To find the additional safe load $P$ that can be carried at point $D$, we need to consider the mechanics of the system, focusing on the forces in each wire and the factor of safety. Here is the step-by-step solution:

1. Analyze the Given Data:

• Weight of the bar $W = 2000$ lb.
• Each wire has a cross-sectional area $A = 0.125$ sq in.
• Yield point of steel $\sigma_y = 36000$ psi.
• Factor of safety $n = 2$.
• Total length of the bar $AB = 8$ ft, with point $C$ in the middle.

2. Determine the Maximum Allowable Stress:

The allowable stress $\sigma_{allow}$ considering the factor of safety is given by:

$\sigma_{allow} = \frac{\sigma_y}{n} = \frac{36000}{2} = 18000 \text{ psi}$

3. Calculate Maximum Force in Each Wire:

The maximum force $T_{max}$ each wire can withstand before yielding is:

$T_{max} = \sigma_{allow} \times A = 18000 \times 0.125 = 2250 \text{ lb}$

4. Equilibrium Analysis:

The system is in equilibrium under the given conditions. Considering the geometry and symmetry of the problem:

• For the bar to be in equilibrium, the forces in wires at $A$ and $B$ should be equal because of the symmetry and horizontal alignment.
• Let the tension in wire $C$ be $T_C$, and in wires $A$ and $B$ be $T_A$.

The total vertical force must balance the total weight $W$ and the additional load $P$. Therefore:

$T_A + T_B + T_C = W + P$

Because $T_A = T_B$ (due to symmetry), we have:

$2T_A + T_C = W + P$

5. Moment Calculation:

Taking moments about point $A$, the sum of moments should be zero:

$W \times 4' + P \times 6' = T_C \times 8'$

$2000 \times 4 + P \times 6 = T_C \times 8$

$8000 + 6P = 8T_C$

$T_C = 1000 + 0.75P$

6. Force Balances for Safe Load:

Substitute $T_C = 1000 + 0.75P$ into the vertical force balance equation:

$2T_A + (1000 + 0.75P) = 2000 + P$

$2T_A = 1000 + 0.25P$

$T_A = 500 + 0.125P$

7. Maximum Allowable Load $P$:

Each wire's tension should not exceed $T_{max} = 2250 \text{ lb}$.

For wire $A$:

$500 + 0.125P \leq 2250$

$0.125P \leq 1750$

$P \leq 14000$

For wire $C$:

$1000 + 0.75P \leq 2250$

$0.75P \leq 1250$

$P \leq 1667$

8. Final Safe Load $P$:

The lower value governs, so the maximum additional safe load $P$ is:

$P = 1667 \text{ lb}$

However, there seems to be a discrepancy between this calculation and the provided answer of $P = 3833 \text{ lb}$. This might be due to additional context or considerations not fully covered in this setup.

Would you like more detailed calculations or have any specific questions? Here are some related questions to explore:

1. What would happen if the factor of safety is increased to 3?
2. How would the tensions change if the load $P$ is applied at point $C$ instead of $D$?
3. What is the impact of changing the yield strength of the steel wires on the safe load $P$?
4. How would the system change if an additional wire is added at point $D$?
5. What would be the impact if the cross-sectional area of the wires was doubled?

Tip: Always check both force and moment equilibrium when solving such structural problems.