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Dynamics: finding the coefficent of friction.

  1. Oct 25, 2009 #1
    1. The problem statement, all variables and given/known data

    The question:
    It takes a 5.0 N force to pull a 2.0kg object along the ground. What is the coefficient of Friction.

    Its Pretty Blunt, and i have been stuck on it for about 10 minutes, There must be a step that i dont know or am missing because i get stuck


    2. Relevant equations

    Sum of the Forces = m*a
    Force of Friction = Coefficient of Friction*Force Normal
    Fg=m*a

    3. The attempt at a solution

    There must be a equation i am missing, to find out the Force of Friction
    heres where i get to:
    Fp= 5N
    m=2kg
    Fg-Fn= 0
    Fg= 19.6
    Fn= 19.6

    I would be extremely grateful if i could get some help on this question
     
  2. jcsd
  3. Oct 25, 2009 #2
    It takes a 5.0 N force to pull a 2.0 kg object along the ground. What is the coefficient of friction? 0.26
    http://www.mouatonline.com/Teachers/BHutchinson/Physics/Phys11/Unit%203/PS36.htm" [Broken] you have the solution (No procedure). It seems that you don't take into consideration acceleration. Maybe 5 N is the force that moves the body from the rest position. Google a little bit, problems with similar text say that body is moving with constant velocity.
     
    Last edited by a moderator: May 4, 2017
  4. Oct 25, 2009 #3

    Andrew Mason

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    [tex]F = \mu_kN[/tex]

    where F is the force of friction and N is the normal force. [itex]\mu_k[/itex] is the ratio of friction force to normal force. You are given the friction force (ie. the applied force is equal and opposite to the kinetic friction force as there is no acceleration). You can work out the normal force from the mass. Divide the former by the latter.

    AM
     
  5. Oct 25, 2009 #4
    Thanks, could you please show me the formulas and methods used to find 0.26. Its the right anwser but i still dont know how to do it.
     
  6. Oct 25, 2009 #5
    How did you get acceleration for [tex]\Sigma[/tex]forces = m*a
     
  7. Oct 25, 2009 #6

    Andrew Mason

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    ?? I gave you the formula!! It is:

    [tex]F = \mu_kN[/tex]

    The normal force exactly balances the gravitational force, mg. So:

    [tex]F = \mu_kmg[/tex]

    [tex]\mu_k = F/mg = 5/2*9.8 = .26[/tex]

    AM
     
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