I don't understand why torques produced by internal forces cancel in the sum $\sum \tau$. My textbook gives the following explanation: due to N3L, if a particle exerts a force on another particle of the body, there will be an equal in magnitude but opposite in direction reaction force. If the line of action of the two forces is the line joining the particles, the lever arm will be equal, and thus the torque produced by each force will be equal and opposite again.

However, what if multiple forces act on the particles (if more than 2 particles are interacting)? Then the torque produced by the net force on one particle might not lie on the same line of action as before and might not have the same magnitude anymore, so how come the torques cancel nonetheless (think a system of three particles, where the first and the second particle interact with the third). [I haven't studied angular momentum yet and only seen torque in 2D, $\tau = Fl = rFsin\phi = F_{tan}r$]

  • $\begingroup$ Somewhere there is an assumption that when more than 2 particles interact, the result is the equivalent of the sum of the result of all the particle pairs interacting. So if each pair has no internal torque, then overall there is no internal torques. $\endgroup$
    – JAlex
    Apr 22 at 13:16

1 Answer 1


The General Case

Let $k$ and $j$ be any two particles, and $\vec F^i_{k\to j}$ be the (internal) force of particle $k$ on $j$. For the total internal torque to vanish, a stronger form of Newton's third law is assumed. Specifically, in addition to $\vec F^i_{k \to j}$ being equal and opposite to $\vec F^i_{j \to k}$, these forces act along the line joining the two particles; that is the two particles can only attract or repel each other. With this assumption, it can be shown that the total internal torque is zero. See an intermediate/advanced mechanics textbook such as Mechanics by Symon, for the total proof; it is rather lengthy and involves summations over all particles of vector cross products ($3{\rm D}$ torques), but I can reproduce it here if you wish. The internal torque on any one particle is not necessarily zero, but the total internal torque defined as the total torque on all particles from internal forces is zero.

Response added based on your comment.

See the figure below. Consider any two particles: $1$ and $2$. Using $O$ as the point about which to evaluate torque, $\vec r_1 \times \vec F^i_{2 \to 1}$ is the torque on particle $1$ from particle $2$, and $\vec r_2 \times \vec F^i_{1 \to 2}$ is the torque on particle $2$ from particle $1$. Using the third law, $\vec F^i_{2\to 1} = -\vec F^i_{1\to 2}$ so summing we have: $(\vec r_2 - \vec r_1 ) \times \vec F^i_{1 \to 2}$. $(\vec r_2 - \vec r_1 ) = \vec r_{12}$ has direction along the line joining point $1$ to point $2$, and if $\vec F^i_{1\to 2}$ acts along this line, as we have required, the cross product in the sum is zero. Similar arguments can be made for the other internal torques so the total internal torque is zero.

As mentioned earlier for the general case, the internal torque on an individual particle is not necessarily zero; for two particles the internal torque on particle 1 from particle 2 is $\vec r_1 \times \vec F^i_{2 \to 1}$. But, the total internal torque, the sum of 1 on 2 plus 2 on 1, is zero. enter image description here

  • $\begingroup$ Is the proof possible with 2D torques as well (see the end of my question)? $\endgroup$
    – Wernher
    Apr 21 at 12:47
  • $\begingroup$ FYI the proof of the cancellation of internal torques (assuming central forces) is part of my answer here. $\endgroup$ Apr 21 at 15:17
  • $\begingroup$ Why is the torque produced by force $\vec{F_{12}}$ $\vec{r_{12}} \times \vec{F_{12}}$ instead of $\vec{r_{2}} \times \vec{F_{12}}$ ? As the point were trying to rotate about is O. $\endgroup$
    – Wernher
    Apr 21 at 15:44
  • $\begingroup$ I added more information to explain this more clearly. $\endgroup$
    – John Darby
    Apr 21 at 23:11

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