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Derive the expression for the work done by the friction force

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  1. Dec 1, 2017 #1
    1. The problem statement, all variables and given/known data
    14.6 The coefficient of kinetic friction between the slider and the rod is μ, and the
    free length of the spring is ##L_0 = b##. Derive the expression for the work done by
    the friction force on the slider as it moves from A to B. Neglect the weight of the slider.

    Fig P14_5_6.png

    2. Relevant equations


    3. The attempt at a solution

    Pytels_Dynamics101.jpg
    I
    am going the right direction with my solution?
    I don't see to be getting the ##-0.1186μkb^2## solution.
     
  2. jcsd
  3. Dec 1, 2017 #2

    BvU

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    ##\mu## stands for the ratio of the friction force and some normal force. What's that normal force as a function of position ?

    Pity you don't give any relevant equations. I don't think the exercise wants you to look at the velocities.
     
  4. Dec 2, 2017 #3
    Pytels_Dynamics103.jpg
    Am going the right direction with the above calculation?
     
  5. Dec 2, 2017 #4

    haruspex

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    It is preferred that you type in your algebra. This makes it easier to comment on specific lines. Or you could number the equations.
    Your FBD omits the force dragging the slider from A to B. You should take acceleration to be negligible. This invalidates your force balance equation for the x direction.
    The NA=P cos(φ) equation is the useful one, but you need to find P as a function of φ and integrate.
     
  6. Dec 4, 2017 #5
    upload_2017-12-4_22-3-44.png

    For the x direction:
    ##ΣFx=0 => F-F_k=0 => F=F_k## (1)

    For the y direction:
    ##ΣFy=0 => N_A-Pcosφ=0 => N_A=Pcos(φ)## (2)

    How do I do that? I am a bit confused...
     
  7. Dec 4, 2017 #6

    BvU

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    How would you calculate ##P## at point B ?
     
    Last edited: Dec 4, 2017
  8. Dec 4, 2017 #7

    haruspex

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    It's the tension in a spring. What determines the value of that?
     
  9. Dec 7, 2017 #8
    P at point B be should be ## -kx## where x is the elongation of the spring.

    ##x = L_B-L_0 = \sqrt {b^2+b^2} - b = 0.414b##

    So P = -0.414kb

    Am I correct? (it is the second problem with springs ever i am trying to solve )
     
  10. Dec 7, 2017 #9

    PeroK

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    I think the problem is a lot more complicated than that. First, you need an expression for the length of the spring as a function of the distance from A to B.
     
  11. Dec 7, 2017 #10
    So to do that I split the distance between A and B to 3 sections. from 0 to 1/3 of b, from 1/3b to 2/3b, and from 2/3b to b.

    At 1/3b ##sinφ = \frac {1b/3} {l} => sinφ = \frac {1b/3} {1.054b} = 0.316## (where ##l = b + L_{1/3b} - L_0##)

    At 2/3b ##sinφ = 0.55##
    At 3/3b ##sinφ = 0.707## that verifies that when collar is at position B the angle φ is 45 degrees.

    So from above we can say that ##l = L_{AB} = \sqrt {b^2+(xb)^2}## where x is between 0 and 1
     
  12. Dec 7, 2017 #11

    PeroK

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    Well, I'd say just use good old Pythagoras:

    ##L = \sqrt{x^2 + b^2}##, where ##0 \le x \le b##

    Now you need to analyse the forces at each point ##x## between ##A## and ##B##.
     
  13. Dec 7, 2017 #12

    BvU

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    Remember the last line in #4: You need ##P## as a function of ##\phi##. I think you do fine for ##\phi## at specific values, but you need a function for all ##\phi## in ##[0,\pi/4]##
     
  14. Dec 9, 2017 #13
    Well using Inkscape, FBD is:

    FBD P14_6.jpg

    in y-axis ##N_A = P_y => N_A = Pcos(φ)##

    in x-axis ##-F_k - P_x < F## (Is this correct?)

    ##P=Pcos(φ) ## (Is this correct?)
     
  15. Dec 9, 2017 #14

    BvU

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    ##\ P_y = P \cos\phi## is right. Both factors change during the trip. So fill in something for ##P##: a function of ##\phi##.
     
  16. Dec 11, 2017 #15
    The only thing i understand is that ##P = -kl##
    I can see that as the angle increases so does P. So P∝φ

    should it be P related to ##1/cos(φ) ? If angle is 0deg the 1/cos(φ) is 1; Then as angle increases so does 1/cos(φ);
     
  17. Dec 11, 2017 #16

    BvU

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    Write it out in full so you have something that you can integrate from ##\phi=0## to ##\phi = \pi/2##....
    To help you: ##k## is given. Now you need ##l(\phi)## -- you did it already, see above
     
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