Screw Jack – Efficiency Formula in Terms of Helix Angle and Friction Angle Which of the following correctly gives the efficiency η of a screw jack in terms of helix angle α and friction angle φ (for raising the load)?

Introduction / Context:
Screw jacks convert a small effort over a long helical path into a large lifting force over a small linear distance. Their performance is captured by efficiency η, which depends on geometry (lead or helix angle α) and friction (expressed via friction angle φ).

Given Data / Assumptions:

• Right-hand square/Acme thread model; helix angle α measured at mean radius.
• Friction angle φ where tan(φ) = μ.
• Raising the load (upward motion).

Concept / Approach:
For a screw jack, the efficiency in raising is the ratio of ideal mechanical advantage to the actual mechanical advantage. Using the force analysis on the screw thread (viewed as an inclined plane wrapped helically), the standard result is η = tan(α) / tan(α + φ) when lifting.

Step-by-Step Solution:
Relate friction to φ via tan(φ) = μ. Model thread as an inclined plane of angle α with friction φ. For raising: efficiency η = tan(α) / tan(α + φ). For lowering (for comparison): η_lower differs; self-locking depends on α and φ.

Verification / Alternative check:
If φ = 0 (no friction), η = tan(α)/tan(α) = 1 (100%), which is consistent with an ideal machine. As φ increases, denominator increases, efficiency drops logically.

Why Other Options Are Wrong:
tan(α − φ)/tan(α): not the raising-load expression; appears in other contexts. tan(α + φ)/tan(α): predicts η > 1 for φ > 0, impossible. tan(φ)/tan(α + φ) and product forms: not supported by thread equilibrium derivation.

Common Pitfalls:
Confusing raising vs lowering formulas. Using lead (l) and mean radius (r) inconsistently when deriving α via tan(α) = l / (2π r).

Final Answer:
η = tan(α) / tan(α + φ)