/*! 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} Q77PE During heavy lifting, a disk bet... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

During heavy lifting, a disk between spinal vertebrae is subjected to a \({\rm{5000 N}}\) compressional force. (a) What pressure is created, assuming that the disk has a uniform circular cross-section \({\rm{2}}{\rm{.00 cm}}\) in radius? (b) What deformation is produced if the disk is \({\rm{0}}{\rm{.800 cm}}\) thick and has Young’s modulus of \({\rm{1}}{\rm{.5 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N/}}{{\rm{m}}^{\rm{2}}}\)?.

Short Answer

Expert verified

(a) The pressure is obtained as: \[{\rm{3}}{\rm{.980 x 1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/}}{{\rm{m}}^{\rm{2}}}\].

(b) Deformation produced in the disk is obtained as: \[{\rm{0}}{\rm{.02123 mm}}\].

Step by step solution

01

Conceptual Introduction

Fluid statics, often known as hydrostatics, is a branch of fluid mechanics that investigates the state of balance of a floating and submerged body, as well as the pressure in a fluid, or imposed by a fluid, on an immersed body.

02

Given Data

The values we have is:

Force Applied is \[{\rm{F = 5000 N}}\].

Radius of disc is \[{\rm{2 cm = 0}}{\rm{.02 m}}\].

The cross-sectional area of disk is:

\(\begin{array}{c}A = \pi {r^2}\\ = 3.14 \times {(0.02)^2}\\ = 1.256 \times 1{0^{ - 3}}{m^2}\end{array}\)

Thickness of disk is:

\[\begin{array}{c}{\rm{L = 0}}{\rm{.800 cm}}\\{\rm{ = 0}}{\rm{.008 m}}\end{array}\]

Young’s modulus of disk is Y = \(1.5 \times 1{0^9}N/{m^2}\)

Pressure created on disk = P = ? N/m

Deformation of disk = l = ? m

03

Evaluating the pressure

(a) Pressure exerted on the disk is given by:

\(\begin{array}{l}P = F/A\\P = 5000/(1.256 \times 1{0^{ - 3}})\\P = 3.980 \times 1{0^6}N/{m^2}\end{array}\)

Therefore, the pressure is: \[{\rm{3}}{\rm{.980 x 1}}{{\rm{0}}^{\rm{6}}}{\rm{ N/}}{{\rm{m}}^{\rm{2}}}\]

04

Solution of part b

(b) Young’s modulus of the disk is given by:

\(\begin{array}{l}Y = \frac{{F/A}}{{l/L}}\\I = \frac{{(F/A)*L}}{Y}\\I = \frac{{(3.980 \times 1{0^6})*0.008}}{{1.5 \times 1{0^9}}}\\l = \frac{{31840}}{{1.5 \times 1{0^9}}}\end{array}\)

\(\begin{array}{l}I = 2.123 \times 1{0^{ - 5}}m\\l = 0.02123mm\end{array}\)

Therefore, the deformation produced in the disk is: \[{\rm{0}}{\rm{.02123 mm}}\].

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!

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

Classical physics is a good approximation to modern physics under certain circumstances. What are they?

(a) An excimer laser used for vision correction emits193 - nm UV. Calculate the photon energy in eV.

(b) These photons are used to evaporate corneal tissue, which is very similar to water in its properties. Calculate the amount of energy needed per molecule of water to make the phase change from liquid to gas. That is, divide the heat of vaporization in kJ/kg by the number of water molecules in a kilogram.

(c) Convert this to eV and compare to the photon energy. Discuss the implications.

(a) Suppose that a person has an average heart rate of 72.0beats/minbeats/ min. How many beats does he or she have in 2.0 y ? (b) In 2.00 y ? (c) In 2.000 y ?

(a) Echo times are measured by diagnostic ultrasound scanners to determine distances to reflecting surfaces in a patient. What is the difference in echo times for tissues that are 3.50 and 3.60cm beneath the surface? (This difference is the minimum resolving time for the scanner to see details as small as 0.100cm, or 1.00mm. Discrimination of smaller time differences is needed to see smaller details.) (b) Discuss whether the period T of this ultrasound must be smaller than the minimum time resolution. If so, what is the minimum frequency of the ultrasound, and is that out of the normal range for a diagnostic ultrasound?

The wavelength of the four Balmer series lines for hydrogen are found to be 410.3, 434.2, 486.3, and 656.5 nm. What average percentage difference is found between these wavelength numbers and those predicted by\[\frac{{\bf{1}}}{{\bf{\lambda }}}{\bf{ = R}}\left( {\frac{{\bf{1}}}{{{\bf{n}}_{\bf{f}}^{\bf{2}}}}{\bf{ - }}\frac{{\bf{1}}}{{{\bf{n}}_{\bf{i}}^{\bf{2}}}}} \right){\bf{?}}\]It is amazing how well a simple formula (disconnected originally from theory) could duplicate this phenomenon.

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.