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

Explanation

To solve the problem step by step, we will analyze the forces acting on the body placed on the inclined plane and derive the relationship between the angle of inclination, the coefficient of friction, and the equilibrium condition.

### Step-by-Step Solution:

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

2. **Determine the Normal Force:**
- The normal force (N) acts perpendicular to the surface of the incline.
- In equilibrium, the normal force balances the perpendicular component of the weight:
\[ N = mg \cos \theta \]

3. **Determine the Frictional Force:**
- The frictional force (f) acts up the incline to counteract the component of the weight acting down the incline.
- The maximum static frictional force can be expressed as:
\[ f = \mu N \]
- Substituting for N, we get:
\[ f = \mu (mg \cos \theta) \]

4. **Set Up the Equilibrium Condition:**
- Since the body is in equilibrium, the frictional force must balance the component of the weight acting down the incline:
\[ mg \sin \theta = f \]
- Substituting for f, we have:
\[ mg \sin \theta = \mu (mg \cos \theta) \]

5. **Simplify the Equation:**
- We can cancel out mg from both sides (assuming m ≠ 0):
\[ \sin \theta = \mu \cos \theta \]

6. **Rearranging the Equation:**
- Dividing both sides by cos θ gives:
\[ \tan \theta = \mu \]

7. **Finding the Angle θ:**
- Therefore, we can express θ in terms of the coefficient of friction:
\[ \theta = \tan^{-1}(\mu) \]

### Final Answer:
The angle of inclination θ is given by:
\[ \theta = \tan^{-1}(\mu) \]