Chaining — sag correction for a suspended tape: A tape of length l and weight W (kg per m) is supported at its ends under a pull P (kg). What is the sag correction to be applied to the measured length?
Correct Answer: c_sag = (W^2 * l^3) / (24 * P^2)
Introduction / Context:When a steel tape is supported only at its ends, it sags under its own weight, making the measured horizontal distance longer than the true straight-line distance. A sag correction must be subtracted to obtain the correct length. This question asks for the standard formula in consistent units.
Given Data / Assumptions:
- Tape length = l (m).
- Tape weight per unit length = W (kg/m) representing weight units consistent with pull P.
- Pull (tension) applied at the ends = P (kg).
- Small-sag approximation and symmetric suspension are assumed.
Concept / Approach:Under uniform weight, the tape assumes a catenary; for small sag, the correction is derived from catenary approximations and proportional to W^2 and l^3, and inversely proportional to P^2. The correction is always positive in value and is subtracted from the measured length (because sag makes the measured length too long).
Step-by-Step Solution:
Model sag with small-deflection catenary assumptions.Apply the standard result: c_sag = (w^2 * L^3) / (24 * T^2).Map variables: w → W, L → l, T → P (consistent units).Therefore, c_sag = (W^2 * l^3) / (24 * P^2), to be subtracted from the measured length.Verification / Alternative check:Surveying texts list this formula; note that increasing tension P or shortening span l lowers the correction rapidly (P^2 and l^3 dependence).
Why Other Options Are Wrong:
- (a) and (d): Wrong dimensional dependence; do not match catenary-based form.
- (c): Inverted dependence on W and P; physically incorrect.
Common Pitfalls:Forgetting that the sag correction is subtractive; mixing force/weight units—keep W and P consistent (both treated as kg-force units here).
Final Answer:c_sag = (W^2 * l^3) / (24 * P^2)