At the top of a wedge of inclination $\theta$, a block is kept. There are no external forces acting on this block in horizontal direction.

Is is possible for this block to fall vertically downwards?

(Image below for illustration/clarity ONLY regarding my question)

enter image description here

I have tried to evaluate the answer using FBDs and no matter how I attempt it, my final answer comes out to be either $\sin\theta = 0$ or $cos\theta = 0$ but for either of those conditions, it wouldn't be an inclined plane.

I have also referred to the following questions asked on this site but I haven't got the answer to my question:

Free body diagram of block on accelerating wedge

A block on inclined plane

block slides on smooth triangular wedge kept on smooth floor.Find velocity of wedge when block reaches bottom

Please notify me in the comments if more clarity is needed anywhere.

  • $\begingroup$ "There is no external force in this system except for the gravitational forces", there is also an external force from the table, no? $\endgroup$ Commented Jun 24, 2022 at 15:47
  • $\begingroup$ @MariusLadegårdMeyer Yes there is. What I meant was there's no external force in the horizontal direction. Thanks for pointing it out. I'll edit the question $\endgroup$
    – Guri
    Commented Jun 24, 2022 at 15:52

1 Answer 1


Can a block set on a inclined plane fall vertically downwards

Only if the mass of the incline, $M$, were zero.

Since the only external forces acting on the combination of the two blocks are vertical (gravity and the normal reaction force of the supporting surface on $M$), the center of mass (COM) of the two block system can only undergo vertical downward motion.

Since block $M$ can only move horizontally to the left in the reference frame of the floor, block $m$ must move downward and to the right in the reference frame of the floor, in such a manner that the horizontal components of the two motions cancel so that the COM moves downward. The greater the mass of $m$ is compared to that of $M$ the smaller the component of the motion of $m$ to the right.

Hope this helps.

  • $\begingroup$ Thank you for your answer. It was a really nice explanation that has cleared my query :D $\endgroup$
    – Guri
    Commented Jun 24, 2022 at 16:53
  • $\begingroup$ You’re welcome. $\endgroup$
    – Bob D
    Commented Jun 24, 2022 at 16:55
  • $\begingroup$ There's another weird edge case where the block can fall straight down: $\theta = 90°$. That's about as realistic as assuming that $M = 0$. $\endgroup$ Commented Jun 28, 2022 at 14:32
  • $\begingroup$ @MichaelSeifert agree. Of course then you can no longer call it an incline as well as no longer call it a wedge $\endgroup$
    – Bob D
    Commented Jun 28, 2022 at 14:37

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