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Block and acceleration of wedge

  1. Sep 15, 2016 #1
    1. The problem statement, all variables and given/known data
    A block rests on a wedge inclined at angle ##\theta##. The coefficient of friction between the block and plane is ##\mu##. The wedge is given a horizontal acceleration ##a##. Assuming that ##\tan \theta > \mu##, find the minimum acceleration from the block to remain on the wedge without sliding

    2. Relevant equations
    Newton's second law
    Law of static friction

    3. The attempt at a solution
    So essentially, I feel as though I have the correct solution, but I am not sure why it is the solution, physically.

    So we have three forces acting on the block. the normal force, friction, and gravity.
    If we define our coordinate system so that the y-direction is up and the x-direction is down, then we have the following equations:

    y-direction: ##N \sin \theta + f \sin \theta - mg = 0##
    x-direction: ##N \sin \theta - f \cos \theta = ma##

    then
    ##N \cos \theta + f \sin \theta = mg##

    Since ##\tan \theta > \mu##
    we know that ##f = \mu N##

    Then
    ##N (\cos \theta + \mu \sin \theta) = mg##
    ##N (\sin \theta - \mu \cos \theta) = ma##

    ##\displaystyle a = \frac{g(\sin \theta - \mu \cos \theta)}{\cos \theta + \mu \sin \theta}##

    I think that this is the expression for the minimum, but I am not sure why this is the minimum. What is the condition that makes this expression the minimum, rather than just some acceleration?
     
    Last edited: Sep 15, 2016
  2. jcsd
  3. Sep 15, 2016 #2

    kuruman

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    In what direction is the acceleration? Horizontal could be to the left or to the right. This will affect whether the friction is up the incline or down the incline. However, there are two coefficients of friction, static and kinetic. Which one is equal to tanθ and which one should appear in the answer? How did you choose your coordinate axes? I cannot tell from the equations you show.
     
    Last edited: Sep 15, 2016
  4. Sep 15, 2016 #3
    The friction in this case would be static. Also, acceleration is to the right, so right is positive and up is positive. The friction will be towards the top of the inclined plane.
     
  5. Sep 15, 2016 #4

    kuruman

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    Which way is downhill for the incline? To the right or to the left?
     
  6. Sep 15, 2016 #5
    To the right
     
  7. Sep 15, 2016 #6

    TSny

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    Did you mean to type "Assuming that ##\tan \theta > \mu## ...?
     
  8. Sep 15, 2016 #7
    Oops, yes.
     
  9. Sep 15, 2016 #8

    Simon Bridge

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    It's not surprising, your working does not seem to involve anything physical. The maths needs to be motivated from your understanding of the physics... this also helps you check your answer.
    You should make a statement at the start about the physical conditions needed for the block to not slide down the wedge (hint: talk in terms of the sum of forces).

    If I define x to be horizontal, and y upwards, and do my working in the lab frame, then to stop the block falling I need to provide an upwards force equal and opposite gravity. That way, provided there was no initial vertical component, the only non-zero acceleration will be horizontal. I can do this by directly pushing upwards on the block, with force mg, and so have no net acceleration at all ... but that option is not available ... instead I am forced to push on the block at an angle... that help?

    This is the same sort of physics as a sailboat tacking at an angle to the wind.
     
  10. Sep 15, 2016 #9

    kuruman

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    I assume that when the problem says "the weight is given a horizontal acceleration a" this means that a horizontal force F is acting on the block. This force is necessary since the block is initially at rest. You need to include that force in your equations because ma is not a force. Also, since the block is moving, which of the two coefficients of friction do you think should appear in the equations? That's why I asked which one is given. Finally, you need to consider that the acceleration has both an x and a y component in the horizontal-vertical coordinate system.
     
  11. Sep 15, 2016 #10

    TSny

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    To understand why your answer gives you the minimum acceleration, you need to understand why you wrote ##f = \mu N## for the friction force. Under what condition does this formula give the static friction force?
     
  12. Sep 16, 2016 #11
    ##f = \mu N## when the static friction force is at a maximum, and relative motion ensues, right?
     
  13. Sep 16, 2016 #12

    TSny

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    Yes
    Relative motion need not happen, but ##f = \mu N## does indicate that the object is "on the verge of slipping".
     
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