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Mecanics 1: friction watch

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    need some help with this q...

    A 40kg mass is in limiting equlibrium when in contact with a rough inclined plane (30 degrees) . Find the coefficient of friction (mu) ?




    well i already know that F = uR


    but i havent a clue about how to logically work out F and R...

    an explanation would also be appreciated.

    mochas gracoious. cheers.
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    Resolve parallel and perpendicular to the plane then solve the simultaneous equations.
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    h its like u draw the picture, ie: the triangle and then u draw the object and its mass going downwards, thing. like the mass ok the object, and when its in limiting equilibrim i think it equals to 0.

    but i could be very wrong bout that.. ok i prob wasnt any help at all then.

    sorry.
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    In limiting equilibrium, the frictional force (parallel to the plane) preventing the particle from moving will be equal to the component of the weight that's parallel to the plane.
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    (Original post by grace_)
    In limiting equilibrium, the frictional force (parallel to the plane) preventing the particle from moving will be equal to the component of the weight that's parallel to the plane.

    Haaa...thats kinda what i meant. lol. but i dont think i said it well enough.

    well at least u were more helpful than me.
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    Forces acting down the plane = 40gsin30

    as it is in equilibrium the forces acting up the plane ie. friction is equal to the forces acting down the plane

    so

    40gsin30 = Fr

    R= 40gcos30

    Fr= u R

    40gsin30 = 40gcos30 x U

    40gsin30/40gcos30 = u

    u = sin30/cos30
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    thanx for the input guyz.
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    if damaster is correct ... sin(30)/cos(30) = 0.5 / (Sqrt(3)/2)) = 1/Sqrt(3) .... = tan(30); way more useful in these forms ;-)
 
 
 
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