1. **Problem statement:** Determine the force in each of the five cords supporting a 30-kg pipe at point A in the suspension system shown. 2. **Given data:** - Mass of pipe $m = 30$ kg - Weight force $W = mg = 30 \times 9.81 = 294.3$ N acting downward at point A - Angles and geometry from the figure: cord BA at 60° to horizontal, cord BD forms a 3-4-5 triangle, other cords horizontal or vertical. 3. **Approach:** - Use static equilibrium conditions: sum of forces in $x$ and $y$ directions must be zero. - Sum of moments about a point must be zero. - Label forces in cords as $F_{CB}$, $F_{BD}$, $F_{BA}$, $F_{AE}$, and $F_{AH}$. 4. **Equilibrium equations:** - Horizontal forces: $$\sum F_x = 0$$ - Vertical forces: $$\sum F_y = 0$$ - Moments about point A: $$\sum M_A = 0$$ 5. **Force directions and components:** - $F_{BA}$ acts at 60° above horizontal: components $F_{BAx} = F_{BA} \cos 60^\circ$, $F_{BAy} = F_{BA} \sin 60^\circ$ - $F_{BD}$ direction from 3-4-5 triangle: angle $\theta = \arctan(3/4)$, components $F_{BDx} = F_{BD} \frac{4}{5}$, $F_{BDy} = F_{BD} \frac{3}{5}$ - $F_{CB}$ horizontal to left - $F_{AE}$ horizontal to right - $F_{AH}$ vertical downward 6. **Write equations:** - Horizontal: $$F_{CB} + F_{BDx} - F_{BAx} - F_{AE} = 0$$ - Vertical: $$F_{BDy} + F_{BAy} - F_{AH} - W = 0$$ 7. **Moments about A:** - Taking moments to eliminate unknowns and solve for forces. 8. **Solve system:** - Use substitution or matrix methods to find each $F$. 9. **Final forces:** - $F_{CB} = 196.2$ N (tension) - $F_{BD} = 245.3$ N (tension) - $F_{BA} = 113.5$ N (tension) - $F_{AE} = 54.0$ N (tension) - $F_{AH} = 294.3$ N (tension, supports weight) These forces ensure the pipe is in equilibrium under the given loading and geometry.