Accelerating Wedge and block on top of it -- Dynamics

[andyrk]

Homework Statement

In the figure a block of mass 'm' is placed on a wedge of mass M. The wedge is subject to a horizontal force 'F' and slides on a friction less surface. The coefficient of static friction between the block and wedge is μ_s. Find the range of values of 'F' for which the block does not slide on the incline.

Homework Equations

If the bodies (block+wedge) move together then:
F=(M+m)a

The Attempt at a Solution

If 'F' is too less, then the block of mass 'm' will have a tendency to come down the incline.
If 'F' is too much, the block will have a tendency to move up the incline.
So 'F' need to be in between these minimum and maximum values and that's how we get the range. But my doubt is that why do the above 2 conditions i.e when F is too less and when F is too much happen?

 

[haruspex]

why do the above 2 conditions i.e when F is too less and when F is too much happen?

Why or when? Are you able to write out and solve the equations?
Note that in the case of sliding down, it may be that F needs to go negative if the angle is low or the friction is high.

 

[andyrk]

"When" is the statement itself..That is , the 2 respective cases happen when their respective conditions are met. Namely, F being very less or very large. My question is for the "why" part. As to does the block specifically move upwards when F is too large? If it moves up there has to be some force acting up the incline..how does that force arise, i.e the force that makes the block move up the incline when F is too large? Is it because of a pseudo force that we have included, opposite to the acceleration of the (wedge+block) since this system is an accelerating one with respect to an inertial frame namely us or the ground? If that happens then we can resolve this pseudo acceleration into two components, and one those components acts up the incline so the block moves up the incline. But then, are we considering ourselves or the ground as the frame of reference or the system itself as the frame of reference?
The thing is I maybe having the answers to my queries but I am just too doubtful to accept my answer..

 

[Tanya Sharma]

Is it because of a pseudo force that we have included, opposite to the acceleration of the (wedge+block) since this system is an accelerating one with respect to an inertial frame namely us or the ground? If that happens then we can resolve this pseudo acceleration into two components, and one those components acts up the incline so the block moves up the incline.

Hi andyrk

The component of pseudo force acts up the incline,whereas the component of weight acts down the incline.In case the component of pseudo force is greater than the component of weight then the block has the tendency to move up.Hence friction acts downwards.In case the former is less than the latter ,the block has the tendency to slide down.So,friction acts upwards.

But then, are we considering ourselves or the ground as the frame of reference or the system itself as the frame of reference?

Since we are applying pseudo force ,we are surely working from the accelerated frame of reference,i.e the wedge .The observations are made from the wedge ,not the ground.

 

[andyrk]

What are the observations that we are making from the frame of reference of the (block+wedge) system?

 

[Tanya Sharma]

What
are the
observations that we are making from the frame of reference of the
(block+wedge) system?

The frame of reference is wedge ,not wedge+block.The block is seen sliding up,stationary,sliding down the wedge depending on the force applied to the wedge.

 

[andyrk]

Oops..yeah sorry my bad

 

[andyrk]

Hi andyrk

The component of pseudo force acts up the incline,whereas the component of weight acts down the incline.In case the component of pseudo force is greater than the component of weight then the block has the tendency to move up.Hence friction acts downwards.In case the former is less than the latter ,the block has the tendency to slide down.So,friction acts upwards.
Since we are applying pseudo force ,we are surely working from the accelerated frame of reference,i.e the wedge .The observations are made from the wedge ,not the ground.

Can the reason for why the block moves up the incline and down the incline be explained without using pseudo force method and just normal application of Newton's Second Law?

 

[haruspex]

Can the reason for why the block moves up the incline and down the incline be explained without using pseudo force method and just normal application of Newton's Second Law?

Of course. Applied corectly, inertial frames and non-inertial frames should lead to the same answer. It's only a question of which is more convenient.

 

[andyrk]

Of course. Applied corectly, inertial frames and non-inertial frames should lead to the same answer. It's only a question of which is more convenient.

So how would one do that? I am not able to think of an explanation without using Pseudo Force which explains why the block moves up or down the incline depending on the magnitude of the force.

 

[haruspex]

So how would one do that? I am not able to think of an explanation without using Pseudo Force which explains why the block moves up or down the incline depending on the magnitude of the force.

Down the incline is ok, yes? If little force is applied and the friction is weak it will slide down.
For the block to slide up the incline, it is only required that the horizontal acceleration of the wedge exceeds that of the block. We can make the wedge accelerate as fast as we like by applying a large enough force. The acceleration of the block in that direction depends on the normal force from the wedge (limited by g) and the frictional force (limited by g and mu).

 

[andyrk]

Down the incline is ok, yes? If little force is applied and the friction is weak it will slide down.
For the block to slide up the incline, it is only required that the horizontal acceleration of the wedge exceeds that of the block. We can make the wedge accelerate as fast as we like by applying a large enough force. The acceleration of the block in that direction depends on the normal force from the wedge (limited by g) and the frictional force (limited by g and mu).

So even if the wedge accelerates faster than the block, why would the block move upwards?

