A disk between vertebrae in the spine is subjected to a shearing force of 825 N. Find its shear deformation, taking it to have a shear modulus of 1.50×109 N/m2. The disk is equivalent to a solid cylinder 0.550 cm high and 3.00 cm in diameter.
A disk between vertebrae in the spine is subjected to a shearing force of 825 N. Find its shear deformation, taking it to have a shear modulus of 1.50×109 N/m2. The disk is equivalent to a solid cylinder 0.550 cm high and 3.00 cm in diameter.
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13) A disk between vertebrae in the spine is subjected to a shearing force of 825 N. Find its shear deformation, taking it to have a shear modulus of 1.50×109 N/m2. The disk is equivalent to a solid cylinder 0.550 cm high and 3.00 cm in diameter.
![**Problem Statement:**
A disk between vertebrae in the spine is subjected to a shearing force of 825 N. Find its shear deformation, taking it to have a shear modulus of \(1.50 \times 10^9 \, \text{N/m}^2\). The disk is equivalent to a solid cylinder 0.550 cm high and 3.00 cm in diameter.
**Solution Box:**
- Shear Deformation: [Input field for answer] m
**Explanation:**
To calculate shear deformation (\(\Delta x\)) of the cylindrical disk, use the formula for shear deformation:
\[
\Delta x = \frac{F \cdot L_0}{A \cdot G}
\]
Where:
- \(F\) is the force applied (825 N)
- \(L_0\) is the original length (height) of the cylinder (0.0055 m)
- \(A\) is the cross-sectional area of the cylinder
- \(G\) is the shear modulus (\(1.50 \times 10^9 \, \text{N/m}^2\))
For a cylinder:
\[
A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.03}{2}\right)^2 \, \text{m}^2
\]
**Credit:**
Question Credit: OpenStax College Physics](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37e1710d-5d92-4416-9037-10beff7be39e%2F5b2f41b4-e2cf-4c95-b041-d7d7d7336fd1%2F8zgxhdv_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A disk between vertebrae in the spine is subjected to a shearing force of 825 N. Find its shear deformation, taking it to have a shear modulus of \(1.50 \times 10^9 \, \text{N/m}^2\). The disk is equivalent to a solid cylinder 0.550 cm high and 3.00 cm in diameter.
**Solution Box:**
- Shear Deformation: [Input field for answer] m
**Explanation:**
To calculate shear deformation (\(\Delta x\)) of the cylindrical disk, use the formula for shear deformation:
\[
\Delta x = \frac{F \cdot L_0}{A \cdot G}
\]
Where:
- \(F\) is the force applied (825 N)
- \(L_0\) is the original length (height) of the cylinder (0.0055 m)
- \(A\) is the cross-sectional area of the cylinder
- \(G\) is the shear modulus (\(1.50 \times 10^9 \, \text{N/m}^2\))
For a cylinder:
\[
A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.03}{2}\right)^2 \, \text{m}^2
\]
**Credit:**
Question Credit: OpenStax College Physics
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