1. Tension Analysis

The main string supports both movable pulleys. Since the string is continuous and pulleys are frictionless, the tension $T$ acts on every block identically.

For the blocks A ($m$), B ($2m$), C ($2m$), and D ($4m$), the equation of motion is: $T – mg = ma$.

2. Constraint Relation

Summing the length of strings and differentiating twice gives the constraint: $\sum a_i = 0$.

$$(\frac{T}{m} – g) + (\frac{T}{2m} – g) + (\frac{T}{2m} – g) + (\frac{T}{4m} – g) = 0$$

Solving for T:

$$T \left( \frac{1}{m} + \frac{1}{2m} + \frac{1}{2m} + \frac{1}{4m} \right) = 4g$$ $$T = \frac{16mg}{9}$$

3. Final Accelerations

Substituting $T$ back into $a = \frac{T}{m} – g$:

Block A ($m$): $a_A = \frac{7g}{9} \uparrow$

Block B ($2m$): $a_B = \frac{g}{9} \downarrow$

Block C ($2m$): $a_C = \frac{g}{9} \downarrow$

Block D ($4m$): $a_D = \frac{5g}{9} \downarrow$