Infinite dry slope, c = 10 kPa, φ = 25°

Data: $c = 10$ kPa, $\phi = 25^\circ$, $z = 5$ m (depth of the failure plane, as given), $\gamma = 16$ kN/m³, $\beta = 10^\circ$, no water.

For a dry infinite slope:

$$F = \frac{c}{\gamma z\sin\beta\cos\beta} + \frac{\tan\phi}{\tan\beta}$$

Step 1: Values

$$\sin10^\circ\cos10^\circ = 0.1710, \quad \tan25^\circ = 0.4663, \quad \tan10^\circ = 0.1763$$

Step 2: Cohesion term

$$\frac{10}{16\times5\times0.1710} = \frac{10}{13.68} = 0.731$$

Step 3: Friction term

$$\frac{0.4663}{0.1763} = 2.645$$ $$F = 0.731 + 2.645 = 3.38$$

$F = 3.38 > 1$, so the slope is stable with a large margin.

If cohesion becomes zero

$$F = \frac{\tan\phi}{\tan\beta} = 2.645$$

The factor of safety drops by about 22% but remains well above 1, so the slope is still stable. The stability then depends only on friction ($\beta=10^\circ<\phi=25^\circ$).

Answer: $F = 3.38$ (stable); with $c = 0$, $F = 2.64$ (still stable).