Question

(II) A 15-cm-long tendon was found to stretch 3.7 mm by a force of 13.4 N. The tendon was approximately round with an average diameter of 8.5 mm. Calculate Young’s modulus of this tendon.

Short answer

The Young’s modulus of the tendon is \(9.6 \times {10^6}\;{\rm{N/}}{{\rm{m}}^2}\).

Step-by-step solution

Young’s modulus

When a force (F) is applied to an object made of a certain material, the length of the object gets changed. This change in length is proportional to the original length (\({l_0}\)) and inversely proportional to the cross-sectional area (A), i.e., \(\Delta l = \frac{1}{E}\frac{F}{A}{l_0}\)

Here,E is the constant of proportionality, termed the elastic modulus or Young’s modulus.

In this problem, Young’s modulus of the tendon is related to the change in length of the tendon, applied force, and cross-sectional area of the tendon by the above expression.

Given information

The original length of the tendon is\(l = 15\;{\rm{cm}} = 15 \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}\).

Force applied on the tendon is F = 13.4 N.

Change in length of the tendon on the application of force is\(\Delta l = 3.7\;{\rm{mm}} = 3.{\rm{7}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{m}}\).

The average diameter of approximately round cross-section of the tendon is\(d = 8.5\;{\rm{mm}} = 8.5 \times {10^{ - 3}}\;{\rm{m}}\).

Let Ebe Young’s modulus of the tendon.

Determination of area of cross-section of the tendon

The area of the cross-section of the tendon is as follows:

\(\begin{array}{c}A = \pi {\left( {\frac{d}{2}} \right)^2}\\ = 3.14 \times {\left( {\frac{{8.5 \times 1{{\rm{0}}^{ - 3}}\;{\rm{m}}}}{2}} \right)^2}\\ = 5.67 \times 1{{\rm{0}}^{ - 5}}\;{{\rm{m}}^2}\end{array}\)

Calculation of Young’s modulus of the tendon

The change in length of the tendon due to force applied on it is given as follows:

\(\Delta l = \frac{1}{E}\frac{F}{A}{l_0}\)

Thus, the expression of Young’s modulus of the tendon can be written as follows:

\(\begin{array}{c}E = \frac{F}{A} \times \frac{{{l_0}}}{{\Delta l}}\\ = \frac{{\left( {13.4\;{\rm{N}}} \right)}}{{\left( {5.67 \times 1{{\rm{0}}^{ - 5}}\;{{\rm{m}}^2}} \right)}} \times \frac{{\left( {{\rm{15}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}} \right)}}{{\left( {3.7 \times {\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{m}}} \right)}}\\ = 9.581 \times {10^6}\;{\rm{N/}}{{\rm{m}}^2}\\ = 9.6 \times {10^6}\;{\rm{N/}}{{\rm{m}}^2}\end{array}\)

Thus, the value of Young’s modulus of the tendon is \(9.6 \times {10^6}\;{\rm{N/}}{{\rm{m}}^2}\).