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Static Equilibrium - Ladder problem with a twist.

  1. Oct 31, 2016 #1
    1. The problem statement, all variables and given/known data
    A board that is of length L = 6.0 m and weight W = 335.7 N rests on the ground and against a frictionless contact at the top of a wall of height h = 2 m (see figure). The board does not move for any value of
    theta.gif greater than or equal to 65 degrees but slides along the floor if
    theta.gif <65 degrees. Find the coefficient of static friction between the board and the floor.

    2. Relevant equations
    c11_6h.gif
    Σt=0
    F(roller)=Fs
    Fn=W
    Fs=Fn(μ)

    Horizontal Forces.
    F(roller) is the horizontal force are the roller point.
    Fs is the force of friction at the point of contact with the ground and direction towards the wall
    Vertical Forces.
    Fn is the normal force from the point of contact.
    W is the force of the rod.

    3. The attempt at a solution

    Fn=335.7
    Fs=335.7μ
    F(roller)=335.7μ

    Σt=0=Fs*d+Fn*d+W*d1-F(rol)*d2

    d=0,[Fs*d+Fn*d]=0

    Σt=0=W*d1-F(rol)*d2

    F(rol)*d2, d2=h

    Fr(2)=W*d1

    W*d1,
    d1= L/2cos65

    [335.7(6/2)(cos65)]/2=Fr

    Fr=335.7μ

    ( [335.7(6/2)(cos65)]/2 )/335.7 =μ

    Am I making a huge mistake with my thought process?
     
    Last edited: Oct 31, 2016
  2. jcsd
  3. Oct 31, 2016 #2

    TSny

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    One of your forces listed as horizontal is not actually horizontal.
     
  4. Oct 31, 2016 #3
    The force of the roller is perpendicular to the to the board, using this I got it to F(rol)cos(25)

    Fn=W*d - F(rol)cos(25)

    But now I am confused what to do.
     
  5. Oct 31, 2016 #4

    TSny

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    OK. Good
    Is this the magnitude of the horizontal component of the force of the roller? If so, that looks right.

    This looks strange. On the left side you have a vertical force. The first term on the right side looks like a torque. The second term on the right looks like a horizontal force.
    You know all the forces acting on the ladder and I'm sure you've carefully drawn a free body diagram. Use the diagram to set up the conditions for static equilibrium.
     
  6. Nov 1, 2016 #5
    2a7hax0.jpg

    F(rol)cos25=Fs=Fnμ
    Fn=W

    Σt=0=w*d1-F(rol)cos(25)d2

    d1=(L/2)sin(65)
    d2=h
    w*d1=F(rol)cos(25)d2
    w*d1/d2=F(rol)cos(25)
    F(rol)cos(25)=Fnμ
    (W(d1))/2=Wμ

    Getting 1.5 out for μ which is ridiculous.

    I feel like my issues lies with the d1 selection. I tried it with cos(65) too and still a wrong answer of .7=μ

    EDIT: I also just noticed my final answer in the first post is the same as this one. Seems like I need to do some more review.
     
  7. Nov 1, 2016 #6

    TSny

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    You've left out one of the torques. Note that F(rol) also has a vertical component.
    [EDIT: You might find it easier to write the expression for the torque due to F(rol) by finding the lever arm for F(rol), rather than finding separate torques for each component of F(rol). But either approach will work.]

    d1 is not quite correct. Make sure you have identified the lever arm for W correctly.

    Otherwise, you are close.
     
    Last edited: Nov 1, 2016
  8. Nov 1, 2016 #7
    3cos(65)W-2tan(25)sin(25)Fr-2cos(25)Fr=0

    335.7(3)cos65=Fr(2tan(25)sin(25)+2cos(25))
    Fr = [335.7(3)cos65]/(2tan(25)sin(25)+2cos(25))
    Fs = Fr*cos25 = Fnμ
    Fn = W-Fr*sin25

    μ=(Fr*cos25)/(W-Fr*sin25)
    μ=.688 ANS

    Thank you for the help it was the correct answer.
     
  9. Nov 1, 2016 #8

    TSny

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    OK, good work!
     
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