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    (Original post by Insanity514)
    Ok so I've some how got the answer..here's how

    First you wanna split the middle part into two triangles.
    The horizontal component of on the ring will therefore be mcos30.
    SInce this is perpendicular to the plane this is the reaction force.
    Now you wanna find the friction thingy on the vertical string not the string on the pully. If the system was not in rest the ring would move up the vertical string (logic).
    Therefore friction acts downwards of the vertical string and force acts up. Now you want the force that acts up the string. This is found from resolving the forces on the top triangle. If you resolve it the verticale force is msin30. m being the mass of that black particle. Since the ring is stationary, mewR=msin30. (mew is coefficient of friction). But you know R is mcos30 as explained above. so Mewmcos30=msin30.
    Cancel the m out from both sides. gives sin30=Mewcos30. Mew=sin30/cos30

    Which gives 0.577.


    Hope it helps and tanks for posting this got me thinkin.
    Excuse spelling mistakes please lol

    Any pictures of the diagram you drew?

    I'm wondering how you split it into two triangles.
    :confused:
    Help please
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    (Original post by pearlypearl)
    How did you rearrange Z(-mgcos60+mg) to get mg(cos60Z+Z) ?
    Shouldn't it be mgZ(1-cos60) ?
    I posted that solution in April, but yeah, you're right, it should.
    In my solution I think I labeled the friction and reaction forces wrong, which is why it doesnt work.

    The Reaction force points horizontally left because the loop in the string is been pulled against the egde of the ring creating a reaction force in the opposite direction. (You assume it's horizontal since the string is about in the middle of the ring on the picture)

    If there was no friction, the ring would slip downwards since the downward forces on the ring (mg) are greater then the upward forces on the ring (mgcos60). Since the system is in eqm, the friction must be acting upwards.

    From this, the reaction force is mgsin60 meaning friction is Zmgsin60. (Resolve horizontally)
    Friction is also mg-mgcos60. (Resolve Vertically)

    mg-mgcos60=Zmgsin60
    1-cos60=Zsin60
    Z=0.577
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    (Original post by Ollie901)
    I posted that solution in April, but yeah, you're right, it should.
    In my solution I think I labeled the friction and reaction forces wrong, which is why it doesnt work.

    The Reaction force points horizontally left because the loop in the string is been pulled against the egde of the ring creating a reaction force in the opposite direction. (You assume it's horizontal since the string is about in the middle of the ring on the picture)

    If there was no friction, the ring would slip downwards since the downward forces on the ring (mg) are greater then the upward forces on the ring (mgcos60). Since the system is in eqm, the friction must be acting upwards.

    From this, the reaction force is mgsin60 meaning friction is Zmgsin60. (Resolve horizontally)
    Friction is also mg-mgcos60. (Resolve Vertically)

    mg-mgcos60=Zmgsin60
    1-cos60=Zsin60
    Z=0.577

    Thanks a lot ! :^_^::^_^:
 
 
 
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