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Frictional Force issue

  1. May 27, 2016 #1
    1. The problem statement, all variables and given/known data
    You pull on a 30 kg box with a horizontal force of 110 N to the right. (i) If there are no other forces acting on this box, what is the resulting acceleration? (ii) If the resulting acceleration of the block is only 1.6 m/s, what is the magnitude and direction of the frictional force acting on the block


    3. The attempt at a solution
    For (i) a- f/m and I got a=3.67

    i have no idea how to solve for (ii) i thought it would f=m times a, but i have no idea where to start/solve any help, please?
     
  2. jcsd
  3. May 27, 2016 #2

    mfb

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    You know m and a, so you can calculate F, the net force acting on the block. This net force has two components.
     
  4. May 27, 2016 #3
    so f= 30 x 1.6=48
    110-48=62N
    correct ?

    subtract because it moving in the negative direction, right?
     
  5. May 27, 2016 #4

    mfb

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    Correct.
     
  6. May 27, 2016 #5
    im also stuck on static frictional force question

    The static co-efficient of friction between a football blocking sled (90 kg) and grass is 0.56. (i) If an80 kg coach stands on top of the blocking sled, what is the maximal static frictional force?

    any hints on the formula to apply?
     
  7. May 27, 2016 #6

    mfb

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    Which formulas do you have for (maximal) static friction?
     
  8. May 27, 2016 #7
    i do not, i was curious to find out which one to use?

    do i have to find the net force?
     
  9. May 27, 2016 #8

    mfb

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    There should be exactly one.
    You have to find the maximal force before block+trainer start to move.
     
  10. May 27, 2016 #9
    i have fs(max) = c0efficient of static friction x normal force x 9.81
     
  11. May 28, 2016 #10

    mfb

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    That part is not correct, even if you add the missing units.
     
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