Question

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 77.5 mm^2 , it's average length is 20 cm, and it's average Young's modulus is 1474 MPa.

Part A

How much tensile stress is required to stretch this muscle by 4.8 % of its length?

Part B

If we model the tendon as a spring what is the force constant?

Part C

If a 75 kg sprinter exerts a force of 15 times his weight on his Achilles' tendon, by how much will it stretch?

Answer 1

l= 20cm zo.20m (a) A = 7.705 mm²=7705x100 m² Y = 1474 MPa = 1474 4100 Pa 408% elongation means strain = se = 4.8 = 0.048 lo 100 Y = stress i Stress=YX strain strain Stress = (1474X106) (0.048). | stres f) = 7608810 Pa Answer (6) F TA = (4.08 X107)(77.5x10-5) F 5487 N a = (strain) (8) = (0.048)( 0.20) : 0.0096 m F=kX 5487 k = // = 2.0096 x= 5.7x105 N/m] Answer (C) F = 15 g = 15(15) 9 8) = 10 25 N F = kx i x = F = 11025 5.7X105 = 1x = 0.019 m 1.9 cm / Answer

Hope this helps..