Why does the big block accelerate?

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Homework Statement:

A block with mass M lies on a slippery horizontal surface. On top of it there is another block with mass m which in turn is attached to an identical block by a string. The string
has been pulled across a pulley situated at the corner of the big block and the second small block is hanging vertically. Initially, the system is held at rest. Find the acceleration of the big block immediately after the system is released. You may neglect friction, as well as masses of the string and the pulley.

Relevant Equations:

$$ \sum F =ma$$
Screenshot_2020_0809_100341.png

Clearly, in the picture I can see that on the small block on top, tension and gravitational force act. Gravity gets balance by the normal force, so tension is the only force causing acceleration. On the block at the side gravity and tension result in vertical acceleration. However, I do not see any force acting on the big block except the normal forces and gravity, which appear to me, to balance out. So, which force am I missing out on that causes the acceleration? And after that how should I go about finding the constraint equations? Any help would be appreciated.

I have also attached the given answer.
 

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  • #2
DaveC426913
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Conceptually:

Where is the centre of mass (approximately) of the entire system at the start?
If mass M did not move, where would the centre of mass of the entire system be (approximately) at the end?
 
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Conceptually:

Where is the centre of mass (approximately) of the entire system at the start?
If mass M did not move, where would the centre of mass of the entire system be (approximately) at the end?
I think I might've read something like this somewhere. Centre of mass at the start would be somewhere between the three masses, but closer to the big block. At the end, if the big block didn't move, it probably would shift somewhere away from that point. But no external force acts on the system, so that shouldn't happen.
Is this reasoning proper sir? Even if it is, I don't know, how it express it on paper in this context.
 
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As gravity on the side block impels the top block toward the pulley, it also impels the pulley toward the top block, and the pulley is affixed to the big block. If you imagine the top block held in place, the side block could not descend without moving the larger block. If you imagine the density of each of the 2 smaller blocks to be much greater than that of the larger block, it's perhaps easier to see that the larger block would move ##-## a lesser-mass large block would move more than a greater-mass top block.
 
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As gravity on the side block impels the top block toward the pulley, it also impels the pulley toward the top block, and the pulley is affixed to the big block. If you imagine the top block held in place, the side block could not descend without moving the larger block. If you imagine the density of each of the 2 smaller blocks to be much greater than that of the larger block, it's perhaps easier to see that the larger block would move ##-## a lesser-mass large block would move more than a greater-mass top block.
Okay, I think I see something here. So, the tension in the string would also act on the big block, to drive it in the opposite direction as the top block, I guess? I think that works. That gets me the answer too. But if there are any nuances, then kindly point them out as well. Thank you!
 
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Okay, I think I see something here. So, the tension in the string would also act on the big block, to drive it in the opposite direction as the top block, I guess? I think that works. That gets me the answer too. But if there are any nuances, then kindly point them out as well. Thank you!
One 'nuance' is that as the descending block descends, it will move in the direction in which the large block moves with the pulley, so that the descent path will be a curve, the horizontal component of which will be in the opposite direction to that of the path of the top block.
 
  • #7
jbriggs444
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One 'nuance' is that as the descending block descends, it will move in the direction in which the large block moves with the pulley, so that the descent path will be a curve, the horizontal component of which will be in the opposite direction to that of the path of the top block.
Note that, as drawn, there is no horizontal force on the descending block which could explain a non-zero horizontal component to its motion. It will begin by going straight down. Meanwhile, the big block is moving left. The two blocks will separate -- the descending block will swing clear.

Once it has swung clear of the big block, the supporting force from tension on the descending block will have a non-zero horizontal component. The descending block will start to move to chase the big block.
 
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Note that, as drawn, there is no horizontal force on the descending block which could explain a non-zero horizontal component to its motion. It will begin by going straight down. Meanwhile, the big block is moving left. The two blocks will separate -- the descending block will swing clear.

Once it has swung clear of the big block, the supporting force from tension on the descending block will have a non-zero horizontal component. The descending block will start to move to chase the big block.
There would be no delay between the onset of the horizontal motions of the top block and the larger block, nor between the pulley beginning to move and the motion of the descending block acquiring a non-zero horizontal component.
 
