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    This is my current question:
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    A horizontal platform is made to move up and down with shm. according to the relationship dtwox over dtsquared (acceleration) = -49x., where x is the vertical displacement of the platform from its mean position at time t.

    Show that any mass placed on the platform will leave the platform if the amplitude of the platform is > 20cm.

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    Do I have to show that at some point the acceleration of the platform is greater than gravity. Is that it?
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    (Original post by maggiehodgson)
    This is my current question:
    ----------------------------------------------------------------
    A horizontal platform is made to move up and down with shm. according to the relationship dtwox over dtsquared (acceleration) = -49x., where x is the vertical displacement of the platform from its mean position at time t.

    Show that any mass placed on the platform will leave the platform if the amplitude of the platform is > 20cm.

    ----------------------------------------------------------------------


    Do I have to show that at some point the acceleration of the platform is greater than gravity. Is that it?
    indeed
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    (Original post by maggiehodgson)
    This is my current question:
    ----------------------------------------------------------------
    A horizontal platform is made to move up and down with shm. according to the relationship dtwox over dtsquared (acceleration) = -49x., where x is the vertical displacement of the platform from its mean position at time t.

    Show that any mass placed on the platform will leave the platform if the amplitude of the platform is > 20cm.

    ----------------------------------------------------------------------


    Do I have to show that at some point the acceleration of the platform is greater than gravity. Is that it?
    The mass leaves the platform if it is no longer in contact with it. At that instant, the normal reaction exerted on the mass becomes 0. Applying Newton II to the mass, supposing that the platform is accelerating downwards at a with the mass in contact, we have:

    mg-R=ma \Rightarrow R=mg-ma

    Then R >0 is the condition for contact which gives us

    R=mg-ma>0 \Rightarrow mg>ma \Rightarrow g>a.

    When a=g then R=0 and the mass leaves the platform. In short: you are (almost) right.
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    (Original post by atsruser)
    The mass leaves the platform if it is no longer in contact with it. At that instant, the normal reaction exerted on the mass becomes 0. Applying Newton II to the mass, supposing that the platform is accelerating downwards at a with the mass in contact, we have:

    mg-R=ma \Rightarrow R=mg-ma

    Then R >0 is the condition for contact which gives us

    R=mg-ma>0 \Rightarrow mg>ma \Rightarrow g>a.

    When a=g then R=0 and the mass leaves the platform. In short: you are (almost) right.
    So, should the question have asked "show that the amplitude is greater than or equal to 20cm" rather than just "greater than 20cm"
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    (Original post by maggiehodgson)
    So, should the question have asked "show that the amplitude is greater than or equal to 20cm" rather than just "greater than 20cm"
    Based on the numbers in the question, then yes, I think that the "or equal to" should be there.
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    (Original post by atsruser)
    Based on the numbers in the question, then yes, I think that the "or equal to" should be there.
    Thanks, I will remember that, I hope.
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    SHMONEY?
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    (Original post by maggiehodgson)
    Thanks, I will remember that, I hope.
    Remember with these mechanics problems that you should start by reasoning from Newton's laws, as I did with the analysis of the reaction. You shouldn't try to remember specific results.

    Anyway, I suspect that I'm being a little picky in pointing out that question isn't quite right.
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    (Original post by atsruser)
    Remember with these mechanics problems that you should start by reasoning from Newton's laws, as I did with the analysis of the reaction. You shouldn't try to remember specific results.

    Anyway, I suspect that I'm being a little picky in pointing out that question isn't quite right.
    I have been mulling over that question/answer since your original reply. The question asks when it will leave the platform. Will it actually leave it or just be on it with no reaction?

    I'm afraid I have these "real life" thoughts that more often than not are incorrect but they do cloud my grasp of mechanics. I hope, therefore, that my question isn't too silly for you.
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    (Original post by maggiehodgson)
    I have been mulling over that question/answer since your original reply. The question asks when it will leave the platform. Will it actually leave it or just be on it with no reaction?
    Well, it's a bit of a philosophical question, and that's why I said that maybe I was being too picky. To my mind though, the only sensible definition for "leave the platform" is that the condition R=0 occurs. That's because once that happens, the mass and the platform no longer "know" about each other - they feel no force from each other, and so from the POV of the platform, say, the mass is no longer there. That is true whether or not the gap between the mass and the platform is 1mm or 1 proton's width.

