Accelerating Wedge and block on top of it -- Dynamics

[haruspex]

The equations are: Ncosθ = mg and Nsinθ = mA. None of these forces are along the incline so how can the block slide up or down the incline?

Further to TSny's reply, we are trying to analyse it in an inertial frame, yes? The incline is accelerating, so for the block to slide up the incline does not mean the block will accelerate in the direction of the incline. It will stay in contact with the incline, certainly, so the question becomes whether the vertical component of the net force on the block is up or down or zero.

 

[andyrk]

Since there is no friction force in these equations, we must still be considering the case of no friction.

When you write Ncosθ = mg, you are assuming at the outset that the block will have no vertical acceleration. In that case, the block can naturally remain in one place on the incline and not slide up or down the incline as the wedge is pushed. But, note that Ncosθ = mg tells you what N must be for this to happen. Using this value of N in your second equation will determine the specific value of A that must occur in this case. From that you can get the specific force F that needs to be applied to the wedge so that the block will not slide up or down the frictionless incline. So, only if you push the wedge with this one specific value of F will the block not slide up or down the incline. If F is greater than this specific force, the block will slide up the incline; if less, it will slide down.

Do you mean to say that if we increase force more than or less than this specific value, the normal reaction would also change accordingly? How is it similar to the scenario when Normal Reaction on the weighing machine inside a moving elevator increases or decreases depending on whether the elevator is moving up or down respectively?

 

[TSny]

Do you mean to say that if we increase force more than or less than this specific value, the normal reaction would also change accordingly?

Yes, N varies as F varies.

How is it similar to the scenario when Normal Reaction on the weighing machine inside a moving elevator increases or decreases depending on whether the elevator is moving up or down respectively?

N varies as the lifting force applied to the elevator varies. That's similar to N varying as F varies for the wedge problem.

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Suppose you did the wedge experiment with gravity switched off and no friction. So, we go to an inertial reference frame deep in outer space. In this frame we start with the wedge and block "floating" at rest and the block is positioned on the incline of the wedge. N is zero while they just sit there in our frame. Now apply F as usual and consider what happens from the point of view of our inertial frame. The block will slide "up the incline". The only force acting on the block is N. N is always oriented perpendicular to the incline. So, the block must accelerate in the direction perpendicular to the incline, yet it "slides up the incline". Try drawing a figure similar to the one in post #23 for this case.

 

[haruspex]

Do you mean to say that if we increase force more than or less than this specific value, the normal reaction would also change accordingly? How is it similar to the scenario when Normal Reaction on the weighing machine inside a moving elevator increases or decreases depending on whether the elevator is moving up or down respectively?

It's not a question of the elevator moving up or down, but accelerating up or down.

 

[andyrk]

It's not a question of the elevator moving up or down, but accelerating up or down.

Yup, I meant accelerating. Sorry for the typo, my bad.

 

[andyrk]

Yes, N varies as F varies.
N varies as the lifting force applied to the elevator varies. That's similar to N varying as F varies for the wedge problem.

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Suppose you did the wedge experiment with gravity switched off and no friction. So, we go to an inertial reference frame deep in outer space. In this frame we start with the wedge and block "floating" at rest and the block is positioned on the incline of the wedge. N is zero while they just sit there in our frame. Now apply F as usual and consider what happens from the point of view of our inertial frame. The block will slide "up the incline". The only force acting on the block is N. N is always oriented perpendicular to the incline. So, the block must accelerate in the direction perpendicular to the incline, yet it "slides up the incline". Try drawing a figure similar to the one in post #23 for this case.

It slides up the incline because by the time it goes up, the wedge has also come forward to to keep in touch with block so that block is seen sliding up, while in reality it was always going vertically straight up.

 

[TSny]

It slides up the incline because by the time it goes up, the wedge has also come forward to to keep in touch with block so that block is seen sliding up, while in reality it was always going vertically straight up.

The block does not move vertically upward. An object that starts at rest and then is acted on by a constant force will move in the direction of the force.

