# Normal force in a compound pendulum (physical pundulum) system?

Consider a compound pendulum pivoted about a fixed horizontal axis, illustrated by the force diagram on the right:

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Okay, I can't figure out where the normal force on the pendlum should point (this force isn't indicated in the diagram).

On the one hand, I think it should point in the same direction as the tension force on the left-side diagram, in order to produce the rotation. But on the other hand, shouldn't the normal force point towards the pendulum rather than away from it? In any case, I know that normal face is perpendicuar to the contact surface, but how does "contact surface" fit into the context of this scenario?

Please help me understand the concepts involved.

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This seems to have been discussed at physics.stackexchange.com/questions/22248/… , but unfortunately I don't really understand the answers(comments) given there. – Ryan Apr 28 '12 at 7:30

## 1 Answer

If you think of the pendulum as swinging on a nail going through a smooth hole near the top, then the tension in the pendulum will pull in the direction of the pendulum's length. The normal force will be opposite to this (your first answer) and will be exerted by the nail on the point of the hole in contact with it. So if the nail has a circular cross section, the force will be exerted at the point of this circle at the "top end" of the diameter which is in the direction of the pendulum's length.

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Thank you!! The original problem in my study material uses the example of a swinging pub sign, and we are asked to assume that the rotation is about the top edge of the sign. Your explanation has helped me realise that this assumption is just to help simplify the distance/length calculation-- but when considering the force digram, it wouldn't make sense to really let the pub sign swing about its edge. – Ryan Apr 28 '12 at 8:05
So in fact the normal force does point toward the compound pendulum rather than away from it. And to further answer my own question: the "contact surface" in this scenario is the arc of the nail's cross-section, so perpendicular to it would be radial from its centre, ie. along the dotted line in the diagram above. So, everything makes sense now. Cheers – Ryan Apr 28 '12 at 8:10
The "normal force" exerted by the nail on the surface of the hole (this surface is part of the pendulum) is in the direction away from the pendulum body (i.e. "upwards" in your diagram). You know this force is in this direction, because if you suddenly pull out the nail, the pendulum will fall. – twistor59 Apr 28 '12 at 8:18
Yes, yep. Doesn't contradict what I last said. Along the dotted line, and pointing diagonally "upwards". – Ryan Apr 28 '12 at 8:34