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Box sliding down an inclined plane

  1. Oct 10, 2010 #1
    The problem statement, all variables and given/known data

    A box is placed on an inclined plane of angle [tex]\theta[/tex]. There is no fiction between the box and the plane nor between the inclined plane and the floor. The mass of the box is [tex] m_1 [/tex] and the mass of the inclined plane is [tex] m_2 [/tex].

    Find the acceleration of the box and that of the inclined plane.


    The attempt at a solution

    It is easy to write down Newton's equations for both the box and the inclined plane.
    Here they are:

    [tex] N_1 \sin \theta = m_1 a_{1x} [/tex]

    [tex] N_1 \cos \theta - m_1 g = m_1 a_{1y} [/tex]

    [tex] - N_1 \sin \theta = m_2 a_{2x} [/tex]

    [tex] N_2 - N_1 \cos \theta - m_2 g = 0 [/tex]

    where [tex] N_1 [/tex] is the contact force between the box and the inclined plane and [tex] N_2 [/tex] is the contact force between the inclined plane and the floor.

    These are four equations in the unknowns [tex] N_1, N_2, a_{1x}, a_{1y}, a_{2x} [/tex].

    How am I supposed to solve them?

    My idea is the following: since the box is sliding down the inclined plane, the motion with respect to the inclined plane is easier. I mean the acceleration of the box relative to the inclined plane must be in the direction of the inclined plane. That is, the vector

    [tex] \vec{a}_{box/plane} = \vec{a}_{box} - \vec{a}_{plane} [/tex]

    should be in the direction of the plane. Thus

    [tex] \tan \theta = \frac{a_{box/plane,y}}{a_{box/palne,x}} = \frac{a_{box,y}-a_{plane,y}}{a_{box,x}-a_{plane,x}} = \frac{a_{1y}}{a_{1x}-a_{2x}}. [/tex]

    Is that right?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 10, 2010 #2

    ehild

    User Avatar
    Homework Helper

    You are on the right track. Go ahead.

    ehild
     
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