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Finding the minimum force F which must be exerted

  1. Nov 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Given: The friction between the block with mass 6 kg and the wedge with mass 18 kg is 0.27 . The surface between the wedge with mass 18 kg and the horizontal plane is smooth (without friction). The acceleration of gravity is 9.8 m/s^ 2. A block is released on the inclined plane (top side of the wedge). What is the minimum force F which must be exerted on the 18 kg block in order that the 6 kg block does not move down the plane?
    29zen9h.jpg

    2. Relevant equations
    Fnet = ma
    and possibly(?): Ff = uFn

    3. The attempt at a solution
    Having the wedge in the problem is confusing me.
    The forces acting on the x axis of the 18 kg block are F applied and Fg parallel
    The forces acting on the x axis of the 6kg block are F frictional and Fg parallel.
    Fnet = ma / Fnet = 0
    Fa + Fg(parallel) = 0
    I think I am solving for Fa (18kg), i'm not sure.
    Fa = Fgsin20
    Fa = (18kg)(9.8m/s2)sin20
    Fa = 60.33 N
    I know the 6kg block plays a role, but i don't know how or where to incorporate it.
    I'm really confused, any help is appreciated.
     
  2. jcsd
  3. Nov 28, 2014 #2

    OldEngr63

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    Gold Member

    Having the wedge in the problem is designed to get you to think.

    Start with FBDs for both the wedge and the 6 kg block and write equations of motion for each AFTER you draw the FBDs.
     
  4. Nov 28, 2014 #3
     
  5. Nov 28, 2014 #4
    upload_2014-11-28_18-22-38.png
    this is my work so far. I know I am looking of F at this point, but I am still not sure where to go from here? (unless my FBDs are incorrect...)
     
  6. Nov 28, 2014 #5

    OldEngr63

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    Gold Member

    Your FBDs do not agree with each other. For example, the Fn you show on the 6 kg block is upward and to the left; the Fn you show on the wedge is straight up. There should be one Fn that is the opposite of the other. You also have not distinguished the fact that there are two normal forces involved. And it goes on...

    I suggest that you try again on the FBDs.
     
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