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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/m²).
  • 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

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?

The compressive strength of our bones is important in everyday life. Young's modulus for bone is about 1.4 \(\times\) 10\(^{10}\) Pa. Bone can take only about a 1.0% change in its length before fracturing. (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 cm\(^2\)? (This is approximately the crosssectional area of a tibia, or shin bone, at its narrowest point.) (b) Estimate the maximum height from which a 70-kg man could jump and not fracture his tibia. Take the time between when he first touches the floor and when he has stopped to be 0.030 s, and assume that the stress on his two legs is distributed equally.

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

An angler hangs a 4.50-kg fish from a vertical steel wire 1.50 m long and 5.00 \(\times\) 10\(^{-3}\) cm\(^2\) in cross-sectional area. The upper end of the wire is securely fastened to a support. (a) Calculate the amount the wire is stretched by the hanging fish. The angler now applies a varying force \(\overrightarrow{F}\) at the lower end of the wire, pulling it very slowly downward by 0.500 mm from its equilibrium position. For this downward motion, calculate (b) the work done by gravity; (c) the work done by the force \(\overrightarrow{F}\), (d) the work done by the force the wire exerts on the fish; and (e) the change in the elastic potential energy (the potential energy associated with the tensile stress in the wire). Compare the answers in parts (d) and (e). \(\textbf{Torques and Tug-of-War.}\) In a study of the biomechanics of the tug-of-war, a 2.0-m-tall, 80.0-kg competitor in the middle of the line is considered to be a rigid body leaning back at an angle of 30.0\(^\circ\) to the vertical. The competitor is pulling on a rope that is held horizontal a distance of 1.5 m from his feet (as measured along the line of the body). At the moment shown in the figure, the man is stationary and the tension in the rope in front of him is \({T_1} =\) 1160 N. Since there is friction between the rope and his hands, the tension in the rope behind him, \({T_2}\) is not equal to \({T_1}\). His center of mass is halfway between his feet and the top of his head. The coefficient of static friction between his feet and the ground is 0.65.

Two people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

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