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Static Friction and Frictional Force Ranking Task

  1. Sep 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Below are six crates at rest on level surfaces. The crates have different masses and the frictional coefficients [given as (μs, μk) ] between the crates and the surfaces differ. The same external force is applied to each crate, but none of the crates move.

    Rank the crates on the basis of the frictional force acting on them.
    Rank from largest to smallest. To rank items as equivalent, overlap them.

    Items:
    (mass in kg, μs, μk)
    A. 250, 0.2, 0.1
    B. 1500, 0.3, 0.1
    C. 500, 0.6, 0.3
    D. 750, 0.6, 0.5
    E. 600, 0.8, 0.5
    F. 750, 0.4, 0.3


    2. Relevant equations

    fs ≤ μs * m * g


    3. The attempt at a solution

    I used the equation fs = μs * m* g and got the answer wrong. Now I am just completely confused. Any help will be appreciated, thanks.
     
  2. jcsd
  3. Sep 21, 2012 #2

    TSny

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    Note the less than or equal sign. The static force of friction could be less than μs*m *g. So, you'll need to bring in another idea to determine what the actual force of friction would be for each block. Hint: Newton's 2nd law.
     
    Last edited: Sep 21, 2012
  4. Sep 21, 2012 #3
    F=ma

    there is no acceleration because the crates do not move, correct? so force would be 0 for all of them?

    still a little confused.
     
    Last edited: Sep 21, 2012
  5. Sep 21, 2012 #4

    TSny

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    The "F" on the left side of the second law represents the net force (i.e., the sum of all the forces acting on the object.)
     
  6. Sep 21, 2012 #5
    Ok, sooo

    F - μkmg = m * a

    there is no acceleration so

    F = μkmg

    So am I supposed to use μk instead of μs?
     
  7. Sep 21, 2012 #6

    TSny

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    No, there is not kinetic friction in this problem since the blocks don't slide. You don't need to use any specific formula for friction. If I push horizontally on a block sitting on a horizontal surface with 5 N of force and it doesn't move, what is the value of the friction force?
     
  8. Sep 21, 2012 #7
    5 N.

    how would I figure out the horizontal force in this case?

    is this equation correct?

    F - μs(m * g) = 0 ; where F is the horizontal force
     
  9. Sep 21, 2012 #8

    TSny

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    Right, 5N. That makes the net force zero which agrees with the second law since we know a = 0.
    In your problem, you don't need to figure out the horizontal applied force. Whatever it is, it's the same applied force for each crate. So, how would the friction forces compare for the various crates? (Again, μs*m * g is not relevant here since it represents only the maximum possible static friction force and you cannot assume that the forces of friction are at the maximum value.)
     
    Last edited: Sep 21, 2012
  10. Sep 21, 2012 #9
    so...
    would it simply be ranked based on the static frictional coefficient?

    the only other known variable is mass.
     
  11. Sep 21, 2012 #10

    TSny

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    Let F represent the applied force (assumed horizontal) and let fs represent the friction force. How would you express the net horizontal force using these symbols?
     
  12. Sep 21, 2012 #11
    Net Force = F - fs

    i think
     
  13. Sep 21, 2012 #12

    TSny

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    Good. What does the net force have to equal according to the 2nd law?
     
  14. Sep 21, 2012 #13
    it has to equal ma

    so... F - fs = m * a
     
  15. Sep 21, 2012 #14

    TSny

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    Yes. And what is the value of the acceleration for the crates?
     
  16. Sep 21, 2012 #15
    0, i think.

    a = F/m ; and net Force, F, is 0
     
  17. Sep 21, 2012 #16

    TSny

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    Well, we know a = 0 because we are told that the crates don't move. So, if a = 0, what does the formula F - fs = ma tell you about the amount of friction force?
     
  18. Sep 21, 2012 #17
    if a = 0 , then F = fs ; which was already determined

    this seems like a loop to me
     
  19. Sep 21, 2012 #18

    TSny

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    Good. So, you should now be able to answer the question. Remember, the problem states that each crate has the same applied force.
     
  20. Sep 21, 2012 #19
    OMG, WOW!

    IT WAS SO EASY! They are all the same?!

    I LOVE YOU!
     
  21. Sep 21, 2012 #20

    TSny

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    Hey hey, now :blushing:. Anyway, good work.:smile:
     
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