/*! 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 63 BIO Tendon-Stretching Exercises.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 63

BIO Tendon-Stretching Exercises. As part of an exercise program, a 75 -kg per- son does toe raises in which he raises his entire body weight on the ball of one foot (Fig. P11.63). The Achilles tendon pulls straight upward on the heel bone of his foot. This tendon is 25 \(\mathrm{cm}\) long and has a a cross-sectional area of 78 \(\mathrm{mm}^{2}\) and a Young's modulus of 1470 MPa. (a) Make a free-body diagram of the person's foot (everything below the ankle joint). You can neglect the weight of the foot. (b) What force does the Achilles tendon exert on the heel during this exercise? Express your answer in newtons and in multiples of his weight. (c) By how many millimeters does the exercise stretch his Achilles tendon?

Short Answer

Expert verified
The Achilles tendon stretches about 1.575 mm, exerting a force of 735 N equal to the person's weight.

Step by step solution

01

Understand the Information Given

We have a person weighing 75 kg. The Achilles tendon has a length of 25 cm, a cross-sectional area of 78 mm data{}^{2} data{}, and a Young's Modulus of 1470 MPa. We are tasked to analyze forces and deformations in the tendon.
02

Free-Body Diagram

In the free-body diagram, the forces acting on the foot include the gravitational force due to the person's weight and the upward force exerted by the Achilles tendon. The gravitational force can be calculated as the person's weight, given by \( F_g = mg = 75 \times 9.8 = 735 \ N \). The upward force is exerted by the Achilles tendon to balance this weight.
03

Calculate the Force Exerted by the Tendon

Since the person stands on the ball of one foot, the entire body weight is balanced by the force exerted by the Achilles tendon. Therefore, the upward force exerted by the tendon, \( F_t \), is equal to the gravitational force: \( F_t = 735 \, N \).
04

Calculate the Stretch in the Tendon

To find the amount by which the Achilles tendon stretches, we use Hooke's Law for materials, which is given by the formula:\[\Delta L = \frac{FL}{AE}\]Where \( F = 735 \ N \) is the force exerted by the tendon, \( L = 0.25 \ m \) is the original length of the tendon, \( A = 78 \times 10^{-6} \ m^2 \) is the cross-sectional area, and \( E = 1470 \times 10^6 \ N/m^2 \) is the Young's modulus. Plugging in these values:\[\Delta L = \frac{735 \times 0.25}{78 \times 10^{-6} \times 1470 \times 10^6} \approx 1.575 \times 10^{-3} \, m = 1.575 \, mm\]Thus, the tendon stretches by approximately 1.575 mm.

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 91影视!

Key Concepts

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

Free-body diagram
A free-body diagram is a simple illustration that shows all the forces acting upon an object. It's like taking a snapshot of the forces around an object to better understand what is affecting it. In this exercise, we're interested in the person's foot as they do a toe raise, specifically the part of the foot below the ankle joint.
When drawing the diagram, we focus on two main forces:
  • The gravitational force, which is the weight of the person exerted downward. This force for a 75 kg person is found by multiplying their mass by the gravitational acceleration, resulting in a force of 735 N.
  • The force exerted by the Achilles tendon, which acts upward. This is because the tendon needs to exert an upward force to counteract the gravitational pull.
The two forces must balance each other out when the person is stationary on one foot. This concept of balancing forces is crucial to understanding many scenarios in mechanics.
Force calculation
Force calculation helps us determine how much force is being exerted in a specific direction. In this problem, we want to find out the force exerted by the Achilles tendon. Since the person raises their whole weight on the ball of one foot, the entire body weight acts as a downward force. Therefore, the tendon must exert an equal force upward to maintain balance.
The downward force is simply the weight of the person, which is calculated using the formula \( F = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (9.8 m/s虏). Given the mass is 75 kg, the force \( F_g \) becomes \( 75 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 735 \, N \).
This means the Achilles tendon must exert exactly 735 N upward. In terms of the person's weight, it carries all of it as they lift themselves.
Material deformation
Material deformation refers to how a material changes shape or extends when a force is applied. The Achilles tendon in this scenario withstands the entire weight of a person, causing it to stretch slightly. The amount of stretch is determined by the material properties, such as its modulus of elasticity (also known as Young's modulus), and the magnitude of the force applied.
Using Hooke's Law, we describe the stretch with the formula \[\Delta L = \frac{FL}{AE}\], where:
  • \( F \) is the force applied (735 N).
  • \( L \) is the original length of the tendon (0.25 m).
  • \( A \) is the cross-sectional area (78 mm虏 converted to m虏).
  • \( E \) is Young's modulus (1470 MPa converted to N/m虏).
When these values are input, the tendon stretches by about 1.575 mm. The ability to predict this deformation helps in designing materials and structures and understanding the limits before a material fails.
Young's modulus
Young's Modulus is a measure of the stiffness of a material. It describes the material's ability to resist deformation under stress. In simpler terms, it's like the material's "springiness". The higher the Young's modulus, the stiffer the material, meaning it will deform less when a given force is applied.
In our exercise, the Young's modulus for the Achilles tendon is given as 1470 MPa. This value indicates the tendon鈥檚 resistance to stretch under the force exerted during toe-raise exercises. By using Young's modulus, we can calculate the precise deformation (or change in length) the tendon undergoes.
This concept is pivotal in fields like mechanical engineering and bio-mechanics, where understanding how materials behave under various forces is crucial for safety and performance considerations.

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

Two people carry a heavy electric motor by placing it on a light board 2.00 \(\mathrm{m}\) long. One person lifts at one end with a force of \(400 \mathrm{N},\) and the other lifts the opposite end with a force of 600 \(\mathrm{N}\) . (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 \(200 \mathrm{N},\) 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 Truck on a Drawbridge. A loaded cement mixer drives onto an old drawbridge, where it stalls with its center of gravity three-quarters of the way across the span. The truck driver radios for help, sets the handbrake, and waits. Meanwhile, a boat approaches, so the drawbridge is raised by means of a cable attached to the end opposite the hinge (Fig. P11.56). The drawbridge is 40.0 \(\mathrm{m}\) long and has a mass of \(18,000 \mathrm{kg} ;\) its center of gravity is at its midpoint. The cement mixer, with driver, has mass \(30,000\) kg. When the drawbridge has been raised to an angle of \(30^{\circ}\) above the horizontal, the cable makes an angle of \(70^{\circ}\) with the surface of the bridge. (a) What is the tension \(T\) in the cable when the drawbridge is held in this position? (b) What are the horizontal and vertical components of the force the hinge exerts on the span?

Bulk Modulus of an Ideal Gas. The equation of state (the equation relating pressure, volume, and temperature) for an ideal gas is \(p V=n R T,\) where \(n\) and \(R\) are constants. (a) Show that if the gas is compressed while the temperature \(T\) is held constant, the bulk modulus is equal to the pressure. (b) When an ideal gas is compressed without the transfer of any heat into or out of it, the pressure and volume are related by \(p V^{\gamma}=\) constant, where \(\gamma\) is a constant having different values for different gases. Show that, in this case, the bulk modulus is given by \(B=\gamma p\)

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?

CP Blo Stress on the Shin Bone. 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 \(\mathrm{cm}^{2} ?\) (This is approximately the cross-sectional 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 the 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 is distributed equally between his legs.
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Famous Physicists

Read Explanation

Nuclear Physics

Read Explanation

Waves Physics

Read Explanation

Fields in Physics

Read Explanation

Physics of Motion

Read Explanation
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.