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Newton laws prob

  1. Aug 5, 2012 #1
    a mass of m=6 kg is free to move without friction on a body with a mass of M=18 kg at an angle of theta=32 deg. M is placed on a friction free table. At the moment t=0 m is released from rest from the top of M at an height of H=4m .

    At what moment does m reach to the table?

    I'm uploading the question and my solution, which is t=1.54 s, while the correct answer should be t=1.08 s. I'd like to know what I did wrong...
     

    Attached Files:

  2. jcsd
  3. Aug 5, 2012 #2

    Simon Bridge

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  4. Aug 5, 2012 #3
    What do you mean "my reasoning"? I can't see why I got the wrong answer and I uploaded a detailed solution... Obviouslly I have a mistake along the way...
     
  5. Aug 5, 2012 #4
    At the end of your attempt, you get the acceleration of m w.r.t. M. However, M itself is accelerated. It is not an inertial frame of reference and I do not see you have made any attempt to account for that.

    Nor do I think it is required. I think all you need is conservation of energy and horizontal momentum.
     
  6. Aug 5, 2012 #5
    But I did account for that, I got the acceleration of m w.r.t M using the Delambre force on m. And relative to M, m passed the distanc L did it not?
     
  7. Aug 5, 2012 #6
    You may have tried, indeed. Problem is, I find it difficult to distinguish [itex]a_M[/itex] from [itex]a_m[/itex] in that picture. Could you post the equations here to avoid any misunderstanding?
     
  8. Aug 5, 2012 #7
    The first 2 equations are on M and in the second one it's aM. and it says: "aM is the accerleration of M relative to the ground!".

    The next equation is the Fy equation on m, and it says:

    N+m*aM*sin(theta)-mgcos(theta)=0 where m*aM*sin(theta) is the Delambre force in the y axis direction of m's coordinate system (drawn at the right side of the picture)

    After that linr I simply put N=(M*aM)/sin(theta) and after some calculations you get an expression for aM.

    Afterwards the Fx equation for m is written and it says:

    m*aM*cos(theta)+mgsin(theta)=m*am where m*aM*cos (theta) is the Delambre force in the x axis direction of m's coordinate system. After I put the expression I got for aM you can get an expression for am, which is the acceleration of m relative to M.

    Hope this helps
     
  9. Aug 5, 2012 #8
    I have followed your derivation and I cannot spot any problem with it.

    Moreover, if your expression for [itex]a_m[/itex] is simplified, it can be seen that in the limit of infinite M it results in [itex]a_m = g sin \theta[/itex], as it should be for the immovable wedge, and in the limit of [itex]\theta → \pi / 2[/itex] it results in [itex]a_m = g[/itex], again as it should be. So I have strong reasons to believe that your solution is correct.
     
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