To solve the problem step by step, we will analyze the forces acting on the rods A and B and apply the principles of equilibrium and stress calculations.
### Step 1: Understand the System
We have two rods, A and B, supporting a rigid bar horizontally. The distance between the two rods is 60 cm. A load of 5000 kg is applied at a distance of 20 cm from rod A.
### Step 2: Calculate the Weight of the Load
The weight (W) of the load can be calculated using the formula:
\[ W = mg \]
Where:
- \( m = 5000 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
Calculating the weight:
\[ W = 5000 \times 10 = 50000 \, \text{N} \]
### Step 3: Determine the Forces in the Rods
Let \( F_A \) be the force in rod A and \( F_B \) be the force in rod B. According to the equilibrium of moments about point A:
\[ F_A \times 20 = F_B \times 40 \]
From this, we can express \( F_A \) in terms of \( F_B \):
\[ F_A = 2F_B \]
### Step 4: Relate the Forces to the Total Weight
The total weight supported by both rods is equal to the weight of the load:
\[ F_A + F_B = 50000 \, \text{N} \]
Substituting \( F_A = 2F_B \) into the equation:
\[ 2F_B + F_B = 50000 \]
\[ 3F_B = 50000 \]
\[ F_B = \frac{50000}{3} \approx 16666.67 \, \text{N} \]
Now substituting back to find \( F_A \):
\[ F_A = 2F_B = 2 \times 16666.67 \approx 33333.33 \, \text{N} \]
### Step 5: Calculate the Stress in Rod B
Given the stress in rod B is \( \sigma_B = 50 \, \text{N/mm}^2 \), we can find the area of rod B using the formula:
\[ \sigma_B = \frac{F_B}{A_B} \]
Rearranging gives:
\[ A_B = \frac{F_B}{\sigma_B} \]
Substituting the values:
\[ A_B = \frac{16666.67}{50} = 333.33 \, \text{mm}^2 \]
### Step 6: Calculate the Stress in Rod A
Using Young's modulus for rod B:
\[ Y_B = \frac{\sigma_B}{\epsilon_B} \]
Where \( \epsilon_B \) is the strain. Since both rods have the same free length and are under the same load conditions, we can relate the stresses and areas using Young's modulus:
\[ \frac{\sigma_A}{Y_A} = \frac{\sigma_B}{Y_B} \]
Rearranging gives:
\[ \sigma_A = \sigma_B \times \frac{Y_A}{Y_B} \]
Substituting the values:
- \( Y_A = 2 \times 10^5 \, \text{N/mm}^2 \)
- \( Y_B = 9 \times 10^4 \, \text{N/mm}^2 \)
Calculating \( \sigma_A \):
\[ \sigma_A = 50 \times \frac{2 \times 10^5}{9 \times 10^4} = 50 \times \frac{20}{9} \approx 111.11 \, \text{N/mm}^2 \]
### Step 7: Calculate the Area of Rod A
Using the formula for stress:
\[ \sigma_A = \frac{F_A}{A_A} \]
Rearranging gives:
\[ A_A = \frac{F_A}{\sigma_A} \]
Substituting the values:
\[ A_A = \frac{33333.33}{111.11} \approx 300 \, \text{mm}^2 \]
### Final Results
- Stress in rod A: \( \sigma_A \approx 111.11 \, \text{N/mm}^2 \)
- Area of rod A: \( A_A \approx 300 \, \text{mm}^2 \)
- Area of rod B: \( A_B \approx 333.33 \, \text{mm}^2 \)
