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Magnitude of Normal Force (Frictionless)

  1. Sep 18, 2008 #1
    1. The problem statement, all variables and given/known data

    "A crate of mass 50 kg is pushed across a frictionless horizontal floor with a force of 100N directed 23.5 degrees below the horizontal. The magnitude of the normal force of the floor on the crate is:"

    2. Relevant equations

    F = ma? Not sure what else would apply.. mainly a conceptual issue here

    3. The attempt at a solution

    I took 100N x cos(23.5) in order to get the horizontal force which turns out to by 91.7N.. not sure what to do from here
     
  2. jcsd
  3. Sep 18, 2008 #2

    Doc Al

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    Staff: Mentor

    To solve for the normal force, analyze the vertical components of the forces acting on the crate. (Three forces act.)
     
  4. Sep 18, 2008 #3
    Well gravity is acting on the crate (9.8 m/s^2 downward), and the crate itself has a normal force exerted in the opposite direction
     
  5. Sep 18, 2008 #4

    Doc Al

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    Staff: Mentor

    Gravity and the normal force are two of the three forces. What's the third vertical component?
     
  6. Sep 18, 2008 #5
    Well I imagine the third component could be sin(23.5 degrees) x 100 = 39.875
     
  7. Sep 18, 2008 #6

    Doc Al

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    Staff: Mentor

    Right. So what must the normal force be to balance this force plus gravity?
     
  8. Sep 18, 2008 #7
    Ah so 9.8 m/s^2 x 50kg = 490N

    And adding that to 39.875N will result in a magnitude of 530N
     
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