B Question about a simple free body diagram

[sysprog]

In this diagram, if the system begins by being held static, and then at time $T$ release of all components occurs, at all times $>T$, until vertical movement stops upon vertical block $m$ making contact with the Normal Force at the base of block $M$, will the lateral acceleration of vertical block $m$ be steadily equal to the lateral acceleration of block $M$, throughout the descent of vertical block $m$?

 

[jbriggs444]

I struggle to understand the diagram.

You have a force $F_1$ which appears to be the force of tension on the upper mass $m$.
You have a force $mg$ which appears to be the force of gravity on the upper mass $m$.
You do not depict any contact force between the upper mass $m$ and the main block $M$.

You have a force labelled mg/(2M+m) which is applied diagonally to both the pulley and the right hand mass $m$. That one makes zero sense.
You do not depict any other forces acting on the pulley.

You have a force labelled $F_2$ acting leftward on the right hand mass $m$. I have no idea what force this is supposed to be.
You do not depict any other forces acting on the right hand mass.

Finally, you depict a normal force $N$ acting upward on the main mass $M$
You do not depict any other forces acting on the main mass.

You are missing a lot of forces. But your question does not concern itself with that...

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

Well, what horizontal forces act on the two blocks?

You do not have any of them on your diagram.

 

[sysprog]

I recognize that the diagram is unconventional and partly incorrect (arrow at upper left is pointing in the wrong direction). What about this diagram?

Wouldn't the large block and the vertically hanging block move horizontally? The horizontal force should be $mg/(2M+m)$, right?

 

[jbriggs444]

I recognize that the diagram is unconventional and partly incorrect (arrow at upper left is pointing in the wrong direction). What about this diagram?

View attachment 295438

Wouldn't the large block and the vertically hanging block move horizontally? The horizontal force should be $mg/(2M+m)$, right?

Things are worse now. You have three free objects on your diagram and a total of four forces depicted.

Which horizontal force are you talking about?

 

[sysprog]

Things are worse now. You have three free objects on your diagram and a total of four forces depicted.

Which horizontal force are you talking about?

$F=mg/(2M+m)$

 

[Lnewqban]

In this diagram, ..., will the lateral acceleration of vertical block $m$ be steadily equal to the lateral acceleration of block $M$, throughout the descent of vertical block $m$?

I would say yes, as both bodies are linked by the pulley and the string.

In real life, if the big body moves towards the left, initially the small body should not move until the horizontal component of the tension in the string (now adopting certain angle respect to the vertical as the pulley moves) makes it accelerate towards the left as well.

But that is an unnecessary complication for this simple problem.
Just consider that the small mass simultaneously descends and moves towards the left at the same rate of acceleration of the big mass, providing a constant tension force in the string.

 

[jbriggs444]

$F=mg/(2M+m)$

You've quantified a force but not identified what it is. What body exerts this force? What body is it exerted on? What manner of force is it?

You are supposed to finish with the free body diagram first and calculate the magnitudes of the various forces afterward. The idea is to use the free body diagram to help you figure out what equations you can write down and solve. It is supposed to be an exhaustive list of the forces acting on a particular body. One normally puts only one free body in a free body diagram. Things get cluttered when you have three bodies.

A proper free body diagram could let you see that there is no leftward force acting on the right-hand mass $m$ and that there is a net leftward force acting on the big mass $M$. It follows that the two will separate at least momentarily. But that does not affect the initial force balance.

You can still write down some equations that will hold exactly for an instant and approximately for a while longer.

 

[PeroK]

