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Good old wedge and block sliding problem

  1. Sep 27, 2005 #1
    The problem is as follows.

    We know the angle of the wedge to the horizontal. Lets say its a given.

    We dont know the mass of the wedge or the mass of the block.

    The planes are frictionless. The mass rests on a "frictionless table".

    There is a force pushing the entire system. The acceleration of the wedge is equal to 5 m/s^2 towards its point along the table.

    What is the relative acceleration of the block down the slope?

    I already turned this problem in. I was able to get an answer that was like [itex]g\sin{\theta}-5\cos{\theta}[/itex] but I was unable to show my complete derivation. I began by seperating the forces, and showing that the forces created acceleration on the block relative to the ground, then took the position that relative to the wedge, the accelerations must have satisfied the wedges triangle. so i got something like [itex]\tan\theta = F_y/F_x [/itex]

    From there its a wash for me. Could someone sort out this a bit for me? I drew my force diagrams, I did the calculations I was supposed to, I just kept getting something a little off. Maybe Ill see it in someones more elegant solution...
  2. jcsd
  3. Sep 29, 2005 #2


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    The two ways are there of thinking.
    If we take the motion of the mass relative to the wedge then, as the wedge is accelerating, is a non inertial frame of reference, we have to consider a pshudoforce = ma (a is the acc. of the wedge) in the derection apposite to the acceleration a. Resolving these forces you will get your answer.


    If you want to consider only real forces then the acceleration of mass in x direction will be ax = a + ar cos@ where ar is the acceleration of mass relative to wedge, as the wedge is accelerating. and that in y direction is ay = ar sin@. You have tocalcutat ar.
    Last edited: Sep 29, 2005
  4. Sep 29, 2005 #3
    Im sorry that I dont follow method 1 very well. I understand that the block and wedge are not in an IFR, and that we must make some sort of comparison to something that is in an IFR with itself, and that the relative motion of each the block and wedge can be described. this is obviously the ground. I went about this and didnt get the right answer. maybe you could specify what you mean about this part

    "consider a pshudoforce = ma (a is the acc. of the wedge) in the derection apposite to the acceleration a. Resolving these forces you will get your answer"

    does this mean that when i plug my forces back into the [itex]\tan\theta[/itex] equation, i would be adding the A_wedge to the A_block, then resolving that? that seems backwards, and still doesnt give me the right solution in the algebra, it only changes the sign of the solution from [itex]9\cos\theta-5\sin\theta[/itex] to [itex]9\cos\theta+5\sin\theta[/itex] when what i really want is [itex]9\sin\theta-5\cos\theta[/itex]. maybe you can help me understand that method more.

    as to method 2. how does [itex]a_y = a_r \sin\theta[/itex] expand to solve for a_r? im not sure i see how to find a solution easily from a single equation with multiple unknowns. a_r is unknown, and a_y is unknown, it involves the normal force and the mass of the block, both of which are also unknown. maybe you can be a bit more specific. i think i know that this leads us to finding two expressions for a_x and a_y and then using them both, but thats more like method 1, as above. i know the language barrier is hard to overcome for some, maybe using more latex would help?
  5. Sep 29, 2005 #4


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    Did a long thinking now enjoy solutions !!! :smile: See the attachment.

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