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Sign dilemma with coefficients of friction

  1. Feb 8, 2017 #1
    Oiy. It is my 6th year teaching physics and early on, I diverged from the textbook (Holt Modern Physics) regarding how they handled "g". To me, it made more sense that "g" was -9.81m/s2 , while the text handled it as "g" = 9.81 m/s2, and negatives are assigned directionally.

    Overall, my approach (I felt) simplified calculations.

    When we do sum of forces I handled it as actually adding forces on each axis and assigning + or - to the force as indicated by cartesian direction (since "g" is negative, this automatically makes Fg a negative number. Life is good!

    If on the y axis a=zero then ΣFy= Fg + Fn=0, therefore Fn is positive. Life is still good!

    When things begin to move on the x and we often default to the acceleration being in the positive direction, then Fk, is a negative. OK, seems reasonable.

    Then we get to μ!
    μk= Fk/Fn.
    When using this to create an expression for Fn, this almost always renders a negative normal force! This makes it difficult to just trust the signs and be careful with the algebra. Sigh.

    I muddle through it with some fudging every year, but I want to do better than that. Help! Should I just go back and train myself to handle it the way the text does? Is there some brilliant 3rd road that I am missing?
    Thanks in advance.
     
  2. jcsd
  3. Feb 8, 2017 #2

    Doc Al

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    I think it's a mistake to have "g" be anything other than the constant 9.81 m/s2.

    The vector Fg is always "down". The magnitude is simply "mg". Whether components are positive or negative just depends on your sign convention, which can vary for convenience.
     
  4. Feb 8, 2017 #3

    vela

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    The reason this doesn't work is because the relationship is between the magnitudes of the normal and friction force. You could work around this problem by pointing out the relationship is really ##\lvert \vec{F}_k \rvert = \mu_k \lvert \vec{F}_N \rvert##.
     
  5. Feb 9, 2017 #4
    Friction, like ALL non-conservative forces, ALWAYS do negative work, steal energy, and slow things down.
     
  6. Feb 9, 2017 #5
    Aha! That's the piece I was missing! I've actually done that, but I thought it was a cheat (and so did my students). Excellent, all is better now. Thanks!
     
  7. Feb 9, 2017 #6
    True, but I am thinking of instances when (from a calculations and signs perspective) the sign on a frictional force can be positive; example- an object sliding down a ramp toward cartesian negative, where Fgx is negative and the net acceleration is negative, but the friction is in opposition to both of those AND Fk is in the cartesian positive direction, so it makes sense to assign a positive sign to the frictional force. (At least that's how I've handled those instances... right? wrong? pragmatic?)

    Thank you!
     
  8. Feb 9, 2017 #7

    Doc Al

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    The sign of the friction force (its component along some axis) can certainly be negative or positive. But as vela points out, that is irrelevant for the definition of coefficient of friction, which uses magnitudes.
     
  9. Feb 9, 2017 #8

    robphy

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    Some conditions might need to be specified to claim that.

    A brick sits on the rough flatbed of a truck, initially at rest.
    The truck is then accelerated with a small acceleration to the right, with the block still sitting on that same spot on the flatbed.
    The person on the street observes that
    1) the brick is being accelerated by the horizontal net force.... the frictional force due to the truck's flatbed---that force points to the right.
    2) the displacement of the brick points to the right
    3) so, the work done by that frictional force is positive.

    ...but this example doesn't get to the heart of the question in this thread.
    As noted, @vela 's reply addresses it.

    Using g = -9.81 m/s^2 might be okay if all you do is solve kinematics and mechanics problems with your y-axis pointing vertically upwards. However, if you do more, i think it is better to emphasize that ##\vec g## is a downward pointing vector, whose magnitude is ##g=|\vec g|=9.81\rm\ m/s^2##, using the usual convention that removing the arrowhead denotes the magnitude. The value of ##-9.81## (with the negative sign) comes from asking for the y-component of ##\vec g## with your upward-pointing y-axis. However, while it may be mathematically convenient for many situations, the physics doesn't force you to use an upward-pointing y-axis or a rightward-pointing x-axis.

    Later, if you study the electric field, you'll see that the electric force ##\vec F_{elec}=q\vec E## is analogous to ##\vec F_{grav}=m\vec g##. (In general, this ##\vec g## refers to the gravitational field due to other masses... Near the surface of the earth, this ##\vec g## reduces to the so-called "uniform gravitational field" of strength 9.81 Newtons/kg.) There is value in maintaining consistency when viewing physics from a wider perspective.
     
  10. Feb 9, 2017 #9
    I second what was mentioned by @Doc Al: better to define g as the magnitude of the gravitational acceleration near earth's surface. Doing so simplifies the most basic introductory exercises by eliminating the negative sign and also can be used to stress the importance of establishing a coordinate system and sticking to it.

    I don't want to derail the conversation, but I take issue with this. I've seen this explanation elsewhere too. Work is only done by a force if the point of application of the force undergoes displacement. This does not happen in the case of static friction.
     
    Last edited: Feb 9, 2017
  11. Feb 10, 2017 #10

    robphy

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    Since the brick's kinetic energy changes, nonzero net work is done on the brick.
    The net work is the work done by the net force on the brick. Note the brick is accelerating.
    Now, draw the free-body diagram of the brick. Which forces on the brick contribute nonzero work to the net work?


    For simplicity, assume the brick is a point-particle. The point of application of the force is the location of the brick. The brick is undergoing displacement.
     
  12. Feb 10, 2017 #11
    I understand the argument and perhaps it is unnecessary to distinguish between connecting forces and ones that 'actually' do work. A related, though perhaps more complicated, question is "what is the work done by static friction when walking". Would you say that static friction does work here?
     
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