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    Attachment 539863539865Name:  IMAG0167.jpg
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Size:  340.5 KB Hi could someone please help me with part c of this question? I managed to work out the normal force in part b which is 1.44. However, I don't get how part c is done. The frictional force would be fr=ur . The new normal force is less than 4.44 in part b, so surely fr would be < 2cosx, meaning the particle stays in equlibrium. However that's not the case, so I really need some help on this. Thanks.

    This is the jan 2002 m1 paper Q7
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    Actually , I might have confused myself a bit here. So since fr is < 0, this mean the particle is will move as fmax is smaller than 2cosx?
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    Your work looks good, since the particle is being pulled at an angle the normal reactionary force has decreased and so has your frictional max.

    Your new frictional max is now (0.45)(1.44) which is 0.648,

    Your force acting parallel to the plane is 2 as you've clearly gotten, since 2 is bigger than 0.648 it should move.

    I think, my M1 is rusty.
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    (Original post by plaguarist)
    Your work looks good, since the particle is being pulled at an angle the normal reactionary force has decreased and so has your frictional max.

    Your new frictional max is now (0.45)(1.44) which is 0.648,

    Your force acting parallel to the plane is 2 as you've clearly gotten, since 2 is bigger than 0.648 it should move.

    I think, my M1 is rusty.
    Yh I think that's right lol I have confused myself with fmax and f. Thanks!
 
 
 
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