(III) A steel cable is to support an elevator whose total (loaded) mass is not to exceed 3100 kg. If the maximum acceleration of the elevator is \({\bf{1}}{\bf{.8}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\), calculate the diameter of the cable required. Assume a safety factor of 8.0.

You know that stress is the ratio of force and cross-sectional area.

For this problem, first, find the tension in the cable. Then, using the formula\({\bf{tensile strength = safety factor \times stress}}\), you will get the diameter.

The maximum load is \(m = 3100\;{\rm{kg}}\).

The safety factor is 8.0.

The acceleration of the elevator is \(a = 1.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\).

Let T be the tension in the wire due to the elevation of the elevator.

Using Newton’s second law,

\(\begin{array}{c}T - mg = ma\\T = m\left( {g + a} \right)\end{array}\).

Then, the force on the cable is

\(\begin{array}{c}F = T\\ = m\left( {g + a} \right)\end{array}\).

You can write that

\(\begin{array}{c}{\rm{tensile strength}} = {\rm{safety factor}} \times {\rm{stress}}\\\frac{{{\rm{tensile strength}}}}{{{\rm{safety factor}}}} = \frac{F}{A}\\\pi \frac{{{d^2}}}{4} = F\left( {\frac{{{\rm{safety factor}}}}{{{\rm{tensile strength}}}}} \right)\\d = \sqrt {\frac{{4m\left( {g + a} \right)}}{\pi }\left( {\frac{{{\rm{safety factor}}}}{{{\rm{shear strength}}}}} \right)} \end{array}\).

Substituting the values in the above equation,

\(\begin{array}{c}d = \sqrt {\frac{{4 \times \left( {3100\;{\rm{kg}}} \right)\left( {9.80\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} + 1.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)}}{\pi }\left( {\frac{{{\rm{8}}{\rm{.0}}}}{{{\rm{500}} \times {\rm{1}}{{\rm{0}}^6}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}}}} \right)} \\ = 2.7 \times {10^{ - 2}}\;{\rm{m}}\\ = 2.7\;{\rm{cm}}\end{array}\)

Hence, the diameter of the cable should be 2.7 cm.