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Block on inclined plane acceleration without rotating coordinates

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data
    I was doing this for fun, but this seemed like the right place to post. Typically inclined plane problems are solved by rotating the coordinate system so the x axis is along the plane. I decided to try solving the problem without rotating the coordinates. In this case there should be acceleration in the x and y direction, and the magnitude of the acceleration vector should return the same value as the problem solved with rotating the coordinates. My attempts at solving the problem with "normal" coordinates have been incorrect and I'd appreciate any insight as to what I'm doing wrong.

    If I need to state the problem than here it is: A block mass m is on an inclined plane at angle [tex]\Theta[/tex]. Find components of the acceleration of the block without rotating the coordinates and also find the magnitude of the acceleration.

    http://dots.physics.orst.edu/graphics/image_maps/inclined_plane.gif [Broken]


    2. Relevant equations
    magnitude of the acceleration of a block down a plane using rotated coordinates: a = g Sin[[tex]\Theta[/tex]]
    magnitude of the normal force: N = mg Cos[[tex]\Theta[/tex]]


    3. The attempt at a solution
    The sum of the forces in the x direction only consist of the x component of the normal force on the block
    [tex]\Sigma[/tex]Fx = Nx= -mgCos[[tex]\Theta[/tex]]Cos[[tex]\Theta[/tex]]
    or Nx=-mgCos2[tex]\Theta[/tex]=max
    The extra cos theta exists because the magnitude of the normal force is mgCos[[tex]\Theta[/tex]]
    From this, ax= -gCos2[tex]\Theta[/tex]
    The sum of the forces in the y direction are the gravitational force and the y component of the normal force
    [tex]\Sigma[/tex]Fy= -mg + mgCos[[tex]\Theta[/tex]]Sin[[tex]\Theta[/tex]]
    From this, ay= -g+gCos[tex]\Theta[/tex]Sin[tex]\Theta[/tex]
    or if you wish to apply the double angle formula:
    ay= -g+1/2gSin[2[tex]\Theta[/tex]]


    Clearly these acceleration components do not give the correct value for the magnitude of the acceleration. A zero degree angle returns an acceleration in the -x direction equal to g. What did I miss?

    PS. I apologize if there are any mistakes or this is hard to read. This is my first post.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 13, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You messed up the calculation of the x & y components of the normal force. Note that the normal force makes an angle Θ with the vertical, not the horizontal.
     
  4. Nov 13, 2009 #3
    Thanks for the reply. Careless errors like that can be so frustrating!
     
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