What Causes the Differences in Newton's Law on an Inclined Plane?

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SUMMARY

The discussion centers on the application of Newton's laws to a box resting on an inclined plane at an angle \(\theta\). Two different coordinate systems yield conflicting results for the normal force \(N\): \(N = mg \sec \theta\) when using a vertical coordinate system and \(N = mg \cos \theta\) when using an inclined coordinate system. The discrepancy arises from the omission of frictional forces in the analysis. A complete free body diagram is essential for resolving these differences and understanding the forces at play.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of inclined plane mechanics
  • Familiarity with free body diagrams
  • Basic trigonometry related to angles and forces
NEXT STEPS
  • Study the derivation of forces on inclined planes using free body diagrams
  • Learn about frictional forces and their role in static equilibrium
  • Explore coordinate transformations in physics problems
  • Investigate the implications of different reference frames in mechanics
USEFUL FOR

Physics students, educators, and anyone interested in classical mechanics, particularly those studying forces on inclined planes and static equilibrium.

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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?
 
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hi

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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