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Friction force

  1. May 28, 2007 #1
    Suppose that an object is moving up an inclined plane, then stops and after moves down. There exists frictional force. How do you represent the frictional force when v=0, when it stops? Or it has a zero value? Thanks.
     
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  3. May 28, 2007 #2
    Frictional force is represented by [tex]F_f = \mu_k*N[/tex]. Neither mu nor N go to zero, so we know that the frictional force is still being applied.

    Even with this being stated, I am still confused. Frictional force does not require a differential equation, does it? Maybe I am just thinking of static friction. Blah!

    Edit: When v=0, the frictional force is equal to [tex]F_f = \mu_s*N[/tex]. This is when the block is at the tipping point of moving (which is is going to do), so I believe this is correct.
     
    Last edited: May 28, 2007
  4. May 28, 2007 #3

    AlephZero

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    Static friction [tex](v = 0)[/tex] and dynamic friction [tex](v \ne 0)[/tex] are different.

    For static friction the friction force can be in any direction, had has magnitude [tex]|F| <= \mu_sN[/tex]. You find the direction and magnitude of the friction force from the static equilibrium of the system (resultant force and moment = 0). If the friction force to maintain equilibrium is bigger than [tex]\mu_sN[/tex] then static equilibrium is not possible.

    For dynamic friction, the magnitude of the friction force is equal to [tex]\mu_dN[/tex] and the direction is opposite to the direction of motion.

    In general the static and dynamic coefficients of friction are not the same, and [tex]\mu_s > \mu_d[/tex].

    For an inclined plane, as the object moves up the plane there is a constant friction force acting down the plane. When the object stops, either there is a static friction force which balances the other forces on the object and it remains stopped, or that is not possible and it starts to slide down. If it slides down the friction force actis up the plane, to oppose the motion.
     
    Last edited: May 28, 2007
  5. May 29, 2007 #4
    Ok, but when body stops, and v=0, how to draw the frictional force if it exists? Pointing downward the plane or upward? Thanks for your interest in help.
     
  6. May 29, 2007 #5

    AlephZero

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    This is no different from drawing a vector to represent any unknown force. If you guess the direction wrong, when you solve the equations the magnitude will come out as a negative number. That's not a problem. The important thing is that you use the SAME direction as "positive" in ALL your equations when you sum forces, take moments, etc.

    As a simple example, it doesn't matter if you consider "weight" to be a force of magnitude +Mg acting downwards, or magnitude -Mg acting upwards. You get the same correct answer either way.

    If the only force on the block is its weight, you can see that if there was no friction, the block would slide down the plane. So the friction force on the block must be acting up the plane to stop that happening.

    But if there was an applied force trying to push the block UP the plane but not actually moving it, the friction force could be in either direction, or zero, depending on the size of the applied force.
     
  7. May 29, 2007 #6
    Ok, but the only forces that act on the body are is weight, normal plane reaction, and if moving, frictional force. My problem is when the body stops in the plane and after inverts its motion. When v=0 there exists a frictional force? If so it points to where, up or down? Note: there is no force applied trying to push the block UP. Thanks again.
     
  8. May 29, 2007 #7

    AlephZero

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    The friction force when it stops is up the plane.

    If you already know the block slides down again, "the direction of the friction force at the instant when it stops" doesn't really matter, because the block is stationary for zero amount of time. When the block changes direction, you can assume the friction force instantly flips from down the plane to up.

    What you are using here is Coulomb's law of friction which is only a model of reality. What "really happens" when the block changes direction is a lot more complicated - but the complicated version is irrelevant for solving this type of problem.
     
  9. May 29, 2007 #8
    Thanks AlephZero!
    Do you know a place in the net where i can learn the complicated version for the problem?
     
  10. May 29, 2007 #9

    AlephZero

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    There isn't "the" complicated version, there are several...

    Try Googling for papers on friction by Menq (mostly written in the 1980s).

    You could also try Google for "macroslip" and "microslip" friction laws.

    Coulomb's law works pretty well for large-scale motions and "small" loads - which is why it's still being used 200 years after Coulomb invented it.

    The theories I mentioned were developed to model friction for problems with small oscillating motion, e.g. friction between vibrating components, or between the parts of a bolted joint when the load on the structure varies with time, etc.
     
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