Infinite slope FoS; water table rises

Derivation (cohesionless soil, no water table)

Take a slice of width $b$ and depth $z$ in an infinite slope inclined at $\beta$. Soil is dry or moist with unit weight $\gamma$ and angle of friction $\phi$. Side forces cancel.

$$W = \gamma z b, \quad N = W\cos\beta, \quad T = W\sin\beta$$

On the failure plane of length $b/\cos\beta$:

$$\sigma = \gamma z\cos^2\beta, \qquad \tau = \gamma z\sin\beta\cos\beta$$

Shear strength: $\tau_f = \sigma\tan\phi = \gamma z\cos^2\beta\tan\phi$

$$\boxed{F = \frac{\tau_f}{\tau} = \frac{\tan\phi}{\tan\beta}}$$

The factor of safety does not depend on depth or unit weight. The slope is stable when $\beta < \phi$, and at $\beta=\phi$ ($F=1$) the slope is at its limit (angle of repose).

Water table rises to the surface

With seepage parallel to the slope and water at the surface, the pore pressure is $u = \gamma_w z\cos^2\beta$, so the effective normal stress falls to $\gamma' z\cos^2\beta$ while the shear stress rises to $\gamma_{sat} z\sin\beta\cos\beta$:

$$F_{sat} = \frac{\gamma'}{\gamma_{sat}}\frac{\tan\phi}{\tan\beta}$$

Since $\gamma'/\gamma_{sat}\approx 0.5$, the factor of safety falls to about half. For example, a slope with $\phi = 35^\circ$ and $\beta = 20^\circ$ has $F = 1.92$ when dry but only about 0.96 when the water reaches the surface, so the slope would fail. This is why heavy rain triggers landslides in sandy slopes.