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$\bullet$ $\bullet$ Tendon-stretching exercises. As part of an exercise program, a 75 $\mathrm{kg}$ person does toe raises in which he raises his entire body weight on the ball of one foot, as shown in Figure $11.43 .$ The Achilles tendon pulls straight upward on the heel bone of his foot. This tendon is 25 $\mathrm{cm}$ long and has a cross-sectional area of 78 $\mathrm{mm}^{2}$ and a Young's modulus of 1470 MPa. (a) Make a free-body diagram of the person's foot (everything below the anklejoint). You can neglect the weight of the foot. (b) What force does the Achilles tendon exert on the heel during this exercise? Express your answer in newtons and in multiples of his weight. (c) By how many millimeters does the exercise stretch his Achilles tendon?

$\delta l=4.36 \mathrm{mm}$

Physics 101 Mechanics

Chapter 11

Elasticity and Periodic Motion

Equilibrium and Elasticity

Periodic Motion

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first question we could drop a free body diagram. Well, hope Boots. So we got that. There's upward force on a hue and a downward force on the toe. And since all body weight was acting on toast So the force on toys assume that you go to the gravity of the person. Okay. And there's ah, point, which is called an ankle joint between the heel and toe F brown. The toe to the ankle joints is tall, bony by centimeter. And find a hell to the angle joins this world 0.6 centimeter. Okay, question be, was asking us was the force on the heel? We're going to little law off the moment. We gotta have times the distance from the heel to the ankle joint C go to the mg times the distance from the toe to the ankle joints. Okay. And we do some calculation. Here we get force Siegel to MGI times 12 point five centimeter over 4.6 centimeter. What is 70? Your worms? I'm sorry to 70 or 75 75. My bit. 75 kilograms times. Now I'm going a meter per second square times trial going by sandy meter over 4.6 centimeter. Okay, this will give us the force is you go to 1997 pointing to a You're Tom's. Okay, The last question was asking us What is the, um, displacement you stretch on his ark you attended. Okay, so we're looking for to Vigata Delta L in this case and we are in on the force is equal to 1997.2 in your times. Okay. And you know the force constants equal to the force to arrive at displacement. It was in the ah, yes. Modules tends a car sessional area over the men off the architect tender, which will give us the changing displacements equal to force. Divided by yes, modules tends cross sectional area over the length off the ark. Intention, which is equal fl over. Why? OK, so delta l is equal to half, which is 1 1919 97 Going to a new town times the end of our key tension, which is 25 centimeter okay and 25 centimeters. Joe going to buy meter. Okay. Over. Why was it's Kiemas 14. 70. Ah m p A, which is 14 70 times tend to about six Pasco. Okay, times The cross sectional area, which is given as 78 millimetre square, was just sandy a past tens of power off negative six meter square. Okay, it was one millimeters. Where is equal to tens? About all the that to six meters. Where? And this will give us the downtown. L is evil Zoo joe 0.0 or three six. Meaner. You become worried. Two millimeter is 4.36 millimeter. Uh, so 4.36 millimeter. Okay. And that's the answer to his question. Thank you.

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