Problem 9 - College Physics

Question

Human hair. According to one set of measurements, the tensile strength of hair is 196 MPa, which produces a maximum strain of 0.40 in the hair. The thickness of hair varies considerably, but let's use a diameter of \(50 \mu \mathrm{m}\). (a) What is the magnitude of the force giving this tensile stress? (b) If the length of a strand of the hair is \(12 \mathrm{~cm}\) at its breaking point, what was its unstressed length?

Step-by-Step Solution

Answer

(a) The force is approximately 0.384 N. (b) The unstressed length is about 8.3 cm.

Step 1: Determine the Cross-Sectional Area of the Hair

The diameter of the hair is given as \(d = 50 \mu m = 50 \times 10^{-6} \text{ m}\). The cross-sectional area \(A\) of the hair, assuming it is circular, can be found using the formula for the area of a circle: \[A = \pi \left(\frac{d}{2}\right)^2\]Substitute the given diameter:\[A = \pi \left(\frac{50 \times 10^{-6}}{2}\right)^2 = \pi (25 \times 10^{-6})^2 \approx 1.96 \times 10^{-9} \text{ m}^2\]

Step 2: Calculate the Force using Tensile Stress

Tensile stress \(\sigma\) is defined as the force \(F\) applied per unit area \(A\). Thus, \(\sigma = \frac{F}{A}\).Given the tensile strength \(\sigma = 196 \text{ MPa} = 196 \times 10^6 \text{ N/m}^2\), we can rearrange to solve for the force:\[F = \sigma \times A = 196 \times 10^6 \times 1.96 \times 10^{-9} \approx 0.384 \text{ N}\]

Step 3: Calculate Unstressed Length of the Hair

Strain \(\epsilon\) is the ratio of change in length (\(\Delta L\)) to the original length \(L_0\). Given \(\epsilon = 0.40\) and the final (stressed) length \(L\) is \(12 \text{ cm} = 0.12 \text{ m}\), use \(\epsilon = \frac{\Delta L}{L_0}\) with \(\Delta L = L - L_0\):\[0.40 = \frac{0.12 - L_0}{L_0}\]Solving for \(L_0\), we get:\[0.40 L_0 = 0.12 - L_0 \Rightarrow 1.40 L_0 = 0.12 \Rightarrow L_0 \approx 0.083 \text{ m} = 8.3 \text{ cm}\]

Key Concepts

Tensile Stress

Tensile stress plays a crucial role when studying the strength of materials like human hair. It refers to the internal forces acting per unit area within a deformable body following the application of external pulling forces.
The formula representing tensile stress is \(\sigma = \frac{F}{A}\), where \(F\) is the applied force and \(A\) is the cross-sectional area.
This concept is important because it helps us understand how materials can withstand different forces without breaking.

• A higher tensile stress indicates a material can support greater forces.
• In our example, the hair with a tensile strength of 196 MPa can endure significant stress before reaching its breaking point.

This knowledge assists engineers in designing and testing materials for various applications, ensuring safety and efficiency.

Strain

Strain is the measure of deformation experienced by an object due to applied stress. It is expressed as a dimensionless ratio of the change in length to the original length.
Mathematically, strain \(\epsilon\) is defined as:\[\epsilon = \frac{\Delta L}{L_0}\]where \(\Delta L\) is the change in length and \(L_0\) is the original length.
For our hair strand, a strain of 0.40 means it extended by 40% of its original length before breaking.

• Strain is useful for determining how much a material can stretch.
• Materials with higher strain ratios can be more ductile and flexible, allowing for greater elongation without failing.

Understanding this concept assists in assessing the elasticity and flexibility of materials under stress.

Cross-Sectional Area

The cross-sectional area is an essential parameter in determining how a material responds to stress. It is the area of an object's cross-section, usually perpendicular to the applied forces.
For cylindrically-shaped materials, like a strand of hair, the formula for the cross-sectional area is:\[A = \pi \left(\frac{d}{2}\right)^2\]where \(d\) is the diameter.
In the exercise, a hair with a diameter of \(50 \mu m\) resulted in a cross-sectional area of approximately \(1.96 \times 10^{-9} \text{ m}^2\).

• Cross-sectional area helps calculate tensile stress, since it is the denominator in the stress formula.
• A larger cross-sectional area generally improves a material's ability to withstand higher stress, given the same force.

Knowing the cross-sectional area is critical when designing objects to ensure they can support anticipated loads without breaking.

Mechanics

Mechanics is the branch of physics dealing with the behavior of objects or materials subjected to forces or displacements. It encompasses concepts like force, motion, stress, and strain, which are fundamental in understanding how materials perform under different conditions.
The principles of mechanics allow us to predict how materials respond to various types of forces, helping optimize their use in practical applications.

• By assessing forces, mechanics helps determine the strength and flexibility of materials.
• Tensile stress and strain are core concepts in examining material performance within mechanical studies.

Incorporating the principles of mechanics into hair tensile testing and other similar experiments ensures accurate analysis and prediction of material behavior under stress.

Engineering Physics

Engineering physics bridges the gap between basic physics and more applied engineering applications. It provides the foundational principles that guide the design, analysis, and improvement of engineering systems.
Through engineering physics, concepts like tensile stress, strain, and cross-sectional area become practical tools used in real-world engineering challenges.

• These concepts aid in designing reliable structures capable of withstanding external forces.
• This interdisciplinary approach ensures that efficient, safe, and innovative engineering solutions are developed.

Engineering physics equips engineers with the knowledge to implement scientific principles effectively in evaluating and improving product and system performance.