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Why is the normal force applied from the corner? How can I find the application point of the normal force?

  • $\begingroup$ I'm afraid some information more is needed because the graphic alone is a bit ambiguous. My guess is that it represents a solid standing on a spinning disk and problem is intended at finding if the object topples at a given angular speed, but that is not he only possible interpretation. $\endgroup$
    – Pere
    May 17 '20 at 13:06
  • $\begingroup$ If it is on the point of toppling, then the resultant torque about the bottom right corner = 0. Consider what happens when the distance between N and the corner increases. $\endgroup$
    – jamie
    May 17 '20 at 13:21
  • $\begingroup$ @jamie That's almost an answer. You should edit your text and post an answer. Comments are for questions about the problem or other clarifying statements. $\endgroup$
    – Bill N
    May 17 '20 at 13:58
  • $\begingroup$ @Pere your guess is right $\endgroup$
    – EmirB.
    May 17 '20 at 14:46

The normal force will be distributed on the base of the solid in a way that it balances all other forces acting in the solid - just the same way that the normal force in any solid at rest balances all the other acting force. By equilibrium in the vertical direction, the magnitude of the normal force equals the weight. Then, with changing angular velocity, the normal force magnitude doesn't change but only its point of application changes - it moves outward as the angular speed increases. However, there is a limit on how far can it move outward: it can move at most to the end of the base, because it's application point must lie in the base.

Then, if the disc is spinning fast enough to be just to topple the solid, the force will be acting in the furthest point in can, as it is shown in the graphic.

  • $\begingroup$ Thanks for help. $\endgroup$
    – EmirB.
    May 17 '20 at 17:33

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