Question about a simple free body diagram

[jbriggs444]

I don't see how it does that. Can you please elaborate?

There is no external horizontal force on the assembly. So its center of mass cannot accelerate horizontally.

Gravity is vertical. No horizontal component.

The normal support force from below is vertical. No horizontal component.

We assume (but are never explicitly told that I recall) that the bottom surface of the big block is free of all friction. No horizontal component.

Any force between top block and big block is internal. It cannot affect the center of gravity.
Any force between the little-blocks, the cord and the pulley is internal. It cannot affect the center of gravity.
Any force between the side block and the big block is internal. It cannot affect the center of gravity.

We do not discuss reactionless drives on Physics Forums.

 

[sophiecentaur]

This again solidifies my prejudice that free-body diagrams are the greatest obstacle to solve mechanical problems.

I never 'did' FBDs explicitly and I often have similar problems when they are called for as a panacea for all mechanical problems. They will (have to) always work but require a lot of care to include everything that counts.

 

[sysprog]

Thanks for that explanation, @jbriggs444 $-$ the reminder in your first sentence made it perspicuous for me. @PeroK, now I see why the top $m$ block has to move, and do so commensurately against the remainder of the system.

 

[sysprog]

We assume (but are never explicitly told that I recall) that the bottom surface of the big block is free of all friction. No horizontal component.

I think that there's a convention by which that's what the row of diagonal marks means.

 

[vanhees71]

I think @PeroK 's analysis with the Lagrange formalism is correct. I'd have a hard time to figure it out with FBDs, but it should be possible.

 

[PeroK]

Newton's laws demand that the Centre of mass of the system does not move horizontally.

In fact, we can use this to get one of the Euler-Lagrange equations directly (without using the Lagrangian). Looking at the horizontal component of the CoM of the system gives us:
$$m(x + X) + (M + m)X = 0$$Hence $$mx + (M + 2m) X = 0$$Differentiating this twice gives us the second E-L equation from my previous post:

$$m\ddot x + (M + 2m)\ddot X = 0$$

To get the solution, we can relate the KE of the system (as expressed previously) with the lost GPE of the side block:
$$mgx = (\frac 1 2 M + m)\dot X^2 + m\dot x^2 + m\dot x \dot X$$Now, we can use the equation relating $x$ and $X$ to get:
$$\dot X = -\frac{m}{M+2m}\dot x$$And use this to eliminate $\dot X$ from our energy equation, giving:
$$gx = \frac{2M + 3m}{2M + 4m}\dot x^2$$We can then differentiate that to give:
$$\ddot x = \frac{M + 2m}{2M + 3m} g$$And, finally, get:
$$\ddot X = -\frac{m}{2M + 3m}g$$

 

[jbriggs444]

I think that there's a convention by which that's what the row of diagonal marks means.

My understanding of those marks is much more mundane. They indicate that the object, the floor in this case, is part of or anchored to the rest of the world. It is an artistic equivalence of ellipses ("...") and says that the object continues past the part that was explicitly drawn.

 

[sophiecentaur]

I needs some confirmation about the falling mass and the large mass. Conservation of momentum says that the large mass will move to the left so will there be any normal force between falling mass and big mass? I don't thinks so; it certainly starts of at zero. The only horizontal force on the large mass will be due to that component of the radial force on the pulley and the only horizontal force on the falling mass will be due to the horizontal component of the vertical string.

I think this has to mean the only solution must be based on energy and not forces. The available energy is mgh and the energy must be shared according to the momentum conservation.

This problem is ideal for Lagrangian mechanics.

I think it may be only soluble that way.

 

[PeroK]

I needs some confirmation about the falling mass and the large mass. Conservation of momentum says that the large mass will move to the left so will there be any normal force between falling mass and big mass? I don't thinks so; it certainly starts of at zero. The only horizontal force on the large mass will be due to that component of the radial force on the pulley and the only horizontal force on the falling mass will be due to the horizontal component of the vertical string.

I think this has to mean the only solution must be based on energy and not forces. The available energy is mgh and the energy must be shared according to the momentum conservation.