 

[andyrk]

The block does not move vertically upward. An object that starts at rest and then is acted on by a constant force will move in the direction of the force.

Then why does the block slide up or down if the applied force is too much or too less?

 

[haruspex]

Then why does the block slide up or down if the applied force is too much or too less?

It goes up (or down), but not vertically - it will also have a horizontal component.

 

[andyrk]

It goes up (or down), but not vertically - it will also have a horizontal component.

Why? Why would it have a horizontal component?

 

[haruspex]

Why? Why would it have a horizontal component?

This thread is in danger of never coming to a conclusion all the while you refuse to get serious about the equations.
There is a normal force between block and wedge.
There is, in general, a frictional force between block and wedge.
The block, in general, may have a horizontal component of acceleration.
The block, in general, may have a vertical component of acceleration.
The wedge has a horizontal acceleration.
(There's no point in considering vertical forces on the wedge since the normal force from the ground will be whatever it needs to be to ensure no acceleration in that direction.)
There is a relationship between the three accelerations, given by the fact that the block maintains contact with the wedge but does not penetrate it.
So, you have five unknowns (two forces, three accelerations) and four equations ($\Sigma F = ma$ in two directions for the block, one direction for the wedge; plus the relationship between the accelerations).
This should allow you to obtain one equation relating the normal and frictional forces, with no other unknowns.
I will only respond to your attempts to pursue that plan.

 

[andyrk]

Yeah! I got that now. Since increasing or decreasing the force affects the normal reaction, it also affects the static friction, making it too low or too high in comparison to component of the weight of the block along the incline. So if the static friction force outweighs the component of weight of the block down the incline, it goes up the incline, and its the other way around, it goes down the incline. And if it is just perfect, the block doesn't slide.

But what happens if the incline is smooth? Have I said enough in this to earn your reply?

 

[TSny]

In the scenario presented at the end of post #33 (no friction or gravity), the block will slide up the incline for any value of the applied force F on the wedge. The block moves up and to the right (in the inertial frame) because the normal force points up and to the right. The block slides "up the incline" even though there is no force component parallel to the incline.

For the case where we also have gravity, the block will still always have a horizontal component of acceleration to the right due to the horizontal component of the normal force. See figure in post #23. The vertical acceleration of the block can now be upward, downward, or zero depending on whether the vertical component of the normal force is greater than, less than, or equal to the weight of the block. Again, when the block slides "up the incline" it does so without there being any force component up along the incline.

I might be wrong, but it seems to me that you are thinking that if the block slides up the incline then there must be a force component up along the incline; i.e., up and to the left. That would be true only if, in the inertial frame, the block had an acceleration in the direction of up and to the left. But it doesn't. In the inertial frame, the block accelerates up and to the right, not up and to the left.

 

[andyrk]

I might be wrong, but it seems to me that you are thinking that if the block slides up the incline then there must be a force component up along the incline; i.e., up and to the left. That would be true only if, in the inertial frame, the block had an acceleration in the direction of up and to the left. But it doesn't. In the inertial frame, the block accelerates up and to the right, not up and to the left.

If there is no force component along the incline then how does it slide on the incline? Shouldn't it stay wherever it was in the beginning? Also, when you are saying that the block accelerates up and to the right, you are assuming it to be a case where there is no gravity, right?

Lastly, see the attachment. It shows all the possible forces acting on the block. Keeping in mind this diagram, explain me, how could the block ever slide up the incline (I think it can slide down the incline though)? I had one doubt though. Is the Nsinθ force a vector sum of N and mgsinθ forces? If yes, then Nsinθ might not always be parallel to the surface on which the wedge is kept and that might lead to interesting results.

 

[TSny]

If there is no force component along the incline then how does it slide on the incline? Shouldn't it stay wherever it was in the beginning?

It slides on the incline because the wedge is "slipping out from under" the block. Imagine letting θ go to zero. This would be like putting the block on a horizontal board. If you push the board to the right and assume no friction between the board and the block, the board will just slip out from under the block while the block doesn't move. The block ends up coming off the back edge of the board even though there was never a force pushing the block toward the back edge of the board.

