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This is an extremely silly and wierd question.

https://en.wikipedia.org/wiki/Varignon%27s_theorem_(mechanics) While reading about Varignon's Theorem in wikipedia I noticed this sentence,

"If many concurrent forces are acting on a body, then the algebraic sum of torques of all the forces about a point in the plane of the forces is equal to the torque of their resultant about the same point."

Well what does "a point" specify here? Can it be any point even outside the object which we are dealing with?

Another thing, lets say we are dealing with 2 concurrent forces about a point C and the two forces are acting on A. Now according to the above theorem shouldn't the above scenario be looked as $$F_1 *AC+F_2*AC=F*CC=F*0=0?$$

Perhaps I misunderstood the theorem. As for the 1st question, I have never thought of any reference point in case of rotational motion other than the center of mass of the object and mostly dealt with uniform shapes.

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    $\begingroup$ Yes, it can be any point. The point that is the centre of rotation is just as important to clearly define as the coordinate system in generel is when you define vectors. $\endgroup$
    – Steeven
    Sep 25 at 18:06
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In general, the sum of vectors gives their resultant. But, for that to be true they must be similar and related: the sum of electric fields at a point, the sum of forces acting on a mass, and the sum of torques about a point. In the case of torques, the sum produces the rate of change of the angular momentum measured relative to that point. Some situations suggest a choice of point. If you have a fixed axle of rotation, the point should be on that axle. For a freely moving mass, the center of mass is a good choice. In a statics problem choose a point that reduces the number of torques.

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  • $\begingroup$ Yeah I understood that torque is also a vector and can be added to another torque if acting onnthe same point but my problem is with the reference point, I have shown an equation above as per that equation is torque always zero at that point? Or is the force infinite for which F*0≠0? $\endgroup$
    – MSKB
    Sep 26 at 5:11
  • $\begingroup$ Ahhh previous comment was a dumb one.......why would a point rotate with respect to itself..... $\endgroup$
    – MSKB
    Sep 26 at 5:12
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Not so silly. Toppling of a cylinder on a block may help set the stage. A torque is a pair of anti-parallel forces of equal magnitude with a sideways displacement applied to an object.

Suppose you have a pulley supported by a fixed axis. Suppose a bunch of forces are applied to the pulley, causing it to spin. The fixed axis supplies a reaction force to each applied force. So each applied force is half of a pair of forces that apply a torque to the pulley.

$$\tau = F d_{perp}$$

Since the magnitudes are equal, you can calculate with either force. All the reaction forces are applied to the same place. So it is easy to ignore the reaction forces and focus on the applied forces.

In cases like this, we can speak of each applied force as generating a torque about the axis.

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One condition is that all the forces lie on a plane, as well as the point of summation. Additionally, if you consider that forces can slide along their axis, it is a requirement they all meet at a point. Let us call this point $\vec{r}_{\rm all}$ and some point each force $\vec{F}_i$ goes through is $\vec{r}_i$.

The above conditions are $$ \sum_i (\vec{r}_i - \vec{r}_{\rm all}) \times \vec{F}_i = \vec{0} \tag{1} $$

The vector cross product of the relative position of each force to the force gives the torque of this force summed at $\vec{r}_{\rm all}$. When the torque is zero, it means the forces all go through this point.

Now with some simple manipulation (1) can be re-stated as

$$ \sum_i \vec{r}_i \times \vec{F}_i = \sum_i \vec{r}_{\rm all} \times \vec{F}_i = \vec{r}_{\rm all} \times \sum_i \vec{F}_i \tag{2} $$

Or as the theorem states, the sum of all the individual torque components (due to each force) equals the torque of the combined force through the common point.

Note that in my above scenario the point of summation is the origin. It is common to use a common origin to sum up torques using $\sum_i \vec{r}_i \times \vec{F}_i$. Without loss of generality, you can translate the points around, and the conclusion will remain the same, so choosing the origin as the point of summation just makes the math easier to see.

As noted by @R.W.Bird in the comments, there is really no need to restrict this problem into a plane and thus the torque vectors being parallel. The only real restriction is that the lines of actions of all the forces meet at a single point.

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  • $\begingroup$ Unless you are looking for a resultant of zero, I can't think of any reason why the vectors representing the torques about a point must be parallel. $\endgroup$
    – R.W. Bird
    Sep 26 at 13:48
  • $\begingroup$ @R.W.Bird - I think you are correct. The more general statement would for a "Pencil of forces" in 3D. If two or more forces meet at a single point, then the net moment would be equivalent to the moment of the resultant. => Without the planar restriction. $\endgroup$ Sep 26 at 15:30

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