A body of mass 10kg is on a rough inclined plane having an inclination of 30° with the horizontal. If coefficient of friction between the surfaces of contact of the body and the plane is 0.5. Find the least force required to pull the body up the plane.

Options

A. 80.5 N
B. 91.45 N
C. 85.4 N
D. 78.4 N

Answer

To solve the problem of finding the least force required to pull a body of mass 10 kg up a rough inclined plane at an angle of 30° with a coefficient of friction of 0.5, we can follow these steps:

### Step 1: Identify the forces acting on the body
The forces acting on the body on the inclined plane include:
1. The gravitational force (weight) acting downwards, \( mg \).
2. The normal force \( N \) acting perpendicular to the inclined plane.
3. The frictional force \( f_k \) acting down the plane, opposing the motion.
4. The applied force \( F \) acting up the plane.

### Step 2: Resolve the weight into components
The weight of the body can be resolved into two components:
- The component parallel to the incline: \( mg \sin \theta \)
- The component perpendicular to the incline: \( mg \cos \theta \)

Given:
- Mass \( m = 10 \, \text{kg} \)
- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)
- Angle of inclination \( \theta = 30° \)

Calculating these components:
- \( mg = 10 \times 9.8 = 98 \, \text{N} \)
- \( mg \sin 30° = 98 \times \frac{1}{2} = 49 \, \text{N} \)
- \( mg \cos 30° = 98 \times \frac{\sqrt{3}}{2} \approx 84.87 \, \text{N} \)

### Step 3: Calculate the normal force
The normal force \( N \) is equal to the component of the weight acting perpendicular to the incline:
\[ N = mg \cos \theta = 98 \times \frac{\sqrt{3}}{2} \approx 84.87 \, \text{N} \]

### Step 4: Calculate the frictional force
The frictional force \( f_k \) can be calculated using the coefficient of friction:
\[ f_k = \mu N = 0.5 \times N = 0.5 \times 84.87 \approx 42.44 \, \text{N} \]

### Step 5: Write the equation for the applied force
The least force \( F \) required to pull the body up the incline must overcome both the gravitational component down the incline and the frictional force:
\[ F = mg \sin \theta + f_k \]
Substituting the values we calculated:
\[ F = 49 + 42.44 \approx 91.44 \, \text{N} \]

### Step 6: Final calculation
Thus, the least force required to pull the body up the inclined plane is approximately:
\[ F \approx 91.45 \, \text{N} \]

### Conclusion
The least force required to pull the body up the plane is \( 91.45 \, \text{N} \).

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