A block of mass 10 kg is placed on the rough horizontal surface. A pulling force F is acting on the block which makes an angle `theta` above the horizontal. If coefficient of friction between block and surface is `4/3` then minimum value of force required to just move the block is `(g = 10 m/s^2)`

Options

A. 80 N
B. 160 N
C. 60 N
D. 120 N

Answer

To solve the problem, we need to find the minimum pulling force \( F \) required to just move a block of mass \( 10 \, \text{kg} \) on a rough horizontal surface, given that the coefficient of friction \( \mu = \frac{4}{3} \) and \( g = 10 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the block, \( m = 10 \, \text{kg} \)
- Coefficient of friction, \( \mu = \frac{4}{3} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Weight of the Block:**
\[ W = mg = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N} \]

3. **Determine the Angle of Friction:**
The angle of friction \( \theta \) can be calculated using:
\[ \theta = \tan^{-1}(\mu) = \tan^{-1}\left(\frac{4}{3}\right) \]
This angle corresponds to the condition when the block just begins to move.

4. **Set Up the Force Equations:**
- In the vertical direction:
\[ N + F \sin \theta = mg \]
Where \( N \) is the normal force.
- In the horizontal direction:
\[ F \cos \theta = f \]
Where \( f \) is the frictional force, given by:
\[ f = \mu N \]

5. **Express Normal Force \( N \):**
From the vertical force balance:
\[ N = mg - F \sin \theta \]

6. **Substitute \( N \) into the Friction Equation:**
The frictional force can be expressed as:
\[ f = \mu (mg - F \sin \theta) \]
Substitute this into the horizontal force equation:
\[ F \cos \theta = \mu (mg - F \sin \theta) \]

7. **Rearranging the Equation:**
\[ F \cos \theta = \frac{4}{3} (100 - F \sin \theta) \]
Simplifying gives:
\[ F \cos \theta + \frac{4}{3} F \sin \theta = \frac{400}{3} \]

8. **Factor out \( F \):**
\[ F \left( \cos \theta + \frac{4}{3} \sin \theta \right) = \frac{400}{3} \]
Thus,
\[ F = \frac{400}{3 \left( \cos \theta + \frac{4}{3} \sin \theta \right)} \]

9. **Substituting \( \theta \):**
Using \( \theta \) calculated from step 3, we can find \( \cos \theta \) and \( \sin \theta \) using trigonometric identities. For \( \tan \theta = \frac{4}{3} \):
- \( \sin \theta = \frac{4}{5} \)
- \( \cos \theta = \frac{3}{5} \)

10. **Final Calculation:**
Substitute \( \sin \theta \) and \( \cos \theta \) into the equation:
\[ F = \frac{400}{3 \left( \frac{3}{5} + \frac{4}{3} \cdot \frac{4}{5} \right)} \]
Simplifying gives:
\[ F = \frac{400}{3 \left( \frac{3}{5} + \frac{16}{15} \right)} = \frac{400}{3 \left( \frac{9}{15} + \frac{16}{15} \right)} = \frac{400}{3 \cdot \frac{25}{15}} = \frac{400 \cdot 15}{75} = 80 \, \text{N} \]

### Conclusion:
The minimum value of the force required to just move the block is \( F = 80 \, \text{N} \).