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Chapter 10: Problem 13

What is the maximum load that could be suspended from a copper wire of length \(1.0 \mathrm{m}\) and radius \(1.0 \mathrm{mm}\) without permanently deforming the wire? Copper has an elastic limit of \(2.0 \times 10^{8} \mathrm{Pa}\) and a tensile strength of \(4.0 \times 10^{8} \mathrm{Pa}\).

Short Answer

Expert verified
Answer: The maximum load that can be suspended from the copper wire without permanently deforming it is approximately 127.9 N.

Step by step solution

01

Calculate cross-sectional area of the wire

To determine the cross-sectional area of the wire, we can use the formula for the area of a circle, which is \(A = \pi r^2\). In this case, the radius of the wire is given as \(1.0\,\text{mm} = 1.0 \times 10^{-3}\,\text{m}\). So the area is: \(A = \pi (1.0 \times 10^{-3}\,\text{m})^2\)
02

Use the elastic limit and cross-sectional area to determine maximum force

The elastic limit of the material is given as \(2.0 \times 10^{8}\,\text{Pa}\). The formula to calculate the force required to reach this stress level can be obtained using the equation: \(Stress = \frac{Force}{Area}\) We can rearrange the equation to find the maximum force: \(Force = Stress \times Area\) Substitute the given values for stress and area: \(Force = (2.0 \times 10^{8}\,\text{Pa}) \times \pi (1.0 \times 10^{-3}\,\text{m})^2\)
03

Calculate the maximum load

To find the maximum load that could be suspended from the wire without permanently deforming it, we can divide the maximum force by the acceleration due to gravity (\(9.81\,\text{m/s}^2\)): \(Load = \frac{Force}{g}\) Substitute the expression for the maximum force and the value of g: \(Load = \frac{(2.0 \times 10^{8}\,\text{Pa}) \times \pi (1.0 \times 10^{-3}\,\text{m})^2}{9.81\,\text{m/s}^2}\) Finally, calculate the maximum load: \(Load \approx 127.9\,\text{N}\) So, the maximum load that could be suspended from a copper wire of length \(1.0\,\text{m}\) and radius \(1.0\,\text{mm}\) without permanently deforming the wire is approximately \(127.9\,\text{N}\).

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Most popular questions from this chapter

A clock has a pendulum that performs one full swing every \(1.0 \mathrm{s}(\) back and forth). The object at the end of the pendulum weighs $10.0 \mathrm{N}$. What is the length of the pendulum?
A \(0.50-\mathrm{kg}\) object, suspended from an ideal spring of spring constant \(25 \mathrm{N} / \mathrm{m},\) is oscillating vertically. How much change of kinetic energy occurs while the object moves from the equilibrium position to a point \(5.0 \mathrm{cm}\) lower?
A 91 -kg man's thighbone has a relaxed length of \(0.50 \mathrm{m},\) a cross- sectional area of \(7.0 \times 10^{-4} \mathrm{m}^{2},\) and a Young's modulus of \(1.1 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) By how much does the thighbone compress when the man is standing on both feet?
What is the period of a pendulum consisting of a \(6.0-\mathrm{kg}\) mass oscillating on a 4.0 -m-long string?
A gibbon, hanging onto a horizontal tree branch with one arm, swings with a small amplitude. The gibbon's CM is 0.40 m from the branch and its rotational inertia divided by its mass is \(I / m=0.25 \mathrm{m}^{2} .\) Estimate the frequency of oscillation.
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