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  1. MHB Apprentice

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    #1


    im stuck on question 8. the equation i got for the wedge is .6R -.5S=ma r stands for relative force of the particle s stands for relative force of the wedge . i got the equation 4mg-R=5m(.6a) for the particle where a stands for acceleration of the wedge. does anyone know where to go from here. are my equations wrong? i can post a picture of my diagram if anyone needs it
    Last edited by markosheehan; December 8th, 2016 at 13:33.

  2. MHB Apprentice

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    #2

    The forces acting on the wedge are mg, friction, S the normal reaction between the wedge and the surface, and R the normal reaction between the wedge and the particle.
    The forces actin on the particle are 5mg and R the normal reaction between the wedge and the particle
    I have called α the angle with tan(α) = 3/4
    If we call the acceleration of the wedge a then the equations of motion for the wedge are
    1) vertically S = mg + R cos(α)
    2) horizontally ma = R sin(α) - 0.5S

    The equations of motion for the particle are
    3) Horizontally 5m(bcos(α) - a) = Rsin(α) since bcos(α) - a is the total acceleration in this direction
    4) Vertically 5mbsin(α) = 5mg - Rcos(α)

    Eliminating S from 1) and 2) gives R = 5ma +2.5mg

    Eliminating b from 3) and 4) gives 4mg - 3ma = R

    Equating the two above expressions for R gives a = 3g/16

    Substituting in 4 then gives b = 3g/4

  3. MHB Apprentice

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    #3 Thread Author
    can equation 3 also be 5m(b-.8a) = 3mg
    and also i dont understand equation 4 5mbsin(α) = 5mg - Rcos(α) should it not be 4mg-R=5m(.6a) are you not supposed to look at the forces horizontal and vertical to the plane

  4. MHB Apprentice

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    #4
    Quote Originally Posted by markosheehan View Post
    can equation 3 also be 5m(b-.8a) = 3mg
    and also i dont understand equation 4 5mbsin(α) = 5mg - Rcos(α) should it not be 4mg-R=5m(.6a) are you not supposed to look at the forces horizontal and vertical to the plane
    To obtain my equations 3) and 4) I looked at the motion vertically and horizontally but you can certainly also look at motion along the plane and perpendicular to the plane. I believe your equations are correct for motion resolved the second way and these should also lead to the same solutions.

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