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Chapter 10: Problem 6

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?

Short Answer

Expert verified
Answer: The abductin ligament is compressed by approximately 0.481 mm.

Step by step solution

01

Calculate the stress

To calculate the stress, we will use the formula \(\sigma = \frac{F}{A}\). We are given the force exerted by the muscles (\(F = 1.5 \, \mathrm{N}\)) and the cross-sectional area of the abductin ligament (\(A = 0.78 \, \mathrm{mm}^2 = 0.78 \times 10^{-6} \, \mathrm{m}^2\)). Now, we can find the stress: \(\sigma = \frac{1.5 \, \mathrm{N}}{0.78 \times 10^{-6} \, \mathrm{m}^2} = 1.923 \times 10^6 \, \mathrm{N / m}^2\)
02

Calculate the strain

Now that we have the stress, we can calculate the strain using the Young's modulus equation, which is \(\sigma = E \cdot \epsilon\): \(\epsilon = \frac{\sigma}{E} = \frac{1.923 \times 10^6 \, \mathrm{N / m}^2}{4.0 \times 10^6 \, \mathrm{N / m}^2} = 0.48075\)
03

Calculate the compression

Now that we have the strain, we can find the compression using the relationship between strain and compression, which is \(\epsilon = \frac{\Delta L}{L}\), where \(\Delta L\) is the compression and \(L\) is the relaxed length: \(\Delta L = \epsilon \cdot L = 0.48075 \cdot 1.0 \, \mathrm{mm} = 0.48075 \, \mathrm{mm}\) The abductin ligament is compressed by approximately \(0.481 \, \mathrm{mm}\).

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