Question

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 teh system and the force by the rod on one of the blocks.

Answer 1

From the free body diagram
$R_1=M_{1}g\cos \theta R_2=M_{2}g\cos \theta T+M_{1}gsin\theta -M_{1}a-\mu R_1=0T-M_{2}g+M_{2}a+\mu R=0$
From equation (iii)
$T+M_{1}g\sin \theta -M_{1}a-cos\theta =0$
From equation (iii)
$T-M_{2}g\sin \theta +\mu M_{2}g\cos \theta =0$
From equation (v)
$g\sin \theta (M_1+M_2)=\mu g\cos \theta (M_1+M_2)\Rightarrow a(M_1+M_2)=g\sin \theta (M_1+M_2)-\mu g\cos \theta (M_1+M_2)\Rightarrow a=g(\sin \theta -\mu \cos \theta )$
$\therefore$ The block (system) has acceleration
$=g(\sin \theta -\mu \cos \theta )$
The force exerted by the rod on one of the blocks is tension.
Tension,
$T=-M_{1}g\sin \theta +M_{1}a+\mu M_{1}g\cos \theta =-M_{2}g\sin \theta +M_1(g\sin \mu -\mu g\cos \theta )+mu{M}_{1}g\cos \theta =0$