/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 A relaxed biceps muscle requires... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A relaxed biceps muscle requires a force of 25.0 N for an elongation of 3.0 cm; the same muscle under maximum tension requires a force of 500 N for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with length 0.200 m and cross-sectional area 50.0 cm\(^2\).

Short Answer

Expert verified
Young's modulus is 333.3 N/m^2 for relaxed tension and 6666.7 N/m^2 for maximum tension.

Step by step solution

01

Gather Known Values

We know the following values for both conditions:- Relaxed tension: Force, \( F = 25.0 \) N.- Maximum tension: Force, \( F = 500 \) N.- Elongation, \( \, \Delta L = 0.03 \) m (converted from 3.0 cm).- Original length of the muscle, \( L = 0.20 \) m.- Cross-sectional area, \( A = 50.0 \times 10^{-4} \) mdue to conversion from \( cm^2 \) to \( m^2 \).
02

Understand Young's Modulus

Young's Modulus \( E \) is calculated using the formula:\[ E = \frac{F \cdot L}{A \cdot \Delta L} \]where \( F \) is the force applied, \( L \) is the original length of the material, \( A \) is the cross-sectional area, and \( \Delta L \) is the change in length (elongation).
03

Calculate Young's Modulus for Relaxed Tension

Substitute the known values into the Young's modulus formula for the relaxed state:\[E_{\text{relaxed}} = \frac{25.0 \, \text{N} \times 0.20 \, \text{m}}{50.0 \times 10^{-4} \, \text{m}^2 \times 0.03 \, \text{m}}\]Calculate to find:\[ E_{\text{relaxed}} = \frac{5.0}{1.5 \times 10^{-2}} = \frac{5.0}{0.015} = 333.3 \, \text{N/m}^2 \]
04

Calculate Young's Modulus for Maximum Tension

Substitute the known values into the Young's modulus formula for the maximum tension state:\[E_{\text{maximum}} = \frac{500 \, \text{N} \times 0.20 \, \text{m}}{50.0 \times 10^{-4} \, \text{m}^2 \times 0.03 \, \text{m}}\]Calculate to find:\[ E_{\text{maximum}} = \frac{100.0}{0.015} = 6666.7 \, \text{N/m}^2 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Elasticity
Elasticity is the property of materials that allows them to return to their original shape after being deformed by an external force. This characteristic is what makes materials like rubber bands snap back after being stretched. In the context of a biceps muscle, elasticity describes the muscle's ability to stretch and recoil. When a muscle is relaxed, it stretches with less force, but under maximum tension, it requires a significantly greater force to achieve the same amount of elongation. This behavior under different forces is a classic example of elasticity at work. A material with higher elasticity will return to its original shape more quickly and with less permanent deformation. Understanding elasticity is crucial for applications like designing sports equipment or medical prosthetics, where materials must endure repeated stress.
Stress-Strain Relationship
The stress-strain relationship is central to understanding material behavior under load. Stress refers to the force applied to a material, expressed per unit area. It is mathematically defined as \( ext{stress} = rac{F}{A} \), where \( F \) is the force applied and \( A \) is the cross-sectional area. Strain, on the other hand, measures the deformation of the material relative to its original length and is expressed as a ratio: \( ext{strain} = rac{ ext{change in length}}{ ext{original length}} \).

In our example, both stress and strain were used to determine how the biceps muscle behaves under different levels of tension. By plotting stress against strain, we create a stress-strain curve, which provides insights into a material's elastic limit and ability to undergo plastic deformation. The slope of the initial, linear portion of this curve is what we refer to when calculating Young's modulus, providing a measure of a material's stiffness.
Material Properties
Material properties such as hardness, tensile strength, and elasticity define how substances respond to external forces. These properties are essential for engineers and scientists when they need to choose materials for specific applications. For example, the muscle tissue in our exercise exhibits different Young's modulus values under relaxed and maximum tension, indicating variable stiffness.

The intrinsic material properties information allows for understanding and predicting how a material will behave in a given situation, reducing the risk of failure in applications like medical implants or athletic equipment. Moreover, knowledge of these properties helps in creating models for analyzing how a muscle or any cylindrical structure might react under different stress conditions. This aids in biologically relevant studies, ensuring safety and efficacy in biological and mechanical systems. By studying these characteristics, professionals can push the limits of what materials can be used for various innovations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A uniform, horizontal flagpole 5.00 m long with a weight of 200 N is hinged to a vertical wall at one end. A 600-N stuntwoman hangs from its other end. The flagpole is supported by a guy wire running from its outer end to a point on the wall directly above the pole. (a) If the tension in this wire is not to exceed 1000 N, what is the minimum height above the pole at which it may be fastened to the wall? (b) If the flagpole remains horizontal, by how many newtons would the tension be increased if the wire were fastened 0.50 m below this point?

A nylon rope used by mountaineers elongates 1.10 m under the weight of a 65.0-kg climber. If the rope is 45.0 m in length and 7.0 mm in diameter, what is Young's modulus for nylon?

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?

A holiday decoration consists of two shiny glass spheres with masses 0.0240 kg and 0.0360 kg suspended from a uniform rod with mass 0.120 kg and length 1.00 m (\(\textbf{Fig. P11.62}\)). The rod is suspended from the ceiling by a vertical cord at each end, so that it is horizontal. Calculate the tension in each of the cords \(A\) through \(F\).

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.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.