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  • #9
PeroK
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Homework Statement:: A block with mass M lies on a slippery horizontal surface. On top of it there is another block with mass m which in turn is attached to an identical block by a string. The string
has been pulled across a pulley situated at the corner of the big block and the second small block is hanging vertically. Initially, the system is held at rest. Find the acceleration of the big block immediately after the system is released. You may neglect friction, as well as masses of the string and the pulley.
Relevant Equations:: $$ \sum F =ma$$

View attachment 267513
Clearly, in the picture I can see that on the small block on top, tension and gravitational force act. Gravity gets balance by the normal force, so tension is the only force causing acceleration. On the block at the side gravity and tension result in vertical acceleration. However, I do not see any force acting on the big block except the normal forces and gravity, which appear to me, to balance out. So, which force am I missing out on that causes the acceleration? And after that how should I go about finding the constraint equations? Any help would be appreciated.

I have also attached the given answer.
Another way to think about forces here: look at the pulley at the top right. Are there any forces acting on the pulley from the tight string?
 
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  • #10
jbriggs444
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There would be no delay between the onset of the horizontal motions of the top block and the larger block, nor between the pulley beginning to move and the motion of the descending block acquiring a non-zero horizontal component.
Yes, I agree. Although first, second and third derivatives of the descending block's horizontal position are zero, I believe that the fourth derivative is leftward and non-zero.
 
  • #11
haruspex
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There would be no delay between the onset of the horizontal motions of the top block and the larger block, nor between the pulley beginning to move and the motion of the descending block acquiring a non-zero horizontal component.
Not so.
At release, there is no horizontal force on the suspended block, so it has only vertical acceleration. The large block does have a net horizontal force, so does have a horizontal acceleration.
The relative acceleration of the two therefore has a horizontal component.
They separate. We can calculate a steady state angle, but starting from vertical the complete motion likely has an oscillatory component.
 
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  • #12
DaveC426913
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I think I might've read something like this somewhere. Centre of mass at the start would be somewhere between the three masses, but closer to the big block. At the end, if the big block didn't move, it probably would shift somewhere away from that point. But no external force acts on the system, so that shouldn't happen.
Is this reasoning proper sir? Even if it is, I don't know, how it express it on paper in this context.
Your reasoning is correct, yes.
I simply wanted to give you a conceptual push toward being sure that M does, in fact, move - so you know you're on the right track.
 
  • #13
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... However, I do not see any force acting on the big block except the normal forces and gravity, which appear to me, to balance out. So, which force am I missing out on that causes the acceleration?...
There is a reaction force at the support of the pulley.
If we decompose it into x and y imaginary axes, ##R_x## would be the force you are missing out.

18r.gif

30913D15-4B1F-44CD-BA20-2C90F07D219C.jpeg
 
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Yes, I agree. Although first, second and third derivatives of the descending block's horizontal position are zero, I believe that the fourth derivative is leftward and non-zero.
At the instant of release, both of the velocity vectors (first derivatives of position with respect to time) on the descending block are zero, the second time derivatives (acceleration) of the vertical and horizontal position of the descending block are both non-zero, and the third and fourth derivatives are (in classical mechanics) inconsequential regarding force.
 
  • #15
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sysprog said:
There would be no delay between the onset of the horizontal motions of the top block and the larger block, nor between the pulley beginning to move and the motion of the descending block acquiring a non-zero horizontal component.
Not so.
So you think that it's not so that there would not be either such delay. For there to be a delay, there would have to be some force in the system that commences prior to another. There isn't. All lateral accelerations in the system at the moment of release are immediate and due to gravity.
At release, there is no horizontal force on the suspended block, so it has only vertical acceleration.
Not so. All of the horizontal acceleration on the large block is also imparted to the pulley, the horizontal acceleration of which immediately causes the descending block to also move horizontally. As little and as much as the pulley moves horizontally, so little and so much does the descending block move horizontally.
The large block does have a net horizontal force, so does have a horizontal acceleration.
The horizontal acceleration on the large block is not independent of the vertical acceleration of the descending block, all the lateral acceleration of the large block is due to the downward acceleration of the descending block, Although the descending block moves in a curve, its path is vertical with respect to the pulley.
The relative acceleration of the two therefore has a horizontal component.
Both the big block and the descending block have a horizontal acceleration, both of which simultaneously result from the vertical force of gravity acting on the system.
They separate.
The descending block remains in contact the large block, because the pulley from which it is descending moves exactly with the large block, and all of the lateral acceleration on the assemblage of the large block and the pulley comes solely from the gravitational force on the descending block.
We can calculate a steady state angle, but starting from vertical the complete motion likely has an oscillatory component.
Just as the block does, the pulley acquires non-zero horizontal velocity immediately. That horizontal acceleration is not due to a system-independent or external force; it's all due to gravity and the effect of the positionings of the blocks and the pulley system. The block continually moves straight down with respect to the pulley. It doesn't move straight down while the pulley independently moves to the left, separating away from it as if not connected to it. That very connection is the only means by which the big block is accelerated laterally. The angle between the path of the descending block and that of the top block is always 90 degrees. The force of the pulley on the block to which it is connected is always at 45 degrees.
 