This problem is ideal for Lagrangian mechanics. Let $x$ be the dispacement of the small top block to the right towards the pullley (which is also the vertical displacement of the side block). And, let $X$ be the displacement of the large block to the left ($X$ will be negative). Then, the displacement of the top block is $x + X$ and the displacement of the side block is $(X, -x)$. We have:
$$T = \frac 1 2 M \dot X^2 + \frac 1 2 m(\dot x + \dot X)^2 + \frac 1 2 m (\dot X^2 + \dot x^2) = (\frac 1 2 M + m)\dot X^2 + m\dot x^2 + m\dot x \dot X$$And $$U = -mgx$$This gives us the Lagrangian:
$$L = T - U = (\frac 1 2 M + m)\dot X^2 + m\dot x^2 + m\dot x \dot X + mgx$$And:
$$\frac{\partial L}{\partial X} = 0, \ \frac{\partial L}{\partial x} = mg, \ \frac{\partial L}{\partial \dot X} = (M + 2m) \dot X + m\dot x, \ \frac{\partial L}{\partial \dot x} = 2m\dot x + m \dot X$$Giving the Euler-Lagrange equations:$$2m\ddot x + m\ddot X = mg, \ \ m\ddot x + (M + 2m)\ddot X = 0$$Which gives the solution:
$$\ddot x = \frac{M+2m}{2M + 3m}g, \ \ \ddot X = -\frac{m}{2M + 3m}g$$Although, I haven't double-checked that answer.

 

[sysprog]

@PeroK but wouldn't $x=0$? Why would the the top $m$ block go anywhere? It seems to me that the pulley and $M$ block move leftward beneath the top $m$ block, and with them the side $m$ block that is hanging vertically directly below the pulley also moves leftward, the vertical descent of that block being the source of the energy transfer for all of the horizontal motion.

 

[Herman Trivilino]

Why would the the top $m$ block go anywhere? It seems to me that the pulley and $M$ block move leftward beneath the top $m$ block, and with them the side $m$ block that is hanging vertically directly below the pulley also moves leftward, the vertical descent of that block being the source of the energy transfer for all of the horizontal motion.

It would help if you did a free body diagram of each of the objects shown in your original drawing. Note that your original drawing is not a free body diagram. Moreover, you don't include a description of the apparatus. Physics is a study of the phenomena, not of drawings in textbooks.

 

[sysprog]

It would help if you did a free body diagram of each of the objects shown in your original drawing. Note that your original drawing is not a free body diagram. Moreover, you don't include a description of the apparatus. Physics is a study of the phenomena, not of drawings in textbooks.

Yes, the first drawing is in incorrect; the second diagram is a free-body diagram; there is no real apparatus involved -- it's an abstraction; your remark about Physics seems to me to be overly restrictive.

 

[jbriggs444]

Yes, the first drawing is in incorrect; the second diagram is a free-body diagram; there is no real apparatus involved -- it's an abstraction; your remark about Physics seems to me to be overly restrictive.

A free body diagram has one free body and depicts all of the forces acting on that body.

Why would the the top $m$ block go anywhere?

It is subject to an unbalanced rightward horizontal pull from the cord running over the pulley. It must move rightward.

A proper free body diagram for the upper block would have made this obvious.

 

[sysprog]

It is subject to an unbalanced rightward horizontal pull from the cord running over the pulley. It must move rightward.

Assuming, as is common for such idealized problems, that the cord is 'massless', it looks balanced to me $-$ the $m$ masses are equal $-$ I think that when the distance between the top $m$ and the pulley decreases, it is due to the leftward movement of the '$M$ + pulley' object, and that the freedom to move of that object is what allows the hanging $m$ block to descend.

 

[A.T.]

the second diagram is a free-body diagram

Free-body means that you draw all bodies separately with all the farces on each, like shown here for a similar scenario (slide 1):

https://slideplayer.com/slide/4164519/

But In your case you you want to include the block M too, nut just the pulley.

 

[sysprog]

But In your case you you want to include the block M too, nut just the pulley.

(Thanks for that to-the-point illustration.) Block $M$ is connected to the pulley, whence the reference to the '$M$ + pulley' object.

 

[robphy]

$F=mg/(2M+m)$

That doesn't have the correct units for a force.

 

[sysprog]

That doesn't have the correct units for a force.

oops $-$ you're right, of course $-$ thanks for pointing that out $-$ I meant to indicate the lateral acceleration $-$ the expression should be $a=mg/(2M+m)$.

 

[PeroK]

@PeroK but wouldn't $x=0$? Why would the the top $m$ block go anywhere? It seems to me that the pulley and $M$ block move leftward beneath the top $m$ block, and with them the side $m$ block that is hanging vertically directly below the pulley also moves leftward, the vertical descent of that block being the source of the energy transfer for all of the horizontal motion.

