A body of mass m is placed on a rough surface with coefficient of friction `mu` inclined at `theta`. If the mass is in equilibrium , then

Options

A. `theta = tan^(-1) ( ( m)/(mu))`
B. `theta = tan^(-1) ( mu ) `
C. `theta = tan^(-) ((mu)/(m))`
D. `theta = tan^(-1) ( 1//mu )`

Answer

To solve the problem, we need to analyze the forces acting on a body of mass \( m \) placed on a rough inclined surface with an angle \( \theta \) and a coefficient of friction \( \mu \). Since the body is in equilibrium, we can set up the equations based on the forces acting on it.

### Step-by-Step Solution:

1. **Draw the Free Body Diagram (FBD)**:
- Start by sketching the inclined plane and the block of mass \( m \) on it.
- The angle of inclination is \( \theta \).
- The weight of the block \( mg \) acts vertically downward.

2. **Identify Forces**:
- The weight \( mg \) can be resolved into two components:
- Perpendicular to the inclined plane: \( mg \cos \theta \)
- Parallel to the inclined plane: \( mg \sin \theta \)
- The normal force \( N \) acts perpendicular to the surface.
- The frictional force \( f \) acts parallel to the surface, opposing the motion.

3. **Set Up Equilibrium Conditions**:
- Since the block is in equilibrium, the sum of forces in both the perpendicular and parallel directions must be zero.
- In the direction perpendicular to the incline:
\[ N = mg \cos \theta \quad \text{(1)} \]
- In the direction parallel to the incline:
\[ f = mg \sin \theta \quad \text{(2)} \]

4. **Relate Frictional Force to Normal Force**:
- The frictional force can also be expressed in terms of the normal force and the coefficient of friction:
\[ f = \mu N \quad \text{(3)} \]

5. **Substitute Normal Force into Friction Equation**:
- Substitute equation (1) into equation (3):
\[ f = \mu (mg \cos \theta) \]
- Now set this equal to the expression from equation (2):
\[ \mu (mg \cos \theta) = mg \sin \theta \]

6. **Cancel Common Terms**:
- We can cancel \( mg \) from both sides (assuming \( m \neq 0 \)):
\[ \mu \cos \theta = \sin \theta \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{\sin \theta}{\cos \theta} = \mu \]
- This can be rewritten as:
\[ \tan \theta = \mu \]

8. **Finding the Angle**:
- To find \( \theta \), take the inverse tangent:
\[ \theta = \tan^{-1}(\mu) \]

### Final Answer:
Thus, the angle \( \theta \) at which the body remains in equilibrium on the inclined plane is:
\[ \theta = \tan^{-1}(\mu) \]