A plank B of mass 4 kg and length 0.9 m lies on a long horizontal table…

Step 1. The friction forces. The table holds up both A and B, so it pushes up on B with $(m_A + m_B)g = 60\ \text{N}$. Once B moves, the table's friction on it is $\mu_2 \times 60\ \text{N} = 6\ \text{N}$. Between A and B the normal force is $m_A g = 20\ \text{N}$, so friction on A can be at most $\mu_1 \times 20\ \text{N} = 6\ \text{N}$.

Step 2. Moving together. If A, B and C move as one with acceleration $a$, then

A is pushed forward only by friction, so it needs $f = m_A a$. This cannot be more than $\mu_1 m_A g$, so $a \le \mu_1 g = 3\ \text{m/s}^2$.

Step 3. The range. Put $a = \mu_1 g$ into the equation of Step 2: $m_C(1 - \mu_1) = (\mu_1 + \mu_2)(m_A + m_B)$, so

Also, nothing moves unless the weight of C beats the table's friction: $m_C > \mu_2(m_A + m_B) = 0.6\ \text{kg}$. So the range is $0.6\ \text{kg} < m_C \le 3.43\ \text{kg}$.

Step 4. Now $m_C = 6\ \text{kg}$. This is more than $3.43\ \text{kg}$, so A slips. Friction on A is kinetic, $6\ \text{N}$ forwards, so

B and C move together, with the string taut. Along their motion act the weight of C, $60\ \text{N}$, forwards; the friction from A on B, $6\ \text{N}$, backwards; and the table's friction, $6\ \text{N}$, backwards:

The tension follows from C alone: $T = m_C(g - a_B) = 6\ \text{kg} \times 5.2\ \text{m/s}^2 = 31.2\ \text{N}$.

Step 5. Falling off. Seen from B, A moves backwards with acceleration $4.8\ \text{m/s}^2 - 3.0\ \text{m/s}^2 = 1.8\ \text{m/s}^2$, starting from rest. It must slide back the whole plank:

In that time A moves forwards over the table by $\tfrac12 \times 3.0\ \text{m/s}^2 \times (1.0\ \text{s})^2 = 1.5\ \text{m}$.