Question

What will be the sag correction for a 100 m tape which is suspended between the ends under a pull of 10 N? The total weight of the tape is 12 N.

• A. 1 m
• B. 6 m
• C. 3 m
• D. 12 m

Hint:

For sag correction: \[C_s = \frac{w^2 L^3}{24P^2}.\] Remember to convert total weight into weight per unit length.

Answer 1

The Correct Option is B

The sag correction for a tape is calculated using the formula:
\[C_s = \frac{w^2 L^3}{24P^2}\]
Where:
 $C_s =$ sag correction,
   $w =$ weight of the tape per unit length,
   $L =$ length of the tape,
   $P =$ applied pull.
Given:
   Total weight of the tape = 12 N,
   Length of the tape, $L = 100 \, \text{m}$,
   Applied pull, $P = 10 \, \text{N}$.
The weight per unit length is:
\[w = \frac{\text{Total weight}}{\text{Length}} = \frac{12}{100} = 0.12 \, \text{N/m}.\]
Substitute these values into the formula:
\[C_s = \frac{(0.12)^2 (100)^3}{24(10)^2}\]
Simplify:
\[C_s = \frac{0.0144 \times 1000000}{24 \times 100}\]
\[C_s = \frac{14400}{2400}\]
Calculate:
\[C_s = 6 \, \text{m}.\]