/*! 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 15 Bone has a Young's modulus of \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Bone has a Young's modulus of \(18 \times 10^{9} \mathrm{~Pa}\). Under compression, it can withstand a stress of about \(160 \times 10^{6} \mathrm{~Pa}\) before breaking. Assume that a femur (thigh bone) is \(0.50 \mathrm{~m}\) long, and calculate the amount of compression this bone can withstand before breaking.

Short Answer

Expert verified
The bone can withstand a compression of about 4.45 mm before breaking.

Step by step solution

01

Understand the Given Values

The Young's modulus \(E\) for bone is given as \(18 \times 10^{9} \mathrm{~Pa}\) and the stress (\(\sigma\)) it can withstand before breaking is \(160 \times 10^{6} \mathrm{~Pa}\). The length (\(L\)) of the femur is 0.50 m.
02

Define the Relevant Equation

The formula to calculate strain (\(\epsilon\)), is derived from the formula of Young's Modulus, \(E = \sigma / \epsilon \), can be rearranged as \(\epsilon = \sigma / E\).
03

Insert the Known values

Substitute the given values into the equation: \(\epsilon = \sigma / E = (160 \times 10^{6} \mathrm{~Pa}) / (18 \times 10^{9} \mathrm{~Pa})\).
04

Calculate the Strain

By doing the division, the strain comes out to be about \(\epsilon \approx 0.0089 \).
05

Calculate Compression

The amount of compression this bone can withstand before it breaks can be obtained by multiplying the total length of the bone with the strain. This yields \(0.50 \, m \times 0.0089 = 0.00445 \, m\) or 4.45 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 91Ó°ÊÓ!

Key Concepts

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

Understanding Strain Calculation
Strain is a measure of deformation representing the displacement between particles in the material body. It's important to know that strain is dimensionless, meaning it has no unit. This is because strain is a ratio of lengths: the change in length divided by the original length.
To compute strain (\(\epsilon\)), you can use the formula derived from Young's Modulus:
  • \(\epsilon = \frac{\sigma}{E}\)
  • \(\sigma\) is the stress applied to the material
  • \(E\) represents Young's Modulus
In our exercise, the known stress (\(160 \times 10^{6} \,\mathrm{Pa}\)) is inserted into the equation together with the Young's Modulus for bone (\(18 \times 10^{9} \,\mathrm{Pa}\)) to find the strain. The resultant strain was calculated to be about \(0.0089\), indicating that the bone slightly elongates under stress before returning to its original shape after the load is removed.
Conducting Stress Analysis
Stress analysis allows us to understand how a material behaves under various forces. In essence, stress (\(\sigma\)) is defined as force exerted over an area (\(F/A\)). Stress is measured in pascals (Pa) in the SI unit system. Key points to consider when working with stress analysis include:
  • Always consider the type of stress (compressive, tensile, or shear)
  • Be aware of the material's ability to handle stress before failure
In the example, the femur can withstand a stress of \(160 \times 10^{6} \,\mathrm{Pa}\). Knowing the bones' stress limit is crucial in preventing excess stress that can lead to fracture or failure.
Exploring Material Properties
Material properties, such as Young's Modulus, are intrinsic qualities that dictate how materials respond to forces. Young's Modulus (\(E\)) specifically measures a material's stiffness, reflecting resistance to elastic deformation. For bone, this modulus is cited as \(18 \times 10^{9} \,\mathrm{Pa}\). Higher values indicate more stiffness, and therefore, less deformation under the same stress.
Understanding these properties helps engineers and scientists predict material behaviors in structures and biological contexts. For example, for the femur, we know how much it can compress without reaching the breaking point, considering its length and the stress it can bear. This insight is invaluable for practical applications, such as designing prosthetics or surgical procedures.

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

Glycerin in water diffuses along a horizontal column that has a cross- sectional area of \(2.0 \mathrm{~cm}^{2}\). The concentration gradient is \(3.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}^{4}\), and the diffusion rate is found to be \(5.7 \times 10^{-15} \mathrm{~kg} / \mathrm{s}\). Determine the diffusion coefficient.

As a first approximation, Earth's continents may be thought of as granite blocks floating in a denser rock (called peridotite) in the same way that ice floats in water. (a) Show that a formula describing this phenomenon is $$ \rho_{g} t=\rho_{p} d $$ where \(\rho_{g}\) is the density of granite \(\left(2.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right), \rho_{p}\) is the density of peridotite \(\left(3.3 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right), t\) is the thickness of a continent, and \(d\) is the depth to which a continent floats in the peridotite. (b) If a continent sinks \(5.0 \mathrm{~km}\) into the peridotite layer (this surface may be thought of as the ocean floor), what is the thickness of the continent?

In a water pistol, a piston drives water through a larger tube of radius \(1.00 \mathrm{~cm}\) into a smaller tube of radius \(1.00 \mathrm{~mm}\) as in Figure \(\mathrm{P} 9.51\). (a) If the pistol is fired horizontally at a height of \(1.50 \mathrm{~m}\), use ballistics to determine the time it takes water to travel from the nozzle to the ground. (Neglect air resistance and assume atmospheric pressure is \(1.00 \mathrm{~atm}\).) (b) If the range of the stream is to be \(8.00 \mathrm{~m}\), with what speed must the stream leave the nozzle? (c) Given the areas of the nozzle and cylinder, use the equation of continuity to calculate the speed at which the plunger must be moved. (d) What is the pressure at the nozzle? (e) Use Bernoulli's equation to find the pressure needed in the larger cylinder. Can gravity terms be neglected? (f) Calculate the force that must be exerted on the trigger to achieve the desired range. (The force that must be exerted is due to pressure over and above atmospheric pressure.)

The viscous force on an oil drop is measured to be equal to \(3.0 \times 10^{-13} \mathrm{~N}\) when the drop is falling through air with a speed of \(4.5 \times 10^{-4} \mathrm{~m} / \mathrm{s}\). If the radius of the drop is \(2.5 \times 10^{-6} \mathrm{~m}\), what is the viscosity of air?

The true weight of an object can be measured in a vacuum, where buoyant forces are absent. A measurement in air, however, is disturbed by buoyant forces. An object of volume \(V\) is weighed in air on an equal-arm balance with the use of counterweights of density \(\rho\). Representing the density of air as \(\rho_{\text {air }}\) and the balance reading as \(F_{g}^{\prime}\), show that the true weight \(F_{g}\) is $$ F_{g}=F_{g}^{\prime}+\left(V-\frac{F_{g}^{\prime}}{\rho g}\right) \rho_{\text {air }} g $$

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.