The value of F at which no friction will act on block on inclined plane , is :

Options

A. `2mumg`
B. `2mumg+2mgtantheta`
C. `2mumg+mgtantheta`
D. `2mumg+mgsintheta`

Explanation

To find the value of the force \( F \) at which no friction will act on a block on an inclined plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block**:
- The weight of the block \( W = mg \) acts vertically downwards.
- The normal force \( N \) acts perpendicular to the inclined plane.
- The component of weight acting parallel to the incline is \( W_{\parallel} = mg \sin \theta \).
- The component of weight acting perpendicular to the incline is \( W_{\perpendicular} = mg \cos \theta \).

2. **Apply the Horizontal Force**:
- A horizontal force \( F \) is applied. This force will have components affecting both the normal force and the frictional force.
- The force \( F \) can be resolved into components along the incline and perpendicular to the incline.

3. **Determine the Normal Force**:
- The normal force \( N \) can be expressed as:
\[ N = mg \cos \theta - F \sin \theta \]

4. **Condition for No Friction**:
- For no friction to act on the block, the force due to gravity parallel to the incline must be equal to the maximum static friction force. The maximum static friction is given by:
\[ f_{\text{max}} = \mu N \]
- Setting the gravitational force parallel to the incline equal to the maximum friction gives:
\[ mg \sin \theta = \mu N \]

5. **Substituting Normal Force**:
- Substitute \( N \) into the friction equation:
\[ mg \sin \theta = \mu (mg \cos \theta - F \sin \theta) \]

6. **Rearranging the Equation**:
- Expanding and rearranging gives:
\[ mg \sin \theta = \mu mg \cos \theta - \mu F \sin \theta \]
- Rearranging further leads to:
\[ mg \sin \theta + \mu F \sin \theta = \mu mg \cos \theta \]

7. **Isolate \( F \)**:
- Factor out \( \sin \theta \):
\[ F = \frac{mg \cos \theta - mg \sin \theta}{\mu \sin \theta} \]
- Simplifying gives:
\[ F = \frac{mg (\cos \theta - \sin \theta)}{\mu \sin \theta} \]

8. **Final Expression**:
- The final expression for the force \( F \) at which no friction acts on the block is:
\[ F = \frac{mg (\cos \theta - \sin \theta)}{\mu \sin \theta} \]