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.

 

[Mister T]

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 $a=mg/(2M+m)$?