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Coefficient of Friction involving weights

  1. Apr 2, 2008 #1
    1. The problem statement, all variables and given/known data
    **picture located as attachment**

    2. Relevant equations
    Tccw= Tcw

    3. The attempt at a solution

    My biggest problem is that I don't know what components are needed to find the coefficient of friction. I am able to find things like the tension/force of the string, but I'm not sure if or how this can used to find the coefficient of friction. This is an extra credit problem, and any help is most welcome! :!!)

    Attached Files:

    • D2.doc
      File size:
      47 KB
  2. jcsd
  3. Apr 3, 2008 #2
  4. Apr 3, 2008 #3


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    Welcome to PF!

    Hi Erickimiscool! Welcome to PF! :cool:

    Can you just briefly describe the picture?

    (It'll be a lot quicker than waiting for your .doc to be approved.)

    Then we can give you a hint as how to go. :smile:

    (the coefficient of friction is usually the sideways frictional force on the object divided by the perpendicular normal force from the object.)
  5. Apr 3, 2008 #4
    Thanks for the welcome! So, I uploaded the picture to ==>


    Don't know if it'll work so I'll describe the picture as best as I can! The picture is representative of a right triangle. There is a piece of string attached to the wall, and attached to the string is a ruler. When the ruler is perpendicular to the wall, a right angle is formed. The string is the hypotenuse, ruler base, and the wall is the last part of the triangle. In addition to the picture is a weight. It is attached to the ruler, so that when the ruler is perpendicular to the wall, it is able to support itself, without any intervention. What I need to find in the problem is the coefficient of friction where the ruler meets the wall, or the 90 degree angle.

    Additional Info:
    -where ruler meets string= 40 degree angle
    - weight is 37.5 N
    - ruler length: 1.015 meters

    ~Hope that makes sense
    Last edited: Apr 3, 2008
  6. Apr 3, 2008 #5


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    You forgot to include the point on the ruler at which the weight is attached.
  7. Apr 3, 2008 #6
    its .25 meters from the right hand side of the triangle
  8. Apr 3, 2008 #7
    I am assuming that the whole state is under equillibrium since nothing else [like angular velocity or anything] has been mentioned.

    In that case, you need to draw a free body diagram. Here there are two unknowns with you: The Tension in the string and the frictional force. So, we can use two conditions, that of transitional equillibrium and rotational equilibrium to calculate these two values.

    also, the tension can be broken into two components, One along the ruler (or parallel to it) and one perpendicular to it. The parallel component will not produce any torque on the ruler, but the perpendicular will contribute fully. However, do note that the parallel component will provide us with the needed Normal force on the ruler, which is necessary for friction to act.

    It is a good idea to compute Torque along the point where atleast one force acts [the weight, the frictional force or the Tension]. When you do so, the force does not give any torque to the ruler and hence it can simplify the problem a bit.

    Get two equations from these conditions, solve it and you shall have your answer. This diagram should help you:


    P.S: IMHO, this is a nice question...
    Last edited: Apr 3, 2008
  9. Apr 3, 2008 #8
    I solved for perpendicular tension and net torque, where do I go from there?
  10. Apr 3, 2008 #9
    Did you get the value of the frictional force?
  11. Apr 3, 2008 #10
    umm how exactly do I do that??
  12. Apr 3, 2008 #11
    Could you please provide the equations you've come up with and the values for the forces you've already computed? I'm not able to comprehend what exactly have you done till now..
  13. Apr 3, 2008 #12
    ok, so I did Tccw=Tcw
    Tsin40(1.015)= 20(.5075)+(37.5)(.25)
    ...=29.93 N and thats the tension in the string

    Tnet= -29.93(sin35)(.25)+ 20(.5075)+ 37.5(.25)= 19.78 N-m
  14. Apr 3, 2008 #13
    What exactly did you do 20(0.5075) for?? The equations i got were:

    For transitional equillibrium:

    T \sin{\theta} - 37.5 - f_s = 0

    and, for rotational equillibrium

    (1.015)\mu T \cos{\theta} + (37.5)(0.35) = 0

    [Net torque calculated along the point where the string is fastened to the ruler]. Using the fact, [itex]f_s = \mu T \cos{\theta}[/itex]

    did you get something like this?
  15. Apr 4, 2008 #14
    for some reason i accidentally inserted a middle force of the ruler :/

    for the most part my calculations were on the right track, but I was totally off on the rotational equilbirum
  16. Apr 4, 2008 #15
    thanks for the help I really appreciated it, I never expected such a fast response ot my question! :)
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