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Kinetic friction free body diagram

  1. Oct 26, 2008 #1
    A Physics 1AL student uses a force P of magnitude 80 N and angle θ = 70 (with respect to the horizontal) to push a 5.0 kg block across the ceiling of her room. The coefficient of kinetic friction between the block and the ceiling is 0.40.

    a) Draw a free body diagram of the system.

    b) What is the magnitude of the block’s
    acceleration?


    This is a question that I can't really figure out. Please help
     
  2. jcsd
  3. Oct 26, 2008 #2

    Kurdt

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    Have you drawn the free body diagram?
     
  4. Oct 26, 2008 #3
    no i dont know how to do it i drew a line which is the ceiling and the box under it but i dont know what to do after that
     
  5. Oct 26, 2008 #4

    Kurdt

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    What are the components of force acting on the box? Put them in the diagram.
     
  6. Oct 26, 2008 #5
    ok well there is a normal force upward and Force due to gravity pointing down right? Plus there is the Static friction which is a horizontal force. However I dont know how to deal with the force being exerted on the box. I figure out that I have to break it down in components but I dont know if the y-component will be added or subtracted for the vertical forces.
     
  7. Oct 26, 2008 #6

    Kurdt

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    The y-component is pushing the box up onto the ceiling isn't it. The weight and normal force from the ceiling act in the same direction toward the floor. You know the normal force will be whatever resultant force is pushing the box into the ceiling and yo also should know how the normal force and the friction force are related.
     
  8. Oct 26, 2008 #7
    I thought Fg and the normal force were pointing opposite directions I guess not. But if it is how you say it is then that means that the normal force will be the same as the y-component because that is the only opposite force right? and I do know that kinetic friction is equal to the coefficient of friction times the normal
     
  9. Oct 26, 2008 #8

    Kurdt

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    Well with the box being on the ceiling the normal force is in the same direction as the weight. It is not just equal to the y-component though. It is equal to the resultant force (i.e. the vector sum of the y-component and the weight).
     
  10. Oct 26, 2008 #9
    oh ok so the sum of vertical components = fn - Fg - 80 x sin70 so that Fn will be equal to the weight + the y component
     
  11. Oct 26, 2008 #10
    hey B good I was about to ask the same question
     
  12. Oct 27, 2008 #11

    Kurdt

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    As long as that is a vector sum then yes. Remember the weight and y-component are in opposite directions.
     
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