A block of mass 10 kg is placed on a rough horizontal surface of coefficient of friction 0.1. A pulling force of 100 N acts on it at `37^(@)` with the horizontal. The frictional force acting on the body is ..... N.

Options

A. 2
B. 4
C. 6
D. 7

Answer

To solve the problem step by step, we will analyze the forces acting on the block and calculate the frictional force.

### Step 1: Identify the given data
- Mass of the block (m) = 10 kg
- Coefficient of friction (μ) = 0.1
- Pulling force (F) = 100 N
- Angle of pulling force (θ) = 37°

### Step 2: Calculate the weight of the block
The weight (W) of the block can be calculated using the formula:
\[ W = m \cdot g \]
Where \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).
\[ W = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \]

### Step 3: Resolve the pulling force into components
The pulling force can be resolved into horizontal (F_x) and vertical (F_y) components:
- Horizontal component:
\[ F_x = F \cdot \cos(θ) = 100 \cdot \cos(37°) \]
Using \( \cos(37°) \approx \frac{4}{5} = 0.8 \):
\[ F_x = 100 \cdot 0.8 = 80 \, \text{N} \]

- Vertical component:
\[ F_y = F \cdot \sin(θ) = 100 \cdot \sin(37°) \]
Using \( \sin(37°) \approx \frac{3}{5} = 0.6 \):
\[ F_y = 100 \cdot 0.6 = 60 \, \text{N} \]

### Step 4: Calculate the normal force (N)
The normal force can be calculated using the equilibrium of vertical forces. The weight of the block is balanced by the normal force and the vertical component of the pulling force:
\[ W = N + F_y \]
Substituting the values:
\[ 100 = N + 60 \]
\[ N = 100 - 60 = 40 \, \text{N} \]

### Step 5: Calculate the maximum static friction force (f_s)
The maximum static friction force can be calculated using the formula:
\[ f_s = μ \cdot N \]
Substituting the values:
\[ f_s = 0.1 \cdot 40 = 4 \, \text{N} \]

### Step 6: Determine the frictional force acting on the block
Since the pulling force's horizontal component (80 N) is greater than the maximum static friction force (4 N), the block will move. Therefore, the frictional force acting on the block will be equal to the maximum static friction force:
\[ f = f_s = 4 \, \text{N} \]

### Final Answer
The frictional force acting on the body is **4 N**.

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