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Newton's Laws of Motion Problem

  1. Mar 31, 2014 #1
    1. The problem statement, all variables and given/known data
    Mass of Block A = 1 kg
    Mass of Wedge B = 2 kg
    Inclination of Wedge is 37°
    and Cos 37° = 4/5
    The wedge is given an acceleration of 5 ms-2
    Find -
    1. Acceleration of Block wrt Wedge
    2. Normal Force
    attachment.php?attachmentid=68172&stc=1&d=1396243029.jpg

    2. Relevant equations

    Newton's Law for a System (hereby referred as NLS)
    ∑fext = m1a1+m2a2......+mnan

    3. The attempt at a solution

    Let the Acceleration of Block wrt Wedge be λ along the inclination.
    In Y axis, there is only one Ext force i.e. The gravity and ∑fnet in Y direction is m1g+ m2g = (m1+ m2)g
    Therefore in Y axis using Newton's Law for a System,
    mAaA + mBaB = (m1+ m2)g
    but in Y axis , aA = 0
    and ab in Y axis = λsinθ + aAcos 90
    = λsinθ + 0
    = λsinθ
    Therefore, mBaB = (m1+ m2)g
    1 (λsin(θ)) = (1 + 2)g
    λ sin 37° = (3)10
    λ(3/5) = 30
    λ = 50
    but using pseudo force, λ = 2 ms-2 which is correct answer in the Answer Key
    Where am I going Wrong?
    Please Help
     

    Attached Files:

  2. jcsd
  3. Mar 31, 2014 #2
    Is wedge B being given an acceleration horizontally to the left or right?

    Assuming then no friction block A will also be sliding down the inclined wedge B.
     
  4. Apr 1, 2014 #3

    haruspex

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    Are you assuming that the blocks are in free fall? I would have guessed the wedge is supported somehow.
     
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