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For example:

There is an object A on the left side of the see-saw that exerts a torque of 10 Nm due to its weight. On the other side of the see-saw, there is an object B that exerts a torque of 20 Nm due to its weight.

According to what I have understood of Newton's Third Law of Rotational Motion, object A exerts a torque of 10 Nm towards you (using right-hand rule) on object B. Object B then reacts equally but oppositely by exerting 10 Nm of torque away from you(again using right-hand rule) on Object A. The same thing happens for object B. It exerts a torque of 20 Nm away from you on Object A. Object A reacts equally but oppositely by exerting a force of 20 Nm towards you on Object B.

Therefore, in total, object B exerts a torque of 30 Nm away from you on object A and object A exerts a torque of 30 Nm towards you on object B. According to my understanding, the see-saw does not undergo angular acceleration and stays at a constant angular velocity, which is obviously wrong!

Where am I going wrong?

Edit: I've been thinking about this, and what happens when two people are arm-wrestling and one is winning? How is the torque that they apply on each other's arms different?

Any help is much appreciated.

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  • $\begingroup$ Object A and Object B are not exerting forces on each other, so how are they exerting torques on each other? It looks like you are completely ignoring the forces between each object and the seesaw (where N3L would actually apply) $\endgroup$ May 17, 2021 at 16:45
  • $\begingroup$ Oh I get it. Object A exerts a torque of 10 Nm on the SEE-SAW, which exerts reacts equally but oppositely by exerting a force of 10Nm on object A. Object B, meanwhile, exerts a torque of 20 Nm on the SEE-SAW, which exerts reacts equally but oppositely by exerting a force of 20Nm on object B. Therefore, the see-saw undergoes angular acceleration. $\endgroup$ May 17, 2021 at 17:02
  • $\begingroup$ Thanks a lot @BioPhysicist $\endgroup$ May 17, 2021 at 17:02
  • $\begingroup$ I've been thinking about this, and what happens when two people are arm-wrestling and one is winning? How is the torque that they apply on each other's arms different? $\endgroup$ May 17, 2021 at 17:38
  • $\begingroup$ To the last sentence: Remember that you don't need different torques to achieve motion (acceleration). $\endgroup$
    – Steeven
    May 18, 2021 at 5:40

1 Answer 1

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To somewhat simplify the problem and not deal with angular quantities, let's say we have the following scenario: You have two blocks of different mass (say m and M) where m is smaller than M. They are both attached to a massless string which is hung over a frictionless ideal pulley.

This would be the same idea but just with forces rather than torques. Now, if you were to apply Newton's Third Law of Action/Reaction, you might say that the block of mass m has a weight of magnitude mg. This would cause the other block to have a reaction of magnitude mg as well. The other block of mass M also has a weight of magnitude Mg. This would cause the first block to have a reaction of Mg. Therefore, both blocks have a net force of (m + M)g and thus nothing will move. There is obviously something wrong here.

The fallacy here is that the weight of the two masses (m and M) do not form an action reaction pair. The action reaction pair to the weight is the force of gravity by the block on the earth. The earth is being pulled up just as hard as the block is being pulled down. We therefore cannot use Newton's Third Law in this case.

Instead, what we do know is that the tension T in the string is the same for both sides. And just as the string is pulling the blocks up, the blocks also pull the string down. There is another action reaction pair.

In a free body diagram on block m, it would look something like:

free body diagram on block m

However, on the string, it would look something like:

free body diagram on string

Therefore the only forces acting on block m are the weight (mg) and the tension (T). There is no reaction force from block M.

If we were to do the math and write a set of equations for both blocks using Newton's Second Law, F=ma with F being the net force and a the shared acceleration of the two blocks, we could solve this system for acceleration and find that the acceleration is indeed non-zero.

This is the same idea with the two objects on the see-saw. The two objects are not directly exerting forces on each other. Therefore they are not exerting torques on each other. Rather, the torque is being applied on the see-saw itself and the force is not the weight of the objects but the normal force between the blocks and the see-saw. Since one block exerts more torque on the see-saw than the other, there is a net torque and thus produces angular acceleration. The see-saw is just like the string and the pulley in the previous scenario.

What about arm-wrestling? Well in this case, both hands are in contact so there is a normal force. This normal force produces a torque because it is a certain distance/radius away from the center of rotation. However, since this normal force forms an action/reaction pair, there is no net force on the system. But the normal force is not the only force in play here (although it is what makes your hands sore). The muscles in the arm also provide an external force on the entire system, thus producing a torque in one direction or another. In this case, the contact between the hands is like the see-saw or the string and the pulley. They do not form an action/reaction pair with the external force of your muscles.

The key takeaway here is to correctly identify the action/reaction pairs. If two objects are determined to be action/reaction but do not share the same interaction, all bets are off!

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  • $\begingroup$ so the torque that the hands exert on each other is due to normal force and cancel out and cause no angular acceleration on the system. But the force that Person A exerts on the pair of arms compared to the force that Person B exerts on the pair of arms is what causes one person to win. Correct? $\endgroup$ May 18, 2021 at 11:34
  • $\begingroup$ Yes that is correct $\endgroup$
    – Luke
    May 18, 2021 at 16:09

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