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A metal wire 3.50 \(\mathrm{m}\) long and 0.70 \(\mathrm{mm}\) in diameter was given the following test. A load weighing 20 \(\mathrm{N}\) was originally hung from the wire to keep it taut. The position of the lower end of the wire was read on a scale as load was added (a) Graph these values, plotting the increase in length horizontally and the added load vertically. (b) Calculate the value of Young's modulus. (c) The proportional limit occurred at a scale reading of 3.34 \(\mathrm{cm} .\) What was the stress at this point?

Short Answer

Expert verified
(a) Plot length vs load graph with test values you collect. (b) Young's modulus \( \approx 5.44 \times 10^9 \, \text{N/m}^2 \). (c) Stress at proportional limit \( \approx 5.19 \times 10^7 \, \text{N/m}^2 \).

Step by step solution

01

Calculate Initial Parameters

First, we need to find the initial values needed for Young's modulus.- Original Length, \( L = 3.50 \) m- Diameter, \( d = 0.70 \) mm = 0.0007 m- Cross-sectional Area, \( A = \frac{\pi d^2}{4} \)Let's calculate the cross-sectional area:\[ A = \frac{\pi (0.0007)^2}{4} \approx 3.85 \times 10^{-7} \text{ m}^2 \]
02

Understand Graph Plotting Requirement

We are asked to plot the graph with "increase in length horizontally and load vertically." However, the exact values for elasticity tests are not provided, so assume measurements were taken as weights were added. Use these generic instructions to guide actual plotting.
03

Young's Modulus Formula

Young's modulus \( E \) can be calculated using:\[E = \frac{\text{Stress}}{\text{Strain}} = \frac{\sigma}{\epsilon}\]Where:- Stress, \( \sigma = \frac{\text{Force}}{\text{Area}} \)- Strain, \( \epsilon = \frac{\Delta L}{L} \)We need to calculate stress and strain before finding Young's modulus.
04

Calculate Stress and Strain at Proportional Limit

Given a scale reading of 3.34 cm for the proportional limit, calculate the new length increase:- Change in Length, \( \Delta L = 3.34 \) cm = 0.0334 m**Stress:**\[\sigma = \frac{\text{Force}}{\text{Area}} = \frac{20}{3.85 \times 10^{-7}} \approx 5.19 \times 10^7 \text{ N/m}^2\]**Strain:**\[\epsilon = \frac{\Delta L}{L} = \frac{0.0334}{3.50} \approx 0.00954\]
05

Calculate Young's Modulus

Using the calculated stress and strain, determine Young's modulus:\[E = \frac{5.19 \times 10^7}{0.00954} \approx 5.44 \times 10^{9} \, \text{N/m}^2\]
06

Determine Stress at Proportional Limit

The stress at the proportional limit was already calculated in Step 4 as:\[\sigma = 5.19 \times 10^7 \, \text{N/m}^2\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

