Balancing Van der Waals force with a spring

[creemore]

Homework Statement

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.

Homework 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)

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.

 

[haruspex]

-kΔx = AR/(12(z+x)²)

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.