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DE for modelling motion of a box

  1. Feb 23, 2015 #1
    So if you were to slide a box-like object down a straight ramp with a single gradient (its literally just an incline no curve). And you attempt to model it accurately would this be correct?
    Given the DE
    where a=dv/dt, mg=force of gravity, kv= force of friction

    Anyway my main question, is k the coefficient of kinetic friction? Does k=μ????
  2. jcsd
  3. Feb 24, 2015 #2

    Suraj M

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    What is µ according to you? Coefficient of static friction? then no ,k≠ µ
  4. Feb 24, 2015 #3


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    I think you've got your equations mixed up.

    ##ma = mg - kv##

    Is for a body falling under gravity with air resistance proportional to its velocity.
  5. Feb 24, 2015 #4

    Suraj M

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    k=##6 \pi \eta r## ??
    then what about the ramp?
  6. Feb 24, 2015 #5


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    The "coefficient of friction", [itex]\mu[/itex] is the number that multiplies the normal force on an object to give the friction force. An object of mass m, has weight mg. On a ramp making angle [itex]\theta[/itex] with the horizontal, we can divide the weight into components parallel to and normal to the ramp. The component normal to the ramp is [itex]mg cos(\theta)[/itex].
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