Answer :
To solve this problem, we need to find the point along the rod where a mass [tex]\(M\)[/tex] should be suspended so that the strains in wires [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are equal. Let's break down the solution into detailed steps:
1. Given Data:
- Length of the rod ([tex]\(L\)[/tex]) = 0.8 meters
- Cross-sectional area of wire [tex]\(A\)[/tex] ([tex]\(A_A\)[/tex]) = 1 mm[tex]\(^2\)[/tex] = [tex]\(1 \times 10^{-6}\)[/tex] m[tex]\(^2\)[/tex]
- Cross-sectional area of wire [tex]\(B\)[/tex] ([tex]\(A_B\)[/tex]) = 1.5 mm[tex]\(^2\)[/tex] = [tex]\(1.5 \times 10^{-6}\)[/tex] m[tex]\(^2\)[/tex]
- Young's Modulus of wire [tex]\(A\)[/tex] ([tex]\(Y_A\)[/tex]) = [tex]\(2.4 \times 10^{11}\)[/tex] Pa
- Young's Modulus of wire [tex]\(B\)[/tex] ([tex]\(Y_B\)[/tex]) = [tex]\(1.6 \times 10^{11}\)[/tex] Pa
2. Establishing the Condition of Equal Strains:
- Strain in wire [tex]\(A\)[/tex] = Strain in wire [tex]\(B\)[/tex]
- Stress in wire [tex]\(A\)[/tex] / Young's Modulus of wire [tex]\(A\)[/tex] = Stress in wire [tex]\(B\)[/tex] / Young's Modulus of wire [tex]\(B\)[/tex]
3. Relating Stress to Forces and Areas:
- Stress is defined as Force per unit area.
- Stress in wire [tex]\(A\)[/tex] = [tex]\(F_A / A_A\)[/tex]
- Stress in wire [tex]\(B\)[/tex] = [tex]\(F_B / A_B\)[/tex]
4. Equal Strains Condition:
[tex]\[ \frac{F_A}{A_A \cdot Y_A} = \frac{F_B}{A_B \cdot Y_B} \][/tex]
5. Solving for the Forces:
- Let’s express the forces [tex]\(F_A\)[/tex] and [tex]\(F_B\)[/tex] in terms of their respective areas and Young's Modulus:
[tex]\[ F_A = A_A \cdot Y_A \][/tex]
[tex]\[ F_B = A_B \cdot Y_B \][/tex]
6. Calculating the Forces:
[tex]\[ F_A = (1 \times 10^{-6} \ \text{m}^2) \times (2.4 \times 10^{11} \ \text{Pa}) = 240000.0 \ \text{N} \][/tex]
[tex]\[ F_B = (1.5 \times 10^{-6} \ \text{m}^2) \times (1.6 \times 10^{11} \ \text{Pa}) = 240000.0 \ \text{N} \][/tex]
7. Establishing the Equilibrium of Moments:
- Let [tex]\(x\)[/tex] be the distance from wire [tex]\(A\)[/tex] to the point where the mass [tex]\(M\)[/tex] is suspended.
- The distance from wire [tex]\(B\)[/tex] to the point where the mass [tex]\(M\)[/tex] is suspended will be [tex]\(L - x\)[/tex].
- Considering the moments about wire [tex]\(A\)[/tex]:
[tex]\[ F_B \cdot (L - x) = F_A \cdot x \][/tex]
8. Solving for [tex]\(x\)[/tex]:
- Substitute [tex]\(F_A\)[/tex] and [tex]\(F_B\)[/tex] into the moment equation:
[tex]\[ 240000.0 \ \text{N} \cdot (0.8 \ \text{m} - x) = 240000.0 \ \text{N} \cdot x \][/tex]
[tex]\[ 192000.0 \ \text{N} \cdot \text{m} - 240000.0 \ \text{N} \cdot x = 240000.0 \ \text{N} \cdot x \][/tex]
[tex]\[ 192000.0 \ = 480000.0 \cdot x \][/tex]
[tex]\[ x = \frac{192000.0}{480000.0} = 0.4 \ \text{m} \][/tex]
Therefore, the mass [tex]\(M\)[/tex] should be suspended at a point [tex]\(0.4 \ \text{meters}\)[/tex] from wire [tex]\(A\)[/tex]. This ensures that the strains in both wires [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are equal.
