Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 66

An approximately round tendon that has an average diameter of \(8.5 \mathrm{~mm}\) and is \(15 \mathrm{~cm}\) long is found to stretch \(3.7 \mathrm{~mm}\) when acted on by a force of \(13.4 \mathrm{~N}\). Calculate Young's modulus for the tendon.

Short Answer

Expert verified
To calculate the Young's modulus of a tendon with a diameter of 8.5 mm, a length of 15 cm, and an elongation of 3.7 mm when a force of 13.4 N is applied, we followed these steps: 1. Calculate the cross-sectional area: A = 56.75π mm². 2. Calculate the stress: Stress = 76.11 N/m². 3. Calculate the strain: Strain = 0.0247. 4. Calculate Young's modulus: Y = 3081.0 N/m² or 3.08 kN/m². Therefore, the Young's modulus of the tendon is approximately 3.08 kN/m².

Step by step solution

01

Calculate the Cross-Sectional Area

To find the stress, we need to calculate the tendon's cross-sectional area. Since the tendon is approximately round, we can use the formula for the area of a circle, which is \(A = \pi r^2\), where \(r\) is the radius. Given the diameter of \(8.5\mathrm{~mm}\), we can find the radius and then the area. First, find the radius: \(r = \frac{\text{diameter}}{2} = \frac{8.5\mathrm{~mm}}{2} = 4.25\mathrm{~mm}\) Next, calculate the cross-sectional area using the radius: \(A = \pi r^2 = \pi (4.25\mathrm{~mm})^2 = 56.75\pi\mathrm{~mm^2}\)
02

Calculate the Stress

Stress is the force acting on a material divided by its cross-sectional area. Using the force of \(13.4 \mathrm{~N}\) and the calculated cross-sectional area, we can find the stress: Stress = \(\frac{\text{Force}}{\text{Area}} = \frac{13.4 \mathrm{~N}}{56.75\pi\mathrm{~mm^2}} = \frac{13.4 \mathrm{~N}}{56.75\pi\mathrm{~mm^2}} \times \frac{1\mathrm{~N/m^2}}{1\mathrm{~N/mm^2}} = 76.11\mathrm{~N/m^2}\)
03

Calculate the Strain

Strain is the change in length of a material divided by its original length. Using the given elongation of \(3.7 \mathrm{~mm}\) and the original length of \(15\mathrm{~cm}\), we can find the strain: Strain = \(\frac{\Delta L}{L} = \frac{3.7\mathrm{~mm}}{15\mathrm{~cm}} = \frac{3.7\mathrm{~mm}}{150\mathrm{~mm}} = 0.0247\)
04

Calculate Young's Modulus

Now that we have values for stress and strain, we can calculate Young's modulus using the formula \(Y = \frac{\text{Stress}}{\text{Strain}}\): \(Y = \frac{76.11\mathrm{~N/m^2}}{0.0247} = 3081.0\mathrm{~N/m^2}\) or \(3.08\mathrm{~kN/m^2}\) The Young's modulus of the tendon is approximately \(3.08\mathrm{~kN/m^2}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

In many problems involving application of Newton's Second Law to the motion of solid objects, friction is neglected for the sake of making the solution easier. The counterpart of friction between solids is viscosity of liquids. Do problems involving fluid flow become simpler if viscosity is neglected? Explain.

The average density of the human body is \(985 \mathrm{~kg} / \mathrm{m}^{3}\) and the typical density of seawater is about \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) a) Draw a free-body diagram of a human body floating in seawater and determine what percentage of the body's volume is submerged. b) The average density of the human body, after maximum inhalation of air, changes to \(945 \mathrm{~kg} / \mathrm{m}^{3}\). As a person floating in seawater inhales and exhales slowly, what percentage of his volume moves up out of and down into the water? c) The Dead Sea (a saltwater lake between Israel and Jordan ) is the world's saltiest large body of water. Its average salt content is more than six times that of typical seawater, which explains why there is no plant and animal life in it. Two-thirds of the volume of the body of a person floating in the Dead Sea is observed to be submerged. Determine the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the seawater in the Dead Sea.

A scuba diver must decompress after a deep dive to allow excess nitrogen to exit safely from his bloodstream. The length of time required for decompression depends on the total change in pressure that the diver experienced. Find this total change in pressure for a diver who starts at a depth of \(d=20.0 \mathrm{~m}\) in the ocean (density of seawater \(\left.=1024 \mathrm{~kg} / \mathrm{m}^{3}\right)\) and then travels aboard a small plane (with an unpressurized cabin) that rises to an altitude of \(h=5000 . \mathrm{m}\) above sea level.

Brass weights are used to weigh an aluminum object on an analytical balance. The weighing is done one time in dry air and another time in humid air, with a water vapor pressure of \(P_{\mathrm{h}}=2.00 \cdot 10^{3} \mathrm{~Pa}\). The total atmospheric pressure \(\left(P=1.00 \cdot 10^{5} \mathrm{~Pa}\right)\) and the temperature \(\left(T=20.0^{\circ} \mathrm{C}\right)\) are the same in both cases. What should the mass of the object be to be able to notice a difference in the balance readings, provided the balance's sensitivity is \(m_{0}=0.100 \mathrm{mg}\) ? (The density of aluminum is \(\rho_{\mathrm{A}}=2.70 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3} ;\) the density of brass is \(\left.\rho_{\mathrm{B}}=8.50 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right)\)

You fill a tall glass with ice and then add water to the level of the glass's rim, so some fraction of the ice floats above the rim. When the ice melts, what happens to the water level? (Neglect evaporation, and assume that the ice and water remain at \(0^{\circ} \mathrm{C}\) during the melting process.) a) The water overflows the rim. b) The water level drops below the rim. c) The water level stays at the top of the rim. d) It depends on the difference in density between water and ice.
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Thermodynamics

Read Explanation

Scientific Method Physics

Read Explanation

Oscillations

Read Explanation

Circular Motion and Gravitation

Read Explanation

Work Energy and Power

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.

Sign-up for free