Question

Answer 1

Calculate the cross-sectional area of cable wire using the relation

\(A_{C}=\frac{\pi}{4} d^{2}\)

Here, \(A_{C}\) is a cross-sectional area of cable and \(d\) diameter of the cable.

Substitute 0.25 in in \(d\)

\(A_{C}=\frac{\pi}{4} \times 0.25^{2}\)

\(A_{C}=0.049 \mathrm{in}^{2}\)

Calculate the tensile load on cable wire using the relation

\(P_{C}=\sigma_{C} \times A_{C}\)

Here, \(P_{C}\) is a tensile load on cable wire, \(\sigma_{C}\) is the working stress in cable wire, and \(A_{C}\) is cross-sectional area cable wire.

Substitute \(0.049 \mathrm{in}^{2}\) in \(A_{W}\) and \(60000 \mathrm{psi}\)

\(P_{C}=60000 \times 0.049\)

\(P_{C}=2945.24 \mathrm{lb}\)

Draw the free-body diagram

Calculate the load acting on the wood pole using force equilibrium equation along the \(\mathrm{y}\) -axis

\(P_{W}=2 P_{C} \sin 60^{\circ}\)

Here, \(P_{C}\) is a tensile load on cable wire, and \(P_{W}\) is a compressive load on a wood pole.

Substitute \(2945.243 \mathrm{lb}\) in \(P_{C}\)

\(P_{W}=2 \times 2945.24 \sin 60^{\circ}\)

\(P_{W}=5101.3 \mathrm{lb}\)

Determine the smallest permissible diameter of the pole using the relation

\(P_{W}=\sigma_{W} \times A_{W}\)

Here, \(P_{W}\) is a compressive load on a wood pole, \(A_{W}\) is the cross-sectional area of the wood pole, and \(\sigma_{w}\) is working compressive stress for wood.

Substitute \(5103.3 \mathrm{lb}\) in \(P_{W}, 200\) psi in \(\sigma_{W}\) and \(\frac{\pi}{4} d_{w}^{2}\)

\(5103.3=200 \times \frac{\pi}{4} d_{w}^{2}\)

\(d_{w}^{2}=32.488\)

\(d_{w}=5.69\) in

Therefore, the smallest permissible diameter of the pole 5.69 in