Difficulty: Medium
Correct Answer: 0.053 m
Explanation:
Introduction / Context:
When measuring distances with a tape carried between supports, the tape sags and the measured length exceeds the true horizontal distance. A sag correction must be subtracted. The magnitude depends on span length, tape weight per unit length, and applied pull. Field practice often supports a long tape at equal intervals to reduce the correction.
Given Data / Assumptions:
Concept / Approach:
Sag makes the measured distance too long. The correction is derived from the catenary approximation for small sag and uniform load. Supporting at intermediate points reduces l and thus the cubic dependence on span length, dramatically lowering the total correction compared to an end-supported 50 m span.
Step-by-Step Solution:
Verification / Alternative check:
If the tape were supported only at its ends (one 50 m span), c = (0.08^2 * 50^3) / (24 * 25) ≈ 1.33 m, showing why intermediate supports are used; the 10 m support interval yields a much smaller, practical correction.
Why Other Options Are Wrong:
Common Pitfalls:
Applying the end-support formula to a multi-support setup; mixing SI units (N) with kgf without consistency; forgetting to subtract the correction from the measured length.
Final Answer:
0.053 m
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