As the block ‘m’ does not slinover M', ct will have same acceleration as that of M'
From the freebody diagrams.
T + Ma – Mg = 0 ...(i) (From FBD – 1) ( T works in the upper direction)
T – M'a – R sin = 0 ...(ii) (From FBD -2)
R sin – ma = 0 ...(iii) (From FBD -3)
R cos – mg =0 ...(iv) (From FBD -4)
Eliminating T, R and a from the above equation, we get M = (M'+m)/cot -1