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 find the minimum value of the force \( F \) required to just move the block, we need to analyze the forces acting on it.

### Step 1: Identify the forces acting on the block
1. **Weight of the block (W)**: This acts downwards and is given by:
\[ W = mg \]
where \( m = 10 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
\[ W = 10 \times 10 = 100 \, \text{N} \]

2. **Normal force (N)**: This acts perpendicular to the surface. It will be affected by the vertical component of the pulling force \( F \).

3. **Frictional force (f)**: This opposes the motion and is given by:
\[ f = \mu N \]
where \( \mu = \frac{4}{3} \) is the coefficient of friction.

### Step 2: Resolve the pulling force \( F \)
Assuming the angle \( \theta \) is above the horizontal:
- The horizontal component of the force \( F \) is:
\[ F_x = F \cos(\theta) \]
- The vertical component of the force \( F \) is:
\[ F_y = F \sin(\theta) \]

### Step 3: Apply Newton's second law in the vertical direction
In the vertical direction, the forces balance out:
\[ N + F_y = W \]
Substituting for \( F_y \):
\[ N + F \sin(\theta) = mg \]
Thus, we can express the normal force \( N \) as:
\[ N = mg - F \sin(\theta) \]

### Step 4: Substitute \( N \) into the frictional force equation
Now substituting \( N \) into the frictional force equation:
\[ f = \mu N = \mu (mg - F \sin(\theta)) \]
Substituting the values:
\[ f = \frac{4}{3} (100 - F \sin(\theta)) \]

### Step 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:
\[ F_x = f \]
Substituting the expressions for \( F_x \) and \( f \):
\[ F \cos(\theta) = \frac{4}{3} (100 - F \sin(\theta)) \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ F \cos(\theta) + \frac{4}{3} F \sin(\theta) = \frac{400}{3} \]
Factoring out \( F \):
\[ F \left( \cos(\theta) + \frac{4}{3} \sin(\theta) \right) = \frac{400}{3} \]
Thus, we can solve for \( F \):
\[ F = \frac{400}{3 \left( \cos(\theta) + \frac{4}{3} \sin(\theta) \right)} \]

### Step 7: Finding the minimum force
To find the minimum value of \( F \), we need to analyze the function \( \cos(\theta) + \frac{4}{3} \sin(\theta) \). The minimum value occurs when \( \theta \) is such that the expression is maximized. This can be done using calculus or by recognizing that the maximum occurs at specific angles.

### Conclusion
The minimum value of the force \( F \) required to just move the block can be calculated using the derived formula, depending on the angle \( \theta \).