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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 isThe 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:Consistently retaining only first order in thermal expansion givesIf the topographic term is retained within the same linear-density column model, it adds $1.52\times 10^{11}\mathrm{N}{\mathrm{m}}^{-1}$, giving $F\simeq 2.75\times 10^{12}\mathrm{N}{\mathrm{m}}^{-1}$. 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.

The ridge-push force in a plate model is the depth integral of the horizontal pressure difference between hot and cold compensated columns. To leading order, the old plate's linear excess density gives $F=g{\int }_0^{a}s\delta \rho (s)ds={\rho }_{m}g\alpha \Delta T{a}^2/6$. Here $s$ is depth below the old seafloor. Retaining the geometrical contribution from its lowered surface adds $g({\rho }_m-{\rho }_w)h^2/2$ within the linear-density model. Force per ridge length has units $\mathrm{N}{\mathrm{m}}^{-1}$, distinct from stress.