/*! 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 80 Between each pair of vertebrae i... [FREE SOLUTION] | 91Ó°ÊÓ

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Between each pair of vertebrae in the spinal column is a cylindrical disc of cartilage. Typically, this disc has a radius of about \(3.0 \times 10^{-2} \mathrm{m}\) and a thickness of about \(7.0 \times 10^{-3} \mathrm{m} .\) The shear modulus of cartilage is \(1.2 \times 10^{7} \mathrm{N} / \mathrm{m}^{2} .\) Suppose that a shearing force of magnitude \(11 \mathrm{N}\) is applied parallel to the top surface of the disc while the bottom surface remains fixed in place. How far does the top surface move relative to the bottom surface?

Short Answer

Expert verified
The top surface moves about \(2.27 \times 10^{-6} \text{ meters}.\)

Step by step solution

01

Understand the Problem

We need to find how far the top surface of a cartilage disc moves relative to the bottom surface when a shearing force is applied. We know the radius, thickness, shear modulus, and the force applied.
02

Identify the Formula for Shear Deformation

The formula for shear deformation is given by \[ \Delta x = \frac{F \cdot L}{A \cdot G} \]where \( \Delta x \) is the displacement, \( F \) is the force, \( L \) is the thickness, \( A \) is the area, and \( G \) is the shear modulus.
03

Calculate the Area of the Disc

The area \( A \) of the top surface of the disc can be calculated using the formula for the area of a circle, \( A = \pi r^2 \). Substituting the given radius, \[ A = \pi \times (3.0 \times 10^{-2})^2 \approx 2.827 \times 10^{-3} \, \text{m}^2 \]
04

Substitute Values into the Formula

Substitute the known values into the shear deformation formula: \[ \Delta x = \frac{11 \times 7.0 \times 10^{-3}}{2.827 \times 10^{-3} \times 1.2 \times 10^{7}} \]
05

Perform the Calculations

Perform the arithmetic calculations: \[ \Delta x = \frac{7.7 \times 10^{-2}}{3.3924 \times 10^{4}} \approx 2.27 \times 10^{-6} \, \text{m} \]
06

Interpret the Result

The top surface of the disc moves approximately \(2.27 \times 10^{-6} \) meters (or \(2.27 \mu m\)) relative to the bottom surface when the force is applied.

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Key Concepts

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

Cartilage Biomechanics
Cartilage is a flexible connective tissue found in various parts of the body. It plays a crucial role in joints, providing a smooth surface for motion and cushioning impact forces.

In the context of the spine, cartilage is integral to the intervertebral discs, which are the soft pads situated between each bone in the spinal column. This makes them important for absorbing shocks and allowing flexibility in the spine.

These discs are made of a unique type of cartilage that can withstand considerable pressure and shear forces. This resilience is partly due to its composition, which includes water, collagen fibers, and proteoglycans.
  • Water provides elasticity and allows cartilage to deform without damage.
  • Collagen fibers add strength and structure.
  • Proteoglycans help trap water molecules, maintaining the disc's shape and allowing it to function effectively under stress.
Understanding the biomechanical properties of cartilage, such as its ability to resist deformation, is essential when analyzing its behavior under different types of forces, including shear forces as demonstrated in the original exercise.
Shear Modulus
The shear modulus, denoted by the symbol \( G \), is a material property that indicates its ability to resist shear deformation. It's defined as the ratio of shear stress to the shear strain in a material.

In mathematical terms: \[ G = \frac{\text{Shear stress}}{\text{Shear strain}} \]

In the context of cartilage in the spinal column, the shear modulus gives us an idea of how the cartilage will react when a force is applied parallel to its surface. It's crucial because it determines the amount of deformation (see "\( \Delta x \)" in the formula mentioned in the original exercise) the material will experience under a given load.

A higher shear modulus means the material is more rigid, while a lower shear modulus indicates it is more easily deformed. In the original problem, the cartilage's shear modulus is given as \( 1.2 \times 10^{7} \, \text{N/m}^{2} \,\). This relatively high value shows that cartilage is quite firm, ensuring that the vertebrae are stable while still allowing for essential movement. Knowing this modulus is vital for designing medical treatments or devices that interact with the spine.
Spinal Column Mechanics
The spinal column is a complex structure that provides support and flexibility to the human body. It consists of vertebrae, intervertebral discs, ligaments, and muscles working together to support weight, protect the spinal cord, and permit a range of motions.

Each intervertebral disc acts as a shock absorber and allows for some movement of the vertebrae. This flexibility is crucial for performing everyday activities that require bending, twisting, and stretching.

The mechanics of the spinal column are finely balanced.
  • The vertebrae are solid bone structures that provide stability.
  • Intervertebral discs of cartilage allow slight movements and absorb pressures through compression and shear motions.
  • Ligaments and muscles surrounding the spine assist in maintaining alignment and allowing controlled movement.
The exercise demonstrates how cartilage discs undergo shear deformation when forces are applied. This form of movement is a critical aspect of spinal mechanics, emphasizing the importance of maintaining healthy cartilage for overall spine function and health.

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Most popular questions from this chapter

A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to the top of the tower and hung down the center of the staircase. However, nothing is available with which to measure the length of the rope. Therefore, at the bottom of the rope a small object is attached so as to form a simple pendulum that just clears the floor. The period of the pendulum is measured to be 9.2 s. What is the height of the tower?

A 75 -kg diver is standing at the end of a diving board while it is vibrating up and down in simple harmonic motion, as indicated in the figure. The diving board has an effective spring constant of \(k=\) \(4100 \mathrm{N} / \mathrm{m},\) and the vertical distance between the highest and lowest points in the motion is \(0.30 \mathrm{m} .\) Concepts: (i) How is the amplitude \(A\) related to the vertical distance between the highest and lowest points of the diver's motion? (ii) Starting from the top, where is the diver located one-quarter of a period later, and what can be said about his speed at this point? (iii) If the amplitude were to double, would the period also double? Explain. Calculations: (a) What is the amplitude of the motion? (b) Starting when the diver is at the highest point, what is his speed one-quarter of a period later? (c) If the vertical distance between his highest and lowest points were changed to \(0.10 \mathrm{m},\) what would be the time required for the diver to make one complete motional cycle?

An archer, about to shoot an arrow, is applying a force of \(+240 \mathrm{N}\) to a drawn bowstring. The bow behaves like an ideal spring whose spring constant is \(480 \mathrm{N} / \mathrm{m} .\) What is the displacement of the bowstring?

A heavy-duty stapling gun uses a 0.140 -kg metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of \(f=2.00 \mathrm{Hz} .\) On this tray is an empty cup. Obtain the coefficient of static friction between the tray and the cup, given that the cup begins slipping when the amplitude of the motion is \(5.00 \times 10^{-2} \mathrm{m} .\)

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