1. Given Data:
- Length of the rod ([tex]\(L\)[/tex]) = 0.8 meters
- Cross-sectional area of wire [tex]\(A\)[/tex] ([tex]\(A_A\)[/tex]) = 1 mm[tex]\(^2\)[/tex] = [tex]\(1 \times 10^{-6}\)[/tex] m[tex]\(^2\)[/tex]
- Cross-sectional area of wire [tex]\(B\)[/tex] ([tex]\(A_B\)[/tex]) = 1.5 mm[tex]\(^2\)[/tex] = [tex]\(1.5 \times 10^{-6}\)[/tex] m[tex]\(^2\)[/tex]
- Young's Modulus of wire [tex]\(A\)[/tex] ([tex]\(Y_A\)[/tex]) = [tex]\(2.4 \times 10^{11}\)[/tex] Pa
- Young's Modulus of wire [tex]\(B\)[/tex] ([tex]\(Y_B\)[/tex]) = [tex]\(1.6 \times 10^{11}\)[/tex] Pa
2. Establishing the Condition of Equal Strains:
- Strain in wire [tex]\(A\)[/tex] = Strain in wire [tex]\(B\)[/tex]
- Stress in wire [tex]\(A\)[/tex] / Young's Modulus of wire [tex]\(A\)[/tex] = Stress in wire [tex]\(B\)[/tex] / Young's Modulus of wire [tex]\(B\)[/tex]
3. Relating Stress to Forces and Areas:
- Stress is defined as Force per unit area.
- Stress in wire [tex]\(A\)[/tex] = [tex]\(F_A / A_A\)[/tex]
- Stress in wire [tex]\(B\)[/tex] = [tex]\(F_B / A_B\)[/tex]
4. Equal Strains Condition:
[tex]\[ \frac{F_A}{A_A \cdot Y_A} = \frac{F_B}{A_B \cdot Y_B} \][/tex]
5. Solving for the Forces:
- Let’s express the forces [tex]\(F_A\)[/tex] and [tex]\(F_B\)[/tex] in terms of their respective areas and Young's Modulus:
[tex]\[ F_A = A_A \cdot Y_A \][/tex]
[tex]\[ F_B = A_B \cdot Y_B \][/tex]
6. Calculating the Forces:
[tex]\[ F_A = (1 \times 10^{-6} \ \text{m}^2) \times (2.4 \times 10^{11} \ \text{Pa}) = 240000.0 \ \text{N} \][/tex]
[tex]\[ F_B = (1.5 \times 10^{-6} \ \text{m}^2) \times (1.6 \times 10^{11} \ \text{Pa}) = 240000.0 \ \text{N} \][/tex]
7. Establishing the Equilibrium of Moments:
- Let [tex]\(x\)[/tex] be the distance from wire [tex]\(A\)[/tex] to the point where the mass [tex]\(M\)[/tex] is suspended.
- The distance from wire [tex]\(B\)[/tex] to the point where the mass [tex]\(M\)[/tex] is suspended will be [tex]\(L - x\)[/tex].
- Considering the moments about wire [tex]\(A\)[/tex]:
[tex]\[ F_B \cdot (L - x) = F_A \cdot x \][/tex]
8. Solving for [tex]\(x\)[/tex]:
- Substitute [tex]\(F_A\)[/tex] and [tex]\(F_B\)[/tex] into the moment equation:
[tex]\[ 240000.0 \ \text{N} \cdot (0.8 \ \text{m} - x) = 240000.0 \ \text{N} \cdot x \][/tex]
[tex]\[ 192000.0 \ \text{N} \cdot \text{m} - 240000.0 \ \text{N} \cdot x = 240000.0 \ \text{N} \cdot x \][/tex]
[tex]\[ 192000.0 \ = 480000.0 \cdot x \][/tex]
[tex]\[ x = \frac{192000.0}{480000.0} = 0.4 \ \text{m} \][/tex]
Therefore, the mass [tex]\(M\)[/tex] should be suspended at a point [tex]\(0.4 \ \text{meters}\)[/tex] from wire [tex]\(A\)[/tex]. This ensures that the strains in both wires [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are equal.
