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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

Equipment to be used in airplanes or spacecraft is often subjected to a shake test to be sure it can withstand the vibrations that may be encountered during flight. A radio receiver of mass \(5.24 \mathrm{kg}\) is set on a platform that vibrates in SHM at \(120 \mathrm{Hz}\) and with a maximum acceleration of $98 \mathrm{m} / \mathrm{s}^{2}(=10 \mathrm{g}) .$ Find the radio's (a) maximum displacement, (b) maximum speed, and (c) the maximum net force exerted on it.
(a) Sketch a graph of \(x(t)=A \sin \omega t\) (the position of an object in SHM that is at the equilibrium point at \(t=0\) ). (b) By analyzing the slope of the graph of \(x(t),\) sketch a graph of $v_{x}(t) .\( Is \)v_{x}(t)$ a sine or cosine function? (c) By analyzing the slope of the graph of \(v_{x}(t),\) sketch \(a_{x}(t)\) (d) Verify that \(v_{x}(t)\) is \(\frac{1}{4}\) cycle ahead of \(x(t)\) and that \(a_{x}(t)\) is \(\frac{1}{4}\) cycle ahead of \(v_{x}(t) .\) (W) tutorial: sinusoids)
A mass-spring system oscillates so that the position of the mass is described by \(x=-10 \cos (1.57 t),\) where \(x\) is in \(\mathrm{cm}\) when \(t\) is in seconds. Make a plot that has a dot for the position of the mass at $t=0, t=0.2 \mathrm{s}, t=0.4 \mathrm{s}, \ldots, t=4 \mathrm{s}$ The time interval between each dot should be 0.2 s. From your plot, tell where the mass is moving fastest and where slowest. How do you know?
The displacement of an object in SHM is given by $y(t)=(8.0 \mathrm{cm}) \sin [(1.57 \mathrm{rad} / \mathrm{s}) t] .$ What is the frequency of the oscillations?
The leg bone (femur) breaks under a compressive force of about $5 \times 10^{4} \mathrm{N}\( for a human and \)10 \times 10^{4} \mathrm{N}$ for a horse. The human femur has a compressive strength of \(1.6 \times 10^{8} \mathrm{Pa},\) while the horse femur has a compressive strength of $1.4 \times 10^{8} \mathrm{Pa} .$ What is the effective crosssectional area of the femur in a human and in a horse? (Note: since the center of the femur contains bone marrow, which has essentially no compressive strength, the effective cross- sectional area is about \(80 \%\) of the total cross-sectional area.)
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