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Balancing Van der Waals force with a spring

  1. Nov 7, 2012 #1
    1. The problem statement, all variables and given/known data

    An AFM tip has a spring constant k, a colloidal bead with radius r is glues onto the cantilever tip. Derive an expression and plot the deformation of the cantilever spring, Δx, as a function of distance the tip of the cantilever is moved towards another bead of the same radius.


    2. Relevant equations

    The cantilever tip can be treated as a Hookean spring. So Fspring = -kΔx

    The non-retarded Van der Waals interaction free energy between two spheres of the same radius is:

    W = -AR/(12D)

    where D is the distance between the spheres, R is the radius of the spheres, and A is the Hamaker constant.

    The Van der Waals force is simply the derivative of the potential w.r.t. D:

    Fvan = AR/(12D2)


    3. The attempt at a solution

    The question seems fairly straight forward. At equilibrium, Fspring = Fv, so

    -kΔx = AR/(12D2)

    we can say D is the distance the cantilever tip is manual moved (z), plus the distance the spring is stretched because of VDW interaction.

    -kΔx = AR/(12(z+x)2)

    For some reason, I can't solve for Δx as a function of z. I feel like I'm missing some trivial step, and would appreciate any help with a solution.

    Thanks in advance.
     
  2. jcsd
  3. Nov 9, 2012 #2

    haruspex

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    Should that read -kΔx = AR/(12(z+Δx)2)?
    If so, can't you just multiply it out to get a cubic? If Δx can be assumed small c.w. z then you can approximate it thus:
    -kΔx = AR/(12z2(1+Δx/z)2) ≈ AR/(12z2(1+2Δx/z)) ≈ AR(1-2Δx/z)/(12z2)
    That now being linear in Δx, solve in the obvious way.
     
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