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Depending on how you fall, you can break a bone easily. The severity of the break depends on how much energy the bone absorbs in the accident, and to evaluate this let us treat the bone as an ideal spring. The maximum applied force of compression that one man's thighbone can endure without breaking is \(7.0 \times 10^{4} \mathrm{N} .\) The minimum effective cross-sectional area of the bone is \(4.0 \times 10^{-4} \mathrm{m}^{2},\) its length is \(0.55 \mathrm{m},\) and Young's modulus is \(Y=9.4 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} .\) The mass of the man is \(65 \mathrm{kg} .\) He falls straight down without rotating, strikes the ground stiff-legged on one foot, and comes to a halt without rotating. To see that it is easy to break a thighbone when falling in this fashion, find the maximum distance through which his center of gravity can fall without his breaking a bone.

Short Answer

Expert verified
The maximum fall distance where the bone absorbs energy without breaking is approximately 0.562 m.

Step by step solution

01

Determine the maximum force the bone can endure

The maximum applied force that the bone can withstand is given as \( F = 7.0 \times 10^{4} \; \mathrm{N} \). This will be compared against the force generated by the fall in subsequent steps.
02

Calculate the pressure on the bone

Pressure can be calculated using the formula: \( \text{Pressure} = \frac{F}{A} \). Substituting the given force and area, we have:\[ \text{Pressure} = \frac{7.0 \times 10^{4} \; \mathrm{N}}{4.0 \times 10^{-4} \; \mathrm{m}^{2}} = 1.75 \times 10^{8} \; \mathrm{N/m^2} \]This value shall be used for equating the stress and strain in later steps.
03

Relate it to Young’s modulus equation

Young's modulus \( Y \) is the ratio of stress to strain and given by:\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\sigma}{\epsilon} \]Substituting \( Y = 9.4 \times 10^{9} \; \mathrm{N/m^2} \) and \( \sigma = 1.75 \times 10^{8} \; \mathrm{N/m^2} \) from above, solve for strain \( \epsilon \).
04

Solve for strain

Using the equation for strain:\[ \epsilon = \frac{\sigma}{Y} \]Substitute the known values into the equation:\[ \epsilon = \frac{1.75 \times 10^{8} \; \mathrm{N/m^2}}{9.4 \times 10^{9} \; \mathrm{N/m^2}} \approx 0.0186 \]
05

Calculate compression (extension) using strain

Strain \( \epsilon \) is also defined as \( \epsilon = \frac{\Delta L}{L} \), where \( \Delta L \) is the change in length (compression in this case) and \( L \) is the original length. Rearranging gives:\[ \Delta L = \epsilon \times L \]Substituting the values, \( L = 0.55 \; \mathrm{m} \):\[ \Delta L = 0.0186 \times 0.55 = 0.01023 \; \mathrm{m} \]
06

Find the potential energy lost while falling

The work done (or potential energy lost) in compressing the bone by \( \Delta L \) is given by:\[ \frac{1}{2} F \Delta L \]Acting as a spring where \[ = \frac{1}{2} \times 7.0 \times 10^{4} \times 0.01023 \approx 357.05 \; \mathrm{J} \] This represents the energy absorbed by the bone without breaking.
07

Calculate the maximum allowable fall distance

Set the potential energy lost equal to the gravitational potential energy (\( mgh \)) absorbed during the fall where \( g = 9.81 \; \mathrm{m/s^2} \), solve for height \( h \):\[ mgh = 357.05 \; \mathrm{J} \]\[ 65 \times 9.81 \times h = 357.05 \; \mathrm{J} \]\[ h = \frac{357.05}{65 \times 9.81} \approx 0.562 \; \mathrm{m} \]

