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There is a nice puzzle about reel of thread. I wonder why the solution works.

The puzzle:

You take a reel with threads on it. Inner radius r, outer radius R. You put it on a table. If you pull the tread under big enough angle Alpha the reel will move in opposite direction. The question is what is the minimal Alpha angle required. No slippage and rolling friction are present.

Image.

The solution:

The solution takes tangency point of the reel and table and says that it won't move. So friction force $F_f = F \cdot \cos(\alpha)$. Then it considers moment of forces about the center of the reel. If $F \cdot R > F_f \cdot r$ the reel will move in the opposite direction.

My question:

The solution is quite simple and clear. But there is one thing I can't understand. If $F_f = F\cos(\alpha)$ and both gravity and normal reaction forces are normal to the table then why does the centre mass of the reel move? The sum of projection of all forces is equal to zero, so it should not move. Where am I wrong here?

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The (static) friction force, once we've started pulling, will provide the force to offset m * a here. So your assumption that the friction force is normal to the surface is wrong.

Typically friction forces are calculated using normal forces, but they're not in the same direction.

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  • $\begingroup$ 1. Ff = F*cos(Alpha) isn't it? 2. friction forces depends on normal forces only when object is moving. $\endgroup$ – klm123 Jun 1 '14 at 7:08
  • $\begingroup$ Is that formula for the friction in the solution itself? You need static friction here, because there is no slippage. $\endgroup$ – Johan Rensink Jun 1 '14 at 8:17
  • $\begingroup$ I do not see your point. We consider static friction. It should be compensated by some forces because the lowest point of the reel doesn't move. What are these forces? Why centre of the reel moves? $\endgroup$ – klm123 Jun 1 '14 at 8:28
  • $\begingroup$ I'm thinking I was wrong initially, I'll post an update as soon as I can. $\endgroup$ – Johan Rensink Jun 1 '14 at 12:26

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