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Incline Friction Problem?

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data

    A 2.00 kg block is held in equilibrium on an incline of angle = 75° by a horizontal force applied in the direction shown in Figure P4.50. If the coefficient of static friction between block and incline is µs = 0.300, determine the following.
    (a) the minimum value of F
    (b) the normal force exerted by the incline on the block

    Illustration of Problem: http://www.webassign.net/sercp/p4-50.gif...

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    For Y Component:
    n-mgcos75=0
    n=mgcos75
    n=2.00*9.8cos75
    n=5.07 N
    It says this is wrong because of: Your answer differs from the correct answer by orders of magnitude.

    For X
    F-fs=0
    F=usn=0
    F=usn
    F=0.300 x 5.07 N = 1.52 N

    Does this look like I did it correctly?
     
    Last edited: Sep 8, 2007
  2. jcsd
  3. Sep 8, 2007 #2

    learningphysics

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    No. You have missing components. The F force has a component in the x direction (parallel to the incline) and y direction (perpendicular to the incline).

    Also the mgsintheta component is missing (component of gravity parallel to the incline)...

    And friction acts upward along the incline... it's preventing the block from sliding down. It works together with F to prevent it from sliding down.
     
  4. Sep 8, 2007 #3

    Doc Al

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    You neglected the Y component of the applied force F.
     
  5. Sep 8, 2007 #4
    Ok, well now I really don't even almost have a clue what to do.

    Of course the Force has x and y components, but if I don't know the force how would I solve to find those?
     
  6. Sep 8, 2007 #5

    learningphysics

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    Resolve F into its components using the angle.
     
  7. Sep 8, 2007 #6
    Would the normal force be equal to mg?

    That is, 19.6 N?
     
  8. Sep 8, 2007 #7
    I am so confused, I don't even know what angle goes with F.

    I can figure out the components of mg, but haven't a clue how to go about finding it for F.
     
  9. Sep 8, 2007 #8

    learningphysics

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    It's the same theta. Just have to be careful about sin vs. cos.

    From the tail of the vector F (the part without the arrow) draw a line perpendicular to the incline. And from where that line intersects the incline draw another line to the head of the vector F (the part with the arrow). See the two components? One of the angles of this triangle is theta.
     
  10. Sep 8, 2007 #9

    learningphysics

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    No. that's not the answer.
     
  11. Sep 8, 2007 #10
    I think I'm just screwed.

    Thanks for the help anyways.
     
  12. Sep 8, 2007 #11

    learningphysics

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    Draw an x - axis parallel to the incline... y -axis perpendicular to the incline... now rotate your picture so that the x-axis is horizontal, and y -axis is vertical... now try to resolve the components of F along the x-axis and y-axis.
     
  13. Sep 8, 2007 #12
  14. Sep 8, 2007 #13

    learningphysics

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  15. Sep 8, 2007 #14
    So should the component parallel to N be pointing up (above F)?
     
  16. Sep 8, 2007 #15

    learningphysics

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    No sorry... it's opposite to N...

    And fs should be upward along the plane... and the other component of F is parallel to fs upward along the plane.
     
  17. Sep 8, 2007 #16
  18. Sep 8, 2007 #17

    learningphysics

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  19. Sep 8, 2007 #18
    Ok, so I drew it and there are 3 angles: 90, 45, and 45.

    Which angle should I be working with?
     
  20. Sep 8, 2007 #19
    Should I get Fsin45 for my x component and Fcos45 for my y-component?
     
  21. Sep 8, 2007 #20
    Ok, I got it. It's n=35.64 N and F=31.66, correct?
     
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