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Box on an inclined plane

  1. Sep 19, 2011 #1
    Suppose a box is sitting, motionless, on an inclined plane which makes an angle [itex]\theta[/itex] with the horizontal. If I'm not mistaken, if one writes out Newton's law in the [itex]y[/itex] direction (if [itex]+y[/itex] is up), one gets [itex]N \cos \theta - mg = 0[/itex]; i.e., [itex]N = mg \sec \theta[/itex]. However, if you ROTATE the coordinate system so that the [itex]x[/itex] axis points along the incline, you get that [itex]N = mg \cos \theta[/itex], which is apparently RIGHT. What's going on here? Why are these two cases different?
  2. jcsd
  3. Sep 19, 2011 #2

    if the box is sitting on an incline plane and motionless, there MUST be a frictional force
    which is keeping it steady. i think you are ignoring that. first draw full free body diagram and
    things will become clearer.
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