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Direction of friction

  1. Jul 8, 2012 #1
    ixFM5.png

    [strike]In this question, the direction of friction (which I assumed correctly) is acting downwards on the wire,[/strike] however:

    bKtet.png

    The friction is acting upwards on the wire, could someone walk me through how I would figure out the direction of the friction force?

    I understand friction is against motion, but I don't see the motion.
     
    Last edited: Jul 8, 2012
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  3. Jul 8, 2012 #2

    cepheid

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    In the first example, a free body diagram for the particle (the black circle) along with the statement that the system is in equilibrium, tells you that T = mg, where T is the tension in the rope. Imposing the requirement of vertical equilibrium for the ring, you see that the downward force on it is mg, but the upward force on it due to the tension in the string is only mgcos(60°) < mg. So, the ring wants to slide downward. In order for the ring to be static, there must be an upward frictional force that keeps it suspended by making up the balance of the vertical force.

    So, the force of friction on the ring due to the wire is upwards. Therefore, by Newton's 3rd Law, the force of friction on the wire due to the ring is downwards, just as you stated in your original post.

    In contrast, for the second problem, the vertical component of the tension in the rope is 3mgcos(30°) > 2mg. So the force pulling upward on the ring is greater than its weight. It wants to slide upward, and it would, were it not for a downward frictional force with the wire that keeps it in place.

    So, the force of friction on the ring due to the wire is downwards. Therefore, by Newton's 3rd Law, the force of friction on the wire due to the ring is upwards, just as you stated in your original post.
     
  4. Jul 8, 2012 #3
    I'm sorry but the bold part, how did you get it less than mg? Also, I've looked at the solutions and for the first 1, they have the friction upwards, and downwards for the second one, so it seems I was wrong.
     
  5. Jul 8, 2012 #4
    For purely geometrical reasons, mgcos60 is less than mg. cepheid stated this in order to then say that the force of friction must be what's holding everything in equilibrium.
     
  6. Jul 8, 2012 #5

    cepheid

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    Yeah, what gnulinger said. cos(any angle) ≤ 1.


    They are talking about the friction acting ON THE RING. You were talking about the friction acting ON THE WIRE. These are in opposite directions according to Newton's 3rd law. I already explained that above. I gave the directions for both of the forces in the action-reaction pair in both examples.
     
  7. Jul 9, 2012 #6
    thanks both, get it now.
     
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