In which of the following cases the friction force between `'A'` and `'B'` is maximum. In all cases `mu_(1)=0.5, mu_(2)=0`.

Options

A.
B.
C.
D.

Explanation

To determine in which case the friction force between blocks A and B is maximum, we can analyze each case step by step using the given coefficients of friction and the forces acting on the blocks.

### Given:
- Coefficient of friction between A and B, \( \mu_1 = 0.5 \)
- Coefficient of friction between B and the surface, \( \mu_2 = 0 \)

### Case Analysis:

**Case 1:**
- Block A (2 kg) is on top of Block B (3 kg).
- A force of 10 N is applied to Block B.
- The total mass \( M = 2 + 3 = 5 \) kg.
- Acceleration \( a = \frac{F}{M} = \frac{10 \, \text{N}}{5 \, \text{kg}} = 2 \, \text{m/s}^2 \).
- The normal force \( N \) acting on block A due to its weight is \( N = mg = 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N} \).
- The friction force \( F_{AB} = \mu_1 N = 0.5 \times 20 \, \text{N} = 10 \, \text{N} \).

**Case 2:**
- Block A (2 kg) is on Block B (3 kg), which is inclined at 37 degrees.
- The normal force \( N = mg \cos(37^\circ) = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times \frac{4}{5} = 8 \, \text{N} \).
- The friction force \( F_{AB} = \mu_1 N = 0.5 \times 8 \, \text{N} = 4 \, \text{N} \).

**Case 3:**
- Block A (2 kg) is on Block B (3 kg), and B is fixed.
- The normal force \( N = mg = 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N} \).
- The friction force \( F_{AB} = \mu_1 N = 0.5 \times 20 \, \text{N} = 10 \, \text{N} \).

**Case 4:**
- There are three blocks (A, B, C) with block A on top of B, and B is on C.
- The total mass \( M = 2 + 3 + 1 = 6 \) kg.
- The acceleration \( a = \frac{10 \, \text{N}}{6 \, \text{kg}} = \frac{5}{3} \, \text{m/s}^2 \).
- The normal force \( N = mg = 2 \, \text{kg} \times 10 \, \text{m/s}^2 = 20 \, \text{N} \).
- The friction force \( F_{AB} = \mu_1 N = 0.5 \times 20 \, \text{N} = 10 \, \text{N} \).

### Summary of Friction Forces:
- Case 1: \( F_{AB} = 10 \, \text{N} \)
- Case 2: \( F_{AB} = 4 \, \text{N} \)
- Case 3: \( F_{AB} = 10 \, \text{N} \)
- Case 4: \( F_{AB} = 10 \, \text{N} \)

### Conclusion:
The maximum friction force between blocks A and B occurs in Case 1, Case 3, and Case 4, where the friction force is \( 10 \, \text{N} \).