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How exactly do we justify that two parallel forces $\vec{\mathbf{F}}_1$ and $\vec{\mathbf{F}}_{2}$ having the same magnitude, acting on a rigid body, do not produce a moment?

Yes, if we consider any point $O$ exactly in between the the lines of action of the forces, the net moment is $0$. Since one of them will produce a clockwise torque and the other will produce an anticlockwise torque of the same magnitude about $O$.

But if we consider some point $P$ on the same side of the two lines of action, then the torques due to both the forces would be either both clockwise or both anticlockwise.

I'm adding a picture to illustrate my point.

enter image description here

In this picture as we see, there is an anticlockwise moment about $P$, due to the two parallel forces having equal magnitude. So, shouldn't there rotational motion in the body, as moment is not zero about all points of the rigid body? However, real life experience says that two parallel equal magnitude forces on rigid body produces no rotation motion!

It would be helpful if someone could help me understand this apparent "paradox".

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    $\begingroup$ "real life experience " . One needs links that demonstrate such assertions. The simplest explanation is that you tend to include the projection of the center of mass in the middle of the line between your two hands. If your point P were a fixed axis, then there would be a torque about this axis. $\endgroup$ – anna v Feb 24 '18 at 13:51
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How exactly do we justify that two parallel forces $\vec{\mathbf{F}}_1$ and $\vec{\mathbf{F}}_{2}$ having the same magnitude, acting on a rigid body, do not produce a moment?

This is not a correct statement unless the lines of action of the two forces are equidistant from the centre of mass of the object. What you have to be careful of in trying to replicate the situation in real life is the presence of friction.
Once there is a frictional force you will not get the expected result which is explained below. Assume that the magnitudes of all the forces shown in the diagram below are the same $F$ and the centre of mass of the object is at position $C$.

enter image description here

In the left had diagram with force $\vec{\mathbf{F}}_{1}$ acting one can add two forces $\vec{\mathbf{F}}_{1\rm a}$ and $\vec{\mathbf{F}}_{1 \rm b}$ whose line of action pass through the centre of mass of the object as shown in the left hand diagram.
You then have a force acting through the centre of mass $\vec{\mathbf{F}}_{1 \rm b}$ and a couple consisting of the two forces $\vec{\mathbf{F}}_{1}$ and $\vec{\mathbf{F}}_{1 \rm a}$ and magnitude $Fd$ acting in an anticlockwise direction.

This result, a translational motion plus a rotational motion, is difficult to replicate unless one does the experiment on an air table etc where the frictional forces have been reduced to a minimum.

Now add a second force $\vec{\mathbf{F}}_{2}$ as in the right hand diagram.
This time there is a force $\vec{\mathbf{F}}_{2 \rm b}$ whose line of action is through the centre of mass of the object $C$ and a couple consisting of forces $\vec{\mathbf{F}}_{2}$ and $\vec{\mathbf{F}}_{2 \rm a}$ whose magnitude is $Fe$ acting in a clockwise direction.

So the net torque clockwise due to the two couples is $Fe - Fd$ which will equal zero, and hence cause no rotational motion, if $e=d$.

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  • $\begingroup$ Thanks. That clears it somewhat. Another question. What if a rigid axis passes through the body (through some point other than COM)? In that case what is the condition for rotational equilibrium? Does net torque about COM have to be 0, or does net torque about that axis have to be 0? $\endgroup$ – aiai12 Feb 24 '18 at 14:21
  • $\begingroup$ @aiai12 That axis will exert a force on the body and also will constrain the body to rotate about that axis. I am not sure as to what you mean by "rotational equilibrium" but if it means no net torque about the axis then the line of action of the external force must pass through the axis. $\endgroup$ – Farcher Feb 24 '18 at 14:30

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