Question

(II) If a compressive force of \({\bf{3}}{\bf{.3 \times 1}}{{\bf{0}}^{\bf{4}}}\;{\bf{N}}\) is exerted on the end of a 22 cm long bone of a cross-sectional area of \({\bf{3}}{\bf{.6}}\;{\bf{c}}{{\bf{m}}^{\bf{2}}}\), (a) will the bone break, and (b) if not, by how much does it shorten?

Short answer

(a) The bone will not break.

(b)The bone will shorten by a length of \(1.3 \times {10^{ - 3}}\;{\rm{m}}\).

Step-by-step solution

Concepts

For this problem, you should use the relation between stress, force, and area, which is\({\bf{stress = }}\frac{F}{A}\).

Explanation

The length of the bone is \({l_ \circ } = 22\;{\rm{cm}} = 0.22\;{\rm{m}}\).

The cross-sectional area of the bone is \(A = 3.6\;{\rm{c}}{{\rm{m}}^{\rm{2}}} = 3.6 \times {10^{ - 4}}\;{{\rm{m}}^{\rm{2}}}\).

The compressive force is \(F = 3.3 \times {10^4}\;{\rm{N}}\).

Calculation for part (a)

You know that the compressive strength of the bone is \(170 \times {10^6}\;{\rm{N/}}{{\rm{m}}^2}\).

Now, the stress on the bone due to the applied force is

\(\begin{array}{c}{\rm{stress}} = \frac{F}{A}\\ = \frac{{3.3 \times {{10}^4}\;{\rm{N}}}}{{3.6 \times {{10}^{ - 4}}\;{{\rm{m}}^{\rm{2}}}}}\\ = 91.7 \times {10^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\end{array}\).

Therefore, the stress on the bone is less than the compressive strength of the bone.

Hence, the bone will not break.

Calculation for part (b)

As the bone does not break, assume that it will shorten by \(\Delta l\) distance.

You know that the Young’s modulus of the bone is \(E = 15 \times {10^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

Then, you get

\(\begin{array}{c}E = \frac{{\frac{F}{A}}}{{\frac{{\Delta l}}{{{l_ \circ }}}}}\\\frac{{\Delta l}}{{{l_ \circ }}} = \frac{F}{{EA}}\\\Delta l = \frac{{F{l_ \circ }}}{{EA}}\end{array}\).

Now, substituting the values in the above equation,

\(\begin{array}{c}\Delta l = \frac{{\left( {3.3 \times {{10}^4}\;{\rm{N}}} \right) \times \left( {0.22\;{\rm{m}}} \right)}}{{\left( {15 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right) \times \left( {3.6 \times {{10}^{ - 4}}\;{{\rm{m}}^{\rm{2}}}} \right)}}\\ = 1.3 \times {10^{ - 3}}\;{\rm{m}}\end{array}\).

Hence, the bone will shorten by a length of \(1.3 \times {10^{ - 3}}\;{\rm{m}}\).