/*! 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 78 The femur is a bone in the leg w... [FREE SOLUTION] | 91Ó°ÊÓ

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The femur is a bone in the leg whose minimum cross-sectional area is about \(4.0 \times 10^{-4} \mathrm{m}^{2} .\) A compressional force in excess of \(6.8 \times 10^{4} \mathrm{N}\) will fracture this bone. (a) Find the maximum stress that this bone can with- stand. (b) What is the strain that exists under a maximum-stress condition?

Short Answer

Expert verified
(a) Maximum stress is \(1.7 \times 10^8 \, \text{N/m}^2\). (b) Strain is \(0.0106\).

Step by step solution

01

Understanding Stress Formula

Stress is defined as the force applied per unit area. It is calculated using the formula: \( \text{Stress} = \frac{\text{Force}}{\text{Area}} \). In this problem, the force is given as \( 6.8 \times 10^4 \, \text{N} \) and the area is \( 4.0 \times 10^{-4} \, \text{m}^2 \).
02

Calculate Maximum Stress

Substitute the given values into the formula to calculate stress: \( \text{Stress} = \frac{6.8 \times 10^4}{4.0 \times 10^{-4}} = 1.7 \times 10^8 \, \text{N/m}^2 \). So, the maximum stress the femur can withstand is \( 1.7 \times 10^8 \, \text{N/m}^2 \).
03

Understanding Strain Formula

Strain is defined as the deformation experienced by a material in the direction of force, divided by its original length. The formula is: \( \text{Strain} = \frac{\text{Change in Length}}{\text{Original Length}} \). However, it can also be related to stress by Young’s modulus, \( E \), where \( \text{Strain} = \frac{\text{Stress}}{E} \).
04

Relate Young’s Modulus to Femur

For human bone (such as the femur), Young’s modulus \( E \) is approximately \( 1.6 \times 10^{10} \, \text{N/m}^2 \). We will use this value to calculate strain given the maximum stress.
05

Calculate Strain Under Maximum Stress

Using the relation \( \text{Strain} = \frac{\text{Stress}}{E} \), substitute the known values: \( \text{Strain} = \frac{1.7 \times 10^8}{1.6 \times 10^{10}} = 0.010625 \). The strain under maximum stress condition is approximately \( 0.0106 \).

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Key Concepts

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

Young's modulus
Young's modulus is an essential concept in material science and engineering. It measures a material's ability to endure length changes under tensile or compressive stress. Essentially, it quantifies how stiff a material is. The formula is given by:
  • Young's modulus, \( E = \frac{\text{Stress}}{\text{Strain}} \)
Young's modulus is typically expressed in Pascals (Pa), with higher values indicating a stiffer material. For example, steel has a considerably higher Young's modulus than rubber, demonstrating its rigidity and resistance to deformation. When dealing with bones, such as the femur, Young's modulus helps determine how much the bone will stretch or compress under force.
In our context, the femur has a Young's modulus of approximately \(1.6 \times 10^{10} \, \text{N/m}^2\). This tells us how the bone responds to the stresses of everyday activities, such as walking and jumping, and helps in understanding its resilience and limits to fracture.
compressional force
Compressional force pertains to forces that tend to squash or compress a material. When applied, it acts perpendicular to the surface, reducing the material's volume. It is crucial to understanding how materials like bone can withstand weight and other pressures.
The compressional force is provided in Newtons (N) and dictates how much stress a material can endure before it deforms or breaks. In the exercise, we see a compressional force as high as \(6.8 \times 10^{4} \, \text{N}\) applied to the femur, which exceeds its fracture point. This shows how bones are designed to handle significant pressure. Understanding compressional forces also helps in designing improvements for safety gear, like helmets and protective gear.
femur bone fracture
A femur bone fracture occurs when the stress applied to the femur surpasses its ultimate strength. This bone, which is the longest and strongest in the human body, plays a critical role in supporting the body's weight. When a force exceeding its maximum endurance pressures it, structural failure happens.
In our exercise, the fracture threshold is at a compressional force greater than \(6.8 \times 10^{4} \, \text{N}\). This fracture often results from high-impact accidents or falls, especially common in the elderly due to reduced bone density. Knowing the fracture limits assists in healthcare, allowing for preventative strategies and the design of surgical devices that can strengthen or replace weak bones.
  • Monitoring bone health through regular check-ups
  • Engaging in weight-bearing exercises to enhance bone strength
  • Ensuring a diet rich in calcium and vitamin D
These measures are essential for reducing fracture risks and maintaining overall bone health.

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Most popular questions from this chapter

An 86.0-kg climber is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a spring with a spring constant of \(1.20 \times 10^{3} \mathrm{N} / \mathrm{m} .\) He accidentally slips and falls freely for \(0.750 \mathrm{m}\) before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

A spring \((k=830 \mathrm{N} / \mathrm{m})\) is hanging from the ceiling of an elevator, and a 5.0-kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at \(a=0.60 \mathrm{m} / \mathrm{s}^{2} ?\)

A square plate is \(1.0 \times 10^{-2} \mathrm{m}\) thick, measures \(3.0 \times 10^{-2} \mathrm{m}\) on a side, and has a mass of \(7.2 \times 10^{-2}\) kg. The shear modulus of the material is \(2.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) One of the square faces rests on a flat horizontal surface, and the coefficient of static friction between the plate and the surface is 0.91 . A force is applied to the top of the plate, as in Figure \(10.29 a .\) Determine (a) the maximum possible amount of shear stress, (b) the maximum possible amount of shear strain, and (c) the maximum possible amount of shear deformation \(\Delta X\) (see Figure \(10.29 b\) ) that can be created by the applied force just before the plate begins to move.

A vertical ideal spring is mounted on the floor and has a spring constant of \(170 \mathrm{N} / \mathrm{m} .\) A \(0.64-\mathrm{kg}\) block is placed on the spring in two different ways. (a) In one case, the block is placed on the spring and not released until it rests stationary on the spring in its equilibrium position. Determine the amount (magnitude only) by which the spring is compressed. (b) In a second situation, the block is released from rest immediately after being placed on the spring and falls downward until it comes to a momentary halt. Determine the amount (magnitude only) by which the spring is now compressed.

A piece of mohair taken from an Angora goat has a radius of \(31 \times 10^{-6} \mathrm{m} .\) What is the least number of identical pieces of mohair needed to suspend a \(75-\mathrm{kg}\) person, so the strain experienced by each piece is less than \(0.010 ?\) Assume that the tension is the same in all the pieces.

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