This is going to be a long post-like question so make sure you have enough time before reading this. And this I got after thinking continuously for about 4 hours on the same question and I want to verify that if my ideology about this question is correct. First, what the question actually is?
Question
Please refer to the diagram in image
Explanation of the Image (You can skip this)
The image of the question is just below, please refer to it for better visualization. There are two blocks A and B. Block A has a mass of $m$ while block two has a mass of $2m$. They are placed on top of each other. The surface between them (Block A and B) has a coefficient of friction $\mu$ and that between Block B and ground is $2\mu$. In front of these two blocks is a wall that has a pulley attached to it (the wall). Now the block A is attached to a string which then runs over the pulley which is attached to the wall. And from that pulley, the string runs over another pulley which is attached to Block B and this string finally terminates getting attached back to the wall ( refer image for better visualization ). Now a force $\overleftarrow{F}$ is applied on the Block A.
Actual Question
What would be the minimum value of $F$ so that the blocks start to move?
My Solution
One Note Before Starting This
Whenever I say a force being "paid by a system", I mean either it gets canceled by another force on the system or it gets exerted on the system. Also, the source of the force mentioned above in the main title of the question refers to the main force $\overleftarrow{F}$ acting on the system.
Now the solution
The most shocking thing I learned while solving this question was that the friction between block A and B will be paid two times all differently by the same force $\overleftarrow{F}$ acting in disguise on the two blocks. This was shocking as I first thought that the friction will be the usual force paid by the two blocks in some ration and not two times. But what I came to figure out was that it will be paid by each block separately and the amount of force of friction paid will be the same for two blocks and not in some ratio and those two will actually be the action-reaction pairs. One way we can thing about this is that the first block let's say starts moving in the left direction then it will exert a force of friction (which would be the reaction pair of the friction on block A) on the second block. Now to overcome this force the second block will pay the same amount of force in opposite direction. Hence the two pairs of force which are action-reaction pairs will be paid differently by each block. Now, these two pairs will be equal to $\mu N$ where $N$ is the Normal force by Block A on B. And $N = mg$. Hence
Friction b/w Block A and B $ = \mu N \\ \rightarrow \mu mg$
Friction b/w Block B and ground $= 2\mu N \\ \rightarrow 2\mu(3m)g \\ \rightarrow 6\mu mg$
For Block A
Now on block A, there is a force $\overleftarrow{F}$ and a friction $\mu mg$ which would be paid completely by this block so the tension $T$ in the string would be
$$ T = F - \mu mg $$ Let's say this equation one.
For Block B
Now there is $2T$ on block B and friction from the ground and surface above and remember that this block will pay its reaction pair separately. Hence
$$ 6\mu mg + \mu mg = 2T $$
Let's say this equation two.
Analyzing equation 1 and 2 we get
$$7\mu mg = 2F - 2\mu mg \\ F = \frac{9\mu mg}{2}$$
Final Question
Now here I want to ask that was this way of thinking correct or was it just a coincidence that I reached this solution? And thank you for spending your precious time on this question to help me out. I appreciate your efforts.