Solve the following :


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.


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)

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

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

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

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



Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now