Solve the following :

Both blocks are connected by rod.

So, there acceleration must be same.

$M_{1} g \sin \theta-T-f f_{1}=M_{1} a$

$f f_{1}=\mu N_{1}=\mu M_{1} g \cos \theta$

Add (1) and (3)

$M_{1} g \sin \theta+M_{2} g \sin \theta-f f_{1}-f f_{2}=\left(M_{1}+M_{2}\right) a$

$\left(M_{1}+M_{2}\right) g \sin \theta-\mu g \cos \theta\left(M_{1}+M_{2}\right)=\left(M_{1}+M_{2}\right) a$

$g \sin \theta-\mu g \cos \theta=a$

$M_{1} g \sin \theta-T-\mu M_{1} g \cos \theta=M_{1}(g \sin \theta-\mu g \cos \theta)$

$\mathrm{T}=0$