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Resolving in the vertical direction and along the direction of the normal force

  1. Apr 15, 2015 #1
    ImageUploadedByPhysics Forums1429110373.315217.jpg
    Ok I know this is simple resolving but I'm a bit confused. If we resolve forces in the vertical y direction (as in the attached image), we obtain Rcos(theta)-W=0 and hence Rcos(theta) = W meaning R = W/cos(theta) [where theta = 18 degrees]

    However, if we resolve forces along the direction of the normal force, R, we obtain R - Wcos(theta) = 0 and hence R = Wcos(theta) [again theta=18]. These 2 results for R appear to contradict so what have I done wrong?

    Cheers
     
  2. jcsd
  3. Apr 15, 2015 #2

    PhanthomJay

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    The first approach as per the solution is correct, since there is no acceleration or component thereof in the vertical direction.

    The second approach is incorrect, because if you resolve forces along the normal direction, the acceleration is not 0 in the normal direction; there is a component of the radial acceleration in that direction, so equilibrium does not apply in that direction normal to the plane.
     
  4. Apr 15, 2015 #3
    Ah I see. With radial acceleration you are referring to an right?
     
  5. Apr 15, 2015 #4

    PhanthomJay

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    Yes, right.
     
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