Say there is a block sliding down an inclined plane that rests on a frictionless table. There is kinetic friction between the block and inclined plane. If the block slides downhill, then the kinetic friction acting on it points uphill. By Newton’s third law, the inclined plane will experience a friction force pointing down hill, in the direction of the block’s velocity/acceleration. Shouldn’t the plane want to move with the block, then? If it does, then wouldn’t the center of mass move with the block and plane too? There’s no friction on the block–plane system, however, so the center of mass should not move, but my analysis claims that it does. Where have I gone wrong?
[Physics] Inclined Plane and Center of Mass
forcesfree-body-diagramfrictionnewtonian-mechanics
Best Answer
You are forgetting the normal force between the block and the wedge (inclined plane). This force has a horizontal component, pushing the wedge to the left as the block slides down to the right.
The 2 forces acting on the wedge due to the block are the normal reaction $N$ and the friction force $F=\mu |N|$ (see diagram). Both have horizontal components : $N\sin\theta$ and $F\cos\theta$ respectively. The net horizontal force acting to the left on the wedge is $N\sin\theta-F\cos\theta=(\tan\theta -\mu)N\cos\theta$.
The condition for the block to start sliding down the incline is $\tan\theta \gt \mu$. So if the block slides to the right then the horizontal force on the wedge is always +ve to the left.
Note that in this situation the normal force is not $N=mg\cos\theta$. The wedge accelerates to the left away from the plane of contact with the block, so the normal force is reduced from this value.