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Newtons' Second Law with Kinetic Friction Problem

  1. Apr 20, 2015 #1
    1. The problem statement, all variables and given/known data

    While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction between the box and the floor is 0.41. The pushing force is directed downward and an angle θ below the horizontal. When θ is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. Find the that value of θ.

    2. Relevant equations
    F = ma, fk = μk Fn

    3. The attempt at a solution
    I have tried working this problem multiple times, I must be missing something in the initial set up. I keep ending up with variables that don't cancel out, or I have given values for. (The only value given is for the coefficient for
    kinetic friction). The solution given is just an angle in degrees.

    I could really use a walk through of the solution to this problem. I'm missing some concept in my initial setup, and I just can't seem to see it.

    Thanks in advance.
    Rich.
     
  2. jcsd
  3. Apr 20, 2015 #2
    What if you try to make a free body diagram, from that I get

    ## \mu_k ( F_{1} sin(\theta) + F_n) < F_1 cos(\theta) ##

    Where ##F_1## is the pushing force, ##F_n## the normal force of the box
     
  4. Apr 20, 2015 #3

    CWatters

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    Out of interest is the book answer about 22 degrees?
     
  5. Apr 20, 2015 #4
    Well with that little info that you have, the ## F_k = \mu_k F_n ##, the friction force divided with the normal force ## \frac{F_k}{F_n} = \mu_k ##, means that the angle will become ##\theta = tan^{-1}(\mu_k)##. I can give you a more detailed explanation in a while.
     
  6. Apr 20, 2015 #5
    The actual answer given is 68o
     
  7. Apr 20, 2015 #6
    Sorry a miscalculation

    ## tan(\theta) = \frac{F_n}{F_k} = \frac{F*sin(\theta)}{F*cos(\theta)} = \frac{1}{\mu_k}##
    From this equation you can see that ##tan(\theta) = \frac{1}{\mu_k} \rightarrow \theta = tan^{-1}(\frac{1}{\mu_k})##
     
  8. Apr 20, 2015 #7
    This seems to be on the right track. The answer (by guessing, and trial and error) looks to be ##\theta = tan^{-1}(1/\mu_k)##.

    The problem is I don't know why. Can you post your Free Body Diagram? I think that is where I made my mistake.

    Thanks.
     
  9. Apr 20, 2015 #8
    That's the same answer I guessed at. But I still don't understand Why? It would help to see the FBD.
     
  10. Apr 20, 2015 #9
    rrhwg6.jpg

    ##F_y=F*sin(\theta)##
    ##F_x=F*cos(\theta)##
    ##F_n-F_y = 0## Forces in y-direction
    Due to the constant velocity the forces in the x-direction ##F_x-F_k = 0##
    ##\theta## is the angle between F and ##F_x##.

    j8gl1f.jpg
    The result force diagram will look like the picture above, from there it is just geometry.
     
  11. Apr 20, 2015 #10
    That looks "Almost" the same as mine. The only difference is that I included a vector for the Weight of the box. That also seems to be where my problem came from.
    Can you tell me why you didn't include the weight of the box?

    I'm pretty certain that is my mistake. I just don't understand why it should be excluded???
     
  12. Apr 20, 2015 #11
    Are you familiar with the concept of limits? I would also like to add that I think that you should include the weight of the box in the FBD, even if the weight of the box ultimately proves negligible.
     
  13. Apr 21, 2015 #12

    CWatters

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    Not "should" but "can"....

    There is a hint in..

    One of the equations contains...

    μk(mg + FSinθ)

    What can you do to simplify that if F is extremely large compared to mg?
     
  14. Apr 21, 2015 #13
    Yes, I'm familiar with the concept of a Limit. Unfortunately, I don't think that would be of any help here.

    This problem is from an Algebra based Intro to Physics text. Cutnell and Johnson, 9th edition to be exact.

    So, I think we are looking for an Algebra based solution.
     
  15. Apr 21, 2015 #14
    If F is extremely large it would make mg too small to matter. But I don't think that a large F is a requirement to the solution, It seems more of a way to emphasize that the angle is the critical issue. That's the way it reads to me, anyway.
    Thoughts??
     
  16. Apr 21, 2015 #15

    "no matter how large the pushing force is" means that F is very large, this means that F >> mg, as a result ##mg+F*sin(\theta) \approx F*sin(\theta)##
     
    Last edited: Apr 21, 2015
  17. Apr 21, 2015 #16
    Ok, I'm going to start fresh with this new way of reading the problem. I'll be back after I give it a try. Thanks for your input.
     
  18. Apr 22, 2015 #17
    In the context of an algebra-based book then I suppose it is reasonable not to analytically evaluate the limit. But you must make a mathematical agreement on closeness in order to solve this problem. There is no simple algebraic solution.

    Or maybe there is. It is certainly beyond me, however.
     
    Last edited: Apr 22, 2015
  19. Apr 22, 2015 #18
    Thanks. I agree. At least I know now I wasn't missing any obvious physics concept I should be learning. That is reassuring.
    Thanks again.
     
  20. Apr 27, 2015 #19
    I reworked the problem with your understanding of the meaning of "extra" information, and approaching it that way everything works out. Thanks for the help.
    Also, I have never encountered a problem where I had to apply information that way to solve the problem. It was very eye opening experience. When it was done, and I was rereading it, it looks almost beautiful in how it leads to the solution. Up until this problem, everything I had encountered was solved with straight formula use, or algebraic manipulation. I feel like I really learned a lot by getting through this problem. I couldn't have done it without your help. :smile::smile::smile: Thanks so much.
     
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