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Question with Normal Force

  1. Jun 26, 2007 #1
    1. The problem statement, all variables and given/known data
    I was doing a problem with normal force on an inclined plane and found the normal force by finding the y-component of the gravity vector. Then I also found that if I find the components of the normal force, I can also get another answer.
    -How are the 2 different equations available for the normal force the same


    2. Relevant equations
    Fg = mg
    Fgy = -mgcos(theta)

    Fny = Fncos(theta)


    3. The attempt at a solution
    Fgy = -mgcos(theta)
    Fn - mgcos(theta) = 0 'movement along plane; not vertically
    Fn = mgcos(theta)

    Fny = Fncos(theta)
    Fncos(theta) - mg = 0 'movement along plane; not vertically
    Fn = mg/cos(theta)

    I got 2 equations for Fn; Fn = mgcos(theta), Fn = mg/cos(theta)...I'm not understanding how the 2 are the same; can anyone help
     
    Last edited: Jun 26, 2007
  2. jcsd
  3. Jun 26, 2007 #2

    andrevdh

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    Homework Helper

    The object is accelerating, [tex]a[/tex], along the direction of the incline.

    This means that in a direction perpendicular to the incline it is not accelerating as you correctly assumed in the derivation of your first formula.

    One can decompose the acceleration along the incline into x- and y- components though. This means that in the x and y direction one cannot assume that the acceleration of the object is zero. So your second formula is invalid.
     
  4. Jun 26, 2007 #3
    I see; thanks. But, in that case, the second equation would work if lets say the object was moving perpendicular to the plane (such as a car on a banked road), right?
     
  5. Jun 26, 2007 #4

    andrevdh

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    Yes, because the resultant acceleration is horizontal and there is then no vertical accelleration component.
     
  6. Jun 26, 2007 #5
    Alright; thanks
     
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