Question: [![enter image description here](https://i.stack.imgur.com/X4cBv.png)](https://i.stack.imgur.com/X4cBv.png)

Here is a picture from [Wikipedia](https://en.wikipedia.org/wiki/Euler%27s_critical_load), a free body diagram used to derive the maximum load in Euler's buckling theory. The moment is calculated around point A. It is claimed that the load $P$ causes a moment $Pw$ around point A, but why is this? In the left picture above, we can see that the load $P$ is applied directly towards point A without offset. As the vertical distance from the point is zero, shouldn't the moment be zero?

I understand that the same force $P$ is also present on the top of the buckle. So let's consider a similar situation:

[![enter image description here](https://i.stack.imgur.com/tovKS.png)](https://i.stack.imgur.com/tovKS.png)

We could imagine the rectangular arch to be any semi-circular object. If we push the ends (marked with the arrows) with equal forces, we have equilibrium in the x-direction. But is there a net moment around either end here? I think not. I tried this in real life with several objects and the ends simply bend towards inside, no rotation. **So what is happening in the buckling case?** Why is there a moment from the force $P$, if its distance is actually zero from the horizontal line?

Answer: 
I think the best way to analyse a compressive loaded column is to suppose that it is already buckled, and see if it stays buckled for a small decrease of the load. 

For an horizontal beam, of length $L$, supported at both sides, with a concentrated load at the centre, the deflection at the centre is a function of the applied momentum, that is a function of the force. Any small decrease of the force results in a small decrease of the momentum, that results in a small decrease of the deflection. And the new equilibrium is reached.

For a buckled column of the same length, loaded in compression with a force $F$, the deflection $d$ at the centre is also a function of the momentum $M$, but $M$ depends on $F$ and $d$. 

Suppose that we decrease $F$ by a $\Delta F$, and we want to calculate the new equilibrium state by an iterative method:

The momentum is now $M = (F-\Delta F)d$. 

But $d$ is a function of the momentum and is now smaller by a $\Delta d$. 

The momentum is then now: $M = (F-\Delta F)(d-\Delta d)$, what leads to a smaller $d$ and so on.

If the iteration converges to a new smaller $d$, the initial situation is stable. But if the initial $d$ is smaller than a threshold, it is possible that the iteration doesn't converge. 

That threshold is the smallest stable $d$ value. And that corresponds to the force $F$ limit for buckling. 

Of course I did not proof that such a threshold exists, it is only a conjecture to explain intuitively why there is a specific $F$ value to start buckling.
Your example won’t really test whether there’s a moment at A, as your finger forms a free joint there. 

Try bending a shape that runs along the left arrow, up, over, back down, and then out along the right arrow. Pushing in on the ends will then show how a moment appears that acts to bend the new joints.