stress and strain
When you exert force on an object, it undergoes deformation. This phenomenon is explained by two fundamental concepts—stress and strain. Stress is defined as the force applied per unit area of the material. Essentially, it's a measure of how much force is acting on a specific area. It's expressed in Newton per square meter (N/m²).Strain, on the other hand, is a measure of deformation. It indicates how much the material has stretched or compressed compared to its original length. It's the ratio of the change in length (\( \Delta L \)) to the original length (\( L \)) and has no unit since it's a ratio of lengths. Relative deformation of a wire under stress can be calculated using the formula:- **Stress (\( \sigma \))**: \( \frac{\text{Force (N)}}{\text{Area (m}^2\text{)}} \)- **Strain (\( \epsilon \))**: \( \frac{\Delta L}{L} \)Understanding these concepts is crucial when studying material properties as they help us predict how materials will behave under different types of forces.
After knowing these values, you can find other important mechanical properties like Young's Modulus.
proportional limit
The proportional limit is a key concept in understanding the elastic behavior of materials. It is a point in the stress-strain curve up to which the material behaves, according to Hooke’s Law, which means stress and strain are directly proportional. This implies that the material will return to its original shape after the load is removed.In our exercise, the proportional limit is noted at a scale reading of 3.34 cm, which represents the increase in length (\( \Delta L \)). Beyond this point, materials often undergo plastic deformation, meaning they won't return exactly to their original configuration.At the proportional limit, materials still remain elastic and follow the formula:- **Linear relationship**: \( \sigma = E \cdot \epsilon \)Here, \( E \) is Young's Modulus, representing material stiffness.Keep in mind that different materials have unique proportional limits, defining their utility depending on the application in engineering and construction domains. Recognizing this limit is vital for ensuring the structural integrity and performance of materials in practical applications.
elasticity test
An elasticity test is often conducted to determine how materials react under stress. During such a test, as explained in our exercise, you measure how different weights affect the material's length. The core objective is to understand the elasticity or the ability of a material to return to its original dimensions after removing the load. The elasticity test often involves: - **Initial Setup**: Measuring the original dimensions of the material such as length and diameter.
- **Applying Load**: Successively adding incremental weights and noting down how much the length of the material changes with each weight.
- **Recording Data**: Observing and recording the load versus the increase in length until the proportional limit is reached. In practice, this test helps in characterizing material behavior, and how much stress it can endure while maintaining the elastic properties. Such tests provide crucial insights, especially when picking materials for construction, manufacturing, and various other industrial applications. They can also highlight the onset of permanent deformation marks when the material returns partly to its original state after the removal of the load.

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Most popular questions from this chapter

Two friends are carrying a 200 -kg crate up a flight of stairs. The crate is 1.25 \(\mathrm{m}\) long and 0.500 \(\mathrm{m}\) high, and its center of gravity is at its center. The stairs make a \(45.0^{\circ}\) angle with respect to the floor. The crate also is carried at a \(45.0^{\circ}\) angle, so that its bottom side is parallel to the slope of the stairs (Fig. P11.71). If the force each person applies is vertical, what is the magnitude of each of these forces? Is it better to be the person above or below on the stairs?

A bookcase weighing 1500 \(\mathrm{N}\) rests on a horizon- tal surface for which the coefficient of static friction is \(\mu_{\mathrm{s}}=0.40 .\) The bookcase is 1.80 \(\mathrm{m}\) tall and 2.00 \(\mathrm{m}\) wide; its center of gravity is at its geo- metrical center. The bookcase rests on four short legs that are each 0.10 \(\mathrm{m}\) from the edge of the bookcase. A person pulls on a rope attached to an upper corner of the bookcase with a force \(\vec{\boldsymbol{F}}\) that makes an angle \(\theta\) with the bookcase (Fig. \(P 11.97 )\) . (a) If \(\theta=90^{\circ},\) so \(\vec{F}\) is horizontal, show that as \(F\) is increased from zero, the bookcase will start to slide before it tips, and calculate the magnitude of \(\vec{\boldsymbol{F}}\) that will start the bookcase sliding. (b) If \(\theta=0^{\circ},\) so \(\vec{\boldsymbol{F}}\) is vertical, show that the bookcase will tip over rather than slide, and calculate the magnitude of \(\vec{\boldsymbol{F}}\) that will cause the bookcase to start to tip. (c) Calculate as a function of \(\theta\) the magnitude of \(\vec{\boldsymbol{F}}\) that will cause the bookcase to start to slide and the magnitude that will cause it to start to tip. What is the smallest value that \(\theta\) can have so that the bookcase will still start to slide before it starts to tip?

A door 1.00 m wide and 2.00 m high weighs 280 N and is supported by two hinges, one 0.50 \(\mathrm{m}\) from the top and the other 0.50 \(\mathrm{m}\) from the bottom. Each hinge supports half the total weight of the door. Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one- fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

A 350 -N, uniform, 1.50 -m bar is suspended horizontally by two vertical cables at each end. Cable \(A\) can support a maximum tension of 500.0 \(\mathrm{N}\) without breaking, and cable \(B\) can support up to 400.0 \(\mathrm{N} .\) You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

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