I think it may be only soluble that way.

We can sort things out by attaching the side block on frictionless rails to the large block. That enforces the constraint that these two blocks move together horizontally. I haven't tried to analyse the forces in this scenario,

If we rely on the cord, then it must swing out at an angle to impart the horizontal force on the side block. And, if that angle is significant, then that changes the GPE in the energy equation. In any case, we have another system parameter (this angle of equilibrium) and the problem has changed somewhat.

 

[vanhees71]

But there are horizontal forces: One is at the mass $m$ lying on the block through the thread connected to the hanging mass $m$ on the right side of the block, and then there's an opposite force of equal magnitude on the block, leading to the horizontal component of the acceleration of the three masses as can be calculated from the analysis in #8.

 

[Mister T]

the second diagram is a free-body diagram;

No, it's not. A free body diagram contains only one body, and it shows all the forces acting on that one body. To analyze a situation with more than one body, you draw a separate free body diagram for each body.

there is no real apparatus involved -- it's an abstraction;

Doesn't matter what you call it. You still need a description of it

 

[Steve4Physics]

If we rely on the cord, then it must swing out at an angle to impart the horizontal force on the side block. And, if that angle is significant, then that changes the GPE in the energy equation. In any case, we have another system parameter (this angle of equilibrium)

Isn't the non-vertical part of the cord critical to the question?

The original question asks:

"… will the lateral acceleration of vertical block $m$ be steadily equal to the lateral acceleration of block $M$, throughout the descent of vertical block $m$?"​

Using $m_{top}$ and $m_{side}$ for the two smaller masses, the question can be paraphrased:

Will the horizontal accelerations of $m_{side}$ and $M$ be equal while $m_{side}$ descends?​

$M$ (and its pulley) accelerate left. The angle to the vertical of the string-section supporting $m_{side}$ increases (from zero).

$m_{side}$ accelerates left - accelerated by the horizontal component of the tilted string-section’s tension.

The gap between $m_{side}$ and $M$ will change (increase from zero as angle increases). This is only possible if $m_{side}$ and $M$ have unequal horizontal accelerations.

Answer to actual question asked: no!

 

[sysprog]

The angle to the vertical of the string-section supporting $m_side$ increases (from zero).

What force could temporarily prevent it from accelerating leftward? Doesn't the '$M$ + pulley' object accelerate leftward only as the $m$ block hanging from it descends? Isn't it that descent that drives the system? How could the hanging block not accelerate leftward immediately? If there is a $\theta$ angle, what would it be? How long would the block hang at this angle with nothing to support it?

side accelerates left - accelerated by the horizontal component of the tilted string-section’s tension.

Why would that acceleration be delayed? For how long? Tilted at what angle? How would it for some non-zero time interval hold its place laterally, and not as it descended accelerate leftward wrt to the ground? Isn't it true that what accelerates it leftward is the leftward acceleration of the pulley, that is driven by the descent, and that via the unchanging tension of the cord, suffices to impart to the hanging block the same lateral acceleration as the rest of the system, other than the top $m$ block, which accelerates commensurately rightward?isn't it true that the expression $a=(mg/(2M+m)$ is correct for the leftward acceleration within the system, and that the $m$ in that expression is the $m$ of the hanging $m$ block, and that for the system as a whole, the acceleration referenced by that expression is commensurate with the rightward acceleration of the top $m$ block?

 

[Lnewqban]

... Why would the top block move, when it has the same mass as the side block? What wrong with "the side block falls, not pulling the top block with it but pushing the large block to the left, via a force at the pulley, the leftward movement of which is transmitted to the descending block via the cord tension"?

This statement suggests to me that you still believe that it is possible for the top block not to move (respect to the ground) while the side block descends, and the big block is induced to move towards the left, both by the action of gravity on the side block.

 

[sysprog]

This statement suggests to me that you still believe that it is possible for the top block not to move (respect to the ground) while the side block descends, and the big block is induced to move towards the left, both by the action of gravity on the side block.

I mistakenly had that notion, until @jbriggs444 and @PeroK successfully disabused me of it a few posts later $-$ thanks for pointing it out.

