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Newton's laws and Friction

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data

    A 4.0 kg toboggan rests on a frictionless icy surface, and a 2.0 kg block rests on
    top of the toboggan. The coefficient of static friction between the block and
    the surface of the toboggan is 0.60, whereas the kinetic friction coefficient is
    0.51. The block is pulled by a 30 N-horizontal force as shown. What are the
    magnitudes and directions of the resulting accelerations of the block and the
    toboggan?

    2. Relevant equations
    Fnet = F-Ffr = ma
    Ffr = μk*Fn

    3. The attempt at a solution
    I found that the box will move ontop of the toboggan because μs *Fn = 11.76 N which is less than 30 N so the box will move.
    F-Ffr = ma
    30N - 10.0N = 2.0 kg (a)
    accel. = 10 m/s^2
    so i know thats the box on the toboggan. How do i find the toboggan on the ground?
    do i just do it as
    F-Fr = (Mb +Mt) *a
    30 - 10 = (2.0 + 4.0) *a
    20 = 6.0 *a
    a = 3.33m/s^2
    does that make sense at all? any help would be great.
    thanks
     
  2. jcsd
  3. Feb 24, 2009 #2

    Delphi51

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    I agree with your a = 10 for the box.
    But isn't the friction force of 10 N the only force on the toboggan? That is the "grip" the pulling system (including the box) has on it. I would apply F = ma to the toboggan with F = 10.
     
  4. Feb 24, 2009 #3
    I dont really understand the "grip" pulling system. I dont understand why the F = 10N. wouldnt the toboggan still move in the same direction as the box because 30N is greater than 10N?
     
  5. Feb 25, 2009 #4
    can someone please help me with this question? i just need to know if this is right or not or if im completely wrong and need to start again
     
  6. Feb 25, 2009 #5
    Have you drawn free body diagrams for these? It might help clear it up if you haven't...
     
  7. Feb 25, 2009 #6
    well im not sure about the FBD of the toboggan. i figured out the block one though. The above calcuations for the toboggan that i did i guess is not right?
    i figured that there would still be the force of 30N west and the force of friction against it, so the total force is 20N west.
    is the above correct?
     
  8. Feb 25, 2009 #7
    The 30N are only working on the block... what is pulling the toboggon forward?
     
  9. Feb 25, 2009 #8

    Delphi51

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    Block: - Friction + 30 = m*a
    Toboggan: Friction = m*b (where b is the acceleration of the tobagan)
    Friction = .51*2*g = 10
    so the block accelerates at 10 m/s^2 and the toboggan accelerates at 2.5.
     
  10. Feb 25, 2009 #9
    does that mean that the toboggan is moving east and the block is moving west? because i need to find the directions of the accelerations as well.
     
  11. Feb 25, 2009 #10

    Delphi51

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    Both move the same way. The way the 30 N pull is exerted. You haven't said which way that is.
     
  12. Feb 25, 2009 #11
    I answer your question with a question of my own:
    What way is the friction between the box and the toboggon acting on the toboggon?
     
  13. Feb 25, 2009 #12
    i guess the friction between the box and toboggan will act in the opposite direction on the toboggan and ground?
    little confused.
    how do i find the directions of the accelerations?
     
  14. Feb 25, 2009 #13

    Delphi51

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    The ground is not involved - frictionless.
    The friction between the box and the toboggan acts one way on the box and the other way on the toboggan. It opposes the motion of the box and causes the motion of the toboggan.
     
  15. Feb 25, 2009 #14
    delphi is right, accel would be in the same direction of friction because the ground is frictionless...
     
  16. Feb 25, 2009 #15
    ok so let me just sure i have this right
    accel box = 10m/s^2 to the left
    accel toboggan = 2.5 m/s^2 to the left???
     
  17. Feb 25, 2009 #16
    *friction acting on the toboggon
     
  18. Feb 25, 2009 #17
    yes, you would be correct
     
  19. Feb 25, 2009 #18
    thank you for the help :eek:)
     
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