Problem

The power transmission cable weighs 10 lb/ft. If the resultant horizontal force on tower B...

The power transmission cable weighs 10 lb/ft. If the resultant horizontal force on tower BD is required to be zero, determine the sag h of cable BC.

Step-by-Step Solution

Solution 1

Calculate the horizontal force \(\left(F_{A B}\right)_{H}\) in cable \(\mathrm{AB}\),

$$ \left(F_{A B}\right)_{H}=\frac{w\left(L_{A B}\right)^{2}}{8 h_{A B}} $$

Here \(w\) is the weight per unit length of the power transmission cable, \(L_{A B}\) is the length of the cable \(A B\) and \(h_{A B}\) is the sag.

Substitute \(10 \mathrm{lb} / \mathrm{ft}\) for \(w, 300 \mathrm{ft}\) for \(L_{A B}\) and \(10 \mathrm{ft}\) for \(h_{A B}\)

$$ \begin{aligned} \left(F_{A B}\right)_{H} &=\frac{10 \times 300^{2}}{8 \times 10} \\ &=11250 \mathrm{lb} \end{aligned} $$

Calculate the horizontal force \(\left(F_{B C}\right)_{H}\) in cable \(A B\),

$$ \left(F_{B C}\right)_{H}=\frac{w\left(L_{B C}\right)^{2}}{8 h_{B C}} $$

Here \(w\) is the weight per unit length of the power transmission cable, \(L_{B C}\) is the length of the cable \(B C\) and \(h_{B C}\) is the sag.

Substitute \(10 \mathrm{lb} / \mathrm{ft}\) for \(w, 200 \mathrm{ft}\) for \(L_{B C}\).

$$ \left(F_{B C}\right)_{H}=\frac{10 \times 200^{2}}{8 \times h_{B C}} $$

Determine the sag \(\left(h_{B C}\right)\) in cable \(B C\) At equilibrium condition

$$ \begin{array}{l} \left(F_{A B}\right)_{H}=\left(F_{B C}\right)_{H} \\ \frac{10 \times 300^{2}}{8 \times 10}=\frac{10 \times 200^{2}}{8 \times h_{B C}} \\ h_{B C}=\frac{400000}{90000} \\ =4.44 \mathrm{ft} \end{array} $$

Therefore, the sag \(\left(h_{B C}\right)\) in cable \(\mathrm{BC}\) is \(4.44 \mathrm{ft}\).

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