Question

The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alternately stretches, as you bring your weight down onto your forward foot, and contracts to push you off the ground. A 70 kg runner has an Achilles tendon that is 15 cm long and has a cross-section area of 110mm2, typical values for a person of this size. a. By how much will the runner's Achilles tendon stretch of the maximum force on it is 8.0 times his weight, a typical value while running? b. What fraction of the tendon's length does this correspond to?

Answer 1

(a)
`\triangle l=5 mm`

(b) 0.033

Step-by-step explanation:

(a)

Force F=mg where m is mass and g is acceleration due to gravity whose value is taken as
`9.81 m/s^(2)`

However, for this case, the maximum force is 8 times the weight of runner hence F=8mg

Assume Young's modulus for tendon is
`0.15*10^(10) N/m^(2)`

Young's modulus is given by

`E=\frac {Fl}{A\triangle l}` and
`\triangle l=\frac {Fl}{EA}` and substituting F with 8mg we obtain
`\triangle l=\frac {8mgl}{EA}`

Where E is young's modulus, l is stretched length and \triangle l is change in length

Substituting m as 70 kg, g as
`9.81 m/s^(2)`, l as 15cm=0.15 m, E as
`0.15*10^(10) N/m^(2)` and A as
`110 m^(2)=0.000110 m^(2)`

`\triangle l=\frac {8*70 Kg*9.81 m/s^(2)*0.15m}{0.15*10^(10) N/m^(2) *0.00011 m^(2)}=0.004994182 m`

`\triangle l=5 mm`

(b)

Strain,
`\epsilon=\frac {\triangle l}{l}` and the fraction of tendon’s length is the ratio of change in length to the stretched length

The fraction of tendon, f is given by

`f=\frac {\triangle l}{l}`. Substituting
`\triangle l` with 0.005m and l with 0.15m we obtain

`\epsilon=f=\frac {0.005}{0.15}=\frac {1}{30}=0.033`

Therefore, fraction of the tendon’s length is 0.033