Imagine a wedge where θ is extremely small. So, the incline is almost horizontal. When you push hard on the wedge to the right, it will be very similar to the case with the horizontal board. The wedge will slip out from under the block so that the block ends up falling off the back edge of the wedge. There was never a force on the block in a direction "up the incline". In this case, the normal force on the block has a small horizontal component so the block actually moves a little to the right as the wedge slips out from under the block. The block is able to rise upward to get to the top of the wedge because the vertical component of the normal force exceeds the weight of the block. So, the block moves both vertically and horizontally as the wedge slips out from under it. That is, the block moves up and to the right.

As you make θ larger, it will be similar. But now you have to push harder on the wedge in order to get the vertical component of the normal force to exceed the weight of the block so that the block can move upward as the wedge slips out from under the block.

Also, when you are saying that the block accelerates up and to the right, you are assuming it to be a case where there is no gravity, right?

No. The picture in post #23 is for the case where there is gravity (but no friction). Note how the block accelerates up and to the right along the faint gray line because that is the direction of the vector sum of the normal force and the gravity force.

Lastly, see the attachment. It shows all the possible forces acting on the block. Keeping in mind this diagram, explain me, how could the block ever slide up the incline (I think it can slide down the incline though)?

Think of the wedge as slipping out from under the block as discussed above.

I had one doubt though. Is the Nsinθ force a vector sum of N and mgsinθ forces? If yes, then Nsinθ might not always be parallel to the surface on which the wedge is kept and that might lead to interesting results.

No. N is the force which the surface of the wedge exerts against the block. It is entirely separate from the force of gravity. Nsinθ is horizontal and Ncosθ is vertical.

 

[haruspex]

If there is no force component along the incline then how does it slide on the incline?

I think you are making the mistake of thinking that it can only slide up the incline if the net force makes an angle between the normal and the vertical. Not so. It will lie between the normal and the horizontal. But that will still lead to an upward acceleration, and since it stayus in contact with the wedge that must mean it slides up the wedge.
If only you would stop shirking getting to grips with the equations you could work all this out for yourself!

 

[andyrk]

If only you would stop shirking getting to grips with the equations you could work all this out for yourself!

Here are the equations. But they make no logical sense as to why would the block move up the incline if the force is too much.
Now when the force is just perfect, then:

Ncosθ = mg​

and​

Nsinθ = mA​

A is the combined acceleration of the whole block + wedge system.
Dividing the 2 equations above we get

A = gtanθ.
So F = (m+M)A = (m+M)gtanθ
​

(This is for the case when the block doesn't slide). For any general case,

A = Nsinθ/m ⇒ F = (m+M)Nsinθ/m ⇒ Fm/((m+M)sinθ) = N
.
​

So N α F. So as F increases, N also increases. And hence Ncosθ ≠ mg anymore.
So

Ncosθ - mg = ma_y,
​

i.e the block has an upwards acceleration. And it also has a horizontal acceleration A as given by the second equation. So, the sum of both of these accelerations. according to me, should lead in direction which is diagonally upwards to the right. Now the thing that confuses me is, how can the block ever go to a direction that is diagonally upwards and towards the left when the net acceleration on the block is diagonally upwards and to the right? It just doesn't make any logical sense at all.

And if N is too less, then the second equation reverses and it becomes: mg - Nsinθ = ma_y, i.e. it has a downwards acceleration. Also it has a horizontal acceleration. So, the vector addition of these two accelerations would give an acceleration in diagonally down direction to the right. So in this case it makes sense that the block would slide down (diagonally down and to the right).

 

[haruspex]

how can the block ever go to a direction that is diagonally upwards and towards the left

It goes up and to the right, but its acceleration to the right will be less than that of the wedge, so, relative to the wedge, it moves up and left.

 

[andyrk]

It goes up and to the right, but its acceleration to the right will be less than that of the wedge, so, relative to the wedge, it moves up and left.

You mean vector addition of a_y and A would be less than A? How can that be possibly true?