If we take $M$ to be large, or the large block to be fixed, then the small masses slide along the top of the block and down the side respectively.

If the block moves, this still happens but the dynamics of the system are complicated by the motion of large block.

$x$ is the displacement of the top block relative to the large block. And $X + x$ is the displacement of the top block relative to the ground. $x = 0$ initially but when released the top block will slide towards the pulley.

 

[sysprog]

I think that the magnitude of the mass of $M$ is irrelevant $-$ if it's not fixed, it moves, however slowly. I think that the '$M$ + pulley' object and the side $m$ block hanging below the pulley move leftward, while the top $m$ block remains stationary.

 

[PeroK]

I think that the magnitude of the mass of $M$ is irrelevant $-$ if it's not fixed, it moves, however slowly. I think that the '$M$ + pulley' object and the side $m$ block hanging below the pulley move leftward, while the top $m$ block remains stationary.

If that were true, then the equations of motion of the system would tell you that. The top block cannot remain stationary. The side block must fall and pull the top block with it.

The only source of KE for the system is if the side block falls under gravity. If this does not happen, then you have a stationary rigid object.

 

[sysprog]

@PeroK, as you stated, the $m$ blocks won't move if the '$M$ + pulley' object is fixed. I don't see why when it's not fixed, that would make the top $m$ block move. What impetus would make it move? Does it depend on the relative masses of the $m$ block and the '$M$ + pulley' object? Or does only the speed, and not the whether, of the rightward motion depend on that? Are you treating the connection between the the top $m$ block and the '$M$ + pulley' object as frictionless? I was treating the cord as massless, and the pulley as frictionless, but not treating the the top $m$ block as frictionless. I thought that the absence of diagonal marks at either of the surfaces there, in a diagram that has them, meant that frictionlessness was not a condition of for that connection.

 

[PeroK]

@PeroK, as you stated, the $m$ blocks won't move if the '$M$ + pulley' object is fixed.

The $m$ blocks will move in any case. If $M$ is fixed, then the problem is simple.

It's only when $M$ is allowed to move that it gets complicated.

 

[sysprog]

Why would they move with $M$ fixed if the cord is massless?

 

[PeroK]

Why would they move with $M$ fixed if the cord is massless?

Gravity pulls the side block down!

 

[sysprog]

But wouldn't friction hold the top block in place?

 

[PeroK]

But wouldn't friction hold the top block in place?

If it does, then you effectively have a rigid body that remains at rest. Nothing moves.

The solution I posted assumed frictionless sliding in all cases.

 

[sysprog]

If the connection of the '$M$ + pulley' object to the normal force is frictionless (as depicted), and the connection between it and the top $m$ block isn't, then wouldn't the '$M$ + pulley' object be accelerated with $a=(2M+m)$, that being either sufficient or insufficient to overcome the friction and thereby cause leftward movement of it and the hanging $m$ block? Wouldn't it have to be sufficient, for the side $m$ block to descend? And if it were sufficient, wouldn't the top $m$ block remain stationary?

 

[PeroK]

If the connection of the '$M$ + pulley' object to the normal force is frictionless (as depicted), and the connection between it and the top $m$ block isn't, then wouldn't the $M$ + pully object be accelerated with $a=(2M+m)$, that being either sufficient or insufficient to overcome the friction and thereby cause leftward movement of it and the hanging $m$ block? And if it were sufficient, wouldn't the top $m$ block remain stationary?

That's physically impossible. Either the side block falls, pulling the top block with it and pushing the large block to the left (via a force at the pulley. Or, the side and top blocks are stuck to the large block in some way, in which case the whole system remains at rest.

You seem to be assuming that an irregularly shaped body will horizontally accelerate spontaneously, in defiance of Newton's laws!

 

[sysprog]

There is friction between the top $m$ block and the '$M$ + pulley' object, which accelerates because of the force on the pulley, but no friction between the side $m$ block and anything. What's wrong with $a=(2M + m)$? 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"?

 

[PeroK]

What's wrong with $a=(2M + m)$? 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"?

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.

 

[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.