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Amount of force to pull a sled up an icy slope

  1. Sep 27, 2011 #1
    1. The problem statement, all variables and given/known data

    You're using a rope to pull a sled at constant velocity across level snow,with coefficient of kinetic friction 0.050 between sled and the snow. The sled is loaded with physics books, giving a total mass of 19 {\rm kg}.

    2. Relevant equations

    F=\mu*n

    3. The attempt at a solution

    I tried solving for the force and the force of friction seperately then subtracting between them... Not sure if that's correct.
    EDIT: I'm sorry I'm on the second part of this question which asks for the same force but at a slope of 30 degrees above horizontal.
     
  2. jcsd
  3. Sep 27, 2011 #2
    Okay, a free body diagram (FBD) would be useful here, if you haven't yet drawn it.

    Remember the equation for friction involves the normal force, not the weight. What would be the normal force when the sled is on a slope?

    When you have the friction force, what other force is there that you need to overcome if you are pulling the sled up a hill?
     
  4. Sep 27, 2011 #3
    The normal force would just be cos30*g right? Since weights not a factor in the friction...?

    The other force beside the frictional is the tension of the rope being pulled which would factor in m*g and cos30?
     
  5. Sep 27, 2011 #4
    I guess my statement came out a little confusing. Weight is a factor in friction only to the point that it contributes to the normal force. So the normal force would be

    [itex] F_N = m g\cos{(30)}[/itex]

    Remember to check your units - g is an acceleration. It needs a mass to become a force.

    And you are applying the tension, but what force are you overcoming? The tension you provided would overcome this force.

    Take a look at your FBD. You should have the friction, the normal force, the tension on the rope and one other force.
     
  6. Sep 27, 2011 #5
    The final force being gravity
     
  7. Sep 28, 2011 #6
    hi

    if the applied force is 30 degrees above the horizontal then the upward component would be
    [tex]F_{app}\sin(30)[/tex]. now in FBD , in the vertical direction there are 3 forces, normal force N is acting on the sled upward , [tex]F_{app}\sin(30)[/tex] is acting on the sled upward and force due to gravity , mg , is acting on the sled downwards. since there is no net force in vertical direction,

    [tex]F_{y-net}=0[/tex]

    use this to solve for the normal force N.
     
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