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Chapter 10: Problem 16

The maximum strain of a steel wire with Young's modulus $2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2},\( just before breaking, is \)0.20 \%$ What is the stress at its breaking point, assuming that strain is proportional to stress up to the breaking point?

Short Answer

Expert verified
Answer: The stress at the breaking point of the steel wire is \(4.0 \times 10^8 \mathrm{N/m^2}\).

Step by step solution

01

Convert strain to decimal form

To find the stress, the first step is to convert the given strain percentage value to decimal form. To do this, simply divide the given percentage value by 100. The strain in decimal form is: $$ \text{strain in decimal} = \frac{0.20\%}{100} = 2.0 \times 10^{-3} $$
02

Calculate stress using Young's modulus and strain

Now that we have the strain value in decimal form, we can use the relationship between stress, Young's modulus, and strain to find the stress at the breaking point. The equation that connects stress, strain, and Young's modulus is: $$ \text{stress} = \text{Young's modulus} \times \text{strain} $$ Substituting the given values, we get: $$ \text{stress} = (2.0 \times 10^{11} \mathrm{N/m^2}) \times (2.0 \times 10^{-3}) $$
03

Calculate the stress at the breaking point

Using the equation from Step 2, we will multiply the Young's modulus and the strain in decimal form to find the stress at the breaking point: $$ \text{stress} = (2.0 \times 10^{11} \mathrm{N/m^2}) \times (2.0 \times 10^{-3}) = 4.0 \times 10^8 \mathrm{N/m^2} $$ So, the stress at the breaking point of the steel wire is \(4.0 \times 10^8 \mathrm{N/m^2}\).

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Most popular questions from this chapter

Christy has a grandfather clock with a pendulum that is \(1.000 \mathrm{m}\) long. (a) If the pendulum is modeled as a simple pendulum, what would be the period? (b) Christy observes the actual period of the clock, and finds that it is \(1.00 \%\) faster than that for a simple pendulum that is \(1.000 \mathrm{m}\) long. If Christy models the pendulum as two objects, a \(1.000-\mathrm{m}\) uniform thin rod and a point mass located \(1.000 \mathrm{m}\) from the axis of rotation, what percentage of the total mass of the pendulum is in the uniform thin rod?
The gravitational potential energy of a pendulum is \(U=m g y .\) (a) Taking \(y=0\) at the lowest point, show that \(y=L(1-\cos \theta),\) where \(\theta\) is the angle the string makes with the vertical. (b) If \(\theta\) is small, \((1-\cos \theta)=\frac{1}{2} \theta^{2}\) and \(\theta=x / L\) (Appendix A.7). Show that the potential energy can be written \(U=\frac{1}{2} k x^{2}\) and find the value of \(k\) (the equivalent of the spring constant for the pendulum).
A pendulum (mass \(m\), unknown length) moves according to \(x=A\) sin $\omega t .\( (a) Write the equation for \)v_{x}(t)$ and sketch one cycle of the \(v_{x}(t)\) graph. (b) What is the maximum kinetic energy?
What is the period of a pendulum consisting of a \(6.0-\mathrm{kg}\) mass oscillating on a 4.0 -m-long string?
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