Solution Found!
This problem returns to the tightrope walker studied in
Chapter 5, Problem 43(choose chapter or problem)
This problem returns to the tightrope walker studied in Example \(4.6\), who created a tension of \(3.94 \times 10^{3} \mathrm{~N}\) in a wire making an angle \(5.0^{\circ}\) below the horizontal with each supporting pole. Calculate how much this tension stretches the steel wire if it was originally \(15 \mathrm{~m}\) long and \(0.50 \mathrm{~cm}\) in diameter.
Equation Transcription:
Text Transcription:
4.6
3.94 times 10^3 N
5.0 degrees
15 m
0.50 cm
Questions & Answers
QUESTION:
This problem returns to the tightrope walker studied in Example \(4.6\), who created a tension of \(3.94 \times 10^{3} \mathrm{~N}\) in a wire making an angle \(5.0^{\circ}\) below the horizontal with each supporting pole. Calculate how much this tension stretches the steel wire if it was originally \(15 \mathrm{~m}\) long and \(0.50 \mathrm{~cm}\) in diameter.
Equation Transcription:
Text Transcription:
4.6
3.94 times 10^3 N
5.0 degrees
15 m
0.50 cm
ANSWER:
Problem 43PE
This problem returns to the tightrope walker studied in Example 4.6, who created a tension of 
Example 4.6:
Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure 4.17
Figure 4.17 The weight of a tightrope walker causes a wire to saf 5.0 degrees. The system of interest here is the wire at which the tightrope walker is standing.
Step by step solution
Step 1 of 2
The change in length due to applied force can be calculated by using the formula below as
Here, 
applied and 