Block a of mass 4 kg is to be kept at rest against a smooth vertical wall by applying a force F as shown in figure. The force required is :- `(g=10m//s^(2))`

Options

A. `40sqrt(2)N`
B. `20sqrt(2)N`
C. `10sqrt(2)N`
D. `15sqrt(2)N`

Answer

To solve the problem of keeping block A of mass 4 kg at rest against a smooth vertical wall by applying a force F, we will follow these steps:

### Step 1: Identify the forces acting on the block
The forces acting on block A are:
1. The gravitational force (weight) acting downwards, \( mg \).
2. The applied force \( F \) at an angle of 45 degrees to the horizontal.

### Step 2: Calculate the weight of the block
Given:
- Mass of the block, \( m = 4 \, \text{kg} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

The weight of the block is calculated as:
\[ mg = 4 \, \text{kg} \times 10 \, \text{m/s}^2 = 40 \, \text{N} \]

### Step 3: Set up the equilibrium condition
Since the block is at rest, the net force acting on it must be zero. We will analyze the vertical forces. The vertical component of the applied force \( F \) must balance the weight of the block.

The vertical component of the force \( F \) can be expressed as:
\[ F_y = F \sin(45^\circ) \]

For equilibrium in the vertical direction:
\[ F_y = mg \]
Substituting the values:
\[ F \sin(45^\circ) = 40 \, \text{N} \]

### Step 4: Solve for the applied force \( F \)
We know that \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \). Substituting this into the equation gives:
\[ F \cdot \frac{\sqrt{2}}{2} = 40 \]

To isolate \( F \), we multiply both sides by \( \frac{2}{\sqrt{2}} \):
\[ F = 40 \cdot \frac{2}{\sqrt{2}} = 40 \sqrt{2} \, \text{N} \]

### Conclusion
The force required to keep block A at rest against the wall is:
\[ F = 40 \sqrt{2} \, \text{N} \]

### Final Answer
The correct answer is \( 40 \sqrt{2} \, \text{N} \).

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