A block of mass 50 kg can slide on a rough horizontal surface. The coefficient of friction between the block and the surface is 0.6. The least force of pull acting at an
angle of `30^(@)` to the upwawrd drawn ve3rtical which causes the block to just slide is

Options

A. 28.3N
B. 215.2N
C. 219.6N
D. 294.2N

Answer

To solve the problem step by step, we will analyze the forces acting on the block and apply Newton's laws of motion.

### Step 1: Identify the forces acting on the block
The forces acting on the block are:
1. Weight (W) acting downwards: \( W = mg \)
2. Normal force (N) acting upwards
3. Applied force (F) acting at an angle of \( 30^\circ \) to the vertical
4. Frictional force (f) acting opposite to the direction of motion

### Step 2: Calculate the weight of the block
Given:
- Mass of the block, \( m = 50 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

The weight of the block is:
\[ W = mg = 50 \, \text{kg} \times 10 \, \text{m/s}^2 = 500 \, \text{N} \]

### Step 3: Resolve the applied force into components
The applied force \( F \) can be resolved into two components:
- Vertical component: \( F \cos(30^\circ) \)
- Horizontal component: \( F \sin(30^\circ) \)

Using the values of cosine and sine:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \sin(30^\circ) = \frac{1}{2} \]

Thus, the components become:
\[ F \cos(30^\circ) = F \cdot \frac{\sqrt{3}}{2}, \quad F \sin(30^\circ) = F \cdot \frac{1}{2} \]

### Step 4: Apply the equilibrium condition in the vertical direction
In the vertical direction, the sum of forces must equal zero for the block to be in equilibrium:
\[ N + F \cos(30^\circ) = W \]
Substituting the weight:
\[ N + F \cdot \frac{\sqrt{3}}{2} = 500 \, \text{N} \]
Rearranging gives:
\[ N = 500 - F \cdot \frac{\sqrt{3}}{2} \quad \text{(Equation 1)} \]

### Step 5: Apply the friction condition in the horizontal direction
The frictional force \( f \) when the block just starts to slide is given by:
\[ f = \mu N \]
Where \( \mu \) is the coefficient of friction. Given \( \mu = 0.6 \):
\[ f = 0.6 N \]

The block starts to slide when the horizontal component of the applied force equals the frictional force:
\[ F \sin(30^\circ) = f \]
Substituting for \( f \):
\[ F \cdot \frac{1}{2} = 0.6 N \]
Substituting Equation 1 into this equation:
\[ F \cdot \frac{1}{2} = 0.6 \left(500 - F \cdot \frac{\sqrt{3}}{2}\right) \]

### Step 6: Solve for F
Expanding and rearranging:
\[ F \cdot \frac{1}{2} = 300 - 0.3 F \sqrt{3} \]
Combining terms:
\[ F \cdot \left(\frac{1}{2} + 0.3 \sqrt{3}\right) = 300 \]
Calculating \( 0.3 \sqrt{3} \approx 0.5196 \):
\[ F \cdot 0.8196 = 300 \]
Thus:
\[ F = \frac{300}{0.8196} \approx 366.0 \, \text{N} \]

### Step 7: Final calculation
Calculating the exact value:
\[ F \approx 366.0 \, \text{N} \]

### Conclusion
The least force of pull acting at an angle of \( 30^\circ \) to the upward vertical that causes the block to just slide is approximately **366.0 N**.