Step 1: Understanding the Concept:
This problem is best solved from the non-inertial frame of the accelerating larger block. In this frame, pseudo forces act on the two smaller blocks in the direction opposite to the acceleration. We must ensure equilibrium for both smaller blocks.
Step 2: Key Formula or Approach:
Pseudo force: \(F_p = m a\).
Friction condition to prevent slipping: \(f \le \mu N\).
Equilibrium equations: \(\sum F_x = 0\) and \(\sum F_y = 0\).
Step 3: Detailed Explanation:
The larger block accelerates rightward with \(a = g/2\).
For the 2 kg block (top, smooth surface):
Pseudo force acts leftward: \(F_{p1} = m_1 a = 2 \times (g/2) = g\).
Tension \(T\) acts rightward. For equilibrium:
\[ T = g \]
For the 1.5 kg block (vertical, rough surface):
Pseudo force acts leftward, pushing it into the wall: \(F_{p2} = m_2 a = 1.5 \times (g/2) = 0.75g\).
The normal reaction is therefore \(N = 0.75g\).
In the vertical direction, gravity pulls down (\(m_2 g = 1.5g\)), while Tension \(T\) and static friction \(f\) pull up.
\[ T + f = 1.5g \]
Substitute \(T = g\):
\[ g + f = 1.5g \implies f = 0.5g \]
To prevent the block from slipping, the required friction must not exceed the maximum available static friction:
\[ f \le \mu N \]
\[ 0.5g \le \mu (0.75g) \]
\[ \mu \ge \frac{0.5}{0.75} = \frac{1/2}{3/4} = \frac{2}{3} \]
Thus, the minimum value is \(\mu = 2/3\).
Step 4: Final Answer:
The minimum value of \(\mu\) is \(2/3\).
