Two masses `M_1 and M_2` are connected by a light rod and the system is slipping down a rough incline of angle `theta` with the horizontal. The friction coefficient at both the contacts is `mu`. Find the acceleration of the system, and the force by the rod on one of the blocks.

Explanation

Fro the free body diagram
`R_1=M_1gcostheta`..i
`R_2=M_2gcostheta`……ii
`T+M_1gsintheta-M_1a-mu R_1=0`……….iii
`T-M_2g+M_2a+muR=0`……..iv
From equation iii,
`T+M_1gsintheta-M_1a-costheta=0` …….iv
from equation iii,
`T-M_2gsintheta+muM_2gcostheta=0` .........v
From equation v
`g sin theta (M_1+M_2)=Mugcostheta(M_1+M_2)`
`-. a(M_1+M_2)=g sintheta (M_1+M_2)`
`-mug costheta (M_1+M_2)`
`gt a=g(sintheta-mucostheta)`
`:. The blcok (system) has acceleration
`=g(sintheta+mucostheta)`
The force exerted by the rod on one of the blocks is tension
`Tenson T=-M_1 sin theta+M_1a+mu M_1g cos theta`
`=M_1g sin theta+M_1(gsinmu-mugcostheta)+muM_1gcostheta=0`