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

[alkaspeltzar]

For example, let's say i have force acting on an object about some pivot. That object will apply an equal force about the same pivot, at equal lever arm, back. So there it seems like if the forces are in line, acting on the same pivot point, there is an equal and opposite torque.

However, if i have two gears, A and B, with A being half the diameter of B, as A drives gears B, there are contact forces between the gears and since there are different lever arms and point of rotations, the torques created are not equal.

SO i am wrong to say that N3L doesn't apply to torque?

Thanks for the clarification.

 

[PeroK]

However, if i have two gears, A and B, with A being half the diameter of B, as A drives gears B, there are contact forces between the gears and since there are different lever arms and point of rotations, the torques created are not equal.

SO i am wrong to say that N3L doesn't apply to torque?

Thanks for the clarification.

Torque is defined relative to a point. There's no reason that the torque about one point should be equal and opposite to the torque about some other point.

 

[alkaspeltzar]

Torque is defined relative to a point. There's no reason that the torque about one point should be equal and opposite to the torque about some other point.

Okay i agree with that. But with that being said, if i compute the torque about a constant point, let's say the center of Gear A, then is there an equal torque about the center of gear A that equals the torque imparted to gear B?

I see a torque applied to Gear b, about its center, at its radius. So there has to be an equal torque to that then right?

Or back to my original question, does N3L not apply here to torque

 

[russ_watters]

Okay i agree with that. But with that being said, if i compute the torque about a constant point, let's say the center of Gear A, then is there an equal torque about the center of gear A that equals the torque imparted to gear B?

I see a torque applied to Gear b, about its center, at its radius. So there has to be an equal torque to that then right?

No, as said, there's no reason for them to be the same. Further, making them different is one of the main functions of gears.

Gears (mechanical advantage) tend to be better described using conservation of energy.

Or back to my original question, does N3L not apply here to torque

At least not in the way you are applying it. It applies at a point where there is a force pair; the contact point between the gears.

 

[alkaspeltzar]

No, as said, there's no reason for them to be the same. Further, making them different is one of the main functions of gears.

Gears (mechanical advantage) tend to be better described using conservation of energy.

At least not in the way you are applying it. It applies at a point where there is a force pair; the contact point between the gears.

SO please correct me if this is wrong: There can be equal forces according to N3L, but only if those forces act at the same radii about the same point will it create equal and opposite torques. TO say there a N3L for torques is not true, esp given the gear example as we need there to be different torques. I am probably trying to make N3L do to much. Agree?

 

[PeroK]

Okay i agree with that. But with that being said, if i compute the torque about a constant point, let's say the center of Gear A, then is there an equal torque about the center of gear A that equals the torque imparted to gear B?

I see a torque applied to Gear b, about its center, at its radius. So there has to be an equal torque to that then right?

Or back to my original question, does N3L not apply here to torque

You need to be careful to stop talking about "the torque imparted to gear B" - in this case by gear A. Strictly speaking, you have a torque about any point for the force gear A applies to gear B. There's no reason to pick out the torque about the centre of gear B as something special. There can't possibly be a fixed relationship between the torque about the centre of object A and torque about the centre of object B, where A and B are arbitrary objects.

In fact, if you think about the vector nature of torque, then the torque about the centre of A and the torque about the centre of B may be equal - but not equal and opposite, as both the forces and the displacement from the centres to the point of contact may be equal and opposite.

 

[Dale]

I disagree. There is a perfectly valid N3 for torque, you are just applying it wrong. You cannot change the axes mid-analysis. All torques are measured about the same axis, which can be any single axis desired.

 

[PeroK]

SO please correct me if this is wrong: There can be equal forces according to N3L, but only if those forces act at the same radii about the same point will it create equal and opposite torques. TO say there a N3L for torques is not true, esp given the gear example as we need there to be different torques. I am probably trying to make N3L do to much. Agree?

The special case is that the torques are equal, but they won't be opposite.

 

[alkaspeltzar]

I disagree. There is a perfectly valid N3 for torque, you are just applying it wrong. You cannot change the axes mid-analysis. All torques are measured about the same axis, which can be any single axis desired.

can you explain then? How does it work in the case with two gears of different sizes where the contact forces between them are equal and opposite. Can you expand on this

 

[alkaspeltzar]

can you explain then? How does it work in the case with two gears of different sizes where the contact forces between them are equal and opposite. Can you expand on this

I am trying to use the axis of Gear A for consistency. IF i use gear A's center, gear A applies a torque to B, so relative to the center of A, how do all the torques balance equal and opposite?

