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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


    3. The attempt at a solution

    Particles motion parallel to the wedge

    (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


    [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


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    Homework Helper

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

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