The problem involves a plank executing vertical simple harmonic motion (SHM) with a mass placed on top. The discussion centers around determining the conditions under which the mass will break off from the plank, particularly focusing on the forces acting on the mass and the timing of its separation during the motion.
The discussion is ongoing, with participants raising questions about the mechanics of the situation, particularly regarding the conditions under which the normal force becomes zero. Some participants are attempting to clarify their understanding of the forces acting on the block and how these relate to its motion.
Participants are considering the implications of the plank's acceleration and its effect on the normal force experienced by the block. There is a recognition of the complexities involved in comparing the accelerations of the block and the plank during SHM.
So when you say "we gradually increase A somehow" you mean increasing the amplitude of the SHM that the whole system (plank+block) is executing? So the break-off point is reached at a certain amplitude? But how can you increase A since it is fixed and if you change A you change the equation of SHM, which I don't think can be done.haruspex said:No, I didn't say that. The minimum normal force, Nmin, is at the top of the movement, agreed? When Nmin>0 and the plank is at the top of its movement the block will have acceleration ω2A, and it will be less than g. Suppose we gradually increase A somehow. As ω2A gets closer to g, Nmin will decrease. At the point where ω2A equals g, Nmin will be zero and the block is on the verge of losing contact. The slightest further increase in A will cause the block to lose contact.
I don't care how we increase it. Maybe we just run hundreds of different experiments, each with a slightly higher A than the one before.andyrk said:So when you say "we gradually increase A somehow" you mean increasing the amplitude of the SHM that the whole system (plank+block) is executing? So the break-off point is reached at a certain amplitude? But how can you increase A since it is fixed?
The equation of SHM is fixed for the problem and so is the amplitude. So how will you go about proving that the block leaves contact/is about to leave contact with the plank even though their accelerations are same at the highest point? This problem has become so complicated.haruspex said:I don't care how we increase it. Maybe we just run hundreds of different experiments, each with a slightly higher A than the one before.
How about rather than increasing A, we see start with small enough ω so that the block maintains contact with the plank, then similar to what haruspex has suggested, we run successive experiments each time with slightly larger ω until we find the minimum value of ω that results in the block losing contact with the plank. Certainly you can't object to increasing ω in this fashion.andyrk said:So when you say "we gradually increase A somehow" you mean increasing the amplitude of the SHM that the whole system (plank+block) is executing? So the break-off point is reached at a certain amplitude? But how can you increase A since it is fixed ...
Rightly said. So the block would leave contact with the plank at a specific ω. But when ω2A = g, then the block is on the verge of losing contact, right? Any slight increase in ω would make the block to lose contact with the plank. But right now the acceleration of the block equals that of the plank, right? So doesn't that happen at all instants, even when the plank is not at its highest point? Or does it not because N is not 0 at all other instants except for the topmost point?SammyS said:How about rather than increasing A, we see start with small enough ω so that the block maintains contact with the plank, then similar to what haruspex has suggested, we run successive experiments each time with slightly larger ω until we find the minimum value of ω that results in the block losing contact with the plank. Certainly you can't object to increasing ω in this fashion.
So finally, N = 0 has nothing to do with different accelerations of the block and the plank. They are the same at the topmost point and all points, yet N = 0 because of the algebraic reasoning we provided earlier. Is that correct?SammyS said:How about rather than increasing A, we see start with small enough ω so that the block maintains contact with the plank, then similar to what haruspex has suggested, we run successive experiments each time with slightly larger ω until we find the minimum value of ω that results in the block losing contact with the plank. Certainly you can't object to increasing ω in this fashion.
Many are here.andyrk said:Anybody there?
By that I mean when the plank and the block are the extreme position of the SHM, i.e. A distance from the mean position.SammyS said:In post # 65, what do you mean by "right now" when you say, "But right now the acceleration ..." .
