Question

a 50 - m tape is used to measure between two points. the average weight of the tape per meter is 0.20 n. if the measured distance is 48.88 m, with the tape supported at the ends only and with a tension of 100 n, find the corrected distance.
a distance of 72.53 ft is recorded using a steel tape supported only at the ends with a tension of 15 lb and weighing 0.016 lb per foot. find the distance corrected for sag.

Explanation:

Step1: Recall sag - correction formula

The sag - correction formula for a tape supported only at the ends is $C_s=-\frac{w^{2}L^{3}}{24T^{2}}$, where $w$ is the weight per unit length of the tape, $L$ is the measured length, and $T$ is the tension in the tape.

Step2: First, convert units for the first case

For the first case:
The measured length $L_1 = 72.53\ ft$. The weight per unit length $w_1=0.016\ lb/ft$ and tension $T_1 = 15\ lb$.
First, substitute the values into the sag - correction formula:
$C_{s1}=-\frac{w_1^{2}L_1^{3}}{24T_1^{2}}$
$w_1^{2}=(0.016)^{2}=0.000256\ lb^{2}/ft^{2}$, $L_1^{3}=(72.53)^{3}=72.53\times72.53\times72.53 = 382979.97\ ft^{3}$
$24T_1^{2}=24\times15^{2}=24\times225 = 5400\ lb^{2}$
$C_{s1}=-\frac{0.000256\times382979.97}{5400}$
$C_{s1}=-\frac{97.042862}{5400}\approx - 0.018\ ft$
The corrected length $L_{c1}=L_1 + C_{s1}=72.53-0.018 = 72.512\ ft$

Step3: For the second case

The measured length $L_2 = 48.88\ m$. The weight per unit length $w_2 = 0.20\ N/m$ and tension $T_2=100\ N$.
Substitute into the sag - correction formula $C_{s2}=-\frac{w_2^{2}L_2^{3}}{24T_2^{2}}$
$w_2^{2}=(0.20)^{2}=0.04\ N^{2}/m^{2}$, $L_2^{3}=(48.88)^{3}=48.88\times48.88\times48.88=116997.97\ m^{3}$
$24T_2^{2}=24\times100^{2}=240000\ N^{2}$
$C_{s2}=-\frac{0.04\times116997.97}{240000}$
$C_{s2}=-\frac{4679.9188}{240000}\approx - 0.0195\ m$
The corrected length $L_{c2}=L_2 + C_{s2}=48.88-0.0195 = 48.8605\ m$

Answer:

For the first case: $72.512\ ft$; For the second case: $48.8605\ m$