I wonder, for a rigid body, if the vector sum of all external forces acting on it is equal to zero ($F_{\rm net} = 0$), does this imply that the vector sum of all external torques acting on it about any point is also equal to zero?


2 Answers 2


does it also leads to ...

No, it doesn't.

A simple Counterexample:

Consider the figure below (the bar $\textrm {AB}$ and forces $F$ are on a plane parallel to $\textrm {xy}$ plane)

enter image description here

We have $\Sigma \vec F=\vec 0$, but, if we calculate vector sum of torques about point $\textrm A$ we will obtain $\Sigma \vec M_A=F (\overline{AB})\vec k\neq \vec 0$ ($\vec k$ is the unit vector of $\textrm z$ direction)

  • 1
    $\begingroup$ In addition to this answer. The requirement for statics in mechanics is that Newton's 1st law and the equivalent for rotation are obeyed: $$\sum \vec F=0\quad \text{ and }\quad \sum \vec M=0$$ No translation and no rotation ($M$ is torque). Compare this with the general non-static cases desscribed by Newton's 2nd law and the equivalent rotational version of the same law: $$\sum \vec F=m\vec a\quad \text{ and }\quad \sum\vec M=I\vec \alpha$$ $\endgroup$
    – Steeven
    Commented Jun 14, 2016 at 9:09
  • $\begingroup$ And on a tangent, another point: zero net force means exactly the same net torque, regardless around which point it is calculated. $\endgroup$
    – LLlAMnYP
    Commented Jun 14, 2016 at 10:35
  • $\begingroup$ why haven't you considered pseudo torque as point A is accelerating $\endgroup$ Commented Aug 26, 2022 at 6:06

The answer is no.

The total torque on the system given in terms of the center of mass coordinates is $$\vec\tau=\vec R_{cm}\times\vec F^{ext}+\sum_i\vec r_i'\times\vec F_i^{ext}$$ where $M$, $\vec R_{cm}$ and $\vec F^{ext}$ are respectively the mass of the system, the center of mass position vector and the total external force on the system. The vector $\vec r_i'$ is the position vector of the particle $i$ with respect to the center of mass and $\vec F_i^{ext}$ is the external force acting on this particle. Therefore, if $\vec F^{ext}=0$, then $$\vec\tau=\sum_i\vec r_i'\times\vec F_i^{ext},$$ which is in general non vanishing. The important result associated is that if the resultant external force vanishes then the torque is independent of the point it is calculated. It has the same value with respect to any point.


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