Question

A steel rod of radius\(R = 15\;{\rm{cm}}\)and length\({l_0}\)stands upright on a firm surface. A 65-kg man climbs atop the rod. (a) Determine the percent decrease in the rod’s length. (b) When a metal is compressed, each atom moves closer to its neighboring atom by exactly the same fractional amount. If iron atoms in steel are normally\(2.0 \times {10^{ - 10}}\;{\rm{m}}\)apart, by what distance did this interatomic spacing have to change in order to produce the normal force required to support the man? [Note: Neighboring atoms repel each other, and this repulsion accounts for the observed normal force.

Short answer

1. The percentage decrease in the length of the rod is\(\left( {4.506 \times {{10}^{ - 6}}} \right)\% \).
2. The decrease in the interatomic spacing of the iron atoms in the steel rod is\(9.012 \times {10^{ - 18}}\;{\rm{m}}\).

Step-by-step solution

Identification of the given data

The cross-sectional radius of a steel rod is\(R = 15\;{\rm{cm}} = 0.15\;{\rm{m}}\).

The length of the rod is\({l_0}\).

The mass of the man is\(m = 65\;{\rm{kg}}\).

The inter-atomic spacing between the iron atoms in the steel is\({d_0} = 2 \times {10^{ - 10}}\;{\rm{m}}\).

The Young’s modulus of steel is\(E = 200 \times {10^9}\;{\rm{N/}}{{\rm{m}}^2}\).

Definition of Young’s modulus

Young’s modulus, E, is the measure of the stiffness of any material, i.e., how easy it is bent or stretched. It is also known as elastic modulus and is defined as the ratio of the stress to the corresponding strain.

\(E = \frac{{{\rm{Stress}}}}{{{\rm{Strain}}}} = \frac{{\frac{F}{A}}}{{\frac{{\Delta l}}{{{l_0}}}}}\) … (i)

Young’s modulus can be used to determine the elongation or compression of any material, as long as the stress is below the proportional limit (less than the yield strength of the material).

(a) Calculation of the percentage decrease in the length of the rod

The strain is defined as the ratio of the deformation over the initial length of any material. So, the fractional change in the length of the rod is the strain.

The force applied on the rod is the weight of the man, i.e., mg.

Using equation (1),

\(\begin{array}{c}\frac{{\Delta l}}{{{l_0}}} = \frac{F}{{AE}}\\ = \frac{{mg}}{{\left( {\pi {r^2}} \right)E}}\\ = \frac{{\left( {65\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{\frac{{22}}{7} \times {{\left( {0.15\;{\rm{m}}} \right)}^2}\left( {200 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^2}} \right)}}\\ = 4.506 \times {10^{ - 8}}\end{array}\)

Thus, the percentage change will be as follows:

\(\begin{array}{c}\frac{{\Delta l}}{{{l_0}}} \times 100 = 4.506 \times {10^{ - 8}} \times 100\\ = \left( {4.506 \times {{10}^{ - 6}}} \right)\% \end{array}\)

(b) Calculation of the decrease in the interatomic spacing of the iron atoms in the steel rod

The fractional change is the same for the steel rod as a whole as for the atoms. Thus, the fractional change in the interatomic spacing of the atoms will be as follows:

\(\begin{array}{c}\frac{{\Delta d}}{{{d_0}}} = \frac{{\Delta l}}{{{l_0}}}\\\frac{{\Delta d}}{{2 \times {{10}^{ - 10}}\;{\rm{m}}}} = 4.506 \times {10^{ - 8}}\\\Delta d = 4.506 \times {10^{ - 8}} \times \left( {2 \times {{10}^{ - 10}}\;{\rm{m}}} \right)\\ = 9.012 \times {10^{ - 18}}\;{\rm{m}}\end{array}\)