 

[Steve4Physics]

What force could temporarily prevent it from accelerating leftward? Doesn't the '$M$ + pulley' object accelerate leftward only as the $m$ block hanging from it descends? Isn't it that descent that drives the system? How could the hanging block not accelerate leftward immediately? If there is a $\theta$ angle, what would it be? How long would the block hang at this angle with nothing to support it?

Why would that acceleration be delayed? For how long? Tilted at what angle? How would it for some non-zero time interval hold its place laterally, and not as it descended accelerate leftward wrt to the ground? Isn't it true that what accelerates it leftward is the leftward acceleration of the pulley, that is driven by the descent, and that via the unchanging tension of the cord, suffices to impart to the hanging block the same lateral acceleration as the rest of the system, other than the top $m$ block, which accelerates commensurately rightward?isn't it true that the expression $a=(mg/(2M+m)$ is correct for the leftward acceleration within the system, and that the $m$ in that expression is the $m$ of the hanging $m$ block, and that for the system as a whole, the acceleration referenced by that expression is commensurate with the rightward acceleration of the top $m$ block?

That’s al lot of interwoven questions, tangled up with some misconceptions. Impractical to answer - but see if this helps...

Consider what’s happening at the pulley. That’s where the action is at. (Some sort of 60's Lagrangian pun intended.)

The tensions in the 2 parts of cord (left of pulley and below pulley) exert a force on the pulley and hence on $M$.

The horizontal component of this force is what causes $M$ to accelerate left.

The natural tendency of $m_{side}$ is to move vertically downwards. But since this occurs while the pulley is moving left, the cord attached to $m_{side}$ gets tilted.

As a result, $m_{side}$ is moving vertically downwards while simultaneously being pulled left by the horizontal component of the cord’s tension.

This is a smooth, continuous process with the gap between $M$ and $m_{side}$ smoothly increasing from the moment of release. (A bit like the gap between a ball and your hand smoothly increasing as soon as you release the ball.)

The fact that the lower part of the cord is tilted makes the problem quite tricky. I wouldn't like to spend time trying to sort out the maths. And this isn't required according to the question as stated in Post #1.

 

[sysprog]

The natural tendency of mside is to move vertically downwards. But since this occurs while the pulley is moving left, the cord attached to $m_{side}$ gets tilted.

How would it "get tilted", when it is the descent of the side $m$ block that is driving the leftward acceleration of the pulley?

As a result, $m_{side}$ is moving vertically downwards while simultaneously being pulled left by the horizontal component of the cord’s tension.

Doesn't that contradict what you just said, that I just quoted? If the leftward acceleration of the descending block is simultaneous with the descent, the cord would not "get tilted", right?

 

[Steve4Physics]

How would it "get tilted", when it is the descent of the side $m$ block that is driving the leftward acceleration of the pulley?

The acceleration of $M$ is being driven by force - the resultant of the tensions acting on the pulley.

$M$ would still accelerate left if you replaced $m_{top}$ and $m_{side}$ by 2 people maintaining the tension in the cord.

Ask yourself:
- does $m_{side}$ accelerate to the left?
- if yes, what force causes its left acceleration?
- if no, what would the path of $m_{side}$ look like?

The potential energy lost by $m_{side}$ as it falls is converted to the kinetic energy of $M$, $m_{top}$ and $m_{side}$. The falling motion provides the kinetic energy; the tensions in the cord provide the force on $M$.

Doesn't that contradict what you just said, that I just quoted? If the leftward acceleration is simultaneous with the descent, the cord would not "get tilted", right?

Not right! If the cord remained vertical there would be ZERO horizontal force available to accelerate $m_{side}$ to the left. That would mean $m_{side}$ would have zero horizontal velocity.

 

[sysprog]

- does $m_{side}$ accelerate to the left?

I think yes.

- if yes, what force causes its left acceleration?

I think that it's the leftward component of the tension on the cord.

- if no, what would the path of $m_{side}$ look like?

I think that there would have be an external rightward force to balance its leftward acceleration. It would then descend vertically from its original position, with the '$M$ + pulley' object moving increasingly away.

