Question

The largest tendon in the body, the Achilles tendon, connects the calf muscle to the heel bone of the foot. This tendon is typically 14.0 cm long, 5.90 mm in diameter, and has a Young's modulus of 1.20×109 Pa . If an athlete has stretched the tendon to a length of 14.9 cm , what is the tension T , in newtons, in the tendon?

Answer 1

Answer:

$T=2109N$

Explanation:

The Young's modulus is defined as the quotient of the stress and the strain in a material. So

$E=\frac{\sigma}{\varepsilon}$

And, how does Young's modulus is related to the tension (in this particular case in the tendon)? We can use the definition of stress and strain:

$\sigma=\frac{F}{S}\\\\\varepsilon=\frac{\Delta L}{L}$,

Where F is the applied force to the material, S is the cross-sectional area of the material, $\Delta L$ is the deformation and L is the original length.

In this case $F=T$, so we can rewrite the Young's modulus expression as follows:

$E=\frac{\frac{T}{S}}{\frac{\Delta L}{L}}$,

now solving for T we get

$T=E(\frac{\Delta L}{L})(S)$.

Before computing the tension we need to compute the cross-sectional area and the deformation.

The cross-sectional area is the cross section of a cylinder (the tendon's form):

$S=\pi(D/2)^{2}$,

where D is de diameter ($D=5.90*10^{-3}m$).

$S=\pi(5.90*10^{-3}/2)^{2}$,

$S=2.73*10^{-5}(m^{2})$.

The Deformation is simply the change in length, so

$\Delta L=14.9cm-14cm=0.9cm=0.009m$.

Now we can easily compute the tension:

$T=(1.20*10^{9})(\frac{0.009}{0.14})(2.73*10^{-5})$,

$T=2109N$.

Answer 2

Answer:

F = 2.1 10³ N

Explanation:

Young's modulus is defined as the tensile strength (pressure) between the unit strain, whose equation is as follows:

    Y = P / (ΔL / L)

Body pressure is defined as the relationship between force and area over which it is applied.

    P = F / A

Let's replace and clear the force

    Y = (F / A) / (DL / L

    Y  ΔL / L = F / A

    F = Y A ΔL / L

Let's calculate the area .   The tendon is shaped like a cylinder, so the area is the circle

    A = π R²

    R = d / 2 = 5.90 10⁻³ / 2

    R = 2.95 10⁻³ m

   

    A = π (2.95 10-3)²

    A = 2,734 10⁻⁵ m²

Let's calculate deformation

   ΔL = 0.149 -0.14

   ΔL = 0.009 m = 9 10⁻³ m

Let's replace and calculate

    F = 1.20 10⁹ 2,734 10⁻⁵ 9 10⁻³ / 0.14

    F = 2.1 10³ N