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Chapter 11: Problem 27

A metal rod that is 4.00 \(\mathrm{m}\) long and 0.50 \(\mathrm{cm}^{2}\) in cross sectional area is found to stretch 0.20 \(\mathrm{cm}\) under a tension of 5000 \(\mathrm{N.}\) What is Young's modulus for this metal?

Short Answer

Expert verified
Young's modulus is \( 2.00 \times 10^{11} \, \text{N/m}^2 \).

Step by step solution

01

Identify Known Values

First, we need to identify the given values from the problem statement: - Original length of the rod, \( L_0 = 4.00 \, \text{m} \)- Cross-sectional area, \( A = 0.50 \, \text{cm}^2 = 0.50 \, \times \, 10^{-4} \, \text{m}^2 \)- Tension force applied, \( F = 5000 \, \text{N} \)- Extension in length, \( \Delta L = 0.20 \, \text{cm} = 0.0020 \, \text{m} \)
02

Understand Young's Modulus

Young's modulus \( E \) is defined as the ratio of tensile stress to tensile strain in a material:\[ E = \frac{\text{Tensile Stress}}{\text{Tensile Strain}} \]Tensile stress is given by \( \frac{F}{A} \) and tensile strain is \( \frac{\Delta L}{L_0} \).
03

Calculate Tensile Stress

The tensile stress \( \sigma \) is calculated using the formula:\[ \sigma = \frac{F}{A} \]Substitute the given values:\[ \sigma = \frac{5000 \, \text{N}}{0.50 \, \times \, 10^{-4} \, \text{m}^2} = 1.00 \, \times \, 10^8 \, \text{N/m}^2 \]
04

Calculate Tensile Strain

Tensile strain \( \varepsilon \) is the ratio of extension in length to the original length:\[ \varepsilon = \frac{\Delta L}{L_0} \]Plug in the values:\[ \varepsilon = \frac{0.0020 \, \text{m}}{4.00 \, \text{m}} = 5.00 \, \times \, 10^{-4} \]
05

Compute Young's Modulus

Using the formula for Young's modulus:\[ E = \frac{\sigma}{\varepsilon} \]Substitute the values:\[ E = \frac{1.00 \, \times \, 10^8 \, \text{N/m}^2}{5.00 \, \times \, 10^{-4}} = 2.00 \, \times \, 10^{11} \, \text{N/m}^2 \]
06

Conclusion

Young's modulus for the metal rod is \( 2.00 \, \times \, 10^{11} \, \text{N/m}^2 \).

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Key Concepts

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

Tensile Stress
Tensile stress refers to the force exerted on a material as it is being stretched or pulled. It is a measure of how much internal force the material is experiencing per unit area when subjected to pulling forces. In the context of our exercise, tensile stress is calculated by dividing the tension force applied to the rod by its cross-sectional area. This relationship is expressed in the formula below:

\[\text{Tensile Stress} \sigma = \frac{F}{A}\]
Here, \(F\) is the applied force, and \(A\) is the cross-sectional area. It's important to remember that tensile stress is expressed in units of pressure, such as pascals (Pa) or newtons per square meter (N/m虏).

Calculating tensile stress helps us understand the external load's impact on the material, which is crucial in assessing whether a material or structure can withstand the applied forces without breaking or deforming.
Tensile Strain
Tensile strain measures how much a material deforms compared to its original length when a force is applied. It is a dimensionless quantity, as it represents the relative change in length. In mathematical terms, tensile strain is expressed as:

\[\text{Tensile Strain} \varepsilon = \frac{\Delta L}{L_0}\]
Here, \(\Delta L\) is the change in length of the material, and \(L_0\) is the original length.

Tensile strain tells us about the elasticity of a material; materials that show small strain for a given stress are considered stiff or rigid, while larger strains correspond to more flexible materials. It also helps in understanding the material's potential for elastic deformation, which is essential for designing structures and materials that need to maintain their shape under stress.
Elasticity
Elasticity is a fundamental property of materials that describes their ability to return to the original shape or size after the removal of forces causing deformation. In other words, an elastic material can undergo a certain amount of deformation and return to its initial form once the load is removed.

Young's modulus is a key factor in determining a material's elasticity. It is the ratio of tensile stress to tensile strain within the elastic limit, where a material's ability to return to its original form is not permanently affected. Mathematically, Young's modulus \(E\) is defined as:

\[E = \frac{\text{Tensile Stress}}{\text{Tensile Strain}}\]
A high Young's modulus indicates a material is stiff and has low elasticity, while a lower modulus suggests that the material is more flexible.

Understanding elasticity is crucial in engineering and design, as it helps predict how materials will behave under different load conditions, ensuring safety and integrity in various applications. Materials are carefully selected based on their elastic properties depending on the requirements of different structural components.

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Most popular questions from this chapter

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?

BIO Leg Raises. In a simplified version of the musculature action in leg raises, the abdominal muscles pull on the femur (thigh bone) to raise the leg by pivoting it about one end (Fig. P11.57. When you are lying. horizontally, these muscles make an angle of approximately \(5^{\circ}\) with the femur, and if you raise your legs, the muscles remain approximately horizontal, so the angle \(\theta\) increases. We shall assume for simplicity that these muscles attach to the femur in only one place, 10 \(\mathrm{cm}\) from the hip joint (although, in reality, the situation is more complicated). For a certain \(80-\mathrm{kg}\) person having a leg 90 \(\mathrm{cm}\) long, the mass of the leg is 15 \(\mathrm{kg}\) and its center of mass is 44 \(\mathrm{cm}\) from his hip joint as measured along the leg. If the person raises his leg to \(60^{\circ}\) above the horizontal, the angle between the abdominal muscles and his femur would also be about \(60^{\circ} .\) (a) With his leg raised to \(60^{\circ},\) find the tension in the abdominal muscle on each leg. As usual, begin your solution with a free-body diagram. (b) When is the tension in this muscle greater: when the leg is raised to \(60^{\circ}\) or when the person just starts to raise it off the ground? Why? (Try this yourself to check your answer.) (c) If the abdominal muscles attached to the femur were perfectly horizontal when a person was lying down, could the person raise his leg? Why or why not?

A moonshiner produces pure ethanol (ethyl alcohol) late at night and stores it in a stainless steel tank in the form of a cylinder 0.300 \(\mathrm{m}\) in diameter with a tight-fitting piston at the top. The total volume of the tank is 250 \(\mathrm{L}\left(0.250 \mathrm{m}^{3}\right) .\) In an attempt to squeeze a little more into the tank, the moonshiner piles 1420 \(\mathrm{kg}\) of lead bricks on top of the piston. What additional volume of ethanol can the moonshiner squeeze into the tank? (Assume that the wall of the tank is perfectly rigid.)

You take your dog Clea to the vet, and the doctor decides he must locate the little beast's center of gravity. It would be awkward to hang the pooch from the ceiling, so the vet must devise another method. He places Clea's front feet on one scale and her hind feet on another. The front scale reads \(157 \mathrm{N},\) while the rear scale reads 89 \(\mathrm{N}\) . The vet next measures Clea and finds that her rear feet are 0.95 \(\mathrm{m}\) behind her front feet. How much does Clear weigh, and where is her center of gravity?

CP 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. (c) If the cable suddenly breaks, what is the magnitude of the angular acceleration of the angular speed of the drawbridge as it breaks? (d) What is the angular speed of the drawbridge as it becomes horizontal?
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