Problem 26 - University Physics with Modern Physics

Question

Find the minimum diameter of a \(50.0-\mathrm{m}\) -long nylon string that will stretch no more than \(1.00 \mathrm{~cm}\) when a load of \(70.0 \mathrm{~kg}\) is suspended from its lower end. Assume that \(Y_{\text {nylon }}=3.51 \cdot 10^{8} \mathrm{~N} / \mathrm{m}^{2}\)

Step-by-Step Solution

Answer

Answer: The minimum diameter of the nylon string that will stretch no more than 1.00 cm when a load of 70.0 kg is suspended from its lower end is 3.52 mm.

Step 1: Convert the given data to the proper units

First, we need to convert the given data to the proper units. The load is given in kg, so we need to convert it to Newtons (N) using the equation \(F = mg\), where \(m\) is the mass and \(g\) is the acceleration due to gravity (9.8 m/s²). The elongation is given in cm, so we need to convert it to meters. \(F = (70.0 ~kg)(9.8 ~m/s^2) = 686 ~N\) \(\Delta L = 1.00 ~cm = 0.0100 ~m\) Now we have: - \(F = 686 ~N\) - \(L = 50.0 ~m\) - \(\Delta L = 0.0100 ~m\) - \(Y = 3.51 × 10^8 ~N/m^{2}\)

Step 2: Find the cross-sectional area (A)

Now we can solve for A by rearranging the elongation equation and plugging in our known values: \(A = \frac{FL}{Y \Delta L} = \frac{(686 ~N)(50.0 ~m)}{(3.51 × 10^8 ~N/m^2)(0.0100 ~m)} = 9.785 × 10^{-6} ~m^2\)

Step 3: Find the diameter of the string

Since the cross-sectional area of the string is circular, we can use the formula \(A = \pi r^2\), where \(r\) is the radius of the string. To find the diameter, we simply need to find the radius and multiply by 2: \(r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{9.785 × 10^{-6} ~m^2}{\pi}} = 0.00176 ~m\) The diameter is twice the radius: \(diameter = 2r = 2(0.00176 ~m) = 0.00352 ~m\)

Step 4: Convert the diameter to the proper units

Finally, let's convert the diameter from meters to millimeters for the final answer: \(diameter = 0.00352 ~m = 3.52 ~mm\) The minimum diameter of the nylon string that will stretch no more than 1.00 cm when a load of 70.0 kg is suspended from its lower end is 3.52 mm.

Key Concepts

Understanding Young's Modulus

Young's Modulus is a fundamental concept in material science that describes how much a material will stretch when a certain force is applied to it. It is often referred to as the "stiffness" of a material because it measures the resistance of a material to linear deformation under stress. Young's modulus (Y) is defined as the ratio of tensile stress (σ) to tensile strain (ε) in the linear elasticity of a material. The formula is expressed as:
\[Y = \frac{σ}{ε}\]where:

• σ = tensile stress, measured in N/m² (Pascals)
• ε = tensile strain, a dimensionless quantity which is the ratio of change in length to the original length

In the context of nylon strings, Young's Modulus helps us understand how the material will behave under different forces. For example, nylon with a higher Young's modulus will stretch less than one with a lower modulus when the same force is applied. It's a crucial factor that helps determine how thin or thick the string must be so that it doesn't stretch more than desired under a specific load.
By knowing the Young's modulus of nylon, which is often provided in problems like this one, various calculations can be carried out to predict the performance of nylon under tension, as seen in the original solution.

Exploring Tensile Stress

Tensile stress is another vital concept that describes the type of stress experienced by materials subjected to pulling forces. It measures how much force is acting on a specific area of the material. In mathematical terms, tensile stress (σ) is calculated using the formula:
\[ σ = \frac{F}{A} \]where:

• F = applied force in Newtons
• A = cross-sectional area in m²

In everyday terms, it's like asking how much pulling force is acting on each square meter of the material. A good analogy is stretching an elastic band: the force you apply to it causes tensile stress across its cross-sectional area.
Relating this to the nylon string problem, by determining the cross-sectional area and applying force due to the suspended load, you can calculate the tensile stress. This calculation helps ensure that the material will not break or stretch too much. Properly managing tensile stress is essential for the durability and performance of materials under force, such as in ropes, cables, and in this case, nylon strings used for specific applications.

Understanding Material Elongation

Material elongation is a key factor in understanding how materials respond when subjected to forces. Elongation refers to the change in length of a material when a force is applied. It's usually expressed as a percentage of the original length, providing a clear idea of how much the material stretches under tension. The formula to calculate elongation (ΔL) is:
\[ ΔL = rac{FL}{AY} \]where:

• F = applied force in Newtons
• L = original length in meters
• A = cross-sectional area in m²
• Y = Young's modulus in N/m²

In practical terms, elongation helps engineers and designers ensure that materials will perform adequately without excessive stretching in their intended application. When it comes to the nylon string problem, the maximum allowable elongation is provided (1 cm), allowing us to use the elongation formula to calculate the needed cross-sectional area to limit stretching within acceptable bounds. Understanding material elongation is thus essential in applications where precision and material stability under mechanical stress are critical factors.