The minimum force required to move a body up an inclined plane of inclination 30° is found to be thrice the minimum force required to prevent if from sliding down the plane. The co-efficient of friction between the body and the plane is -

Options

A. `(1)/(sqrt(3))`
B. `(1)/(2sqrt(3))`
C. `(1)/(3sqrt(3))`
D. `(1)/(4sqrt(3))`

Answer

To solve the problem, we need to analyze the forces acting on a body on an inclined plane and relate them to the coefficient of friction. Let's break down the solution step by step:

### Step 1: Understand the Forces Acting on the Body
When a body is on an inclined plane, the weight of the body (mg) acts vertically downward. This weight can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

Where \( \theta \) is the angle of inclination (30° in this case).

### Step 2: Define the Forces for Moving Up the Plane
Let \( F \) be the minimum force required to move the body up the incline. The frictional force \( f \) acts down the incline when moving up, and it can be expressed as:
\[ f = \mu N \]
Where \( N \) is the normal force, which is equal to \( mg \cos \theta \).

Thus, the force balance when moving up can be expressed as:
\[ F = f + mg \sin \theta \]
Substituting for \( f \):
\[ F = \mu (mg \cos \theta) + mg \sin \theta \]

### Step 3: Define the Forces for Preventing Sliding Down
Let \( F' \) be the minimum force required to prevent the body from sliding down. In this case, the frictional force acts up the incline. The force balance can be expressed as:
\[ F' + f = mg \sin \theta \]
Substituting for \( f \):
\[ F' + \mu (mg \cos \theta) = mg \sin \theta \]
Rearranging gives:
\[ F' = mg \sin \theta - \mu (mg \cos \theta) \]

### Step 4: Relate the Two Forces
According to the problem, the force required to move the body up is three times the force required to prevent it from sliding down:
\[ F = 3 F' \]

### Step 5: Substitute and Rearrange
Substituting the expressions for \( F \) and \( F' \):
\[ \mu (mg \cos \theta) + mg \sin \theta = 3 \left( mg \sin \theta - \mu (mg \cos \theta) \right) \]

### Step 6: Simplify the Equation
Expanding and simplifying:
\[ \mu (mg \cos \theta) + mg \sin \theta = 3mg \sin \theta - 3\mu (mg \cos \theta) \]
Combining like terms:
\[ \mu (mg \cos \theta) + 3\mu (mg \cos \theta) = 3mg \sin \theta - mg \sin \theta \]
\[ 4\mu (mg \cos \theta) = 2mg \sin \theta \]

### Step 7: Cancel Common Terms
Dividing both sides by \( mg \) (assuming \( m \neq 0 \) and \( g \neq 0 \)):
\[ 4\mu \cos \theta = 2 \sin \theta \]

### Step 8: Solve for the Coefficient of Friction
Rearranging gives:
\[ \mu = \frac{2 \sin \theta}{4 \cos \theta} = \frac{\sin \theta}{2 \cos \theta} = \frac{1}{2} \tan \theta \]

### Step 9: Substitute the Angle
Substituting \( \theta = 30° \):
\[ \tan 30° = \frac{1}{\sqrt{3}} \]
Thus,
\[ \mu = \frac{1}{2} \cdot \frac{1}{\sqrt{3}} = \frac{1}{2\sqrt{3}} \]

### Final Answer
The coefficient of friction \( \mu \) between the body and the plane is:
\[ \mu = \frac{1}{2\sqrt{3}} \]