What is the tension force in the wire shown below if it stretches 0.35 cm when tightened if young's modulus for the wire is 2 × 1011 N/m² 2.5 m 0.4 cm a) 1374 N b) 1978 N c) 2693 N d) 3517N
What is the tension force in the wire shown below if it stretches 0.35 cm when tightened if young's modulus for the wire is 2 × 1011 N/m² 2.5 m 0.4 cm a) 1374 N b) 1978 N c) 2693 N d) 3517N
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![## Problem Statement: Calculation of Tension Force in a Wire
### Given Data:
- **Length of the Wire (L):** 2.5 m
- **Diameter of the Wire:** 0.4 cm (Radius = 0.2 cm)
- **Extension of the Wire (ΔL):** 0.35 cm
- **Young's Modulus (E):** \(2 \times 10^{11} \, \text{N/m}^2\)
### Question:
What is the tension force in the wire shown below if it stretches 0.35 cm when tightened if Young's modulus for the wire is \(2 \times 10^{11} \, \text{N/m}^2\)?
### Options:
a) 1374 N
b) 1978 N
c) 2693 N
d) 3517 N
### Diagram Explanation:
The diagram provided illustrates a wire of length 2.5 meters, with one end fixed and the other end where the tension force is applied causing the extension. The diameter of the wire is 0.4 cm, and it shows a circular cross-section to emphasize the diameter for calculation of the area.
### Solution Steps:
1. **Calculate the Cross-Sectional Area (A) of the Wire:**
Given radius (r) = 0.2 cm = 0.002 m
\[
A = \pi r^2 = \pi (0.002)^2 = \pi \times 4 \times 10^{-6} \approx 1.2566 \times 10^{-5} \, \text{m}^2
\]
2. **Use Hooke's Law for Stress and Strain:**
\[
\text{Strain} = \frac{\Delta L}{L} = \frac{0.35 \, \text{cm}}{250 \, \text{cm}} = \frac{0.0035}{2.5} = 0.0014
\]
\[
\text{Stress} = E \times \text{Strain} = 2 \times 10^{11} \times 0.0014 = 2.8 \times 10^8 \, \text{N/m}^2
\]
3. **Calculate the Tension](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbeae604a-2e4b-47ef-907a-313aaf16b611%2F4c9daeb6-d4e8-4df6-add3-3d4692906b93%2F2rox1ep_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Problem Statement: Calculation of Tension Force in a Wire
### Given Data:
- **Length of the Wire (L):** 2.5 m
- **Diameter of the Wire:** 0.4 cm (Radius = 0.2 cm)
- **Extension of the Wire (ΔL):** 0.35 cm
- **Young's Modulus (E):** \(2 \times 10^{11} \, \text{N/m}^2\)
### Question:
What is the tension force in the wire shown below if it stretches 0.35 cm when tightened if Young's modulus for the wire is \(2 \times 10^{11} \, \text{N/m}^2\)?
### Options:
a) 1374 N
b) 1978 N
c) 2693 N
d) 3517 N
### Diagram Explanation:
The diagram provided illustrates a wire of length 2.5 meters, with one end fixed and the other end where the tension force is applied causing the extension. The diameter of the wire is 0.4 cm, and it shows a circular cross-section to emphasize the diameter for calculation of the area.
### Solution Steps:
1. **Calculate the Cross-Sectional Area (A) of the Wire:**
Given radius (r) = 0.2 cm = 0.002 m
\[
A = \pi r^2 = \pi (0.002)^2 = \pi \times 4 \times 10^{-6} \approx 1.2566 \times 10^{-5} \, \text{m}^2
\]
2. **Use Hooke's Law for Stress and Strain:**
\[
\text{Strain} = \frac{\Delta L}{L} = \frac{0.35 \, \text{cm}}{250 \, \text{cm}} = \frac{0.0035}{2.5} = 0.0014
\]
\[
\text{Stress} = E \times \text{Strain} = 2 \times 10^{11} \times 0.0014 = 2.8 \times 10^8 \, \text{N/m}^2
\]
3. **Calculate the Tension
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