The problem involves determining the magnitudes of tensions $T_2$ and $T_3$ such that the resultant force of $10,000 \, \text{lb}$ acts vertically downward. Here’s how we can approach the solution step by step:

Step 1: Break forces into components

The system is in equilibrium, so the sum of the forces in both the horizontal ($x$) and vertical ($y$) directions must equal zero. We will decompose the tensions $T_1$, $T_2$, and $T_3$ into their respective horizontal and vertical components using the given slopes.

Tension $T_1$:

• Given slope: $4:3$, the ratio of vertical to horizontal components.
  • Vertical component: $\frac{4}{5} T_1 = \frac{4}{5} \times 5000 = 4000 \, \text{lb}$
  • Horizontal component: $\frac{3}{5} T_1 = \frac{3}{5} \times 5000 = 3000 \, \text{lb}$

Tension $T_2$:

• Given slope: $3:4$ (horizontal:vertical).
  • Vertical component: $\frac{3}{5} T_2$
  • Horizontal component: $\frac{4}{5} T_2$

Tension $T_3$:

• Given slope: $5:12$ (horizontal:vertical).
  • Vertical component: $\frac{12}{13} T_3$
  • Horizontal component: $\frac{5}{13} T_3$

Step 2: Apply equilibrium conditions

Since the system is in equilibrium, we have two main conditions:

1. Sum of horizontal components equals zero: $T_1 \text{(horizontal)} + T_2 \text{(horizontal)} = T_3 \text{(horizontal)}$ Substituting the components: 3000 + \frac{4}{5} T_2 = \frac{5}{13} T_3 \tag{1}

2. Sum of vertical components equals the downward resultant: $T_1 \text{(vertical)} + T_2 \text{(vertical)} + T_3 \text{(vertical)} = 10,000 \, \text{lb}$ Substituting the components: 4000 + \frac{3}{5} T_2 + \frac{12}{13} T_3 = 10,000 \tag{2}

Step 3: Solve the system of equations

Now we solve the system of equations (1) and (2) to find the values of $T_2$ and $T_3$.

Let me calculate these values.The solution to the system of equations is:

• $T_2 = -476.19 \, \text{lb}$ (This negative value indicates that the assumed direction for $T_2$ may be incorrect. It likely points in the opposite direction).
• $T_3 = 6809.52 \, \text{lb}$

Thus, the tension $T_3$ is approximately $6809.52 \, \text{lb}$, while the tension $T_2$, considering its negative value, should be re-evaluated for its direction, suggesting the cable could be in compression rather than tension.

Would you like a detailed breakdown of the interpretation of these results or have any further questions?

Here are 5 related questions to deepen understanding:

1. Why does a negative value for $T_2$ indicate a change in direction for the force?
2. How would the problem change if the resultant force was not vertical?
3. What would happen if the slopes of the cables were different?
4. How can we visualize the forces in terms of vector components?
5. How does the equilibrium condition affect structures in real-world engineering?

Tip: Always check the direction of forces when solving equilibrium problems, as the sign can provide important insights into the physical behavior of the system.