We have three blocks A, B, and C of equal mass $m$.

• Block C: Slides horizontally on the frictionless floor.
• Block B: Is constrained against the right vertical face of C. It moves horizontally with C but can slide vertically.
• Block A: Is a hanging mass that drives the system.

String Path: The string comes from the fixed pulley, passes over the pulley mounted on the top-right corner of C, and connects down to Block B.

Let $x_C$ be the horizontal position of block C (and pulley).
Let $y_A$ be the downward position of block A.
Let $y_{B/C}$ be the vertical position of block B relative to the pulley on C.

The total length of the string $L$ is constant. The segments are:

1. Horizontal segment between pulleys: $X_{fixed} – x_C$
2. Vertical segment supporting A: $y_A$
3. Vertical segment supporting B: $y_{B/C}$

$$ L = (X_{fixed} – x_C) + y_A + y_{B/C} $$ Differentiating twice with respect to time (acceleration): $$ 0 = -a_C + a_A + a_{B/C} $$ $$ a_C = a_A + a_{B/C} \quad \text{…(1)} $$

Let $T$ be the tension. All masses are $m$.

For Block A (Hanging):
$$ mg – T = m a_A \quad \text{…(2)} $$

For Block B (Vertical motion):
B moves vertically relative to C. $$ mg – T = m a_{B/C} \quad \text{…(3)} $$

For Block B (Horizontal motion):
B moves to the right with acceleration $a_C$. The force providing this is the Normal force $N$ from C acting on B. $$ N = m a_C \quad \text{…(4)} $$

For Block C (Horizontal motion):
Forces on C in x-direction:

• Tension $T$ acting on the pulley (pulling right).
• Normal force $N$ from B acting on C (pushing left, Newton’s 3rd law).

$$ T – N = m a_C \quad \text{…(5)} $$

Comparing (2) and (3), forces are identical ($mg-T$). Thus: $$ a_A = a_{B/C} $$ Substitute into (1): $$ a_C = a_A + a_A = 2a_A $$ Substitute $N$ from (4) into (5): $$ T – m a_C = m a_C \Rightarrow T = 2 m a_C $$ Using $a_C = 2 a_A$: $$ T = 4 m a_A $$ Substitute $T$ back into (2): $$ mg – 4 m a_A = m a_A $$ $$ mg = 5 m a_A \Rightarrow a_A = \frac{g}{5} $$ Since $g = 10 \, \text{m/s}^2$: $$ a_A = 2 \, \text{m/s}^2 $$ Then $a_C = 2(2) = 4 \, \text{m/s}^2$. And $a_{B/C} = 2 \, \text{m/s}^2$.