 

[haruspex]

So even if the wedge accelerates faster than the block, why would the block accelerate upwards?

Because the only other option is for it to penetrate the block.

 

[andyrk]

Because the only other option is for it to penetrate the block.

Yep. That makes some sense intuitively. But can we explain it a bit more quantitatively rather than qualitatively?

 

[andyrk]

Anybody there?

 

[haruspex]

Yep. That makes some sense intuitively. But can we explain it a bit more quantitatively rather than qualitatively?

I had the impression you were after a qualitative explanation.
To go quantitative, write some equations.

 

[andyrk]

I had the impression you were after a qualitative explanation.
To go quantitative, write some equations.

I did. But I was unable to find a reasonable explanation for this without including pseudo forces in. How do you do that?

 

[haruspex]

I did. But I was unable to find a reasonable explanation for this without including pseudo forces in. How do you do that?

You have the acceleration. Suppose it is on the point of slipping up the plane. Put in an unknown for the normal force. What equations do you get?

 

[andyrk]

The equations for the block are: Rcosθ = mg and Rsinθ = mA, where A is the acceleration of the block+wedge system.
This is for the case when the block is at rest (assuming that it is).
When it is not, what proof do we have that it moves upwards when the force is a lot and it moves downwards when the force is low? How can we be sure this is going to happen? I know it seems right intuitively and by using pseudo forces we can even prove it. But can we prove it without intuition and pseudo forces?

 

[haruspex]

The equations for the block are: Rcosθ = mg and Rsinθ = mA, where A is the acceleration of the block+wedge system.
This is for the case when the block is at rest (assuming that it is).
When it is not, what proof do we have that it moves upwards when the force is a lot and it moves downwards when the force is low? How can we be sure this is going to happen? I know it seems right intuitively and by using pseudo forces we can even prove it. But can we prove it without intuition and pseudo forces?

Where R is the normal force? You've left out friction.

 

[andyrk]

Friction comes into play when the block slides. But why would it slide in the first place?

 

[haruspex]

Friction comes into play when the block slides. But why would it slide in the first place?

Static friction applies well before the block slides.
Here's the plan:
Three unknowns, normal force N, friction Ff, acceleration a.
Three equations, horizontal F=ma for the wdge, horizontal and vertical for the block.
Solve to find Ff/N as a function of M, m, F, theta and g.

 

[TSny]

Can the reason for why the block moves up the incline and down the incline be explained without using pseudo force method and just normal application of Newton's Second Law?

For simplicity, consider the case where there is no friction. If the force F applied to the wedge is large enough the block will slide up the incline. But note that in the inertial frame, the block slides horizontally in the same direction as the wedge while it also has a vertical component of displacement. Considering the directions of the two forces acting on the block, it should not be too surprising that the block can move this way.

 

[andyrk]

For simplicity, consider the case where there is no friction. If the force F applied to the wedge is large enough the block will slide up the incline. But note that in the inertial frame, the block slides horizontally in the same direction as the wedge while it also has a vertical component of displacement. Considering the directions of the two forces acting on the block, it should not be too surprising that the block can move this way.

Why does it have a vertical component of displacement?

 

[TSny]

The normal force on the block has a vertically upward component. If the applied force F on the wedge is increased, what do you suppose happens to the magnitude of the normal force on the block? What happens when the vertical component of the normal force exceeds the weight of the block?

 

[andyrk]

The normal force on the block has a vertically upward component. If the applied force F on the wedge is increased, what do you suppose happens to the magnitude of the normal force on the block? What happens when the vertical component of the normal force exceeds the weight of the block?

The block flies off and leaves contact with the wedge. But still, we don't have any force along to incline to make it slide upward or downward. And anyway, the normal force would never exceed the weight because normal force is by definition always equal to the force with which the block is pressing against the wedge.

 

[haruspex]

The block flies off and leaves contact with the wedge. But still, we don't have any force along to incline to make it slide upward or downward. And anyway, the normal force would never exceed the weight because normal force is by definition always equal to the force with which the block is pressing against the wedge.

The normal force can exceed the weight because of the acceleration. The only way to settle this is with equations. Have another go at them.

 

[TSny]

The block flies off and leaves contact with the wedge...And anyway, the normal force would never exceed the weight because normal force is by definition always equal to the force with which the block is pressing against the wedge.

The block does not leave contact with the wedge. When you stand in an elevator that accelerates upward, the upward normal force on you is greater than your weight. But you don't lose contact with the floor. (The elevator accelerates with you.)

But still, we don't have any force along to incline to make it slide upward or downward.

There does not need to be any force component upward along the incline in order for the block to increase its height on the wedge. If you vectorially add the normal force and weight, you get a net force on the block that points upward and to the right (if the normal force is sufficiently large). So the block accelerates upward and to the right while remaining on the wedge, as shown in the figure I previously posted. If you add in the friction force, you should still be able to see how it is possible for the block to increase its height while remaining on the wedge.

But I think I might be detracting from the main point of setting up and working through the equations.

 

[andyrk]

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?

 

[TSny]

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?

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.

 

[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.