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The following text is from the book "Concepts of Physics" by Dr.H.C.Verma, from the chapter "Rotational Mechanics", under the topic "Torque of a Force about the Axis of Rotation":

If there are more than one forces $\vec{F_1},\vec{F_1},\vec{F_1},\dots$ acting on a body, one can define the total torque acting on the body about a given line.

To obtain the total torque, we have to get separately the torques of the individual forces and then add them.

$$\vec{\Gamma_{}}=\vec{r_1}\times\vec{F_1}+\vec{r_2}\times\vec{F_2}+\dots$$

You may be tempted to add the forces $\vec{F_1},\vec{F_1},\vec{F_1},\dots$ vectorially and then obtain the torque of resultant force about the axis. But that won't always work. Even if $\vec{F_1}+\vec{F_2}+\dots=0$, $\vec{r_1}\times\vec{F_1}+\vec{r_2}\times\vec{F_2}+\dots$ may not. However, if the forces act on the same particle, one can add the forces and then take the torque of the resultant.

(Emphasis mine)

The statement "But that won't always work" made me think of different cases where finding the net force to find the net torque would work or would not work. One of the cases for which I was unable to determine whether this would work or not is as follows:

If the forces $\vec{F_1},\vec{F_1},\vec{F_1},\dots$ act on different particles of a body in such a way that their lines of action intersect at a common point, is it possible to find the net torque by finding the net force $\vec{F_{}}$ then using $\vec{\Gamma_{}}=\vec{r_{}}\times\vec{F_{}}$ where $\vec{r_{}}$ is the position vector of the common point of the lines of action of the forces?

Please Note: The question - Net torque on an object when when all the forces pass through a common point is related to that of mine, but it doesn't address whether we could use the alternate method of finding the net torque. The question/answer there uses the conventional (always working) method to determine the net torque.

Community
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Vishnu
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2 Answers2

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But that won't always work. Even if $\vec{F_1}+\vec{F_2}+\dots=0$, $\vec{r_1}\times\vec{F_1}+\vec{r_2}\times\vec{F_2}+\dots$ may not.

Clearly this is true. If the sum of the forces being zero always meant the sum of the moments (torques) is also zero, you would be able, for example, to determine static equilibrium requirements for beams based only on the sum of the forces being zero. That is not the case. The exception is for concurrent forces, i.e., forces that intersect at a common point.

Varignon's theorem states that the sum of the moments produced by any two concurrent forces with respect to a point is equal to the moment produced by the resultant with respect to the same point. Though initially stated for two concurrent forces, it applies to any number of concurrent forces.

So yes, if the forces are concurrent, it is possible to use the net force (resultant) and the vector $r$ from the point or axis about which you desire the total torque (moment) to the point where the concurrent forces intersect to find the total torque (moment).

Hope this helps.

Vishnu
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Bob D
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Even if $\vec{F_1}+\vec{F_2}+\dots=0$, $\vec{r_1}\times\vec{F_1}+\vec{r_2}\times\vec{F_2}+\dots$ may not.

One example, consider such lever system : enter image description here

Weights cancels each other : $$ W_U + W_D = 0, $$ however, net toque is $$ \tau = WL_1 - W(L_1 + L_2) = -WL_2. $$ So it is not zero and directed downwards, given that lever will not break at point A

Agnius Vasiliauskas
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  • Thank you for your answer. The diagram looks cute. I feel that you didn't notice what I was asking. But it was really useful. Thanks :) – Vishnu Nov 23 '19 at 13:06
  • As I have understood, you was asking for cases where $\tau_{net}$ is different from $F_{net} r_{avg}$. I gave you one example of this case. If this answer is useful to you - give +1 point ;-) – Agnius Vasiliauskas Nov 23 '19 at 13:21