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Help with basic forces problem

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  1. Jul 12, 2016 #1
    1. The problem statement, all variables and given/known data
    A block is at rest on a ramp on a table with no friction between the surfaces. The ramp has an incline of 30 degrees. The mass of the block is 4kg, and the mass of the wedge is 6kg. What magnitude force can be applied to the opposite side of the ramp such that both the ramp and the block accelerate without the block sliding?

    2. Relevant equations
    F = ma
    m * gravity * sin(theta) = ma
    normalForce = mg cos(theta)
    forceByGravity = mgsin(theta) = ma
    Force(block in x direction) = 4(-a) //negative a because it's going left.
    Force(ramp in x direction) = 6(-a)

    3. The attempt at a solution
    My first thought was to figure out the acceleration the block would have without a push from the ramp (though since the block starts at rest I'm not sure how relevant this information is):
    mgsin(theta) = ma
    (4)(9.8)(sin30) = (4)a
    a = 4.9 m/s^2
    and that the force from the ramp would match up to have the same acceleration (i.e. F = ma = 6*(-4.9) ), but I think there's something conceptual about this that I'm not taking into account because I'm unable to explain exactly why that would work. Should the push force act similarly to friction in this model? What should I try to do?
     
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  3. Jul 12, 2016 #2

    haruspex

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    That approach certainly seems doubtful.
    Why not just apply the usual ΣF=ma to each body?
     
  4. Jul 13, 2016 #3
    Maybe you did not realize that it is a static problem. The sum of all the forces it is not ## \sum F=m \cdot a ## but ## \sum F=0##.
     
  5. Jul 13, 2016 #4

    haruspex

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    Reread the question.
     
  6. Jul 13, 2016 #5
    I understand that the ramp moves with acceleration creating a force that must neutralize the force (weight) which makes the block slide down. So the sum of forces acting on the block must be zero.

    - Force one: From the acceleration applied on the ramp.
    - Force two. From gravity on the block.
    - Result: Block does not slide down because the sum of the forces is zero.

    If the block was sliding down (or up) with some acceleration I would then use ##F= m \cdot a##, but not if it is quiet or with constant velocity.

    Where is my mistake?
     
  7. Jul 13, 2016 #6

    haruspex

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    The wedge is accelerating, the block is staying still relative to the wedge, and the sum of the forces on the block is zero... is that what you are saying? You might care to think about that some more.
     
  8. Jul 13, 2016 #7
    I don't get your point, sorry. It's ok. It may be the language, English is not my mother tongue.
     
  9. Jul 13, 2016 #8
    The block is actually accelerating at the same rate as the wedge ,therefore it has an acceleration equal to the wedge.
     
  10. Jul 13, 2016 #9

    haruspex

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    ... and therefore the net force on it is not zero.
     
  11. Jul 14, 2016 #10
    You are putting the reference system on the table, I am putting it on the block. The equation is the same.
     
  12. Jul 14, 2016 #11

    haruspex

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    Are you saying you are using a non-inertial reference frame?
     
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