Problem 56 - Physics

Question

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 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?

Step-by-Step Solution

Answer

The top surface moves approximately \(2.28 \times 10^{-5}\) meters relative to the bottom surface.

Step 1: Identify Known Values

In this problem, we are given the following values: - Radius of the disc, \( r = 3.0 \times 10^{-2} \ m \)- Thickness of the disc, \( t = 7.0 \times 10^{-3} \ m \)- Shear modulus of cartilage, \( G = 1.2 \times 10^{7} \ N/m^2 \)- Shearing force, \( F = 11 \ N \).

Step 2: Calculate Disc Area

The area of the cylindrical disc's top surface is given by the formula for the area of a circle, \( A = \pi r^2 \). Substitute the disc's radius:\[ A = \pi \times (3.0 \times 10^{-2})^2 \].Calculate this value to find the area.

Step 3: Calculate Disc Area Value

Plug in the radius and calculate:\[ A = \pi \times (3.0 \times 10^{-2})^2 \approx 2.827 \times 10^{-3} \ m^2 \].

Step 4: Use Shear Deformation Formula

The shear deformation formula is given by \( \Delta x = \frac{F \cdot t}{G \cdot A} \), where- \( \Delta x \) is the relative displacement,- \( F \) is the force applied,- \( t \) is the thickness,- \( G \) is the shear modulus of the material,- \( A \) is the area.Substitute the known values into this formula.

Step 5: Substitute Values into the Formula

Substitute \( F = 11 \ N \), \( t = 7.0 \times 10^{-3} \ m \), \( G = 1.2 \times 10^{7} \ N/m^2 \), and \( A = 2.827 \times 10^{-3} \ m^2 \) into the formula:\[ \Delta x = \frac{11 \times 7.0 \times 10^{-3}}{1.2 \times 10^{7} \times 2.827 \times 10^{-3}} \].

Step 6: Calculate Final Displacement

Calculate the expression:\[ \Delta x = \frac{11 \times 7.0 \times 10^{-3}}{1.2 \times 10^{7} \times 2.827 \times 10^{-3}} \approx 2.28 \times 10^{-5} \ m \].

Key Concepts

Shear Modulus

The shear modulus, also known as the modulus of rigidity, is a property that describes how a material responds to shear stress. This modulus is critical in understanding how materials like cartilage behave when forces are applied parallel to their surface.
The shear modulus is a measure of the stiffness of a material.

• A high shear modulus indicates a material is difficult to deform, meaning it's very stiff.
• A low shear modulus means the material is more easily deformed, indicating resilience.

In the given problem, the shear modulus of the cartilage disc is specified as \(1.2 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}\). This means for every unit of shear stress applied, the material will deform to a certain extent. The value here suggests cartilage is quite firm, resisting deformation under typical forces. This resistance is crucial to providing stability in the spinal column.

Cylindrical Disc

A cylindrical disc is a solid, round object with a specific height and circular base. In our context, cartilage discs are found between vertebrae in the spine, serving as cushions and providing flexibility. These discs have both a thickness and a radius, determining their size and shape.
For the exercise:

• The radius is given as \(3.0 \times 10^{-2} \mathrm{~m}\)
• The thickness is \(7.0 \times 10^{-3} \mathrm{~m}\)

This shape is crucial in determining the surface area, which is calculated as \(A = \pi r^2\).
This surface area plays a key role in how the disc responds to applied forces, as seen in the provided formula for calculating deformation.

Shearing Force

Shearing force refers to a force that acts parallel or tangential to a surface. Such forces can cause materials to deform by sliding layers over each other, much like spreading a deck of cards across a table. In our exercise, a shearing force of \(11 \mathrm{~N}\) is applied to the top surface of the cylindrical disc, while the bottom remains fixed.
When a shearing force is applied:

• It causes one part of the material to move in the direction of the force.
• The bottom layer remains stationary, creating a shear effect.

The significance of this force lies in determining how much the top surface of the disc will shift relative to the bottom. As demonstrated in the solution, it plays a vital role in calculating the amount of deformation a cylindrical disc will undergo.

Relative Displacement

Relative displacement refers to the change in position of the top surface of the cylindrical disc compared to its bottom due to the applied shearing force. In simpler terms, it is how much the upper part of the disc slides over the stationary lower part when force is applied. In this exercise, the formula \(\Delta x = \frac{F \cdot t}{G \cdot A}\) is used to compute the displacement. Here, \(\Delta x\) indicates the actual measure of how far the top surface moves, with:

• \(F\) being the force applied.
• \(t\) signifying the thickness of the disc.
• \(G\) representing the shear modulus.
• \(A\) showing the area of the surface.

Through substitution of these known values, one can calculate the displacement, contributing to our understanding of how forces affect structural parts like spinal discs.