The plank is in static equilibrium, so the net force and net torque on it must be zero. Let $F_A$ and $F_B$ be the upward support forces from supports A and B, respectively.

For vertical force equilibrium:

$$\sum F_y = F_A + F_B - Mg - mg = 0$$

For rotational equilibrium, we sum the torques about pivot A:

$$\sum \tau_A = F_B \cdot L - Mg \cdot \frac{L}{2} - mg \cdot \frac{L}{4} = 0$$

Solving the torque equation for $F_B$:

$$F_B L = \frac{MgL}{2} + \frac{mgL}{4}$$ $$F_B = \frac{Mg}{2} + \frac{mg}{4}$$

Substituting the given values:

$$F_B = \frac{(6.0 \text{ kg})(10 \text{ m/s}^2)}{2} + \frac{(4.0 \text{ kg})(10 \text{ m/s}^2)}{4} = 30 \text{ N} + 10 \text{ N} = 40 \text{ N}$$

Now, use the force equilibrium equation to find $F_A$:

$$F_A = (M+m)g - F_B$$ $$F_A = (6.0 \text{ kg} + 4.0 \text{ kg})(10 \text{ m/s}^2) - 40 \text{ N} = 100 \text{ N} - 40 \text{ N} = 60 \text{ N}$$

By Newton's third law, the forces exerted by the plank on the supports are equal in magnitude to the forces exerted by the supports on the plank.