1. **State the problem:** We have a measured distance between points A and B as 308.955 units at field conditions with temperature -21°C and applied tension 27 lbs. Points A and B differ in elevation by 3.635 units (B is lower). We need to find the correct horizontal distance between A and B. 2. **Given data:** - Original length $L_0 = 30.037$ at 20°C and 20 lbs tension - Cross-sectional area $A = 0.009$ - Modulus of elasticity $E = 29 \times 10^6$ - Weight per unit length $W = 2.00$ lbs - Field temperature $T_f = -21^\circ C$ - Field tension $P_f = 27$ lbs - Measured length $L_m = 308.955$ - Elevation difference $h = 3.635$ 3. **Calculate temperature change:** $$\Delta T = T_f - 20 = -21 - 20 = -41^\circ C$$ 4. **Calculate change in length due to temperature:** Assuming linear expansion coefficient $\alpha$ is unknown, but since not given, we focus on tension and weight effects. 5. **Calculate elongation due to tension change:** Initial tension $P_0 = 20$ lbs, field tension $P_f = 27$ lbs Change in tension $\Delta P = P_f - P_0 = 7$ lbs Elongation $\Delta L = \frac{L_0 \Delta P}{A E} = \frac{30.037 \times 7}{0.009 \times 29 \times 10^6} = \frac{210.259}{261000} \approx 0.000805$ units 6. **Calculate elongation due to weight:** Weight per unit length $W = 2$ lbs, length $L_0 = 30.037$ Total weight $W_t = W \times L_0 = 2 \times 30.037 = 60.074$ lbs Elongation due to weight $\Delta L_w = \frac{W_t L_0}{2 A E} = \frac{60.074 \times 30.037}{2 \times 0.009 \times 29 \times 10^6} = \frac{1804.4}{522000} \approx 0.003455$ units 7. **Total elongation:** $$\Delta L_{total} = \Delta L + \Delta L_w = 0.000805 + 0.003455 = 0.00426$$ 8. **Corrected length at field conditions:** $$L_c = L_0 + \Delta L_{total} = 30.037 + 0.00426 = 30.04126$$ 9. **Calculate horizontal distance:** Measured distance $L_m = 308.955$ Elevation difference $h = 3.635$ Horizontal distance $d = \sqrt{L_m^2 - h^2} = \sqrt{308.955^2 - 3.635^2} = \sqrt{95436.5 - 13.21} = \sqrt{95423.29} \approx 308.91$ **Final answer:** The correct horizontal distance between points A and B is approximately **308.91** units.