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Rolling ramp displacement due to weigh of block

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Basically there's a ramp; its inclined plane has length "a", the base has length "b" and its height has length "c". The ramp is stationary, but it has wheels under it. The ramp weighs 120N.

    The center of mass of the ramp is located at 2/3 of "b" on the horizontal axis and 1/3 of "c" in the vertical axis.

    Then a block is placed on top of the ramp (1). It weighs 80N.

    Given that the inclination of the ramp is α; sin α = 0.6 and "a" = 15m. How much will the ramp be displaced when the block reaches the end of the ramp (2)?

    Also, disregard any friction and assume the block is a point mass.

    2. Relevant equations



    3. The attempt at a solution

    I have a couple of questions to begin with..

    Does the ramp get displaced only because it's specified a center of gravity? Like if the whole ramp was a point mass; would it not be displace at all? Then what happens if I place the center of gravity of a block on top of the center of mass of a ramp?

    Also; what role does the center of mass play in this particular problem?
     

    Attached Files:

  2. jcsd
  3. Nov 11, 2012 #2

    mfb

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    Every object has a center of gravity, it does not matter whether it is given to you or not. If the ramp was a point-mass, the object could not slide down the ramp and the setup would be pointless (;)).

    That point has no special meaning in the setup. The sliding distance would change, and the displacement of the ramp would change, but the setup stays similar.

    Momentum conservation ensures that the center of mass of the whole setup stays at the same horizontal position. This can be used to solve the problem.
     
  4. Nov 11, 2012 #3
    Okay, so the center of mass of the ramp and the block won't change along the x direction because there is no external forces acting on the block/ramp system horizontally; correct?

    If the x coordinate of the center of mass is always gonna be 4b/5, can I say that:

    [tex] \frac{m_{block}x_{block}+m_{ramp}x_{ramp}}{m_{block}+m_{ramp}}=\frac{4b}{5}[/tex]

    How do I continue from here?

    EDIT: sorry the x coordinate of the center of mass is [itex] \frac{4b}{5} [/itex]

    Actually, I still can't solve the problem.. Is the fact that the block is on an inclined changes something in the expression

    [tex] \frac{m_{block}x_{block}+m_{ramp}x_{ramp}}{m_{block}+m_{ramp}}[/tex]

    where I would take [itex] m_{block} [/itex] to be 80/9.8 and [itex] m_{ramp} [/itex] to be 120/9.8.
     
    Last edited: Nov 11, 2012
  5. Nov 11, 2012 #4

    mfb

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    Right

    You know the distance between "ramp center of mass" and "block center of mass" after the block slided down. This leaves one unknown (the position of the ramp, for example) and one equation (the known center of mass).

    The factor of 9.8 is the same for both objects, it does not change anything.
     
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