If the block maintains contact with the plank, then they have the same acceleration at all positions.andyrk said:By that I mean when the plank and the block are the extreme position of the SHM, i.e. A distance from the mean position.
But you didn't answer my question. I was asking that N = 0 (block on the verge of leaving contact) even though the acceleration of the block and plank is the same and not different. So am I correct on this part or not? And if they are not different then can you explain how?SammyS said:If the block maintains contact with the plank, then they have the same acceleration at all positions.
I could just leave it at that, but ...
Even in the limiting case in limiting in which Aω2 = g, they maintain contact at all positions, because there would be no position at which the normal force would need to be negative to keep them in contact.
Wow! 52 minutes and no answer to an ill-formed request. It's not a complete sentence nor a complete question.andyrk said:Why doesn't anyone reply until I post a message like this? :/
The post explains it itself. N = 0 at the top, but the acceleration of the plank is not greater than the block. They are equal. And if they are not equal, i.e. the acceleration of the plank is greater than the block can you prove it?SammyS said:What is it precisely that you are trying to ask in Post # 71?
Not me, I've given up. The problem is not complicated at all, and I'm at a loss to understand why it still baffles you. Take your objection to my considering varying A. It strikes me as so inane as to be deliberately obtuse. Byeee...andyrk said:Anybody there?
I think not. It was incomprehensible.andyrk said:The post explains it itself.
Yes. Provided that you are referring to the case in which Aω2 = g.N = 0 at the top, but the acceleration of the plank is not greater than the block. They are equal.
If Aω2 > g, then at some point the block loses contact unless it is fastened to the plank, in which case the plank can exert downward (i.e. negative) force on the block.And if they are not equal, i.e. the acceleration of the plank is greater than the block can you prove it?
The fact that you are mentioning something like acceleration of the plank being greater than the block, confuses me. The solution I am looking at doesn't say anything like that. It just simply puts N = 0 in the equation mg - N = mω2A to get the value of ω. But you say that this equation holds only when the block is in contact? And N can't go beneath 0 since that can't happen. N = 0 means contact lost or about to lose contact (but hasn't yet). So are you saying the same thing? It hasn't yet? But will? How will it lose contact if accelerations are not different? And for this you then begin your story of comparing accelerations of the block with that of the plank involving greater than (leaves contact completely) or less than equal to (stays in touch). Which I am simply not able to comprehend! But I just showed you that the accelerations are equal, so why do involve greater than or less than equal to cases, even though they are not required and meaningless?haruspex said:The value of g is fixed. Give the plank enough amplitude and its acceleration at the top of the cycle will exceed g. The block will have less downward acceleration than the plank and lose contact with it. The normal force cannot be negative.
haruspex said:The value of g is fixed. Give the plank enough amplitude and its acceleration at the top of the cycle will exceed g. The block will have less downward acceleration than the plank and lose contact with it. The normal force cannot be negative.
You quote haruspex but I believe he recently said he no longer participates in this thread.andyrk said:The fact that you are mentioning something like acceleration of the plank being greater than the block, confuses me. The solution I am looking at doesn't say anything like that. It just simply puts N = 0 in the equation mg - N = mω2A to get the value of ω. But you say that this equation holds only when the block is in contact? And N can't go beneath 0 since that can't happen. N = 0 means contact lost or about to lose contact (but hasn't yet). So are you saying the same thing? It hasn't yet? But will? How will it lose contact if accelerations are not different? And for this you then begin your story of comparing accelerations of the block with that of the plank involving greater than (leaves contact completely) or less than equal to (stays in touch). Which I am simply not able to comprehend! But I just showed you that the accelerations are equal, so why do involve greater than or less than equal to cases, even though they are not required and meaningless?
Basically, you say that N = 0 then acceleration of block and plank is g. Now you say that acceleration of plank is increased somehow (by increasing ω).. But then we would again have to write the equation mg - N = mω2A with new ω and put N = 0 again so as to get the new value of ω at which the block leaves contact at the topmost point.