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Value of friction with constant velocity

  1. Dec 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A block of mass "m" is pulled across a rough table by a string at an angle "θ" above the horizontal at a constant velocity. What is an equation for the magnitude of the kinetic friction force, fk?

    The answer that I was given was μk(mg-T*cos(θ))

    2. Relevant equations
    Fk=Fn*uk
    mg*cos(theta)=magnitude of force in x direction



    3. The attempt at a solution

    I just thought it would be mg*cos(theta)=uk*Fn

    Since there is no acceleration, the frictional force in the positive x direction must equal the force in the negative x direction...so, T*cos(theta), which is the same in this situation as uk*Fn.
    Why that weird answer? Anybody?
     
  2. jcsd
  3. Dec 15, 2015 #2

    SteamKing

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    For a block of mass m and acceleration due to gravity g, what is Fn?
     
  4. Dec 15, 2015 #3
    Oh, I see now...there are three vertical forces acting on the object. I forgot about the vertical y component of the pull. Thanks, SteamKing. And, it should be uk(mg-T*sin(theta)). not T*cos(theta). Appreciate it.
     
  5. Dec 15, 2015 #4

    SammyS

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    You are NOT being asked for the coefficient of friction. You're being asked for the magnitude of the kinetic friction force.

    μk does not need to be used here.
     
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