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 $$4.44 \mathrm{ft}$$.