A body of mass 10kg is on a rough inclined plane having an inclination of `30^(@)` with the horizontal. If co-efficient of friciton between the surface of contact of the body and the plane is 0.25. Find the least force required to pull the body up.

Options

A. 70.2 N
B. 80.6 N
C. 60.5 N
D. 90.4 N

Answer

To find the least force required to pull a body of mass 10 kg up a rough inclined plane at an angle of 30 degrees with a coefficient of friction of 0.25, we can follow these steps:

### Step 1: Identify the forces acting on the body
1. **Weight (W)**: The weight of the body can be calculated using the formula:
\[ W = mg \]
where \( m = 10 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).
\[ W = 10 \times 10 = 100 \, \text{N} \]

### Step 2: Resolve the weight into components
2. **Weight component parallel to the incline (W_parallel)**: This component acts down the incline and can be calculated as:
\[ W_{\text{parallel}} = W \sin(\theta) = 100 \sin(30^\circ) = 100 \times \frac{1}{2} = 50 \, \text{N} \]

3. **Weight component perpendicular to the incline (W_perpendicular)**: This component acts perpendicular to the incline and can be calculated as:
\[ W_{\text{perpendicular}} = W \cos(\theta) = 100 \cos(30^\circ) = 100 \times \frac{\sqrt{3}}{2} = 50\sqrt{3} \, \text{N} \approx 86.6 \, \text{N} \]

### Step 3: Calculate the normal force (N)
4. The normal force \( N \) is equal to the perpendicular component of the weight:
\[ N = W_{\text{perpendicular}} = 50\sqrt{3} \, \text{N} \]

### Step 4: Calculate the frictional force (F_friction)
5. The frictional force \( F_{\text{friction}} \) can be calculated using the coefficient of friction \( \mu \):
\[ F_{\text{friction}} = \mu N = 0.25 \times 50\sqrt{3} \approx 0.25 \times 86.6 \approx 21.65 \, \text{N} \]

### Step 5: Calculate the total force required to pull the body up (F)
6. The total force \( F \) required to pull the body up the incline is the sum of the force needed to overcome the gravitational pull down the incline and the frictional force:
\[ F = W_{\text{parallel}} + F_{\text{friction}} = 50 + 21.65 \approx 71.65 \, \text{N} \]

### Step 6: Round the answer
7. The least force required to pull the body up the incline is approximately:
\[ F \approx 71.65 \, \text{N} \quad \text{(can be rounded to 72 N)} \]

### Final Answer
The least force required to pull the body up the inclined plane is approximately **71.65 N**.

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