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Friction/ Newtons Laws

  1. Jan 8, 2014 #1
    ImageUploadedByPhysics Forums1389194813.781348.jpg

    Hi I am stuck with a question which I have attached a photo of, it is one involving friction! I have done a few of these questions already and find it is when I am defining forces etc it's where I go wrong, not the easy computational part:(

    I also have attached my attempt at the solution and have drawn my free body diagram as clear as possible:)


    Any help would be great:) thanks


    ImageUploadedByPhysics Forums1389195068.087240.jpg
     
  2. jcsd
  3. Jan 8, 2014 #2
    I see no mistake in your attempt so far.
     
  4. Jan 8, 2014 #3
    Excellent I shall continue onwards and reply shortly:)
     
  5. Jan 8, 2014 #4
    ImageUploadedByPhysics Forums1389195983.885812.jpg


    Ok I went on and done this to find the acceleration but the model answer is 3.71 m/s^2
     
  6. Jan 8, 2014 #5
    In your computation you assumed that static friction will not be strong enough to prevent motion. Part (a) of the problem required you to determine whether this was the case.
     
  7. Jan 8, 2014 #6
    It is not strong enough
     
  8. Jan 8, 2014 #7
    It is the two fractions which are confusing me, kinetic friction is the friction between two moving bodies and static is between two non moving bodies would there be a case where both would have to come into a computation?
     
  9. Jan 8, 2014 #8
    Can you demonstrate that?
     
  10. Jan 8, 2014 #9
    I did not see #7 when I wrote #8. Please explain how you determined that static friction was not strong enough to prevent motion. That will help clarify what you confusion really is.
     
  11. Jan 8, 2014 #10
    So for the block to move mg sinθ must be greater than static friction?
     
  12. Jan 8, 2014 #11
    Sorry F cosθ
     
  13. Jan 8, 2014 #12
    Yes, the horizontal component of force must be greater than the maximal static friction. The maximal static friction is given by ## \mu_s N ##.
     
  14. Jan 8, 2014 #13
  15. Jan 8, 2014 #14
    I agree with your result, but it is not correct to write ## F \cos \theta - \text{fr} = 0 ##. You should write ## F \cos \theta - \text{fr} > 0 ##.
     
  16. Jan 8, 2014 #15

    CWatters

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    Check what value you used for μ.

    I put 0.2 into your earlier equation and got 3.7 as the answer.
     
  17. Jan 8, 2014 #16
    Oh! Haha I think I've got it now lol
     
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