To solve this problem, we will address both the stress and the elongation conditions separately, and find the maximum allowable load $P$ that satisfies both conditions.

Given data:

• Diameter of cable $d = 5.8 \, \text{mm}$
• Young's Modulus $E = 179 \, \text{GPa} = 179 \times 10^9 \, \text{Pa}$
• Maximum stress $\sigma_{\text{max}} = 203 \, \text{MPa} = 203 \times 10^6 \, \text{Pa}$
• Maximum elongation $\Delta L_{\text{max}} = 6.6 \, \text{mm} = 6.6 \times 10^{-3} \, \text{m}$
• Length of cable $L = 5 \, \text{m}$

Step 1: Stress condition

The stress in the cable is given by the formula: $\sigma = \frac{P}{A}$ where:

• $\sigma$ is the stress,
• $P$ is the load,
• $A$ is the cross-sectional area of the cable.

The cross-sectional area of the cable is the area of a circle: $A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{5.8 \times 10^{-3}}{2} \right)^2$

The maximum stress is 203 MPa, so: $\sigma_{\text{max}} = \frac{P}{A}$ Thus, the maximum allowable load due to stress is: $P_{\text{stress}} = \sigma_{\text{max}} \times A$

Step 2: Elongation condition

The elongation $\Delta L$ of the cable is related to the load by Hooke's Law: $\Delta L = \frac{P L}{A E}$ Rearranging for $P$: $P_{\text{elongation}} = \frac{\Delta L_{\text{max}} A E}{L}$

Step 3: Calculation of $P_{\text{stress}}$ and $P_{\text{elongation}}$

1. Calculate the cross-sectional area $A$:

$A = \pi \left( \frac{5.8 \times 10^{-3}}{2} \right)^2 = 2.654 \times 10^{-5} \, \text{m}^2$

2. Maximum load due to stress:

$P_{\text{stress}} = 203 \times 10^6 \times 2.654 \times 10^{-5} = 5383.5 \, \text{N}$

3. Maximum load due to elongation:

$P_{\text{elongation}} = \frac{6.6 \times 10^{-3} \times 2.654 \times 10^{-5} \times 179 \times 10^9}{5} = 7854.6 \, \text{N}$

Step 4: Determine the maximum allowable load

The maximum load $P_{\text{max}}$ is the smaller of the two values, since both conditions must be satisfied: $P_{\text{max}} = \min(P_{\text{stress}}, P_{\text{elongation}}) = \min(5383.5, 7854.6) = 5383.5 \, \text{N}$

Thus, the maximum allowable load is:

$P_{\text{max}} = 5383.5 \, \text{N}$

Final Answer:

The maximum load allowed is 5383.5 N.

Would you like further clarification or details on any of the steps? Here are a few related questions for exploration:

1. How would the maximum load change if the diameter of the cable increased?
2. How does Young's Modulus affect the elongation of the cable?
3. What would happen if the maximum stress were increased beyond 203 MPa?
4. How does the length of the cable affect its elongation for a given load?
5. What is the relationship between force and stress in different materials?

Tip: When dealing with stress and elongation problems, always ensure the units are consistent across all calculations (e.g., convert diameters to meters, pressures to Pascals).