/*! 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 94 CP Blo Stress on the Shin Bone. ... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The maximum force is 42,000 N, and the maximum jump height is about 4.1 meters.

Step by step solution

01

Understanding Young's Modulus and Stress

Young's Modulus \((E)\) is a measure of a material's ability to withstand changes in length when a force is applied. Stress \(\sigma\) on the bone can be calculated using the formula \(\sigma = \frac{F}{A}\), where \(F\) is the force applied and \(A\) is the cross-sectional area of the bone.
02

Calculate Maximum Stress Before Fracture

The maximum stress that bone can withstand before fracturing is the product of Young's Modulus and the maximum strain: \(\sigma_{\text{max}} = E \times \text{strain}_{\text{max}} = 1.4 \times 10^{10} \, \text{Pa} \times 0.01 = 1.4 \times 10^{8} \, \text{Pa}\).
03

Determine Maximum Force on Tibia

Using the maximum stress \(\sigma_{\text{max}} = 1.4 \times 10^8 \, \text{Pa}\) and the cross-sectional area \(A = 3.0 \, \text{cm}^2 = 3.0 \times 10^{-4} \, \text{m}^2\), calculate the maximum force: \(F = \sigma_{\text{max}} \times A = 1.4 \times 10^8 \, \text{Pa} \times 3.0 \times 10^{-4} \, \text{m}^2 = 4.2 \times 10^4 \, \text{N}\).
04

Estimate Maximum Jump Height

The impulse \( J = \Delta p = F \times \Delta t \) is the change in momentum. Assuming each leg takes half the force, \(F = \frac{4.2 \times 10^{4}}{2} = 2.1 \times 10^{4} \, \text{N}\). With \( m = 70 \, \text{kg}\), calculate \( v = \frac{F \times \Delta t}{m} = \frac{2.1 \times 10^4, \text{N} \times 0.030, \text{s}}{70, \text{kg}} = 9, \text{m/s}\) approximately. Use \( v^2 = 2gh \) to find height: \( h = \frac{v^2}{2g} = \frac{9^2}{2 \times 9.81} \approx 4.1, \text{m}\).

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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 a fundamental property of materials that describes their ability to deform under stress. It is denoted by the symbol \( E \), and it essentially tells us how stiff a material is. The value of Young's Modulus for bone is approximately \(1.4 \times 10^{10} \) Pascal (Pa). This means bones can withstand significant force without changing much in length. For example, the slight stretch or compression that happens when you apply force to a bone falls under the purview of Young's Modulus. When we talk about stress in terms of Young's Modulus, it's about the amount of deformation (strain) that a material can endure in relation to the force (stress) applied to it. The formula to represent stress is given by \( \sigma = \frac{F}{A} \), where \( F \) is the force applied, and \( A \) is the cross-sectional area.
Compressive Strength of Bones
The compressive strength of bones refers to their ability to withstand loads that tend to reduce size. This is crucial in contexts where bones experience squeezing forces. Human bones are incredibly strong in compression because of their dense, robust structure. Compressive strength is particularly important for bones subjected to large forces, like those in the legs when jumping or landing. It's measured by determining the maximum stress a bone can handle before it fractures. As mentioned in the bone stress problem, the bone can tolerate only about a 1% change in its length before fracturing. The calculation of maximum stress is done using the formula \( \sigma_{\text{max}} = E \times \text{strain}_{\text{max}} \), which helps determine the risk of fracture under compressive forces.
Maximum Force on Tibia
The tibia, also known as the shin bone, is one of the most robust bones in the human body. Calculating the maximum force it can withstand is crucial, especially in dynamic activities like jumping. To determine the maximum force the tibia can handle, use the maximum stress it can endure before fracturing. With the maximum stress \( \sigma_{\text{max}} = 1.4 \times 10^8 \, \text{Pa} \) and the cross-sectional area of the tibia \( A = 3.0 \, \text{cm}^2 = 3.0 \times 10^{-4} \, \text{m}^2 \), we find the force with the equation \( F = \sigma_{\text{max}} \times A \). Applying these values: \( F = 1.4 \times 10^8 \, \text{Pa} \times 3.0 \times 10^{-4} \, \text{m}^2 = 4.2 \times 10^4 \, \text{N} \). This shows the tibia can withstand significant force, supporting activities like running or jumping.
Impulse and Momentum
Impulse is a concept that links force and the change in momentum. It can be expressed using the formula \( J = \Delta p = F \times \Delta t \), where \( J \) is the impulse, \( \Delta p \) is the change in momentum, \( F \) is the force, and \( \Delta t \) is the time over which the force is applied. In the context of the bone stress problem, this formula helps to estimate the maximum jump height a person can achieve without breaking the tibia. Considering the landing force distributed between two legs, each leg takes half of \( 4.2 \times 10^4 \, \text{N} \), resulting in \( 2.1 \times 10^4 \, \text{N} \). By determining the speed at impact using \( v = \frac{F \times \Delta t}{m} \), where \( m = 70 \, \text{kg} \), and using the result in the energy conservation equation \( v^2 = 2gh \), we can ascertain the maximum height achievable safely during a jump.
Fracture Strain
Fracture strain refers to the amount of deformation necessary for a material, like bone, to fracture. It is a critical measure in determining how much a bone can elongate or compress before breaking. For bones, the fracture strain is relatively low, at around 1%. This means the bone can only tolerate a tiny change in length before failing. The fracture strain plays a vital role in estimating the limitations of bone stress and understanding the conditions leading to potential fractures. In the example problem, it helps to define the maximum changes the tibia can undergo, ensuring the calculations for forces and potential jumps remain within the bone's limits to prevent injury.

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Most popular questions from this chapter

A \(0.120-\mathrm{kg}, 50.0\) -cm-long uniform bar has a small 0.055 -kg mass glued to its left end and a small \(0.110-\) kg mass glued to the other end. The two small masses can each be treated as point masses. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?

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