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Accelerating Wedge Problem

  1. Mar 3, 2016 #1
    upload_2016-3-2_22-13-29.png

    A block of mass m rests on a frictionless wedge that has an inclination of θ and mass M. Find the acceleration of the wedge M to the right such that the block m remains stationary relative to the wedge

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

    a, [itex] \theta [/itex], M, m


    2. Relevant equations


    3. The attempt at a solution
    Draw a FBD centered on m. There is the force due to gravity, the normal force due to contact between M and m. There is also a force from M that imparts to m. Am I missing anything else?
     
  2. jcsd
  3. Mar 3, 2016 #2

    haruspex

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    You already listed a normal force from M on m. What is this other force?
     
  4. Mar 3, 2016 #3
    I would find the component of acceleration (due to gravity) acting on m down the slope first. Then consider the component in direction of a.
     
  5. Mar 4, 2016 #4
    I realize that there is a force that from M that imparts to m due to M's acceleration. I'm not sure what the direction and magnitude of this force. Is the direction of this force just parallel to the movement of M towards the right? So how many force vectors are there altogether for the FBD? 3?
     
  6. Mar 4, 2016 #5
    The only force the plane exerts on m is the normal force.
    Have you tried resolving the normal force into its components and relating these
    components to the motion of the block?
     
  7. Mar 4, 2016 #6

    haruspex

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    M accelerates to the right. If it were to exert no force on m then m would move inside M. A normal force between surfaces is that force which opposes interpenetration. The force that M 'imparts' to m due to M's acceleration is the normal force.
     
  8. Mar 11, 2016 #7
    I just realized that there should be a force parallel to the plane that points up along the slope? Is that true?

    The y component of the normal force is mg cos theta and the x component is mg sin theta
     
  9. Mar 11, 2016 #8

    haruspex

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    What would supply such a force? The slope is frictionless.
    Don't confuse normal force with gravitational force. Are you defining the Y direction as vertical or normal to the wedge slope?
    Whichever you choose, try to answer this question: if the normal force is N, what is its component in the Y direction?
    (To answer this you do not need to consider gravity or mass or acceleration.)
     
  10. Mar 11, 2016 #9
    The force that points up along the slope is probably s fictitious force. But that's beyond the scope of the course I'm taking.

    I'm defining the y direction as vertical to the wedge slope. It's component in the y direction is simply cos theta.
     
  11. Mar 11, 2016 #10

    haruspex

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    Yes.
    No, it's either 'vertical' or 'normal to the slope'; which?
    What direction does the normal force act in? (The clue is in the name.)
     
  12. Mar 11, 2016 #11
    Alright, the direction is vertical.
     
  13. Mar 11, 2016 #12

    haruspex

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    OK, we can agree that the component of the normal force N in the y direction is N cos(theta), yes?
    What other forces act in the y direction?
    What is the acceleration of the block in the y direction?
    What equation does that give you?
     
  14. Mar 12, 2016 #13
    There's gravity (mg) that acts in the y direction. Since the block is static, then the equation should be N cos(theta) - mg = 0
     
  15. Mar 12, 2016 #14

    haruspex

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    Well, it's not static, but you are right that it has no vertical acceleration, so yes, that is the right equation.
     
  16. Mar 12, 2016 #15
    So is the equation for the x direction N sin(theta) = a?
     
  17. Mar 12, 2016 #16

    haruspex

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    Yes.
     
  18. Mar 12, 2016 #17
    so a = mg tan(theta)?
     
  19. Mar 12, 2016 #18

    haruspex

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    Yes.
     
  20. Mar 12, 2016 #19
    Thanks, I think that was the answer.
     
  21. Mar 12, 2016 #20
    Actually, the answer is a = tan (theta), since I forgot to cancel out the m. Close enough.
     
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