/*! 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 2 A 91 -kg man's thighbone has a r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 91 -kg man's thighbone has a relaxed length of \(0.50 \mathrm{m},\) a cross- sectional area of \(7.0 \times 10^{-4} \mathrm{m}^{2},\) and a Young's modulus of \(1.1 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) By how much does the thighbone compress when the man is standing on both feet?

Short Answer

Expert verified
Question: Calculate the compression in length experienced by a man's thighbone when he is standing on both feet, given the following information: the man's mass is 91 kg, the length of the thighbone is 0.50 m, the cross-sectional area of the thighbone is \(7.0 \times 10^{-4} \mathrm{m}^{2}\), and the Young's modulus of the thighbone is \(1.1 \times 10^{10} \mathrm{N} / \mathrm{m}^{2}\). Answer: The thighbone compresses by approximately \(2.898 \times 10^{-5}\) meters when the man is standing on both feet.

Step by step solution

01

Calculate the weight of the person

To calculate the weight of the person, we can use the formula: \(W = m \times g\) where W is the weight, m is the mass of the person (91 kg), and g is the acceleration due to gravity (9.81 m/s²). \(W = 91 \mathrm{kg} \times 9.81 \mathrm{m/s²} = 892.71 \mathrm{N}\)
02

Calculate the force exerted on one leg

The force exerted on one leg is half of the person's weight as the weight is equally distributed on both legs. Hence, the force on one leg is: \(F = \frac{W}{2} = \frac{892.71 \mathrm{N}}{2} = 446.355 \mathrm{N}\)
03

Calculate the stress

Stress is calculated by dividing the force exerted on the bone by its cross-sectional area. Here, Stress = \(\frac{F}{A}\) Stress = \(\frac{446.355 \mathrm{N}}{7.0 \times 10^{-4} \mathrm{m}^{2}} = 637650 \mathrm{N/m^{2}}\)
04

Calculate the strain

Using the formula for Young's modulus, we have: \(Y = \frac{\text{Stress}}{\text{Strain}}\) Here, we want to find the strain, so rearrange the formula: Strain = \(\frac{\text{Stress}}{Y}\) Strain = \(\frac{637650 \mathrm{N/m^{2}}}{1.1 \times 10^{10} \mathrm{N} / \mathrm{m}^{2}} = 5.796818 \times 10^{-5}\)
05

Calculate the compression

Now that we have the strain, we can calculate the compression in the thighbone. Strain is the ratio of the compression (ΔL) to the initial length (L). Strain = \(\frac{\Delta L}{L}\) Here, we want to find the compression, so rearrange the formula: \(\Delta L = \text{Strain} \times L\) \(\Delta L = 5.796818 \times 10^{-5} \times 0.50 \mathrm{m} = 2.898409 \times 10^{-5} \mathrm{m}\) The thighbone compresses by approximately \(2.898 \times 10^{-5}\) meters when the man is standing on both feet.

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!

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

Resilin is a rubber-like protein that helps insects to fly more efficiently. The resilin, attached from the wing to the body, is relaxed when the wing is down and is extended when the wing is up. As the wing is brought up, some elastic energy is stored in the resilin. The wing is then brought back down with little muscular energy, since the potential energy in the resilin is converted back into kinetic energy. Resilin has a Young's modulus of $1.7 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}$ (a) If an insect wing has resilin with a relaxed length of \(1.0 \mathrm{cm}\) and with a cross-sectional area of \(1.0 \mathrm{mm}^{2},\) how much force must the wings exert to extend the resilin to \(4.0 \mathrm{cm} ?\) (b) How much energy is stored in the resilin?
A cart with mass \(m\) is attached between two ideal springs, each with the same spring constant \(k\). Assume that the cart can oscillate without friction. (a) When the cart is displaced by a small distance \(x\) from its equilibrium position, what force magnitude acts on the cart? (b) What is the angular frequency, in terms of \(m, x,\) and \(k,\) for this cart? (W) tutorial: cart between springs) (IMAGE NOT COPY)
(a) Sketch a graph of \(x(t)=A \sin \omega t\) (the position of an object in SHM that is at the equilibrium point at \(t=0\) ). (b) By analyzing the slope of the graph of \(x(t),\) sketch a graph of $v_{x}(t) .\( Is \)v_{x}(t)$ a sine or cosine function? (c) By analyzing the slope of the graph of \(v_{x}(t),\) sketch \(a_{x}(t)\) (d) Verify that \(v_{x}(t)\) is \(\frac{1}{4}\) cycle ahead of \(x(t)\) and that \(a_{x}(t)\) is \(\frac{1}{4}\) cycle ahead of \(v_{x}(t) .\) (W) tutorial: sinusoids)
An ideal spring has a spring constant \(k=25 \mathrm{N} / \mathrm{m}\). The spring is suspended vertically. A 1.0 -kg body is attached to the unstretched spring and released. It then performs oscillations. (a) What is the magnitude of the acceleration of the body when the extension of the spring is a maximum? (b) What is the maximum extension of the spring?
The ratio of the tensile (or compressive) strength to the density of a material is a measure of how strong the material is "pound for pound." (a) Compare tendon (tensile strength \(80.0 \mathrm{MPa}\), density $1100 \mathrm{kg} / \mathrm{m}^{3}\( ) with steel (tensile strength \)\left.0.50 \mathrm{GPa}, \text { density } 7700 \mathrm{kg} / \mathrm{m}^{3}\right)$ which is stronger "pound for pound" under tension? (b) Compare bone (compressive strength \(160 \mathrm{MPa}\), density $1600 \mathrm{kg} / \mathrm{m}^{3}$ ) with concrete (compressive strength $\left.0.40 \mathrm{GPa}, \text { density } 2700 \mathrm{kg} / \mathrm{m}^{3}\right):$ which is stronger "pound for pound" under compression?
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.