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Particle on a slope with friction question

  1. Feb 18, 2015 #1
    Hi guys! Im really struggling with this poorly written question. Any help would be appreciated!



    a 3.5 kg box is on a slope. What is the minimum angle which will cause the box to slip if μ (friction) =0.6? If the plane is tilted to an angle of θ above this slipping value what is the acceleration on the box when θ is 10 degrees and -10 degrees (below). How long will it take the box to stop if θ = -10 degrees with an initial velocity of 2 m/s and with θ = 10 degrees how far will the box slide in that time starting from rest?
     
  2. jcsd
  3. Feb 18, 2015 #2

    PeroK

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    Can you try the first part about finding the minimum angle?
     
  4. Feb 18, 2015 #3

    Svein

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    This is standard mechanics. m*g is straight down, the friction is dependent on the force normal to the slope and the accelerating force is parallel to the slope.
     
  5. Feb 18, 2015 #4
    If i was given the angle i can easily calculate the forces parallel and perpendicular to the slope ie mgcosθ for perpendicular resultant force. I just havent a clue how to go about this question! Dynamics are not my forte.
     
  6. Feb 19, 2015 #5

    Svein

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    Yes - just continue...
     
  7. Feb 19, 2015 #6

    haruspex

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    Be careful about reusing symbols with different meanings. I assume here you mean theta as the angle of an arbitrary slope, but in the question it is the additional angle of the slope beyond a critical angle.
     
  8. Feb 20, 2015 #7
    I can get you started.
    The point at which it breaks stasis is when the forces up and down the incline are equal.
    So:
    Gravitational force = friction force
    ( m * g * sine ( incline angle ) ) = ( m * g * cosine ( incline angle ) * friction coefficient )
    Transpose for friction coefficient
     
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