System Analysis First, we analyze the two blocks A and B as a single system of total mass $M = m_A + m_B$. The system moves horizontally with a common acceleration $a$. Applying Newton's second law in the horizontal direction:

$$F - f_k = (m_A + m_B)a$$

The total kinetic friction force is $f_k = \mu N_{total} = \mu(m_A+m_B)g$.

$$F - \mu(m_A+m_B)g = (m_A+m_B)a$$

This gives the acceleration of the system:

$$a = \frac{F}{m_A+m_B} - \mu g$$

[Q1] Interaction Force between A and B Let $N_{AB}$ be the magnitude of the normal force exerted by A on B. This force is perpendicular to the contact surface. We analyze the forces on block B. The forces acting on block B are:

1. Weight $m_B g$ (downward).
2. Normal force from the ground $N_B$ (upward).
3. Kinetic friction from the ground $f_B = \mu N_B$ (leftward).
4. Interaction force from A, $N_{AB}$. This force has a horizontal component $N_{AB}\sin\alpha$ (rightward) and a vertical component $N_{AB}\cos\alpha$ (downward).

Applying Newton's second law to block B:

• Vertical equilibrium ($\sum F_y = 0$):

$$N_B - m_B g - N_{AB}\cos\alpha = 0 \implies N_B = m_B g + N_{AB}\cos\alpha$$

• Horizontal motion ($\sum F_x = m_B a$):

$$N_{AB}\sin\alpha - f_B = m_B a \implies N_{AB}\sin\alpha - \mu N_B = m_B a$$

Substitute the expression for $N_B$ into the horizontal equation:

$$N_{AB}\sin\alpha - \mu(m_B g + N_{AB}\cos\alpha) = m_B a$$

Rearranging to solve for $N_{AB}$:

$$N_{AB}(\sin\alpha - \mu\cos\alpha) = m_B(a + \mu g)$$

From the system analysis, we have $a + \mu g = \frac{F}{m_A+m_B}$. Substituting this in:

$$N_{AB}(\sin\alpha - \mu\cos\alpha) = m_B \left(\frac{F}{m_A+m_B}\right)$$

The interaction force is:

$$N_{AB} = \frac{m_B F}{(m_A+m_B)(\sin\alpha - \mu\cos\alpha)}$$

[Q2] Allowed Range for F For the blocks to move together as described, three conditions must be met:

1. The blocks must accelerate: $a > 0$.

   $$ \frac{F}{m_A+m_B} - \mu g > 0 \implies F > \mu(m_A+m_B)g $$

   This sets the minimum force required.

2. The blocks must remain in contact: $N_{AB} > 0$. Since $F$, $m_A$, and $m_B$ are positive, this requires the denominator of the $N_{AB}$ expression to be positive:

   $$ \sin\alpha - \mu\cos\alpha > 0 \implies \tan\alpha > \mu $$

   This is a necessary condition on the system parameters for this type of motion to be possible.

3. Block A must not lift off the ground: The normal force on A from the ground, $N_A$, must be non-negative ($N_A \ge 0$). We analyze the vertical forces on block A. The interaction force from B on A, $N_{BA}$, is equal in magnitude to $N_{AB}$ and acts with an upward vertical component $N_{BA}\cos\alpha$. Vertical equilibrium for block A:

   $$N_A - m_A g + N_{BA}\cos\alpha = 0 \implies N_A = m_A g - N_{AB}\cos\alpha$$

   The condition $N_A \ge 0$ implies:

   $$m_A g \ge N_{AB}\cos\alpha$$

   Substitute the expression for $N_{AB}$:

   $$m_A g \ge \frac{m_B F \cos\alpha}{(m_A+m_B)(\sin\alpha - \mu\cos\alpha)}$$

   Solving for $F$ gives the maximum allowed force:

   $$F \le \frac{m_A g (m_A+m_B)(\sin\alpha - \mu\cos\alpha)}{m_B \cos\alpha}$$ $$F \le \frac{m_A(m_A+m_B)g}{m_B}(\tan\alpha - \mu)$$

   (Note: The normal force on B, $N_B = m_B g + N_{AB}\cos\alpha$, is always positive, so B never lifts off.)

Combining the conditions on $F$, the allowed range is:

$$\mu(m_A+m_B)g < F \le \frac{m_A(m_A+m_B)g}{m_B}(\tan\alpha - \mu)$$