A cubical block rests on a plane of `mu= sqrt(3)` . The angle through which the plane be inclined to the horizontal so that the block just slides down will be -

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

To solve the problem of determining the angle at which a cubical block will just start to slide down an inclined plane with a coefficient of friction (μ) equal to √3, we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
When the block is placed on the inclined plane, two main forces act on it:
- The gravitational force (weight) acting downwards, which can be represented as \( mg \).
- The normal force acting perpendicular to the surface of the inclined plane.

### Step 2: Resolve the Gravitational Force
The gravitational force can be resolved into two components:
- A component parallel to the incline: \( mg \sin \theta \)
- A component perpendicular to the incline: \( mg \cos \theta \)

### Step 3: Write the Expression for Normal Force
The normal force (N) acting on the block is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

### Step 4: Determine the Maximum Static Friction
The maximum static friction force (F_friction) that can act on the block before it starts sliding is given by:
\[ F_{friction} = \mu N = \mu (mg \cos \theta) \]

### Step 5: Set Up the Condition for Sliding
The block will just start to slide when the component of the gravitational force parallel to the incline equals the maximum static friction force:
\[ mg \sin \theta = \mu (mg \cos \theta) \]

### Step 6: Simplify the Equation
We can cancel \( mg \) from both sides (assuming \( mg \neq 0 \)):
\[ \sin \theta = \mu \cos \theta \]

### Step 7: Divide Both Sides by \( \cos \theta \)
This gives us:
\[ \tan \theta = \mu \]

### Step 8: Substitute the Value of μ
Given that \( \mu = \sqrt{3} \):
\[ \tan \theta = \sqrt{3} \]

### Step 9: Find the Angle θ
To find θ, we take the arctangent (inverse tangent) of both sides:
\[ \theta = \tan^{-1}(\sqrt{3}) \]

### Step 10: Calculate θ
From trigonometric values, we know:
\[ \tan 60^\circ = \sqrt{3} \]
Thus,
\[ \theta = 60^\circ \]

### Final Answer
The angle through which the plane can be inclined to the horizontal so that the block just slides down is \( 60^\circ \).

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