Equilibrium Conditions Under the Principle of Transmissibility I'm basically asking a question that's already been asked here, but it wasn't answered well. I have the same textbook as this guy, and in fact, it does claim that "the conditions of equilibrium or motion of a rigid body remain unchanged" regardless of where a force acts, so long as it is along the same line of action. My understanding of equilibrium is that it requires both net force and net moment to equal zero. An image such as the following one clearly shows that the moment of the rigid body is not the same at any point along the line of action. It will spin either clockwise or counter-clockwise depending on where it is contacted.

Additionally, I'm not certain the translational motion, ignoring the moment, is the same either. As seen in this next image, depending on where I apply force to the L-shaped object it will either spin violently or be pulled fairly linearly.

Is my understanding of the book wrong or is it just incorrect? Thank you.
 A: The answer in the question that you cite is correct, but it's a bit short so maybe that makes it less helpful. Also note I'm assuming that you meant 'torque' where you wrote 'moment.'
This is one of those tricky situations where intuition can actually be a bit misleading when you first glance at the problem. So you might find it helpful to actually make some little paper/cardboard cutouts and experiment. In your first image, it looks like the line of action passes through the center of mass of the rectangle. In this case, $\vec{r}\times \vec{F} = 0$, so the object actually doesn't spin (remember that $\vec{r}$ is the vector from the object's center of mass to the point where the force is applied). If the line of action is displaced to the left, then the object will rotate clockwise, no matter where the force is applied.

It seems like your intuition for which way the object should rotate is based on the situation where the force is applied only in the vertical direction, rather than at a diagonal as pictured. And indeed, if you set a book on a table and try this experiment by pushing it with your finger along the line of action, it will confirm your intuition if your finger slips while you push it. But if your finger slips, then you're applying less force in the horizontal direction than you intended, so this is not the experiment that you want to do! If instead you tape a piece of string to the book and pull, then you will find that the book does not rotate (or only rotates a bit, since the setup won't be perfect).
Another thing that can mess with your intuition is the fact that the line of action changes as the object moves. In these problems, you're interested in what happens to the object when everything is oriented as pictured. As soon as the object moves a little bit, the line of action may change relative to the center of mass. This gives another reason why you might think that the left and right cases are quite different. The left case is like hanging the book by a string from the top (which is stable), while the right is like trying to balance the book on your finger (which is not). But although this means that the dynamics after $t=0$ will be quite different, the dynamics at $t=0$ are identical. I think this line of reasoning can explain your confusion about the L-shaped object.
Also for those (like me) who had never seen this thing about line of action before, here's a quick proof: If you apply a force at a different point along the line of action, the torque in this case is $\vec{\tau} = (\vec{r} + \vec{r}')\times \vec{F} = \vec{r}\times\vec{F} + \vec{r}'\times\vec{F}$, where $\vec{r}'$ is the difference between the original and new points of action. Because $\vec{r}'$ is parallel to $\vec{F}$, the second term is zero and the torque is unchanged.
