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    Hi just wondering if someone could explain this to me?

    Finished the statics chapter just over a week ago and currently going through review exercise 2. Just wondered why they could resolve to find Mew (friction coefficient) separately for each ring? Is it because the friction coefficient is always the same for both rings?

    If that is the case and they instead asked me to find Friction and gave me the friction coefficient could I resolve for one ring or would I have to do so for the whole system?

    Teaching myself m1 so any explanation would be great thanks!!
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    (Original post by AJA1994)
    Hi just wondering if someone could explain this to me?

    Finished the statics chapter just over a week ago and currently going through review exercise 2. Just wondered why they could resolve to find Mew (friction coefficient) separately for each ring? Is it because the friction coefficient is always the same for both rings?

    If that is the case and they instead asked me to find Friction and gave me the friction coefficient could I resolve for one ring or would I have to do so for the whole system?

    Teaching myself m1 so any explanation would be great thanks!!
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    :five: I'm self-teaching the same thing and found trouble with the review exercise. When are your exams btw? January or June?
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    (Original post by Reshyna)
    :five: I'm self-teaching the same thing and found trouble with the review exercise. When are your exams btw? January or June?
    My exams are in June, have you tried UKMathsTeacher on YouTube? He was great help understanding before I started the exercises.
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    (Original post by AJA1994)
    My exams are in June, have you tried UKMathsTeacher on YouTube? He was great help understanding before I started the exercises.
    No I haven't but thanks. I'll try them.
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    (Original post by AJA1994)
    Hi just wondering if someone could explain this to me?

    Finished the statics chapter just over a week ago and currently going through review exercise 2. Just wondered why they could resolve to find Mew (friction coefficient) separately for each ring? Is it because the friction coefficient is always the same for both rings?

    If that is the case and they instead asked me to find Friction and gave me the friction coefficient could I resolve for one ring or would I have to do so for the whole system?

    Teaching myself m1 so any explanation would be great thanks!!
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Views: 39
Size:  490.5 KB
    I don't have time to help now. Please tag me if you haven't received a reply by the end of today and I'll take a look.
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    (Original post by notnek)
    I don't have time to help now. Please tag me if you haven't received a reply by the end of today and I'll take a look.
    Okay I'll quote you? Don't know how to tag I'm new. Thanks again
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    Taking account of (a) symmetry, (b) the fact that the ring C is smooth, and (c) the whole system being in equilibrium, you can say that the string tension acting on ring A is the same as the string tension acting on ring B. In terms of forces, once you've worked out the string tension in part (a), you know everything you need to know about the forces acting on both the rings A and B. Therefore you can look at just one of them in isolation to work out the coefficient of friction. The question implies that there is just one coefficient of friction.

    Pretty much the same applies if they had given the coefficient of friction and asked for the force due to friction.
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    (Original post by old_engineer)
    Taking account of (a) symmetry, (b) the fact that the ring C is smooth, and (c) the whole system being in equilibrium, you can say that the string tension acting on ring A is the same as the string tension acting on ring B. In terms of forces, once you've worked out the string tension in part (a), you know everything you need to know about the forces acting on both the rings A and B. Therefore you can look at just one of them in isolation to work out the coefficient of friction. The question implies that there is just one coefficient of friction.

    Pretty much the same applies if they had given the coefficient of friction and asked for the force due to friction.
    So under what circumstances can I look at a particle in isolation for future questions? When there is symmetry?
    I thought tension was equal due to the fact that the string was light? (UkMathsTeacher)
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    (Original post by AJA1994)
    So under what circumstances can I look at a particle in isolation for future questions? When there is symmetry?
    I thought tension was equal due to the fact that the string was light? (UkMathsTeacher)
    It's difficult to give a universal answer to that, but once you've reached the stage where you know all the forces acting on a particle (or all the forces bar one, when you already know the coefficient of friction), you can then focus on that particle to work out whatever is the remaining unknown.

    Of course you will often have to refer to other parts of the overall system in determining all the forces acting on a particular particle.

    Things will get more complicated with some systems, where you can end up equations for two points in the system and still have two unknowns. You then have to employ elimination or (more likely) substitution to get rid of one of the unknowns and then solve for the remaining one.
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    (Original post by old_engineer)
    It's difficult to give a universal answer to that, but once you've reached the stage where you know all the forces acting on a particle (or all the forces bar one, when you already know the coefficient of friction), you can then focus on that particle to work out whatever is the remaining unknown.

    Of course you will often have to refer to other parts of the overall system in determining all the forces acting on a particular particle.

    Things will get more complicated with some systems, where you can end up equations for two points in the system and still have two unknowns. You then have to employ elimination or (more likely) substitution to get rid of one of the unknowns and then solve for the remaining one.
    I hear you mate, thanks a lot for the time and detailed answers!
 
 
 
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