[Q1] Case: $m_A > \mu m_B$ (System Accelerates)

The condition $m_A g > \mu m_B g$ indicates that the gravitational force on mass A is greater than the maximum friction force on mass B, so the system will accelerate. Let $a$ be the magnitude of the acceleration of both masses and $T$ be the tension in the string.

1. Equations of Motion: We apply Newton's second law to each object. Assume A accelerates downwards and B accelerates to the right.

   • For mass A (vertical motion): The net force is the weight minus the tension.

     $$m_A g - T = m_A a \quad (1)$$

   • For mass B (horizontal motion): The net force is the tension minus the kinetic friction force, $f_k$. The normal force on B is $N = m_B g$, so the kinetic friction is $f_k = \mu N = \mu m_B g$.

     $$T - \mu m_B g = m_B a \quad (2)$$

2. Solve for Acceleration ($a$): Add equations (1) and (2) to eliminate the tension $T$.

   $$(m_A g - T) + (T - \mu m_B g) = m_A a + m_B a$$ $$g(m_A - \mu m_B) = (m_A + m_B)a$$ $$a = \frac{m_A - \mu m_B}{m_A + m_B}g$$

3. Solve for Tension ($T$): Substitute the expression for $a$ back into equation (2).

   $$T = m_B a + \mu m_B g$$ $$T = m_B \left( \frac{m_A - \mu m_B}{m_A + m_B}g \right) + \mu m_B g$$ $$T = \frac{m_A m_B g - \mu m_B^2 g + \mu m_B g(m_A + m_B)}{m_A + m_B}$$ $$T = \frac{m_A m_B g - \mu m_B^2 g + \mu m_A m_B g + \mu m_B^2 g}{m_A + m_B}$$ $$T = \frac{m_A m_B g(1 + \mu)}{m_A + m_B}$$

[Q2] Case: $m_A \leq \mu m_B$ (System at Rest)

1. Condition for Static Equilibrium: The system is initially at rest. The driving force attempting to cause motion is the weight of A, $W_A = m_A g$. The maximum static friction force that resists the motion is $f_{s,max} = \mu_s N = \mu_s m_B g$. Assuming the coefficient of static friction $\mu_s$ is at least as large as the given kinetic friction coefficient $\mu$, the condition for the system to remain at rest is $W_A \leq f_{s,max}$. The problem condition is $m_A \leq \mu m_B$, which is equivalent to $m_A g \leq \mu m_B g$. This means the driving force is less than or equal to the maximum available friction force. Therefore, the system remains in static equilibrium.

2. Acceleration and Tension:

   • Since the system does not move, its acceleration is zero. $$a = 0$$
   • With $a=0$, we analyze the forces on mass A in equilibrium. $$m_A g - T = m_A(0) = 0$$ $$T = m_A g$$