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Math Help - normal contact force with angles

  1. #1
    Senior Member furor celtica's Avatar
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    normal contact force with angles

    A small smooth ring R, of mass 0.6 kg, is threaded on a light inextensible string. One end of the string is attached to the fixed point A and the other end is attached to a ring B, of mass 0.2 kg, which is threaded on a fixed rough horizontal wire which passes through A (see diagram (excuse the obviously unequal angles)). The system is in equilibrium, with B about to slip and with the part AR of the string making an angle of 60 to the wire.
    a. Explain, with reference to the fact that ring R is smooth, why the part BR of the string is inclined at 60 to the wire.
    b. Show that the normal contact force between B and the wire has magnitude 5 N.
    c. Find the coefficient of friction between B and the wire.

    For a. I was unsure of what wording to use exactly, and answered the following:
    a. Ring R being smooth and ring B being on the same horizontal plane as fixed point A, ring R will have slipped to the midpoint of the string: angles ABR and RAB are therefore equal.
    Angle RAB being given as 60, this means that the part BR of the string is inclined at 60 to the wire.

    For b. however I have several problems. Firstly, of how the 6N weight of R is distributed over both string segments (RB and AR). I would assume that the each segment carries half the load, but Im unsure of how to prove this. Secondly, even taking 3cos30 N (half of Rs weight resolved for the slope of the string) as the tension in the string segment BR, I end up with normal contact force = 2 (weight of B) + 3cos30(cos30) = 4.25 N

    Any suggestions?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Re: normal contact force with angles

    b. Why did you take 3cos(30)cos(30)?

    What is the component of the weight of ring R downwards?
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