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$\bullet$ Achilles tendon. 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 13 times a person's weight. According to one set of experiments, the average area of the Achilles tendon is $78.1 \mathrm{mm}^{2},$ its average length is $25 \mathrm{cm},$ and its average Young's modulus is 1474 MPa. (a) How much tensile stress is required to stretch this muscle by 5.0$\%$ of its length? (b) If we model the tendon as a spring, what is its force constant? (c) If a 75 kg sprinter exerts a force of 13 times his weight on his Achilles tendon, by how much will it stretch?

a) $7.37 \times 10^{7} \mathrm{Pa}$b) $4.604776 \times 10^{5} \mathrm{N} / \mathrm{m}$c) 2.08 $\mathrm{cm}$

Physics 101 Mechanics

Chapter 11

Elasticity and Periodic Motion

Equilibrium and Elasticity

Periodic Motion

Cornell University

University of Washington

Hope College

McMaster University

Lectures

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$\bullet$$\bullet$ Muscles…

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in part maybe want to figure out what the stresses. So recall that the Youngs module us is to find a stress or strain. Yeah, sub stress over strange. And now we can determine that. Eh? 5% allegation which is given in the problem 5% in elongation implies that delta L over l non Izzy go 0.5 since delta l is the change in length and Ellen honesty original length So 5% elongation implies that the ratio is equal to this. And so this is just the stream, and we know Young's module is so we can find the stress by solving for it. So stress is equal to Young's module US time strain and this is equal to the strain are giving young small angeles So we can solve this to get 7.4 times 10 to the seventh past gowns. And that's the answer to party in part B. We want to figure out what the spring constant is if this could be modelled as a spring. And so now we recall that the stress is also written as FT over the cross sectional area. We can solve this for ft. since we just solve for the stress and we know eh? And we get the FT is equal to 5.76 times 10 to the third Nunes. This will be the force of a plugging into who hooks law and where to find the spring constant. We also need X, the change in length because we eventually want to plug this into this Got here to self. Okay, so we need X to do that ex In this case is the change in life from the context of hooks along. And we found in party that delta L is just equal to the original length times the strain which is 0.5 so we can plug in the original life. And when get 1.257 years or if we wanted meters, this is 1.25 times 10 to minus two meters is now We can take this value backs in this volume f and plugging into this to find the value of K. We do that we get that K is 4.6 times 10 to the fifth news premiere. And that's answered to part me and parsley. We want to find the stretch of the Achille still intending when it's under a force of 13 times, the athletes wait. And so in this case, our force is going to be equal to 13 times the weight, which is MG. We plug in the mass value they give us. This turns out to be 9,555 Nunes, and then we can take this value of F. Plug it into this equation here with our known value of K to figure out what exodus since Kei will not change from situation to situation. This kid's a value of X. It is 2.8 centimetres. So that's how much the Achilles tendon stretches under this force, if it could be modeled by a spring.

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