Find out acceleration at center of mass of the given system.

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Solution

The net force on this system is weight of the block $3 M$
$\Rightarrow {\to F}_{N e t} = M_{C O M} {\to a}_{C O M} (M_{C O M} = M + 3 M)$
$\Rightarrow 3 M g = 4 M \times {\to a}_{C O M}$
$\Rightarrow {\to a}_{C O M} = \frac{3}{4} g$

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Find out acceleration at center of mass of the given system.

The net force on this system is weight of the block 3M⇒→FNet=MCOM→aCOM(MCOM=M+3M)⇒3Mg=4M×→aCOM⇒→aCOM=34g

The net force on this system is weight of the block $3 M$
$\Rightarrow {\to F}_{N e t} = M_{C O M} {\to a}_{C O M} (M_{C O M} = M + 3 M)$
$\Rightarrow 3 M g = 4 M \times {\to a}_{C O M}$
$\Rightarrow {\to a}_{C O M} = \frac{3}{4} g$