A body of mass 10 kg is placed on rough surface, pulled by 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=10 ms^(-2)`)

Options

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

Explanation

To solve the problem, we need to find the minimum force \( F \) required to move a body of mass \( 10 \, \text{kg} \) on a rough surface when the force is applied at an angle of \( 30^\circ \) above the horizontal. The angle of friction is also \( 30^\circ \).

### Step 1: Identify the forces acting on the body
- The weight of the body \( W = mg = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N} \) acts downwards.
- The normal force \( N \) acts perpendicular to the surface.
- The applied force \( F \) can be resolved into two components:
- Horizontal component: \( F \cos(30^\circ) \)
- Vertical component: \( F \sin(30^\circ) \)

### Step 2: Write the equations for vertical forces
The vertical forces must balance for the body to move:
\[ N + F \sin(30^\circ) = mg \]
Substituting the known values:
\[ N + F \cdot \frac{1}{2} = 100 \]
Thus, we can express the normal force \( N \):
\[ N = 100 - \frac{F}{2} \]

### Step 3: Determine the frictional force
The frictional force \( f \) can be expressed using the normal force and the coefficient of friction. The coefficient of friction \( \mu \) is given by:
\[ \mu = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
The frictional force \( f \) is:
\[ f = \mu N = \frac{1}{\sqrt{3}} N \]
Substituting for \( N \):
\[ f = \frac{1}{\sqrt{3}} \left(100 - \frac{F}{2}\right) \]

### Step 4: Set up the equation for horizontal forces
For the body to just start moving, the horizontal component of the applied force must equal the frictional force:
\[ F \cos(30^\circ) = f \]
Substituting the known values:
\[ F \cdot \frac{\sqrt{3}}{2} = \frac{1}{\sqrt{3}} \left(100 - \frac{F}{2}\right) \]

### Step 5: Solve for \( F \)
Now we can solve for \( F \):
\[ F \cdot \frac{\sqrt{3}}{2} = \frac{100}{\sqrt{3}} - \frac{F}{2\sqrt{3}} \]
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 force \( F \) required to move the body is \( 50 \, \text{N} \).

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