Angle of friction from limiting horizontal push on a horizontal surface A 50 kgf block rests on a horizontal plane. A horizontal force of 18 kgf just causes impending motion. Determine the angle of friction φ (where tan φ = μ).

Introduction / Context:
At impending motion on a horizontal plane, the frictional force reaches its limiting value μN. Converting the limiting push to the friction coefficient and then to the angle of friction is common in design of wedges, clamps, and handling equipment where stick–slip must be controlled.

Given Data / Assumptions:

• Normal reaction N equals the weight W on a horizontal plane.
• Limiting horizontal force P at impending motion equals friction μN.
• Given W = 50 kgf, P = 18 kgf; use kgf consistently (force units).

Concept / Approach:

At the threshold of motion: P = μ * N. For a horizontal plane, N = W. Thus μ = P / W. The angle of friction φ satisfies tan φ = μ. Compute μ and take inverse tangent to obtain φ in degrees and minutes.

Step-by-Step Solution:

Verification / Alternative check:

Check by resolving along motion: Required horizontal push at limiting state equals μW. Using μ = tan φ ensures consistency with the angle interpretation from the friction cone.

Why Other Options Are Wrong:

Other listed angles correspond to μ values not equal to 0.36. For example, 17°48′ → μ ≈ 0.32; 21°48′ → μ ≈ 0.40, which do not match P/W.

Common Pitfalls:

Confusing kg (mass) with kgf (force); rounding too early; using sin or cos instead of tan for the friction angle.

Final Answer:

19° 48′