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Friction of a sliding box on a slope

  1. Jun 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A loaded penguin sled weighing 75 N rests on a plane inclined at angle θ = 21° to the horizontal (see the figure). Between the sled and the plane, the coefficient of static friction is 0.30, and the coefficient of kinetic friction is 0.18. (a) What is the minimum magnitude of the force F , parallel to the plane, that will prevent the sled from slipping down the plane? (b) What is the minimum magnitude F that will start the sled moving up the plane? (c) What value of F is required to move the sled up the plane at constant velocity?

    http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c06/fig06_29.gif

    2. Relevant equations
    fss Fn is the magnitude of the maximum static force.
    fkk Fn is the kinetic friction force.

    3. The attempt at a solution
    http://s1294.photobucket.com/user/jhoversten/media/20150626_193750_zpsvqv6akeu.jpg.html [Broken]

    i can see why my attempt at solving a is not working. thanks
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jun 26, 2015 #2

    Nathanael

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    You did it right, except you miscalculated FT and FN

    The angle is 21 degrees not 20 degrees.
     
  4. Jun 26, 2015 #3
    opps thanks
     
  5. Jun 26, 2015 #4
    im getting 4.646 N but this is still wrong.
     
  6. Jun 26, 2015 #5

    Nathanael

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    That's not what I get. Did you correct both FT and FN?
     
  7. Jun 26, 2015 #6
    oh yea i forgot to correct the right side of my equation. thanks
     
  8. Jun 26, 2015 #7
    for part b; i set up and equation identical to the equation in part a except is used μk and got an answer 14.276 N. And this would be the force required to start the sled moving up the ramp after the static force was applied. Im not sure how to set it up

    i added the force from part a to part b and that was not good.
     
  9. Jun 26, 2015 #8
    i initiall thought it would be the same answer as part a, because once the block is broken free from gravitational forces and static frictional force it should be ready to move with any increase in force applied right?
     
  10. Jun 26, 2015 #9

    Nathanael

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    μk does not come into play yet. You want to find the force F which puts it on the verge of slipping upwards, so you will still want to use μs

    It's similar to part a but it's not the same:
    Friction opposes the motion of the block; in part a, the block was about slip downwards. In part b, the block is about to slip upwards.
     
  11. Jun 26, 2015 #10
    oh right friction is a non conservative force so it works in both directions.
     
  12. Jun 26, 2015 #11
    Thanks Nathanael
     
  13. Jun 26, 2015 #12

    Nathanael

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    No problem
     
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