# Why does the Block have an Acceleration Only Downwards

1. Sep 1, 2016

### decentfellow

1. The problem statement, all variables and given/known data
Block $A$ of mass $m$ is placed over a movable wedge of mass $m$. Both the block and the wedge are placed over a fixed incline plane. Assuming all the surfaced to be smooth, calculate the displacement of the block $A$ in the ground frame reference in $1s$.

2. Relevant equations
$$\vec{F}=m\vec{a}$$

3. The attempt at a solution
Before beginning, the question that I posed above is a solved example, and I am just confused about the motion of the block $A$. So here goes.

The solution says that "Let acceleration of wedge in ground's frame is $a$ down the plane. The acceleration of block $A$ will be $a\sin\theta$ vertically downward"

Now, my problem is with the very first line of the solution, how can the block only have an acceleration down the plane, though in the solution, to justify the first line it is stated that "From FBD of A it is clear that Block A cannot accelerate horizontally, i.e. in $x-$direction because there is no force acting on the block $A$ in $x-$direction, hence block $A$ can accelerate in $y-$ direction only."

Though the explanation seems right in its place but before seeing the solution, my line of thought regarding the question was that as the block is constrained to move with the movable wedge hence it has an acceleration in the the direction along the plane. I don't know why but the explanation in the solution is just not sinking in.

Last edited: Sep 1, 2016
2. Sep 1, 2016

### Staff: Mentor

I guess that the crux of the problem. Everything is assumed to be frictionless, so block A is not constrained to move with the wedge. As the wedge slides down the inclined plane, from the point of view of block A the "floor" is sliding from under it and moving down. Only the latter leads to any motion of block A.

3. Sep 1, 2016

### PeroK

One of the reasons to do a FBD is to isolate the forces on each component. If there is no friction between A and the wedge, then the only force on A is gravity, hence it must move vertically.

If there were friction between A and the wedge that would be a different matter.

4. Sep 1, 2016

### decentfellow

After removing the constraint of them moving together, i.e. not considering the friction force which I was considering while visualizing the problem(unkowingly), and now that I have removed it I again have a doubt that doesn't it happen that as the wedge slides down the incline it so happens that the block looses contact with the movable wedge and moves along the incline, the only problem is that that the length of the platform on which the block stands is not given, so isn't the question incomplete.

5. Sep 1, 2016

### PeroK

If you moved the wedge fast enough, then the block would lose contact. But, the acceleration of the wedge is limited by gravity, so it can't out-accelerate the block.

6. Sep 1, 2016

### decentfellow

But in the (non-inertial)reference frame of the movable wedge there would be a pseudo force in the $+ve x-$ direction making it accelerate in that direction w.r.t the movable wedge so after some time when the block would have traversed the length of the wedge's platform on which it stands wouldn't it have partially lost contact with the wedge and have made contact with the fixed incline.

7. Sep 1, 2016

### Staff: Mentor

I guess that's why the problem asks for displacement for the first second, so that the block doesn't reach the inclined plane.

8. Sep 1, 2016

### decentfellow

Yeah, you are right I think I am jumping to conclusions before even attempting it.

9. Sep 1, 2016

### PeroK

Yes, of course the block will eventually fall off the side of the wedge!