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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. $$ \begin{array}{cc}{\text { Added Load (N) }} & {\text { Scale Reading (cm) }} \\\ \hline 0 & {3.02} \\ {10} & {3.07} \\ {20} & {3.12} \\ {30} & {3.17} \\\ {40} & {3.22} \\ {50} & {3.27} \\ {60} & {3.32} \\ {70} & {4.27}\end{array} $$ (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
Young's Modulus is found from the linear portion of the stress-strain graph. The stress at the proportional limit is calculated using the corresponding force and cross-sectional area.

Step by step solution

01

Convert the scale readings to wire elongation

First, determine the actual elongation of the wire. The original scale reading is 3.02 cm (with no added load). To find how much the wire stretches, subtract 3.02 cm from each subsequent scale reading. This gives the elongation when different loads are added.
02

Set up the graph

Using the elongation values from Step 1 for the x-axis and the added load values for the y-axis, create a graph. This will plot the increase in length (elongation in cm) against the added load in Newtons.
03

Calculate Young's Modulus

Young's Modulus, E, is calculated using the formula:\[ E = \frac{\text{Stress}}{\text{Strain}} \]Stress is the force per unit area, \( \sigma = \frac{F}{A} \), where \( F \) is the force and \( A \) is the cross-sectional area. The area \( A \) can be calculated from the diameter: \( A = \pi \left( \frac{d}{2} \right)^2 \). Strain is the relative deformation, \( \text{Strain} = \frac{\Delta L}{L_0} \), where \( \Delta L \) is the change in length and \( L_0 \) is the original length. Use these relationships to find the modulus between data points for the linear part of the graph (ignoring the point where the reading is 4.27 cm, which indicates a deviation from linearity).
04

Calculate the stress at the proportional limit

The proportional limit is reached when the scale reads 3.34 cm. The elongation here is 3.34 cm - 3.02 cm = 0.32 cm = 0.0032 m. Stress (\( \sigma \)) is calculated as:\[ \sigma = \frac{F}{A} \]The force corresponding to this elongation is determined from the graph or known data points just before 3.34 cm. Use the area calculated in Step 3 to find the stress at this point.

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Key Concepts

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

Understanding Stress and Strain
Stress and strain are fundamental concepts in physics that help describe how materials deform under forces. When a force is applied to an object, it can change shape, shorten, or elongate. This change is described in terms of stress and strain.

Stress (\(\sigma\)) is the force applied to a material per unit cross-sectional area. It is measured in pascals (\(\text{Pa}\)) and is calculated using the equation: \[\sigma = \frac{F}{A}\]where \(F\) is the force in newtons and \(A\) is the cross-sectional area in square meters.

Strain, on the other hand, is a measure of deformation and is the ratio of the change in length to the original length. It is a dimensionless quantity, expressed as: \[\text{Strain} = \frac{\Delta L}{L_0}\]where \(\Delta L\) is the change in length and \(L_0\) is the original length.

The relationship between stress and strain is often linear in the initial stages, which is a crucial principle in material science. This relationship is represented by Young's Modulus, an essential parameter for evaluating the elasticity of materials. Recognizing the stress and strain relationship helps in predicting how materials will behave under different loading conditions.
Exploring the Proportional Limit
The proportional limit is a pivotal point on the stress-strain graph where the linear relationship between stress and strain ends. Up to this limit, stress and strain have a directly proportional relationship, meaning the material behaves elastically — like a stretched spring that returns to its size when the load is removed.

Beyond this point, the material starts to deform plastically, and the deformation is not fully recoverable. In our exercise, the proportional limit occurred at a scale reading of 3.34 cm. At this point, the elongation was 0.32 cm, equivalent to 0.0032 m.

To calculate the stress (\(\sigma\)) at this limit, the force just before this elongation is needed. Using the known relationship between stress and force, and deriving the cross-sectional area from the given diameter, we find: \[\sigma = \frac{F}{A}\]

This stress value at the proportional limit allows us to assess the maximum elastic stress the material can withstand.
Graphing in Physics: Visualizing Data
Graphs are invaluable tools in physics and other sciences, allowing for clear visualization of relationships between variables. In this exercise, graphing the elongation of the wire against the added load provides a visual representation of the stress-strain relationship.

Starting with plotting elongation on the x-axis and force on the y-axis, a linear portion is usually evident for metals below the proportional limit. This indicates a constant modulus of elasticity, helping to determine Young’s Modulus — a measure of stiffness.

The key points to ensure in graphing include:
  • Accurately labeling axes with correct units.
  • Marking data points clearly.
  • Drawing a best-fit line to highlight the linear region effectively.


Such graphs not only provide a clear way to determine critical points like the proportional limit but also offer insights into a material's mechanical properties and behavior under various loads. This understanding is crucial for the fields of engineering and material science.

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Most popular questions from this chapter

A uniform drawbridge must be held at a \(37^{\circ}\) angle above the horizontal to allow ships to pass underneath. The drawbridge weighs \(45,000 \mathrm{N}\) and is 14.0 \(\mathrm{m}\) long. A cable is connected 3.5 \(\mathrm{m}\) from the hinge where the bridge pivots (measured along the bridge) and pulls horizontally on the bridge to hold it in place. (a) What is the tension in the cable? (b) Find the magnitude and direction of the force the hinge exerts on the bridge.

A nonuniform fire escape ladder is 6.0 \(\mathrm{m}\) long when extended to the icy alley below. It is held at the top by a frictionless pivot, and there is neghigible frictional force from the icy surface at the bottom. The ladder weighs 250 \(\mathrm{N}\) , and its center of gravity is 2.0 \(\mathrm{m}\) along the ladder from its bottom. A mother and child of total weight 750 \(\mathrm{N}\) are on the ladder 1.5 \(\mathrm{m}\) from the pivot. The ladder makes an angle \(\theta\) with the horizontal. Find the magnitude and direction of \((a)\) the force exerted by the icy alley on the ladder and (b) the force exerted by the ladder on the pivot. (c) Do your answers in parts (a) and (b) depend on the angle \(\theta ?\)

A uniform ladder 5.0 \(\mathrm{m}\) long rests against a frictionless, vertical wall with its lower end 3.0 \(\mathrm{m}\) from the wall. The ladder weighs 160 \(\mathrm{N}\) . The coefficient of static betwoen the foot of the ladder and the ground is \(0.40 .\) A man weighing 740 \(\mathrm{N}\) climbs slowly up the ladder. Start by drawing a free-body diagram of the ladder. (a) What is the maximum frictional force that the ground can exert on the ladder at its lower end? (b) What is the actual frictional force when the man has climbed 1.0 \(\mathrm{m}\) along the ladder? (c) How far along the ladder can the man climb before the ladder starts to slip?

A steel cable with cross-sectional area 3.00 \(\mathrm{cm}^{2}\) has an elastic limit of \(2.40 \times 10^{8}\) Pa. Find the maximum upward acceleration that can be given a \(1200-\mathrm{kg}\) elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of 90.8 \(\mathrm{N}\) is applied perpendicular to each end. If the diameter of the wire is \(1.84 \mathrm{mm},\) what is the breaking stress of the alloy?

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