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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?

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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. enter image description here

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.

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  • $\begingroup$ Excellent answer, but I suggest that you remove the disclaimer and put mg cos$\theta$ back in there for N. The force which drives acceleration isn't reduced by the acceleration. In this case that force is (tan$\theta$-$\mu$)N cos$\theta$, so N doesn't change. You may be confusing this with the change in the normal force exerted on a scale by a person in an elevator accelerating downward. But in that case the acceleration of the scale isn't driven by the normal force exerted on it. $\endgroup$ – D. Ennis Oct 15 '16 at 20:31
  • $\begingroup$ Thank you for the suggestion. However, I think this situation is like the accelerating elevator. There is a component of acceleration perpendicular to the plane of contact, so the normal reaction is affected. $\endgroup$ – sammy gerbil Oct 15 '16 at 21:01
  • $\begingroup$ I see it as more like me pushing my car, and the car pushing back. As the car accelerates it pushes back with an equal force. I could put a bathroom scale between me and the car and read the force. By contrast, in the elevator I push down on the stationary scale with my weight, but when it and the elevator accelerate away from me the scale reads less because my weight isn't causing the acceleration. $\endgroup$ – D. Ennis Oct 15 '16 at 21:54
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Treat the plane and the block as a system

When you will treat the block and the inclined plane as a system you find that the only forces which are acting on the system are in the vertical direction. therefore the center of mass of the system will accelerate ( if it does ) in a vertical line only. also the net force vector in this case will point downwards so the center of mass will accelerate downwards.

or to simplify the center of mass won't be accelerated in the horizontal direction.

The inclined plane does want to move with block

The friction which acts on the inclined plane definitely opposes the relative motion between them but what you forgot to count I think is the normal form the smooth friction-less table. The table may be smooth but it can still exert a normal force on any body that presses it.

Also as a representation of forces you may look at the diagram I have attatchedFree Body Diagram of system (Friction force not included)

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  • $\begingroup$ I can't understand this answer very well, the way you write is unclear. $\endgroup$ – Suzu Hirose Oct 15 '16 at 1:46

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