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>> No.12309351 [View]
File: 6 KB, 250x171, inclined plane.jpg [View same] [iqdb] [saucenao] [google]
12309351

Quick sanity check.
Ignoring [math]f[/math], the resultant force on the block is [math]mg \sin \theta[/math], right? [math]mg \cos \theta[/math] and [math]N[/math] cancel out.
The real issue is, basically, that if a penny is spinning frictionlessly in a cone, then the resultant force needs to point to the cone's vertex, so it eventually falls in, right?

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