IB MYP 4-5 Physics- Stretching and compressing materials- Study Notes - New Syllabus
IB MYP 4-5 Physics-Stretching and compressing materials- Study Notes
Key Concepts
- Stretching and compressing materials
Stretching and Compressing Materials
Stretching and Compressing Materials
When forces are applied to a material, they can cause it to change shape by stretching (tensile deformation) or compressing (compressive deformation). This change in shape is directly related to the size of the applied force and the properties of the material.
- Stretching – Occurs when the material is subjected to a pulling (tensile) force, increasing its length.
- Compressing – Occurs when the material is subjected to a pushing (compressive) force, reducing its length.
Concepts:
- When a force is applied, particles in the material are displaced from their equilibrium positions.
- The deformation is elastic if the material returns to its original shape after the force is removed, and plastic if the shape change is permanent.
- The relationship between the applied force and the resulting extension or compression is given by Hooke’s Law, within the elastic limit:
$\mathrm{F = kx}$
where:
F = \text{Applied force (N)}
k = \text{Spring constant (N/m)}
x = \text{Extension or compression (m)}
- The stiffness of a material is measured by its spring constant \( k \). Larger \( k \) means the material is harder to stretch or compress.
Stress, Strain, and Young’s Modulus (for more advanced analysis):
Stress:
\(\text{Stress} = \dfrac{\text{Force}}{\text{Cross-sectional area}} = \dfrac{F}{A}\)
Strain:
\(\text{Strain} = \dfrac{\text{Change in length}}{\text{Original length}} = \dfrac{\Delta L}{L_0}\)
Young’s Modulus:
E = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{F L_0}{A \Delta L}
Elastic and Plastic Behavior:
- Elastic deformation: Reversible: material returns to original shape.
- Plastic deformation: Irreversible: permanent change in shape after removing the force.
Example:
A steel spring has \( k = 250 \ \text{N/m} \). How much will it extend if a 10 N force is applied?
▶️ Answer/Explanation
Using Hooke’s Law: \( F = kx \)
\( x = \dfrac{F}{k} = \dfrac{10}{250} = 0.04 \ \text{m} = 4 \ \text{cm} \)
\(\boxed{4\ \text{cm}}\) extension
Example:
A rubber block has \( k = 500 \ \text{N/m} \). What compression occurs when a 25 N force is applied?
▶️ Answer/Explanation
\( x = \dfrac{F}{k} = \dfrac{25}{500} = 0.05 \ \text{m} = 5 \ \text{cm} \)
\(\boxed{5\ \text{cm}}\) compression
Example:
A metal wire of original length \( 2 \ \text{m} \) and diameter \( 2 \ \text{mm} \) is stretched by \( 1 \ \text{mm} \) when a \( 50 \ \text{N} \) force is applied. Find the stress, strain, and Young’s modulus.
▶️ Answer/Explanation
Step 1: Cross-sectional area: \( A = \pi r^2 = \pi (0.001)^2 = 3.14 \times 10^{-6} \ \text{m}^2 \)
Step 2: Stress: \( \sigma = \dfrac{F}{A} = \dfrac{50}{3.14 \times 10^{-6}} = 1.59 \times 10^{7} \ \text{Pa} \)
Step 3: Strain: \( \epsilon = \dfrac{\Delta L}{L_0} = \dfrac{0.001}{2} = 5 \times 10^{-4} \)
Step 4: Young’s Modulus: \( E = \dfrac{\sigma}{\epsilon} = \dfrac{1.59 \times 10^7}{5 \times 10^{-4}} = 3.18 \times 10^{10} \ \text{Pa} \)
\(\boxed{\text{Stress} = 1.59 \times 10^{7} \ \text{Pa}, \ \text{Strain} = 5\times 10^{-4}, \ E = 3.18 \times 10^{10} \ \text{Pa}}\)