Which direction does the normal contact force act?

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SUMMARY

The normal contact force acting on a ring threaded on a fixed, rough, horizontal curtain pole is directed upwards, perpendicular to the surface of the pole. In this scenario, the ring, with a mass of 0.3 kg, is subjected to a downward tension of 2.5 N from a string making an angle A with the horizontal, where tan A = 3/4. The equilibrium condition necessitates that the normal force counteracts the downward tension, ensuring that all forces acting on the ring are balanced.

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Gughanath
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A ring of mass 0.3kg is threaded on a fixed, rough, horizontal curtain pole. A light inextensible string is attached to the ring. The string and the pole lie in the same vertical plane. The ring is pulled downwards by the string which makes an angle "A" to the horizontal, where tan A=3/4. The tension in the string is 2.5N. The ring is in equilibrium.

This question doesn't seem that difficult, but in which direction does the normal contact force act?
 
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The contact force between the ring and the rod has two components: the component parallel to the rod, which is the friction; the component perpendicular to the rod, which is the normal force. If I picture the setup correctly, the normal force on the ring points up.
 


The normal contact force in this scenario would act in the upward direction, perpendicular to the surface of the curtain pole. This is because the ring is in equilibrium, meaning that all forces acting on it must be balanced. Since the tension in the string is pulling the ring downwards, the normal contact force must be equal and opposite to counteract this downward force and keep the ring in place on the pole.
 

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