A block of dimensions `axxaxx2a` is kept on an inclined plane of inclination `37^(@)`. The longer side is perpendicular to the plane. The co`-` efficient of friction between the block and the plane is `0.8.` By numerical analysis find whether the block will topple or not.

Explanation

To determine whether the block will topple or not, we can analyze the forces and torques acting on the block placed on the inclined plane. Here’s a step-by-step solution:

### Step 1: Understand the Setup
The block has dimensions \( a \times a \times 2a \) and is placed on an inclined plane at an angle of \( 37^\circ \). The longer side (length \( 2a \)) is perpendicular to the inclined plane.

### Step 2: Identify Forces Acting on the Block
1. **Weight (W)**: The weight of the block acts downward and is given by \( W = mg \), where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity.
2. **Normal Force (N)**: The normal force acts perpendicular to the inclined plane.
3. **Frictional Force (F)**: The frictional force acts parallel to the inclined plane and is given by \( F = \mu N \), where \( \mu = 0.8 \) is the coefficient of friction.

### Step 3: Calculate the Components of Weight
The weight of the block can be resolved into two components:
- **Perpendicular to the incline**: \( W_{\perp} = mg \cos(37^\circ) \)
- **Parallel to the incline**: \( W_{\parallel} = mg \sin(37^\circ) \)

Using trigonometric values:
- \( \sin(37^\circ) \approx 0.6 \)
- \( \cos(37^\circ) \approx 0.8 \)

Thus:
- \( W_{\perp} = mg \cdot 0.8 \)
- \( W_{\parallel} = mg \cdot 0.6 \)

### Step 4: Determine the Normal Force
The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = mg \cos(37^\circ) = mg \cdot 0.8 \]

### Step 5: Calculate the Frictional Force
The frictional force can be calculated as:
\[ F = \mu N = 0.8 \cdot (mg \cdot 0.8) = 0.64 mg \]

### Step 6: Analyze the Torque
To determine if the block will topple, we need to analyze the torques about the edge of the block that is in contact with the inclined plane.

1. **Torque due to \( W_{\parallel} \)** (which tries to topple the block):
- This force acts at a distance of \( a \) from the pivot (the edge of the block).
- Torque \( \tau_{\parallel} = W_{\parallel} \cdot a = (mg \cdot 0.6) \cdot a \)

2. **Torque due to \( W_{\perp} \)** (which stabilizes the block):
- This force acts at a distance of \( \frac{a}{2} \) from the pivot (the center of mass of the block).
- Torque \( \tau_{\perp} = W_{\perp} \cdot \frac{a}{2} = (mg \cdot 0.8) \cdot \frac{a}{2} \)

### Step 7: Set Up the Torque Equation
For the block to be in equilibrium (not toppling), the clockwise torque must equal the counterclockwise torque:
\[ mg \cdot 0.6 \cdot a = mg \cdot 0.8 \cdot \frac{a}{2} \]

### Step 8: Simplify the Equation
Cancelling \( mg \) and \( a \) from both sides:
\[ 0.6 = 0.4 \]
This shows that the block will not remain in equilibrium.

### Step 9: Conclusion
Since the torque due to the weight component parallel to the incline is greater than the torque due to the weight component perpendicular to the incline, the block will topple.

### Final Answer
**The block will topple.**

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