# Balancing Van der Waals force with a spring

1. Nov 7, 2012

### creemore

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. Nov 9, 2012

### haruspex

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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