    I'm afraid I have these "real life" thoughts that more often than not are incorrect but they do cloud my grasp of mechanics. I hope, therefore, that my question isn't too silly for you.
    Far from silly - they are precisely the kinds of thoughts that make mechanics tricky. Physicists get around them by learning to rewrite vague statements like "leave the platform" into precise ones that can be analysed according to Newton's laws.

    One other point: the question is one of those silly ones that work with some precise value of g (9.8 m/s^2 or something). Really the value for the amplitude should be given in terms of the variable g.
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    (Original post by atsruser)
    Well, it's a bit of a philosophical question, and that's why I said that maybe I was being too picky. To my mind though, the only sensible definition for "leave the platform" is that the condition R=0 occurs. That's because once that happens, the mass and the platform no longer "know" about each other - they feel no force from each other, and so from the POV of the platform, say, the mass is no longer there. That is true whether or not the gap between the mass and the platform is 1mm or 1 proton's width.


    Far from silly - they are precisely the kinds of thoughts that make mechanics tricky. Physicists get around them by learning to rewrite vague statements like "leave the platform" into precise ones that can be analysed according to Newton's laws.

    One other point: the question is one of those silly ones that work with some precise value of g (9.8 m/s^2 or something). Really the value for the amplitude should be given in terms of the variable g.
    Thank you again. Your reply has put my mind at rest and the phrase about the platform and mass not "knowing" about each other makes perfect sense to me.
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    (Original post by maggiehodgson)
    Thank you again. Your reply has put my mind at rest and the phrase about the platform and mass not "knowing" about each other makes perfect sense to me.
    My pleasure. Incidentally, in mechanics problems, I find that it's often useful to try and put yourself mentally in place of the objects, and think about what they would feel.

    For example, imagine a pulley problem, with 3 masses, and masses A and B on one side connected by a rope, mass B on the bottom, and C on the other connected by a rope to mass A over the pulley. Here, people sometimes get confused about the forces on mass A, and start saying that mass A feels the weight of mass B.

    However, that's nonsense. Mass A cannot feel the weight of another body. It doesn't have extrasensory powers allowing it to sense the force of gravity on another body. Instead, if you were mass A, you would feel a pull from the rope connecting to you mass B, and a pull from the mass connecting you to mass C. So those are the only forces affecting your motion. They are the forces that we would draw on a free body diagram for A.

    Similarly if you were the platform in your original problem, once the reaction force has fallen to zero, you have no way of knowing precisely where the mass is: from your POV, it no longer exists. You, as a platform or a mass, can only "know" about objects that are directly in contact with you, and exerting a force on you (such as via ropes connected to you, or masses sitting on you)
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    (Original post by atsruser)
    My pleasure. Incidentally, in mechanics problems, I find that it's often useful to try and put yourself mentally in place of the objects, and think about what they would feel.

    For example, imagine a pulley problem, with 3 masses, and masses A and B on one side connected by a rope, mass B on the bottom, and C on the other connected by a rope to mass A over the pulley. Here, people sometimes get confused about the forces on mass A, and start saying that mass A feels the weight of mass B.

    However, that's nonsense. Mass A cannot feel the weight of another body. It doesn't have extrasensory powers allowing it to sense the force of gravity on another body. Instead, if you were mass A, you would feel a pull from the rope connecting to you mass B, and a pull from the mass connecting you to mass C. So those are the only forces affecting your motion. They are the forces that we would draw on a free body diagram for A.

    Similarly if you were the platform in your original problem, once the reaction force has fallen to zero, you have no way of knowing precisely where the mass is: from your POV, it no longer exists. You, as a platform or a mass, can only "know" about objects that are directly in contact with you, and exerting a force on you (such as via ropes connected to you, or masses sitting on you)
    It sounds a good idea. I'll try it. Thanks for all the effort you have put into these replies.
 
 
 
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