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Min Force to keep from Slipping!

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data

    A block weighing 70.6 N rests on a plane inclined at 26.9o to the horizontal. The coefficients of static and kinetic friction are 0.23 and 0.14 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping?

    Picture Include!


    2. Relevant equations

    F=m*a

    us*Fn = static friction


    3. The attempt at a solution

    I split force of gravity into its components. (Y DIRECTION). I found the Fn to equal mg cos theta

    decided that for the X DIRECTION to do

    F - usFn - mg sin theta = m * a.... then I made A = 0... which made the [ ...left side... = 0 ]

    Then I solved for F... which gave me 48.1799 N.

    But unfortunately this is wrong. Any Ideas? Thank you!
     

    Attached Files:

  2. jcsd
  3. Sep 29, 2011 #2
    More of my attempt....
    __________________
    Y Direction:

    Sum F = M * A
    Fn-mg = 0
    Fn = mgcosTheta <---- I got this from breaking gravity into it's components.

    _________________

    X Direction:

    Sum F = M * A
    F - Static Friction - mgSinTheta = 0
    F = uS*Fm - mgsinTheta

    _________________

    Can you please give me some direction? Thank you!
     
  4. Sep 29, 2011 #3
    First equation is OK.

    Now you were asked for the minimum force. Hence the block will tend to slip DOWNWARDS. Therefore frictional force is UP the plane.
    If you correct the 2nd equation you are OK.
     
  5. Sep 29, 2011 #4
    F - (0.23)(70.6cos(26.9)) - (70.6)sin(26.9)) = 0

    F - Kintic Friction - mgSinTheta = 0

    ^^ Is that what you meant?

    Thank you.
     
  6. Sep 29, 2011 #5
    what is Fm ? kinetic friction doesn't come into the picture
     
  7. Sep 29, 2011 #6
    I think Fn = W.... so in the beginning when they said that it's W = 70.6 that means that Fn = 70.6?
     
  8. Sep 29, 2011 #7
    I thought you ment Fn... I'm not sure what the variable Fm stands for
     
  9. Sep 29, 2011 #8
    No that is not what I meant.
    The first minus must be changed into plus since friction is up in same direction as the applied force F.
    All is at rest so kinetic friction is out.
     
  10. Sep 29, 2011 #9
    the downward component of the weight is more than the maximum static frictional force...
    so you will need to apply the amount of force in the upward direction, which is the difference between the two
     
  11. Sep 29, 2011 #10
  12. Sep 29, 2011 #11
  13. Sep 29, 2011 #12
    Correct!
     
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