A rough inclined plane has its angle of inclination equal to `45^(@)` and `mu=0.5`. The magnitude of the least force in kg wt parallel to the plane required to move a body of 4 kg up the plane is

Options

A. `3sqrt(2)`
B. `2sqrt(2)`
C. `sqrt(2)`
D. `1//sqrt(2)`

Answer

To solve the problem, we need to determine the least force required to move a body of mass 4 kg up a rough inclined plane with an angle of inclination of 45 degrees and a coefficient of friction (μ) of 0.5.

### Step 1: Identify the forces acting on the body
When the body is on the inclined plane, the following forces act on it:
1. Weight (W) of the body acting vertically downward.
2. Normal force (N) acting perpendicular to the surface of the incline.
3. Frictional force (f) acting parallel to the incline, opposing the motion.

### Step 2: Calculate the weight of the body
The weight (W) of the body can be calculated using the formula:
\[ W = m \cdot g \]
Where:
- \( m = 4 \, \text{kg} \) (mass of the body)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 4 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 39.2 \, \text{N} \]

### Step 3: Resolve the weight into components
The weight can be resolved into two components:
1. Perpendicular to the incline: \( W_{\perp} = W \cdot \cos(\theta) \)
2. Parallel to the incline: \( W_{\parallel} = W \cdot \sin(\theta) \)

For \( \theta = 45^\circ \):
\[ W_{\perp} = 39.2 \cdot \cos(45^\circ) = 39.2 \cdot \frac{1}{\sqrt{2}} \approx 27.8 \, \text{N} \]
\[ W_{\parallel} = 39.2 \cdot \sin(45^\circ) = 39.2 \cdot \frac{1}{\sqrt{2}} \approx 27.8 \, \text{N} \]

### Step 4: Calculate the normal force
The normal force (N) is equal to the perpendicular component of the weight:
\[ N = W_{\perp} = 27.8 \, \text{N} \]

### Step 5: Calculate the frictional force
The frictional force (f) can be calculated using the formula:
\[ f = \mu \cdot N \]
Where \( \mu = 0.5 \):
\[ f = 0.5 \cdot 27.8 \approx 13.9 \, \text{N} \]

### Step 6: Determine the total force required to move the body up the incline
The total force (F) required to move the body up the incline must overcome both the parallel component of the weight and the frictional force:
\[ F = W_{\parallel} + f \]
\[ F = 27.8 + 13.9 \approx 41.7 \, \text{N} \]

### Step 7: Convert the force into kg weight
To convert the force from Newtons to kg weight, we use the relation:
\[ 1 \, \text{kg wt} = 9.8 \, \text{N} \]
Thus, the force in kg wt is:
\[ F_{\text{kg wt}} = \frac{41.7}{9.8} \approx 4.25 \, \text{kg wt} \]

### Final Answer
The magnitude of the least force required to move the body of 4 kg up the plane is approximately **4.25 kg wt**.

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