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Body's Acceleration

  1. Apr 23, 2006 #1
    I have a question in my mind. If the body is supposed to slide on with mass m on a fixed triangular platform with angle [tex]\theta[/tex]. Ignoring friction and with g as gravitational acceleration, what will be the acceleration of the body if it slide on the platform?
     
  2. jcsd
  3. Apr 23, 2006 #2

    Doc Al

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    Consider the forces acting on the mass parallel to the surface (since that's the only direction the mass can move).
     
  4. Apr 23, 2006 #3
    In short, the acceleration of the body is the same as the gravitational pull of the earth([tex]9.8 m/s^2[/tex])disregarding the friction on the platform?How about the angle?Do I need to disregard it?
     
  5. Apr 24, 2006 #4

    Hootenanny

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    Do you mean there is a body on an incline? As in this image;

    [​IMG]

    ~H
     
  6. Apr 24, 2006 #5

    Doc Al

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    9.8 m/s^2 is the acceleration of a freely falling body (neglecting air resistance)--it is not the acceleration of an object sliding down an incline.

    Most certainly not; the angle is key. As I had asked before, identify the forces acting parallel to the surface. (Hoot gave you a good diagram to refer to.) Then you can apply Newton's 2nd law to find the acceleration.
     
  7. Apr 24, 2006 #6
    Meaning the acceleration of the body is -mgsin[tex]\theta[/tex] by disregarding the friction of the platform. My answer would be negative because it is going downward?
     
  8. Apr 24, 2006 #7

    Hootenanny

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    That would be the force acting down the slope. Simply apply Newton's law, F=ma to find the acceleration.

    ~H

    edit: sorry for jumping in on you Doc, I didn't realise you were still online.
     
    Last edited: Apr 24, 2006
  9. Apr 24, 2006 #8

    Doc Al

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    That's the force, not the acceleration. (What is the acceleration?)
    The acceleration is down the incline. Depending upon the sign convention used, that could be negative. Perhaps you only need the magnitude of the acceleration.
     
  10. Apr 24, 2006 #9
    So the acceleration of the body from Newton's law, all forces from x axis would be -mgsin[tex]\theta[/tex]=[tex]ma_x[/tex] if we disregard friction. So the magnitude of the acceleration is a=gsin[tex]\theta[/tex]. Am i correct now?
     
  11. Apr 24, 2006 #10

    Doc Al

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    Now you've got it.

    (No problem, Hoot. The more, the merrier! :smile: )
     
  12. Apr 24, 2006 #11
    This question would be interesting. What if the platform moves with an acceleration of a on the horizontal plane, the body stands still on the slope of the moving stand. What is the acceleration a of the stand?
     
  13. Apr 24, 2006 #12

    Hootenanny

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    Can you figure it out yourself? What would be the forces involed? Direction is important for this question.

    ~H
     
  14. Apr 24, 2006 #13

    Doc Al

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    You'd attack this problem by identifying all the forces on the body, then applying Newton's 2nd law. In this case you'd be solving for the conditions which make the vertical acceleration zero. (Write separate equations for the horizontal and vertical components.)
     
  15. Apr 24, 2006 #14
    I am sorry because I still catching up on these forces acting upon the body.
     
  16. Apr 24, 2006 #15
    I want to quote the image. Should it be -mgcos[tex]\theta[/tex]?
     
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