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Chapter 10: Problem 7

A 0.50 -m-long guitar string, of cross-sectional area $1.0 \times 10^{-6} \mathrm{m}^{2},\( has Young's modulus \)Y=2.0 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}$ By how much must you stretch the string to obtain a tension of \(20 \mathrm{N} ?\)

Short Answer

Expert verified
Answer: The guitar string should be stretched by 5.0 micrometers to achieve the desired tension.

Step by step solution

01

Identify the relevant formula

We will use the definition of Young's modulus, which is given by: $$Y = \frac{F}{A} \frac{\Delta L}{L}$$ where \(Y\) is Young's modulus, \(F\) is the force (tension) applied, \(A\) is the cross-sectional area, \(\Delta L\) is the change in length, and \(L\) is the original length. We'll need to solve for \(\Delta L\).
02

Rearrange the formula for the desired variable

We want to find the change in length \(\Delta L\). First, we'll isolate the term \(\frac{\Delta L}{L}\): $$\frac{\Delta L}{L} = \frac{F}{Y \cdot A}$$ Now, we can solve for \(\Delta L\) by multiplying both sides by \(L\): $$\Delta L = L \cdot \frac{F}{Y \cdot A}$$
03

Insert given values into the formula

Now, we'll plug in the given values for the length \(L = 0.50\,\text{m}\), the force \(F = 20\,\text{N}\), Young's modulus \(Y = 2.0 \times 10^{9}\,\frac{\text{N}}{\text{m}^{2}}\), and the cross-sectional area \(A = 1.0 \times 10^{-6}\,\text{m}^{2}\): $$\Delta L = 0.50\,\text{m} \cdot \frac{20\,\text{N}}{(2.0 \times 10^{9}\,\frac{\text{N}}{\text{m}^{2}}) \cdot (1.0 \times 10^{-6}\,\text{m}^{2})}$$
04

Calculate the change in length

Perform the calculation: $$\Delta L = 0.50\,\text{m} \cdot \frac{20\,\text{N}}{(2.0 \times 10^{9}\,\text{N/m}^{2}) \cdot (1.0 \times 10^{-6}\,\text{m}^2)} = 5.0 \times 10^{-6} \,\text{m}$$
05

Interpret the result

The change in length required to obtain a tension of 20 N is \(5.0 \times 10^{-6}\, \text{m}\). This means the guitar string should be stretched by 5.0 micrometers to achieve the desired tension.

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Most popular questions from this chapter

The ratio of the tensile (or compressive) strength to the density of a material is a measure of how strong the material is "pound for pound." (a) Compare tendon (tensile strength \(80.0 \mathrm{MPa}\), density $1100 \mathrm{kg} / \mathrm{m}^{3}\( ) with steel (tensile strength \)\left.0.50 \mathrm{GPa}, \text { density } 7700 \mathrm{kg} / \mathrm{m}^{3}\right)$ which is stronger "pound for pound" under tension? (b) Compare bone (compressive strength \(160 \mathrm{MPa}\), density $1600 \mathrm{kg} / \mathrm{m}^{3}$ ) with concrete (compressive strength $\left.0.40 \mathrm{GPa}, \text { density } 2700 \mathrm{kg} / \mathrm{m}^{3}\right):$ which is stronger "pound for pound" under compression?
A steel beam is placed vertically in the basement of a building to keep the floor above from sagging. The load on the beam is $5.8 \times 10^{4} \mathrm{N},\( the length of the beam is \)2.5 \mathrm{m},$ and the cross- sectional area of the beam is \(7.5 \times 10^{-3} \mathrm{m}^{2}\) Find the vertical compression of the beam.
Abductin is an elastic protein found in scallops, with a Young's modulus of \(4.0 \times 10^{6} \mathrm{N} / \mathrm{m}^{2} .\) It is used as an inner hinge ligament, with a cross-sectional area of \(0.78 \mathrm{mm}^{2}\) and a relaxed length of \(1.0 \mathrm{mm} .\) When the muscles in the shell relax, the shell opens. This increases efficiency as the muscles do not need to exert any force to open the shell, only to close it. If the muscles must exert a force of $1.5 \mathrm{N}$ to keep the shell closed, by how much is the abductin ligament compressed?
A pendulum of length \(120 \mathrm{cm}\) swings with an amplitude of $2.0 \mathrm{cm} .\( Its mechanical energy is \)5.0 \mathrm{mJ} .$ What is the mechanical energy of the same pendulum when it swings with an amplitude of \(3.0 \mathrm{cm} ?\)
A marble column with a cross-sectional area of \(25 \mathrm{cm}^{2}\) supports a load of \(7.0 \times 10^{4} \mathrm{N} .\) The marble has a Young's modulus of \(6.0 \times 10^{10} \mathrm{Pa}\) and a compressive strength of $2.0 \times 10^{8} \mathrm{Pa} .$ (a) What is the stress in the column? (b) What is the strain in the column? (c) If the column is \(2.0 \mathrm{m}\) high, how much is its length changed by supporting the load? (d) What is the maximum weight the column can support?
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