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Chapter 11: Problem 27

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

Short Answer

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

Step by step solution

01

Understand the Problem

We are asked to find Young's modulus, which is a measure of the stiffness of a material. The formula for Young's modulus is given by \( E = \frac{F}{A} \times \frac{L_0}{\Delta L} \) where \( E \) is Young's modulus, \( F \) is the force applied, \( A \) is the cross-sectional area, \( L_0 \) is the original length, and \( \Delta L \) is the change in length.
02

Identify Given Values

From the problem statement, we have:- Force \( F = 5000 \text{ N} \)- Original Length \( L_0 = 4.00 \text{ m} \)- Cross-sectional Area \( A = 0.50 \text{ cm}^2 = 0.50 \times 10^{-4} \text{ m}^2 \) (converted to square meters)- Change in Length \( \Delta L = 0.20 \text{ cm} = 0.0020 \text{ m} \) (converted to meters).
03

Substitute the Values into the Formula

Substitute the values into the formula: \[ E = \frac{5000 \text{ N}}{0.50 \times 10^{-4} \text{ m}^2} \times \frac{4.00 \text{ m}}{0.0020 \text{ m}} \].
04

Calculate the First Fraction

First, calculate the stress (force per unit area): \[ \sigma = \frac{5000 \text{ N}}{0.50 \times 10^{-4} \text{ m}^2} \] \[ \sigma = 1.00 \times 10^8 \text{ N/m}^2 \].
05

Calculate the Second Fraction

Next, calculate the strain (change in length per original length): \[ \epsilon = \frac{0.0020 \text{ m}}{4.00 \text{ m}} = 0.0005 \].
06

Complete the Calculation for Young's Modulus

Now calculate Young's modulus using the values from Steps 4 and 5: \[ E = 1.00 \times 10^8 \text{ N/m}^2 \times \frac{1}{0.0005} \]. Evaluate this to obtain: \[ E = 2.00 \times 10^{11} \text{ N/m}^2 \].
07

Conclusion

The value of Young's modulus for the metal is \( 2.00 \times 10^{11} \text{ N/m}^2 \). This suggests that the metal is quite stiff, which is typical for metals like steel.

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Key Concepts

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

Stress
Stress is a measure of the internal forces that develop within a material when an external force is applied. In simple terms, stress can be thought of as the pressure experienced by the material. It is defined as the force applied divided by the cross-sectional area over which the force acts. Mathematically, you can express stress (\( \sigma \)) using the formula:
\[ \sigma = \frac{F}{A} \]where \( F \) represents the applied force, and \( A \) represents the cross-sectional area.
  • Unit: Stress is measured in Pascals (Pa), where 1 Pascal equals 1 Newton per square meter (N/m虏).
  • Application: Knowing the stress helps predict whether a material will withstand the applied forces or fail.
Stress gives a fundamental insight into the mechanical behavior of materials and is critical in designing structural components.
Strain
Strain is related to the deformation a material undergoes under stress. When a force is applied to an object, it may change its shape or size. Strain quantifies this change and is defined as the ratio of the change in length (\( \Delta L \)) to the original length (\( L_0 \)) of the object. Mathematically, strain (\( \epsilon \)) is given by:
\[ \epsilon = \frac{\Delta L}{L_0} \]Strain is a dimensionless quantity, meaning it has no units. It simply represents how much a material deforms in response to stress.
  • Types of Strain: There are mainly two types of strains: tensile strain (stretching) and compressive strain (squeezing).
  • Elastic vs Plastic Strain: Elastic strain is reversible, while plastic strain is permanent.
Understanding strain is crucial for assessing how materials deform and behave under various conditions.
Elasticity
Elasticity describes a material's ability to return to its original shape and size after the external forces are removed. Think of it like a spring that stretches when pulled but returns to its original shape when released. The extent to which an object can stretch or compress and still return to its original size determines its elasticity.
  • Hooke's Law: This principle states that, within the elastic limit, the amount of stretch or compression (strain) is directly proportional to the applied load (stress). Mathematically, it is expressed as \( \sigma = E \cdot \epsilon \)
  • Elastic Limit: The maximum extent to which a material can return to its original shape after the applied stress is removed. Beyond this limit, permanent deformation occurs.
Elasticity is vital for understanding how structures flex under loads and is critical in ensuring materials perform safely under stress.
Mechanical Properties
Mechanical properties indicate how a material reacts to applied forces, including tension, compression, and bending. They define the material's behavior and capabilities under mechanical loads and conditions and can be specific to types of applications, such as construction or manufacturing.
  • Young's Modulus: This is a significant mechanical property that measures a material's stiffness, indicating how much it will deform under stress. It's defined as the ratio of stress to strain.
  • Tensile Strength: The maximum stress a material can withstand while being stretched before breaking.
  • Ductility and Brittleness: Ductility describes a material's ability to deform under tensile stress. Brittleness refers to a lack of ductility, where a material breaks rather than deforms.
Understanding these properties is crucial for selecting materials for specific applications, ensuring both functionality and safety in their end use.
Material Deformation
Material deformation refers to the change in shape, size, or volume of a material due to applied forces. In engineering, it is vital to understand the limits up to which a material can deform.
  • Elastic Deformation: Temporary shape change that is recoverable upon the removal of stress.
  • Plastic Deformation: Permanent change, where the material doesn't return to its original shape after the stress is removed.
  • Factors Influencing Deformation: These include material composition, temperature, and type of applied stress.
Understanding material deformation helps in designing components that can withstand external forces without failing or suffering permanent damage.

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Most popular questions from this chapter

Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of 400 N, and the other lifts the opposite end with a force of 600 N. (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs 200 N, with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

A vertical, solid steel post 25 cm in diameter and 2.50 m long is required to support a load of 8000 kg. You can ignore the weight of the post. What are (a) the stress in the post; (b) the strain in the post; and (c) the change in the post's length when the load is applied?

A door 1.00 m wide and 2.00 m high weighs 330 N and is supported by two hinges, one 0.50 m from the top and the other 0.50 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 uniform, 7.5-m-long beam weighing 6490 N is hinged to a wall and supported by a thin cable attached 1.5 m from the free end of the beam. The cable runs between the beam and the wall and makes a 40\(^\circ\) angle with the beam. What is the tension in the cable when the beam is at an angle of 30\(^\circ\) above the horizontal?

In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm 3.80 cm from the elbow joint. Assume that the person's hand and forearm together weigh 15.0 N and that their center of gravity is 15.0 cm from the elbow (not quite halfway to the hand). The forearm is held horizontally at a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the forearm. (a) Draw a free-body diagram for the forearm, and find the force exerted by the biceps when the hand is empty. (b) Now the person holds an 80.0-N weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is 33.0 cm from the elbow. Draw a free-body diagram for the forearm, and find the force now exerted by the biceps. Explain why the biceps muscle needs to be very strong. (c) Under the conditions of part (b), find the magnitude and direction of the force that the elbow joint exerts on the forearm. (d) While holding the 80.0-N weight, the person raises his forearm until it is at an angle of 53.0\(^\circ\) above the horizontal. If the biceps muscle continues to exert its force perpendicular to the forearm, what is this force now? Has the force increased or decreased from its value in part (b)? Explain why this is so, and test your answer by doing this with your own arm.
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