Power screws — in a screw jack, the effort required to lower a load W is expressed in terms of helix angle α and friction angle φ as:

Introduction / Context: For power screws, raising and lowering efforts differ due to friction. When lowering a load in a screw jack, the applied effort must overcome the difference between frictional resistance (φ) and the geometric helix angle (α).

Given Data / Assumptions:

• Lead (helix) angle = α.
• Friction angle = φ, where tan φ = μ.
• Load = W; square/Acme threads modeled with equivalent friction.

Concept / Approach: The equilibrium on the screw thread (viewed as an inclined plane wrapped around the cylinder) gives the tangential effort required. For lowering under control (no overhauling), the effort is P = W tan(φ − α). Self-locking occurs when φ > α; if α ≥ φ, the screw may overhaul (run down).

Step-by-Step Solution:

Verification / Alternative Check: For raising, P_raise = W tan(α + φ); swapping sign for lowering yields P_lower = W tan(φ − α).

Why Other Options Are Wrong:
P = W tan(α − φ) — Sign and order do not match lowering case.
P = W tan(α + φ) — Formula for raising, not lowering.
P = W cos(α + φ) — Dimensionally incorrect for effort ratio.

Common Pitfalls: Confusing raising and lowering expressions; neglecting self-locking criterion φ > α.

Final Answer: P = W tan(φ − α).