A 100 m tape is suspended between the ends under a pull of 200 N. If the weight of the tape is 30 N, the correct distance between the tape ends will be nearly

Question

Question

This question was previously asked in

ESE Civil 2020: Official Paper

Answer 1

Concept:

Sag correction is given by:

\(C = \frac{{{W^2}l}}{{24{P^2}{}}}\)

Where,

l = the length of the tape (in metres) suspended between supports

P = Pull applied in kg or N

W = weight of the tape in kg or N

C = Sag correction in metres for length (l)

Calculation:

Correction, \(C = \frac{{{W^2}l}}{{24{P^2}{}}}\)

\(= \frac{{{{30}^2} \times 100}}{{24 \times {{200}^2} {}}}\)

= 0.09375 m

Correct distance = 100 – 0.09375 = 99.90 m