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    Two poles are fixed, one horizontal and one vertical, 1.2m apart. The horizontal pole is smooth, the vertical pole is rough. A ring of weight 50n fits round the vertical pole. A chord is attached to the ring, passes over the horizontal pole and carries a block of weight 130N at its end.

    If the ring is at the same level as the horizontal pole, it is in limiting equilibrium. Calculate the coefficient of friction between the ring and the vertical pole.


    Just dont get questions like this, I worked out the resultant force which didn't help me as I got mew=1,



    any help?
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    Look at the forces on the ring. We already have 130N to the right (tension) and 50N down (weight). The ring is in equilibrium so we must have 130N to the left (normal contact force) and 50N up (this can only be friction). So we know normal contact force and friction (which we are told is limiting) so we can calculate mu.
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    (Original post by ttoby)
    Look at the forces on the ring. We already have 130N to the right (tension) and 50N down (weight). The ring is in equilibrium so we must have 130N to the left (normal contact force) and 50N up (this can only be friction). So we know normal contact force and friction (which we are told is limiting) so we can calculate mu.
    thanks, feel like such an idiot now, just when contact force concerns things on vertical poles, i get confused
 
 
 
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