  • #16
haruspex
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For there to be a delay, there would have to be some force in the system that commences prior to another.
I'm only claiming a delay (at least) in the horizontal acceleration components of the suspended mass and the large block reaching the same magnitude. That is enough for the descending mass to lose contact with the large block.
As I wrote, we can find the equilibrium angle for steady accelerations. I get that if the suspended mass trails at angle theta from the vertical then ##s=\sin(\theta)## satisfies ##2Ms=m(1-s)^2##.
If there were no detachment s would be zero.

If you disagree, do the math.

Edit: perhaps a more cogent path is to have the descending mass trapped in a smooth vertical track so that it cannot separate from the large block. Compute the horizontal force the large block must exert via the track on the descending mass. You will find it is not zero.
 
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  • #17
jbriggs444
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At the instant of release, both of the velocity vectors (first derivatives of position with respect to time) on the descending block are zero, the second time derivatives (acceleration) of the vertical and horizontal position of the descending block are both non-zero.
If you do a force balance on the descending block, you will see that this claim regarding horizontal acceleration is not correct. You cannot have horizontal acceleration without a horizontal force. No such force is present.
 
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  • #18
jbriggs444
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All lateral accelerations in the system at the moment of release are immediate and due to gravity.
This statement lacks clarity.

It would be correct to say that the lateral acceleration of the descending block after any finite delay is non-zero. In this sense the lateral acceleration is "immediate".

It would not be correct to say that the lateral acceleration of the descending block at the start of the scenario is non-zero. It is zero. In this sense the lateral acceleration is not "immediate".
 
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sysprog said:
All lateral accelerations in the system at the moment of release are immediate and due to gravity.
This statement lacks clarity.

It would be correct to say that the lateral acceleration of the descending block after any finite delay is non-zero. In this sense the lateral acceleration is "immediate".

It would not be correct to say that the lateral acceleration of the descending block at the start of the scenario is non-zero. It is zero. In this sense the lateral acceleration is not "immediate".
I didn't say " at the start of the scenario"; I said "at the moment of release". The problem statement says "Initially the system is held at rest. Find the acceleration of the big block immediately after the system is released." I think that the difference between "at the moment of" (with "are immediate") and "immediately after" doesn't make my statement "unclear".
 
  • #20
haruspex
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I didn't say " at the start of the scenario"; I said "at the moment of release". The problem statement says "Initially the system is held at rest. Find the acceleration of the big block immediately after the system is released." I think that the difference between "at the moment of" (with "are immediate") and "immediately after" doesn't make my statement "unclear".
At the instant of release there is no lateral force on the suspended block so no lateral acceleration.
Any nonzero time later, no matter how small, there will be a lateral acceleration of the suspended mass, but only because there is now a nonzero distance between it and the large block. The two have to separate in order for the suspended block to be pulled sideways at all.
 
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  • #21
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I'm only claiming a delay (at least) in the horizontal acceleration components of the suspended mass and the large block reaching the same magnitude. That is enough for the descending mass to lose contact with the large block.
As I wrote, we can find the equilibrium angle for steady accelerations. I get that if the suspended mass trails at angle theta from the vertical then ##s=\sin(\theta)## satisfies ##2Ms=m(1-s)^2##.
If there were no detachment s would be zero.