 

[PeroK]

I am trying to use the axis of Gear A for consistency. IF i use gear A's center, gear A applies a torque to B, so relative to the center of A, how do all the torques balance equal and opposite?

In general, if the displacement from your reference point (centre of gear A) to the contact point is $\vec r$ and the N3L pair at the point are $\vec F_1 = - \vec F_2$, then$$\vec \tau_1 = \vec r \times \vec F_1 = - \vec r \times \vec F_2 = - \vec \tau_2$$

 

[A.T.]

I am trying to use the axis of Gear A for consistency. IF i use gear A's center, gear A applies a torque to B, so relative to the center of A, how do all the torques balance equal and opposite?

Relative to the same reference point, two equal but opposite forces with the same point of application will always produce equal but opposite torques. This is because:

$$\vec r \times -\vec F = - (\vec r \times \vec F)$$

 

[alkaspeltzar]

Relative to the same reference point, two equal but opposite forces with the same point of application will always produce equal but opposite torques. Do the math.

I have done the math. Let's say A radius is 1. Let's say B radius is 5. IF we apply a torque of 10lb-inches to A and there is a contact force of 10 lbs between the gears, A has zero torque, B has 50 lb-inches. Where is the equal and opposite torque of B which has 50?

 

[alkaspeltzar]

In general, if the displacement from your reference point (centre of gear A) to the contact point is $\vec r$ and the N3L pair at the point are $\vec F_1 = - \vec F_2$, then$$\vec \tau_1 = \vec r \times \vec F_1 = - \vec r \times \vec F_2 = - \vec \tau_2$$

i agree with this. But gear B has torque, where does its equal and opposite balance come from? If you see the math over i did, gear B would feel a torque of 50 in lbs. How does this balance then relative to center of A

 

[russ_watters]

I am trying to use the axis of Gear A for consistency. IF i use gear A's center, gear A applies a torque to B, so relative to the center of A, how do all the torques balance equal and opposite?

You have a force applied to gear A, which presumably is opposed by an equal and opposite torque around its shaft/point A. They sum to zero.

Where is the equal and opposite torque of B which has 50?

It's your setup, you need to tell us. But presumably its being applied to the shaft.

If you see the math over i did, gear B would feel a torque of 50 in lbs. How does this balance then relative to center of A

Again: it doesn't. You say you are trying to be consistent, but you aren't. I think it is because the thing you really want to do here breaks the consistency and isn't possible. You can't equate the torques around A and B in the way you are trying to.

 

[A.T.]

IF we apply a torque of 10lb-inches to A and there is a contact force of 10 lbs between the gears, A has zero torque,

This is not how Newtons 3rd Law works. It has nothing with the net force or net torque on an object. It applies to an interaction between two objects. But you are using three objects: gear A, gear B and "we".

You have two torque interactions here, and the N3 applies to each of them :
torque_by_us_on_A = -torque_by_A_on_us
torque_by_B_on_A = -torque_by_A_on_B

 

[PeroK]

But gear B has torque,

That statement is meaningless. Torque applies about a point.

 

[russ_watters]

can you explain then? How does it work in the case with two gears of different sizes where the contact forces between them are equal and opposite. Can you expand on this

I think @Dale is saying that when you sum the torques around a point (and they sum to zero), that's an application of N3.

 

[alkaspeltzar]

I think @Dale is saying that when you sum the torques around a point (and they sum to zero), that's an application of N3.

Dale And Russ, can you confirm this? IF i sum the torque about A, i agree, but if i look for equal and opposite which is what started this thread than that's different.

Here is an example. Assume Gear A has radius 1, Gear B has radius 5. Gear A and B are mounted the Earth and there is a motor providing the 10lbs-inches of torque to gear B. Steady state, no acceleration of the system. No friction. Torque B requires leave A with zero, so its speed is constant

 

[alkaspeltzar]

I think @Dale is saying that when you sum the torques around a point (and they sum to zero), that's an application of N3.

This is what my physics books says. It says that thru N3L, internal torques on a system should sum to zero. But doesn't say there is a N3L strictly for torque

 

[russ_watters]

This is what my physics books says. It says that thru N3L, internal torques on a system should sum to zero. But doesn't say there is a N3L strictly for torque

That second sentence doesn't parse/mean anything as far as I can tell.

 

[alkaspeltzar]

That second sentence doesn't parse/mean anything as far as I can tell.

