Find value of pulling force F for which block just moves. Given coefficient of friction for surface is `mu`.

Answer

To find the value of the pulling force \( F \) for which the block just moves, we need to analyze the forces acting on the block.

### Step-by-Step Solution:

1. **Identify the Forces**:
- The weight of the block acting downwards: \( mg \) (where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity).
- The normal force \( N \) acting upwards.
- The pulling force \( F \) applied at an angle \( \theta \) to the horizontal.
- The frictional force \( f \) opposing the motion.

2. **Resolve the Pulling Force**:
- The horizontal component of the pulling force is \( F \cos \theta \).
- The vertical component of the pulling force is \( F \sin \theta \).

3. **Apply Newton's Second Law in the Vertical Direction**:
- The sum of the vertical forces must equal zero for the block to be in equilibrium vertically:
\[ N + F \sin \theta = mg \]
- Rearranging gives:
\[ N = mg - F \sin \theta \]

4. **Calculate the Frictional Force**:
- The frictional force \( f \) is given by:
\[ f = \mu N \]
- Substituting for \( N \):
\[ f = \mu (mg - F \sin \theta) \]

5. **Set Up the Equation for Motion**:
- For the block to just start moving, the horizontal component of the pulling force must equal the frictional force:
\[ F \cos \theta = f \]
- Substituting for \( f \):
\[ F \cos \theta = \mu (mg - F \sin \theta) \]

6. **Rearranging the Equation**:
- Expanding the right side:
\[ F \cos \theta = \mu mg - \mu F \sin \theta \]
- Bringing all terms involving \( F \) to one side:
\[ F \cos \theta + \mu F \sin \theta = \mu mg \]
- Factoring out \( F \):
\[ F (\cos \theta + \mu \sin \theta) = \mu mg \]

7. **Solve for \( F \)**:
- Finally, solving for \( F \):
\[ F = \frac{\mu mg}{\cos \theta + \mu \sin \theta} \]

### Final Answer:
The value of the pulling force \( F \) for which the block just moves is:
\[ F = \frac{\mu mg}{\cos \theta + \mu \sin \theta} \]