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Kinetic and Static Friction

  1. Feb 27, 2008 #1
    [SOLVED] Kinetic and Static Friction

    I've done so many of these but for some reason can't figure this one out. Any help iwll be greatly appreiciated. Here is the problem:

    A book of mass M = 0.55 kg rests on a table where the coefficient of static friction μs = 0.48. What is the minimum horizontal force needed to move it?

    So I draw my free body diagram and get this


    where N is normal force, P is horizontal force

    P = 2.59 N This part is correct

    Now for the part I can't solve.

    A different force of F = 3 N now acts on the book at an angle of θ = 30 deg above the surface of the table. What is the acceleration of the book if the coefficient of kinetic friction μk = 0.25?

    I've tried this part two different ways. First I checked for movement

    Fy+mg = N
    f = μs*N
    Fx - f = ma

    -.712 N = ma since friction force > applied force, the book doesn't move so acceleration is 0. However, they don't like that answer.

    So now I assumed the book must already be in motion when the force is applied which is why they give me a kinetic friction coefficient.

    .874 N = ma
    a = 1.59 m/s^2

    They don't like this answer either. Can anyone helpshed some light on what I"m doing wrong? Thanks so much in advance.
  2. jcsd
  3. Feb 27, 2008 #2

    Doc Al

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    Staff: Mentor

    Check your signs (and the normal force). Does the applied force act upwards at 30 degrees?
    Last edited: Feb 27, 2008
  4. Feb 27, 2008 #3
    You should not put m*a and static friction force in the same equation. Even if you
    get the right result, it is conceptually wrong, since static friction is present only if there is no motion.

    Exact answer is:

    If the book did not move when the force was applied, then it won't start to move.
    If the book moved when the force was applied, then it accelerates with a (the value you got in second point).
  5. Feb 27, 2008 #4
    Well, I know that ma = 0 from no acceleration and the book being at rest. I see your point and will use 0 from now on.
  6. Feb 27, 2008 #5
    How I am reading it is the force acts downwards. In my picture down and to the right. Since there is no acceleration in the y direction, Fy+mg = N. Fy and mg act down and N acts up. Is this correct?
  7. Feb 27, 2008 #6
    Why do you read it like that? IF that were the case, then what you typed would be right, but it's not
  8. Feb 27, 2008 #7

    Doc Al

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    Staff: Mentor

    In that case, what's the maximum static friction versus the horizontal component of the applied force? (Related to your first answer.)

    Lacking a diagram, and in order to have some kind of acceleration :wink:, I interpreted the force as upwards:
  9. Feb 27, 2008 #8
    Also, you should differ between maximum friction force and real friction force.
    Real friction force f is the one you get from Fx-f=0. Maximum friction force fmax is the one you get from fmax=k*N. The required condition for no movement is:

  10. Feb 27, 2008 #9
    Thank you for everyone your help. The applied force is acting up as you suggested.

    "A different force of F = 3 N now acts on the book at an angle of θ = 30 deg above the surface of the table." I can see this being read two ways. What it tells us for sure is that the angle is 30 deg when measured from the horizontal. However, does wording such as, "above the surface of the table" always refer to the direction of the force? Since no picture is given you'd think they would try to clarify a bit. Anyway, thanks again.
  11. Feb 27, 2008 #10
    I don't know why you guys are acting like it's so ambiguous unless the meaning of "above" has changed drastically since last I checked
  12. Feb 27, 2008 #11
    Well, if I was standing and pushing "down" on the book at a 30 deg angle, it is my understanding the force would still be acting from above the surface of the table. If this is not the case then the wording is fine. Ambiguous or not, I'm happy for the help.

    The wording in question:
    "A different force of F = 3 N now acts on the book at an angle of θ = 30 deg above the surface of the table."
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