The potential energy lost by $m_{side}$ as it falls is converted to the kinetic energy of $M$, $m_{top}$ and $m_{side}$.

And to that of the '$M$ + pulley' object, right? I think that leftward acceleration is imparted to the side $m$ block as it descends, just as it is to the '$M$ + pulley' object. This appears to me to contradict the following statement that you made in post #51:

The natural tendency of ##m_{side} is to move vertically downwards. But since this occurs while the pulley is moving left, the cord attached to gets tilted.

The falling motion provides the kinetic energy; the tensions in the cord provide the force on $M$.

The energy transfer via the connection of the cord to the pulley results in leftward acceleration of the '$M$ + pulley' object and the side $m$ block, right?

Not right! If the cord remained vertical there would be ZERO horizontal force available to accelerate $m_{side}$ to the left. That would mean $m_{side}$ would have zero horizontal velocity.

I think that the cord remains vertical wrt the pulley, but not wrt the ground, because of its leftward acceleration.

 

[berkeman]

So, after 3 pages of discussion, can I please put the word "simple" in the thread title in quotes? Please, please? When I first clicked into this thread I thought it was simple, until I saw that the supporting surface was frictionless...

 

[jbriggs444]

I think that it's the leftward component of the tension on the cord.

But there can be no leftward component of tension in the cord unless the large block $M$ has already moved away from the side block $m$.

If you were hoping to argue that the two will not separate using an argument based on the fact that they do separate, you have some explaining to do.

 

[sysprog]

But there can be no leftward component of tension in the cord unless the large block $M$ has already moved away from the side block $m$.

I think that the leftward component of tension in the cord is what drives all the leftward acceleration in the system.

If you were hoping to argue that the two will not separate using an argument based on the fact that they do separate, you have some explaining to do.

I had no such hope. I think that they will not separate, but that idea is based is based on the idea that the leftward component of the tension on the cord is based on the balance of tensions between the top $m$ block and the descending $m$ block. I think that the descent of that block is the only driver of accelerations in the system.

 

[sysprog]

So, after 3 pages of discussion, can I please put the word "simple" in the thread title in quotes? Please, please? When I first clicked into this thread I thought it was simple, until I saw that the supporting surface was frictionless...

Please as far as I'm concerned feel free to remove that word altogether.

 

[jbriggs444]

I think that the leftward component of tension in the cord is what drives all the leftward acceleration in the system.

We are talking about the acceleration of the right-hand side block and trying to get you to explain why you think it stays in contact with the middle block. The tension of the cord on the right hand block acts in a vertical direction unless the two blocks separate. You are arguing that they do not separate. So your own argument destroys itself.

It is fairly easy to see that if the side block is positioned against the side of the large block that the system will evolve so that the two become separated. It is also fairly easy to see that if the side block is positioned very far away, the system will evolve so that it closes the gap. It seems clear that there is an equilibrium position at some angle in the middle. Based on this, one might expect some sort of oscillation. A damped oscillation, almost certainly. The hard part is deducing a formula for that oscillation as a function of time.

 

[sysprog]

It appears to me that you're saying that something like this (presumably with a smaller angle) must occur:

Wouldn't there have to be an external rightward force on the descending $m$ block for that to occur?

 

[jbriggs444]

Wouldn't there have to be an external rightward force on the descending $m$ block for that to occur?

No. The middle block moves leftward away from the right hand block which stays more or less in place.

The middle block is subject to a leftward force from the pulley. The right hand block is subject to no such force -- until the cord becomes sufficiently non-vertical anyway.

 

[sysprog]

Why would it do that? Why wouldn't there have to be an external force holding it in place? Wouldn't something like this make more sense?

 

[jbriggs444]

Why would it do that? Why wouldn't there have to be an external force holding it in place?

Absent a horizontal force, the right side block stays in place. You have not identified a leftward force acting on the right side block.

There are exactly two forces that act on the right side block:

1. Gravity. Downward. Its horizontal component is zero.
2. Tension. Upward. Unless the right side block is separated from the middle block, its horizontal component is zero.