 

[haruspex]

You mean vector addition of a_y and A would be less than A? How can that be possibly true?

No. When the block slides up the wedge its horizontal acceleration is less than that of the wedge. You can no longer use A for both.

 

[andyrk]

No. When the block slides up the wedge its horizontal acceleration is less than that of the wedge. You can no longer use A for both.

How can you be so certain that the horizontal acceleration of the wedge and the block will be different? Can you prove it?

 

[haruspex]

How can you be so certain that the horizontal acceleration of the wedge and the block will be different? Can you prove it?

It's not up to me to prove it. You can't prove they're the same, so in the equations you must allow them to be different, then find out if they are.

 

[andyrk]

It's not up to me to prove it. You can't prove they're the same, so in the equations you must allow them to be different, then find out if they are.

Simply put, from what I have gathered, I have come to the conclusion that I don' t think that you can explain why the block slides up or down without involving pseudo forces in. If you know the reasoning behind something, then why don't you just say it? I am not able to figure it out myself, so there's no point in waiting and creating more confusions. And I don't even think that we can ever prove that without using pseudo forces. That is because, look at the example of a man standing on a rotating wheel. If there is no friction between his feet and the wheel, with time the man automatically goes towards the rim of the wheel due to pseudo force. There isn't any other explanation as to why this happens without using pseudo forces. Same is with the case with the wedge and block, according to me. The block slides up or down purely because of pseudo force and nothing else.

 

[haruspex]

Simply put, from what I have gathered, I have come to the conclusion that I don' t think that you can explain why the block slides up or down without involving pseudo forces in. If you know the reasoning behind something, then why don't you just say it? I am not able to figure it out myself, so there's no point in waiting and creating more confusions. And I don't even think that we can ever prove that without using pseudo forces. That is because, look at the example of a man standing on a rotating wheel. If there is no friction between his feet and the wheel, with time the man automatically goes towards the rim of the wheel due to pseudo force. There isn't any other explanation as to why this happens without using pseudo forces. Same is with the case with the wedge and block, according to me. The block slides up or down purely because of pseudo force and nothing else.

In you are standing on a rotating disk then lose friction, you will move in a straight line, at a tangent to the circle you were moving in before losing friction. You will observe this as moving towards the outer rim of the disk. This is standard Newtonian mechanics.
Pseudoforces only exist in non-inertial frames. Analysis in inertial frames and different non-inertial frames, done correctly, all lead to the same answer, therefore pseudoforces cannot be a necessary part of the explanation.
The only way you will see this is to develop the right equations and see where they lead. E.g. do not assume the horizontal accelerations are the same. I am happy to work with you on that, but don't see any other way to make progress.

 

[andyrk]

In you are standing on a rotating disk then lose friction, you will move in a straight line, at a tangent to the circle you were moving in before losing friction. You will observe this as moving towards the outer rim of the disk. This is standard Newtonian mechanics.
Pseudoforces only exist in non-inertial frames. Analysis in inertial frames and different non-inertial frames, done correctly, all lead to the same answer, therefore pseudoforces cannot be a necessary part of the explanation.
The only way you will see this is to develop the right equations and see where they lead. E.g. do not assume the horizontal accelerations are the same. I am happy to work with you on that, but don't see any other way to make progress.

Okay, so if the accelerations are different then then just the second equation changes and becomes Nsinθ = ma_x While the horizontal acceleration of the wedge and block is A. How does this account for the fact that the block slides up or down the incline depending on F?

 

[haruspex]

Okay, so if the accelerations are different then then just the second equation changes and becomes Nsinθ = ma_x While the horizontal acceleration of the wedge and block is A. How does this account for the fact that the block slides up or down the incline depending on F?

No, you cannot write A for the horizontal acceleration of 'wedge and block', since they are different.
If the wedge has the greater horizontal acceleration then the block must move up to allow the wedge to advance under it. If the block has the greater then it will descend the face of the wedge as the wedge falls behind. You can write an equation relating the three accelerations and the angle of the slope.