1. **Problem Statement:** A rigid bar ABC is supported by two aluminum cables CE and BD, each with diameter $d = \frac{1}{2}$ inches and modulus of elasticity $E = 10000$ ksi. A downward force $P = 5$ kips is applied at point C. We need to find: (a) The extensions of cables CE and BD. (b) The stresses in cables CE and BD. 2. **Given Data and Conversions:** - Diameter $d = 0.5$ in - Force $P = 5$ kips - Modulus $E = 10000$ ksi - 1 ft = 0.3048 m (not directly needed here) - 1 ksi = 6.89476 MPa (for unit consistency if needed) 3. **Calculate Cross-Sectional Area of Cables:** $$ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.5}{2}\right)^2 = \pi (0.25)^2 = \pi \times 0.0625 = 0.19635 \text{ in}^2 $$ 4. **Determine Cable Lengths and Geometry:** - Vertical distances between cables and base are 5 ft each. - Angle between AB and horizontal is 40°. - Assume cables CE and BD are horizontal, so their lengths correspond to horizontal distances from points C and B to walls E and D respectively. 5. **Equilibrium and Force Distribution:** - The bar is rigid and supported by two cables CE and BD. - The force $P$ acts downward at C. - Let tensions in cables CE and BD be $T_{CE}$ and $T_{BD}$. 6. **Moment Equilibrium about point A:** - Taking moments about A to solve for tensions. - Let distances be: - $AB = L_1$ - $AC = L_2$ - Using geometry and given angles, calculate $L_1$ and $L_2$ or use given distances. 7. **Calculate Tensions:** - Sum of vertical forces: $T_{CE} + T_{BD} = P = 5$ kips - Sum of moments about A: $T_{CE} \times h_{CE} + T_{BD} \times h_{BD} = P \times h_P$ - Using vertical distances $h_{CE} = 10$ ft (two 5 ft segments), $h_{BD} = 5$ ft, and $h_P = 10$ ft (height of C from A). 8. **Solve for $T_{CE}$ and $T_{BD}$:** $$ \begin{cases} T_{CE} + T_{BD} = 5 \\ 10 T_{CE} + 5 T_{BD} = 5 \times 10 = 50 \end{cases} $$ Multiply first equation by 5: $$ 5 T_{CE} + 5 T_{BD} = 25 $$ Subtract from second: $$ (10 T_{CE} + 5 T_{BD}) - (5 T_{CE} + 5 T_{BD}) = 50 - 25 $$ $$ 5 T_{CE} = 25 \Rightarrow T_{CE} = 5 \text{ kips} $$ Then: $$ T_{BD} = 5 - 5 = 0 \text{ kips} $$ 9. **Calculate Extensions:** - Extension formula: $$ \delta = \frac{T L}{A E} $$ - Lengths $L_{CE}$ and $L_{BD}$ correspond to horizontal cable lengths; assume both cables are 5 ft = 60 in long. - Convert $E$ to psi: $E = 10000$ ksi = $10^7$ psi. Calculate extension of CE: $$ \delta_{CE} = \frac{5000 \text{ lb} \times 60 \text{ in}}{0.19635 \text{ in}^2 \times 10^7 \text{ psi}} = \frac{300000}{1.9635 \times 10^6} = 0.1528 \text{ in} $$ Calculate extension of BD: $$ \delta_{BD} = 0 \text{ in} \quad \text{(since } T_{BD} = 0) $$ 10. **Calculate Stresses:** - Stress formula: $$ \sigma = \frac{T}{A} $$ Stress in CE: $$ \sigma_{CE} = \frac{5000}{0.19635} = 25475 \text{ psi} = 25.475 \text{ ksi} $$ Stress in BD: $$ \sigma_{BD} = 0 \text{ psi} $$ **Final Answers:** - (a) Extensions: $\delta_{CE} = 0.153$ in, $\delta_{BD} = 0$ in - (b) Stresses: $\sigma_{CE} = 25.5$ ksi, $\sigma_{BD} = 0$ ksi