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The situation looks as follows: situation

The incline is moving rightwards with a constant acceleration
a=g

The forces on the picture are: mg - weight, N - normal force, F - inertial force (already shown), since it'll be easy to analise the problem in a noninertial frame.

The question is: what's the trajectory of the block, i.e. how the block will be moving?

I will show my hypothesis; please tell me if it's right or not:

I've calculated the vector sum of the forces acting on the block when it's on the incline (bold letters symbolize vectors, bold i and j are unit vectors):

net force

Let's say theta = 30 degrees, and the incline is infinitely long. In this scenario, the net force looks like this:
situation

The net force is such, that the block will actually loose contact with the incline. But then the normal force will disappear and the net force will become:
situation

This force brings the block back down. So my hypothesis is, that for a theta equal to 30 degrees the block will be "jumping" down. Is it right?

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  • $\begingroup$ It appears you've equated $N$ with $mg \cos \theta$. Could you explain to me why you've done this? (That wasn't meant to sound accusatory, I'm just uncertain as to what your logic is). $\endgroup$ – gj255 Jan 10 '14 at 22:18
  • $\begingroup$ Image Sorry for no vectors and lame signs, but I think it's easy to read. As you can see on the image: alfa + beta = 90 degrees Then the angle between mg and F1 is also alfa. cos alfa = F1/mg, then F1 = mgcos alfa. Since N is the reaction force to F1, its value has to be equal to the value of F1. Thus, N = F1 = mgcos alfa. $\endgroup$ – neverneve Jan 11 '14 at 23:05
  • $\begingroup$ So your mistake is saying that because N is the reaction force to F1, its value has to equal F1. This would only be true if the block was not moving. The reason we so frequently set N = F1 is because it's very rare that the block has any motion in the direction perpendicular to the plane. In this case, however, there is going to be motion in this direction. So by Newton's second law, $F = ma$, if the acceleration in this direction is non-zero then the net force in this direction will be non-zero. In other words, the forces won't cancel out. $\endgroup$ – gj255 Jan 12 '14 at 11:48
  • $\begingroup$ Thank you for your comments, I think I haven't explained the first image in the question enough. I thought of it as a picture of the instant the F force starts acting - so it's basically a normal, incline situation (where N cancels F1) with the F force added. The net force in this direction will surely by non-zero, but in my opinion N is equal to F1 and the force causing the motion in this direction is F * sin(theta). $\endgroup$ – neverneve Jan 12 '14 at 12:01
  • $\begingroup$ I don't think this logic is sound. If you have your force $F$ turned off and the block stationary, then indeed the equation you gave will be correct. But when the force F is applied, there is going to be some motion --- some acceleration, and the equation you derived before --- $N = mg \cos \theta$ --- is going to stop applying immediately. Instead there will be some complicated equation corresponding to motion in this perpendicular direction. Your argument seems to rest on $N = mg \cos \theta$ holding for a short while --- enough for the block to lift off --- which it doesn't. $\endgroup$ – gj255 Jan 12 '14 at 12:46
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NOTE: The blockquotes only apply with a gradual increase in acceleration starting from $0\ \text m\text s^{-2}$.

The trajectory that it makes depends almost entirely on the coefficient of friction between the two surfaces. This is because the 'net' normal force will become less and less decreasing friction until the force: $\vec F = mg\sin\theta$ is larger than the static friction force: $\vec F = \vec N\mu$, after which the block starts sliding down.

Now, the block will leave the ground if the acceleration, $\vec a_x$, is large enough. Using simple trigonometry it is found that the object leaves the ground when: $$\frac{\vec a_x}{\sin\theta} > \vec g\cos\theta$$ Since the maxima of $\sin x *\cos x$ is $0.5$ , the object will instantly leave the incline. (Because $\sin\theta$ and $\cos\theta$ are positive in the first quadrant and the accelerations are of the same magnitude).

If the static friction is large, the trajectory will look like the block jumps to a lower level. (This is where your last diagram is completely wrong, because the net force acts down - The only force acting on it then is $mg$). If on the other hand, the coefficient of friction is low, the lock will slide before it finally releases, and thus make a much longer jump. Friction affects how much horizontal velocity it gains; i.e. low friction will cause the block to 'move away' from the incline much faster, because its velocity compared to that of the incline (which is also accelerating) is a lot lower.

Note that the block will always catch up with the incline eventually, because you defined $|\vec a|$ to be equal to $|\vec g|$, and $$\vec g>\vec a\sin\theta$$
So essentially what happens is that every time the block touches the incline it leaves again.

Also do not speak about the "inertial force"; simply call it friction. The only forces initially acting on the block are the: normal force, $\vec N=m\vec g\cos\theta$; the weight, $\vec W=m\vec g$; the force along the incline due to its weight, $\vec F=m\vec g\sin\theta$; and lastly friction, which is equal the previous force until that is greater than $\mu \vec N$.

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  • $\begingroup$ I assumed that the frame of reference is moving along with the incline, with a rightward acceleration, so I think the block should be subject to intertial force wherever it is. And isn't the inertial force in this case different than friction? Friction acts up along the incline, and the inertial force acts leftwards. I didn't mention friction because I thought of it as an impediment and wanted to merge into the equations later. Also, could you expand on the trigonometry, that told you when the block leaves the ground? Thank you very much. $\endgroup$ – neverneve Jan 11 '14 at 22:30
  • $\begingroup$ @neverneve I fixed my answer because parts of it were wrong, because I didn't realize that you specified constant acceleration. The only forces that are acting on the block are specified in the last paragraph, and while the object is in the air, the only force is $m\vec g$. Your last diagram for example is completely wrong because the net force acts downwards. Anyway, I have thought about this too much so I'll check back in a few hours. $\endgroup$ – Ruben Jan 12 '14 at 0:29
  • $\begingroup$ Thank you very much, I think I understand now (I'm also content, that my prediction of the block "jumping" down is correct). The only bit I'm not sure of is $$\frac{\vec a_x}{\sin\theta} > \vec g\cos\theta$$. I've also calculated that whatever the value of the F force, the block will leave the ground, but I'm curious how did you come to this inequality. $\endgroup$ – neverneve Jan 12 '14 at 12:10
  • $\begingroup$ @neverneve Sorry it seems one of my comments never posted =/. Anyway, I did this by copying the incline and rotating it 90 degrees clockwise. This way the base is $\vec a_x$, and so the hypotenuse becomes $\frac{\vec a}{\sin\theta}$. Equating this with the normal of the block divided by its mass, $\vec g\cos\theta$, gives that 'equation'. So for most values for acceleration the block will leave the incline. $\endgroup$ – Ruben Jan 12 '14 at 12:18

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