A body of mass 10 kg is at rest on an iinclined plane of angle `30^(@)` . What are its components parallel and perpendicular to the plane ?

Explanation

To find the components of the gravitational force acting on a body of mass 10 kg resting on an inclined plane at an angle of 30 degrees, we can follow these steps:

### Step 1: Identify the forces acting on the body
The only force acting on the body is the gravitational force (weight) which can be calculated using the formula:
\[ F_g = mg \]
where:
- \( m = 10 \, \text{kg} \) (mass of the body)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

### Step 2: Calculate the gravitational force
Substituting the values into the formula:
\[ F_g = 10 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N} \]

### Step 3: Resolve the gravitational force into components
The gravitational force can be resolved into two components:
1. **Component parallel to the incline** (\( F_{\parallel} \))
2. **Component perpendicular to the incline** (\( F_{\perpendicular} \))

Using trigonometric functions:
- The component parallel to the incline is given by:
\[ F_{\parallel} = mg \sin(\theta) \]
- The component perpendicular to the incline is given by:
\[ F_{\perpendicular} = mg \cos(\theta) \]

### Step 4: Calculate the parallel component
Substituting the values for \( \theta = 30^\circ \):
\[ F_{\parallel} = 98 \, \text{N} \times \sin(30^\circ) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ F_{\parallel} = 98 \, \text{N} \times \frac{1}{2} = 49 \, \text{N} \]

### Step 5: Calculate the perpendicular component
Now, calculate the perpendicular component:
\[ F_{\perpendicular} = 98 \, \text{N} \times \cos(30^\circ) \]
Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ F_{\perpendicular} = 98 \, \text{N} \times \frac{\sqrt{3}}{2} \]
Calculating this gives:
\[ F_{\perpendicular} = 49 \sqrt{3} \, \text{N} \approx 84.87 \, \text{N} \]

### Final Result
- The component of the gravitational force parallel to the incline is \( 49 \, \text{N} \).
- The component of the gravitational force perpendicular to the incline is approximately \( 84.87 \, \text{N} \).