We have a system where a single inextensible cord connects the blocks.

• Blocks A ($M_1$) and B ($M_2$): Move horizontally towards each other.
• Blocks C ($m_1$) and D ($m_2$): Hang vertically and move downwards.

The cord path is: Start at C $\rightarrow$ Over pulley on A $\rightarrow$ Across to pulley on B $\rightarrow$ Down to D.

Let $T$ be the tension in the cord. The coordinate system defines the positive x-direction to the right.

For Block A (moves right):
Note: Since C hangs from the string passing over A, C moves horizontally with A. We treat the horizontal acceleration of the system (A+C). However, the tension acts horizontally only on A via the pulley axis. $$ T = (M_1 + m_1) a_A \implies a_A = \frac{T}{M_1 + m_1} $$ (Note: We include $m_1$ in the horizontal inertia because C is suspended from A and must accelerate horizontally with it).

For Block B (moves left):
Similarly for B and D. $$ T = (M_2 + m_2) a_B \implies a_B = \frac{T}{M_2 + m_2} $$ (Magnitude is $a_B$, direction is left).

For Block C (moves down):
$$ m_1 g – T = m_1 a_C \implies a_C = g – \frac{T}{m_1} $$

For Block D (moves down):
$$ m_2 g – T = m_2 a_D \implies a_D = g – \frac{T}{m_2} $$

The total length of the string is constant. The slack created by A moving right ($x_A$) and B moving left ($x_B$) is taken up by C dropping ($y_C$) and D dropping ($y_D$). $$ \sum \Delta \text{Length} = 0 $$ Decrease in horizontal length = Increase in vertical lengths $$ a_A + a_B = a_C + a_D $$

Substitute the acceleration expressions into the constraint: $$ \frac{T}{M_1 + m_1} + \frac{T}{M_2 + m_2} = \left(g – \frac{T}{m_1}\right) + \left(g – \frac{T}{m_2}\right) $$ Group terms with $T$: $$ T \left[ \frac{1}{M_1+m_1} + \frac{1}{M_2+m_2} + \frac{1}{m_1} + \frac{1}{m_2} \right] = 2g $$ Let $K = \left[ \frac{1}{M_1+m_1} + \frac{1}{M_2+m_2} + \frac{1}{m_1} + \frac{1}{m_2} \right]$. Then $T = \frac{2g}{K}$.