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Newton's 2nd Law with Friction

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A man standing in an elevator that is moving with a constant acceleration holds a 3-kg block B between two other blocks in such a way that the motion of B relative to A and C is impending. Knowing that the coefficients of friction between all surfaces are μs = 0.30 and μk = 0.25, determine (a) the acceleration of the elevator if it is moving upward and each of the forces exerted by the man on blocks A and C has a horizontal component equal to twice the weight of B, (b) the horizontal components of the forces exerted by the man on blocks A and C if the acceleration of the elevator is 2.0 m/s2 downward

    2rzammx.png

    2. Relevant equations

    newton's 2nd law
    sum of the forces in x = 0
    sum of the forces in y = ma

    3. The attempt at a solution

    I am stuck on part a. I tried drawing a FBD of block B to find the acceleration of the elevator but it didn't work out.
     
  2. jcsd
  3. Oct 9, 2009 #2

    rl.bhat

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    When the lift is moving up with an acceleration a, the net acceleration is ( g + a)
    When the lift is moving down with an acceleration a, the net acceleration is ( g - a).
    In part (a), find the net frictional force on B and equate it to the apparent weight of B to find a.
     
  4. Oct 9, 2009 #3

    tiny-tim

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    Hi Oblivion77! :wink:
    It should have done :confused: … F = ma should give you a.

    Show us your full calculations, and then we can see what went wrong. :smile:
    Surely the acceleration of B is always a? :confused:
     
  5. Oct 12, 2009 #4
    Hi there,
    I did the work assuming that block B was not moving i.e. force < UsN where N was the force applied horizontal by blocks A and C and got answers for both parts A, and B.
    How would I know that block B was not moving in which case i would use Uk instead of Us. My prof said I would have to assume it was not moving and test to see if F <= UsN if not then F= UkN, but don't know how I would do that in this case.

    Thanks!
     
  6. Oct 13, 2009 #5

    tiny-tim

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    Hi mannie! :smile:

    (are you the same person as Oblivion77? have a mu anyway: µ and try using the X2 tag just above the Reply box :wink:)

    Just calculate F as the vertical force necessary to keep B stationary …

    then if F ≤ µsN, the available force (of static friction) is sufficient … if not, then it isn't, and some other force would be needed (and of course that other force would have to make up the difference, not between F and µsN, but between F and µkN). :smile:
     
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