A triangular block of mass M with angle 30º, 60º, 90º rests with its 30º– 90º side on a horizontal smooth fixed table. A cubical block of mass m rests on the 60º – 30º sideofthe triangular block. What horizontal acceleration a must M have relative to the stationary table so that m remains stationary with respect to the triangular block [M = 9 kg, m = 1 kg]

Options

A. `2.8 m//s^(2)`
B. ` 5.6 m//s^(2)`
C. `8.4 m//s^(2)`
D. Zero

Answer

To solve the problem, we need to find the horizontal acceleration \( a \) that the triangular block of mass \( M \) must have so that the cubical block of mass \( m \) remains stationary with respect to the triangular block.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a triangular block with angles 30º, 60º, and 90º. The 30º-90º side is on a smooth horizontal table.
- A cubical block of mass \( m \) rests on the 60º-30º side of the triangular block.

2. **Identify Forces Acting on the Cubical Block**:
- The weight of the cubical block \( m \) acts downward, which is \( mg \) (where \( g \) is the acceleration due to gravity).
- There is a pseudo force acting on the cubical block due to the acceleration \( a \) of the triangular block. This pseudo force acts horizontally to the left with magnitude \( ma \).

3. **Setting Up the Free Body Diagram (FBD)**:
- The cubical block experiences a gravitational force \( mg \) acting downwards.
- The pseudo force \( ma \) acts horizontally to the left.
- The block is in equilibrium with respect to the triangular block, meaning the forces must balance.

4. **Resolving Forces**:
- The gravitational force can be resolved into components along the incline of the triangular block:
- The component of the weight acting parallel to the incline (down the slope) is \( mg \sin(30^\circ) \).
- The component of the pseudo force acting perpendicular to the incline is \( ma \cos(30^\circ) \).

5. **Balancing Forces**:
- For the cubical block to remain stationary with respect to the triangular block, the forces must balance:
\[ mg \sin(30^\circ) = ma \cos(30^\circ) \]
- Substituting \( \sin(30^\circ) = \frac{1}{2} \) and \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ mg \cdot \frac{1}{2} = ma \cdot \frac{\sqrt{3}}{2} \]

6. **Substituting Known Values**:
- Given \( m = 1 \, \text{kg} \) and \( g \approx 9.8 \, \text{m/s}^2 \):
\[ 1 \cdot 9.8 \cdot \frac{1}{2} = 1 \cdot a \cdot \frac{\sqrt{3}}{2} \]
- Simplifying gives:
\[ 4.9 = a \cdot \frac{\sqrt{3}}{2} \]

7. **Solving for \( a \)**:
- Rearranging the equation:
\[ a = \frac{4.9 \cdot 2}{\sqrt{3}} = \frac{9.8}{\sqrt{3}} \approx 5.66 \, \text{m/s}^2 \]

### Final Answer:
The required horizontal acceleration \( a \) that the triangular block must have is approximately \( 5.66 \, \text{m/s}^2 \).