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Free-Body Diagram

  1. Sep 26, 2005 #1
    I was trying to figure out how to get the outward normal vector to the surface of a ramp inclined θ degrees from the horizontal. Say that a block of mass "m" is on the surface and the surface is frictionless. When I draw the free-body diagram, I come up with a downward force of <0,-mg>. To calculate the force in the direction of the incline, I first want to find the normal vector to add to <0,-mg>. Is this the correct way to do this? It looks like |n|=|g|cosθ, but I cannot find the coordinates of this vector. Any ideas?

    Edit: Working it out, I came up with the following for the coordinates of the outward normal vector:

    x = |g|cosθsinθ
    y = |g|cos2θ

    Are these correct?
    Last edited: Sep 26, 2005
  2. jcsd
  3. Sep 26, 2005 #2

    Doc Al

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    Staff: Mentor

    You got it--except for a missing mass (you left out the "m" in your equations). (Note that it's often more useful to use coordinates parallel and perpendicular to the incline surface, rather than vertical and horizontal components.)
    Last edited: Sep 26, 2005
  4. Sep 26, 2005 #3
    Oh yeah, the m! Anyways, thanks for the response :smile:.
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