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

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The discussion centers on the application of Newton's laws to a box at rest on an inclined plane. Two different coordinate systems yield conflicting results for the normal force, with one suggesting N = mg sec θ and the other N = mg cos θ. The discrepancy arises from the neglect of frictional forces, which are essential for maintaining equilibrium on the incline. A complete free body diagram is recommended to clarify the forces at play. Understanding these forces is crucial for accurately applying Newton's laws in this scenario.
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Suppose a box is sitting, motionless, on an inclined plane which makes an angle \theta with the horizontal. If I'm not mistaken, if one writes out Newton's law in the y direction (if +y is up), one gets N \cos \theta - mg = 0; i.e., N = mg \sec \theta. However, if you ROTATE the coordinate system so that the x axis points along the incline, you get that N = mg \cos \theta, 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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