General sag correction formula for a suspended tape: If a pull F (kgf) is applied at the tape ends, l is the span between supports (m), and w is the tape weight per meter (kgf/m), which expression gives the sag correction to be subtracted from the measured length?

Introduction / Context:A tape suspended between supports forms a curve under its own weight. Because the actual path is longer than the horizontal projection, a sag correction must be applied. Correct formula selection is essential for precise linear measurement in surveying and construction layout.

Given Data / Assumptions:

• Uniform weight per unit length w (kgf/m).
• Pull F (kgf) is steady and the same at both ends.
• Span between supports is l (m); small-sag approximation holds.
• Correction c is to be subtracted from the measured distance.

Concept / Approach:Under uniform load, the tape approximates a catenary. For small sag, the excess length over the horizontal projection is given by c = (w^2 * l^3) / (24 * F^2). The correction grows rapidly with span (proportional to l^3) and decreases with the square of the applied pull. Using consistent gravitational units (kgf and meters) keeps the expression dimensionally coherent for field calculations.

Step-by-Step Solution:

Verification / Alternative check:Compare with multi-support case by summing corrections for each equal span; end-supported 50 m spans yield much larger corrections than 10 m spans, matching practice.

Why Other Options Are Wrong:

• (w * l^2) / (8 * F) is the mid-sag (dip), not the length correction.
• Expressions with F^2 in the numerator or wrong powers of l do not match catenary-derived relationships.

Common Pitfalls:Mixing SI Newtons with kilogram-force without conversion; forgetting the correction is always subtractive.

Final Answer:c = (w^2 * l^3) / (24 * F^2)