Infinite slope — stability number: For an infinite slope of inclination i with soil internal friction angle φ (cohesion may be considered separately), the stability number Sa defined as c / (γ z) required for limiting equilibrium is:

Introduction / Context:
For an infinite slope, the shear stress on a plane parallel to the slope and the corresponding normal stress vary linearly with depth. In a frictional–cohesive soil, cohesion provides additional resistance when the gravitational shear demand exceeds frictional resistance. The required cohesion per unit overburden is often expressed as a stability number, Sa = c / (γ z).

Given Data / Assumptions:

• Infinite slope with inclination i.
• Soil has internal friction angle φ and cohesion c (to be determined for limiting equilibrium).
• Unit weight γ and depth to the plane of interest z.
• Dry or drained condition with no pore pressure; seepage effects are neglected here.

Concept / Approach:
On a plane at depth z parallel to the slope: shear stress τ = γ z sin i cos i and normal stress σ = γ z cos^2 i. Frictional resistance available = σ tan φ = γ z cos^2 i tan φ. The cohesion required to exactly satisfy equilibrium at limiting state must balance the excess of τ over frictional resistance:

Step-by-Step Solution:

Verification / Alternative check:
If i = φ (pure friction slope at limiting state), Sa = 0, which is consistent: no cohesion is needed. If i < φ, Sa is negative (no cohesion required), again consistent with stability.

Why Other Options Are Wrong:

• Forms with sin^2 i do not arise from τ − σ tan φ when using τ = γ z sin i cos i and σ = γ z cos^2 i.
• tan i / tan φ is the inverse of a friction-only factor of safety, not the stability number defined here.

Common Pitfalls:
Confusing friction-only factor of safety FS = (tan φ / tan i) with cohesion requirement; neglecting pore pressure which would modify σ and hence Sa.

Final Answer:
(tan i − tan φ) cos^2 i