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Key Concepts

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

Stress and Strain
When we speak about stress and strain in physics, we're describing how materials respond to forces. **Stress** is the internal force experienced by an object when it is subjected to an external force. Think about the pressure applied. In mathematical terms, stress is defined as the force applied per unit area of the object. The formula is:
  • \( \text{Stress} = \frac{F}{A} \)
where \( F \) is the force applied, and \( A \) is the cross-sectional area. For example, if you push on a wall, the force you apply causes stress all over the area where your hand touches the wall.
**Strain**, on the other hand, is about how much the object deforms due to stress. It’s a measure of the change in shape or size compared to its original form. Strain is calculated by this formula:
  • \( \text{Strain} = \frac{\Delta L}{L} \)
where \( \Delta L \) is the change in length, and \( L \) is the original length. Strain is a way to quantify how much something stretches or compresses when you apply stress.
Understanding how stress and strain work helps engineers and scientists design materials that can withstand various forces without breaking.
Compression in Physics
Compression refers to the action of pushing or pressing something until it becomes more compact. In physics, it usually involves the application of stress that decreases the volume of a material. This process is crucial to comprehend how objects behave under force.
The thighbone in our problem is subject to compression when the body falls and presses against the ground. The **compression** it experiences is closely tied to the strain calculated for such conditions. Here’s how it works:
  • Strain \( \epsilon \) is directly proportional to the compression \( \Delta L \) as \( \epsilon = \frac{\Delta L}{L} \).
  • In our exercise, the bone acts similar to a spring. When it compresses, it absorbs kinetic energy from the fall scenario until reaching the fracture point, if the force is too great.
Compression calculations let us predict how much a bone or material will shorten under stress. This is critical for understanding the limits of a material's resilience to applied forces.
Moreover, materials often have a limit to how much stress in compression they can endure before failing, which is known as the **yield strength** in engineering.
Bone Fracture Mechanics
Bone fracture mechanics deals with understanding why and how bones break. Bones act like natural springs, exhibiting properties of elasticity and strength. However, they have a failure limit beyond which they fracture.
In our exercise, the maximum force the thighbone can endure is calculated. This provides insight into the dynamics of bone fractures:
  • The **Young's Modulus** of the bone helps us determine how much stress it can sustain in relation to how much it can strain.
  • If the bone undergoes a force greater than \( 7.0 \times 10^{4} \; \mathrm{N} \), it risks breaking. This is especially a concern during sudden impacts, like a fall.
The sophisticated structure of bones, a balance of rigidity and pliability, allows them to absorb impacts to a certain extent. However, understanding bone fracture mechanics is crucial in fields like sports medicine and orthopedics to help prevent injuries and promote healing.
By studying the interplay of forces, stress and strain in bones, scientists can determine the conditions under which bones are likely to break, and find ways to mitigate these risks, such as improved protective gear or techniques for fall prevention.

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Most popular questions from this chapter

A 75 -kg diver is standing at the end of a diving board while it is vibrating up and down in simple harmonic motion, as indicated in the figure. The diving board has an effective spring constant of \(k=\) \(4100 \mathrm{N} / \mathrm{m},\) and the vertical distance between the highest and lowest points in the motion is \(0.30 \mathrm{m} .\) Concepts: (i) How is the amplitude \(A\) related to the vertical distance between the highest and lowest points of the diver's motion? (ii) Starting from the top, where is the diver located one-quarter of a period later, and what can be said about his speed at this point? (iii) If the amplitude were to double, would the period also double? Explain. Calculations: (a) What is the amplitude of the motion? (b) Starting when the diver is at the highest point, what is his speed one-quarter of a period later? (c) If the vertical distance between his highest and lowest points were changed to \(0.10 \mathrm{m},\) what would be the time required for the diver to make one complete motional cycle?

A 15.0 -kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of \(5.00 \mathrm{m} / \mathrm{s}\) in \(0.500 \mathrm{s}\). In the process, the spring is stretched by \(0.200 \mathrm{m} .\) The block is then pulled at a constant speed of \(5.00 \mathrm{m} / \mathrm{s}\), during which time the spring is stretched by only \(0.0500 \mathrm{m} .\) Find \((\mathrm{a})\) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

A vertical spring with a spring constant of \(450 \mathrm{N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is unstrained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by \(2.5 \mathrm{cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm}\) ) above the compressed spring was the block dropped?

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring is \(174 \mathrm{N} / \mathrm{m} .\) The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2 . The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2 .

When an object of mass \(m_{1}\) is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of \(12.0 \mathrm{Hz} .\) When another object of mass \(m_{2}\) is hung on the spring along with the first object, the frequency of the motion is \(4.00 \mathrm{Hz} .\) Find the ratio \(m_{2} / m_{1}\) of the masses.

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