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Homework Help: Coefficient of friction force

  1. Sep 27, 2004 #1
    The coefficients of friction between a 1kg mass and a surface are ms=.54 and mk=.46. Assuming the only other force acting on this block is that due to gravity, what is the force of friction (give magnitude and direction) on block:

    -if the block is at rest and the surface is horizontal?
    would this just be f=(ms)(N), and then N would equal (mass of block *gravity)?
  2. jcsd
  3. Sep 27, 2004 #2
    I believe so, yes: [itex]F_{F}=\mu_{s}F_{N}[/itex], where [itex]\mu_{s}[/itex] is the coefficient of static friction.
  4. Sep 28, 2004 #3
    ok so the Force of friction would be F=(.54)(1kg)(9.8m/s^2)=5.3 N to the left?

    also, if the block is at rest and the surface is inclined at 17.5 degrees, would the force then be f=(Ms)(mgcos(theta)), where theta equals 17.5?
  5. Sep 28, 2004 #4


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    Is there any force on the block other than gravity ? If there isn't, your answer would NOT be correct. Where did you get "left" from ?
  6. Sep 28, 2004 #5
    Well we're told to give magnitude and direction, but since the block isn't moving I guess there is no direction, so all I would give is the magnitude.
  7. Sep 28, 2004 #6
    if the block is at rest and there are no forces other than gravity, then there would be no friction because its not acting against anything
  8. Sep 28, 2004 #7


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    Always draw a free-body diagram for problems like this.

    If there is any net (resultant) force on the body, it will accelerate in the direction of that force. For a body to be stationary, there must be NO net force.
  9. Sep 28, 2004 #8


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    static friction

    The correct expression is
    [itex]F_{static} \leq \mu_{s}F_{N}[/itex].
    Interpretation: "The static fricition force is whatever it needs to be to prevent sliding between the surfaces... up to some maximum strength [itex]F_{static,max} = \mu_{s}F_{N}[/itex]."

    Generally, [itex]F_{static} [/itex] is less than [itex]\mu_{s}F_{N}[/itex] unless "the object is just about to start sliding", when it is equal to [itex]\mu_{s}F_{N}[/itex].
  10. Sep 28, 2004 #9


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    Static friction is a limit, it does not have a definite value: [tex]F_s\leq{\mu_sN}[/tex].
    Its value is just enough to keep the body in rest till the resultant of the other forces would exceed [tex]\mu_sN[/tex].

    If the body is in rest on a horizontal surface than there are two forces normal to the support, one is gravity and the other is the normal force. These two forces cancel each other. The friction would counteract against a horizontal pull or push, but there is no such force now. So the force of static friction is zero.

    If the body is in rest on an incline the normal force is [tex]F_N=mg\cos(\phi ) \mbox{ and the force along the incline is } F_i = mg\sin(\phi)-F_s [/tex]
    If the body doesn't move [tex] F_i=0 \mbox{ so } F_s=mg\sin(\phi)[/tex].
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