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Calculate the coefficient of static & kinetic friction

  1. Mar 5, 2016 #1
    1. The problem statement, all variables and given/known data
    You are trying to slide a sofa across a horizontal floor. The mass of the sofa is 200 kg, and you need to exert a force of 350 N to make it begin to move.
    a) Calculate the coefficient of static friction between the floor and sofa
    b) after it starts moving, the sofa reaches a speed of 2.0m/s after 5.0s. Calculate the coefficient of kinetic friction between the sofa and the floor.

    Fa= 350N; m=200kg

    2. Relevant equations
    Fnet=ma; Fnet=Fa-Fs; Fnet= Fa-u*m*9.8

    3. The attempt at a solution: I tried but i got 35.7 for the static coeffcieint which is why i didn't tried b) please help.
     
  2. jcsd
  3. Mar 5, 2016 #2

    SteamKing

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    How is the static coefficient of friction defined? This is one formula which you should have included in Section 2 above.
     
  4. Mar 5, 2016 #3
    This is what I did.
    Fnet=0
    Fnet=ma
    Fnet=Fa-Fs
    so
    ma=Fa-u*m*9.8 both masses cancel out
    0=350-u9.8
    -350/9.8=-u
    35.7=u
     
  5. Mar 5, 2016 #4

    SteamKing

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  6. Mar 5, 2016 #5
    so u=f/N is the answer you are indicating to me? where f is static friction and n is the normal force. But the thing is i am only given force applied in the question not force of satic friction. But still this is something I tried again and got right answer but not sure if why the net force would be zero i just tried it randomly and got correct answer.
    Fnet= Fa-Fs
    0= 350-u(200)(9.8)
    -350/1960=-u
    0.178=u
    but this is just an answer for part a of the question i also need help with b)
     
  7. Mar 6, 2016 #6

    SteamKing

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    The sofa is not going to move by itself.

    The force of 350 N which is applied to the sofa is required to overcome the static friction, so one can reasonably conclude that f = 350 N.

    The sofa has a mass of 200 kg, and since it rests presumably on earth, one can calculate the normal force created by the weight of the sofa on the floor.

    Your equation for Fnet makes no sense because Fa and Fs are acting perpendicular to one another.

    As far as Part b is concerned, what calculations have you done to solve for the coefficient of kinetic friction?
     
  8. Mar 6, 2016 #7
    In my view it made sense because static friction goes against the motion which is force applied the attempted motion, and I don't understand how Fs could be upwards because it is a horizontal motion so help me how i should come up with the equation to calculate the coefficient of static friction. And by the way what f represents in your comment f=350, do u mean fnet=350 or fa=350, because it already says in the question that fa=350.
    as for part b) this is what I did:
    a= (vf-vi)/t
    = 2/5
    = 0.4
    Fa= ma
    = 200*o.4
    = 80Newtons
    Fnet=ma
    Fnet=Fa-Fs
    ma= 80-umg both masses cancel out
    a= 80- u 9.8
    (0.4-80)/9.8 = -u
    8.1=u but this wrong answer book answer is o.14
     
    Last edited: Mar 6, 2016
  9. Mar 6, 2016 #8

    SteamKing

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    The coefficient of static friction is the ratio of the friction force to the normal force acting on a object. The static friction in the case of the sofa is that force which resists the motion of the sofa across the floor. The normal force on the sofa is created by the weight of the sofa acting on the floor. For this particular problem, the friction force and the normal force are acting perpendicular to one another. You can't calculate a meaningful Fnet using only the scalar magnitudes of these two forces.

    Also, you don't seem to be aware that the coefficients of friction, whether static or kinetic, have values which are usually less than 1.

    If ma = 80 - μ m g, you cannot arbitrarily cancel m on both sides of the equation without calculating what 80 / m is. That's just simple algebra.

    Again, you are using this weird formula to calculate the coefficient of kinetic friction which didn't work for calculating the coefficient of static friction. The force of friction and the normal force are acting perpendicular to one another, so the Fnet you have calculated is meaningless.

    Remember, the sofa does not move by itself. You are still pushing with a force of 350 N after the sofa begins to move. Hint: once the sofa gets moving, what is the net force acting on the sofa in the horizontal direction?
     
  10. Mar 6, 2016 #9

    haruspex

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    As I read it, Belma is using Fs for static friction force and Fa for applied force.
    .... maximum frictional force ...
     
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