A particle is resting on a rough inclined plane with inclination `alpha` The angle of friction is `lamda`. The particle will be at rest if and only if

Options

A. `alphagtlamda`
B. `alphagelamda`
C. `alphalelamda`
D. `alphaltlamda`

Answer

To determine the conditions under which a particle remains at rest on a rough inclined plane, we need to analyze the forces acting on the particle and apply the concept of friction. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Particle
The forces acting on the particle on the inclined plane include:
- The gravitational force (weight) acting downwards, which can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \alpha \)
- Parallel to the incline: \( mg \sin \alpha \)
- The normal force \( N \) acting perpendicular to the inclined surface.
- The frictional force \( F_f \) acting parallel to the incline, opposing the motion.

### Step 2: Apply the Condition for Equilibrium
For the particle to remain at rest, the net force acting on it must be zero. This gives us two conditions:
1. The sum of forces perpendicular to the incline must be zero.
2. The sum of forces parallel to the incline must also be zero.

### Step 3: Analyze the Forces Perpendicular to the Incline
The normal force \( N \) balances the perpendicular component of the weight:
\[ N = mg \cos \alpha \]

### Step 4: Analyze the Forces Parallel to the Incline
The frictional force \( F_f \) can be expressed in terms of the normal force and the angle of friction \( \lambda \):
\[ F_f = \mu N = \mu (mg \cos \alpha) \]
Where \( \mu = \tan \lambda \) (the coefficient of friction).

For the particle to remain at rest, the frictional force must be equal to the component of the gravitational force acting down the incline:
\[ F_f \geq mg \sin \alpha \]

### Step 5: Set Up the Inequality
Substituting the expression for \( F_f \):
\[ \mu (mg \cos \alpha) \geq mg \sin \alpha \]

### Step 6: Simplify the Inequality
Dividing both sides by \( mg \) (assuming \( mg \neq 0 \)):
\[ \mu \cos \alpha \geq \sin \alpha \]

### Step 7: Substitute the Coefficient of Friction
Since \( \mu = \tan \lambda \), we can substitute this into the inequality:
\[ \tan \lambda \cos \alpha \geq \sin \alpha \]

### Step 8: Rearranging the Terms
Rearranging gives us:
\[ \cos \alpha \tan \lambda \geq \sin \alpha \]
This can be rewritten as:
\[ \sin \alpha \leq \tan \lambda \cos \alpha \]

### Step 9: Final Condition
Thus, the condition for the particle to remain at rest on the inclined plane is:
\[ \sin \alpha \leq \mu \cos \alpha \]
or equivalently,
\[ \tan \alpha \leq \tan \lambda \]

### Conclusion
The particle will be at rest if and only if:
\[ \alpha \leq \lambda \]