A body of mass 2 kg is lying on a rough inclined plane of inclination 30º. Find the magnitude of force parallel to incline needed to make the block move (a)up the incline (b)down the incline. Coefficient of static friction = 0.2

Options

A. 13N,5N
B. 13N,13N
C. 13N,0N
D. 5N,13N

Answer

To solve the problem step by step, we will analyze the forces acting on the body on the inclined plane and calculate the required force to move the block up and down the incline.

### Given Data:
- Mass of the body (m) = 2 kg
- Inclination angle (θ) = 30°
- Coefficient of static friction (μ) = 0.2
- Acceleration due to gravity (g) = 10 m/s² (approx.)

### Step 1: Calculate the weight of the body.
The weight (W) of the body can be calculated using the formula:
\[ W = m \cdot g \]
\[ W = 2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 20 \, \text{N} \]

### Step 2: Resolve the weight into components.
The weight can be resolved into two components:
1. Perpendicular to the incline (Normal force, N)
2. Parallel to the incline (Force due to gravity along the incline)

The components are given by:
- Normal force (N):
\[ N = W \cdot \cos(θ) = 20 \, \text{N} \cdot \cos(30°) \]
\[ N = 20 \, \text{N} \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{N} \]

- Force parallel to the incline (F_parallel):
\[ F_{\text{parallel}} = W \cdot \sin(θ) = 20 \, \text{N} \cdot \sin(30°) \]
\[ F_{\text{parallel}} = 20 \, \text{N} \cdot \frac{1}{2} = 10 \, \text{N} \]

### Step 3: Calculate the maximum static friction force.
The maximum static friction force (F_friction) can be calculated using:
\[ F_{\text{friction}} = μ \cdot N \]
\[ F_{\text{friction}} = 0.2 \cdot (10\sqrt{3}) \]
\[ F_{\text{friction}} = 2\sqrt{3} \, \text{N} \]

### Step 4: Calculate the force needed to move the block up the incline.
To move the block up the incline, the applied force (F_applied) must overcome both the gravitational force acting down the incline and the frictional force:
\[ F_{\text{applied}} = F_{\text{parallel}} + F_{\text{friction}} \]
\[ F_{\text{applied}} = 10 \, \text{N} + 2\sqrt{3} \, \text{N} \]

Calculating \( 2\sqrt{3} \):
\[ 2\sqrt{3} \approx 3.46 \, \text{N} \]
Thus,
\[ F_{\text{applied}} \approx 10 \, \text{N} + 3.46 \, \text{N} \approx 13.46 \, \text{N} \]

### Step 5: Calculate the force needed to move the block down the incline.
To move the block down the incline, the applied force must overcome the frictional force acting upward and the component of the weight acting down the incline:
\[ F_{\text{applied}} = F_{\text{friction}} - F_{\text{parallel}} \]
Since the block will naturally tend to slide down due to gravity, we can express:
\[ F_{\text{applied}} = F_{\text{parallel}} - F_{\text{friction}} \]
\[ F_{\text{applied}} = 10 \, \text{N} - 2\sqrt{3} \, \text{N} \]
\[ F_{\text{applied}} \approx 10 \, \text{N} - 3.46 \, \text{N} \approx 6.54 \, \text{N} \]

### Final Answers:
- (a) The force needed to move the block up the incline is approximately **13.46 N**.
- (b) The force needed to move the block down the incline is approximately **6.54 N**.