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B Motion analysis of an accelerating wedge and a block

  1. Jun 12, 2017 #1
    I don't understand the motion of an accelerating wedge and a block. I'd really appreciate if you make me understand the motion in both an inertial and a non inertial reference frame.

    Here's a figure I have made, a0 is the acceleration on the wedge with respect to an inertial frame, towards right.
    sfoMZDk.jpg
     
    Last edited: Jun 12, 2017
  2. jcsd
  3. Jun 12, 2017 #2
    Is ##a_0## given? If so then the mass M of the wedge is not needed
     
  4. Jun 12, 2017 #3
    Yes , a0 is towards right. I apologise, I made a mistake earlier by saying a0 is towards left.
     
  5. Jun 12, 2017 #4
    Let's start from inertial viewpoint. Write down the Newton Second Law for the briсk
     
  6. Jun 12, 2017 #5
    EnuXycR.jpg
    Let's take the angle of incline as θ. Applying Newton's Second Law to the block;
    ∑Fx = mgsinθ = ma (Where a is acceleration with respect to ground, which we take to be inertial.)
    ∑Fy = N - mgcosθ = 0 ⇔ N = mgcosθ

    Am I correct?
     
  7. Jun 12, 2017 #6
    Stop! do not be so quickly. First we must write the Newton second Law in vector form: ##m\boldsymbol a=m\boldsymbol g+\boldsymbol T##; here ##\boldsymbol a## is an acceleration of the brick relative to an inertial frame. This was the first step of the solution. Now you can expand this vector equality by any coordinate frame you want. Let's introduce a frame ##Oxy## such that ##O## is the highest angle of the wedge and the axis ##OX## goes down along the slope; Oy is directed upwards perpendicular to the slope.
    Then ##\boldsymbol T=T\boldsymbol e_y;##
    ##\boldsymbol g=..##
    ##\boldsymbol a=##
    go on
     
  8. Jun 12, 2017 #7
    no
     
  9. Jun 12, 2017 #8
    Why not? Those are the only forces I see.
     
  10. Jun 12, 2017 #9

    jbriggs444

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    Partly. You seem to be adopting a coordinate system where "x" is along an axis rotated θ up from the horizontal and where "y" is along the axis at right angles to that. [This coordinate system is notionally fixed in space. It does not move with the wedge]. Against that coordinate system, only g has any component along the x axis. We may immediately determine the x component of the acceleration of the block. You have done that and correctly computed ##ma_x## for the block.

    You have not computed ##m\vec{a}## for the block.

    You have not justified the claim that N = mg cos θ. Why should that be so?
     
  11. Jun 12, 2017 #10
    Doesn't the block stay in contact with wedge through out the motion? Let's see, the wedge moves to right with acceleration a0, and at the same time, the block moves down with an acceleration of mgcosθ. I thought the normal force stayed same. I must admit, this motion is confusing me.
     
  12. Jun 12, 2017 #11

    jbriggs444

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    The normal force does not stay the same. The block does not move down with an acceleration mg cosθ (as it would if the normal force were zero).

    Is ##a_0## given? Or is is something that you wish to compute?
     
  13. Jun 12, 2017 #12
    Ah, yes, that makes sense. I am new to Newtonian mechanics, so I am likely to make mistakes. Can you tell me why normal force does not remain the same?
     
  14. Jun 12, 2017 #13

    jbriggs444

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    The wedge is accelerating (or is allowed to accelerate -- you still have not said which). If ##\vec{A}## is the acceleration of the wedge, what is Ay? If Ay is the acceleration of the wedge in the y direction, what is ay -- the acceleration of the block in the y direction?

    Write down a force balance for the block in the y direction. ∑Fy = may
     
  15. Jun 12, 2017 #14
    The wedge is accelerating towards right. Ay should be 0, as the wedge is not acceleration upward or downward. What do you mean by "Ay is the acceleration of the block"? Didn't you just say A is the acceleration on the wedge?
     
  16. Jun 12, 2017 #15

    jbriggs444

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    You rotated your coordinate axes. "right" has a component in the direction of negative y.
    I editted that post a few times. It should read better now.
     
  17. Jun 12, 2017 #16
    Do I have to choose the same coordinate system for the wedge that I chose for block? If I do so, the component of acceleration in the Y direction will be Asinθ.
     
  18. Jun 12, 2017 #17

    jbriggs444

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    You were trying to figure out the normal force. For normal force, what matters is the acceleration of the block in the direction perpendicular to the surface.

    You persist in not clarifying whether the acceleration of the wedge is a given fixed constant or whether it is something to be computed. I will assume that it is a given fixed constant, A0. In that case, the acceleration of the block and of the wedge in the y direction is -A0sin θ.

    Edit: fixed the sign. This is a "downward" acceleration.

    With that information in hand, can you now expand ∑Fy = may with all relevant forces?
     
  19. Jun 12, 2017 #18
    The acceleration on the wedge which you denoted by A, is a constant, and it is towards right.
    With that information in hand, can you now expand ∑Fy = may with all relevant forces?[/QUOTE]
    The acceleration on the wedge which you denoted by A, is a constant, and it is towards right.
    Applying Newton's second law to the wedge:
    ∑Fy = N = MAsinθ
     
  20. Jun 12, 2017 #19

    jbriggs444

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    You forgot the sign. That term should be -mA0 sinθ.
    You also missed the force of gravity.

    Edit: It is good practice to be consistent with your variable names...

    Having used lower case "m" for the mass of the block, you should continue to use "m" and not switch to upper case "M". That way we can reserve "M" for the mass of the wedge.

    Having used "A0" for the rightward acceleration of the wedge, you should not shift to an un-subscripted "A".

    At one point in this thread, I had used ##\vec{A}## to refer to the acceleration of the wedge as a vector. More properly, I should have used ##\vec{A_0}##.
     
  21. Jun 12, 2017 #20
    jPWOIFm.jpg
    Then there is also normal force from the ground on the wedge, let it be N'. I hope you can see the figure, and thank you for trying to help me. I really appreciate it.
    All right, let's use A0 for the acceleration of the wedge. Let's apply Newton's second law to wedge;
    ∑Fy = N'cosθ - Mgcosθ - N = -MA0sinθ
     
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