B Question about a simple free body diagram

[sysprog]

What's wrong?

It disobeys the physical constraint that the two small blocks are linked by a cord. That puts a fundamental constraint on your system, which was reflected in my Lagrangian.

There is no horizontal external force on the system, so it cannot spontaneously move to the left. Any motion of the large block and side block must be offset by motion of the top block to the right.

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

I think that the energy transfer from the descent of the side $m$ block gets imparted to the leftward movement $-$ that the centroid of the system can and must move leftward if the side $m$ block descends. I mistyped this in my most recent posts, omitting the $mg/$, so, correctly this time, what's wrong with $a=mg/(2M+m)$?

 

[PeroK]

The energy transfer from the descent of the side $m$ block gets imparted to the leftward movement. The centroid of the system can and must move leftward if the side $m$ block descends. I mistyped this in my most recent posts, omitting the $mg/$, so, correctly this time, what's wrong with $a=mg/(2M+m)$?

That defies Newton's laws of motion in the absence of an external horizontal force.

 

[sysprog]

That defies Newton's laws of motion in the absence of an external horizontal force.

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

 

[vanhees71]

This again solidifies my prejudice that free-body diagrams are the greatest obstacle to solve mechanical problems. As demonstrated above, Hamilton's principle rules, and math is your friend ;-)).

 

[sysprog]

@vanhees71, it seems to me that LaGrangian analysis isn't necessary for solving this $-$ what's wrong with leftward $a=mg/(2M+m)$?

 

[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.

 

[Herman Trivilino]

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?