If you disagree, do the math.
The given answer is simply ##mg/(2M + m)##. Your introduction of delay requires that an extra-systemic force be applied laterally to the large block. There is in the problem as stated no such external impetus by which any ##\theta## angle between the large block and the descending block would be introduced.
Edit: perhaps a more cogent path is to have the descending mass trapped in a smooth vertical track so that it cannot separate from the large block.
Compute the horizontal force the large block must exert via the track on the descending mass. You will find it is not zero.
Assuming that you mean that the trapping keeps the descent vertical with respect to the pulley, the added apparatus is superfluous; the track may be disregarded as ornamentation attached to the large block.
At the instant of release there is no lateral force on the suspended block so no lateral acceleration.
Any nonzero time later, no matter how small, there will be a lateral acceleration of the suspended mass, but only because there is now a nonzero distance between it and the large block. The two have to separate in order for the suspended block to be pulled sideways at all.
The large block is pushed sideways (and inconsequentially downward), before and after the release from static condition, by the action of the pulley in response to the string tension. The descending block has lateral acceleration only as soon as it begins to descend. The required non-zero distance corresponding to the non-zero leftward distance traveled by the large block and pulley assembly is that of the vertical difference between the static position of the descending block and the new position thereof. All of the energy in the system that upon/after release begins/continues to move anything comes from the transformation of the potential energy of the block on the right (the descending block) into kinetic energy.
 
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  • #22
haruspex
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Assuming that you mean that the trapping keeps the descent vertical with respect to the pulley, the added apparatus is superfluous;
That is just your assumption, and it is incorrect, as you will discover if you DO THE MATH.
 
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  • #23
jbriggs444
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I didn't say " at the start of the scenario"; I said "at the moment of release". The problem statement says "Initially the system is held at rest. Find the acceleration of the big block immediately after the system is released." I think that the difference between "at the moment of" (with "are immediate") and "immediately after" doesn't make my statement "unclear".
Nothing there clarifies anything.

If t=0 is the moment of release and if ##a_x(t)## is the leftward lateral acceleration of the descending block at time t then ##a_x(0) = 0##.

No one is disagreeing with a claim that ##a_x(t)## > 0 for small ##t## > 0.
 
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  • #24
etotheipi
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If you do assume the suspended block is always in contact with the larger one then it's not so bad, let ##y##, the height of the suspended mass, and ##x##, the position of the centre of mass of the larger box, be two generalised coordinates. You would have$$\mathcal{L} = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) + \frac{1}{2}m(\dot{x} - \dot{y})^2 - mgy$$Then for ##x##, $$M\ddot{x} + m\ddot{x} + m(\ddot{x} - \ddot{y}) = (M+2m)\ddot{x} - m\ddot{y}= 0$$and for ##y##$$m\ddot{y} - m(\ddot{x} - \ddot{y}) = 2m\ddot{y} - m\ddot{x}= -mg$$You can see that the first equation is just momentum conservation in the ##x## direction and the second equation is the vertical NII on the suspended mass. You can solve that for ##\ddot{x}## if you like, you will see that ##\ddot{x} < 0##

But what if the suspended block is no longer in constant contact, as others have mentioned? I think it will be helpful to redefine ##y## as the height of the suspended mass w.r.t. the pulley. Also let the horizontal displacement of the suspended block to the larger block be ##\varepsilon##. If the length of the string is ##l## and the horizontal position of the little block on the top w.r.t. the pulley is ##\beta## (s.t. ##\beta < 0## during the motion), then $$\sqrt{\varepsilon^2 + y^2} - \beta = l \implies \frac{\varepsilon \dot{\varepsilon} + y\dot{y}}{\sqrt{\varepsilon^2 + y^2}} - \dot{\beta}= 0$$ $$\mathcal{L} = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m \left(\dot{x} + \frac{\varepsilon \dot{\varepsilon} + y\dot{y}}{\sqrt{\varepsilon^2 + y^2}} \right)^2 + \frac{1}{2}m\left((\dot{x} + \dot{\varepsilon})^2 + \dot{y}^2 \right) - mgy$$Luckily as is to be expected this reduces to the first lagrangian in the case ##\varepsilon = 0##. Maybe the system of 3 el equations can be solved numericwlly, if I haven't made a mistake? It would be interesting to see if we do indeed get the oscillatory motion...
 
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I think that is incorrect. For one thing, transform into the accelerating frame of the large block, then on the suspended mass there will be a rightward pointing inertial force, weight, and the tension force. The effective weight is not vertical, so we will obtain some sort of oscillatory motion that is always instantaneously about the equilibrium position defined by the instantaneous value of the acceleration of the large block.

Maybe if I am feeling up to it later I will try to plot the solutions from the euler lagrange equations.
You are apparently visualizing a leftward movement of the big block and pulley that is somehow independent of and so not immediately connected to the downward and leftward movement of the descending block.
 

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