Problem 6 - College Physics

Question

\(\bullet\) Stress on a mountaineer's rope. A nylon rope used by mountaineers elongates 1.10 under the weight of a 65.0 \(\mathrm{kg}\) climber. If the rope is 45.0 \(\mathrm{m}\) in length and 7.0 \(\mathrm{mm}\) in diameter, what is Young's modulus for this nylon?

Step-by-Step Solution

Answer

Young's modulus for the nylon rope is approximately 6.78 × 10^8 N/m².

Step 1: Understand the problem and the known values

We need to calculate Young's modulus of the rope. We know the elongation (1.10 m), the mass of the climber (65.0 kg), the original length of the rope (45.0 m), and the diameter (7.0 mm). Young's modulus is calculated using the formula: \( E = \frac{F/A}{\Delta L/L_0} \).

Step 2: Calculate the cross-sectional area

The rope's diameter is 7.0 mm, which is 0.007 m. We calculate the cross-sectional area \( A \) using \( A = \pi r^2 \), where \( r = \frac{d}{2} = 0.0035 \) m. Hence, \( A = \pi \times (0.0035)^2 \approx 3.85 \times 10^{-5} \, \text{m}^2 \).

Step 3: Calculate the force exerted by the climber

The force \( F \) is the weight of the climber, calculated as \( F = m \cdot g \), where \( m = 65.0 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \). Thus, \( F = 65.0 \times 9.81 \approx 637.65 \, \text{N} \).

Step 4: Apply the Young's modulus formula

We can now use the formula \( E = \frac{F/A}{\Delta L/L_0} \). Substitute \( F = 637.65 \, \text{N} \), \( A = 3.85 \times 10^{-5} \, \text{m}^2 \), \( \Delta L = 1.10 \, \text{m} \), and \( L_0 = 45.0 \, \text{m} \). The ratio is \( \frac{F}{A} = \frac{637.65}{3.85 \times 10^{-5}} \approx 16556948 \, \text{N/m}^2 \). The fractional change in length is \( \frac{1.10}{45.0} \approx 0.0244 \).

Step 5: Calculate Young's modulus

Now divide the stress by the strain: \( E = \frac{16556948}{0.0244} \approx 6.78 \times 10^{8} \, \text{N/m}^2 \). Thus, Young's modulus of the nylon is \( 6.78 \times 10^8 \, \text{N/m}^2 \).

Key Concepts

Stress

When a force is applied to a material, it experiences stress. Stress is defined as the force exerted per unit area of the material. In our given problem, the stress on the nylon rope used by the mountaineer is due to the weight of the climber. Formulaically, stress \( \sigma \) is calculated using:

• \( \sigma = \frac{F}{A} \)

where \( F \) is the force exerted, which in this case is the weight of the climber, and \( A \) is the cross-sectional area of the rope. The weight is calculated as the product of mass and gravitational acceleration, here 65 kg multiplied by 9.81 m/s², resulting in approximately 637.65 N. Breaking this down further:

• The diameter of the rope is 7.0 mm, converted into meters (0.007 m) for calculation.
• The area \( A \) is calculated using the formula for the cross-sectional area of a circle: \( A = \pi (r^2) \), where \( r \) is half the diameter.
• Plug the force and area into the stress formula to determine the stress experienced by the rope.

Understanding stress helps in knowing how much force a material can withstand before deforming.

Strain

Strain is a measure of how much a material deforms under stress. It's defined as the ratio of change in length to the original length of the material. For the nylon rope, strain \( \varepsilon \) is computed by:

• \( \varepsilon = \frac{\Delta L}{L_0} \)

where \( \Delta L \) is the change in length (elongation) of the rope, and \( L_0 \) is the original length of the rope.

• In our example, the rope elongates by 1.10 m, and its original length is 45.0 m.
• The strain, therefore, is \( \frac{1.10}{45.0} \approx 0.0244 \).

This dimensionless quantity provides insight into how much a material stretches. It’s crucial for determining the usability and durability of the material under different forces. Knowing the strain helps engineers and scientists predict how a material will behave under load.

Nylon Rope

Nylon is a type of synthetic polymer known for its strength and elasticity, and is widely used in applications like climbing ropes. It possesses desirable properties for this use:

• High strength: Nylon ropes are strong enough to resist substantial amounts of load without breaking, providing security to climbers.
• Elasticity: The material allows for some stretch, absorbing impacts and reducing stress transmitted to the climber during a fall.
• Durability: Resists environmental damage, such as sunlight and moisture, which makes it ideal for outdoor use.

These properties make nylon a preferable choice for applications where both flexibility and strength are essential traits. Understanding these characteristics can help individuals appreciate why nylon is selected over other materials for demanding applications.

Material Properties

Material properties are intrinsic characteristics that define how materials respond to external forces and environmental conditions. They include:

• Elastic Modulus: Also known as Young's modulus, it defines the ability of a material to withstand changes in length under lengthwise tension. It is a measure of stiffness, calculated using stress divided by strain.
• Tensile Strength: The maximum stress that a material can withstand while being stretched or pulled before breaking. It's an indication of the material's durability and capability.
• Ductility: The ability to deform under tensile stress, critical for understanding potential elongation without failure.
• Resistance to wear and environmental factors: This includes how well a material stands up to elements like heat, UV rays, and moisture.

In the context of the mountaineering problem, knowing the elastic modulus of nylon illustrates its robust nature and flexibility. This helps predict how the rope will perform under stress, ensuring the climber's safety.