(II) (a) What is the minimum cross-sectional area required for a vertical steel cable from which a 270-kg chandelier is suspended? Assume a safety factor of 7.0. (b) If the cable is 7.5 m long, how much does it elongate?

For this problem, you should use the relation between stress, tensile strength, and safety factor, which is\({\bf{tensile strength = safety factor \times stress}}\).

The mass of the cylinder is \(m = 270\;{\rm{kg}}\).

The safety factor is 7.0.

The length of the cable is \({l_ \circ } = 7.5\;{\rm{m}}\).

The tensile strength of steel is \({\rm{500}} \times {\rm{1}}{{\rm{0}}^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}\).

Let A be the area of the cable and \(\Delta l\) be the extended length of the cable.

Part (a)

The force on the cable is \(F = mg\).

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

Now, the stress on the cable due to the applied force is:

\(\begin{array}{c}{\rm{tensile strength}} = {\rm{safety factor}} \times {\rm{stress}}\\\frac{{{\rm{tensile strength}}}}{{{\rm{safety factor}}}} = \frac{F}{A}\\A = F\left( {\frac{{{\rm{safety factor}}}}{{{\rm{tensile strength}}}}} \right)\\A = mg\left( {\frac{{{\rm{safety factor}}}}{{{\rm{tensile strength}}}}} \right)\end{array}\).

Now, substituting the values in the above equation,

\(\begin{array}{c}A = \left( {270\;{\rm{kg}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {\frac{{{\rm{7}}{\rm{.0}}}}{{{\rm{500}} \times {\rm{1}}{{\rm{0}}^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}} \right)\\ = 3.7 \times {10^{ - 5}}\;{{\rm{m}}^{\rm{2}}}\end{array}\).

Hence, the cross-sectional area is \(3.7 \times {10^{ - 5}}\;{{\rm{m}}^{\rm{2}}}\).

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

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

Then,

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

Now, substituting the values in the above equation,

\(\begin{array}{c}\Delta l = \frac{{\left( {270\;{\rm{kg}}} \right) \times \left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \left( {7.5\;{\rm{m}}} \right)}}{{\left( {200 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right) \times \left( {3.7 \times {{10}^{ - 5}}\;{{\rm{m}}^{\rm{2}}}} \right)}}\\ = 2.7 \times {10^{ - 3}}\;{\rm{m}}\end{array}\).

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