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Block Falling off of Wall

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem consists of a block that starts off on a ledge. It has an initial vertical imperfection ie it is leaning over at the top such that when the top anchors are cut it will fall off of the ledge. I need to determine how far the block falls from the wall and the velocity of the block when it hits the ground 45 feet below to calculate an impulse.


    2. Relevant equations
    I think the relevant equations are the relative general plane motion about a translational and a rotational axis. Neglecting the sliding of the block off of the ledge as it moves outward.


    3. The attempt at a solution
    I had a FBD and then I started to use rigid body motion about a fixed axis but then concluded that would not work. I have been stuck for two weeks and this forum is my last hope. Thanks

    Scott
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 30, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Scott! Welcome to PF! :wink:

    (i'm not sure i understand exactly what the set-up is, but …)

    if the block is balanced on its edge, on the edge of the shelf, and it starts to rotate about that edge until it loses contact,

    then you need to use conservation of energy (both linear and rotational) to find the speed of the centre of mass just as it loses contact …

    from then on, it's freely falling, so of course use the standard constant acceleration equations. :smile:
     
  4. Apr 30, 2009 #3
    Yes, the block is just barely off the edge of the table, and it starts out with an initial tilt, being held back by wires at the top. The wires are then cut and the block is allowed to move off of the table. Could you possibly start me up with the conservation equations, I think I know which ones to start using, however it has been a while since I have used any of those principles. Thanks for your reply


    Scott
     
  5. May 1, 2009 #4

    tiny-tim

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    Well, the initial equation is just mgh + 1/2 Iω2 = constant, where I is the moment of inertia about the edge of the table, and ω is the angular velocity.
     
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