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Tension and friction on an angle

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    An adult is pulling two small children in a sleigh over level snow. The sleigh and children have a total mass of 47kg. The sleigh rope makes an angle of 23degrees with the horizontal. The coefficient of kinetic friction between the sleigh and the snow is 0.11. Calculate the magnitude of the tension in the rope needed to keep the sleigh moving at a constant velocity.


    2. Relevant equations
    Fk=μFn
    Fg= mg
    ƩFx=ma(x)
    ƩFy=ma(y)


    3. The attempt at a solution

    Fn≠Fg
    Fn= Fgcosθ
    = mgcosθ
    = (47kg)(9.8m/s^2)cos23
    Fn= 424N

    Fg=mg
    =(47kg)(9.8m/s^2)
    Fg=461N

    Fk=μFn
    = (0.11)(424N)
    Fk=46.64N

    If it is to move at a constant velocity, there would be no acceleration
    ƩFx=ma(x)
    Ft+ (-Fk)=ma(x)
    Ft=ma(x)+Fk
    but if there is no acceleration then a=0
    Ft=Fk
    But, that's not the case.

    Ideas?
     
  2. jcsd
  3. Oct 5, 2011 #2
    Why don't you first compute the horizontal force needed to keep the sled in motion? You have the mass and you have gravitational constant. These mutliplied by the kinetic friction coefficient yields the force needed to keep it in motion with no acceleration.

    Then determine the resultant force on the rope. It must be such that when resolved into components the horizontal component is sufficient to maintain sled motion. Therefore, it will be larger than the horizontal force needed to maintain constant speed.
     
  4. Oct 5, 2011 #3
    μmg= Fs does it not?
    By finding the Fs I can determine the minimum horizontal force needed for the sled to move, but μ is for kinetic friction, I didn't think I could use that for static friction.
     
  5. Oct 5, 2011 #4
    Your problem says nothing about getting the sled started. It only asks for the tension to keep it moving at constant speed.

    "An adult is pulling ......"
     
  6. Oct 5, 2011 #5
    Ok, I'm just a little confused as to what you were trying to say before. μmg, where does that equation come from?
    (0.11)(47kg)(9.8m/s^2)
    =50.6N
    and what does that give me? is that the horizontal component of Ft?
     
  7. Oct 5, 2011 #6
    Weight multiplied by coefficient of friction gives the force needed to overcome friction and keep moveing the block at constant speed. It would be the horizontal component of the tension in the rope. So 0.11*47*9.8 = 50.6N. That's the horizontal force needed to keep constant speed.

    Not to complicate matters, but the mg force that is multiplied by friction coefficient MUST be normal to the surface. So if the sled were on a slope things would be different because the force normal to a sloped surface is less than what it is if the surface is horizontal such as with this problem.
     
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