nicely ...

I have given you all the N3L pairs, for all your diagrams. Nowhere in there are torque_by_A_on_B and torque_by_GROUND_on_B a N3L pair.

 

[alkaspeltzar]

Yes, it does.

The point of N3 is the conservation of momentum. The point of the torque version is the conservation of angular momentum. Angular momentum requires the specification of one single axis about which all angular momenta are measured. That is just how angular momentum works.

Dale, can we talk or expalin where the n3l pair is for the torque B has on it? I get the point about being around a single axis, but do I just ignore what is happening to hear B

 

[alkaspeltzar]

I have given you all the N3L pairs, for all your diagrams. Nowhere in there are torque_by_A_on_B and torque_by_GROUND_on_B a N3L pair.

I asked though what about gear B, do we just ignore the torque about B as it is around an axis not related to the problem?

 

[Dale]

Dale, can we talk or expalin where the n3l pair is for the torque B has on it? I get the point about being around a single axis, but do I just ignore what is happening to hear B

No, you don't ignore what is happening to gear B. You must include all torques on all objects in the system. But you must fix the axis of interest. Which axis do you want to know if there is torque about? Then you calculate all torques about that point.

 

[A.T.]

I asked though what about gear B, do we just ignore the torque about B as it is around an axis not related to the problem?

"About B" makes no sense. Do you mean about the center of B? You can use any point, as long it's the same for both torques of an N3L pair.

 

[PeroK]

"About B" makes no sense. Do you mean about the center of B? You can use any point, as long it's the same for both torques of an N3L pair.

This misconception, and the OP's refusal to address it, has persisted for 60 posts now.

 

[alkaspeltzar]

No, you don't ignore what is happening to gear B. You must include all torques on all objects in the system. But you must fix the axis of interest. Which axis do you want to know if there is torque about? Then you calculate all torques about that point.

i want to know the torques about the center of A

 

[russ_watters]

"About B" makes no sense. Do you mean about the center of B? You can use any point, as long it's the same for both torques of an N3L pair.

Sorry, I think I added that: I've been calling the mounting point at the center of gear B, "Point B".

Nowhere in there are torque_by_A_on_B and torque_by_GROUND_on_B a N3L pair.

This might be another interpretation by me that isn't stated, but I interpreted the new drawing to show "Object B" is really a wheel and a gear, connected by a shaft -- but with the scenario morphing I don't know for sure. If that's the case, then a point on the shaft (Point B) has two equal and opposite torques applied to it. If "Object B" is truly one object (a wheel and a gear at the same time?), then it doesn't even need to be mounted on a shaft at all.

 

[A.T.]

i want to know the torques about the center of A

Then why do you keep asking about the torques "about B". Are you confusing "torque about ..." and "torque on ..."?

 

[Dale]

i want to know the torques about the center of A

View attachment 276890

So the torque of the ground force at A is 0, the torques at the gear are 10 in lb and -10 in lb and the torque at the center of B is 60 in lb. So the net torque on the system is 60 in lb about the center of A.

 

[alkaspeltzar]

Then why do you keep asking about the torques "about B". Are you confusing "torque about ..." and "torque on ..."?

Gear A DRIVES Gear B right?

There are contact forces between the two yes? Equal and opposite.

I see a torque arm of 5 inches, with a force of 10 lbs acting on gear B about its center. If there is a N3L for torque, i am trying to understand where does one account for this?

If i take all the torques about the center of gear A, i can see how angular momentum balances, but still Gear B has a torque applied to it, and i don't see where its equal and opposite is?

Perhaps someone ccould sketch on paper and show me what i am missing to help with all the 'about this or that"

 

[alkaspeltzar]

I think I see it now.. sorry I only had to sketch it like 30 times to see it. Sorry everyone. Let me sleep on it

 

[A.T.]

Gear A DRIVES Gear B right?

This is completely irrelevant to the application of N3L.

i want to know the torques about the center of A

I see a torque arm of 5 inches, with a force of 10 lbs acting on gear B about its center.

Make up your mind about which point you take the torques.

 

[alkaspeltzar]

This is completely irrelevant to the application of N3L.
Make up your mind about which point you take the torques.

there's a difference between being rude and trying to understand something you're confused about, I'm sorry I just don't get it and it took me a while to sketch it out and I think now I see how I have to apply Newton's third law to the single point and look at the torques there. Again I'm sorry it's not like I have a classroom to sketch us out in

 

[alkaspeltzar]

It's been almost 20 years since I've been in a physics classroom sometimes the easiest things and the fundamental things are the toughest to remember even though I can do the work on a daily basis so please bear that in mind I'm not trying to be stubborn