MY tipler physics book has a statement in it, that says "Thru the fact and use of Newtons third law, all internal torques of a system must sum to zero if ang. mom. is conserved". What the book doesn't say, is it doesn't say for every torque there is an opposite and equal torque.

 

[russ_watters]

MY tipler physics book has a statement in it, that says "Thru the fact and use of Newtons third law, all internal torques of a system must sum to zero if ang. mom. is conserved". What the book doesn't say, is it doesn't say for every torque there is an opposite and equal torque.

"torques must sum to zero" sounds like "equal and opposite" to me.

 

[A.T.]

...it doesn't say for every torque there is an opposite and equal torque.

If the force producing the torque obeys N3, then so does the torque.

If the force producing the torque doesn't obey N3, then neither does the torque.

 

[Dale]

can you explain then? How does it work in the case with two gears of different sizes where the contact forces between them are equal and opposite. Can you expand on this

You can pick anyone axis you like (it does not need to be the center of either gear). Calculate both torques about that one axis. The sum will be zero.

 

[alkaspeltzar]

What torque does Gear B have applied to it in my example? 50 inch-lbs right? So where does it have an equal and opposite torque. Can someone explain? I am clearly not seeing how the 50 inch lbs of torque applied to B has its equal and opposite. Perhaps someone can draw it on my diagram?[/QUOTE]

 

[alkaspeltzar]

You can pick anyone axis you like (it does not need to be the center of either gear). Calculate both torques about that axis. The sum will be zero.

Dale, i agree. If i pick Gear A axis and sum the torques about A, they do sum to zero. But that doesn't mean there is a N3L for torque does it? IF so, if gear B in my example has 50 inches pounds on it, where does its equal and opposite torque get applied?

 

[alkaspeltzar]

I provided a drawing. IF someone could mark up the 3 law pairs for torque on a two gear system i would be most greatful. I don't see my mistake and the sad face Imogi and tidbits are not helping me. I appreciate it, but if for every torque there should be an equal and opposite, then we cannot just ignore the torque on B

 

[russ_watters]

What torque does Gear B have applied to it in my example? 50 inch-lbs right? So where does it have an equal and opposite torque. Can someone explain? I am clearly not seeing how the 50 inch lbs of torque applied to B has its equal and opposite. Perhaps someone can draw it on my diagram?

As I said above, it's your problem and you need to specify what's in it. You specified two equal/opposite torques for A, but you didn't for B. So you tell us. Is gear B connected to a shaft that has an applied torque?
[edit] Oh; you said there's a motor at B. The motor provides the other torque.

 

[A.T.]

You have 3 torque interactions here, and the N3 applies to each of them, when taken around the same point :

torque_by_GROUND_on_A = -torque_by_A_on_GROUND
torque_by_GROUND_on_B = -torque_by_B_on_GROUND
torque_by_B_on_A = -torque_by_A_on_B

 

[PeroK]

Dale, i agree. If i pick Gear A axis and sum the torques about A, they do sum to zero. But that doesn't mean there is a N3L for torque does it? IF so, if gear B in my example has 50 inches pounds on it, where does its equal and opposite torque get applied?

Let's say you were teaching a junior student about Newton's laws and they would not accept that force had a direction. They kept talking about force as a magnitude only. And, no matter how many times you told them that force was a vector and had a magnitiude and direction, they just kept repeating the same mistake by considering only the magnitude of all the forces.

What would you do?

Because, really, as far as persuading you that a torque is always relative to a point, I'm out of ideas. Sorry. Your stubbornness has defeated me.

 

[alkaspeltzar]

As I said above, it's your problem and you need to specify what's in it. You specified two equal/opposite torques for A, but you didn't for B. So you tell us. Is gear B connected to a shaft that has an applied torque?
[edit] Oh; you said there's a motor at B. The motor provides the other torque.

Here is the problem. A is the driver gear. It has a motor providing 10 inch lbs to it. B is the driven gear.

I see a 10 lb force acting at B, which is 5 away. THat is 50 inch lbs. Let's assume B is a wheel touching the ground

 

[russ_watters]

Here is the problem. A is the driver gear. It has a motor providing 10 inch lbs to it. B is the driven gear.

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.

 

[A.T.]

Here is the problem. A is the driver gear. It has a motor providing 10 inch lbs to it. B is the driven gear.

You have 2 torque interactions here, and the N3 applies to each of them, 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

 

[Dale]

If i pick Gear A axis and sum the torques about A, they do sum to zero. But that doesn't mean there is a N3L for torque does it?

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.