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Can someone please help! Equilibrium/friction

  1. Sep 28, 2007 #1
    [bQuestion Details:

    I am having trouble answering this question.

    The drawing shows a block against a vertical wall to its right. The force F is pointed at the lower left hand corner of the block at 40.0 deg. left of -y-axis


    The weight of the block in the drawing is 88.9N. The coefficient of the static friction between the block and the vertical wall is 0.560. a)What minimum force F is required to prevent the block from sliding down the wall? b) What minimum force F is required to start the block moving up the wall?


    Here is my attempt at it--

    a) Fs(max)=(0.560)(88.9N)=49.8N
    Fy=88.9N-49.8N=39.1N
    Fx=(39.1N)/(cos(40))=51.0N
    sqrt{39.1^2+51.0^2}=F
    F=64.3N

    b) 88.9N +49.8 N=138.7N=Fy
    Fx=(138.7N)/(cos(40))=181.06N
    sqrt{138.7^2+181.06^2}=F
    F=228N

    But it doesn't seem right!
     
  2. jcsd
  3. Sep 28, 2007 #2

    Astronuc

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    Staff: Mentor

    This is incorrect.

    The friction between the block and wall is not dependent on the weight, which is pointing downward with gravity and parallel with the wall.

    The friction is related to [itex]\mu[/itex]Fx, which is perpendicular or normal to the wall.

    a) Weight acts down, Fy acts up, and Fs acts _________?

    b) Weight acts down, Fy acts up, and Fs acts _________?
     
  4. Sep 28, 2007 #3
    a) Fs acts up
    b)Fs acts down
    The problem is I'm not sure how to get Fs from the normal force in the x direction from the wall.
     
  5. Sep 29, 2007 #4

    Astronuc

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    Staff: Mentor

    Sure you do!

    One simply resolves F into its x and y components.

    Fx = F sin [itex]\theta[/itex], where [itex]\theta[/itex] is the angle with the y-axis or vertical. If one selected the angle with the x-axis or horizontal, then one would use cos instead of sin.

    then Fy = F cos [itex]\theta[/itex]

    We know [itex]\mu[/itex] = 0.56 and we know the angle [itex]\theta[/itex] = 40°, and we know how to find Fs.

    Write the equations for the forces expressed in parts a and b.
     
  6. Sep 29, 2007 #5
    a)Fs=[tex]\mu[/tex]sFn
    Fn=F(cos40)
    Fs=(.560)(F(cos40))
    Fs-mg+F(sin40)=0
    F(0.560*cos(40)+sin(40))=88.9N
    F=82.9N

    b)-Fs-mg+F(sin(40))=0
    F(sin(40)-(.560*cos(40))=88.9N
    F=415.4N

    answer for b doesn't seem right.
     
  7. Sep 29, 2007 #6
    You post the same question twice?!?!

    I believe I gave you a good answer:
    Here
     
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