= Solution
<Ridge push> is a horizontal <pressure> <force> integrated with depth, equivalently a difference in gravitational <potential energy> per horizontal <area>. It is not the weight of a slab multiplied by the surface slope at one chosen point.
For an explicit compensated-column derivation, let $y$ measure depth below the ridge seafloor; the old seafloor is at $y=h$. Set $b=\rho_m\alpha\Delta T$. The hot column has greater <pressure>. Its <pressure> excess over the old column is
$$
\Delta p(y)=\begin{cases}
g(\rho_m-\rho_w)y,&0<y<h,\\
\dfrac{gb}{2a}(a-s)^2,&y=h+s,\quad0<s<a,\\
0,&y>h+a.
\end{cases}
$$
The second line follows by integrating $\delta\rho(s)=b(1-s/a)$ upwards from the compensation level. The two expressions agree at $y=h$ because $(\rho_m-\rho_w)h=ba/2$. Integrating this positive <pressure> excess gives the <ridge-push force in a plate model>, directed away from the ridge:
$$
F=\int\Delta p\,dy
=\frac{\rho_mg\alpha\Delta T a^2}{6}
+\frac{g(\rho_m-\rho_w)h^2}{2}.
$$
Consistently retaining only first order in <thermal expansion> gives
$$
\boxed{F\simeq2.60\times10^{12}\,\mathrm{N\,m^{-1}}.}
$$
If the topographic term is retained within the same linear-density column model, it adds $1.52\times10^{11}\,\mathrm{N\,m^{-1}}$, giving \b[$F\simeq2.75\times10^{12}\,\mathrm{N\,m^{-1}}$]. This is a useful small correction, not a claim that all other second-order material effects have been included. Both values use the intended hot initial plate, not the inconsistent printed zero-age profile.
\Image[/past-exam-of-the-physics-of-the-earth-as-a-planet-course-of-the-university-of-cambridge/2013/paper-1-pep-plate-forces.png]
{title=Hot-ridge and old-plate columns, and their compensated pressure difference in the intended cooling model}
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