Simple beginning physics, mindfart

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The discussion focuses on a physics problem involving three blocks connected by pulleys, where the challenge lies in calculating the acceleration of the system. The user is struggling with the forces acting on the central block, particularly how to incorporate tension and friction into their calculations. They have created free-body diagrams but are confused about the different tensions on either side of the middle block. A response suggests that in a single massless string, the tension remains constant, which can simplify the analysis of forces along the x-axis. Understanding this concept is crucial for solving the problem effectively.
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Homework Statement



2 blocks are suspended on the sides of a table, over pulleys, connected to a central block on the table. From the left to the right, the masses are 4kg, 2kg and 6kg. There is a friction coefficient of 0.12 on the middle block. Find the acceleration.



Homework Equations



F=ma, of course


The Attempt at a Solution



For some reason, I'm having trouble with the center block. I have drawn free-body diagrams for each object, and at the center, I'm having trouble with determining the x-axis forces.

I have marked friction in the opposite direction of the acceleration, but can't seem to figure out how to factor in Tension, as there should be tension on both sides of the object.

This, I know is a very simple thing, but for some strange reason, I'm hung up on it.
 
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Show your equations. Do not forget that there are different tensions in the ropes at the opposite sides of the middle block.

ehild
 
In a single massless string, tension is same. Use it to get forces in x axis
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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