The force required to just move a body up an inclined plane is double the force the required prevent it from sliding down. If phi is angle of friction and theta is the angle which incline makes with the horizontal then,

Options

A. `tantheta=tanphi`
B. `tantheta=2tanphi`
C. `tantheta=3tanphi`
D. `tanphi=2tantheta`

Explanation

To solve the problem, we need to analyze the forces acting on a body placed on an inclined plane and derive the relationship between the angles of friction (φ) and the incline (θ).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Body:**
- When the body is about to move up the incline, the forces acting on it include:
- The gravitational force component down the incline: \( mg \sin \theta \)
- The frictional force acting downwards: \( f = \mu_s N = \mu_s (mg \cos \theta) \)
- When the body is about to slide down, the frictional force acts upwards.

2. **Set Up the Equations:**
- For the body moving upwards (force \( F_1 \)):
\[ F_1 = mg \sin \theta + f = mg \sin \theta + \mu_s (mg \cos \theta) \]
- For the body sliding down (force \( F_2 \)):
\[ F_2 = mg \sin \theta - f = mg \sin \theta - \mu_s (mg \cos \theta) \]

3. **Given Condition:**
- According to the problem, the force required to move the body up the incline is double the force required to prevent it from sliding down:
\[ F_1 = 2F_2 \]

4. **Substituting the Expressions:**
- Substitute \( F_1 \) and \( F_2 \) into the equation:
\[ mg \sin \theta + \mu_s (mg \cos \theta) = 2(mg \sin \theta - \mu_s (mg \cos \theta)) \]

5. **Simplifying the Equation:**
- Distributing the 2 on the right side:
\[ mg \sin \theta + \mu_s (mg \cos \theta) = 2mg \sin \theta - 2\mu_s (mg \cos \theta) \]
- Rearranging gives:
\[ mg \sin \theta + \mu_s (mg \cos \theta) + 2\mu_s (mg \cos \theta) = 2mg \sin \theta \]
- Combine like terms:
\[ mg \sin \theta + 3\mu_s (mg \cos \theta) = 2mg \sin \theta \]

6. **Isolate Terms:**
- Move \( mg \sin \theta \) to the right:
\[ 3\mu_s (mg \cos \theta) = 2mg \sin \theta - mg \sin \theta \]
- This simplifies to:
\[ 3\mu_s (mg \cos \theta) = mg \sin \theta \]

7. **Canceling \( mg \):**
- Since \( mg \) is common in all terms, we can cancel it out:
\[ 3\mu_s \cos \theta = \sin \theta \]

8. **Substituting for \( \mu_s \):**
- Recall that \( \mu_s = \tan \phi \):
\[ 3 \tan \phi \cos \theta = \sin \theta \]

9. **Using Trigonometric Identity:**
- Divide both sides by \( \cos \theta \):
\[ 3 \tan \phi = \tan \theta \]

10. **Final Relation:**
- Thus, we have the relationship:
\[ \tan \theta = 3 \tan \phi \]

### Final Answer:
The relationship between the angles is:
\[ \tan \theta = 3 \tan \phi \]