Please identify the force that causes the right side block to move leftward without having it move away from the middle block since you claim that it never does so.

Wouldn't something like this make more sense?

Excuse me. You are contemplating a wire frame constructed specifically to make the apparatus do something that it was not designed to do?! No. That is most definitely not what I am talking about. Do not be silly.

Also, please use your words. Diagrams are nice, but verbal descriptions of what the diagram is intended to depict are helpful.

 

[sysprog]

Absent a horizontal force, the right side block stays in place. You have not identified a leftward force acting on the right side block.

What about the leftward tension of the top $m$ block? Doesn't that act on the hanging $m$ block via the cord?

There are exactly two forces that act on the right side block:

1. Gravity. Downward. Its horizontal component is zero.
2. Tension. Upward. Unless the right side block is separated from the middle block, its horizontal component is zero.

Isn't the leftward tension of the top block a third force? How would such separation introduce a horizontal component? Wouldn't it have to have been there from the outset?

Please identify the force that causes the right side block to move leftward without having it move away from the middle block since you claim that it never does so.

Again, the leftward tension of the top $m$ block, right?

Excuse me. You are contemplating a wire frame constructed specifically to make the apparatus do something that it was not designed to do?! No. That is most definitely not what I am talking about. Do not be silly.

Haven't you been saying that the right block doesn't move leftward when the '$M$ + pulley' object does? If that were so, then what would counteract the leftward tension of the top block impelling the right side $m$ block leftward ?

I attempted an illustration of an external force to do that counteracting, and you seem to dismiss that as my being silly, but I don't seee a good reason why it's not a reasonable depiction, not of the situation as originally postulated, but of the situation as it would, as far as I know, have to be for the right side $m$ block to remain laterally stationary while the '$M$ + pulley' object moved leftward $-$ can you please say why some kind of external restraint would not be necessary to keep the right block from remaining vertical wrt to the pulley, rather than wrt the ground?

 

[PeroK]

I attempted an illustration of an external force to do that counteracting, and you seem to dismiss that as my being silly, but I don't seee a good reason why it's not a reasonable depiction, not of the situation as originally postulated, but of the situation as it would, as far as I know, have to be for the right side $m$ block to remain laterally stationary while the '$M$ + pulley' object moved leftward $-$.

Let's look at the initial motion as several stages:

1) The side mass falls a small distance vertically under gravity.

2) it pulls the top mass a small distance to the right.

3) An internal force at the pulley moves the large block a very small distance to the left.

Note that the CoM (horizontal component) of the system has not changed and a gap has opened up between the large block and the side block.

Now, we look at the next short time interval.

1) The side block falls a little further, but is also pulled to the left by the angled cord.

2) The top block moves a bit more to the right.

3) The large block is pushed a bit more to the left.

Note that if the block is large, we soon get to the scenario where the leftwards motion of the side block equals or exceeds the small leftward motion of the large block. There must be a small equilibrium angle, which could be calculated.

I think I have to point out that your mechanics in this thread is wild and woolly and represents more of an abuse than a use of Newton's laws!

I suggest you take a step back and really try to understand what I've just written. Don't jump in with "that can't be right because ...". We've spent enough time between us arguing against your version of Newton's laws. The time has come for you to focus on the correct analyses in this thread and really try to understand why we are right.

 

[Steve4Physics]

Hi @sysprog. You have hit a ‘blind-spot’, so forget about the original problem for a moment. Think about this one:

You hang a pendulum (small ball on a string) from the roof of your car, and accelerate left.

Q1. Is the ball in equilibrium? I.e. do the forces on the ball balance?

Q2. What individual forces act the ball? If the resultant of these forces is non-zero, in what direction does it act?

Q3. What happens to the string?

 

[vanhees71]

But there are horizontal forces: One is at the mass lying on the block, and then

So, after 3 pages of discussion, can I please put the word "simple" in the thread title in quotes? Please, please? When I first clicked into this thread I thought it was simple, until I saw that the supporting surface was frictionless...

