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Mass on a wedge. Relative acceleration.

  1. Jul 16, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle of mass M is on a wedge of mass 2M. The wedge is smooth, and is inclined at 30* to the horizontal. The system is released from rest. Find the acceleration of the wedge, and find the acceleration of the particle relative to the wedge.


    2. Relevant equations

    F=ma

    3. The attempt at a solution



    Particles motion parallel to the wedge

    F=ma
    (mg)sin30 = m[f - (a)cos30]
    mg = 2mf - ma[itex]\sqrt{3}[/itex]


    Particles motion perpendicular to the wedge

    F = ma
    (mg)cos30 - R = m(asin30)
    mg[itex]\sqrt{3}[/itex] -2R = ma
    R = [itex]\frac{m(g\sqrt{3} - a)}{2}[/itex]


    Wedges Motion

    F = ma
    Rsin30 = 2ma
    R = 4ma




    Substitution

    [itex]\frac{m(g\sqrt{3} - a)}{2}[/itex] = 4ma

    mg[itex]\sqrt{3}[/itex] - ma 8ma

    g[itex]\sqrt{3}[/itex] - a = 8a

    g[itex]\sqrt{3}[/itex] = 9a

    a =[itex]\frac{g\sqrt{3}}{9}[/itex]




    The answer at the back of the book is [itex]\frac{g}{3\sqrt{3}}[/itex]
     
  2. jcsd
  3. Jul 16, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Your solution is the same as that of the book. (Write 9 as 3*√3*√3 and simplify by √3.)

    ehild
     
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