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Person dragging a bow with a rope - Calculate friction

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    The question says
    "A person is dragging a box with the mass of 60kg by a rope. The person begins to drag the box at 4 m/s [F]. All of a sudden the person sees on coming traffic. The person calculates that they have 3 seconds to get out of the way. If at the end of the 3 seconds the box is moving 10 m/s [F] and the coefficient of friction between the box and the ground is 0.7, find the force which the person is pulling the box."

    my biggest problem is I don't know what a coefficient is.

    2. Relevant equations
    im not to sure what equations i would need to use. My guess would be d=(vi)(t) + 1/2 (a)(t^2)


    3. The attempt at a solution
    To find out the force which the person is pulling the box would we still use
    F= m*a

    and we would need to figure out the displacement first with d=(vi)(t) + 1/2 (a)(t^2)
     
  2. jcsd
  3. Dec 6, 2009 #2
    Re: Forces

    The coefficient of friction, [tex]\mu[/tex] is the constant relating the frictional force between two surfaces to the normal contact force between the two surfaces.
    [tex]F_{friction} = \mu N[/tex]​

    The velocity of the box increases from 4m/s to 10m/s in 3s. I presume that should be enough for you to obtain the average force exerted on the box during this time period using the standard F=ma?
     
  4. Dec 6, 2009 #3
    Re: Forces

    so μ is 0.7?
     
  5. Dec 6, 2009 #4
    Re: Forces

    yep.
     
  6. Dec 6, 2009 #5
    Re: Forces

    How would we begin this question
    my known's are
    mass= 60kg
    vi=4 m/s [F]
    v2=10m/s [F]
    t= 10s
    friction=0.7

    but i dont know what equation to use to find the applied force
     
  7. Dec 6, 2009 #6
    Re: Forces

    you have time, v0 and vf. from there, you must solve for acceleration, then plug it into F = ma , which will be your net force. From there, you must consider all the different forces and single out the pulling force. (Hint: the pulling force is not all that is affecting the pull.. what does friction do?)
     
  8. Dec 6, 2009 #7
    Re: Forces


    friction is not 0.7, that's the coefficient of friction. Friction is the coefficient of friction multiplied by the normal force
     
  9. Dec 6, 2009 #8
    Re: Forces

    ok, and the normal force is equal to the mass right?
    so then it would be
    0.7 x 60
    which would then be
    42?
     
  10. Dec 6, 2009 #9
    Re: Forces

    the normal force is what's keeping the object "up", it is the sum of vertical components.. so in this example, you only have the force of gravity or the weight pushing it down to the ground, so the normal force is equal to the weight
     
  11. Dec 6, 2009 #10
    Re: Forces

    Alright. So then the normal force is 60, because the weight is 60kg.
    and then friction would be, 0.7x60=42
    is that right?
     
  12. Dec 6, 2009 #11
    Re: Forces

    no, mass and weight are two different things. I think that you should revise your concepts
     
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