The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme activities, such as sprinting, it can be subjected to forces as high as 15 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is 77.3 mm2 , its average length is 30 cm , and its average Young's modulus is 1474 MPa .
How much tensile stress is required to stretch this muscle by 5.00% of its length?

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CarliReifsteck CarliReifsteck · Answer 1 · 2019-09-05T09:53:16+02:00

Answer:

The tensile stress is 73.7 MPa.

Explanation:

Given that,

Area = 77.3 mm²

Length = 30 cm

Young's modulus = 1474 MPa

We need to calculate the stretched length

Muscle is stretched by 5.0 % of its length

Stretched length x= 5.0% of L

[tex]x=0.05 L[/tex]

We need to calculate the tensile stress

Using Hooke's law

[tex]Y=\dfrac{stress}{strain}[/tex]

Where, Y = Young's modulus

[tex]stress = Y\times strain[/tex]

[tex]stress =Y\times\dfrac{x}{L}[/tex]

Put the value into the formula

[tex]stress = 1474\times10^{6}\times\dfrac{0.05 L}{L}[/tex]

[tex]stress =73700000=7.37\times10^{7}\ Pa[/tex]

[tex]stress=73.7\ MPa[/tex]

Hence, The tensile stress is 73.7 MPa.