As shown in the figure, the friction coefficient between the board...

Let $T$ be the tension in the rope, which is equal to the force exerted by the man on the rope.

Let us consider the man and the board as a single system. The external forces acting on this system in the vertical direction are:
1. Total weight of the man and the board, $(M+m)g$, acting downwards.
2. Normal reaction $N$ from the floor, acting upwards.
3. Tension $T$ from the rope pulling the man upwards.

For vertical equilibrium of the system, we have:
$N + T = (M+m)g$
$\Rightarrow N = (M+m)g - T$

In the horizontal direction, the rope passes around the pulley on the floor and pulls the board to the right with a force equal to the tension $T$. To prevent slipping, the static friction $f$ between the board and the floor must act to the left and balance this pulling force.

For horizontal equilibrium, we have:
$f = T$

The condition for the board not to slip on the floor is that the required friction must be less than or equal to the maximum possible static friction:
$f \le \mu N$

Substituting the expressions for $f$ and $N$ into the inequality:
$T \le \mu [(M+m)g - T]$

Solving for $T$:
$T \le \mu(M+m)g - \mu T$
$T + \mu T \le \mu(M+m)g$
$T(1 + \mu) \le \mu(M+m)g$
$T \le \dfrac{\mu(M+m)g}{1 + \mu}$

Therefore, the maximum force the man can exert on the rope without the board slipping is $\dfrac{\mu(M+m)g}{1 + \mu}$.