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Friction Force between the box and the floor?

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data
    A heavy box (200.0 kg) is sitting still on a warehouse floor, despite the efforts of a worker to push it (with a force of 300.0 N). If the coefficient of static friction between the box and the floor is 0.6, what is the friction force between the box and the floor? What is the frictional force between the box and the floor if another worker puts another box (also 200.0 kg) on top of the first box while the first worker continues to pull with 300.0 N?


    2. Relevant equations
    Not sure


    3. The attempt at a solution
    ok, so this problem deals with static friction. What is the equation for the friction force? I am assuming for the first part of the problem, you plug in the force and mass. Also, If you simply double the the weight in kg, doesnt the force quadruple?
     
  2. jcsd
  3. Sep 30, 2009 #2
    Draw a free body diagram for this problem and include all of the relavent forces.

    Friction force equals normal force times coefficiant of friction.
     
  4. Sep 30, 2009 #3
    so would that be 9.8 m/s2 times 0.6?
     
  5. Sep 30, 2009 #4
    Not quite. Hint: F=ma
     
  6. Sep 30, 2009 #5
    so 200kg(9.8) = 1960.. times that by 0.6 and you get 1176?
     
  7. Sep 30, 2009 #6

    tiny-tim

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    Welcome to PF!

    Hi IneedHelp:-)! Welcome to PF! :wink:
    No, Fr = µN = µmg, so if you double m, then you double the maximum static friction force also. :smile:
     
  8. Sep 30, 2009 #7
    Yes! There is your answer for the first part. :biggrin:
     
  9. Sep 30, 2009 #8

    PhanthomJay

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    The problem is asking for the friction force between the box and floor when the worker pushes on it with a 300 N force (presumably in a horizontal direction), and it doesn't move. Using one of Newton's laws will help to solve for it.
     
  10. Sep 30, 2009 #9
    ok so if you double the mass, then you double 1176 and that is the second part of the answer correct?
     
  11. Sep 30, 2009 #10

    PhanthomJay

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    No. You are not trying to determine in part 1 or 2 the max available static friction force. You are trying to determine the actual friction force. Hint: Use Newton 1.
     
  12. Sep 30, 2009 #11
    ah so would that mean the force is 0, bc there is no acceleration? thats the only thing i can come up with out of the first law clue you are giving me
     
  13. Oct 1, 2009 #12

    tiny-tim

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    No, it means the net force (the sum of all the forces) is 0. :wink:

    So what are all the forces? :smile:
     
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