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Dynamics Boxes 2

  1. Apr 15, 2014 #1
    Please see picture.
    Here is my method. Is it good?
    T=m2g
    m2g=m1a
    a=m2g/m1
    -n+Fpush=Ma. (n=normal force of m2 on M).
    n=m2a=(m2)^2g/(m1a)

    -n+Fpush=Ma.
    Fpush=Ma+(m2)^2g/m
     

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  2. jcsd
  3. Apr 15, 2014 #2

    SammyS

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    No.

    There is a horizontal force of T exerted to the left on M via the pulley.

    There is also a downward force of T exerted on M, but that doesn't come into play here. It would if there was friction opposing the movement of M.

    attachment.php?attachmentid=68689&d=1397609466.png
     
  4. Apr 15, 2014 #3
    Is the answer of the book wrong :O ?
     
  5. Apr 15, 2014 #4

    SammyS

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    Their result looks correct.
     
  6. Apr 15, 2014 #5
    Perfect, I will come back to it!
    What about this problem? I solved the force values:
    I get -(5/8)M^2G/r^2 for the first attraction of the big mass and M^2G/(2r^2) for the attraction of the small mass.

    Where did they get 5/16 and 1/4?
     

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  7. Apr 15, 2014 #6

    SammyS

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    They merely used proportions.

    Gravitational force is directly proportional to mass & inversely proportional to the square of distance.

    attachment.php?attachmentid=68693&d=1397616243.png
     
  8. Apr 15, 2014 #7
    Ok perfect!

    Coming back to the previous question:
    Would the equation for M be:

    -T+Fpush-n=Ma?

    Why is there a T? :O Is the pulley not frictionless?
     
  9. Apr 15, 2014 #8

    SammyS

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    Yes, the equation is
    -T+Fpush-n=Ma .​
    Furthermore, T = m1a as well as m2 g .

    Also, n = m2a .

    Plugging those into -T+Fpush-n=Ma and solving for Fpush gives
    Fpush = (M + m1 + m2)a​
    in agreement with the book.

    Yes, the pulley is frictionless, but there is thension T to the left on it an tension T downward on it. Both of these are transmitted to block M.
     
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