91Ó°ÊÓ

Given that a 75-kg person does toe raises in which he raises his entire body weight on the ball of one foot. The Achilles tendon pulls straight upward on the heel bone of his foot. This tendon is 25 cm long and has a cross-sectional area of 78 mm² and a Young’s Modulus of 1470 MPa.

Mass of the person, m = 75 kg

Initial length of the tendon, ${\mathrm{I}}_0=25\mathrm{cm}$

Area of cross-section, $A=78×{10}^{-6}{m}^2$

Young’s Modulus, $Y=78×{10}^{-6}\mathbf{Pa}$

Torque $\mathrm{Ï„}=FI$

Where Fis force exerted and l is moment arm.

The formula for young’s Modulus is $Y=\frac{T/A}{∆l/l_0}$

Where

T is force exerted

A is area of cross section

$I_0$is initial length

$∆l$is change in length

Weight of the person, $n=\left(75\mathbf{kg}\right)\left(9.8\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\right)$.

The diagram is as follows:

Let T be the force exerted by tendon on the heel.

Applying first condition for equilibrium

$\begin{array}{r}T\left(4.6\mathbf{cm}\right)−n\left(12.5\mathbf{cm}\right)=0\\ T=\left(\frac{12.5\mathbf{cm}}{4.6\mathbf{cm}}\right)\left(75\mathbf{kg}\right)\left(9.8\mathrm{m}/{\mathbf{s}}^2\right)\\ T=2000\mathrm{N}\end{array}$

It is 2.72 times his weight.

Hence, force exerted by tendon on the heel is 2000 N which is 2.72 times his weight.

The formula for young’s Modulus is $Y=\frac{T/A}{∆l/l_0}$

Where

T is force exerted

A is area of cross section

$I_0$is initial length

$∆l$is change in length

The formula implies

$∆l=\left(\frac{T}{YA}\right)l_0$

So

$\mathrm{Δ}I=\frac{2000\mathrm{N}}{\left(1470×{10}^6\mathbf{Pa}\right)\left(78×{10}^{−6}{\mathrm{m}}^2\right)}\left(25\mathrm{cm}\right)=4.4\mathrm{mm}$

Hence, tendon stretches by 4.4 mm.