A 30-m steel tape weighing 0.7 kg has a cross-sectional area of 0.03 cm2 and a standard length of
30.005 m when fully supported, at 20 °C, and with 5 kg of tension.
a. A line on a smooth level ground is measured with the tape under a tension of 10 kg at a temperature
of 20 °C and is recorded as 1242.823 m long. Compute the length of the line applying the corrections.
E = 2,100,000 kg/cm2.
b. This same tape is used to measure a distance recorded as 985.423 m over level terrain with a tape
supported at the two ends and 5 kg of tension (temperature = 20 °C). Calculate the correction due
to sag for this distance.

We are tasked with calculating two corrections for measurements made using a steel tape. First, we'll compute the corrected length for a line measured under a different tension, and second, we'll calculate the sag correction for a different measurement.

### Part A: Correcting for Tension

Given dataWe are tasked with calculating two corrections for measurements made using a steel tape. First, we'll compute the corrected length for a line measured under a different tension, and second, we'll calculate the sag correction for a different measurement.

### Part A: Correcting for Tension

Given 
- Length of the steel tape = 30 m
- Weight of the tape = 0.7 kg
- Cross-sectional area = \(0.03 \, \text{cm}^2 = 0.000003 \, \text{m}^2\)
- Standard length of the tape = 30.005 m
- Temperature = 20°C (no temperature correction needed)
- Initial tension = 5 kg
- New tension = 10 kg
- Measured length = 1242.823 m
- Young’s modulus (E) = \(2,100,000 \, \text{kg/cm}^2 = 2.1 \times 10^9 \, \text{kg/m}^2\)

#### Steps:
1. **Find the elongation due to tension**:
   The elongation correction for tension can be found using the formula:

   \[
   C_T = \frac{(T_2 - T_1) \times L_m}{A \times E}
   \]
   where:
   - \(T_1\) = original tension = 5 kg
   - \(T_2\) = new tension = 10 kg
   - \(L_m\) = measured length = 1242.823 m
   - \(A\) = cross-sectional area = \(0.03 \, \text{cm}^2 = 0.000003 \, \text{m}^2\)
   - \(E\) = Young's modulus = \(2.1 \times 10^9 \, \text{kg/m}^2\)

Let's compute this correction.

#### Part B: Sag Correction

Given data for part B:
- Measured distance = 985.423 m
- Tape is supported only at two ends.
- Weight of the tape per unit length = \( \frac{0.7}{30} = 0.0233 \, \text{kg/m}\)
- Tension applied = 5 kg
- Standard temperature = 20°C (no temperature correction needed).

#### Steps:
1. **Sag correction**:
   The correction due to sag for a tape supported at two ends can be calculated using the formula:

   \[
   C_s = -\frac{w^2 L_m^3}{24 T^2}
   \]
   where:
   - \(w\) = weight of the tape per unit length = 0.0233 kg/m
   - \(L_m\) = measured length = 985.423 m
   - \(T\) = tension = 5 kg

Now, I'll compute the corrections for both parts.

### Part A: Correction for Tension

Let me compute the correction due to tension first.The correction due to the change in tension for Part A is \( C_T = 0.986 \, \text{m} \).

To find the corrected length of the line, we add this correction to the recorded measurement:

\[
L_{\text{corrected}} = L_m + C_T = 1242.823 + 0.986 = 1243.809 \, \text{m}
\]

Thus, the corrected length of the line is \( 1243.809 \, \text{m} \).

---

### Part B: Sag Correction

Now, let's compute the sag correction for Part B.The sag correction for Part B is \( C_s = -868.301 \, \text{mm} \) or \( -0.868 \, \text{m} \).

This correction reduces the measured length due to the sag of the tape. Therefore, the corrected distance would be:

\[
L_{\text{corrected}} = L_m + C_s = 985.423 - 0.868 = 984.555 \, \text{m}
\]

Thus, the corrected distance accounting for sag is \( 984.555 \, \text{m} \).

---

Would you like further details or have any other questions?

Here are 5 related questions for deeper understanding:
1. How would the correction change if the tape were fully supported across its entire length?
2. What effect does temperature have on the measurement correction?
3. Can this method be used to measure vertical distances? Why or why not?
4. How would a change in the cross-sectional area of the tape affect the correction?
5. What is the importance of accurate tape calibration in large-scale projects?

**Tip:** Always consider both sag and tension corrections when measuring long distances with tapes to maintain high accuracy.

A 30-m steel tape weighing 0.7 kg has a cross-sectional area of 0.03 cm² and a standard length of 30.005 m when fully supported, at 20 °C, and with 5 kg of tension. 
a. A line on a smooth level ground is measured with the tape under a tension of 10 kg at a temperature of 20 °C and is recorded as 1242.823 m long. Compute the length of the line applying the corrections. E = 2,100,000 kg/cm². 
b. This same tape is used to measure a distance recorded as 985.423 m over level terrain with a tape supported at the two ends and 5 kg of tension (temperature = 20 °C). Calculate the correction due to sag for this distance.

This detailed solution explains how to apply corrections to steel tape measurements for tension and sag. Using the elongation correction formula and sag correction formula, learn to compute adjustments for measurements under varying tensions and unsupported distances. Perfect for engineering students or professionals working with precise distance measurements.

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