Determine the mass M of the hanging block shown in the figure that...

Let the acceleration of the system be $a$ to the right. For the smaller block of mass $m$ to not slip over the triangular block, it must move with the same horizontal acceleration $a$ as the triangular block.

Considering the free body diagram of the smaller block $m$ from the ground frame, the forces acting on it are its weight $mg$ downwards and the normal reaction $N$ perpendicular to the inclined surface.

Resolving the normal reaction $N$ into horizontal and vertical components:
Vertical equilibrium: $N \cos \theta = mg$
Horizontal acceleration: $N \sin \theta = ma$

Dividing the two equations, we get the required acceleration for no slipping:
$\tan \theta = \dfrac{a}{g} \implies a = g \tan \theta$

Now, considering the entire system of masses $(M' + m)$ moving horizontally and $M$ moving vertically downwards, the common acceleration $a$ is given by the net pulling force divided by the total mass:
$a = \dfrac{Mg}{M + M' + m}$

Equating the two expressions for $a$:
$\dfrac{Mg}{M + M' + m} = g \tan \theta$

Canceling $g$ and rearranging to solve for $M$:
$M = (M + M' + m) \tan \theta$
$M - M \tan \theta = (M' + m) \tan \theta$
$M(1 - \tan \theta) = (M' + m) \tan \theta$

$M = \dfrac{(M' + m) \tan \theta}{1 - \tan \theta}$

Dividing the numerator and the denominator by $\tan \theta$:
$M = \dfrac{M' + m}{\cot \theta - 1}$