Question:
Find the acceleration of the block of mass M in this figure. The coefficient of friction between the two blocks is $\mu_1$ and that between the bigger block and ground is $\mu_2$.
Doubt:
I have understood the question, but the only thing that I am confused with is the tension on the bigger block due to the pulley.
Why will there be a tension in right and downward direction due to the pulley? Kindly help.
It seems to me there are several problems with the force diagrams. First of all, the normal force on the small block is to the right, not to the left, and the "inertial force" ma is to the left, not to the right (when acceleration is to the right, the effective inertial force is to the left, if you are using the constraint that all the forces add up to zero). Making arrows point the wrong way would normally simply result in them coming out with minus signs for their solved magnitudes, which is confusing but not necessarily wrong, but the normal force R on the big block is pointed in the correct direction (to the left), so the signs would not always come out correct with this inconsistency. (They do in this case because as you can see, R = ma, so it doesn't matter which way either one points when you are just getting their magnitudes.) Also, we should assume the big block has the same rightward acceleration a as the little block, so there should be an effective inertial force Ma to the left on the big block. That mistake would give the wrong answer.
By the way, we don't normally include the "inertial force" ma in a force diagram, as it is not a real force, but you can imagine it is a force if you want to make all the forces add up to zero, which is a fine way to get the right answer to questions like this. But you do need the correct force diagram. It's actually a pretty cute problem for testing if you understand force diagrams, I've never seen one this complicated.
The answer to why there are two more T forces, one to the right and one down, is that one of the pulleys is attached to the big block. The pulley is apparently assumed to have zero mass, so must not have any net force on it. The rope applies T to the right and also down, so the big block must apply a force T to the left and also up, and the action-reactions to that are what you see in the diagram.
