A body of mass 10kg is placed on rough surface pulled by a force F making an angle of `30^@` above the horizontal . If the angle of friction is also `30^@`, then the minimum magnitude of force F required to move the body is equal to (take , g= `10ms^(-2)`)

Options

A. 100N
B. `50sqrt(2)N`
C. `100 sqrt(2) N`
D. 50N

Explanation

To solve the problem step by step, we will analyze the forces acting on the body and apply the principles of equilibrium and friction.

### Step 1: Identify the Given Data
- Mass of the body, \( m = 10 \, \text{kg} \)
- Angle of applied force, \( \theta = 30^\circ \)
- Angle of friction, \( \phi = 30^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Weight of the Body
The weight \( W \) of the body can be calculated using the formula:
\[ W = m \cdot g = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \]

### Step 3: Draw the Free Body Diagram (FBD)
In the FBD, we have:
- The applied force \( F \) at an angle of \( 30^\circ \) above the horizontal.
- The weight \( W = 100 \, \text{N} \) acting downwards.
- The normal force \( N \) acting upwards.
- The frictional force \( f \) acting opposite to the direction of motion.

### Step 4: Resolve the Applied Force into Components
The applied force \( F \) can be resolved into horizontal and vertical components:
- Horizontal component: \( F_x = F \cos(30^\circ) \)
- Vertical component: \( F_y = F \sin(30^\circ) \)

### Step 5: Apply the Equilibrium Condition in the Vertical Direction
Since the body is not moving vertically, we can apply the equilibrium condition:
\[ N + F_y = W \]
Substituting the values:
\[ N + F \sin(30^\circ) = 100 \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ N + \frac{F}{2} = 100 \]
Thus, we can express \( N \) as:
\[ N = 100 - \frac{F}{2} \]

### Step 6: Apply the Equilibrium Condition in the Horizontal Direction
The maximum frictional force \( f_{\text{max}} \) is given by:
\[ f_{\text{max}} = \mu N \]
Where \( \mu = \tan(\phi) = \tan(30^\circ) = \frac{1}{\sqrt{3}} \).

Now, applying the equilibrium condition in the horizontal direction:
\[ F_x = f_{\text{max}} \]
Substituting the expressions:
\[ F \cos(30^\circ) = \mu N \]
Substituting for \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ F \cdot \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}} \left(100 - \frac{F}{2}\right) \]

### Step 7: Solve for \( F \)
Now, we can solve the equation:
\[ F \cdot \frac{\sqrt{3}}{2} = \frac{100}{\sqrt{3}} - \frac{F}{2\sqrt{3}} \]
Multiplying through by \( 2\sqrt{3} \) to eliminate the fractions:
\[ F \cdot 3 = 200 - F \]
Rearranging gives:
\[ 3F + F = 200 \]
\[ 4F = 200 \]
\[ F = 50 \, \text{N} \]

### Conclusion
The minimum magnitude of the force \( F \) required to move the body is \( 50 \, \text{N} \).

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