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What is the maximum force of static friction for the block?

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data
    A string is tied to a 4.4 kg block and 120g hanging bucket. Students add 20g washers one at a time to the bucket. The student are unaware that the coefficient of static friction for the block on the table is 0.42.
    A) what is the maximum force of static friction for the block?
    B) how many washers can the students add to the bucket without moving the block?

    2. The attempt at a solution
    a)FssFn
    = (0.42)(4.4x9.8)
    Fs= 18.1N

    b) Ft-Fs=ma
    Ft-18.1=4.4a
    Ft=4.4a+18.1
    Ft=1.6N

    Ft-Fg=m(-a)
    4.4a+18.1=1.176=0.12a
    4.52a=16.924
    a=3.74m/s/s

    I don't know what the next step after this would be. Right now I know that the hanging mass has a tension of 1.6N so another 16.5N can be added to the tension force to reach the maximum static force.

    I also tried this:
    Fg=mg
    16.5=(0.12+0.02x)(9.8)
    16.5=1.176+0.196x
    15.324=0.196x
    x=78 washers

    The correct answer is 86 washers, but I don't know how to get to this answer.
     
    Last edited: Oct 27, 2013
  2. jcsd
  3. Oct 27, 2013 #2

    NascentOxygen

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    Staff: Mentor

    Should be 18.1=(0.12+0.02x)(9.8)

    The friction is overcome by the (bucket + washers). The string just conveys the force, the frictionless pulley changes the direction of the force. Acceleration is 0 until the friction is overcome.

    You should be able to do it now. :smile:
     
  4. Oct 27, 2013 #3

    cepheid

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    The thing that confuses me about your solution is this bit in red here. How did you come up with this number of 1.6 N? It seems like you assumed a value for 'a' (that was wrong).

    The solution is much simpler than you are making it. You are correct that the only two horizontal forces acting on the block are static friction, and the force due to the tension in the rope. If the tension force is *just about* to overcome static friction, what must be true about the value of Ft as compared to Fs, and what is the value of 'a' in that case? (EDIT: NascentOxygen already gave you the answer to that last question about 'a'). The answer to these questions tells you all you need to know.
     
  5. Oct 27, 2013 #4
    I still do not understand why Fg=18.1. I know that the hanging mass must have a maximum net force of 18.1N without the block moving. However, shouldn't Ft-Fg=18.1 because those are the two forces acting on the hanging mass. Or is Fg=18.1 because the acceleration is zero, therefore so is the net force and as a result Ft= -Fg and they both cancel each other out which doesn't cause the object to move. Is that correct?
     
  6. Oct 27, 2013 #5

    cepheid

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    Yes.
     
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