Consider the system consisting of Block A ($m$), Block B ($3m$), and Wedge C ($2m$). Since all surfaces are frictionless and the external forces (gravity and normal force from the floor) act vertically, there is no external horizontal force on the system.

Therefore, the horizontal position of the Center of Mass (CM) remains unchanged: $$ \Delta X_{CM} = \frac{m_A \Delta x_A + m_B \Delta x_B + m_C \Delta x_C}{m_A + m_B + m_C} = 0 $$ $$ m_A \Delta x_A + m_B \Delta x_B + m_C \Delta x_C = 0 $$

Let $x$ be the displacement of the wedge to the left. We define right as positive ($+$) and left as negative ($-$).

• Wedge C ($2m$): Displacement $\Delta x_C = -x$.
• Block B ($3m$): Constrained to move horizontally with the wedge. Displacement $\Delta x_B = -x$.
• Block A ($m$):
  • Since the string is inextensible and Block B moves down by $h$, Block A must move up the incline relative to the wedge by a distance $h$.
  • Horizontal component of this relative motion is $h \cos \theta$ (to the right).
  • Net horizontal displacement: $\Delta x_A = -x + h \cos \theta$.

Substitute these displacements into the center of mass equation:

$$ m(-x + h \cos \theta) + 3m(-x) + 2m(-x) = 0 $$

$$ -mx + mh \cos \theta – 3mx – 2mx = 0 $$

$$ mh \cos \theta – 6mx = 0 $$

$$ x = \frac{h \cos \theta}{6} $$

Given values:

• Vertical fall of B: $h = 27 \text{ cm}$
• Inclination: $\cos \theta = 2/3$

$$ x = \frac{27 \cdot (2/3)}{6} = \frac{18}{6} = 3 \text{ cm} $$