The problem is "simple" when using the right tools, which obviously is the action principle. Analyzing it with free-body diagrams (for me the most complicated subject in mechanics I can think of) is at least tricky if not impossible.

 

[PeroK]

The problem is "simple" when using the right tools, which obviously is the action principle. Analyzing it with free-body diagrams (for me the most complicated subject in mechanics I can think of) is at least tricky if not impossible.

We have a fairly simple solution for the case where the side block is constrained to move with the large block. But, not for the case where the side block may swing away from the large block. That looks quite complicated.

 

[jbriggs444]

What about the leftward tension of the top $m$ block? Doesn't that act on the hanging $m$ block via the cord?

No. That is not how tension and pulleys work.

What matters for the effect of tension on the side block is the direction of the cord where it contacts the side block. One can look at the tension in the last millimeter of [ideal] cord and the direction of the last millimeter of [ideal] cord and determine the resulting force applied at its point of attachment.

The cord between top block and pulley is horizontal. It exerts a horizontal force at its attachment point on the top block.

The cord between pulley and bottom block is vertical. It exerts a vertical force at its attachment point on the side block.

The interaction with the pulley is more interesting. The cord exerts a uniform radial force along a 90 degree arc of the ideal pulley's surface. The net of this is a force whose magnitude is $\sqrt{2}$ times tension at an angle 45 degrees below the horizontal.

Of course, Newton's third law still applies. The cord is subject to a force from its two attachment points and from the pulley. Since it is an ideal cord and since this means that it is massless, the forces on the cord must sum to zero. And indeed they do. We have a leftward force of magnitude T plus a downward force of magnitude T plus an up-and-to-the-rightward force of magnitude $\sqrt{2}T$. They sum to zero. Life is still good and the laws of physics still hold.

The side block is not subject to any magical horizontal force transmitted through a vertical section of cord. That would be a shear force. Ideal strings, threads, ropes and cords do not do shear.

Edit to add...

I've never done the coursework on continuum mechanics where this stuff might be discussed, but @Dale and @Doc Al have let enough leak through over the years that I have a basic grasp.

In first year physics one treats tension as a force that has particular behaviors. It is always parallel to the cord. It is uniform over the length of an ideal cord. It is transmitted unchanged in magnitude over ideal pulleys.

It turns out that there is a deeper and mathematically correct way to think about tension instead. It is not a force at all. It is a particular condition of stress in the cord. It is given by the stress tensor.

Expressed in component form, a tensor is a matrix. For stress within a three-dimensional body it is a 3 by 3 matrix. Its component values are force components per unit area at a point within the volume of the body. There are three directions for the force components. And there are three orthogonal planes on which a directed area can be measured. So there are nine components total.

For a cord under uniform pure tension along its length and a coordinate system aligned nicely with the cord, a pure tension of $T$ results in a stress tensor of
$$\left[ {\begin{array}{ccc}
-T & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array} } \right]$$For an ideal cord, this tensor value will be present at every point within the volume of the cord. [As it passes over an ideal pulley, the stress will change. For a pulley aligned with the coordinate system and a 90 degree turn, a different diagonal element will then have the $-T$ term].

If you perform a tensor multiplication of the stress tensor times a directed area, you get a vector for the force transmitted by the material across that area. For a tensor that corresponds to tension in a cord, it will turn out that no matter how you slice across the cord, the product will be a vector force parallel to the cord at the point of the cut.

That is the mathematics that goes behind the statement that the direction of the last millimeter of cord is what matters.

More generally, one can use the tensor formalism to do beautiful things like computing the power transmitted through a shaft or belt as a surface integral of local stress times incremental area times local velocity over a chosen slice.

 

[PeroK]

For the record, the tension in the cord (for the variation of this problem for which we have a solution) can be calculated from the acceleration of the top block:
$$T = m(\ddot x + \ddot X) = \frac{M + m}{2M + 3m}mg$$And, confirmed by using the acceleration of the side block:
$$(mg - T) = m\ddot x \ \Rightarrow \ T = m(g - \ddot x) = (1 - \frac{M + 2m}{2M + 3m})mg = \frac{M + m}{2M + 3m}mg$$