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If I have a ring threaded onto a fixed rough vertical rod, and it is in equilibrium, in which direction does the normal force point? Why? Isn't there contact with the rod from all directions?

Here's the question for clarity: enter image description here

Also, am I correct in saying the friction force goes vertically upwards, opposing the direction in which the ring would go if it were released?

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  • $\begingroup$ To be honest I don't understand the configuration in your question. You mean a horizontal rod? $\endgroup$ – pglpm Sep 30 at 12:13
  • $\begingroup$ Is the ring a press fit which would have normal force all around, or a loose fit which would only have normal force at the top? $\endgroup$ – Adrian Howard Sep 30 at 12:25
  • $\begingroup$ I added an image of the figure they gave me. There's a string pulling the ring on the rod. $\endgroup$ – Trying Sep 30 at 12:28
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The usual assumption in questions like this is that the ring is a loose fit on the rod. So when a string exerts a sideways force on the ring, the ring is only in contact with the rod at one point. The normal force on the ring then acts in the plane of the rod and the string, perpendicular to the rod. Friction acts along the line of the rod, in a direction that opposes the direction in which the ring would move in the absence of friction.

Although one can certainly imagine scenarios in which this assumption is not realistic, it is the default convention for such questions.

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