An inclined plane is inclined at an angle `theta` with the horizontal. A body of mass m rests on it, if the coefficient of friction is `mu`, then the minimum force that has to be applied to the inclined plane to make the body just move up the inclined plane is

Options

A. ` mg sin theta`
B. `mu mg cos theta`
C. `mu mg cos theta-mg sintheta`
D. `mu mgcostheta+mg sintheta`

Answer

To solve the problem of finding the minimum force that must be applied to an inclined plane to make a body of mass \( m \) move up the plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body**:
- The weight of the body \( W = mg \) acts vertically downward.
- This weight can be resolved into two components:
- Perpendicular to the incline: \( W_{\perp} = mg \cos \theta \)
- Parallel to the incline: \( W_{\parallel} = mg \sin \theta \)

2. **Determine the Normal Force**:
- The normal force \( N \) acting on the body is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]

3. **Calculate the Frictional Force**:
- The frictional force \( f \) that opposes the motion of the body up the incline is given by:
\[ f = \mu N = \mu (mg \cos \theta) \]

4. **Set Up the Equation for Motion**:
- For the body to just start moving up the incline, the applied force \( F \) must overcome both the gravitational component pulling it down the incline and the frictional force. Therefore, we can write:
\[ F = W_{\parallel} + f \]
- Substituting the expressions for \( W_{\parallel} \) and \( f \):
\[ F = mg \sin \theta + \mu (mg \cos \theta) \]

5. **Simplify the Expression**:
- Factor out \( mg \):
\[ F = mg \left( \sin \theta + \mu \cos \theta \right) \]

### Final Result:
The minimum force \( F \) that must be applied to the inclined plane to make the body just move up the inclined plane is:
\[ F = mg \left( \sin \theta + \mu \cos \theta \right) \]