To determine in which case the friction force between A and B is maximum, we will analyze each case step by step.
### Given:
- Coefficient of friction between A and B, \( \mu_1 = 0.5 \)
- Coefficient of friction between B and the surface, \( \mu_2 = 0 \)
- Mass of A = 2 kg
- Mass of B = 3 kg
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)
### Case Analysis:
**Case A:**
1. **Calculate Normal Force (N)**:
\[
N = \text{mass of A} \times g = 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N}
\]
2. **Calculate Maximum Frictional Force (F_max)**:
\[
F_{\text{max}} = \mu_1 \times N = 0.5 \times 20 \, \text{N} = 10 \, \text{N}
\]
3. **Determine Acceleration (a)**:
\[
a = \frac{F_{\text{net}}}{\text{total mass}} = \frac{10 \, \text{N}}{5 \, \text{kg}} = 2 \, \text{m/s}^2
\]
4. **Apply Newton's Second Law on B**:
\[
10 \, \text{N} - F = 3 \, \text{kg} \times 2 \, \text{m/s}^2
\]
\[
10 - F = 6 \implies F = 4 \, \text{N}
\]
**Result for Case A**: Frictional force = 4 N
---
**Case B:**
1. **Calculate Normal Force (N)**:
\[
N = \text{mass of A} \times g \times \cos(37^\circ) = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times \frac{4}{5} = 16 \, \text{N}
\]
2. **Calculate Maximum Frictional Force (F_max)**:
\[
F_{\text{max}} = \mu_1 \times N = 0.5 \times 16 \, \text{N} = 8 \, \text{N}
\]
3. **Determine Force due to Gravity**:
\[
F_{\text{gravity}} = \text{mass of A} \times g \times \sin(37^\circ) = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times \frac{3}{5} = 12 \, \text{N}
\]
4. **Since 12 N > 8 N, the frictional force will be limited to F_max**:
\[
F = 8 \, \text{N}
\]
**Result for Case B**: Frictional force = 8 N
---
**Case C:**
1. **Calculate Normal Force (N)**:
\[
N = 10 \, \text{N}
\]
2. **Calculate Maximum Frictional Force (F_max)**:
\[
F_{\text{max}} = \mu_1 \times N = 0.5 \times 10 \, \text{N} = 5 \, \text{N}
\]
3. **Determine Force due to Gravity**:
\[
F = 20 \, \text{N} \text{ (downward)}
\]
4. **Since 20 N > 5 N, the frictional force will be limited to F_max**:
\[
F = 5 \, \text{N}
\]
**Result for Case C**: Frictional force = 5 N
---
**Case D:**
1. **Calculate Normal Force (N)**:
\[
N = 2 \, \text{kg} \times g = 20 \, \text{N}
\]
2. **Calculate Maximum Frictional Force (F_max)**:
\[
F_{\text{max}} = \mu_1 \times N = 0.5 \times 20 \, \text{N} = 10 \, \text{N}
\]
3. **Determine Acceleration (a)**:
\[
a = \frac{10 \, \text{N}}{6 \, \text{kg}} = \frac{5}{3} \, \text{m/s}^2
\]
4. **Apply Newton's Second Law on B**:
\[
T - F = 3 \, \text{kg} \times \frac{5}{3} \implies F = 5 \, \text{N}
\]
**Result for Case D**: Frictional force = \( \frac{10}{3} \, \text{N} \approx 3.33 \, \text{N} \)
---
### Conclusion:
The maximum frictional force occurs in **Case B**, where the frictional force is **8 N**.
