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Friction on lab table

  1. Oct 21, 2004 #1
    Q.
    A physics student is pushing a lab table across the floor. The table as a mass of 36 kg. The coeffient of static friction between the table and the floor is μs = 0.21 and the coefficient of kinetic friction is μk = 0.16.

    a) She initially pushes with a force of 64 N. What is the magnitude of the force of friction between the table and the floor?


    b) She now pushes with a force of 69 N. What is the magnitude of the force of friction between the table and the floor?


    I know that |Fs fric.| = Us . N, and

    |Fk fric. | = Uk . N

    For the first part I did as follows:

    |Fk fric. | = Uk . N
    We know the mass of the student so we can calculate the weight. N=mg.

    Then Fk fric. , would be Uk.N

    Similarly the second part too, but the answers are wrong. Don't know what is wrong with my approach.

    Help Requested!
    Thanks
     
  2. jcsd
  3. Oct 21, 2004 #2

    Doc Al

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    First thing to figure out: Does it move?

    A good thing to know about static friction: [itex]\mu N[/itex] is the maximum value it can be. It will be whatever it needs to be to prevent sliding---up to that maximum value.
     
  4. Oct 22, 2004 #3
    I tried several things but still the answer is wrong.

    Don't even have the slightest idea how to get started? Can anybody out there help me! :cry:

    Here is what is did !

    divided applied force by mass of table. , found the acc. Tried to use that in the formula . F = Uk .N

    Still the answer is wrong.!! :confused:
     
  5. Oct 22, 2004 #4

    arildno

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    Have you tried Doc Al's suggestions yet?
     
  6. Oct 22, 2004 #5
    Yeah, I do now, that UN is the max. value the table can hold, but the given value is a lot lesser. I really don't know if the table is moving or not.
     
  7. Oct 22, 2004 #6

    arildno

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    Can you please state the value of MAXIMAL static friction?
     
  8. Oct 22, 2004 #7
    Us = 0.21 in the question.

    Us, max. may be Us . Fn
    (0.21) . (36kg)*(9.81)
    = 74.16 N
     
  9. Oct 22, 2004 #8

    arildno

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    Yes, so what does that tell you?
    In particular, will the block be at rest or moving?
     
  10. Oct 22, 2004 #9
    I think the block would be at rest. That's all I now.
     
  11. Oct 22, 2004 #10

    arildno

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    Correct, because it is possible to have a (non-maximal) static friction of value 64N, right?

    So, is it possible to have a static friction force of 69N, so that the block will also remain at rest in the next example?
     
  12. Oct 22, 2004 #11
    No, it is not possible for both parts a and b, the answers all simply the applied forces in the question. Thanks, for guiding me step by step for this problem.

    Thanks, very much!!!!!!!!!!!!
     
  13. Oct 22, 2004 #12
    But , now I have a problem with the following parts:

    c) She now pushes with a force of 80 N. What is the magnitude of the force of friction between the table and the floor?

    d) She now pushes with a force such that the table moves with a velocity of 1.4 m/sec. What is the magnitude of the force of friction between the table and the floor?

    e) She now pushes with a force such that the table accelerates at 0.9 m/sec2. What is the magnitude of the force of friction between the table and the floor?
     
  14. Oct 22, 2004 #13
    Yeah, Found answers to part c and d ,

    Since the table is moving now we can use Uk to find out the force of friction.

    (.16) * (36) * ( 9.81 ) = 56.05 for part c, The applied force is greater than the MAXIMAL Static friction, which was 74.16 N

    and for part d too the answer is 56.05 N
     
  15. Oct 22, 2004 #14
    I am trying to figure out part e.

    The tables accelarates at 0.9 m/s2. How would we calculate the force of friction now.

    we know Uk is 0.16.
     
  16. Oct 22, 2004 #15
    I am trying to figure out part e.

    The tables accelarates at 0.9 m/s2. How would we calculate the force of friction now.

    we know Uk is 0.16.

    The answer again is 56.50 ,same as c and d. Finally got it all, took more than 2 hrs. to do this problem....

    THanks,
    Naeem
     
  17. Oct 22, 2004 #16
    A problem

    !. Masses m1 and m2 rest on a table 1.4 meters above the floor and are attached to m3 via very light string and a frictionless pulley as shown above. The coefficient of static friction between the m2 and the table is μs = 0.21 and their coefficient of kinetic friction is μk = 0.16. m2 = 48 kg and m3 = 24 kg.

    m1 is on top of m2 on the table in the picture, and m3 is hanging by means of a pulley in the -ve y direction. There is a chord which connects m2 via a pulley which also connects m3.

    Tried to apply Newtons laws to block 3 but no success so far.

    They are asking for.

    a) What is the minimum mass that m1 can have to keep the two blocks from sliding off the table?

    b) m1 is completely removed. What will be the acceleration of m3?

    Thanks,
     
  18. Oct 22, 2004 #17
    Hi, folks

    I was able to figure out part b)

    Here's what I did:

    Applied newtons laws as follows:

    for Mass m3 :

    +Ft + (24)(-9.81) = (24)a ----- ( 1 )


    applied newtons laws for m2

    Ft - Ff = (48)a

    Ft - ( .16 ) * (48 ) * ( 9.81 ) = (24)a ------ (2)

    Solved 1 and 2 and got accelaration of m3 as -2.22 which is correct

    My ideas was that if the system of m1 and m2 is moved to the right assuming it as the +ve x directon , m3 would move in the -ve y direction. so a for both the equations would be a = +a2 = -a3

    So, how can I figure out part a ) , how can I find the mass of m1. How can I apply Newtons Laws to m1.?????????


    Please help!!!!!!!




    Thanks,
     
  19. Oct 22, 2004 #18

    Doc Al

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    Hints for part a: Treat m1 and m2 as a single object. What's the tension in the cord? What must the friction force be on m2?
     
  20. Oct 22, 2004 #19
    I dont know if I am right, but here is what I did for part a .

    Ft-Ffric = ( m1 + m2 ) a

    Ft= ( m1 + m2 ) * a + Ffric

    = ( m1 + 48 ) * (-2.22 ) + Us ( m1 + 48 ) * (-9.81)

    That's all I now. Plz. help
     
  21. Oct 23, 2004 #20
    Here is what I did right now!

    For m1 not to slide, max horizontal force forward is Fs=uS*m1*g=m1*a
    So, the system acceleration cannot exceed the max acceleration for m1

    a = uS*g

    For the system

    m3*g - uK(m1+m2)g = (m1+m2+m3)a = (m2+m1+m3)(uS*g)

    rearrange, and collect like terms in m3

    m3*g - uS*m3*g = uS(m2 + m1)g + uK(m2 + m1)g

    common factor of g leaves

    m3(1 - uS) = (uS + uK)(m2 + m1)

    and, finally

    m3 = [(uS+uK)(m2+m1)]/(1-uS)

    I tried to use this equation to find minimum mass for m1 but doesn't work.

    Can anbody tell me what is wrong to the approach
     

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